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2877 Answers 2877

81 82
84 85

Implement Or

or is a dubious esoteric programming language. The author is unknown but it is known that fungot (a chatbot) is currently learning or but it remains unknown how it does it and where it receives the required material to learn or. Check out this IRC log to see fungot revealing part of the language.

Instruction reference

It is only known that a space followed by an f pushes false to the stack. The false value may by any falsy value of your choosing. Since this language does not support output, in order to make verifying answers possible, you should output the resulting stack in the end of the program.

Test cases

Expects one false to the stack


Expects 2 falses on the stack

m f ma f

Expects 1 false on the stack

 f a lf 

Expects 2 falses on the stack

a f fa


This is , so shortest answer in bytes wins.


  • Is this clear enough?
  • I haven't found a duplicate, but anything?
  • Tags are code-golf, string and interpreter. Anything else?
  • Any further feedback?
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  • \$\begingroup\$ Duplicate of this. \$\endgroup\$ – user85052 Oct 21 '19 at 14:01
  • \$\begingroup\$ Duplicate of this. \$\endgroup\$ – user85052 Oct 21 '19 at 14:02
  • \$\begingroup\$ Duplicate of this. \$\endgroup\$ – user85052 Oct 21 '19 at 14:03

It was found that this challenge is a dupe. if you would like to see the challenge, please look in the revision history.

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  • \$\begingroup\$ My explanation is Because LangName. Now I can halve my bytecount! \$\endgroup\$ – user85052 Oct 26 '19 at 1:01
  • \$\begingroup\$ I can simplify this into just a cheating quine by not using a code block at all. \$\endgroup\$ – user85052 Oct 26 '19 at 1:26
  • \$\begingroup\$ @A_, fixed. The challenge should be more loophole-free now. \$\endgroup\$ – Lyxal Oct 26 '19 at 1:59
  • 1
    \$\begingroup\$ Duplicate or at least very near duplicate of codegolf.stackexchange.com/questions/98812/… \$\endgroup\$ – Unrelated String Oct 26 '19 at 3:55

Make a karaoke machine

Given a list of lines and a set time (in seconds) which those lines appear, output those lines after that specific amount of seconds.

Example input:

0 Hello there!
5 This line should come up after 5 seconds.
10 This line should come up 5 seconds after that one.
20 There's a 10-second delay here.

Every line will be preceded by an integer, separated from the rest of the line by a string. The number indicates when the line should come up - the first line should be outputted instantly, followed by the next few lines like so:

Hello there! (output instantly)
(wait 5 seconds)
This line should come up after 5 seconds.
(wait 5 seconds)
This line should come up 5 seconds after that one.
(wait 10 seconds)
There's a 10-second delay here.


  • You're guaranteed that the lines are in order - i.e. this doesn't happen:
    5 Hello there!
    0 Oops, this is meant to happen before the first line.
  • You're also guaranteed that no two lines share the same time.

Winning criteria:

This is , so shortest code in bytes wins!

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  • 1
    \$\begingroup\$ No problem in particular. However I'd expect that many solution will have unbounded drift when the number of lines increases, so you may ban that (w. r. t. the system clock if there's one, for example) plus, I suppose that lines are short enough such that printing them takes a negligible time. \$\endgroup\$ – user202729 Oct 28 '19 at 10:59
  • \$\begingroup\$ I found several related challenges: 1 2 3. None of them were quite the same, but I think they all suffered from some of the problems that user202729 mentioned. Most of them still went well, but I think being more rigorous won't hurt. \$\endgroup\$ – FryAmTheEggman Oct 28 '19 at 15:15

Interpolate between binary images

While there are various ways to interpolate between images with a large color space where for each color there are many similar other colors, it is not so straightforward to do that in a (heavily) quantized color space where you only have few colors available. We now take that to an extreme:

Given two black and white image of the same size as well as a number \$N\$, output a sequence of \$N\$ images that "interpolate" between the two input pictures.

The goal here is coming up with an "interpolation" algorithm that allows you to smoothly transition between two images. But how you define this smoothness is up to you. The more aesthetically pleasing the better.


  • "black and white" here means the images consist only of the colors black and white, no grays.
  • Validity criteria:
    • The first image in the sequence must be the first input image, and the last image in the sequence must be the second image.
    • All images must have the same size and consist of only black and white pixels. Furthermore the output must be deterministic in the sense that if you repeatedly call the program with identical inputs, you must get identical outputs.
  • Please post the sequence of images as an animation in your answer, for any \$N\$ you choose.


To be defined...

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  • \$\begingroup\$ popularity contests are at this point considered almost considered "historical". You need a clear scoring criterion for a valid challenge, which in your case means defining smoothness as something better than just "It's up to you" \$\endgroup\$ – AlienAtSystem Nov 19 '19 at 12:10
  • \$\begingroup\$ @AlienAtSystem The popularity contests need validity criteria, but no scoring criteria - we discussed this quite thoroughly about 2 years ago. The problem is that you cannot force people to vote according to a specific way, but you have to design the contest in a way that the desired output is something people will like anyway. It is not really historical, it is just rare that good popularity contests are written. Check out the discussions linked here. \$\endgroup\$ – flawr Nov 19 '19 at 13:19

Mahjong bot

It seems there isn't a challenge about mahjong bot...

Basic idea is a simplified mahjong rule:

Rules for beginners

Mahjong game use 136 mahjong tiles (similar to poker cards). There are 4 players in a game.


There are 136 tiles in total.

There are 3 simple suits. Tiles numbered from 1 to 9 in each simple suit. There are 4 identical copies of each tiles in simple suits. That is \$3\times 9\times 4=108\$ tiles. These 3 simple suits are donated as m, p, s. For example, m1 means a tile with number 1 in m suit.

There are also 7 honors tiles. And there are also 4 copies of each honors tile. That is \$7\times 4=28\$ tiles. These tiles are donated as z1 ~ z7. Don't be confused with m, p, s, honors tails are not numbered. Donating as z1 ~ z7 just make it easy.


  1. All 136 tiles are shuffled first;
  2. Each player receives 13 tiles at the beginning (These tiles are kept secret to others); Starting from the first player...
  3. The player draw a tile (secret to others), and then
    • Claim win, or
    • Kong, or
    • Discard a tile
  4. After someone discard a tile, others may
    • Claim win, or
    • Pong, or
    • Kong, or
    • Do nothing (pass)
  5. If anyone win, the game ends;
  6. If anyone Pong, the player have to discard a tile and continue to step 4;
  7. If anyone Kong, the player draw a tile and continue to step 3;
  8. Otherwise, next player draw a tile and the game goes on to step 3;
  9. When someone try to draw a tile but there are no more tiles left, the game draw (no winner).


A player can claim win if tiles in his hand meet requirements (known as self-pick). A player may also declare win if others discard the last tile he need to complete the hand (known as feed). In case two or three players claim win after a single discard, only the one closest player win.

A winning hand must consist of four melds and a pair.

Melds includes Pongs, Kongs, or Chows:

  • Pong: A set of three identical tiles.
    • Example: m1m1m1, p8p8p8, z3z3z3
  • Kong: A set of four identical tiles. (Read more details in Kong section)
    • Example: m1m1m1m1, p8p8p8p8, z3z3z3z3
  • Chows: A meld of three suited tiles in sequence. Only simple suits may in a Chows.
    • Example: m1m2m3, p5p6p7, s7s8s9
    • Invalid: m8m9m1 (no wrap), m1p2s3 (no mixed suit), z1z2z3 (no honors tails)

A pair is two identical tiles.

Any tile may only be used in one Meld / Pair. It should not be reused.


When someone discards a stail, other players may teal the tile to complete a Pong.

  • The player claim Pong;
  • The player got the discarded tile;
  • The player expose two more identical tiles to others and place all these three tiles in the front;
  • The player must discard a tile, and the game continue;


When a player

  • Collect all 4 identical tiles in his hand;
  • Collect 3 identical tiles, and others discard the last one;
  • Collect the last one, while the other 3 are already exposed as Pong by him

The player may claim Kong. When someone claim a Kong

  • Expose all 4 tiles to others, and put them in the front;
  • Draw another tile;
  • The game goes on;

Tiles used in exposed Pong / Kong may not be changed or discarded later. When declare win, tiles used in exposed Pong / Kong must not be reused in other Chows.


  • If a player win by self-pick: The winner got 3 points, others lost 1 point;
  • If a player win by feed: The winner got 3 points. The one who discard that card lost 3 points. Others neither got or lost points.
  • In case a player try some illegal moves: The player lost 9 points. All others got 3 points. And the game ends;
  • Once the game draw by run out of tiles, no points got or lost.

Rules if you already know mahjong

  • Use 136 tiles, no bonus tiles (花牌, flowers) included;
  • 13 / 14 tiles each player at beginning, 4 players;
  • Win only if 4 meld (面子, set), and 1 pair (对, 頭, head, eye);
    • no seven pairs (七对, 七対子), thirteen orphans (十三幺, 国士無双), nagashi mangan (流し満貫), Knitted Straight (组合龙), ect.
  • Pong (碰, ポン, pon), Kong (槓, カン, kan) exists, but not Chow (吃, チー, chii);
  • Both self-pick (自摸, ツモ), feed (食糊, 出铳, ロン) exits; But not robbing a Kong (搶槓)
  • All winning hands worth 1 point (分, 点), no yaku (役, pattern), faan (番, han), fu (符), starting dealer (莊, 庄) calculated;
    • self-pick: winner +3, others -1
    • feed: winner +3, loser -3
    • Replacement tile (嶺上牌, rinshanpai) after open Kong (大明槓) count as self-pick
    • A bot try to act some invalid moves lose the game, which -9, and others +3
  • At most one winner each game (盘, 局). No Sichuan rule, nor multiple feed;
  • No need to claim riichi (立直, リーチ), no furiten (振聴 or フリテン), no abortive draws (途中流局, tochuu ryuukyoku), no not-waiting penalty (ノーテン罰符), no joker tiles (百搭);
  • Starting dealer changed every game no matter who won the last game or draw;


The player who got most points after certain times of games win.


