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3371 Answers 3371

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Dollar Bill Auction

  • \$\begingroup\$ I would recommend removing the EmoWolf example, it's not a valid solution by your own rules and I think calling it an "example" could mislead users to believe otherwise. \$\endgroup\$ – Kamil Drakari Apr 10 '18 at 19:34
  • \$\begingroup\$ @KamilDrakari how is it invalid? \$\endgroup\$ – RamenChef Apr 10 '18 at 21:25
  • 1
    \$\begingroup\$ Your Additional Rules indicate that Standard Loopholes are forbidden. EmoWolf is literally the example for one of the standard loopholes. \$\endgroup\$ – Kamil Drakari Apr 10 '18 at 23:26
  • 4
    \$\begingroup\$ remove floating point arithmetic; if bids only go up by 5 cents, floating point arithmetic only introduces rounding errors to discrete data. \$\endgroup\$ – Destructible Lemon Apr 10 '18 at 23:32
  • \$\begingroup\$ I don't see any check that I'm not making a negative bid. \$\endgroup\$ – Peter Taylor Apr 11 '18 at 8:28
  • \$\begingroup\$ @DestructibleLemon I think I'll do that. \$\endgroup\$ – RamenChef Apr 11 '18 at 13:11
  • \$\begingroup\$ @PeterTaylor if your bid is negative the bidding ends because you didn't top your opponent's bid. \$\endgroup\$ – RamenChef Apr 11 '18 at 13:13
  • \$\begingroup\$ @RamenChef, but is my negative bid subtracted from my balance? \$\endgroup\$ – Peter Taylor Apr 11 '18 at 14:11
  • \$\begingroup\$ @PeterTaylor no, it uses your previous bid. \$\endgroup\$ – RamenChef Apr 11 '18 at 17:42
  • \$\begingroup\$ Suggestions have been implemented. \$\endgroup\$ – RamenChef Apr 11 '18 at 17:58
  • \$\begingroup\$ I'll post this soon if nobody objects. \$\endgroup\$ – RamenChef Apr 16 '18 at 16:49
  • 2
    \$\begingroup\$ Honestly, this is rather boring. There's very few strategies, so this will quickly devolve into meta-gaming \$\endgroup\$ – Jo King Apr 17 '18 at 11:08
  • \$\begingroup\$ You're Analyst bot is promoting bad behavior. What happens if you have another bot that also uses reflection to run newAuction? They will infinite loop each other. I highly recommend disallowing reflection. \$\endgroup\$ – Nathan Merrill Apr 17 '18 at 19:11
  • \$\begingroup\$ You haven't mentioned how you are combining auctions. "Bots are scored based on the profit they made during those auctions." isn't nearly specific enough. \$\endgroup\$ – Nathan Merrill Apr 17 '18 at 19:15
  • \$\begingroup\$ Running 2 auctions per pair is a bad idea, the second game provides you with no benefit: it either gives the same result, or you have no idea who actually won between the two players. \$\endgroup\$ – Nathan Merrill Apr 17 '18 at 19:24

Operate on a subset of a register

Note: You do NOT need to know anything about quantum mechanics or quantum computing to understand this challenge. I've tried to include sufficient background for any old code golfer to realize what's going on.

Motivation (much of this can probably be skipped)

In quantum computing, the equivalent of a bit is a qubit. Like a bit, a qubit will have a state of OFF or ON (0 or 1) when observed, but unlike a bit, a qubit can exist in both states simultaneously while unobserved. Trippy.

What's even trippier is that they don't exist in simple superpositions (40% chance to observe 0, 60% for 1)—you can have two qubits that will behave the same when observed, but not when interacting with other qubits. Taking this into account, a convenient representation of a qubit's state is a complex vector, with bases |0> and |1>. A qubit with state (1+0i)|0> + (0+0i)|1> will always show 0 when observed, as will (0+1i)|0> + (0+0i)|1> or (0.6+0.8i)|0> + (0+0i)|1>. What matters here is the squared magnitude of each component—that's the probability of observing the corresponding basis vector. Noting this, (-0.5+0.5i)|0> + (-0.5-0.5i)|1> will "choose" randomly between the two possible states. This ability to have multiple states at once allows qubits to store much, much more information than an equal number of bits.

Now, since a qubit is a two-dimensional vector (as far as we're concerned), an operation on a qubit can be represented as a 2x2 matrix. For example, the identity:

[[1+0i  0+0i]
 [0+0i  1+0i]]

will take any qubit to an equivalent one, while the Hadamard transform, represented by the matrix:

[[1+0i  1+0i]
 [1+0i -1+0i]]

will turn any pure state (P(0) == 0 or 1) into an equal superposition of the two. Check out this challenge for generation of this matrix, generalized to n qubits! These matrices multiply a qubit's vector state to yield a new state.

A qubit on its own is little more useful than a bit. However, they can be put into registers with others, so a "qubyte" can store 0, 255, or any arbitrary superposition of the numbers between.

I won't go into depth on entanglement here, but I'll say that it is a property of these qubits that makes storing quantum registers as lists of qubits not work at all (and makes quantum computers outperform classical ones). It works much better to treat a register much like a qubit and give it a vector representation, this time with a basis vector for each observable state. For example, two (1+0i)|0> + (0+0i)|1> qubits together form a (1+0i)|00> + (0+0i)|01> + (0+0i)|10> + (0+0i)|11> register. Note that the probability distributions of outcomes are identical for the two representations—in this case, both qubits will and up as 0 in both representations.

Important stuff again

These registers are nice and all, but it's difficult (and often impossible—try it yourself on (-0.5+0.5i)|00> + (0+0i)|01> + (0+0i)|10> + (-0.5-0.5i)|11>!) to extract the state of a single qubit in a register from the register's vector—that is, you can have a register in a state unformable by individual qubits. Fortunately, to apply, say, the Hadamard transform to qubit 2 of a register, we can construct a matrix by which we can multiply the entire register but change only the one qubit. Neat!

Let's say we have an operation m on one qubit defined as:

[[x y]
 [z w]]

and a three-qubit register of qubits a|0> + b|1>, c|0> + d|1>, and e|0> + f|1>. This register has a vector form (treating the first of these qubits as the least significant) of:

eca|000> + ecb|001> + eda|010> + edb|011> + fca|100> + fcb|101> + fda|110> + fdb|111>

Defining g := ax + by and h := az + bw, we see that the result of applying m to a|0> + b|1> is g|0> + h|1>. What we want is a matrix that, when multiplied by the register state above, yields the same but with a and b replaced by g and h, respectively:

ecg|000> + ech|001> + edg|010> + edh|011> + fcg|100> + fch|101> + fdg|110> + fdh|111>

Let's look at a matrix that does just this:

[[x y 0 0 0 0 0 0]     [[eca]     [[ecax + ecby]     [[ecg]
 [z w 0 0 0 0 0 0]      [ecb]      [ecaz + ecbw]      [ech]
 [0 0 x y 0 0 0 0]      [eda]      [edax + edby]      [edg]
 [0 0 z w 0 0 0 0]  \/  [edb]  ——  [edaz + edbw]  ——  [edh]
 [0 0 0 0 x y 0 0]  /\  [fca]  ——  [fcax + fcby]  ——  [fcg]
 [0 0 0 0 z w 0 0]      [fcb]      [fcaz + fcbw]      [fch]
 [0 0 0 0 0 0 x y]      [fda]      [fdax + fdby]      [fdg]
 [0 0 0 0 0 0 z w]]     [fdb]]     [fdaz + fdbw]]     [fdh]]

How nice. Of course, we can operate of qubit 1 or 2 (0-indexed) with a modified version of the very same matrix:

[[x 0 y 0 0 0 0 0]
 [0 x 0 y 0 0 0 0]
 [z 0 w 0 0 0 0 0]
 [0 z 0 w 0 0 0 0]
 [0 0 0 0 x 0 y 0]
 [0 0 0 0 0 x 0 y]
 [0 0 0 0 z 0 w 0]
 [0 0 0 0 0 z 0 w]]

[[x 0 0 0 y 0 0 0]
 [0 x 0 0 0 y 0 0]
 [0 0 x 0 0 0 y 0]
 [0 0 0 x 0 0 0 y]
 [z 0 0 0 w 0 0 0]
 [0 z 0 0 0 w 0 0]
 [0 0 z 0 0 0 w 0]
 [0 0 0 z 0 0 0 w]]

These are the same as before, but with the 2x2 squares spread out a bit and overlapping. Verification of these matrices is left as an exercise for the reader.

You can also perform multi-qubit operations (like the controlled-NOT gate) on just part of a register in much the same fashion. Performing on qubits 0 and 2 of a 3-qubit register the following operation:

[[a b c d]
 [e f g h]
 [i j k l]
 [m n o p]]

can be accomplished with yet another 8x8 matrix:

[[a b 0 0 c d 0 0]
 [e f 0 0 g h 0 0]
 [0 0 a b 0 0 c d]
 [0 0 e f 0 0 g h]
 [i j 0 0 k l 0 0]
 [m n 0 0 o p 0 0]
 [0 0 i j 0 0 k l]
 [0 0 m n 0 0 o p]]

This can be generalized to an operation on m qubits of a register of n qubits, where 0 ≤ m ≤ n; in fact, that's what this challenge'll do!

Your task (at long last!)

Write a program or function. It must:

  • Take as input a nonnegative integer n.

  • Take as input a list of m distinct integers from 0 to n (inclusive).

  • Take as input a 2mx2m matrix of integers that represents the operation to be performed on the qubits of an n-qubit register indicated (in order) by the aforementioned list.

  • Return as output a 2nx2n matrix of integers that represents an operation on the entire register equivalent to applying the given matrix to the indicated qubits.

Test cases

n = 4
qubits = {2}
Operation: [[1  2]
            [3  4]]
Return: [[1 0 0 0 2 0 0 0 1 0 0 0 2 0 0 0]
         [0 1 0 0 0 2 0 0 0 1 0 0 0 2 0 0]
         [0 0 1 0 0 0 2 0 0 0 1 0 0 0 2 0]
         [0 0 0 1 0 0 0 2 0 0 0 1 0 0 0 2]
         [3 0 0 0 4 0 0 0 3 0 0 0 4 0 0 0]
         [0 3 0 0 0 4 0 0 0 3 0 0 0 4 0 0]
         [0 0 3 0 0 0 4 0 0 0 3 0 0 0 4 0]
         [0 0 0 3 0 0 0 4 0 0 0 3 0 0 0 4]
         [1 0 0 0 2 0 0 0 1 0 0 0 2 0 0 0]
         [0 1 0 0 0 2 0 0 0 1 0 0 0 2 0 0]
         [0 0 1 0 0 0 2 0 0 0 1 0 0 0 2 0]
         [0 0 0 1 0 0 0 2 0 0 0 1 0 0 0 2]
         [3 0 0 0 4 0 0 0 3 0 0 0 4 0 0 0]
         [0 3 0 0 0 4 0 0 0 3 0 0 0 4 0 0]
         [0 0 3 0 0 0 4 0 0 0 3 0 0 0 4 0]
         [0 0 0 3 0 0 0 4 0 0 0 3 0 0 0 4]]

n = 3
qubits = {0, 2}
Operation: [[1  2  3  4 ]
            [5  6  7  8 ]
            [9  10 11 12]
            [13 14 15 16]]
Return: [[1  2  0  0  3  4  0  0 ]
         [5  6  0  0  7  8  0  0 ]
         [0  0  1  2  0  0  3  4 ]
         [0  0  5  6  0  0  7  8 ]
         [9  10 0  0  11 12 0  0 ]
         [13 14 0  0  15 16 0  0 ]
         [0  0  9  10 0  0  11 12]
         [0  0  13 14 0  0  15 16]]

n = 3
qubits = {2, 0}
Operation: [[1  2  3  4 ]
            [5  6  7  8 ]
            [9  10 11 12]
            [13 14 15 16]]
Return: [[1  3  0  0  2  4  0  0 ]
         [9  11 0  0  10 12 0  0 ]
         [0  0  1  3  0  0  2  4 ]
         [0  0  9  11 0  0  10 12]
         [5  7  0  0  6  8  0  0 ]
         [13 15 0  0  14 16 0  0 ]
         [0  0  5  7  0  0  6  8 ]
         [0  0  13 15 0  0  14 16]]
Wow, things get weird.

Operation: 2^m x 2^m identity matrix
Return: 2^n x 2^n identity matrix

n = 5

qubits = {3, 0, 4, 2, 7}

Operation: Pastebin

Return: [Pastebin] (TODO)


As usual, standard loopholes apply. Builtins are allowed, but try to include a solution sans builtin as well.

You can 1-index or reverse the list if desired.

I/O is flexible.

This is , so the shortest solution (in bytes) in each language wins!

Sandboxy stuff

Anything worth mentioning that I missed?

Is there another application of this that might be easier to explain and scare fewer people off?


  • \$\begingroup\$ If I understood correctly, the part "it's difficult to extract the state of a single qubit in the register from the register's vector." means that most "registers" cannot be formed by (multiple) single qubits, correct? \$\endgroup\$ – user202729 Apr 22 '18 at 13:29
  • \$\begingroup\$ @user202729 That is correct. \$\endgroup\$ – Khuldraeseth na'Barya Apr 22 '18 at 16:23
  • \$\begingroup\$ Is there a way to rearrange the post so that the task is on top and the explanation comes later? \$\endgroup\$ – JayCe Jun 11 '18 at 3:53

Bot Wars at the Auction

I have seen KOTH challenges where the bots fight each other, and I have seen KOTH challenges where the bots are in an auction. So I came up with this:

The Challenge:

You must build a javascript bot that will fight other bots using weapons purchased in an auction. These items are: swords, bows, and shields. You can also purchase healing and quivers. Your function is called once per "Turn", and only 1 weapon action and 1 move action can be called in each turn. You start with 200 health, and 2000 coins. Fight in arena with each bot in its own 5 by 5 square. For example, 16 bots would be a 20 by 20 arena.


During an auction, 2 items will be sold per surviving bot. In the initial auction, 3 items will be sold for each bot. It is up to the bots how they will bid: some bots may spend lots on a cool sword, while others could choose to buy healing in large amounts. There is an auction every few hundred turns. Items start at cost 1.


After the auction, your bots will be pitted against each other. With swords to strike nearby enemies, bows to attack distant bots, and shields to defend yourself, what could go wrong? If you find yourself low on health, you can heal (Which takes between 6 and 8 turns).


As input, your function receives an object filled with functions that can be used to control your bot. This object is also at window.bots["yourbot"].auction or window.bots["yourbot"].fight.


//within window.bots["yourbot"]
mode: integer, //0=Dead, 1=Auction, 2=Fight
items: object, //All items owned
coins: integer,
health: integer,
auction: {
    itemlist: array, //Array of item objects
    placeBid(coins), //Bid on item
    incrementBid(amount), //Bids (current + amount)
    item: {
        current: integer, //Current highest bid
        bidder: string, //Current highest bidder
        type: integer, //1=Sword, 2=Bow, 3=Quiver, 4=Shield, 5=Healing
        power: integer, //Item power (Attack amount, healing amount, or block percentage)
        units: integer //Only for healing and quivers; amount of turns taken to heal, or amount of arrows in quiver
fight: {
    botList: array, //List of bots
    surrenderTo(bot), //Surrenders to a specific bot
    move: {
        north(), //Y+
        east(), //X+
        south(), //Y-
        west(), //X-
    use: {
        swordAttack(bot OR [x,y]), //Attacks a bot, but only if it is adjacent to attacker
        bow(bot OR [x,y]), //Shoots bow at bot, bot must be within 5 spaces
        shield(), //Raises shield for 5 turns
        heal(sor) //Uses least powerful healing or takes input sort function
    util: {
        getNearest(), //Returns nearest bot
        getBot(name), //Returns a specific bot by name
        atLocation([x,y]), //Returns bot at a certain location
        gridSize(), //Returns the side length of the square arena
        getPosition(), //Returns array [x,y] of caller bot
        see([x,y]), //Returns true or false, depending on if there is a bot in a certain location

//within other bot nearestBot(), getBot(), botList, or atLocation(x, y)
location: array, //[x,y]
name: string, //Bot's name
items: object //All objects owned by bot

//within items
type: integer, //1=Sword, 2=Bow, 3=Quiver, 4=Shield, 5=Healing
power: integer, //Item power (Attack amount, healing amount, or block percentage)
units: integer //Only for healing and quivers; amount of turns taken to heal, or amount of arrows in quiver

How it Works:

You will start in an auction. With your 2000 coins, you can bid on items using the functions placeBid(price) or incrementBid(amount). If nobody bids for 5 turns, the item is sold. Once the items are all gone, the fight begins. The initial auction focuses mainly on swords, bows, and shields. During the fight, you can attack players using simple methods like bow(nearestPlayer()), or more complex methods involving mapping the area with see([x,y]) and atLocation([x,y]). When the fight round is over, another auction starts, which mainly focuses on healing and quivers. When the fight begins again, you will be able to heal (Make sure you shield right before doing so!). After that fight, another auction will commence. This will repeat until one person remains, or all other bots use surrenderTo(bot). Inflicting damage on another bot gets you 1 coin per point, and you get a 500 coin bonus for killing a bot. If a bot surrenders to you, you earn 100 coins. Store information that you bot may use in window.bots[yourBotName]. Note that a copy of this object is passed to your bot, but storing a varibale in this object will only affect it locally. Coordinates range from [0,0] in the corner to [max,max] in the opposite corner, max being round(22.36 * sqrt(botCount)).

