This "sandbox" is a place where Code Golf users can get feedback on prospective challenges they wish to post to main. This is useful because writing a clear and fully specified challenge on your first try can be difficult, and there is a much better chance of your challenge being well received if you post it in the sandbox first.

Sandbox FAQ


To post to the sandbox, scroll to the bottom of this page and click "Answer This Question". Click "OK" when it asks if you really want to add another answer.

Write your challenge just as you would when actually posting it, though you can optionally add a title at the top. You may also add some notes about specific things you would like to clarify before posting it. Other users will help you improve your challenge by rating and discussing it.

When you think your challenge is ready for the public, go ahead and post it, and replace the post here with a link to the challenge and delete the sandbox post.


The purpose of the sandbox is to give and receive feedback on posts. If you want to, feel free to give feedback to any posts you see here. Important things to comment about can include:

  • Parts of the challenge you found unclear
  • Comments addressing specific points mentioned in the proposal
  • Problems that could make the challenge uninteresting or unfit for the site

You don't need any qualifications to review sandbox posts. The target audience of most of these challenges is code golfers like you, so anything you find unclear will probably be unclear to others.

If you think one of your posts requires more feedback, but it's been ignored, you can ask for feedback in The Nineteenth Byte. It's not only allowed, but highly recommended! Be patient and try not to nag people though, you might have to ask multiple times.

It is recommended to leave your posts in the sandbox for at least several days, and until it receives upvotes and any feedback has been addressed.


Search the sandbox / Browse your pending proposals

The sandbox works best if you sort posts by active.

To add an inline tag to a proposal, use shortcut link syntax with a prefix: [tag:king-of-the-hill]. To search for posts with a certain tag, include the name in quotes: "king-of-the-hill".

  • \$\begingroup\$ What if I posted on the sandbox a long time ago and get no response? \$\endgroup\$
    – None1
    Commented May 15 at 14:05
  • \$\begingroup\$ @None1 If you don't get feedback for a while you can ask in the nineteenth byte \$\endgroup\$
    – mousetail
    Commented May 29 at 13:27

4705 Answers 4705

129 130
132 133

Create a non-renewable world

In Minecraft, we can split all possible items into two categories: renewable and non-renewable. All non-renewable items are simply items that are not renewable.

Minecraft worlds are not infinite, meaning that, for some resources, there are only a finite number available. This number is often huge, but it is finite. For example, diamond ore can only be obtained by mining blocks that are generated when the world is created, meaning that there is only a finite number of them in any given world.

However, some resources can be obtained infinitely. For example, as Raid Captains will spawn indefinitely, it is possible to obtain an arbitrarily large number of Ominous Banners, dropped when you kill a Raid Captain. Or, as you can create an infinite water source, and you can get infinite iron by killing Iron Golems, it is possible to get an infinite number of Water Buckets.

Any resource for which it is possible to obtain an arbitrary amount of is called renewable, and there are a lot of these. Any item that is not renewable is non-renewable. There are 40 of these that are obtainable in a classic Minecraft 1.16 survival world. They are:

A Netherite Tool               Netherite Armor                Ancient Debris                 Block of Diamond              
Block of Netherite             Coal Ore                       Cobweb                         Conduit                       
Dead Bush                      Diamond                        Diamond Ore                    Dragon Egg                    
Dragon Head                    Elytra                         Emerald Ore                    Enchanted Golden Apple        
Enchantment Table              Gilded Blackstone              Gold Ore                       Heart of the Sea              
Iron Ore                       Jukebox                        Lapis Lazuli Ore               Large Fern                    
Lava Bucket                    Lodestone                      Nether Gold Ore                Netherite Ingot               
Netherite Scrap                Iron/Gold/Diamond Horse Armor  Pigstep Music Disc             Quartz Ore                    
Redstone Ore                   Shulker Box                    Shulker Shell                  Snout Banner Pattern          
Sponge                         Tall Grass                     Thing Banner Pattern           Wet Sponge              

Your task is to get as many of these 40 items as you can. A Netherite Tool means any of a Netherite Sword, Pickaxe, Axe, Shovel or Hoe. Netherite Armor is any of a Netherite helmet, Chestplate, Leggings or Boots. Iron/Gold/Diamond Horse Armor means that any of these three are acceptable. Leather Horse Amor is not.

You should create a vanilla Minecraft Java Edition world, in 1.16.5, in Hardcore difficulty. You then have until you die in that world to collect all 40 of these items, or as many as you can. You should provide a double chest containing 1 of each of these and nothing else.