Later, it should be in Python 3 as it is popular here.

Sandbox: Is the rule given above suitable? (Since no one want to change rules after writing controller...)

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  • \$\begingroup\$ 1. How will you organize the games between multiple bots, e.g. a tournament or a league? 2. A regular mahjong match usually consists of multiple games (east-wind match or south-wind match). I think you'll need to specify how long each match will be (single-game matches will be fine though). 3. After four kans, no one can declare another kan, right? 4. Just a matter of preference: when you explain the mahjong rules, using pure English terms might be easier to read. \$\endgroup\$ – Bubbler Oct 31 '19 at 9:40
  • \$\begingroup\$ The simplified rules themselves are fine I guess. I can think of a general strategy: go for riichi-style, try to maximize the number of remaining winning tiles, go for a pon when I have multiple pairs, and kan whenever possible. I don't think I can set up a reliable defense because no riichi and no furiten. I just can't imagine how to write these into code... \$\endgroup\$ – Bubbler Oct 31 '19 at 9:47
  • \$\begingroup\$ @Bubbler 1. Run Combination(n, 4) times (Assuming n > 4, otherwise maybe I need to duplicate every bots); 2. Maybe 100 games should be fine. (Starting dealer always change, so nothing east-wind / south-wind would include.) Running only 16 games seems not enough. (This will depends on how many answers then). 3. The 5th, 6th kan are also valid, since we do not include abortive draws nor dora here; 4. I will describe the rules for beginners later. And it would be in English. Current rule is for players who already know how to play. \$\endgroup\$ – tsh Oct 31 '19 at 10:06
  • \$\begingroup\$ @Bubbler I comment these words in brackets since I cannot find any stander translations. And translations vary from one to another (for example, all Chow / Chii / Chi spelling are exists... \$\endgroup\$ – tsh Oct 31 '19 at 10:11

Draw a hexagonal grid n×m of ascii hexagons

 /    \
/      \
\      /

with numbers somewhere within each of them, such as numbers goes in snail order starting in the center (but without empty hexagons):

  ____        ____        ____        ____        ____        ____
 /    \      /    \      /    \      /    \      /    \      /    \
/      \____/      \____/      \____/      \____/      \____/      \
\      /    \      /    \      /    \      /    \      /    \      /
 \____/      \____/      \____/  25  \____/      \____/      \____/
 /    \      /    \      /    \      /    \      /    \      /    \
/      \____/      \____/  26  \____/  24  \____/      \____/      \
\      /    \      /    \      /    \      /    \      /    \      /
 \____/      \____/  27  \____/  11  \____/  23  \____/      \____/
 /    \      /    \      /    \      /    \      /    \      /    \
/      \____/  28  \____/  12  \____/  10  \____/  22  \____/      \
\      /    \      /    \      /    \      /    \      /    \      /
 \____/      \____/  13  \____/  3   \____/  9   \____/  40  \____/
 /    \      /    \      /    \      /    \      /    \      /    \
/      \____/  29  \____/  4   \____/  2   \____/  21  \____/      \
\      /    \      /    \      /    \      /    \      /    \      /
 \____/      \____/  14  \____/  1   \____/  8   \____/  39  \____/
 /    \      /    \      /    \      /    \      /    \      /    \
/      \____/  30  \____/  5   \____/  7   \____/  20  \____/      \
\      /    \      /    \      /    \      /    \      /    \      /
 \____/      \____/  15  \____/  6   \____/  19  \____/  38  \____/
 /    \      /    \      /    \      /    \      /    \      /    \
/      \____/  31  \____/  16  \____/  18  \____/  37  \____/      \
\      /    \      /    \      /    \      /    \      /    \      /
 \____/      \____/  32  \____/  17  \____/  36  \____/      \____/
 /    \      /    \      /    \      /    \      /    \      /    \
/      \____/      \____/  33  \____/  35  \____/      \____/      \
\      /    \      /    \      /    \      /    \      /    \      /
 \____/      \____/      \____/  34  \____/      \____/      \____/
 /    \      /    \      /    \      /    \      /    \      /    \
/      \____/      \____/      \____/      \____/      \____/      \
\      /    \      /    \      /    \      /    \      /    \      /
 \____/      \____/      \____/      \____/      \____/      \____/
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  • 4
    \$\begingroup\$ 1. Can you provide examples for smaller various n,ms? Especially, I want to see examples for odd and even widths. 2. Should the numbers be distinct in every cell? 3. Assuming the answer to 2 is true, What should we do if the number doesn't fit in a single line inside the hexagon (i.e. 7 digits or higher)? \$\endgroup\$ – Bubbler Nov 11 '19 at 4:17
  • \$\begingroup\$ Maybe we should assume n,m to be odd. \$\endgroup\$ – Alexey Burdin Nov 12 '19 at 18:17

Monokeyed Words

(Based on https://what-if.xkcd.com/75/)

Old mobile phones had letters and numbers assigned to the same key, as follows:

  • 1
  • 2abc
  • 3def
  • 4ghi
  • 5jkl
  • 6mno
  • 7pqrs
  • 8tuv
  • 9wxyz
  • 0 [space]

From any given list of words or phrases (arbitrary number of inputs, formed of 0-9 a-z and space), find the word which has the most consecutive characters on the same key, and output the key and the count (in any reasonable format)

for example:

  • [nonmonogamous],[qwerty],[false] -> 7, 6 (seven times on the 6 key - nonmonogamous)
  • [tutu],[cat],[mouse] -> 4, 8 (four times on the 8 key - tutu)
  • [cab],[mon],[tin],[tom] -> 3,2 (cab) or 3,6 (mon) - either answer is valid
  • [#],[!!!],[is this valid?] -> any output or error - doesn't fit valid inputs. 2,4 would also be acceptable, for hi in is this valid?
  • bacmon, habitat, fringe, be, test, valid -> 3,2 or 3,6 as it's a tie between those keys
  • 01452301146 -> 2,1 (two in a row on the 1 key)

The output format isn't fixed, so you could also output [3][6] or 3.6 or 6,3 - anything you like, as long as it's consistent which is the key and which the number of repetitions. You could also use one of the letters to represent the key, if you wanted - e.g. instead of 3,6 you could output 3,m

This is , usual exclusions etc. apply

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  • \$\begingroup\$ I don't think I understand what you mean by "wordlists." You don't appear to want us to check a lexicon (as 9wxyz9wxyz isn't a word in any language I know). If you just mean an input which is a list of strings then you don't need to do anything besides say that - the default rules will handle that no problem. Aside from that, this seems to have some "fluff" in that I think you'd be better off asking for just a summary (i.e. the longest chain of what button) of one input word or just finding the maximal word(s) from a list. Once you have one, getting the other is more or less trivial. \$\endgroup\$ – FryAmTheEggman Nov 11 '19 at 20:25
  • \$\begingroup\$ Thanks, that makes sense. I was originally wanting to do exactly the same as the link (i.e. look up based on a whole dictionary), but I changed my mind part way through. As for the "fluff", yes OK I can see that; I'll edit my question \$\endgroup\$ – simonalexander2005 Nov 12 '19 at 9:17
  • \$\begingroup\$ I think changing your mind halfway has happened to all of us at least once :) That said, I find your edits somewhat confusing - you seem to have changed it to just be one word in the new test cases but have kept the old ones. \$\endgroup\$ – FryAmTheEggman Nov 12 '19 at 20:50
  • \$\begingroup\$ @FryAmTheEggman yes, the program should be able to take an aribtrary number of inputs \$\endgroup\$ – simonalexander2005 Nov 13 '19 at 9:35


Cornhole is a game in which players take turns throwing bags filled with corn kernels(or other fillers) onto a wooden board with a hole in the far end. If a bag goes through the hole, it counts as 3 points. If a bag stays on top of the board without falling off it counts as 1 point.

Score is taken at the end of each round in which the two players throw 4 bags (alternating). Only the net score is calculated.

The first to exactly 21 points is the winner. If a player were to score points in a round that would bring them to exceed 21 points, the points for that round are subtracted from the players score.


  1. Given an array of scores from multiple rounds of Cornhole, determine the current score of the game.


The input will be an array of one or more rounds of scores for each bag throw by both players alternating. (you may choose any 3 distinct symbols to cover the scenario of misses, 1 point, or 3 points)


  • In this scenario
    • Player A missed the first shot, Player B missed their first shot, Player A landed the bag on the board for their second shot, Player B missed their second shot, Player A made it in the hole for their third shot, Player B made it in the hole for their third shot, Player A landed on the board for their fourth shot, and Player B landed on their board for their fourth shot.
    • In the next round of play, Player A missed the first shot, Player B missed their first shot, Player A landed the bag on the board for their second shot, Player B missed their second shot, Player A made it in the hole for their third shot, Player B made it in the hole for their third shot, Player A landed on the board for their fourth shot, and Player B landed on their board for their fourth shot.


The score of the Cornhole match as an array of two numbers. From the input above, the output would be:



 Insert multiple samples of scoring, including one example of a player scoring over 21 and losing points

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SandChallenges for Proposed box [draft]

In this challenge you should simulate how "challenges" interact with each other. (You need to output how many iterations have the "challenges" done.)

Symbols used in the documentation

SandChallenges is played on a 3x3 grid.

  • X represents a challenge. It challenges the grid next to its position.

  • represents an empty sandpile. This is what the "challenges" move in their actions.

  • . represents moving sandpiles and is only used in this documentation. Input will never contain those characters; it instead will only contain X and .