Function Specifications:

    Inputs amount of coins to set the bidding at
    If lower than current highest bid or another bid in that turn, returns 1
    If input is invalid, returns 2
    Will set auction.item.current to coins and auction.item.bidder to your bot name
    Inputs amount of coins to increase bidding by
    Adds to highest bid in last turn
    If lower than current highest bid or another bid in that turn, returns 1
    If input is invalid, returns 2
    Will add amount to auction.item.current and auction.item.bidder to your bot name
    Will move bot 1 space Y+
    Returns 1 if bot is at far North edge of arena
    Will move bot 1 space X+
    Returns 1 if bot is at far East edge of arena
    Will move bot 1 space Y-
    Returns 1 if bot is at far South edge of arena
    Will move bot 1 space X-
    Returns 1 if bot is at far West edge of arena
    Returns the nearest bot as of the last turn
    Read information: location, name, items
    Inputs a string, containing the name of a bot
    If bot does not exist, returns 1
    If bot is dead, returns 2
    If input is invalid, returns 3
    Returns bot as of last turn
atLocation([x,y]) or atLocation(x,y)
    Inputs an array containing coordinates, or two integer coordinates
    Returns bot as of last turn
    If no bot exists in that spot, returns 0
    Invalid input returns 1
    Returns an integer containing the side length of the arena
    Returns an array [x,y] containing the position of the bot
    Takes bot as input
    Will set mode to 0 on your bot
    Gives bot surrendered to 100 coins
    Returns 1 if input is invalid
    Returns 2 if bot is dead or inexistant
    Returns true if bot is in location
    Returns false if input is invalid or no bot exists in location
swordAttack(bot OR [x,y])
    Takes either a bot or [x,y] coordinates as input
    Does as much damage as your best sword
    Will return 1 if input is invalid
    Will return 2 if bot is dead/nonexistant, or if coordinates are out of bounds
    Will return 3 if you don't have a sword
    Will return 4 if bot/coordinates are not within 1 space of your bot
bow(bot OR [x,y])
    Takes either a bot or [x,y] coordinates as input
    Doesas much damage as your best bow
    Will return 1 if input is invalid
    Will return 2 if bot is dead/nonexistant, or if coordinates are out of bounds
    Will return 3 if you don't have a bow, and 5 if you have no arrows
    Will return 4 if bot/coordinates are not within 20 spaces of your bot
    Blocks as much damage as your best shield as a percentage
    Lasts for 5 turns (Includes turn activated)
    Will return 1 if you don't have a shield
    Uses your weakest heal, or you can specify a sort function to define the order they are used
    Cannot move for 6-8 turns (Selected randomly)
    Will return 1 if you don't have heal

Item Specifications:

    The sword is used to attack bots within 1 space
    Has set damage amount (Geometric distribution 10-100)
    Is found in:
        Auction 1 = 35%
        Auction 2 = 30%
        Auction 3 = 15%
        Auction 4 = 5%
    The bow is used to remotely attack bots within 20 spaces
    Has set damage amount (Geometric distribution 4-40)
    Is found in:
        Auction 1 = 30%
        Auction 2 = 20%
        Auction 3 = 10%
        Auction 4 = 5%
    Quivers are purchased with arrows in them, used for shooting bows
    Starts with specific amount of arrows (Geometric distribution 5-20)
    Is found in:
        Auction 1 = 7.5%
        Auction 2 = 10%
        Auction 3 = 25%
        Auction 4 = 40%
        Other auctions = 45%
    Shields are used to defend against attack
    Starts with a specific block percentage (Geometric distribution 5%-75%)
    Can be used while healing
    Is found in:
        Auction 1 = 20%
        Auction 2 = 30%
        Auction 3 = 20%
        Auction 4 = 10%
    Heal is used to regenerate during battle
    Heal level is geometric distribution 20-80
    Is found in:
        Auction 1 = 7.5%
        Auction 2 = 10%
        Auction 3 = 30%
        Auction 4 = 40%
        Other auctions = 55%

Example bot:

function readyToFight(obj) {
    var items = obj.items;
    var types = [];
    var coins = obj.coins;
    for (var i = 0, n; i < items.length; i++) {
        n = items[i];
    if (obj.mode == 1) {
        obj = obj.auction;
        if (obj.item.type == 1) {
            if (types.indexOf(1) != -1) {
                for (var i = 0, n; i < items.length; i++) {
                    n = items[i];
                    if (n.type == 1 && n.power < obj.item.power) {
                        var bid = true;
                if (n.item.price < coins) {
            } else {
                if (n.item.price < coins) {
        } else if (obj.item.type == 2) {
            if (types.indexOf(2) != -1) {
                for (var i = 0, n; i < items.length; i++) {
                    n = items[i];
                    if (n.type == 2 && n.power < obj.item.power) {
                        var bid = true;
                if (n.item.price < coins) {
            } else {
                if (n.item.price < coins) {
        } else if (obj.item.type == 3) {
            if (n.item.price < n.item.units * 2) {
        } else if (obj.item.type == 4) {
            if (types.indexOf(4) != -1) {
                for (var i = 0, n; i < items.length; i++) {
                    n = items[i];
                    if (n.type == 4 && n.power < obj.item.power) {
                        var bid = true;
                if (n.item.price < coins) {
            } else {
                if (n.item.price < coins) {
        } else if (obj.item.type == 5) {
            if (n.item.price < n.item.power * 5) {
    } else if (obj.mode == 2) {
        hp = obj.health;
        obj = obj.fight;
        if (types.indexOf(2) != -1) {
        } else if (obj.health < 50 && types.indexOf(5) != -1) {
        } else {
            //Be scared, or implement better fighting techniques!
    } else {
        //Bot can still see the arena and auction when dead, but cannot fight or bid


Standard Loopholes prohibited, javascript only

  • 1
    \$\begingroup\$ Seems like a solid updated version of the previous idea. I think the "The Challenge" section uses the word "function" too much, and should be updated with slightly less technical language. For example, "Your bot can make one Move action and one Combat action each turn". Additionally, the fight object doesn't separate its functions into movement and attacks. I would recommend sub-objects, for example fight.move.north() and fight.attack.sword(x,y). And if some of those functions are supposed to be neither movement nor attacks, fight.util.getNearest() or the like. \$\endgroup\$ – Kamil Drakari Apr 23 '18 at 15:52
  • \$\begingroup\$ Okay, got it. Sub objects for fight: move, attack, and util \$\endgroup\$ – Redwolf Programs Apr 23 '18 at 15:54
  • 1
    \$\begingroup\$ I did use instead of attack, so that shields and heal fit under the same category \$\endgroup\$ – Redwolf Programs Apr 23 '18 at 16:00
  • \$\begingroup\$ Next, only having the option to use the weakest healing item doesn't seem right to me. While non-consumable equipment makes sense to always use the best, and there's only one "trait" so best is never in question, healing would often want to use "largest heal that isn't overheal" or "heal that takes the least time". I suspect it wasn't done that way to simplify the runner, but only allowing smallest heal seems restrictive. \$\endgroup\$ – Kamil Drakari Apr 23 '18 at 16:02
  • \$\begingroup\$ Perhaps add an optional parameter for a sort function? Like function(a,b){return a.power - b.power} or function(a,b){return (a.power / a.units) - (b.power / b.units)} \$\endgroup\$ – Redwolf Programs Apr 23 '18 at 16:05
  • \$\begingroup\$ And then the heals are used in the order they are listed \$\endgroup\$ – Redwolf Programs Apr 23 '18 at 16:06

Single-Digit Representations of Natural Numbers

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Dr. Inder Taneja is a Mathematics Professor at the Federal University of Santa Catarina who wrote and published "Single Digit Representations of Natural Numbers". In this paper, he describes each natural number from 0 to 1000 in terms of each of the digits 1 through 9 in as few digits as possible and using only the five basic arithmetic operations.

For example, the entry for "282" written only using the digit "3" looks like this:

3 + 3 * (3 * (3 ^ 3 + 3) + 3)

Similarly, "466" written using the digit "7" looks like this:

7 * 77 + 77 / 7 - 77 - 7

This is also described in the first half of this video on the Numberphile youtube channel.

This challenge will ask you to write a program (or function, class, etc.) which takes two inputs and outputs a valid mathematical expression according to the rules used by Dr. Taneja. Your score will be calculated based on the size of your code and the quality of your outputs.


Write a program (or function, class, etc.) which takes two integers as input. The first of these will be an integer from 0 to 1000, inclusive. The second number will be a digit from 1 to 9, inclusive.

The program shall output an expression of the first input as a series of arithmetic operations on numbers whose digits are exclusively the digit specified as the second input.

In addition to parentheses ( and ), the following arithmetic operators are allowed in your output:

  • addition (represented by the +)
  • subtraction (represented by the -)
  • multiplication (represented by the *)
  • division (represented by the /)
  • power (represented by the ^)

Numbers, as stated before, must be composed entirely of the digit specified by the second input, but they can be any number of digits. Numbers can be positive or negative. Number tokens cannot contain any characters besides the digit specified and, if negative, a - at the beginning.

Any form and amount of whitespace, including no whitespace at all, is allowed around each token in your output, where "token" refers to a number, operator, or parentheses. Your program's output may not contain any other characters.

The order of operations are evaluated in the usual manner.


Dr. Taneja claims that his answers use "as few digits as possible"; unfortunately, his paper does not show how to prove that a solution is truly minimal. In the spirit of preserving that goal, however, the quality of your solutions will contribute to your score.

The quality of an output is measured, as Dr. Taneja says, by the number of digits used. For example, consider the output for "466" and "7":

7 * 77 + 77 / 7 - 77 - 7

This has a score of 1 + 2 + 2 + 1 + 2 + 1 = 9. Smaller is better.

Your code's score is:

score = 2 * <worst_scoring_output> + <solution_bytecount>

The lowest scoring valid solution wins.

You may determine your worst output either by test battery (trying every input) or logical proof. The following shell script can be used to compute your worst-scoring output, assuming your program is an executable which takes input as command-line arguments and provides output to stdout on a single line. When executing this script, provide the path to your program executable as a command-line argument.

// TODO: write a shell script which could compute your bonus score


Input and output may use any convenient method. This includes writing a function which takes the inputs as arguments to the function and returns or prints a string as described.

This is , so the size of your code in bytes forms a significant part of your score. The standard loopholes are forbidden, this includes parsing the original paper to "look up" Dr. Taneja's solution for a combination of inputs.

Reasonable adjustments to the output format are allowed; for instance: if your language has a native Expression object for mathematical expressions but its toString() (or equivalent) method represents the power operator as two asterisks rather than a caret, you may use the native representation.

Similarly, if it is easier for you to format your output as an equation rather than a one-sided expression, you may do so. For example, you may output <number> = <expression> or <expression> = <number> rather than simply <expression> (where the number is the first input to the program and the expression is the output as defined in the challenge section). Note that in this case, only the digit count of your expressions contribute to your score.

Example Input and Output

Your outputs may vary, but are still valid provided all explicitly-stated rules are respected.


282 3


3 + 3 * (3 * (3 ^ 3 + 3) + 3)


0 7


7 - 7


1000 9


999 + 9 / 9


466 7


7 * 77 + 77 / 7 - 77 - 7

  • \$\begingroup\$ Welcome to the site! I suggest including test-battery as a winning criterion here—in fact, this is almost exactly the challenge described in the tag wiki! \$\endgroup\$ – Khuldraeseth na'Barya Apr 22 '18 at 21:15
  • \$\begingroup\$ @Scrooble thank you! do you mean anything besides just tagging it as test-battery? i've added (to the top) a list of the tags i intend to use for this challenge. \$\endgroup\$ – Woodrow Barlow Apr 22 '18 at 21:23
  • \$\begingroup\$ Nope, that's all I meant! You can use [tag:tag-name] for the nice-looking tag links. \$\endgroup\$ – Khuldraeseth na'Barya Apr 22 '18 at 21:37
  • \$\begingroup\$ Are you allowed to concatenate, i.e. 36 = 3 + 33? (I vote no, but it's quite common to allow that.) \$\endgroup\$ – Nathaniel Apr 23 '18 at 4:20
  • \$\begingroup\$ @Nathaniel concatenation is not allowed as an operator in the output; however, your output is allowed to have multi-digit numbers as long as all of the digits are the digit specified. \$\endgroup\$ – Woodrow Barlow Apr 23 '18 at 13:52
  • \$\begingroup\$ I'd prefer a version that does not include the bonus. In fact, I think that it should instead be the primary scoring method, e.g. "each submission's score is the total number of digits used across all inputs". Bonus points in Code Golf tend to quickly become either ignored or mandatory \$\endgroup\$ – Kamil Drakari Apr 23 '18 at 21:02
  • \$\begingroup\$ @KamilDrakari i really like that suggestion. i was worried it would be hard to find a bonus point value that doesn't end up getting ignored because it isn't worth the extra bytes. \$\endgroup\$ – Woodrow Barlow Apr 23 '18 at 22:27
  • \$\begingroup\$ @KamilDrakari what do you think of your score being the score of your program + your bytecount, to preserve the golf aspect? \$\endgroup\$ – Woodrow Barlow Apr 23 '18 at 22:31
  • \$\begingroup\$ If you just score it the way I mentioned, adding up all digits used, then I think the golfing would just be way too inconsequential in comparison. Maybe "highest number of digits + byte count"? It's not that different from the current scoring, but it avoids weird cutoffs and negative points so it's more likely that both parts matter. \$\endgroup\$ – Kamil Drakari Apr 24 '18 at 4:16
  • \$\begingroup\$ @KamilDrakari i've re-written the scoring rules. please let me know what you think. i think it strikes a nice balance between golfing and quality outputs. \$\endgroup\$ – Woodrow Barlow Apr 24 '18 at 14:34
  • \$\begingroup\$ @WoodrowBarlow Looks good. I think having that script prepared that will calculate the score is definitely a good idea since manually testing all 9000 input possibilities would be tedious. It would also be nice if that script, in addition to outputting the score (or just "highest_scoring_output" since having it count bytes as well might be problematic), also said which input pair is responsible. \$\endgroup\$ – Kamil Drakari Apr 24 '18 at 14:39
  • \$\begingroup\$ The problem space of expression-building has been fairly well explored, and there are a few questions which this might be considered to duplicate. I personally would vote to close as a duplicate of codegolf.stackexchange.com/q/32085/194 : adding exponentiation as a supported operator and a loop to find the smallest k are both trivial changes IMO. \$\endgroup\$ – Peter Taylor Apr 25 '18 at 9:58
  • \$\begingroup\$ Wait… just how does the length of my code affect my score? \$\endgroup\$ – Nissa Apr 25 '18 at 13:22
  • \$\begingroup\$ @StephenLeppik your score is: 2 * [worst_scoring_output] + [length_of_code]. lower score is better. is that not clear? \$\endgroup\$ – Woodrow Barlow Apr 25 '18 at 13:52

ASCII-betical order

The full printable ASCII set (chars 32-126) in ascending order looks like this:


Your task is to create a program or function which takes in an integer n and outputs the nth item of an OEIS sequence. The catch: your score is the length of your code, plus thrice the number of ASCII chars in your code not included in the longest strictly increasing substring.


A valid strictly increasing substring of your code is defined as so:

  • Each character within the substring, except the first, must have a higher ASCII value than the previous.

Other rules:

  • The sequence may be 0- or 1-indexed.
  • The sequence you choose may not be a constant sequence, such as A000012.
  • Entries may only include printable ASCII and newlines. Newlines are ignored when calculating the longest increasing substring.


This JavaScript function outputs A000027 (the natural numbers):


The longest increasing substring of characters is =>[]^n. Therefore, the score is 7 + 3 × (7 - 6) = 10.