In order to verify attempts, you should provide the save file of your world, as a zip file.

The user who gets the most of these 40 items before dying wins. If multiple users get all 40 items, the user who gets all 40 within the fewest in-game days, which can be found by pressing F3.


  • You may not change the game rules of the world from the default, nor may you change any other option during world creation aside from the seed. To be fully clear, you may change exactly 3 things from the default values after clicking "Create New World":
    • The name of the world
    • The gamemode, switching from Survival to Hardcore
    • Optionally, the seed, which may be any value of your choosing
  • While in the world, you may not enable commands. This is typically done by pausing the game, opening the world to LAN and enabling commands, but, regardless of how you may do it, enabling commands is completely forbidden.
  • You may not use any mods, datapacks or anything else that can modify the default behaviour of the game. Texture packs are perfectly fine.
  • I will be testing and confirming each world submitted, so you should not include anything in the save file that isn't generated by the world. You may not edit the save file directly, only by interacting with the world
  • Items are allowed to be enchanted, renamed or otherwise legitimately modified in game

Additionally, as of snapshot 21w20a, there are 18 new non-renewable resources (ignoring products from Deepslate and Cobbled Deepslate), and Lava Buckets are noew renewable:

Block Of Raw Copper        Block Of Raw Gold         
Block Of Raw Iron          Budding Amethyst          
Calcite                    Cobbled Deepslate         
Copper Ore                 Deepslate                 
Deepslate Diamond Ore      Deepslate Gold Ore        
Deepslate Iron Ore         Deepslate Lapis Lazuli Ore
Deepslate Redstone Ore     Raw Copper                
Raw Gold                   Raw Iron                  
Spore Blossom              Tuff     

This brings the total number of non-renewable items in the latest version to 57

I will offer a +500 bounty to a submission which also includes a save file for a world which:

  • Is a vanilla Minecraft Java Edition snapshot 21w20a Hardcore world
  • Meets the requirements for the main challenge
  • Has 3 chests with all 57 non-renewable items, as listed above


  • Is this clear enough?
  • Are there any ways this can be exploited, so that users can trivialise the task? How can I prevent those exploits?
  • Is it a good challenge?
  • Tags are , and (meta discussion). Any suggestions?
  • Minecraft is not free, thus limiting those who can compete. Is this a significant problem?
  • Any further feedback?
  • \$\begingroup\$ I don't think minecraft being not free is a relevant problem here \$\endgroup\$
    – Wezl
    Commented May 27, 2021 at 18:32
  • \$\begingroup\$ I'm sure I saw a way to renew Spore Blossoms in one of the recent snapshots \$\endgroup\$
    – pxeger
    Commented May 27, 2021 at 18:35
  • 2
    \$\begingroup\$ There are, frankly, trivial ways to exploit this system. For it to be more or less uncheatable you would need a speedrun-style verification system requiring full screen recording and ideally livestreaming for the whole thing. I don't think it's a good fit for the site, to be honest, because there's really nothing code-related about it (Code Golf and Coding Challenges), and because it's too complex to implement a reliable verification system. You could try proposing it as a category extension on speedrun.com, though \$\endgroup\$
    – pxeger
    Commented May 27, 2021 at 18:41
  • 4
    \$\begingroup\$ Honestly this might be better as an off-TNB hosted competitive event where a bunch of people can stream (to minimize cheating) and attempt to get [as many of] these resources [as quickly as possible], if we want this to be a thing for people in this community to do for fun. In any case, this was a cool idea initially, but I do agree with pxeger that there isn't really anything in scope for CGCC about this. \$\endgroup\$
    – hyper-neutrino Mod
    Commented May 27, 2021 at 18:52
  • \$\begingroup\$ Is Bedrock allowed? \$\endgroup\$
    – emanresu A
    Commented May 28, 2021 at 5:30
  • \$\begingroup\$ If you do make it a TNB event (or even if you don't?), maybe the scoring system should take into account the rarity of each item. So coal ore scores 1 point and enchanted golden apple scores 10 or something \$\endgroup\$
    – pxeger
    Commented May 28, 2021 at 13:11
  • \$\begingroup\$ Question - are we allowed to start with anything, as long as we get the items from the world itself? \$\endgroup\$ Commented May 28, 2021 at 16:04

for (a.length of a)

A recent conversation in The Nineteenth Byte about language design brought about discussion of an "interesting" snippet of JavaScript code:

for (a.length of a) console.log(a)

This loops over the elements of a, but assigns each element to the length of a, and prints out a on each iteration.