A closer look at X challenges

There are sandpiles between those "challenges". When a "challenge" is active, it pushes sand towards the next grid. The next grid is described as the following:

  • The next grid is usually the grid on the right.
  • If the grid is on the right edge, the next grid becomes the first grid of the next line.
  • If the grid is on the last line, the next grid becomes the first grid of the first line.

The next grid of A is B, the next grid of C is D, and the next grid of I is A.

Another rule: they can get buried by active sandpiles. This makes them fail to function.

Test cases

Spaces are replaced with semicolons for readability problems.


-> (Affecting the next cell)



X..    X..    X..           X..
;;; -> .;; -> ..; -> ... -> ...
;;;    ;;;    ;;;           ...

After that the "challenge" gets buried by active sandpiles, which results in the following state:


This box had went through 9 changes before it stops.

XXX    ...
XXX -> ...
XXX    ...

This means that the challenge had went through 1 change before it stops. All of them push active sand and buries themselves simultaneously.

X;;    X.;
;X; -> ;X.
;;X    ;;X

The bottom-right challenge shoots active sand, which buries the active challenge in the same iteration:

..;    ...    ...    ...
;X. -> ;X. -> .X. -> ...
;;X    .;X    ..X    ...

This goes through 4 iterations.

X;;    X.;    X..    ...
X;; -> X.; -> X.. -> ...
X;;    X.;    X..    ...

This goes though 3 iterations.

X;X    X.X    ...    ...
;X; -> .X. -> ... -> ...
;X;    ;X.    .X.    ...

This goes through 3 iterations.

X;X    ..X    ...
;X; -> .X. -> ...
X;X    X.X    ...
This goes through 2 iterations.


  • Input/output can be given by any convenient method.
  • You can print it to STDOUT or return it as a function result.
  • Either a full program or a function are acceptable.
  • Any amount of extraneous whitespace is permitted, provided the characters line up appropriately.
  • Standard loopholes are forbidden.
  • This is so the shortest code wins.


  • Is this clear enough? Any better wordings for sections?
  • Is this a duplicate?
  • Tags are and and . Any suggestions?
  • Any further feedback?
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  • \$\begingroup\$ As posed the challenge appears too simple. The wrapping method of the grid means you can just save it as a 1d object and reshape it for the output alone. There is no reason why the challenge should be restricted to a 3x3 grid either. Also, you're not clearly stating what you want the output to be. And lastly, that pun for the sake of pun is horrible and will just confuse people. \$\endgroup\$ – AlienAtSystem Nov 18 '19 at 18:48
  • \$\begingroup\$ I mostly agree with the other comment. This seems to be a lot of fluff for "find the longest substring of semicolons in the input." It may even wind up being a dupe because of that - I was going to suggest having them propagate both ways, but I'm pretty sure that is a duplicate as well! If you allow the grid to be any size and make the sandboxes propagate in more directions you may have a good base to build off of. \$\endgroup\$ – FryAmTheEggman Nov 20 '19 at 18:59

Sandbox reviewers, thanks for checking this out. I posted this question (after a few revisions) as a code challenge and pretty much no one likes it. Please give some feedback as to why it's not appropriate, clear, fun, or whatever other criticism you have. I think it's a very interesting problem and I'd love to see some other approaches to it along with some interesting language choices :).

I think the main criticism is how to measure accuracy. I've added a script to calculate a score between the competition video and whatever video contestants submit. Please tell me why that is insufficient, or unfair, unclear, etc.

Again, any feedback for improvement is welcome! I really want to see some interesting solutions :).

Thanks again for reviewing!

TITLE: Reconstruct a video where the frames' pixels have been shuffled

Given a video where all the pixels of every frame have been scrambled with some image scrambling function T(frame), reconstruct the original. T only operates on one frame at a time and has no persistent state between invocations. It's just a constant random map that shuffles pixels to new locations each frame. T is constant and performs the same permutation each frame.

Essentially you have to recover all of the spatial information that's been removed.

Prepare to bake your noodle golfers, this is a doozy!


Since the spatial mapping needs to be determined, the scrambled video source needs to be something with significant motion. An example would be an episode of your favorite sitcom. Lots of colors, lots of movement between and around frames.

Simple example of the scrambling function T(frame):

SourceImage = [44,22,33,55]

T(SourceImage) = [33,44,22,55]

T maps {0:2, 1:0, 2:1, 3:3}

This example has the image in a 1D array. For the problem however, the image would be a 2D RGB image.


  • You are allowed to process the entire video to determine Ti.
  • Ti does not have to be a 100% perfect inverse, which my not even be possible depending on the input.
  • No loopholes. Your program should at a minimum take the example scrambled video as input ('scrambled_city.mp4' below) and output the unscrambled video that is as close to the original as possible ('city.mp4').
  • After producing the output video, calculate your score with the provided average MSE python script below.
  • Submit your unscrambled program source, your average MSE score and your calculated unscrambled video file, see Submission below.
  • Providing your source code and unscrambled video lets the hivemind confirm your submission is honest.


The answer which most accurately reconstructs the video is the winner. Accuracy will be calculated as the avgerage of Mean Squared Error (MSE) of all hidden source and corresponding reconstructed frames, nothing fancy, and should be as close to 0 as possible.


  • Unscrambling a randomly shuffled image by itself is impossible. A video is different, it's possible to some degree.
  • Feel free to post any algorithms already known to accomplish this.
  • This problem is very difficult so if you can generate any unscrambled images that don't look like pure garbage consider yourself a success.

Competition Videos

Hidden source video 'city.mp4': Full Hidden Source Video Your algorithm should work without access to this! Frame 250

Scrambled video (seed 9001) 'scrambled_city.mp4': Full Scrambled Video - 2GB Frame 250 - Scrambled

source webpage for other videos to use

Python code to generate test cases for you:

This will take a video (infile) and output it's scrambled version (outfile).

Dependency: opencv2+ (pip3 install opencv-python)

import cv2

infile = 'city.mp4'
outfile = 'scrambled_city.avi'
seed = 9001

def scramble(image):
    copy = image.copy()
    # reset the seed each frame so we have constant transform function.
    return copy

# Open video, read first frame
vidcap = cv2.VideoCapture(infile)
success,image = vidcap.read()

# Create video writer with same dimensions
h,w,layers = image.shape
size = (w,h)
outvideo = cv2.VideoWriter(outfile,cv2.VideoWriter_fourcc(*'MPEG'), 30, size, True)
print( 'videosize: %s\n'%repr(size))

count = 0
while success:
    scrambled = scramble(image)
    #cv2.imwrite("frame%d.jpg" % count, scrambled)


    print( 'frame: %d   %s'%(count,str(success)) , end='\r') 
    count += 1
    success,image = vidcap.read()


Python code to calculate avg Mean Square Error over two videos

import cv2
import numpy as np

infile1 = 'city.mp4'
infile2 = 'scrambled_city.avi'

# https://www.pyimagesearch.com/2014/09/15/python-compare-two-images/
def mse(imageA, imageB):
    # the 'Mean Squared Error' between the two images is the
    # sum of the squared difference between the two images;
    # NOTE: the two images must have the same dimension
    err = np.sum((imageA.astype("float") - imageB.astype("float")) ** 2)
    err /= float(imageA.shape[0] * imageA.shape[1])
    # return the MSE, the lower the error, the more "similar"
    # the two images are
    return err

# Open video, read first frame
vid1 = cv2.VideoCapture(infile1)
success1,image1 = vid1.read()
vid2 = cv2.VideoCapture(infile2)
success2,image2 = vid2.read()
count = 0
sumMSE = 0
while success1 and success2:
    sumMSE += mse(image1,image2)
    count += 1
    print( 'frame: %d    avgMSE=%f'%(count,sumMSE/float(count)) , end='\r') 
    success1,image1 = vid1.read()
    success2,image2 = vid2.read()

print("\n[Frames: %d] Avg MSE = %f"%(count,sumMSE/float(count)))

Example output avgMSE of city.mp4 and scrambled_city.avi:

[Frames: 5113] avg MSE = 18079.870445918445

Example output avgMSE of city.mp4 and city.mp4:

[Frames: 5113] avg MSE = 0.000000


When you're ready to submit your solution, please score your submission against the example video and for fun include unscrambled versions of the three test cases below.

First (required):

  • Your solution code. No code provided is disqualification.
  • Submit the avg MSE for your reconstructed 'scrambled_city.mp4' and the first example video, 'city.mp4', you calculate from the provided python script.
  • Your unscrambled video file (suggestion: host on youtube or G-drive).
  • You do not need to include your calculation of T or Ti, the visual media you provide (step above) will be enough to prove your algorithm works and score it.

~~ Lowest score wins. Good luck! ~~

Second (optional for fun): You can submit your completely unscrambled videos OR you can submit the unscrambled frames 250,500,750,1000 for all 3 of the test cases below. It will be pretty clear if you solved it or not. They're very recognizable scenes.