This function does exactly the same thing:


The longest increasing substring is =>[]^_abcdefghijklmnopqrstuvwxyz. The score is 59 + 3 × (59 - 32) = 140.

This function also does the same thing:


The longest increasing substring is !"#$%&'()*/=>[]^n. Therefore, the score is 20 + 3 × (20 - 17) = 29.

The lowest score wins.

Sandbox questions

  • Is the scoring system sufficient?
  • Is there a better option for the challenge?
    • Output n'th item in some OEIS sequence
    • Output first n items of some OEIS sequence
    • Output the score of the input
    • etc.
  • What about languages with extended code pages? Should only the ASCII chars be counted, or should the bytes of the program be considered instead?
  • Any other concerns?
  • \$\begingroup\$ I like this version much better. Some ideas: maybe force to choose a non costant oeis sequence? I'm not sure about scoring, seems that standard golfing without caring about increasing ascii can still get a pretty good score... Rule to use only ascii chars in code seems to just be a limit to golfing languages. For the title, i liked the term ASCIIbetical order used in a comment to the previous proposal .) \$\endgroup\$ – Leo Dec 1 '16 at 23:41
  • \$\begingroup\$ @Leo I like the non-constant-sequence idea. I've modified the scoring system to make it a little easier to use (see examples), but I still don't like it because if you need to use that variable name 3 times, it's not worth it, and if you only need it once, it's doubly worth it. Perhaps something taking into account the number and length of individual runs would work better, but I have no idea how to work that out... \$\endgroup\$ – ETHproductions Dec 2 '16 at 4:49
  • \$\begingroup\$ Would something like f n = if ( length "#$% ... |}~" == 92) then n else 0 be valid? For golfing languages the length check won't be much overhead. \$\endgroup\$ – nimi Dec 2 '16 at 15:18
  • \$\begingroup\$ I'd prefer the scoring to work like this (which would be simpler): length, plus three times the number of characters in your code which are not part of the longest substring. That'd prevent people putting the whole ASCIIbet in a string and using a checksum (or length check) to make sure it hadn't been changed. In other words, it'd would encourage programs which were mostly formed of increasing ASCII, rather than programs which contained a lot of increasing ASCII (i.e. favour proportion over length). \$\endgroup\$ – user62131 Dec 6 '16 at 11:27
  • \$\begingroup\$ @nimi Sorry, I missed your comment. I suppose that would be valid. The scoring system would have to account for that. \$\endgroup\$ – ETHproductions Dec 6 '16 at 17:22
  • \$\begingroup\$ @ais523 I like that idea; however, as it's currently written, that would give the first two examples scores of 16 and 140, respectively. I'd like to give the second example a much lower score, if possible, but I'm not sure how to do this, and it may not be a good idea. What do you think? \$\endgroup\$ – ETHproductions Dec 6 '16 at 17:28
  • \$\begingroup\$ I think it's a bad idea. The second example isn't actually any different from the first in terms of AST, it just has a bunch of junk characters written somewhere (in this case, a variable name) that the program doesn't parse them (with an equal number of characters outside to balance). \$\endgroup\$ – user62131 Dec 6 '16 at 17:32
  • \$\begingroup\$ @ais523 OK, I've changed the scoring algorithm. I've also removed the "you can't be able to remove any chars in the substring" rule for now, as it now seems superfluous. \$\endgroup\$ – ETHproductions Dec 6 '16 at 17:47
  • \$\begingroup\$ The problem with this scoring is that the winner would probably be a 2 byte program in a golfing language scoring 2 points (or a 1 byte program scoring 1 point, I bet someone will find it). \$\endgroup\$ – Leo Dec 7 '16 at 16:37
  • \$\begingroup\$ @Leo: I guess even 0 bytes is possible by using a language that takes implicitly input, puts in on the stack (does nothing for 0 bytes) and outputs the stack when the program stops. This is sequence A000027, the natural numbers . \$\endgroup\$ – nimi Dec 7 '16 at 19:42
  • \$\begingroup\$ @Leo Yes, that's a very good point. It's so hard to come up with a good formula.... Do you have any more ideas? \$\endgroup\$ – ETHproductions Dec 7 '16 at 21:44
  • \$\begingroup\$ @ETHproductions Maybe (length of longest increasing sequence)^2/(length of whole program) ? Highest score wins, max would be 95 for a program composed by exactly the whole sequence \$\endgroup\$ – Leo Dec 8 '16 at 8:33
  • \$\begingroup\$ @Leo The problem with that would be it goes back to rewarding solutions containing the whole string with a length check... (For example, n=> !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~"[90]?m:n in JavaScript scores 84.346) \$\endgroup\$ – ETHproductions Dec 9 '16 at 3:25
  • 1
    \$\begingroup\$ A separate comment: you should replace "ASCII chars" with "bytes". That would allow more languages to compete. \$\endgroup\$ – Nathaniel Apr 27 '18 at 14:12
  • 1
    \$\begingroup\$ As a follow-on to my previous comment about the identity function being shorter in some languages, as things stand I could submit a zero-byte answer in GolfScript. When a zero-byte answer is possible, the question has a serious problem. \$\endgroup\$ – Peter Taylor Apr 27 '18 at 18:00

The Derby Stakes

The Derby is a horse race run (near) annually since 1780 near the town of Epsom and is the middle leg of the Triple Crown. The race covers one mile, four furlongs, and six yards. Over the years it has had as few as 4 and as many as 34 entries. The fastest winning time was in 2010 at 2m31.33, but I'm old-fashioned and like horse race results only to the nearest fifth of a second. So let's call it 2:31 2/5.

The task

We will ID horses by number from 0 or 1 (your choice) up to the number of horses in the race (could be anything from 4 to 34).

The task is to take a list of the average horse speeds in miles per hour and produce a report of the race. The report has:

  1. the winning horse ID and its winning time;
  2. the other horse IDs in their order of finish and the difference between each time and the time of the horse that finished ahead of it.


  1. the times are to the nearest fifth of a second. There can be ties, which must be listed in random order;
  2. some horses might not finish the race, in which case their average speed is 0.


The input will have between 4 and 34 non-negative numbers (average speeds) rounded to one decimal place, in any convenient format for your language. The horse ID is just the index (starting at 0 or 1) of their average speed.

The output can be any convenient format, as long as the winner is "first" (in the sense of top-most or left-most) and the other horses are listed in a way that makes their finishing order clear.

The times have to be rounded to the nearest fifth of a second, but otherwise can be represented however you like, such as the number of seconds, or some format like 2:31.4 or whatever. If the time is in "seconds flat", as they say, either 2:32.0 or 2:32 (for example) is fine.

For horses that don't finish the race, no time difference really makes sense, so anything that just indicates they didn't finish is fine. But the do have to appear in the report, and if more than one horse does not finish they have to be listed randomly.

The point with the output is that it needs to be clear, but the format itself is not so important.

You also need to state in your answer how you have ensured that ties are output in a random order.

Test case

31.8, 0, 32.9, 32.1, 30.8, 31.8, 31.1, 33.1  

8 2:43.6
3 0:01.0
4 0:04.0
1 0:01.6
6 0:00.0
7 0:03.8
5 0.01.8

"I made the ranking of ties random by using `rank(x, ties.method = "random")` in R."

I chose a particular time format and used "DNF" for the horse that didn't finish, but you don't need to adhere to these.

I will make more test cases if this challenge seems interesting to anyone.

This is . Standard loopholes forbidden.

  • \$\begingroup\$ I assume .5 is expected to be rounded to .6 ? \$\endgroup\$ – JayCe Jun 11 '18 at 2:12
  • \$\begingroup\$ Do the times get rounded to 1/5 of a second before calculating the differences, or are the differences rounded to 1/5 of a second? \$\endgroup\$ – JayCe Jun 11 '18 at 2:20
  • \$\begingroup\$ Also you might want (or not) to clarify how many miles is 1 mile, 4 furlongs and 6 yards. Or it could be another code-golf challenge if it doesn’t already exist :) \$\endgroup\$ – JayCe Jun 11 '18 at 2:22

How far can I throw a ball up the hill?

Given an initial velocity, an angle to throw the ball, and the angle of a hill, determine how far the ball will go before it hits the ground due to gravity.

Refer to the image below. V=initial velocity vector, a1=angle of hill, a2=angle ball is thrown at, d=distance ball will travel

Highly detailed image


  • The velocity the ball is thrown at (m/s)
  • The angle from level ground that the ball is thrown at (radians or degrees)
  • The angle from level ground the hill is at


  • The number of meters the ball travels before hitting the ground


  • Gravity will pull the ball downwards at 9.81 m/s^2
  • Due to floating point rounding, the answer only need be correct within 0.1 of my test case examples
  • The input angle can be between 0 and pi/2 radians (or 90 degrees)
  • An angle of 0 means the ball is thrown horizontally across the ground, which hits the ground at 0 meters
  • An angle of pi/2 radians (or 90 degrees) means the ball is thrown vertically and hits the ground at 0 meters
  • The angle of the hill will always be less or equal than the angle the ball is thrown at
  • This is code golf, fewest bytes wins

Related to task 3

Test cases

pi/4  (20, 0.7853981633974483, 0) = 40.8
pi/5  (20, 0.6283185307179586, 0) = 38.8
pi/6  (10, 0.5235987755982988, 0) = 8.8
pi/3  (10, 1.0471975511965976, 0) = 8.8
pi/2  (5,  1.5707963267948966, 0) = 0.0
0     (5,  0,                  0) = 0.0


I'm not sure what standard rules are used when accounting for floating point errors, is allowing a .1 margin of error fine?

Test cases for angles > 0 to come.

  • \$\begingroup\$ I think allowing both degrees and radians would be better. \$\endgroup\$ – Esolanging Fruit Apr 30 '18 at 6:20
  • \$\begingroup\$ @PeterTaylor Added a bit more to the challenge. \$\endgroup\$ – aoemica May 1 '18 at 6:23
  • \$\begingroup\$ I think that in light of the changes, some of the points need to be updated. Point 3 talks about the input angle (singular), but may no longer be necessary at all. Point 4 also looks out of date. With respect to floating point errors, that's tricky to do well. IMO the best approach is to specify a range of input values and an error (relative or absolute) which must be obtained within that range; but to verify that the requirements are reasonable may require numerical analysis. \$\endgroup\$ – Peter Taylor May 1 '18 at 7:34
  • \$\begingroup\$ While this is mathematically interesting, it does not allow for very much solution variation. \$\endgroup\$ – Beefster May 1 '18 at 16:03
  • \$\begingroup\$ 1. Regarding floating point inaccuracies, you can visit this meta post to get some inspiration and guidance. 2. I think physics is a good fit here. 3. In its current form, if posted, the challenge is very likely to be closed as Unclear what you are asking, since you don't include a way to compute that distance. I suggest adding this to the challenge \$\endgroup\$ – Mr. Xcoder May 1 '18 at 18:37
  • \$\begingroup\$ I don't think such challenges (where there is a formula, and deriving the formula is the main part of the challenge) have many interesting solutions. Explicit form. (cc @Mr.Xcoder ) \$\endgroup\$ – user202729 May 2 '18 at 12:18
  • \$\begingroup\$ You should make it clear that the output is the horizontal length traveled (instead of the arc length) \$\endgroup\$ – user202729 May 2 '18 at 12:19
  • \$\begingroup\$ @user202729 I myself know how to solve the challenge in its current form, however I just pointed out that a larger audience might not receive this too well. Perhaps this challenge isn’t really suited for our format on PPCG. (P.S. the length is not horizontal, it’s along the inclined plane.) \$\endgroup\$ – Mr. Xcoder May 2 '18 at 12:21
  • \$\begingroup\$ @Mr.Xcoder For the current test cases they're equivalent anyway. \$\endgroup\$ – user202729 May 2 '18 at 12:26


A typical stack-based lanaguage uses nilads, monads, dyads, and occasionally operations with larger numbers of arguments. As each operation is processed, the values are stored on a stack, so that if you want to apply an operation to the result of other operation(s), you simply invoke those operations (and predecessor operations if necessary), at which point their results will end up on the stack ready for the operation you provide next.

Queuez is a fictional language which works differently: instead of the results being stored on a stack, they are stored in a queue, so you have to be careful how you manage your order of operations. For example, let's imagine that digits are nilads and - and / are dyads and you want to calculate ((a / c) - f) / ((b / d) - e). In a stack-based language you would write this as ac/f-bd/e-/, however in Queuez you need to write this as acbd/f/e--/. Explanation:

Command Stack
a       a
c       a c
/       a/c
f       a/c f
-       a/c-f
b       a/c-f b
d       a/c-f b d
/       a/c-f b/d
e       a/c-f b/d e
-       a/c-+f b/d-e
/       (a/c-f)/(b/d-e)
Command Queue
a       a
c       a c
b       a c b
d       a c b d
/       b d a/c
f       b d a/c f
/       a/c f b/d
e       a/c f b/d e
-       b/d e a/c-f
-       a/c-f b/d-e
/       (a/c-f)/(b/d-e)

Please write a program or function that converts a program written in a fictional stack-based language into the equivalent Queuez. The actual nilads, monads and dyads of the two lanaguages will be arbitrarily mapped to three sets of at least 10 printable ASCII characters of your choice (you are not required to use letters or mathematical symbols); your answer only has to output the correct rearrangement based on which characters you have chosen to represent nilads, monads and dyads. You can assume that there will be one value left on the stack/queue at the end of the program. Normal rules apply.

If you prefer, you can reverse the dyad argument order, but this must apply both to the original stack-based language and the output. For instance, for the input edb/-fca/-/ you would need to output dbcae/f/--/:

Command Stack
e       e
d       d e
b       b d e
/       b/d e
-       b/d-e
f       f b/d-e
c       c f b/d-e
a       a c f b/d-e
/       a/c f b/d-e
-       a/c-f b/d-e
/       (a/c-f)/(b/d-e)
Command Queue
d       d
b       b d
c       c b d
a       a c b d
e       e a c b d
/       b/d e a c
f       f b/d e a c
/       a/c f b/d e
-       b/d-e a/c f
-       a/c-f b/d-e
/       (a/c-f)/(b/d-e)
  • \$\begingroup\$ By "ASCII" did you mean "characters" or "symbols"? (can we use SBCS?) \$\endgroup\$ – user202729 May 5 '18 at 14:52
  • \$\begingroup\$ I think this needs to be narrowed from "Make up a language with the operators of your choice, and then do the real task on the basis of that language". That has the potential to move the golfing from the program to the definition of the task, e.g. by making all the dyads commute in order to simplify the transformations, or making all the monads no-ops so that they just serve to rearrange the queue, or having some operators which build lists and an eval dyad which evaluates a string over a list in a stack-based way, or ... \$\endgroup\$ – Peter Taylor May 5 '18 at 17:18
  • \$\begingroup\$ @PeterTaylor Sorry, that was not the intent of the challenge; thank you for pointing out that it needed clarification. Hopefully this new wording will indicate that the meaning of the operators is unknown and therefore the work lies in finding the appropriate rearrangement. \$\endgroup\$ – Neil May 5 '18 at 20:37
  • \$\begingroup\$ @user202729 You don't know the underlying SBCS of the language because you don't need to know, all you know is that it has been mapped to three sets of printable ASCII characters of your choice. \$\endgroup\$ – Neil May 5 '18 at 20:39
  • \$\begingroup\$ @Neil Then call them "bytes". \$\endgroup\$ – user202729 May 6 '18 at 2:00

Multidimensional orthodiagonal steps

Step further in step generation.
Navigation in 2d matrix is common, but something uncommon is even more insteresting.
Now I'll ask you to develope shortest solutions to generate all possible steps in N-dimensional matrix.


Your code takes positive integer number N > 0 as input.
Your code must output all possible steps in matrix of N dimensions. In other words, you need to output coordinates of all cells that touch (0;...;0) cell in any way.