Note: The youtube preview is terribly distorted. The downloaded files are much better quality.

video 1

video 2

video 3 - HARD

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  • \$\begingroup\$ The issue I see is the ambiguity of the sentence "T only operates on one frame at a time and has no persistent state between invocations." If T does not perform the same permutation on all frames, the challenge is simply impossible. In addition, the test cases are all ridiculously huge. 2GB of data for a simple test case is way too much. \$\endgroup\$ – AlienAtSystem Nov 18 '19 at 18:55
  • 2
    \$\begingroup\$ I don't think this one is solvable even if T did perform the same permutation on all frames. Its like the Auditors atomizing paintings to try and figure out why they're beautiful. Also holy hell that test case is literally the worst: a collection of short, unconnected, sequences that are only a few seconds long at best. How is any algorithm supposed to find any patterns in that? \$\endgroup\$ – Draco18s no longer trusts SE Nov 19 '19 at 4:53
  • \$\begingroup\$ T is constant. I clarified the phrasing. Ok I'll add a second smaller video. Maybe 100MB? It's difficult to get smaller because the compression algorithms don't like it when the frames are noisy. Compressing the scrambled version too much would actually make it impossible. \$\endgroup\$ – extracrispy Nov 19 '19 at 8:30
  • \$\begingroup\$ duplicate \$\endgroup\$ – Beefster Nov 19 '19 at 22:49
  • \$\begingroup\$ @Beefster I think you have referred to a similar but different problem. In that problem there is a simple reversible algorithm and you (as in the person reading this) get to decide how to map/unmap the values. However, in this problem the only thing you are presented with is a scrambled video and the knowledge that it's scrambling function is constant. You don't know anything else. Now try your best to unscramble the video. \$\endgroup\$ – extracrispy Nov 20 '19 at 6:02
  • \$\begingroup\$ @extracrispy are you going for a cops-and-robbers then? \$\endgroup\$ – Beefster Nov 20 '19 at 20:29
  • \$\begingroup\$ Adding a sentence is not the same as clarifying. The misleading information is still the first thing people read. And I think before asking people to perform a task like this, you should consider how you would go about it yourself. Here is a very small (32x32x32) and easy (3d perlin noise: it's smooth in all dimensions) test case for you. Show me that you could unshuffle it successfully. \$\endgroup\$ – AlienAtSystem Nov 21 '19 at 10:10

How long should this song last?

Enter the world of sheet music. A composition (the musical piece, which may or may not be a song) is divided into bars. The length of a bar is defined by the time signature. The time signature states how the bar is divided into beats, and what length of note carries the beat.

Note lengths are always powers of 2. 4 means a quarter note, 2 a half note, 8 an eighth note (or a quaver if you're a snob), etcetera. A half note (2) is twice the length of a quarter note (4), which is itself twice the length of a quaver (8), and so on.

A time signature may look like this: 3/4. The 4 means that the quarter note carries the beat, and the 3 means that there are 3 of them in one bar. 3/2 means there are three half-notes in a beat, 7/8 means there are seven quavers, and so on.

Now, the actual speed at which a piece is to be performed depends on the tempo. That is usually expressed in beats per minute (bpm). The tempo also defines the note carrying the beat (usually the same as the one in the time signature but not always). So, you can have the time signature be 8=150, meaning there are 150 quavers in a minute (in the sheet music it would be notated ♪=150).

Both tempo and time signature can change throughout the composition.


Use the following format for your input (or something very similar). It is a list of events:


This is the simplest form. It is a list of integer-string pairs (you're obviously free to go with string-string pairs if it makes your program simpler). The integer defines at which bar the event happens (starting from 1), and the string defines what happens there. If it is in the form of x/y, then it is a new time signature. If it is in the form of x=y, you have a new tempo. Lastly, an empty string designates the end of the score (exclusive, so the above example has 520 bars).

With changes, the format may look like this:

[46, "4=155"],
[67, "5/4"],
[68, "4/4"],

The output of the program should be the duration of the entire piece in "xm ys" (where x is the number of minutes and y is the number of seconds. You can leave out the "ys" part if there's no spare seconds, but it is not necessary).

This is , so shortest code wins!

Important note!

Real artists do not follow the tempo exactly. Only beginners use a metronome to match the exact number of seconds as notated; more experienced musicians know to dynamically speed up or slow down depending on the mood, their personal preference, etcetera. Therefore, it is perfectly acceptable for your answer to be up to 34% higher or lower than the "correct" answer. Also, the minimum length of a composition is 2 minutes and 30 seconds.

Test cases



The time signature has 4 quarter notes a bar, and 520 bars, so 4*520=2080 quarter notes. There's 120 quarter notes per minute, so 2080/120=17.333 minutes, or 17m 20s.


[46, "4=155"],
[66, "5/4"],
[76, "6/8"],

For the first 45 bars, there's 4 quarter notes a bar, so that's 45*4=180 quarter notes. Now the tempo is 120 8th notes per minute, which is 60 quarter notes per minute, meaning the first bit lasts 180/60=3 minutes.

Then a tempo change: from bar 46 to 67 there's 20 bars of 4/4 (thus 4*20=80 quarter notes), and at 155 quarter notes per minute you add 155/80=1.94 minutes = 1m 56s. Total is 4m 56s.

Then a time signature change: 10 bars of 5 quarter notes per bar = 50 quarter notes. 155/50 = 3.1 minutes = 3m 6s. Total is 8m 2s.

Then another time signature change. 6 eighth notes per bar for 76 bars is 76*6=456 eighth notes. Tempo is still 155 quarter notes per minute, which is 310 eighth notes per minute. 456/310=1.471 minutes = 1m 28s. Total comes down to 9m 39s.

3. (The third movement of Shostakovich's second piano concerto, you get 5 bytes off if you listen to this while programming (not really but more people need to listen to Shosty dammit))

[1, "2/4"],
[1, "4=176"],
[75, "7/8"],
[102, "6/8"],
[103, "7/8"],
[106, "3/8"],
[107, "7/8"],
[109, "2/4"],
[112, "3/4"],
[113, "2/4"],
[116, "3/4"],
[117, "2/4"],
[120, "3/4"],
[121, "2/4"],
[124, "3/4"],
[125, "2/4"],
[155, "7/8"],
[160, "2/4"],
[175, "7/8"],
[180, "2/4"],
[181, "6/8"],
[182, "2/4"],
[186, "7/8"],
[188, "2/4"],
[222, "7/8"],
[225, "2/4"],
[286, "7/8"],
[308, "9/8"],
[309, "7/8"],
[314, "2/4"],
[317, "3/4"],
[318, "2/4"],
[321, "3/4"],
[322, "2/4"],
[325, "3/4"],
[326, "2/4"],
[329, "3/4"],
[330, "2/4"],
[356, ""]

Calculation is too long to show here, but it comes down to 4m 47s. The video is 5m 24s, proving my point that this is not an exact rule but rather a guideline.



Does this look like an interesting puzzle? Any tags I miss? Is the +/-34% allowance large enough to matter, or should it be more?

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  • \$\begingroup\$ The "+- 34%" and "at least 2:30 min" are unneccessary for the challenge. They may be relevant for music, but not for programming. I doubt that this allowance will allow you to shave bytes off the task, given that it's straightforward parsing + arithmetic otherwise. \$\endgroup\$ – AlienAtSystem Nov 21 '19 at 11:31
  • 1
    \$\begingroup\$ @AlienAtSystem I want to make the allowance relevant, both because it is musically appropriate, and because it would elicit different kinds of answers that approximate the solution. I suppose the arithmetic needs to be more complex before approximate solutions can be made with shorter programs? How about adding accelerando events that change the tempo over time? \$\endgroup\$ – KeizerHarm Nov 21 '19 at 11:41
  • \$\begingroup\$ Or maybe I should make a hilariously long list of possible events (e.g. fermata changing the length of a single note, all the different types of ritardando with slightly different slowing down behaviours, etc.) and make it part of the challenge to sort out which ones are relevant to get close enough to the solution? \$\endgroup\$ – KeizerHarm Nov 21 '19 at 11:44
  • 2
    \$\begingroup\$ I'm not an expert golfer, so I can't say at what level of complexity an approximation gets shorter than just the straightforward calculation. I think the site consensus is for approximative approaches to use precision scoring (How many out of this long list of test cases do you get accurately), because otherwise there is the hidden condition of "for all possible valid inputs", which is hard to prove for approximation algorithms. \$\endgroup\$ – AlienAtSystem Nov 21 '19 at 11:51
  • \$\begingroup\$ @AlienAtSystem Very well, I could generate a list of 100 "compositions" and make the % of properly solved cases part of the score. I guess that would mean adding a different tag? I couldn't find one for "approximation" or direct synonyms of that. \$\endgroup\$ – KeizerHarm Nov 21 '19 at 11:55
  • 1
    \$\begingroup\$ If you go for the fraction of cases correct method, the tag you want is test-battery. \$\endgroup\$ – FryAmTheEggman Nov 21 '19 at 19:46
  • \$\begingroup\$ So, we wont include 𝄆 repeat sign 𝄇 in this challenge. Am I right? \$\endgroup\$ – tsh Nov 27 '19 at 3:35
  • \$\begingroup\$ @tsh Not at this stage. And if I do, it definitely won't be using the unicode token. \$\endgroup\$ – KeizerHarm Nov 27 '19 at 10:38

Deathmatch Football (Soccer)


2 Teams each of 11 Players and 3 Bank Players compete in a match of 90 minutes to find out who's the best. But... it wouldn't be deathmatch with no casualties, so after scoring, there is a chance to die. Better think twice before you shoot...


Each Team is represanted as a String of 14 chars (11 + 3). Every char az-AZ is unique to the whole match. So 2 teams could look like this

Team1: {"a","c","f","g","j","k","A","D","E","H","I","k","n","o"} //bank: k, n, o
Team2: {"b","d","e","h","i","l","B","C","F","G","J","l","m","P"} //bank: l, m, P

The Game:

The game lasts for 90 (+3) minutes. Every minute each team has one chance to score a goal. The player who's shooting gets randomly selected. The chance of a player to score a goal is the byte value of the char.

Example: "z" has a value of 122, so he has a chance of 122% (>=100%) means he will definetely score

If he scores he will die with a chance of his byte value / 2

Example: "z" has a value of 122, so he has a chance of 61% to die.

Special Events:

  • At minute 45, 60 and 75 a random player from the bank comes into the team (teams are not limited to 11 players, if no one has died yet)
  • If there is a draw at minute 90 OR both teams have less than 5 players, the game lasts for additional 3 minutes (90 + 3)
  • If a team has 0 players the game is over

Result and Rating:

Once a match is over. Display both teams, the final score, and the top scorer including goals.