Input: 1


Input: 2


Input 3: Output:

(0,1,0)   (0,1,1)    (0,1,-1)
(0,-1,0)  (0,-1,1)   (0,-1,-1)
(1,0,0)   (1,0,1)    (1,0,-1)
(-1,0,0)  (-1,0,1)   (-1,0,-1)
(1,1,0)   (1,1,1)    (1,1,-1)
(1,-1,0)  (1,-1,1)   (1,-1,-1)
(-1,1,0)  (-1,1,1)   (-1,1,-1)
(-1,-1,0) (-1,-1,1)  (-1,-1,-1)


  1. Standart loopholes are disalowed
  2. Input number is always integer and always greater than zero
  3. Output order is not relevant
  4. Output is flexible. Coordinates just need to be distinguishable
  5. This is code-golf, so shortest answer in bytes wins
  • \$\begingroup\$ Some typos: insterestinginteresting, developedevelop, StandartStandard, disaloweddisallowed \$\endgroup\$ – wastl May 19 '18 at 9:04

Fridge magnet substitution (posted)

  • \$\begingroup\$ You should add 6 = 9, 7 = L, and C = U. If you're willing to have nuance like intransitivity and directionality, you could add Z = 2 (but not Z = R) H = I (but not H = 1), 7 = T (But not T = L), P => D (D doesn't really work for P, but P acts like a lower case d if you flip it over), Q => O (sorta works if you flip over Q), T => I (also flipped), V => C/U \$\endgroup\$ – Beefster May 14 '18 at 23:20
  • \$\begingroup\$ Good additions! I thought of having directionality, but I wanted to keep the commutation that message A->B implies B->A. I'll add the other ones though. \$\endgroup\$ – maxb May 15 '18 at 7:00

Treasure Map Drawing Bot

You're organizing a treasure hunt for your friends. To conduct things more easily, you want to draw a map of all locations where you hid the precious objects.


A string consisting of the locations of the objects separated by newlines. Each location is represented by two non-negative integer coordinates describing the position in the field by an x- and y-coordinate, 0 0 being the upper left corner. Example:

1 2
3 0
0 1


Your function or program should be able to construct a map denoting every given location with an x where the mark is found in row y + 1 and column x + 1 in the output. Unmarked locations are represented with a . The map also consists of a frame where the corners are +s, the vertical lines are |s and the horizontal lines are -s. Map for the input example given above:

|   x|
|x   |
| x  |

Possible Test Cases

"0 0"

"0 10
 5 5
 10 0"
|          x|
|           |
|           |
|           |
|           |
|     x     |
|           |
|           |
|           |
|           |
|x          |


"0 0
 0 2
 2 0"
|x x|
|   |
|x  |

Of course, this is , meaning that the solution with the lowest byte count wins!

  • 1
    \$\begingroup\$ You may wish to clarify that the (0,0) position is the upper left corner. This is common in computer graphics in my experience, but not any other field. If that isn't correct, then your test cases seem to be wrong (and you should clarify where the (0,0) position is anyway). \$\endgroup\$ – Kamil Drakari May 14 '18 at 17:21
  • \$\begingroup\$ Thanks, did that. \$\endgroup\$ – racer290 May 14 '18 at 17:57

Is it a table?


Take an bitmap (in .bmp, 2D array, string containing line break, etc.), check whether it's a table, i.e. the outer border is a rectangle, and each separated (edge-connected) empty part in it is a rectangle.

Equivalently, each space area, including the outside one, has exactly 4 edges.

You can't assume there is or is no space outside of the table.

Test cases

True samples:

#           #
#           #
#     #     #
#     #     #
#         #
#         #
#         #
#   #     #
#####     #
# # #######
# # #     #
##   #   ##
##   #   ##
##### #####
######   ##
######   ##

Note that in this test case the smallest space area (with 1 space) is a rectangle, although it doesn't have a rectangular border with #.

False Samples:

#     #   #
#     #   #
#     #####
#         #

In this test case the lower-left empty region is a concave hexagon.

#    #     #
#    #     #

The outer border is not a rectangle.

#          #
#          #
#   ####   #
#   #  #   #
#   ####   #
#          #

The outer empty region has a "hole" inside.

Winning criteria

Shortest code win.

  • \$\begingroup\$ This is massively unspecified, and based on the test cases I'm not even sure why the truthy cases are tables and some of the falsey cases not.. As stated in the Sandbox description: "Write your challenge just as you would when actually posting it.", which this clearly isn't.. I'm assuming you just had this challenge as a idea and don't have the time yet to fully specify it, in which case it would be better to save it in a local Notepad doc until you have time to specify it, before posting it here. I do the same pretty often. \$\endgroup\$ – Kevin Cruijssen May 25 '18 at 11:40
  • \$\begingroup\$ @KevinCruijssen the whole one is a rectangle, and each separated part is a rectangle \$\endgroup\$ – l4m2 May 25 '18 at 11:43
  • \$\begingroup\$ Why is the third test cases truthy? Do we look at the # for the outer rectangle, but the spaces for the inner ones? If either both # this third truthy case would be false; and if both space (somewhat), I could imagine the second falsey case to be true. \$\endgroup\$ – Kevin Cruijssen May 25 '18 at 11:51
  • \$\begingroup\$ @KevinCruijssen Also Notepad doesn't help wording, so your suggestion is bad \$\endgroup\$ – l4m2 May 25 '18 at 12:43
  • \$\begingroup\$ I meant creating the answer with correct formatting here, then save it in Notepad++ until it's fully specified (which is what I do). Alternatively you can keep it open in your browser tab until it's fully specified before clicking the "post"-button. Either way, it clearly isn't "Write your challenge just as you would when actually posting it." right now.. That's all I wanted to state. It does look like an interesting challenge when it is fully specified, though. So you'll have my future upvote when it is. \$\endgroup\$ – Kevin Cruijssen May 25 '18 at 12:52
  • \$\begingroup\$ I think you should prove clearly how [it clearly isn't "Write your challenge just as you would when actually posting it."] @KevinCruijssen \$\endgroup\$ – l4m2 May 25 '18 at 12:56
  • \$\begingroup\$ I don't understand this. Why is the third truthy test case truthy? Why are the first and third falsey test cases falsey? \$\endgroup\$ – AdmBorkBork May 25 '18 at 13:28
  • \$\begingroup\$ @AdmBorkBork its outer border is a 8*11 rect, and there are 4 empty rect parts in it, 3 2*3 and a 1*1. First false test contain an empty part with 6 edges, 3rd one with 8 edges \$\endgroup\$ – l4m2 May 25 '18 at 14:00
  • \$\begingroup\$ So this is also a truthy example? \$\endgroup\$ – user202729 May 29 '18 at 5:00
  • \$\begingroup\$ @user202729 True. Another true case I think maybe necessary is ### (no space area)? \$\endgroup\$ – l4m2 May 29 '18 at 8:00

Formic Functions 2: Hierarchies

     Watch live | Active answers | Add new answer | Chat room | Source code | Leaderboard

               This is an open-ended challenge. New answers and updates are always welcome.

We've seen the territorial highland ants wander the lands in search of food. There have also been talks of aggressive forest ants lurking down below whilst feasting on fungi. These ants, however, are unlike any you've seen before. Collecting food is their goal, as always, but colonies don't need a queen - all ants of this subfamily are capable of creating offspring. There's a catch, though. New ants have less strength than their parent, unless the parent ant spends a large amount of food on them. Ants are capable of killing others weaker than them, and weaker ants are worth less, so your ants need to make careful choices.

Watching the game

Because this competition is made in JavaScript, you can see what other players have made already directly in your browser. Just click on one of the links above - you'll figure out the rest.


To keep the comment section clean, I highly encourage you to use the dedicated chat room for questions and discussion.


No official tournaments available yet.


Here are a couple of images taken at the end of a game that should entice you:

No screenshots available yet.

You can look at more simply by running games in your browser. The controller also allows you to zoom in so you can track the ants' decisions that result in the large-scale patterns you ultimately see.


A value range is indicated by a number followed by two dots and another number. "0..7" means "between 0 and 7, inclusive". A different type of range is indicated by a number followed by a plus sign. "1024+" means "greater than or equal to 1024".


Your task is to provide a JavaScript class that implements a getAction(view) function and optionally a constructor. All code provided by you will be run under strict mode.

The constructor is called with no arguments. It must take no more than one second to run. It must be side-effect free and must consistently result in an identical object after every run.

The getAction(view) function (henceforth known as the ant function) is called like this: yourConstructedObject.getAction(someViewArray);. It must also be side-effect free and must consistently return an object, and provide the same output every time if provided the same input.

At the start of every game, your ant function is granted 1 second of reserve time. It is also granted an additional 10 milliseconds of time every time it is called. Time measurement starts right before the function call and ends upon return. That time is then taken away from the reserve time. Your function must not run out of the reserve time.

The constructed object and any temporary data combined must not consume more than 64 MB of memory at any time.

Here's a template for you to fill in:

class Entry { // TODO: Change the "Entry" to whatever you want, preferably your entry name. Remember that this is a JavaScript class name, so you can't use some characters.
  constructor() {
    // TODO: Optionally fill this in.

  getAction(view) {
    // TODO: Fill this in.


The arena is a toroidal (edge wrapping) grid of hexagonal cells arranged in a rhombus of side length 1500. Initially, all cells have the color 0 and 0.1% have food placed on them, for a total of 2250 pieces of food. Up to 12 entries are randomly chosen every game and have their corresponding ants (one per entry) spawned in random locations, with no more than one ant occupying any cell. The initial turn order of ants is also randomized.


The game lasts 32768 turns, which are processed sequentially. Every turn, in an unchanging order, each ant has its corresponding function called with the local state of the arena passed as an argument. In other words, your ant function is called separately for every ant you control. The action chosen by the ant function is then taken immediately - other ants see the arena changes even during the same turn.



Each ant carries the following properties with them:

  • Position: Implementation-defined.
    • Ants do not have access to their global position.
  • Tier: An integer representing the tier of an ant.
    • Equal to 0 for all ants not spawned by other ants.
    • Doesn't change over an ant's lifespan.
    • The lower its tier, the less valuable an ant is.
    • There is no theoretical limit on how high or low tier can be.
  • Age: An integer representing how old an ant is.
    • Initially 0 for all ants.
    • After every action an ant takes its age goes up by one.
  • Food: An integer representing the amount of food an ant is carrying.
    • Initially 0 for all ants.
    • An ant may carry an unlimited amount of food.
  • Memory: An integer in the range 0..15 representing the memory of an ant.
    • Initially equal to 0 for all ants not spawned by other ants.
    • An ant may change it as part of any action.
    • Holds no meaning for the controller.
  • Owner: Implementation-defined.
    • Allows ants to see whether another one is a friend or not.


Each ant sees 6 cells in its neighborhood, as well as the cell it is currently occupying. Each cell contains the following properties:

  • color: An integer in the range 0..7 representing the current color of a cell.
  • food: A boolean representing whether or not a piece of food is present on a cell.
  • ant: An object if an ant is present on a cell, undefined otherwise. If present, the object has the following properties:
    • tier: Maps to Tier.
    • age: An integer in the range 0..3 representing an ant's age on a logarithmic scale. 0 actual age maps to 0, 1..31 actual age maps to 1, 32..1023 actual age maps to 2 and 1024+ maps to 4. [Sandbox note: I'm torn about whether or not the ants should have access to this information.]
    • [Sandbox] Alternative for or complement to age: older: A boolean representing whether or not an ant's age is higher that of the currently active ant.
    • food: An integer in the range 0..3 representing the amount of stored food by an ant on a logarithmic scale. 0 actual food maps to 0, 1..3 actual food maps to 1, 4..15 actual food maps to 2 and 16+ actual food maps to 3.
    • memory: Maps to Memory.
    • friend: A boolean representing whether or not an ant is under control of the same entry as the currently active ant.

Every turn sight is passed to your ant function as an array of 7 objects, arranged in this order:

 0 1
5 6 2
 4 3

The view is randomly rotated by a multiple of 60 degrees every time your ant function is called. This means that it's impossible to extract a consistent sense of direction without clever use of the environment.

Accessing one's own state is done by retrieving the ant object of the 7th element of the array, like this: const me = view[6].ant;


Each ant must perform an action every turn. The possible actions are as follows:

  • Move: Move to the desired cell.
    • Format: {cell: 0..6, action: 0}
    • When moving onto a cell with food on it, an ant automatically picks up the piece of food present.
    • When moving onto a cell with an ant on it, of the two ants only one remains. If the still ant's tier value is lower than that of the moving ant, then the still ant dies and is permanently removed from the game. Otherwise, the unfortunate fate awaits the moving ant instead. In either case, the remaining ant gains all the food of the ant that died and one more.
    • Staying still is a valid action and may hold merit in certain situations.
  • Paint: Change the color of the desired cell to the desired color.
    • Format: {cell: 0..6, action: 1, color: 0..7}
    • Changing the color of a cell to the same one is a valid action.
  • Place food: Put a piece of food onto the desired cell.
    • Format: {cell: 0..6, action: 2}
    • Placing a piece of food onto a cell with an ant on it results in that ant immediately picking it up.
    • Placing food costs food. An ant may only place a piece of food if it has at least one food stored, and one food is taken away from it.
    • Placing food onto own cell is a valid action, but placing food onto a cell with a piece of food already on it is not.
  • Spawn: Create a new ant under control of the same entry on the desired cell.
    • Format: {cell: 0..6, action: 3, tier: -1..+1, state: 0..15}
    • A parent ant must choose to spend 1, 4 or 16 food on the new ant by specifying a tier equal to -1, 0 or +1 respectively.
    • A parent ant may only spawn an ant of the desired tier if it has the necessary amount of food to do so, and that amount is taken away from it.
    • Spawning results in an ant of a tier equal to that of its parent plus the tier value specified.
    • A parent ant must choose the initial memory of the new ant by specifying a state. [Sandbox note: I'm considering not letting parents set the initial memory of offspring.]
    • The new ant is inserted into the turn order right before its parent.
    • Spawning an ant counts as movement for the new ant, and is resolved as such.
    • Spawning an ant onto own cell is a valid action and may hold merit in certain situations.
  • Memorize: Replace the currently active ant's memory with the desired number.
    • Format: {memory: 0..15}
    • Memorization is not an action itself. It must be appended to one of the four previously mentioned actions.
    • An ant may choose to leave this field undefined. No memory change occurs then. [Sandbox note: Torn about this as well.]

Format refers to the JavaScript object that must be returned by your ant function to perform a specific action.

Example outputs:

  • {cell: 0, action: 0} // Move to cell 0.
  • {cell: 6, action: 0} // Do nothing.
  • {cell: 6, action: 1, color: 0} // Set own cell color to 0.
  • {cell: 3, action: 2} // Put a piece of food onto cell 3. Invalid if there is a piece of food on cell 3.
  • {cell: 2, action: 3, tier: -1, state: 3} // Spawn an ant 1 tier lower than that of your own with initial memory equal to 3 on cell 2 at the cost of 1 food. Invalid if you have no food.
  • {cell: 5, action: 3, tier: +1, state: 11, memory: 14} // Spawn an ant 1 tier higher than that of your own with initial memory equal to 11 on cell 5 at the cost of 16 food whilst committing 14 to memory. Invalid if you have less than 16 food.
  • {cell: 6, action: 3, tier: 0, state: 0} // Spawn an ant with tier equal to that of your own with initial memory equal to 0 on your own cell at the cost of 4 food. A conflict occurs and the newly spawned ant dies, giving you a piece of food. Effectively, ou lost 3 pieces of food. Invalid if you have less than 4 food.
  • {cell: 6, action: 3, tier: +1, state: 0} // Spawn an ant 1 tier higher than that of your own with initial memory equal to 0 on your own cell at the cost of 16 food. A conflict occurs and you die, giving the newly spawned ant a piece of food (and any additional food you might've had). Effectively, you upgraded yourself at the cost of 15 food. Invalid if you have less than 16 food.
  • {cell: 3, action: 0, color: 5, test: true, memory: 1} // Move to cell 3 whilst committing 1 to memory. Values ofcolorandtestare irrelevant. You may choose to include unrelated properties - they are simply ignored. [Sandbox note: Maybe the controller shouldn't allow color in this case? This looks like it might be a good breeding ground for hard-to-detect bugs.]

Example invalid outputs (not exhaustive):

  • {cell: 7, action: 0} // Error: Value of "cell" outside required range 0..6.
  • {cell: 3} // Error: "action" left undefined.
  • {action: 1, color: 4} // Error: "cell" left undefined.
  • {cell: 4, action: 3, tier: -1} // Error: "state" left undefined.


At the end of each game, every entry participating in that game is evaluated according to this formula:


n is the number of ants belonging to the entry currently being evaluated
Ti is the tier of the ith ant belonging to the entry currently being evaluated
Fi is the amount of food the ith ant belonging to the entry currently being evaluated is holding

Or as pseudocode:

var evaluation = 0;
for (int i = 0; i < entry.ants.count; i++) {
  var ant = entry.ants[i];
  evaluation += Math.Pow(2, ant.tier) + Math.Pow(2, ant.tier - 1) * ant.food;

[Sandbox note: Should I count food as score or not?]

The final score of each entry is equal to the amount of entries whose evaluation was lower than that of the currently considered entry. This means that the highest score an entry can get is 11.


A tournament is nothing more than a series of individual games. At any point of a tournament the score of each entry is equal to the average of all of its scores.

Official tournaments

Official tournaments will be run by me on the latest version of Chrome on my personal computer, which at the time of writing is an AMD FX-8350, every time a new entry is posted or an existing one receives a meaningful edit. The leaderboard will be updated once the 1st place becomes consistent between 6 subsets of played games (which gives a probability of 96.875% that the first place won't change).

As stated before, there is no permanent winner. This means that no checkmark will be awarded to any answer, ever. I will continue to run new tournaments for as long as is practical.


Each submission must follow this general format:

Entry Title

[Optional text and pictures]

Code block with your JavaScript class

[Optional text, pictures and code blocks]

Not adhering to the above format may result in the submission not being properly picked up by the controller.

Explanations and pictures are highly encouraged, though not necessary. Making your entry pretty and well-documented will entice me (and probably a lot of people) to upvote it.


Your entry will be disqualified if it is found to not adhere to the specification correctly. Most of the time problems will be caught by the controller and reported, but some rules can't be checked programmatically, so they'll be enforced by hand. I reserve the right to disqualify an entry manually if it breaks the rules or if I subjectively believe it has not been made with fair competition in mind. I hope I haven't left any loopholes (and therefore won't have to exercise this power), but if I did then this rule prevents them from potentially ruining the entire challenge.

Disqualification during a game results in all of your ants being immediately and permanently removed from the turn order. Disqualification during a tournament invalidates all games in which your entry has participated and terminates the active game (as it would've been invalid anyway). Disqualification during an official tournament, aside from having the effects of a regular tournament disqualification, prohibits your entry from taking part in any future official tournament until meaningfully edited.

This is not supposed to be an additional challenge - helpful error messages along with your ant function's output and the input that disqualified it will be attached to the disqualification notice. I will also paste these messages into a comment informing you of a disqualification should your entry be disqualified during an official tournament.

Multiple entries and editing

You may provide multiple entries, provided that they do not team up against the others. As long as each entry is working solely towards its own victory, you are permitted to tailor your strategy to take advantage of others' weaknesses. You may also edit your answers whenever you choose. It is up to you whether you post a new entry or edit an existing one; just don't flood the game with nearly identical variations. If you make a variation of another person's entry, remember to give them credit by linking to their entry from your own.

Example entries


This entry demonstrates the minimal amount of code needed for something to be considered valid. It returns the same output regardless of input.

class Randant {
  getAction() {
    return {cell: 0, action: 0};

Because the orientation is random every time, Randant will perform a random walk instead of going straight.

Despite being very simple, this is a perfectly valid submission. It won't ever get disqualified either, because moving is guaranteed to be a valid action.

See Smart Randant on how one could go about improving this entry.

Smart Randant

This entry demonstrates basic mechanics that ought to be commonly used between more advanced entries. It is an improvement of Randant's general design. It depends on input this time around.

class SmartRandant {
  getAction(view) {
    const me = view[6].ant; // Get data about myself

    function wrap(number, cycleLength = 6, negativeSafety = 1) { // Defining functions inside the getAction(view) function is fine
      return (number + cycleLength * negativeSafety) % cycleLength; // Wraps numbers (allows for cyclic array access)

    function stronger(a, b) { // Checks if ant a is stronger than ant b
      return a && (!b || a.tier > b.tier);

    view.forEach((cell, i) => { // Modifying the view is perfectly acceptable (here we do it to store the index alongside each cell)
      cell.index = i;

    const dirs = view.slice(0, 6);
    const safe = dirs.filter((cell, i) => !stronger(dirs[cycle(i - 1)].ant, me) && !stronger(dirs[cycle(i + 1)].ant, me) && stronger(me, cell.ant)); // Create an array of cells that are safe to move to
    // TODO: Also consider attacking other ants that are at the same strength level as us by spawning a stronger ant on their face

    if (safe.length) { // If we've got some safe spaces to move to...
      const victims = safe.filter((cell) => cell.ant && stronger(me, cell.ant)); // Prioritize killing other ants

      if (victims.length) { // If we've got someone to kill...
        const target = victims.reduce((prev, next) => stronger(next.ant, prev.ant) ? next : prev); // Find the strongest victim and target it

        return {cell: target.index, action: 0};

      const food = safe.filter((cell) => cell.food); // Otherwise prioritize grabbing food

      if (food.length) { // If we've got some food to grab...
        return {cell: food[0].index, action: 0} // Just grab any piece
        // TODO: Grab the piece that has the most pieces of food next to it

      return {cell: safe[0].index, action: 0} // Otherwise move to any safe cell

    return {cell: 6, action: 0}; // If we don't have a safe cell to move to, just stay still and hope for the best
    // TODO: Handle this situation better

As you can see, Smart Randant is a lot smarter than his counterpart. Smart Randant dropped his suicidal tendencies by finding safe cells before moving. Note that despite not being suicidal, he may still be killed by particularly cunning entries. He's also more aggressive - he'll gladly attack opponents weaker than him and will grab nearby food. Finally, if he's got anywhere to go, he will go there. Otherwise, he'll just stay still, hoping for the best.

You may have also noticed that Smart Randant still has a lot to learn. You can edit your entries to your heart's content, so W.I.P. submissions are allowed, or even encouraged.


This entry demonstrates smart usage of the environment for the purpose of fighting randomness. It travels in a straight line, leaving behind a path. It also utilizes and expands Smart Randant's basic framework.



The chat room is active already and is a better place for extended discussion than the comment section. I'd also love to dicuss some preliminary ideas for potential strategies to be used in future entries (though let's try to restrict this dicussion to nothing more than interesting isolated systems rather than full submissions - I'm not particularly interested in posting a nearly-solved challenge).

Most links are broken for now. The controller, for example, doesn't exist yet.

I'm aware that I'll have to make (or find) a custom number library that can handle nearly arbitrarily small and large powers of 2 for scoring (I've managed to design a system which spawns progressively less valuable ants every 2 turns at no cost, leading to a score equal to about 1 + 2e-16384).



Hexarun (stylized as Hexarun!) is a simultaneous game with complete information. Hexarun is intended for a minimum of three (3) players.


Up to twelve (12) players start on a regular hexagonal toroidal board with a number placed on each hexagon. The objective of the game is to collect the numbers by navigating the board. Players have complete knowledge of the board and the participating players. The number on a hexagon is reduced by the amount being collected. The game ends when all hexagons become zero or if the game becomes stale. The player whose collection has the highest sum wins the game.

Each game consists of multiple turns. Before the first turn, players choose their own starting location. During each turn, players move within their vicinity at the same time. If a player is alone in a numbered hexagon, this player collects the entire number. If multiple players move into a single hexagon simultaneously, they split the number. Numbers are not exchanged between players during each game.

Game specifications

  1. Regular hexagonal toroidal board: Each edge of the board measures N=2*P hexagons where P is the number of players on the board. The numbers on the board are placed in concentric rings. The numbers on the edge of the board are 1, and each successive inner ring follows off the OEIS sequence A002024 which starts with 1, 2, 2, 3, 3, 3, 4, 4, 4, 4. As a final touch, increase the number in the center by 1 if it is not unique on the board. [Example Board (N=6)]

  2. A game ends when...

    a. Hexagons become zero: Numbers on all hexagons are zero.

    b. Game becomes stale: Scores do not change during 6*P consecutive turns.

  3. Legal moves...

    a. Run!: Moves into one of the six (6) neighboring hexagons.

    b. Nothing: Does literally nothing.

  4. Number collection:

    a. Collect: At the end of a turn, a player alone on a hexagon scores s=h points, where h is the number on the hexagon at the beginning of the turn.

    b. Split: At the end of a turn, n>1 players standing on a hexagon scores s=h/n points, where h is the number on the hexagon at the beginning of the turn, and / is integer division. The number on the hexagon at the end of the turn becomes h-s*n, a.k.a. the remainder.

    Example: At the end of a turn, a total of 3 players are on a hexagon with number 10, then each gets 3 points and the hexagon becomes 1.

How to play


Tournament rules



  • \$\begingroup\$ Note to self: Implement using davidje13/koth-webplayer. \$\endgroup\$ – Frenzy Li May 31 '18 at 9:37
  • 2
    \$\begingroup\$ 2*P scaling means that if there are more players, there's more proportional space for each player. In other words, there's more and more empty space between players. (As an extreme example, if you have 2 players, each player will have 8 squares each. If you have 100 players, each player will have 1600 squares each) \$\endgroup\$ – Nathan Merrill May 31 '18 at 12:57
  • 1
    \$\begingroup\$ The center of the board is more valuable, but because you are using that OEIS sequence, with big boards, there are tons of squares in the middle that will have the same number. I think you were aiming to have fights for the middle, but how I see it, there's plenty for everybody. \$\endgroup\$ – Nathan Merrill May 31 '18 at 13:02
  • \$\begingroup\$ Finally, winning this game is primarily done by predicting where people are going to go. This is problematic, because as it stands, the way to win is to identify bots based on movement, which leads to one-upping. \$\endgroup\$ – Nathan Merrill May 31 '18 at 13:05
  • \$\begingroup\$ @NathanMerrill I meant to say that P is the number of players on the board. If up to 12 players play at the same time, the edge size is up to 24. I'm still not entirely sure about that sequence, and it'll take some testplay to see. \$\endgroup\$ – Frenzy Li May 31 '18 at 13:10
  • 2
    \$\begingroup\$ Right. If you have 12 players, then the edge size is 24. 24*24 = 576, which is 48 squares per player. With 100 players, it is 200*200 = 40000, which is 400 squares per player. What you really want is X*sqrt(P), where X is the number of squares per player. \$\endgroup\$ – Nathan Merrill May 31 '18 at 14:31
  • \$\begingroup\$ @NathanMerrill I'm trying to think in your way. The total number of squares for the hex follows A003215. For edge size 24 (number of hexagons on a side of the hexagonal board), there are 24^3-(23)^3=1657 hexagons, so that's approximately 69 hexagons per player. \$\endgroup\$ – Frenzy Li Jun 1 '18 at 0:39
  • \$\begingroup\$ 69 is miscalculated, 138 should be the answer. 138 hexagons should be plenty. \$\endgroup\$ – Frenzy Li Jun 1 '18 at 0:55
  • \$\begingroup\$ Let us continue this discussion in chat. \$\endgroup\$ – Nathan Merrill Jun 1 '18 at 1:15

Base 18 & Decimal traffic light

Given a number n<180, output the n/10 part and the n%10 part in 7-seg.

7-seg shapes:

 000         222   333         555   666   777   888   999   AAA         CCC         EEE   FFF   GGG
0   0     1     2     3 4   4 5     6         7 8   8 9   9 A   A B     C         D E     F     G     H   H
0   0     1     2     3 4   4 5     6         7 8   8 9   9 A   A B     C         D E     F     G     H   H
0   0     1     2     3 4   4 5     6         7 8   8 9   9 A   A B     C         D E     F     G     H   H
             222   333   444   555   666         888   999   AAA   BBB         DDD   EEE   FFF         HHH
0   0     1 2         3     4     5 6   6     7 8   8     9 A   A B   B C     D   D E     F     G   G H   H
0   0     1 2         3     4     5 6   6     7 8   8     9 A   A B   B C     D   D E     F     G   G H   H
0   0     1 2         3     4     5 6   6     7 8   8     9 A   A B   B C     D   D E     F     G   G H   H
 000         222   333         555   666         888   999         BBB   CCC   DDD   EEE         GGG
  • Either led on or (led off&background) should be constant, and the left characters mean the other state, or
  • At least two of your background, led on and led off should be constant, and the left characters mean the left state.

  • \$\begingroup\$ add the tag ascii-art \$\endgroup\$ – Beefster Jun 5 '18 at 18:46
  • 5
    \$\begingroup\$ 1. Related, related, related, and plenty more. I wouldn't be surprised if this were closed as a dupe. 2. What does this have to do with traffic lights or base 18? 3. I don't understand the last two bullet points. Are they saying that I can choose to use the same character for on and off and essentially just always output 88? 4. Why is the shape for 0 the same as the shape for 8? \$\endgroup\$ – Peter Taylor Jun 6 '18 at 8:50
  • \$\begingroup\$ @PeterTaylor 1. There are lots of related, but none of these are 2. video.baomihua.com/v/18443141 , base 18 because the next letter "i" can be mistaken with "1" 3. clarified somehow, though I don't know which choice you use 4. fixed \$\endgroup\$ – l4m2 Jun 6 '18 at 11:01

Convincingly Fake Compression of Random Data

As you may know, it is impossible to write an compression method that takes strings of length \$n\$ and returns strings of length \$n - 1\$. This can be proven by a simple argument: Each compression function must be a bijection, because otherwise the compressed strings can not unambiguously be uncompressed. However, just considering bit strings, there are \$2^n\$ strings of length \$n\$, but only \$2^{n-1}\$ strings of length \$n-1\$, thus no such bijection can exist.

In other words, for each attempt to write such a compression function, there exist strings which cannot be compressed.

The task for the cops in this challenge is to write a function which looks like it can compress arbitrary fixed-length strings, and the task of the robbers is to find a string for which the compression function fails.


I'm not yet convinced if this will actually work as a challenge, mainly if it is feasible for the cops to write a submission which is not easily crackable. It might be easier for large \$n\$ so robbers can not brute force, so I could make the cops winning criterion about getting the safe submission with the lowest \$n\$. Any ideas for a better winning criterion?

Also do you think this will work as a challenge?

  • 2
    \$\begingroup\$ Might be too easy for a challenge. There's a 50% probability that you find a string which fails. \$\endgroup\$ – Erik the Outgolfer Jun 26 '18 at 14:17
  • 2
    \$\begingroup\$ Looks feasible to me. It may be good to fix the input into binary string, lower-case alphabet, or just ASCII and require the robbers to find two strings that "compress" to the same object. \$\endgroup\$ – JungHwan Min Jun 26 '18 at 14:53
  • \$\begingroup\$ @EriktheOutgolfer yes, but with a well-crafted cop submission, it would be difficult to know that a string wasn't "compressed" unless the robber finds two strings that end up with the same string or writes a "decompression" function. \$\endgroup\$ – JungHwan Min Jun 26 '18 at 14:56
  • \$\begingroup\$ @JungHwanMin Just run it and measure its length? \$\endgroup\$ – user202729 Jun 26 '18 at 15:20
  • 1
    \$\begingroup\$ Anyway I think this is only feasible if the cops solution takes a very long time to run even for a single input, for most/all of the counterexamples. \$\endgroup\$ – user202729 Jun 26 '18 at 15:24
  • \$\begingroup\$ @user202729 What I meant to say is that it is possible that there exists a "compressed" string that corresponds to only one "uncompressed" string, so it would be "compressed" when you feed it to the cop program. With a good cop program, it may be hard for robbers to find specific examples of strings that have the same "compressed" variation. \$\endgroup\$ – JungHwan Min Jun 26 '18 at 16:15
  • \$\begingroup\$ @JungHw Pick random strings of length n and compress it, the expected number of runs until you crack the solution is 2. \$\endgroup\$ – user202729 Jun 26 '18 at 16:21
  • \$\begingroup\$ @user202729 That doesn't make sense. Assuming that we use binary strings and exactly two strings compress to one string, the probability of getting no collision after picking 3 strings is 1 * (2^n - 2)/(2^n - 1) * (2^n - 4)/(2^n - 2), which should be close to 1 with large enough n. \$\endgroup\$ – JungHwan Min Jun 26 '18 at 16:31
  • 1
    \$\begingroup\$ Perhaps the word "fail" has to be clearer. The challenge sounds possible if "failing" means "producing colliding outputs" instead of something on the lines of "crashing." With binary strings, Mathematica tells me that the expected value of trials needed to find colliding strings is ~946 with length 10, and ~1.05e+6 with length 20. \$\endgroup\$ – JungHwan Min Jun 26 '18 at 16:46
  • \$\begingroup\$ @JungHwanMin Shouldn't generating a random string, compressing it, decompressing the result and checking whether this equals the original string yield a collision with probability 0.5? \$\endgroup\$ – Laikoni Jun 26 '18 at 21:14
  • \$\begingroup\$ @Laikoni then the robber would need to create a decompression function. Maybe I misunderstood the challenge, but I thought the only requirement for the cop was to include a compression algorithm. Having some obscure compression function that make "decompression" very challenging would make this challenge feasible. \$\endgroup\$ – JungHwan Min Jun 26 '18 at 21:28
  • \$\begingroup\$ I suggest changing the challenge specification to my misinterpretation of the challenge specs. ;-) The robber would need to come up with either a decompression function or a counterexample to the compression algorithm (i.e. prove that the compression is not bijective). \$\endgroup\$ – JungHwan Min Jun 26 '18 at 21:40
  • 1
    \$\begingroup\$ @JungHwanMin Thanks for the suggestions! The problem with the cop providing only a compression function is that they could submit any hash function, because that's what those fake compression functions actually are, and trying to prevent cops from submitting known-to-be-collision-resistant hash functions probably brings us into dupe-distance to this cnr challenge. \$\endgroup\$ – Laikoni Jun 26 '18 at 22:10
  • \$\begingroup\$ So the cops must write a decompression function too? \$\endgroup\$ – user202729 Jun 27 '18 at 1:57
  • \$\begingroup\$ @user202729 That's what I originally envisioned, but at the moment that's up for debate in order to make the challenge feasible for the cops. \$\endgroup\$ – Laikoni Jun 27 '18 at 8:08

Progrqmming Puzzle andf Co9de Golf |


We all make mistakes. You, me, everyone. But not computers. They aren't making any mistake. Not even a single typo.
Time to change this injustice.