Team1: {"a","f","A","D","p"}
Team2: {"b","e","h","P"}
Score: 20:18 // So Team1 won
Top Scorer: A=>5

The code with less bytes is the winner.

Extra Notes:

You are free to choose the form of the output as long as you can clearly see the remianing teams, who won and the top scorer. No loopholes, do i need to say that?

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Multiplicative Digital Root

The Digital root of a number is found by iteratively summing its digits until you end up with a single number (e.g. 99 -> 18 -> 9)

The multiplicative digital root is found by iteratively multiplying the digits (e.g. 99 -> 81 -> 8)

The Challenge

Print out the digital root of all numbers 0..99 inclusive. Note that this is based on https://oeis.org/A031347, and the first 10 numbers (0-9) are treated without leading 0s.




In any sensible format:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 
0, 2, 4, 6, 8, 0, 2, 4, 6, 8, 
0, 3, 6, 9, 2, 5, 8, 2, 8, 4, 
0, 4, 8, 2, 6, 0, 8, 6, 6, 8, 
0, 5, 0, 5, 0, 0, 0, 5, 0, 0, 
0, 6, 2, 8, 8, 0, 8, 8, 6, 0, 
0, 7, 4, 2, 6, 5, 8, 8, 0, 8, 
0, 8, 6, 8, 6, 0, 6, 0, 8, 4,
0, 9, 8, 4, 8, 0, 0, 8, 4, 8,

Sandbox comments

  • this is the only related challenge I could find, but it doesn't ask the same thing.

  • Would this be better/more interesting as a "calculate the multiplicative digital root of the input" challenge?

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  • 1
    \$\begingroup\$ no, because 0,1..9 are single-digit numbers. 00, 01..09 would make the first row zero, but based on the OEIS sequence (oeis.org/A031347) I want to treat them as shown \$\endgroup\$ – simonalexander2005 Nov 25 '19 at 11:14
  • \$\begingroup\$ Ah ok, I misread it as a multiplication table instead of just a sequence of 0..n. Ignore my now deleted comment. So we just output the first 100 numbers in the (0-based) sequence? Would outputting the 1-based sequence be allowed (1..100 instead of 0..99)? If not np, but I could save a byte in my prepared solution if it would be allowed. :) \$\endgroup\$ – Kevin Cruijssen Nov 25 '19 at 11:48
  • \$\begingroup\$ Sorry yeah, I've just formatted it that way for clarity. No, 0-99 was the challenge I had in mind. Alternatively would this be a better challenge if you had to calculate the multiplicative digital root given an input (arbitrary-length integer)? Or should I just propose both challenges? \$\endgroup\$ – simonalexander2005 Nov 25 '19 at 12:52
  • \$\begingroup\$ Either is fine by me, but not both. If the single input would already exist as challenge, this ranged one would simply use those answers with a 0..99 map around it (which is also what I do in my current answer in my language of choice: calculating the multiplicative digital root of an input is 3 bytes; adding a range 0..99 map around it is 4 more bytes). \$\endgroup\$ – Kevin Cruijssen Nov 25 '19 at 12:58
  • \$\begingroup\$ The challenge you linked to is the same except that it doesn't require wrapping the thing in a for loop. I believe this is a duplicate. \$\endgroup\$ – my pronoun is monicareinstate Nov 25 '19 at 13:40
  • \$\begingroup\$ @mypronounismonicareinstate it wouldn't be a for loop, but rather recursion (or a while), right? Anyway, the challenge as written here currently asks for the whole set - so I'd say that's substantially different to the linked challenge \$\endgroup\$ – simonalexander2005 Nov 25 '19 at 13:47
  • \$\begingroup\$ I personally see it as closely related instead of a dupe @mypronounismonicareinstate, although I agree parts of the answers could be reused. That other challenge asks: remove all zeroes, and take the product of the remaining digits. Whereas this challenge asks: for the numbers in the range 0..99; reduce by taking the product of its digits until a single digit remains. In 05AB1E for example, that other challenge is 0KSP and this challenge is т<ÝεΔSP. The SP part (split to digits, take product) is similar. Both challenges are rather trivial in most languages, though - even non-golf langs \$\endgroup\$ – Kevin Cruijssen Nov 25 '19 at 14:35

Double an infinitely long number

You are given a non-negative real number strictly less than 0.5 as an endless stream of digits. Output twice of it in an endless stream of digits.

The exact format is flexible. You may separate digits by any reasonable separator, or don't use separators. You may prepend something representing "0" or "0.", or just omit the integral part. The input and output don't have to be in the same format. But everything must be in base 10, from the most significant digit to less significant digits.

Your code doesn't have to print anything immediately. But if each digit in the input would be given in a finite amount of time, and it is in a state allowing output in your chosen I/O method (so that if you are using a generator, you may assume some code is repeatedly accessing the generator), each digit in the output should also be printed or returned in a finite amount of time.

Shortest code wins.

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  • 1
    \$\begingroup\$ I presume the stream of digits is given from largest to smaller place? I think this would be cleaner if we didn't have to deal with a decimal point, say by saying the input starts after the decimal point and is less than 0.5. \$\endgroup\$ – xnor Nov 25 '19 at 10:10
  • \$\begingroup\$ Would it be a valid submission to give a stateful function that takes in a digit, and outputs some a string of digits, so that repeatedly calling it with digits from the stream gives the desired output if concatenated? Or does our code have to handle the read-and-output loop itself? \$\endgroup\$ – xnor Nov 27 '19 at 1:30
  • \$\begingroup\$ @xnor Not sure. I'll accept it if it becomes a standard I/O method for reading/writing lists. \$\endgroup\$ – jimmy23013 Nov 27 '19 at 10:52

Product of Shuffle Algebra

Compute the shuffle product of two elements of a Shuffle Algebra.


  1. An Alphabet is a set of symbols. For this challenge it is \$A = \{a,b,c,\ldots,z\}\$. Let us use bold letters for variables that denote elements of the alphabet.
  2. A Word is a tuple of of arbitrary size where each entry is an element of our alhpabet. Let \$W\$ be the set of words. Instead of writing a word as \$(p,p,c,g)\$ we just omit the unnecessary symbols and write it as \$ppcg\$. We write the empty word as \$\varepsilon = ()\$. Let us use lower case greek letters (\$\alpha, \beta, \gamma, \ldots\$) for variables that denote words.
  3. The shuffle algebra \$S\$ (simplified for the purpose of this challenge) is a set of \$\mathbb Z\$-linear combinations of words. This means that each element consists of a few words which each have an integer associated with them (the coefficients), which are then formally summed up. Here are some examples: $$\begin{array}[l] \\ a \\ b \\ 5\varepsilon = 5\\ a + (-5)b = a-5b \\ a + 2a = 3a \\xy - 3code +5golf +5se \end{array}$$ Note that terms (that is a word with it's coefficient) where the coefficient is zero are considered as \$0\$ and are not listed. Let us use uppercase letters form the greek alphabet (\$ \Gamma, \Delta, \Phi, \Psi, \ldots \$) letters for variables denoting an element of the shuffle algebra.
  4. We can add elements of the shuffle algebra just as you'd expect: The coefficients of equal words are just summed. for instance let $$\begin{align*} A &:= aq + 3bccd + 5cg \\B &:= -3bccd + 6cg + 5qq .\end{align*}$$ Then $$\begin{align*} A+B &= aq + 0bbcd + 11cg + 5qq \\&= aq + 11cg + 5qq.\end{align*}$$
  5. We multiply elements of the shuffle algebra using the shuffle product denoted by the symbol \$⧢\$. We can recursively define it as follows:

    A. First we define the shuffle product for words: if one of the words is empty (=\$\varepsilon\$) we have $$\alpha ⧢ \varepsilon = \varepsilon ⧢ \alpha = \alpha.$$ If both are nonempty, we split them up into a one letter suffix and a leading word. So let us write the two words \$\alpha\$ and \$\beta\$ as $$\alpha = \varphi \mathbf{a} \qquad \beta = \psi \mathbf{b}$$ Then the product is defined via the recursion $$\alpha ⧢ \beta = (\varphi ⧢ \beta)\mathbf{a} +(\alpha ⧢ \psi)\mathbf{b}.$$ This is where the "shuffle" in the name is coming from: The shuffle product results in all possible riffle shuffles of the two words. Consider the following examples: $$\begin{align*}ab ⧢ xy = abxy + axby + xaby + axyb + xayb + xyab\end{align*}$$

    B. Two shuffle algebra elements \$\Phi = \sum_i f_i \varphi_i \$ and \$\Psi = \sum_j g_j \psi_j \$ (where \$f_i, g_j \in \mathbb Z, \varphi_i, \psi_j \in W\$) are multiplied as follows: $$\Phi ⧢ \Psi = \sum_i \sum_j \underbrace{(f_i \cdot g_j)}_{\in \mathbb Z} \underbrace{(\varphi_i ⧢ \psi_j)}_{\in S}$$


  • The inputs can each be taken as a string or as a list of pairs where each pair is represents one term (coefficient and word) of the element, or alternatively as two lists of the same length or other similar formats.
  • The output format must match the input format.
  • The output must be reduced: No two terms should share the same word.