Your task will be to take a string as input, and display the string character by character, and with a cursor, as if someone was typing it. But there is a twist : sometimes, the "entered" character will be wrong. In that case there is a little pause, then the last entered character (which is supposed to be wrong) is deleted.


  • Input and output will contain only printable ASCII characters.
  • The cursor should be displayed as |, preceded by a space. This will always be displayed, from the empty string to the exact input displayed.
  • Each "iteration" (new character) is separated by a 0.25s pause (It takes time to write right). You can have a marge error of 0.05s, meaning the pause have to be lower than 0.2s, and lower than 0.3
  • Each character has exactly 1/10 chance to be wrong (that means it will be anything but the right character). In that case you should add a 1s pause ("Wait, did l really made a mistake?"), then delete this character, add a 0.25s pause (with a marge error of 0.05s), then continue.




Possible output

Note : I am planning to add a gif to show what the code should do. For now, consider every line as a gif frame

H |          // 0.25s pause
He |         // 0.25s   "
Hel |        // 0.25s   "
HelG |       // 1s      "
Hel |        // 0.25s   "
Hell |       // 0.25s   "
Hello |      // 0.25s   "
Hello! |     // done

Note for sandbox

  • Is this challenge good enough ?
  • Is this challenge already exist ?
  • Is this challenge clear enough ?
  • \$\begingroup\$ Are those typos in the body also intentional? e.g. whitch \$\endgroup\$ – wastl Jun 22 '18 at 12:48
  • \$\begingroup\$ @wastl Except for the title, this is just me being bad at writing english \$\endgroup\$ – The random guy Jun 22 '18 at 12:52
  • \$\begingroup\$ @AdmBorkBork the wrong character can be anything, except for the character it should have be. In my example the wrong character is a G, it could have been a a, a 7, a #, but not a l since it is the correct character. \$\endgroup\$ – The random guy Jun 22 '18 at 14:28
  • \$\begingroup\$ Is it acceptable if the "wrong" character is not random? For example, if a typo on a always becomes b. It might also be a good idea to restrict what characters need to be handled, e.g. "Input will, and output must, consist of only printable ASCII characters" \$\endgroup\$ – Kamil Drakari Jun 22 '18 at 15:53
  • \$\begingroup\$ @KamilDrakari as long as the "wrong" character is actually wrong, it's ok. And good idea for the input/output restrictions. \$\endgroup\$ – The random guy Jun 22 '18 at 20:04
  • \$\begingroup\$ Can we use \b to remove the previous character? How accurate does the delay have to be (0.24 seconds instead of 0.25?). Same question with randomness. Can we have an initial delay before output? Enough is spelt with a g. \$\endgroup\$ – Jo King Jun 27 '18 at 2:35
  • \$\begingroup\$ @JoKing as long as the output shows that the character is removed, you can use \b. i'd say delay will have an error margin of 0.05s (basically 0.2s<delay<0.3s). However the randomness has to be exactly 1/10. I'll edit the question to add those criterias later. \$\endgroup\$ – The random guy Jun 27 '18 at 10:42
  • \$\begingroup\$ You say each character has a 1/10 chance of being wrong? Does this mean answers have to behave non-deterministically, or could a deterministic method be used as long as 1 out of every 10 characters is wrong. \$\endgroup\$ – Wheat Wizard Jun 28 '18 at 13:27
  • \$\begingroup\$ @CatWizard Answer have to behave non deterministically : the total number of wrong characters will only rely to randomness. \$\endgroup\$ – The random guy Jun 29 '18 at 16:34
  • \$\begingroup\$ how are you going to measure exact timing on people's solutions? \$\endgroup\$ – don bright Jul 7 '18 at 2:16
  • \$\begingroup\$ @donbright with a chronometer ? More seriously, the point here is to get aproximately 4 characters in a second (unless there is a wrong one) \$\endgroup\$ – The random guy Jul 9 '18 at 9:30
  • \$\begingroup\$ i dont know how you can measure it, since computers run at different speeds, and OSes are not realtime OSes. but i like it enough anyways. \$\endgroup\$ – don bright Jul 11 '18 at 1:02

Role reversal

This is related to an old question, but is different enough that answers should be quite different to the older one.

You are given a sentence referring to two different people. Return the sentence with the roles reversed.

For example, for the input I will give you a kiss., you should return You will give me a kiss.

There will always be exactly two people referred to in the sentence, and they will be referred to with different pronouns. Here's a table of the pronouns that might be used (pronouns in a row refer to the same person. Pronouns in a column can be switched with one another to reverse a role).

 I         me       my           mine         myself
 you       you      your         yours        yourself     
 she       her      her          hers         herself
 he        him      his          his          himself      
 they      them     their        theirs       themselves 

Capitalisation matters (the first letter of the sentence should be capitalised, the pronoun "I" should always be capitalised, and no other pronoun is capitalised when not starting a sentence).

Words/punctuation not appearing in the pronoun table above shouldn't be changed. A pronoun word shouldn't be replaced if it appears as a substring of another word (e.g. 'history' shouldn't he changed). None of the inputs will be contractions using pronouns (so there won't be any inputs with "I'm" or "you're" etc).

Input/output examples:

in: She gave them hers!
out: They gave her theirs!

in: He will eat me if I don't eat him.
out: I will eat him if he don't eat me.

in: Get it for them yourself!
out: Get it for you themselves!

in: I think I am going to see him tomorrow.
out: He think he am going to see me tomorrow.

in: I am not interested in history, is he?
out: He am not interested in history, is I?


  • This is code golf so the shortest answer wins.
  • Standard loopholes are banned.
  • \$\begingroup\$ Any last thoughts before I post this? \$\endgroup\$ – LangeHaare Jul 19 '18 at 14:21

Use Japt Shortcuts

Japt is the PPCG Language of the Month for July, and I'm excited to try it out! In accordance with the Tips post I should use the Unicode Shortcuts. However, my keyboard seems to be lacking those important characters like "upside-down exclamation point" needed for optimum golfiness. Please write me a program to change horrible, verbose monstrosities like â m@VgUb==X into pristine, optimal code like â £VgUb¥X!


Given Japt code as input, output the same code making maximal use of the Unicode Shortcuts Japt supports.


  • Answers must support inputs containing any combination of valid Japt characters [Sandbox: is there a list of these? In particular, many languages need to use the NULL byte to indicate end of input, does Japt support NULL bytes in the middle of code?].
  • Input may be in any reasonable format (string, list of chars, etc.)
  • Output may be in any reasonable format, and does not need to be the same as the one used as input (e.g. "input as string => output as list of chars" is fine)
  • If multiple shortcuts are possible, use the one that replaces the most characters. For example, === should be replaced by not ¥=.
  • Your code does not need to handle ambiguous situations. For example, ==== could be shortened to ¶= or , so behavior is undefined if such a string shows up in the input.
  • Only shortcuts available in Japt 1.4.5 (most recent version at time of posting) need to be handled
  • Non-unicode shortcuts like _ and @ don't need to be handled
  • I've replaced easy-to-miss trailing spaces with where I found them. Those should be the literal space character ' ' when running tests or replacements. [Sandbox: Seriously, is there some better way to do this?]

Test Cases

?OvUf\l m_c %H} qV):0 => "?OvUf\l ®c %HÃqV):0"
Ov"y m_î íZ c p0} "p2␠ => Ov"y ®î íZ c p0Ã"²
=== => ¶
ñgJ òXYZ{XgJ <YgJ } mg mg␠ => ñÌòÈÌ<YÌÃmÎmÎ
w å+ m@Vå+ m+S+Xw} c => w å+ £Vå+ m+S+XwÃc

[Sandbox: The first two test cases are grabbed from some real Code Golf answers here and here. Should I replace them with something else?]

For reference, here is the full list of Unicode Shortcuts to be supported. It can also be found on the Japt Interpreter. Note that some shortcuts end in a space. [Sandbox: should I format this differently?]

¡   Um@
¢   Us2␠
£   m@
¤   s2␠
¥   ==
¦   !=
§   <=
¨   >=
©   &&
ª   ||
«   &&!
¬   q␠
®   m_
¯   s0,
°   ++
±   +=
²   p2␠
³   p3␠
´   --
µ   -=
¶   ===
·   qR␠
¸   qS␠
¹   )␠
º   ((
»   (((
¼   .25
½   .5
¾   .75
À   !==
Á   >>>
à   }␠
Ä   +1
Å   s1␠
Æ   o@
Ç   o_
É   -1
Ê   l␠
Ë   mDEF{D
Ì   gJ␠
Í   n2␠
Î   g␠
Ð   $new Date($
Ñ   *2
×   r*1␠
  • \$\begingroup\$ Nice idea! You can include trailing spaces by using <pre><code> ... </code></pre> instead of indenting, though they'll only be visible if the text is selected if you do it that way. (Also, Ê, Ì, and Í want their s back ;-) ) \$\endgroup\$ – ETHproductions Jul 7 '18 at 4:49

A note about this meta post:

I just have one example right now, but I will have three in the final.

I thought it would be interesting to have a problem about something I know a bit about. Right now it seems a bit mathy, but I wanted to ground the problem on something real. It feels more 'real-life' if you need to understand the spec in addition to golfing. The problem is I don't want it to seem like homework. Another problem is the actual computation that needs to take place isn't actually that hard once you understand the simplifications of the problem.

Let me know what you think.



Calculate the strain tensor and volume percent change of a cube given its material properties and stress tensor.


Common Terms

Stress Strain

  • Strain: ε, The amount of elongation per unit length, Units: \$\frac{in}{in}\$

  • Normal Stress: σ, The amount of force per unit area perpendicular to the cross section, Units: \$\frac{lbs}{in^2} = psi\$

  • Shear Stress: τ, The amount of force per unit area parallel to the cross section, Units: \$\frac{lbs}{in^2} = psi\$

  • Young's Modulus: E, The relationship between stress and strain: \$σ = Eε\$, Units: psi


  • Poisson's Ratio: ν, The relationship between strain in different directions. For a uniaxial bar: \$ε_{22} = -ν ε_{11}\$, Units: unitless

  • Index Notation: A short form for tensors written with subscripts \$i,j,k,l\$ to denote which element within the tensor. The number of subscripts the tensor has indicates what order it is. \$σ_{ij} \equiv \tensor{σ}\$ (Second Order)

  • Kroniker Delta: \$δ_{ij}\$, has the value of 1 if i=j, otherwise its value is 0. Index Notation for the Identity Tensor.

$$ δ_{ij} = \left[\begin{array}{ccc} δ_{11} & δ_{12} & δ_{13}\\ δ_{21} & δ_{22} & δ_{23}\\ δ_{31} & δ_{32} & δ_{33}\\ \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{array}\right] = I $$


A tensor in this context can be thought of as three directional components for each of the three positive faces of the cube.

Tensor on Cube

Although a Stress Tensor is not a matrix, it can be represented in matrix form by a 3x3 or other matrices:

$${\tensor{σ}} = \left[\begin{array}{ccc} σ_{xx} & σ_{xy} & σ_{xz}\\ σ_{yx} & σ_{yy} & σ_{yz}\\ σ_{zx} & σ_{zy} & σ_{zz}\\ \end{array}\right] = \left[\begin{array}{ccc} σ_{11} & σ_{12} & σ_{13}\\ σ_{21} & σ_{22} & σ_{23}\\ σ_{31} & σ_{32} & σ_{33}\\ \end{array}\right] = \left[\begin{array}{c} σ_{11}\\ σ_{12}\\ \vdots \\ σ_{33}\\ \end{array}\right] = \left[\begin{array}{cccc} σ_{11} & σ_{12} & \dots & σ_{33}\\ \end{array}\right] $$

If a Tensor and a Kroniker Delta share the same indices, they are combined.

$$ σ_{ij} δ_{kj} = σ_{ik} $$

If a Tensor has repeating indices, then it is taken as a zero order tensor and the indices are summed.

$$ σ_{ij} δ_{ij} = σ_{ii} = \sum_{i=1}^{3} σ_{ii} \equiv tr(\tensor{σ}) = σ_{11} + σ_{22} + σ_{33} $$

Stress Strain Relationship

The relationship in one dimension is \$σ=Eε\$. For three dimensions we can use Hook's law to find the relationship between the strain tensor and the stress tensor as follows:

$$ σ_{ij} = C_{ijkl} ε_{kl} $$

This general case would need \$3^4 = 81\$ independent material properties to calculate the strain tensor. If we assume the cube has a symetric \$\tensor{σ}\$, symetric \$\tensor{ε}\$, is Elastic, Isotropic, Linear, and Homogeneous, then we only need two independent material properties: (Young's modulus: E, Poisson's Ratio: ν) or (Lamé modulus: λ, Shear modulus: μ). We can use either pair of values, but for this example it is much easier to calculate the strain tensor from the stress tensor using Young's Modulus and Poisson's Ratio.

And so we can simplify the stress tensor by what we know.

$${\tensor{σ}} = \left[\begin{array}{ccc} σ_x & σ_{xy} & σ_{xz}\\ & σ_y & σ_{yz}\\ Sym & & σ_z\\ \end{array}\right] $$

And our new relationship is:

$$ ε_{ij} = \frac{1+ν}{E} σ_{ij} - \frac{ν}{E} σ_{kk} δ_{ij} $$

Calculating the Dilation

Volumetric Change

The Volumetric Strain can be found by calculating the trace of the strain tensor for very small values of strain. This is because for small values \$ε^3 \ll ε^2 \ll ε\$.

$$ {\frac {ΔV}{V_0}} \approx tr(\tensor{ε}) $$

Putting it all together

Therefore, in summary we can calculate the strain tensor with the following:

Using Index Notation:

$$ ε_{ij} = \frac{1+ν}{E} σ_{ij} - \frac{ν}{E} σ_{kk} δ_{ij} $$

Using Matrix Notation:

$$ ε_{ij} = \frac{1+ν}{E} \left[\begin{array}{ccc} σ_{11} & σ_{12} & σ_{13}\\ σ_{12} & σ_{22} & σ_{23}\\ σ_{13} & σ_{23} & σ_{33}\\ \end{array}\right] - \frac{ν}{E} tr(\tensor{σ}) \left[\begin{array}{c} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{array}\right] $$

Then calculate the volumetric change:

Index Notation

$$ {\frac {ΔV}{V_0}} = ε_{ij} δ_{ij} = ε_{ii} $$

Matrix Notation

$$ {\frac {ΔV}{V_0}} = tr(\tensor{ε}) $$


One positive long: Young's Modulus (Usually in the range 100 psi - 100,000,000 psi)

One positive decimal: Poissons's Ratio (Usually in the range 0.01 - 0.5)

Array of signed decimals for Stress Tensor (9 values)


Array of Strain Tensor, to at least five significant figures (same format as input)

Percent volume change of cube, to at least five significant figures


(simple example for now)

Young's Modulus: 29,000,000 psi
Poisson's Ratio: 0.30
Stress Tensor: [[50000,      0,     0]
                [    0, -10000,     0]
                [    0,      0, 25000]] psi
Strain Tensor: [[0.00157,        0,       0]
                [      0, -0.00112,       0]
                [      0,        0, 0.00045]] in/in
Dilation: 0.08966%


IO is flexible

  • Stress Tensor input can be any size array, or string

    • Input type and size must be the same as output
  • No formatting or units required

This is , least number of bytes for each language wins

  • \$\begingroup\$ Looks interesting. However, because all answers are publicly viewable, challenges where the main difficulty is the implementation (instead of understanding the challenge) is preferred. Looking at the input/output it appears that the implementation itself is probably complex enough to be interesting. \$\endgroup\$ – user202729 Jul 8 '18 at 10:58
  • \$\begingroup\$ this is very fascinating. IMHO the writing could be adjusted for the audience. if they are like me, i dont really understand stress and strain, and what they mean in laymans terms. also i think it could be simplified a lot. take for example the first graph. what are the dashed lines? why are they numbered 1 and 2? what is 0.2%? why is there a cylinder there, and what does it mean? is any of that graph required to solve the puzzle? \$\endgroup\$ – don bright Jul 11 '18 at 1:42
  • \$\begingroup\$ @donbright Thank you, this is helpful. I will explain the fundamentals of stress and strain more, and annotate the pictures (which are directly from wikipedia). The graphs are not required to solve the puzzle, but they may be useful to understand the concepts. I have them to explain, but they are probably not as helpful to a layman. I will fix it up in the next few days. Btw, what do you you think of the explanations from the tensors section onwards? \$\endgroup\$ – WretchedLout Jul 12 '18 at 5:13
  • \$\begingroup\$ I think it's "Kronecker delta" not "Kroniker delta". Also what is \$\varepsilon_{11}\$ and \$\varepsilon_{22}\$? \$\endgroup\$ – user202729 Jul 13 '18 at 2:54
  • \$\begingroup\$ You should probably describe what is a tensor too. (yes I can read Wikipedia and it's unambiguous but it would be more useful if a post contains all information) \$\endgroup\$ – user202729 Jul 13 '18 at 2:57
  • \$\begingroup\$ still hoping for this one to become a challenge!!!! \$\endgroup\$ – don bright May 27 '19 at 23:46

Haferman Carpet

Given nonnegative integer input \$n\$ output the \$n\$th iteration of the Haferman carpet.