Haskell script for generating Examples

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  • \$\begingroup\$ Might you consider just having the challenge be to multiply two words? When thinking about how I'd do this challenge, I'm imagining implementing the shuffle product, than iterating over pairs of monomials and multiplying them and their coefficients, then reducing by combining like terms. I don't know anything about the shuffle algebra, so I don't know if there's something that lets you work with elements in some more special way. \$\endgroup\$ – xnor Sep 14 '19 at 20:10
  • \$\begingroup\$ @xnor My thought was that if we just take two words for the input, the output will still be a general element (e.g. aa ⧢ ab = 3*aaab +2*aaba +1*abaa), so you would already have to implement the simplification (collecting the the terms with the same words), so the additionl step of multiplying two general elements wouldn't be a lot more compilcated. - Would you suggest also removing the simplification step, so that the output would just be a list? (e.g. aaab,aaab,aaba,aaab,aaba,abaa for the input from above) I think this would on the one hand be a lot easier, but also less of a challenge. \$\endgroup\$ – flawr Sep 14 '19 at 20:31
  • \$\begingroup\$ Hmm, I hadn't considered that a riffle product of monomials would already produce repeats. I don't know what I'd suggest. For what it's worth, I tried searching for a duplicate of generating all possible riffles and didn't find one, though I have a memory of having done something like it on anarchy. \$\endgroup\$ – xnor Sep 14 '19 at 20:37

Walk N Spaces on a Game Board

This challenge is inspired by Touhou Cannonball, a mobile game that includes a board game element similar to Mario Party, except for the fact that movement isn't restricted to one particular direction.

A Touhou Cannonball board.


On your turn, a six-sided dice is rolled to determine how many spaces to move. For each of those spaces, you can move to any adjacent space, as long as you do not backtrack to the space you were on in a previous move.

For example, if you roll a 2 and start from space [2] in the following graph, you can move left twice to space [0] or right twice to space [4], but you cannot move once left and once right to land on space [2] again, because you traveled across the same edge in two consecutive moves.


You can move to a previously visited space, as long as you aren't "backtracking" to that space -- if you roll a 4 on the following graph starting from space [0], you can land on [0] by going in a circle around the graph with the move [0]->[1]->[2]->[3]->[0].

^    ^
|    |
v    v

Some graphs may have some directed edges. If your graph is [0]-->[1], you can move from [0] to [1], but not the other way around.

The Challenge

Given any appropriate directed graph structure (incidence matrix, adjacency list, etc.), a starting position, and a number of spaces to move, output all possible positions you can move to.


Example inputs are zero-indexed in the format starting position, number of spaces, adjacency list. I will also have the visual graph included for better visualization. (More to be added later)

Input: 2, 5, [ [1,2], [0,3], [3,4], [1,6], [2,5], [4,6], [3,5] ]
^       ^
|       |
|       v
^       ^
|       |
v       v

Output: [1,2,5]
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How fair are my dice?

(Inspired by a dream I recently had regarding Mario Party. This is not fully fleshed-out yet.)

Given a pair of dice, e.g., [[1,1,2,2,3,3], [1,2,3,4,5,6]], output how "closely" they resemble a pair of standard 1-6 dice (i.e., [[1,2,3,4,5,6], [1,2,3,4,5,6]]) in terms of distribution of numbers when rolled.

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  • \$\begingroup\$ Is the input always a pair of six-sided dice or can they have a variety of face counts? Sorted? How would [[2,3,4,5,6,7],[2,3,4,5,6,7]] compare considering it'd have the same distribution curve? \$\endgroup\$ – Veskah Dec 2 '19 at 16:20
  • \$\begingroup\$ You need to flesh out details on how this should be measured. Do I go by simulation? Do I base the distribution on the total of each pair of rolls? What is "closely"? Sum of squared error? \$\endgroup\$ – Beefster Dec 6 '19 at 19:56

A game of putting lines through dots.

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Rate your brevity

This challenge is to produce a script that will accept a string as input, and generate a score for it using the rules defined below. Your score is the return value of your program given its own source as input.

The first scoring system

Each additional unique character has an escalating penalty to your score.

The first unique character is worth -1
The second unique character is worth -2
The third unique character is worth -3

The second scoring system

Each repetition of a given character incurs a further escalating penalty to your score.

The first usage of a given character incurs no additional penalty
The second usage of a given character incurs an additional penalty of -1
The third usage of a given character incurs an additional penalty of -2
The fourth usage of a given character incurs an additional penalty of -3


  • The code abc would have a score of -6
  • The code aabc would have a score of -7
  • The code aaaabc would have a score of -12


  • Your score starts at 0
  • The highest score for a given language wins.


The below Stack snippet will parse a string according to the rules defined above and generate the score.

function testScore() {
  var score = 0,
      counter = 0,
      input = document.getElementById('code').value,
      instances = {},
  input = input.split('');

  for(x = 0; x < input.length; x++) {
    if(typeof instances[input[x]] == 'undefined') {
      instances[input[x]] = 0;
      score -= ++counter;

  keys = Object.keys(instances);
  for(x = 0; x < keys.length; x++) {
    for(y = 0; y < instances[keys[x]]; y++) {
      score -= y;

  alert('Your score is:\n' + String(score));
  return false;
<form onsubmit="return testScore(this);">
    <textarea id="code" style="width: 100%; height: 150px;"></textarea>
    <button type="submit">Generate score</button>

| |
  • \$\begingroup\$ I'd appreciate some pointers on what tags to use for this one - it's similar to code golf but isn't quite code golf. \$\endgroup\$ – Scoots Dec 11 '19 at 11:48
  • \$\begingroup\$ Similar but a slightly different scoring system \$\endgroup\$ – Veskah Dec 11 '19 at 13:02
  • \$\begingroup\$ This is two challenges: First, reading your own source code. Second, creating the score for an input string. In addition, the condition of "Parse yourself" is a non-observable requirement. In this case, it's definetly bad because it forces the two separate challenges approach onto people. I understand why you want to exclude print -45, but right now, the whole quine-except-not-really thing doesn't seem helpful. \$\endgroup\$ – AlienAtSystem Dec 13 '19 at 14:09
  • \$\begingroup\$ @AlienAtSystem Respectfully I disagree with your assertion that reading the source code is a challenge; I'm very explicit that any means of getting your source code in is acceptable, even copy/pasting it as a command line argument. The parsing of your own code - I thought it might encourage people to golf in a way they may not be used to due to the escalating penalties involved in the scoring system. \$\endgroup\$ – Scoots Dec 13 '19 at 15:36
  • 1
    \$\begingroup\$ If you don't want the source code reading to be part of the challenge, then don't write things like it is. Really just say Takes a string as input and outputs the uniqueness score and Your score is the return value of your program when given its own source as input. \$\endgroup\$ – AlienAtSystem Dec 14 '19 at 9:13
  • \$\begingroup\$ @AlienAtSystem I take your point, your phrasing is definitley better than what I originally had. \$\endgroup\$ – Scoots Dec 14 '19 at 11:32

Nash equilibrium of 2-player 3-choice zero-sum game


In a 2-player single finite game, a Nash equilibrium (NE) is a pair of strategies chosen by both players where neither has an incentive to deviate from their own strategy. For example, here is a payoff matrix of two players \$ A,B \$ where each player has two possible pure strategies \$ X,Y \$:

$$ \matrix{ A \backslash B & X & Y \\ X & 3 \backslash 3 & 0 \backslash 0 \\ Y & 0 \backslash 0 & 2 \backslash 2 } $$

In this game, if \$ A \$ chooses \$ X \$, \$ B \$ should also choose \$ X \$ because it will give \$ B \$ the best outcome. Likewise, if \$ B \$ chooses \$ X \$, \$ A \$ should also choose \$ X \$. Therefore, \$ (X,X) \$ is a NE. We can observe \$ (Y,Y) \$ is also a NE; although it gives less outcome than \$ (X,X) \$ for both players, there is no reason to select \$ X \$ when the opponent selects \$ Y \$.

But a strategy can also be a mixed strategy, i.e. choosing one of the pure strategies with some chance. Let's take a mixed strategy \$ S = \frac25X+\frac35Y \$. If \$ A \$ takes \$ S \$, \$ B \$'s expected outcome is always \$ \frac65 \$, regardless of \$ B \$'s strategy (either pure or mixed). In this case, \$ B \$ has zero incentive to deviate from whatever strategy \$ B \$ is already taking. The same can be said for \$ B \$ taking \$ S \$, so \$ (S, S) \$ is also a NE.

Nash's existence theorem states that every game has at least one Nash equilibrium, either pure or mixed. Note that a game may have many (possibly infinitely many) NEs.


In this challenge, we consider 2-player 3-choice zero-sum games, where each player has three possible pure strategies, and the sum of the two players' outcomes is always zero. The payoff matrix of such a game might look like this:

$$ \matrix{ A \backslash B & R & P & S \\ R & 0 \backslash 0 & -1 \backslash 1 & 1 \backslash -1 \\ P & 1 \backslash -1 & 0 \backslash 0 & -1 \backslash 1 \\ S & -1 \backslash 1 & 1 \backslash -1 & 0 \backslash 0 } $$

Since it is a zero-sum game, we can omit the payoffs of \$ B \$:

$$ \matrix{ A & R & P & S \\ R & 0 & -1 & 1 \\ P & 1 & 0 & -1 \\ S & -1 & 1 & 0 } $$

This game is equivalent to Rock-Paper-Scissors, and the only NE is, as we all know, mixed \$ (\frac13R + \frac13P + \frac13S,\frac13R + \frac13P + \frac13S) \$.

The challenge is to find at least one of the Nash equilibria, given the payoff matrix for player \$ A \$. Remember that a Nash equilibrium is a pair of \$ (\text{strategy of } A,\text{strategy of } B) \$.

For an outline of finding a mixed-strategy NE, refer to this Math.SE question. Also related: why some games don't have mixed-strategy NE.

Input and output

The input is a 3-by-3 matrix of integers which represent the payoff matrix for player \$ A \$. You can assume the magnitude of each integer won't exceed 100, and you can take the numbers in any order of your choice.

The output is a pair of strategies in the form of \$ (a_XX+a_YY+a_ZZ, b_XX+b_YY+b_ZZ) \$ where \$ a_i \$ and \$ b_i \$ are the chances of selecting one of the strategies. You don't need to format the output; outputting the 6 numbers \$ a_X,a_Y,a_Z,b_X,b_Y,b_Z \$ is fine. If you want to output a pure strategy, you can do something like \$ (1, 0, 0) \$.

Scoring and winning criterion

Standard rules apply. The shortest code in bytes wins.

Test cases

Matrix: (classic RPS)
  0  -1   1
  1   0  -1
 -1   1   0
NE: A's strategy = (1/3, 1/3, 1/3), B's strategy = (1/3, 1/3, 1/3)
Matrix: (scored RPS)
  0  -5   1
  5   0  -2
 -1   2   0
NE: A's strategy = (2/8, 1/8, 5/8), B's strategy = (2/8, 1/8, 5/8)
Matrix: (dumb game, pure strategy NE example)
  0   1   2
 -1   0   0
 -2   0   0
NE: A's strategy = (1, 0, 0), B's strategy = (1, 0, 0)
Matrix: (asymmetric game example, infinitely many NEs)
  2  -1   3
  0   3  -1
 -2  -3   0
NE: A's strategy = (1/2, 1/2, 0),
    B's strategy = any triple (x, y, z) that satisfies
                   x + 2z = 2y, x + y + z = 1, 0 <= x,y,z <= 1
Matrix: (asymmetric game example, one mixed NE)
  2  -1   4
  0   3  -1
 -2  -3   0
NE: A's strategy = (1/2, 1/2, 0), B's strategy = (2/3, 1/3, 0)


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This is a repost of the underspecified challenge.


Remove all duplicates from one list of integers. A list is simply a sequence of connected values that allows the same values to be stored at different positions in this sequence.

If an item was found to be the same value as another item in this list, keep the first occurence of the item and remove the second (third and so on...) occurence of the item.


You may write a function or full program that takes a list as an input and returns a remove-duplicated list as an output.


  • This is a challenge, so shortest answer wins!
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Join two file paths

A file path consists of letters a-z separated by slashes /. Your task is to join two paths as follows:

  • If the second path doesn't start with /, concatenate them with exactly one / in between.
  • If the second path starts with /, just output it.

For example,

/User/Desktop   temp/file   ->  /User/Desktop/temp/file
/User/Desktop/  temp/file   ->  /User/Desktop/temp/file
/User/Desktop   /temp/file  ->  /temp/file
/User/Desktop/  /temp/file  ->  /temp/file
relative/path   more/path/  ->  relative/path/more/path

You may assume the inputs don't contain two consecutive slashes, but they may start or end with a slash. Each input with be nonempty and not just a slash. You may use backslash \ in places of slash /.

Sandbox: Is this too boring? Too likely a built-in?

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String Blockify™ a Hexagon

an obvious rip-off of Hexagonify™ a String Block

What is Hexagonification?

Hexagonification is a transformation that creates a hexagon with 3 copies of a rectangle block™, each skewed to the right by 30 degrees and then rotated by 0, 120 and 240 degrees respectively, as shown in the following image. A triangle hole may appear in the middle, but that isn't a great deal.

diagram of hexagonification


Write a program or function that receives a hexagonified block of string™ as an input and outputs the original string block™. diagram of string blockification The input and output formats are flexible. You may receive a single string, a list of lines or a 2D array as input, and output a single string or a list of lines. You may have leading/trailing spaces on each line and leading/trailing newlines, provided that the block is properly formed. See Sample IO for how the blocks™ should be positioned.

Sample IO


   3 x 3
  e s q u
 u r a r e
3 q a a s 3
 x s r q x
  3 e u 3




   l o n g e r . . . . .
  . r e c t a n g u l a r
 r . b l o c k . . . . . .
. a .                 b r l
 . l .               l e o
  . u .             o c n
   . g .           c t g
    . n k         k a e
     r a c       . n r
      e t o     . g .
       g c l   . u .
        n e b . l .
         o r . a .
          l . r .




        v b
       8 e l
      x l r o
     2 a   t c
    k c     i k
   c i       c 2 
  o t         a x
 l r           l 8
b e l a c i t r e v
 v 8 x 2 k c o l b  




 1 . l i n e
e           1
 n         .
  i       l
   l     i
    .   n
     1 e




     e .
    n   l
   i     i
  l       n
 .         e
1 e n i l . 1




Your score will be the area of the source code as a String Block™. Assuming your code can just fit inside a rectangle of width \$w\$ and height \$h\$, the score will be:

$$w \cdot h$$

Winning Criteria

The submission with the lowest score in each language wins.

| |
  • \$\begingroup\$ The obvious problem here is that the reverse challenge is "cluttered" with the information being available three times. The solutions won't profit from the structure being a full hexagon, one of the three skewed rectangles is enough. \$\endgroup\$ – AlienAtSystem Dec 22 '19 at 16:47

Joyous Kwanzaa!

Kwanzaa is an annual week-long celebration of African culture and history, and takes place on December 26 to January 1. During Kwanzaa it is traditional to light the seven candles of a kinara, one on each day. The middle candle is lit first, followed by the rest of the candles from left to right.


Your challenge will be to write a program or function that when given ASCII art (in any convenient format) of candles on a kinara (as specified below), some of which may be lit, will return or output a ASCII art of what the kinara would look like on the next day.


A kinara will contain of seven even columns of equal length with equal spacing between them, representing the candles. The candles will be constructed with some consistent printable, non-whitespace ASCII character.

Some of the candles might have a flame above them, which will be represented by a single printable, non-whitespace ASCII character, possibly the same as the candle's character. If no candles are lit, there will be an empty line above the candles.

Here is an example of a kinara ASCII art:

*     *
| | | | | | |
| | | | | | |


The input will be a kinara with seven candles, 0-6 of them lit according to the rule (middle candle first, then the rest from left to right).

The output will be a kinara with candles and flames identical to the input kinara, but with one more lit candle than the input. The extra candle will be lit according to the rule. If no candles are lit in the input, you may use any character for the flame.

You may assume the input is valid according to the specification given. You may assume the input has no trailing or leading whitespace in any of the lines, but trailing or leading whitespace is acceptable in the output as long as the kinara has the proper shape.



*     *
| | | | | | |
| | | | | | |

* *   *
| | | | | | |
| | | | | | |


** *



5 5 5 5
6 6 6 6 6 6 6
6 6 6 6 6 6 6

5 5 5 5 5
6 6 6 6 6 6 6
6 6 6 6 6 6 6


               # pretend this is an empty line
q q q q q q q

q q q q q q q


  • No standard loopholes

  • Shortest code in bytes wins


  • Is this too complicated? If so how could I simplify it?

  • Is the wording unclear?

  • Is this too similar to an existing challenge? I found another challenge which involves creating a menorah, but this one takes candles as input instead of a date.

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Hit the most Balloons with one Arrow

You got 1 arrow to hit as much balloons as possible. A common way to solve this, is to mark the tangent points of every balloon and project the angles on the source's position. As seen in the following image this has been down already. The task is now to find the optimal and angle-centered solution to hit the most Balloons.


The Input:

A json array of length 360 (for every 1° one entry) representating the number of Balloons hit at this angle.

JSON: https://pastebin.com/raw/Rd8g7Z6J

The Task:

Find an algorithm that solves the task with the following requirements:

  • return the number of most hits
  • return the max range of angles with the most hits
  • return the the centered angle of this range
  • there are no format restrictions how you are returning the values, it could be 3 prints or an array... whatever, as long as it is clear to see whats the hits, range and center

Solution of given JSON:

  • Most Hits: 6
  • Range: 358 - 0
  • Center: 359


Remember its a circle, so you need to check if the range crosses the 360°/0° mark!!! A range can be like 300° - 10° for example.

For simplicity the angles are 0° - 359° as 360° would be 0°.

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  • 3
    \$\begingroup\$ That's a lot of words and a picture wasted because your challenge is really just "Find the longest max-value sequence in the array", which is probably a duplicate. I think the challenge would be a lot better if you actually had to calculate this angle distribution from an input of balloon coordinates and a size. \$\endgroup\$ – AlienAtSystem Dec 23 '19 at 6:29

Bunda-Gerth parser for abstract APL-family language


APL has four kinds of tokens: arrays, functions, monadic operators (mops) and dyadic operators (dops).1) Let's denote them A (arrays), F (functions), M (mops), and D (dops) respectively. Usually when programming in APL, we use a mental model like this:

  • Strand notation: A A ... A -> A
  • Monadic function application: F A -> A
  • Dyadic function application: A F A -> A
  • Mop with left operand: (A|F) M -> F
  • Dop with both operands: (A|F) D (A|F) -> F
  • Mops and dops bind first from left to right (producing functions), and then function applications are evaluated from right to left

But J.D.Bunda and J.A.Gerth proposed another way to parse APL expressions. In this method, each (ordered) pair of tokens is assigned a binding strength and a resulting token. Below is the translation of the above mental model. Note that some token pairs do not bind at all, and we have a new token type AF, which stands for a function with bound left argument.

Binding strength | Token pair(s) -> result
               5 | A A -> A        # Strand notation
               4 | D (A|F) -> M    # Dop + Right operand -> Mop
               3 | (A|F) M -> F    # Left operand + Mop  -> Function
               2 | A F -> AF       # Array + Function    -> Left-bound function
               1 | (F|AF) A -> A   # Function + Array    -> Array

Then the actual parsing proceeds as follows: given a stream of tokens,

  1. Evaluate the binding strengths between adjacent tokens.
  2. Select the token pair whose binding strength is the rightmost local "peak", i.e. the x y pair in w x y z satisfying wx < xy >= yz. w and/or z can be empty. Note that the leftmost pair is bound first in A A ... A.
  3. Group the pair and produce a new token as specified in the transition table.
  4. Repeat from 1 until single token is left (successful parse) or no more reduction is possible (syntax error).


Given a list of tokens, output the order of binding (refer to the I/O section for specification) using the transition table and algorithm described above.

Input and output

The input is a non-empty list of A/F/M/D tokens. You may use any number or single character to represent a token. You may assume that the input will always parse successfully, and the input won't contain any parentheses (since it isn't specified in the table above).