Constructing the carpet

  • The zeroth iteration is 1.
  • When going from the \$n\$th to the \$(n+1)\$th iteration, replace each \$1\$ with the pattern [[0,1,0],[1,0,1],[0,1,0]] and each \$0\$ with the pattern [[1,1,1],[1,1,1],[1,1,1]].

Test cases

0 [[1]]

1 [[0,1,0],[1,0,1],[0,1,0]]

2 [[1,1,1,0,1,0,1,1,1],[1,1,1,1,0,1,1,1,1],[1,1,1,0,1,0,1,1,1],

Standard I/O stuff


Is this a duplicate? I will make the rules more explicit later.

  • 2
    \$\begingroup\$ Talking about replacing a scalar with a 2D array confused me for quite a while. Having the test cases laid out in a way which obscures the 2D pattern of the carpet also doesn't help. \$\endgroup\$ – Peter Taylor Jul 13 '18 at 8:43
  • \$\begingroup\$ As for dupes: these are probably the most closely related questions: two about the Sierpinski carpet and one which is general enough to draw that one and this one. I think it's borderline whether or not this adds something new to the site. \$\endgroup\$ – Peter Taylor Jul 13 '18 at 8:47

Syllabification and classification (Venpa)


Venpa is a form of classical Tamil poetry, based upon certain rules of metric prosody. The complete rules form a context-free grammar, the task here is to test for a tiny subset of those rules.


  • Vowels: a, A, i, I, u, U, e, E, y, Y, o, O, W
    • small case represents "short" vowels, CAPS represent "long" vowels
    • y represents the y sound in "my".
    • W represents the "ou" sound in "mouse", and always comes in capital (long) form.
  • Consonant: everything else
    • the consonants can be in small case or capitals, whichever is convenient.

Word Segmenting

Words in the input are seen as composed of segments (which roughly correspond to the usual notion of a syllable). A segment divide occurs where:

  • two or more consonants occur together, tr, rpp, etc.
  • a long (capital) vowel occurs in the word, pO, W, mY etc.
  • two vowels have occurred since the last segment's end (or the word's beginning).

sanjIvi has segments san, jI, vi.
Yvar has Y,var.
vigadakavi has viga,daka,vi.
vAnily has vA,nily.
padAgytAngiha has padA, gytAn, giha
radagajaturagapadAdi has rada, gaja, tura, gapa, dA, di

(Consonants at the end of segments make no difference after segmentation and can be ignored - gytAn=gytA, var=va.)

As you can see, the segments are of two types:

  1. those with a single vowel, san, Y, vA, called a straight (S) segment
  2. those with two vowels, viga, nily, padA called a sequence (Q) segment

So the segment patterns of the above words are:

  sanjIvi              =              san + jI + vi          = SSS
  Yvar                 =                 Y + var             = SS
  vigadakavi           =            viga + daka + vi         = QQS
  vAnily               =                vA + nily            = SQ
  padAgytAngiha        =           padA + gytAn + giha       = QQQ
  radagajaturagapadAdi = rada + gaja + tura + gapa + dA + di = QQQQSS

Permitted patterns

The rules for words allowed in a Venpa are:

  1. a word should have one to three segments
  2. a one or two segment word can have any pattern (S, Q, SQ, SS, QS, QQ)
  3. a three segment word must end in a straight (S) segment

So out of the above, SSS, SS, QQS, and SQ are allowed, QQQ and QQQQSS are not.


Given an input word in the above letters, output its segment pattern if it's an allowed word, or a false-y value if it's not an allowed word.


  • Will be a single word composed of the characters mentioned under the Letters heading above
    • (Small case w will not be present anywhere in the text (neither as a vowel nor as a consonant))


  • The segment pattern of the input word, if it's an allowed word
    • can use any two distinct unambiguous characters to represent S segments and Q segments
  • A false-y value in your language if it's not an allowed word
  • Output is undefined for invalid input (i.e. has characters other than those specified)

Test cases

=> QQS
=> SS
=> falsey (QQQ not allowed)
=> SQ
=> SQS
=> falsey (more than 3 segments)
=> SSS
=> Q
  • 1
    \$\begingroup\$ fascinating. im probably very unusual, but i would like it even better if it was actual poetry instead of 'p' everywhere, maybe even in tamil script (although that would probably disqualify some languages) \$\endgroup\$ – don bright Jul 7 '18 at 2:37
  • 1
    \$\begingroup\$ Yeah, I intended to just remove some extraneous complexity, but somehow having input in this form makes it slightly less interesting even to me (and I'm not sure it removes a meaningful level of complexity anyway, relative to the rest of the task). I'll change it to use the actual consonants if anyone else here also feels this way. (The last two test case inputs are from actual poetry by the way, just with the consonants replaced.) \$\endgroup\$ – sundar - Remember Monica Jul 7 '18 at 18:54
  • 1
    \$\begingroup\$ Using Tamil script would also be great, but my concern with that is mainly that it might make testing and debugging much harder for people who can't read the script. I can mitigate some of that by providing exhaustive test cases, but I'm not sure it's worth it. \$\endgroup\$ – sundar - Remember Monica Jul 7 '18 at 18:58
  • \$\begingroup\$ yeah i think i might be a little unusual in my tastes, i still like the puzzle... \$\endgroup\$ – don bright Jul 7 '18 at 20:08
  • 1
    \$\begingroup\$ @donbright I've updated the post to use different consonants (corresponding to actual words). I like this version a lot better, but let's see how it's received. \$\endgroup\$ – sundar - Remember Monica Jul 20 '18 at 14:18
  • \$\begingroup\$ I wrote a program that solves the examples you gave. my first pass was 60 lines of Rust code. i think it is a really good puzzle, great balance between an interesting puzzle and many ways to possibly reduce code size. very interesting. \$\endgroup\$ – don bright Dec 29 '18 at 14:10
  • \$\begingroup\$ i would like some more examples though. feel like my program might be wrong on some inputs. \$\endgroup\$ – don bright Dec 29 '18 at 14:11

DRAW me a picture: A QBasic metagolf challenge

The DRAW command in QBasic takes a string argument, consisting of instructions for moving the cursor and drawing line segments, and produces the appropriate line segments on the screen. The syntax of the instructions is very terse--perfect for a challenge!

The challenge

Write a program or function which:

  • Given a 2-D array of on and off pixels, representing a black-and-white image,
  • Generates a string that, when passed to QBasic's DRAW command, will draw that image on the screen,
  • While keeping the generated string as short as possible.

More about DRAW

Your program may use the following DRAW instructions:

(more details pending)

  • U - draw line upwards
  • D - draw line downwards
  • L - draw line to the left
  • R - draw line to the right
  • E - draw line diagonally up and to the right
  • F - draw line diagonally down and to the right
  • G - draw line diagonally down and to the left
  • H - draw line diagonally up and to the left
  • B - meta-instruction: prepend to any instruction to move the cursor accordingly but not draw the line
  • N - meta-instruction: prepend to any instruction to draw the line but not move the cursor

(examples + pictures pending)

The following instructions are outside the scope of this challenge and may not be used (even if they would improve your score): C, P, S, M, X, A, and TA.

Output requirements

Conceptually, your program's output will be substituted for the ... in the following QBasic program:

SCREEN 9        ' Graphics mode, 640 x 350 pixels
DRAW "B M 0,0"  ' Set drawing cursor to top left corner
DRAW "..."

(If the length of your output exceeds any limits on line or string literal length, it may be split across multiple DRAW commands in such a way that the instructions are preserved.)

The program will then be run, and the output image compared to your program's input. Where the input array has a 1, the output image must have a white pixel; where the input array has a 0, the output image must have a black pixel. The portion of the screen outside the input array's dimensions must be entirely black pixels.

Practically speaking, I will probably write a verification script in some other language, just to make testing easier.


Standard I/O methods apply. Output is case-insensitive. Input array dimensions will not exceed 640 x 350. (more rules pending)

Test cases

(test cases pending)


Your submission's score is the sum of the lengths of its outputs on these test cases. In the case of a tie, the earlier submission wins.

Note: this challenge is probably a variation on the Traveling Salesman Problem, meaning that an optimal solution will take exponential time. In order to receive a score, your submission must complete all test cases, which means that you'll need to take a sub-optimal approach.

Sandbox questions:

  • What's a good number of test cases?
  • Should I instead score submissions on a second, hidden set of test cases to prevent overfitting? Or should the hidden test cases be the (first) tiebreaker?
  • Is the implicit requirement "must complete all test cases before you can post it" enough of a bound on long execution times, or should I add a specific execution-time limit?
  • \$\begingroup\$ I would suggest explaining in what way the draw command is given its path length. \$\endgroup\$ – Jonathan Frech Aug 3 '18 at 22:25

Note: this is my first time posting, so I need help fleshing out the details. I'm aware there are plenty of Roman Numeral problems, but this is somewhat different.

When in Rome, count as Romans do!

This problem is inspired by this website, which published the following diagram:

enter image description here

This diagram shows us that the longest Roman Numeral expression under 250 is that of 188, which requires 9 numerals to express.

The standard symbols used to express most Roman Numerals are the following: {I, V, X, L, C, D, M}. In this challenge, your goal is to, given an positive integer n, compute the number of valid Roman Numeral representations that can be composed through concatenating n of the standard symbols.

Then, your program must output the result of this computation modulo 3997 (to prevent answers from getting too long) in Roman Numerals!

Rules for Roman Numeral Expressions

Roman Numerals originally only had "additive" pairing, meaning that numerals were always written in descending order, and the sum of the values of all the numerals was the value of the number.

Later on, subtractive pairing, the use of placing a smaller numeral in front of a larger in order to subtract the smaller from the larger, became commonplace to shorten Roman Numeral expressions. Subtractive pairs cannot be chained, like the following: IXL. This is considered invalid.

The following are the modern day rules for additive and subtractive pairing.

  1. Only one I, X, and C can be used as the leading numeral in part of a subtractive pair.
  2. I can only be placed before V and X in a subtractive pair.
  3. X can only be placed before L and C in a subtractive pair.
  4. C can only be placed before D and M in a subtractive pair.
  5. Other than subtractive pairs, numerals must be in descending order
  6. M, C, and X cannot be equalled or exceeded by smaller denominations.
  7. D, L, and V can each only appear once.
  8. Only M can be repeated 4 or more times.

Test Cases

Input: 1
Output: VII

More to be added.

Sandbox Users

Thoughts on this problem? I know it is really badly formatted but I thought the concept was cool. Thanks for the help!

  • 1
    \$\begingroup\$ So basically, find how many Roman numbers there are with n numerals? The modulo 3997 is probably not necessary, since the maximum for each n is only 8**n (and valid roman numerals are going to consist of a small fraction of that) \$\endgroup\$ – Jo King Aug 8 '18 at 5:54
  • \$\begingroup\$ Alright, I'll remove the modulo part. Does the rest seem plausible as a concept? \$\endgroup\$ – Don Thousand Aug 8 '18 at 12:36
  • \$\begingroup\$ The reason for my inclusion of the modulus is because I want to ensure the answer can be outputted without using any numerals beyond those I've provided in the Standard Set \$\endgroup\$ – Don Thousand Aug 8 '18 at 15:12
  • \$\begingroup\$ This needs clearer specification on what counts as a valid Roman numeral. E.g. are IC and XCIX both valid expressions for 99? Is MMMMMMMM a valid 8-letter numeral? \$\endgroup\$ – Peter Taylor Aug 8 '18 at 16:38
  • \$\begingroup\$ I added more details to the text. \$\endgroup\$ – Don Thousand Aug 8 '18 at 17:10
  • 2
    \$\begingroup\$ Now that you've posted this to the main site, please add a link to the challenge, edit out the body and delete the post. Thank you! \$\endgroup\$ – Mr. Xcoder Aug 12 '18 at 11:19

Most distinct Turing-complete character subsets

(Inspired by Fewest (distinct) characters for Turing Completeness)


In any language you choose, find the greatest number of distinct and disjoint subsets of characters allowed in that language (i.e. no individual character is in more than one of the subsets), each of which separately makes the language Turing-complete.


JS (2): eval()"\u0123456789bcdf, []+=` (see answer to linked question).


Scoring is by total number of distinct Turing-complete subsets found. Higher scores are better. In case of a tie, the answer with the fewest total characters used across all subsets wins.


Execution of arbitrary code is not required, only Turing completeness.

Explanations of why each of your subsets are Turing complete are highly encouraged.

In case this was unclear, whitespace characters are still counted as characters.

Sandbox notes:

Should I include some stipulation forbidding languages such as Unary which don't care about the particular characters used?
What is unclear about this specification? Where could I give a better/more complete explanation?

  • \$\begingroup\$ I personally wouldn't exclude unary. I recall a previous challenge attempted to exclude all "symbol-independent" languages and it was almost as much of a mess as attempts to ban builtins. Sure Unary will "win" with a score of 256, but that doesn't mean no interesting answers will be posted. \$\endgroup\$ – Kamil Drakari Aug 14 '18 at 18:30
  • \$\begingroup\$ You’ll need to define if spaces are counted as characters. \$\endgroup\$ – JayCe Aug 14 '18 at 20:05
  • \$\begingroup\$ @JayCe done.⠀⠀⠀ \$\endgroup\$ – praosylen Aug 14 '18 at 20:21
  • 2
    \$\begingroup\$ Are you sure you want to use characters instead of bytes? \$\endgroup\$ – Jo King Aug 15 '18 at 4:44

Chain Classification


Objective: Write a program (whose index in the chain of answers is \$n\$) which, when given any program with index \$i\$ (\$ 1 \le i \le n\$), outputs \$i\$.

(The program may do anything else given any other string input, including but not limited to: crashing, erorring, returning other numbers, sending an email to google support, and simulating the universe.)


  • Your program may be in any language that has not appeared in the answer chain yet.
  • You may output to STDOUT, STDERR, return as a function value, etc. Any reasonable method of output.
  • You may take input from STDIN, command line arguments, function parameters, etc. Any reasonable method of input.
  • You must output 1 for the first answer, 2 for the second, etc. Any other form of indexing is not allowed.
  • You must use base 10 when outputting.
  • You may not use the internet in any way, particularly to scrape the answers to this question.
  • No person may answer twice in a row.
  • No person may answer within 1 hour of their previous answer.
  • Languages which differ by version are considered distinct. Thus, Python 2 and Python 3 can both be part of the chain.
  • Languages which differ by compiler or interpreter are not considered distinct. So, Python 3 (Cython) and Python 3 are equivalent.

Answer format

# N. Language



Try It Online links are appreciated, as well as links for the language itself.

You must include how your program performs input and output.


The first answer is:

1. Alumin


Try it online! Input and output through STDIN and STDOUT.