For the output, the order of binding represents in which order each gap between the tokens is closed:

Given the tokens:       A A F M A
Assign each gap an ID:   w x y z
Binding strength:        5 2 3 0
Rightmost local peak:       F M
Bind them first:        A A (F M) A     -> y = 1
Bind A-A next:         (A A) (F M) A    -> w = 2
Bind A-F next:        ((A A) (F M)) A   -> x = 3
Bind AF-A last:      (((A A) (F M)) A)  -> z = 4

Given the tokens:       A A F M A
The order of binding:    2 3 1 4  (the answer)

Note that, if the input contains L tokens, the answer is always a permutation of 1..L-1 (or 0..L-2 if you choose zero-based numbering).

Scoring and winning criterion

Standard rules apply. The shortest code in bytes wins.

Test cases

Input:  A A A A A
Output:  1 2 3 4
Input:  A F F A A F F A
Output:  6 7 5 2 3 4 1
Input:  A  F D A A  M  A A D F M F A
Output:  11 9 8 7 10 12 3 4 2 5 6 1


  1. An actual implementation of APL has some special tokens, e.g. dot ., jot , and index notation [x]. We ignore them here for simplicity.
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  • \$\begingroup\$ Maybe an easy answer format is simply inserting the order of binding? A F F A A F F A((A F) (F (((A A) F) (F A))))((A5F)7(F6(((A1A)2F)4(F3A)))) and maybe even returning the permutation vector 5 7 6 1 2 4 3? \$\endgroup\$ – Adám Dec 17 '19 at 8:42
  • \$\begingroup\$ @Adám That could work, but I don't think it's any easier to produce than a fully structured output. Also, alternative algorithms can produce the same tree with different binding order. \$\endgroup\$ – Bubbler Dec 17 '19 at 23:27
  • \$\begingroup\$ It will be really hard to verify with the broad output allowance. Maybe just require fully parenthesising? \$\endgroup\$ – Adám Dec 17 '19 at 23:55
  • \$\begingroup\$ @Adám IMO it's not a problem, as long as the answers specify which output format they're using. \$\endgroup\$ – Bubbler Dec 18 '19 at 0:16
  • \$\begingroup\$ On second thought, I decided to emphasize the algorithm itself (over the resulting parse tree). \$\endgroup\$ – Bubbler Dec 26 '19 at 2:04

Potentially Prime Punch-Card Patterns


Let us define a punch-card to be a set of 'open' and 'closed' holes which can slide over the positive integer number line:

   |   ___     ___     ___   | -->
   |  |   |   |   |   |   |  |
1  |  | 3 |   | 5 |   | 7 |  |  9   10  11  12  13  14
   |  |___|   |___|   |___|  |
   |_________________________| -->

When the first open hole is on the integer \$ p \$, then this punchcard (as it slides across the number line) yields the integers \$p, p+2,\$ and \$p+4\$. So, we can represent this card as the set \$\{0, 2, 4\}\$.


Given a punch-card we may wonder whether it is possible to slide the card to such a position that only prime numbers are under the 'open' holes.

Clearly, for \$\{0, 2, 4\}\$, we can position the punch-card to make 3, 5, and 7 visible. However, this is the only solution. Taking each element modulo 3 yields \$\{0, 2, 1\}\$, which contains every possible remainder under division by 3, so one visible integer must be divisible by 3. But we only want primes, and so 3 itself must be visible; this leaves a finite number of possible positions, of which only one is a valid solution.

More generally, if there exists a prime \$q\$ such that every integer \$0 \le n < q\$ appears in the punch-card set modulo \$q\$, then the punch-card is inadmissible and has a finite number of positions where all visible integers are prime: as we are bounded by the condition that \$q\$ must be visible. We can place \$q\$ in each hole in turn, and perform primality tests to count the valid solutions.

However, if there exists no such \$q\$, then the punch-card is admissible and we cannot assume there are finite solutions; the K-Tuple Conjecture in fact hypothesizes that every admissible punch-card can assume infinitely many positions where all visible integers are prime.

The Challenge

Your task is to write a program or function which, given a list of ordered positive integers representing a punch-card set, outputs the number of positions the punch-card can assume where all visible integers are prime. If the set is admissible, then give a distinct output such as -1, null, Inf - anything that is not a non-negative integer.

Test Cases

Coming soon.


There is a related challenge - testing for admissible sequences. However, I believe this is distinct enough to be a duplicate as rather than being a simple , if a set is admissible this program will have to then try different positions of the punch-card and count the valid solutions; whereas the previous challenge considers admissible sets in isolation, this challenge applies it to prime numbers.

| |
  • \$\begingroup\$ Given that the admissibility test is already a challenge, how about you split off the two-challenge and special case part: The input can be assumed to be an admissible sequence, the task is simply finding the positions resulting in valid solutions. \$\endgroup\$ – AlienAtSystem Dec 28 '19 at 5:54
  • \$\begingroup\$ @AlienAtSystem i just feel as though that it is much less interesting to golf, and returning the distinct different output would open more interesting golfing opportunities. \$\endgroup\$ – FlipTack Dec 28 '19 at 6:09
  • \$\begingroup\$ Handling special cases is very rarely interesting to golf, for a quite simple reason: Checking if the special case is present requires bytes. Being told the simple case is the case saves those bytes. \$\endgroup\$ – AlienAtSystem Dec 29 '19 at 20:29
  • \$\begingroup\$ @AlienAtSystem but for there to be more golfing opportunities, don't there have to be more bytes to golf? \$\endgroup\$ – FlipTack Dec 29 '19 at 20:37
  • \$\begingroup\$ Not really. The issue is that for multiple task challenges, the byte count is 90% of the time optimal_bytes(task1)+optimal_bytes(task2). Often, one of the tasks is considerably longer to golf, to the point where optimizing the other task is almost irrelevant because it's so small compared to the other. Therefore, the site consensus is to split challenges into their individual components as much as possible, and especially avoid input validation, because it's boring and requires lots of bytes. \$\endgroup\$ – AlienAtSystem Jan 1 at 8:29
  • \$\begingroup\$ But it's not input validation in the sense of there being erroneous or invalid inputs: it's just that the answer to the question "how many prime positions does this have?" may be 0,1,2..., and also (probably) infinity. \$\endgroup\$ – FlipTack Jan 1 at 11:04

How many ways can I count on n?

By using addition of natural numbers {1, 2, 3...} and multiplication of natural numbers larger than 1, we can reach the same outcome in several ways. For example 4 = 2 x 2 but also: 4 = 2 + 1 + 1. Using normal mathematical operator precedence, there are actually 6 ways to express 4 and 9 ways to express 5.

Since addition is commutative, a + b is counted as the same solution as b + a. The same holds for multiplication. So 7 = 1 + 2 x 3 is the same solution as 7 = 3 x 2 + 1.

The (trivial) solution n (using no addition or multiplication) is also counted as one of the solutions of n.

Multiplication with 1 is forbidden, because you can do this infinitely.


For a given input n, output c(n), which is defined as the number of ways n can be uniquely expressed using zero, one or more additions and multiplications of natural numbers, using normal mathematical operator precedence and without multiplication with 1.


  • Input and output are integers (your program should at least support input and output in the range of 1 up to 32767)
  • Invalid input (0, floats, strings, negative values, etc.) may lead to unpredictable output, errors or (un)defined behaviour.
  • Default I/O rules apply.
  • Default loopholes are forbidden.
  • This is , so the shortest answers in bytes wins

Final note

I can think of several ways to approach this problem. I think it's interesting to see how the different approaches impact the length of the solution.

Sandbox questions

  • Please let me know if this task/problem is stated clear enough.
  • Please let me know if there are any loopholes that I should cover in the question.
  • I intentionally omitted more examples (because I hope contesters will think about solutions rather than just reproduce an OEIS sequence; although I'm not sure if this question had an OEIS sequence).
  • I will tag this question with .
| |

Morse with Binary

For every input string (in ASCII, containing only lowercase alphabetic characters), output the alphabetic string converted from Morse code to Binary code.

In this case we are assuming that . is 0 and - is 1.

How to do the conversion

For reference, here are a few Morse code libraries:

a .-
b -...
c -.-.
d -..
e .
f ..-.
g --.
h ....
i ..
j .---
k -.-
l .-..
m --
n -.
o ---
p .--.
q --.-
r .-.
s ...
t -
u ..-
v ...-
w .--
x -..-
y -.--
z --..

What's amazing about those codes is that the code length never exceeds 4 codes.

And now we are trying to cofuse binary with Morse code. First let's convert sample to morse code:

... .- -- .--. .-.. .

And consider the whole thing as a binary:

.... ---. --.. -...

Here is an alternative table for mapping the characters:

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

The result of this operation, after conversion, is aomi. You are never going to end up with a letter past p.

| |
  • 1
    \$\begingroup\$ What if there is a group of characters remaining? Input: e, Output: ? \$\endgroup\$ – Element118 Jan 1 at 5:23

Quine Generator Generator... of any* length!

A Quine Generator Generator... is a quine when the input is empty or 0. Otherwise, for any other sufficiently large positive integer input, it should print a Quine Generator Generator... of that specified length (in the same language, same options).

Let S be the (most likely theoretically infinite) set of Quine Generator Generator... you can generate from your initial Quine Generator Generator....

Your score is the smallest N such that any Quine Generator Generator... in S can generate a Quine Generator Generator... of length N or longer (for reasonably sized N).

Input can be from standard input or as an argument of a function.

Lowest score wins.

Sandbox Meta:

Typically, code is scored by bytes, but now the length would depend on how that is interpreted, since it is included in the description of the question. Hence would it be a problem to put: "You may choose to define characters in terms of bytes or characters (should they differ) for purposes of the program length and scoring purposes."

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