  • \$\begingroup\$ An answer \$a_n\$ may also recognize every answer \$a_i\$, where \$1\le i<n\$, and, if it doesn't, output \$n\$. \$\endgroup\$ – Erik the Outgolfer Aug 17 '18 at 19:09
  • \$\begingroup\$ @EriktheOutgolfer Yup. Do I need to edit anything to reflect that? \$\endgroup\$ – Conor O'Brien Aug 17 '18 at 19:19
  • \$\begingroup\$ No, just a note that answers don't need to be generalized quines. \$\endgroup\$ – Erik the Outgolfer Aug 17 '18 at 19:20
  • \$\begingroup\$ @EriktheOutgolfer I'm not sure what would indicate to someone that the answers have to be generalized quines. \$\endgroup\$ – Conor O'Brien Aug 17 '18 at 20:11
  • \$\begingroup\$ This is more like the comments that often appear below challenges for slight hints (usually posted by others). \$\endgroup\$ – Erik the Outgolfer Aug 17 '18 at 20:12
  • \$\begingroup\$ What do the rules 5 and 8 ensure? \$\endgroup\$ – ბიმო Aug 19 '18 at 16:23
  • \$\begingroup\$ @BWO Rule 5 is for clarity; rule 8 is for variety. \$\endgroup\$ – Conor O'Brien Aug 21 '18 at 6:27
  • \$\begingroup\$ Ok, I don't see a point in 5 but I don't have a strong opinion on it either and it doesn't harm the challenge. However rule 8 seems rather arbitrary, for one I don't think it will be more variable if a user is required to wait some 10 minutes before submitting. Even if in the mean-time another user will answer, I don't think it would become more variable: Just because it's the same user doesn't mean it will be nearly the same answer. But that's just my opinion, in the end it's your call. \$\endgroup\$ – ბიმო Aug 21 '18 at 14:19

King of the Grid

There's a side-project of mine, simply titled My Grid Game, that's been sitting on lefttwix.com for a while. It was only a matter of time before it went automatic.

Grid game image

The rules of the game are simple:

  • Each player plays as a colour.
  • Each player starts with a City (C) 2 spaces in from opposite corners.
  • Cities spawn soldiers, they start with 10 health.
  • Soldiers will all march in the same direction as the other soldiers of the same colour. They will spawn 1 at a time (denoted by the number 1).
  • Players can change the direction of their entire army.
  • When soldiers can't go any further (either because they reach the edge of the grid or a friendly city), they join each other, to become larger groups of soldiers. (Numbers 2 - 9)
  • When 10 soldiers group up, they form a new city.
  • When soldiers meet their enemies, they kill off the enemies. The largest group of soldiers survives. (e.g. if a 5 meets a 3, it survives, but is reduced to 2).
  • When soldiers meet an enemy city, they reduce the health of that city by the size of the group.
  • When cities lose health, it can not be regained.
  • When cities or soldiers run out of health, they are destroyed.

With the API enabled

My experimental API branch is enabled at https://ajfaraday.github.io/grid_game/

The coloured boxes in the grey box correspond to the 2 players.

There's 3 ways to choose a bot for a player:

  • Select 'Code on Page' to write the code in place. The game will stop when the code is invalid.
  • Choose 'Gist ID' and paste the id of a Github Gist onto the page.
  • Select one of the existing bots.

The buttons are self-explanitory.

  • Start Start the game running
  • Stop Halt the game
  • Reset Reset the game with the same bots

The API:

You can interact with the player by making these calls:

  • api.towardsX() Change direction horizontally towards the opponent.
  • api.awayX() Change direction horizontally away from the opponent.
  • api.towardsY() Change direction vertically towards the opponent.
  • api.awayY() Change direction vertically away from the opponent.
  • api.random_direction() Change to a random direction.
  • api.turn() Returns the numbered turn, starting at zero.

Writing a bot:

  • Only use the API calls to interact with the game.
  • It does not need to be defined as a function or any other wrapping code.
  • jQuery is available.
  • Your code will be called once every turn.
  • Please, only one direction call per iteration.

Example bot:

if (api.turn() < 400) {
  if ((api.turn() % 2) == 0) {
  } else {
} else {
  if ((api.turn() % 20) == 0) {

The Rules

Answers should include:

  • A name for your bot.
  • The bot's code.
  • A Github Gist ID.



Note: This is the part I'm least clear about. Please do clear up how this is to be done. I can think of 3 approaches.

(league approach)

For each pair of entries:

  • I'll try it against each other entry, best of 3.
  • The winner will get a point.
  • After a week(?) I'll finalise the results and declare a winner.

(tournament approach)

  • After a week(?) I'll set up a tournament bracket with all the entries.
  • I'll try a best of 3 for each pair in the tournament.
  • The winner wins.

(Challenge the champ approach)

  • All entries get added to a list.
  • When a new entry is added, it is at the bottom of the list.
  • It will then challenge the entry immediately above it (best of 3).
  • If it wins, it moves 1 up the list, then challenges the next one and so on.
  • After 1 week(?) the tournament closes and the top of the ladder wins.


  • Which competition approach is best?
  • Is the Gist code distribution approach okay?
  • How long should the competition last?

Thanks for looking over this!

  • \$\begingroup\$ Won't new cities only be possible on the edges then? \$\endgroup\$ – fəˈnɛtɪk Aug 31 '18 at 21:40
  • \$\begingroup\$ @fəˈnɛtɪk Not quite, you can also build cities against your existing cities. \$\endgroup\$ – AJFaraday Aug 31 '18 at 21:46
  • \$\begingroup\$ So you should mention that soldiers cannot move over friendly cities \$\endgroup\$ – fəˈnɛtɪk Aug 31 '18 at 22:18
  • \$\begingroup\$ 1. What happens when a soldier hits an opponent's city. 2. What's the number of a spawned soldier? 3. I'd recommend doing a round-robin, randomly pairs, or using an ELO system and having similarly scored players play each other. 4. It seems simpler to me to have the method simply return an integer for the direction (N/E/S/W) than making these API calls. 5. Don't require a gist for bots. 6. Where do cities start? \$\endgroup\$ – Nathan Merrill Sep 1 '18 at 3:53
  • \$\begingroup\$ 7. Does the bot get the state of the map? 8. Are you going to have a maximum number of turns? 9. Despite all my feedback, I really like this challenge. Definitely ping me in chat if you want some help along the way. \$\endgroup\$ – Nathan Merrill Sep 1 '18 at 4:07
  • \$\begingroup\$ @NathanMerrill 1, 2, 6, Edited. 3. Thanks for the suggestions. 4. The relative directions are to accommodate a bot starting in either corner. It will rotate it's strategy. Perhaps both options, if we can find a data structure for the state of the grid. 5. Gists were an idea for quick deployment. You seem strongly against it. What issues does it present? 7. I considered this, but it seems it'll be a large, difficult-to-consume data structure. How would you like to see this data presented? 8. I wasn't planning to, although stale mates are quite common. 9. Glad you like this one! \$\endgroup\$ – AJFaraday Sep 1 '18 at 8:34
  • \$\begingroup\$ 4. Right. I'd propose you simply do NESW, but rotate the map when displaying it to the player (so it appears that both players are on the west side). 5. SE needs to be the authority, gists can disappear, and it's extra work players shouldn't need to do (IMO). 7. You need the map to make this a game of strategy. Give it as a 2D array of {count: 4, isCity: false} objects (or maybe a mixed type array?). 8. Right, you need to somehow detect stalemates (and then score them too) \$\endgroup\$ – Nathan Merrill Sep 1 '18 at 12:32

The Input String, But Every Time It Says The First Word, It's Recursion

You wanted to create the ultimate replacement remix, but didn't see a way to add more recursion that ends naturally. Then you saw The Entire Bee Movie But Every Bee Is Replaced With The Entire Bee Movie Without Bees. Unlike the other replacement remixes, all the inserted copies were from the same text. It was the answer - at each recursing, you could remove the word being replaced, making each copy shorter than the one before it and causing the recursion to eventually end. Your master plan was complete, all that was left was to write the program to fulfill it.

The Expansion Function

Let's define some function \$F\$ on a sequence of words. Words are sequences of characters not containing whitespace, which are separated by whitespace.

define \$F(S)\$:

if \$S\$ is empty, return nothing

let \$X\$ be the first word in \$S\$

let \$Y\$ be \$S\$ with all instances of the whole word \$X\$ removed

let \$Z=F(Y)\$

return \$S\$ with all instances of the whole word \$X\$ replaced by \$Z\$

Your task is to implement \$F\$.


Input the sequence of words in some form. You may choose any input method.


Output the sequence of words in some form. You may choose any output method.

Only the words in the output count for correctness. If you output as a string, the leading, trailing, and separator whitespace can be anything.


This is code golf, so shortest code in bytes wins.


Small case


b o o k k e e p e r


r r p r r r r p r r e e p e r r r p r r r r p r r e e p e r k k e e p e r o o k k e e p e r

See the recursion:


value of S
result of F


p r

e e p e r
r r p r r

k         k         e e p e r
r r p r r r r p r r e e p e r

o                             o                             k k e e p e r
r r p r r r r p r r e e p e r r r p r r r r p r r e e p e r k k e e p e r

b                                                                         o o k k e e p e r
r r p r r r r p r r e e p e r r r p r r r r p r r e e p e r k k e e p e r o o k k e e p e r

Large case


peter piper picked a peck of pickled peppers a peck of pickled peppers peter piper picked if peter piper picked a peck of pickled peppers wheres the peck of pickled peppers peter piper picked


218854 words

Hello Sandbox

This section will not appear when the challenge is posted to the main site.

I can't seem to get the quote Markdown right, it might just be a parser bug where it thinks the quote continues even if there are no >s.

I originally put a description of \$F\$ as an implementation in pseudocode, but I'm considering finding a more mathematical description which tells you less about how to go about implementing it yourself. Still though, there is a challenge in optimizing this for tiny code.

  • \$\begingroup\$ I think the title is a bit confusing (as not everyone knows about bee movie or somewhat similar). Maybe "Recursive replacement" a better one? \$\endgroup\$ – Shieru Asakoto Sep 10 '18 at 4:37
  • \$\begingroup\$ Oh btw does "occurrence" means the whole word or being a part of another word is also counted? \$\endgroup\$ – Shieru Asakoto Sep 10 '18 at 4:41
  • \$\begingroup\$ Only a whole word counts. I'll go rewrite that to be more clear. As for the title, you're right that what the challenge is is more important than the inspiration, but I'm not quite sure what wording would be good. I'm thinking something like The Input String, But Every Time It Says This Word, It's Recursion which would state the premise while also referencing the meme. \$\endgroup\$ – EPICI Sep 10 '18 at 5:02
  • \$\begingroup\$ Oh I didn't know about the meme either. The Input blah blah blah would be better for others who don't know about the meme I think, but if you want to retain the meme that's fine. \$\endgroup\$ – Shieru Asakoto Sep 10 '18 at 5:05
  • 2
    \$\begingroup\$ The introductory paragraph primes the reader to expect a very different challenge to the one given: firstly because it sets up a scenario of infinite recursion, and secondly because it sets up a scenario where elements of the input are replaced by elements which are not in the input. IMO it would be better to remove that introduction and use it instead for a challenge which implements the scenario described and asks for the nth word in the limit string (although this may be a dupe, so check for that first). \$\endgroup\$ – Peter Taylor Sep 10 '18 at 8:15
  • \$\begingroup\$ I don't follow how the input creates the output. Could you show a step-by-step example of how each level of recursion works? \$\endgroup\$ – Beefster Sep 11 '18 at 22:39
  • \$\begingroup\$ I added more details to the "bookkeeper" example, it should explain where each part is coming from, but it may still be too brief. // As for the misleading introduction, I attempted another rewrite, but I think it may need some more revisions before I get it right. \$\endgroup\$ – EPICI Sep 12 '18 at 0:51
  • \$\begingroup\$ I think that the current I/O rules are just a little too restrictive. I would allow other inputs and outputs than "space delemited string", such as "list of words". \$\endgroup\$ – Kamil Drakari Sep 13 '18 at 19:48
  • \$\begingroup\$ After a bit of thought, I relaxed the IO to allow other formats and methods. I figured you'll have some work to do either way, so I might as well let you choose the methods that result in the least bytes. I'm not too worried about dedicated data structures as an IO format, since that's probably already covered as a common loophole. \$\endgroup\$ – EPICI Sep 13 '18 at 22:20

Triangular snake

You're given a triangular field:

Triangular field

It has 4 ports.

You're also given five different pieces from A to E:


Each piece has a little piece of path inside it.

Your basic goal is to build paths between ports. There are some rules:

  • Each cell of the field can be occupied by 0 or 1 pieces.
  • A port can either point to an empty cell or be an endpoint of a path. That is, this is illegal: enter image description here
  • The path starts with a port and ends with a port.
  • The path must not be broken up into pieces, i.e. every edge of a piece through which a path goes must not touch an empty cell or a border without a port.

This is an example of a legal path:

enter image description here

This path can be represented as "CBEDEAD" or "DAEDEBC".


There is no input.


Your ultimate task is to output all the possible paths. Your output must not contain illegal paths. You can output the paths in any readable way. Order, repetitions and letter case don't matter.

  • 1
    \$\begingroup\$ Aren't A en E the same piece? \$\endgroup\$ – Kroppeb Sep 17 '18 at 15:12
  • \$\begingroup\$ Yes they are, what a blunder. \$\endgroup\$ – decorator-factory Sep 17 '18 at 15:16
  • \$\begingroup\$ It's sad you can only post comments with 50 reputation, otherwise I would have noticed that already :) \$\endgroup\$ – decorator-factory Sep 17 '18 at 15:16
  • \$\begingroup\$ @int6h Out of interest ... What is the connection? \$\endgroup\$ – Jonathan Frech Sep 18 '18 at 21:12

Iterated Prisoner's Trilemma

Please send me problems or else I'm gonna post this on main.

Prisoner's dilemma ... with three choices. Crazy, huh?

Here's our payoff matrix. Player A on the left, B on the top

A,B| C | N | D
 C |3,3|4,1|0,5
 N |1,4|2,2|3,2
 D |5,0|2,3|1,1

The payoff matrix is engineered so that it's best for both players to always cooperate, but you can gain (usually) by choosing Neutral or Defection.

Here's some (competing) example bots.

# turns out if you don't actually have to implement __init__(). TIL.

class AllC:
    def round(self, _): return "C"
class AllN:
    def round(self, _): return "N"
class AllD:
    def round(self, _): return "D"
class RandomBot:
    def round(self, _): return random.choice(["C", "N", "D"])

class Grudger:
    def __init__(self):
        self.history = []
    def round(self, last):
            if(self.history.count("D") > 0):
                return "D"
        return "C"

class TitForTat:
    def round(self, last):
        if(last == "D"):
            return "D"
        return "C"

Your bot is a Python3 class. A new instance is created for every game, and round() is called each round, with your opponent's choice from last round (or None, if it's the first round)

If we get enough entries for the result to be statistically significant, there's a 50 rep bounty for anyone who can beat Tit For Tat.


  • Round count: [REDACTED]
  • Standard loopholes disallowed.
  • No messing with anything outside your class.
  • \$\begingroup\$ I have no idea how to embed tags in a post. \$\endgroup\$ – SIGSTACKFAULT Sep 28 '18 at 15:56
  • 1
    \$\begingroup\$ [tag:something] should do the trick. \$\endgroup\$ – Alion Sep 28 '18 at 16:24
  • \$\begingroup\$ thanks @Alion <3 \$\endgroup\$ – SIGSTACKFAULT Sep 28 '18 at 16:28
  • 5
    \$\begingroup\$ Way too little substance. You should add a few paragraphs describing the prisoner's trilemma, the challenge, or anything else tangentially related. \$\endgroup\$ – RamenChef Sep 29 '18 at 0:22
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    \$\begingroup\$ Don't do a pretty payoff matrix image. Use text. That way you don't exclude people with limited vision who rely on screen readers. \$\endgroup\$ – Peter Taylor Oct 1 '18 at 15:09
  • \$\begingroup\$ How many rounds per game? And why does N/N suck so much? \$\endgroup\$ – Jo King Oct 10 '18 at 8:13
  • \$\begingroup\$ @JoKing Haven't decided the round count, and probably will never tell. Don't want defect-on-the-last-round bots. No idea why I did that; if it causes problems i'll change it. \$\endgroup\$ – SIGSTACKFAULT Oct 10 '18 at 12:33
  • \$\begingroup\$ Maybe a random amount of rounds within a certain range? Also, wouldn't it be better for the second argument to be a list of all your opponents moves, rather than the latest one? \$\endgroup\$ – Jo King Oct 10 '18 at 12:42
  • \$\begingroup\$ You can do watever you want within the class, so if you need to remember the entire history, that's your problem. \$\endgroup\$ – SIGSTACKFAULT Oct 10 '18 at 12:45
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