  • When an array's length is assigned, it is resized to the length, with excess elements being removed from the end, and additional elements filled with undefined
  • When an array's length is assigned to undefined, an error occurs
  • The loop will halt if the number of iterations is ever greater than a.length

Depending on the contents of a, this produces some wild results.

TODO: finish this post. Posted without finishing to test if Redwolf Program's Sandbox Posts bot is working :)


Given a black-box function f(x) which take a value x and output true for \$p(x)\$, an unknown continious monotone function(not knowing even whether it's increasing), probable; and false otherwise. Output an infinite sequence \$a_n\$ such that \$\lim_{n\rightarrow \infty}p(a_n)=0.5\$. You can assume that the result exists.

Reasonable I/O allowed. Shortest code win.

A possible solution:

for i=1..infty
    S = [i/2] * i*i*2
    for j=0..i*i*2-1
        for k=1..i
            if f(j/i-i)
    print minPos([t*t for t in S])/i-i

Try It Online!

  • \$\begingroup\$ So we are given two functions f and p and need to define an infinite sequence? \$\endgroup\$
    – Quelklef
    Commented Mar 26, 2021 at 1:55
  • \$\begingroup\$ You're not given p, and you're to find a such that p(a)=0.5 \$\endgroup\$
    – l4m2
    Commented Mar 26, 2021 at 5:58
  • \$\begingroup\$ If we are not given p, how are we supposed to find a? Do you want us to generate an a that works for every p? \$\endgroup\$
    – Quelklef
    Commented Mar 26, 2021 at 6:51
  • \$\begingroup\$ @Quelklef Try some a and adjust till probable is 0.5 \$\endgroup\$
    – l4m2
    Commented Mar 26, 2021 at 7:19
  • 2
    \$\begingroup\$ I completely do not understand. Is it possible you could provide an example solution in your problem? That may help others confused like me. \$\endgroup\$
    – Quelklef
    Commented Mar 26, 2021 at 15:13
  • \$\begingroup\$ Say, \$ p\left(x\right) = \frac{4\pi + 2\arctan\left(x\right)}{5\pi} \$. How can you find out such a sequence \$ a_n \$ let \$ \lim_{n\to\infty}p(a_n)=0.5 \$ \$\endgroup\$
    – tsh
    Commented Mar 29, 2021 at 8:44
  • \$\begingroup\$ Even if we can safely assume that there is a value \$x\$ that \$p\left(x\right)=0.5\$, I don't think anyone can provide sequence would even convergent with a non-zero probable. \$\endgroup\$
    – tsh
    Commented Mar 29, 2021 at 8:51

duplicate :(


Convert Lambda Calculus to SKI-notation

Lambda calculus is a formal system in mathematical logic for expressing computation based on function abstraction and application using variable binding and substitution.

However, it's possible to translate pure lambda calculus into SKI-notation. This can be preformed by building an AST for the lambda calculus, converting it through multiple substitution steps where it contains a mixture of lambda calculus and SKI-notation, until a pure SKI-notation expression is reached.

As this is code golf, there is some flexibility on your I/O so that you may be able to accept the lambda calculus and output the SKI-notation as a tree but the examples will use a string-based notation.

  • (c1), where c1 is an arbitrary expression that does not contain a lambda (represented here by >), does not require further substitution.
  • (c1(c2)), where c1 and c2 are arbitrary expressions, can be substituted with (s1(s2)), where s1 and s2 are the result of performing any necessary substitutions on c1 and c2.
  • (x1>x1) can be substituted with I.
  • (x1>c1) (where c1 can be an arbitrary expression as long as it does not contain a reference to x1) can be substituted with (Ks1), where s1 is the result of performing any necessary substitutions on c1.
  • (x1>(x2>x1)) can be substituted with K.
  • (x1>(x2>e1)) (where e1 is an arbitrary expression that is depends on but is not x1) can be substituted by performing substitutions on (x2>e1) resulting in s2 and then performing substitutions on (x1>s2).
  • (x1>c1(e2)) (where c1 can be an arbitrary expression as long as it does not contain a reference to x1) can be substituted with (s1(s2)), where s1 and s2 are the result of performing any necessary substitutions on c1 and (x1>e2).
  • (x1>e1(e2)) can be substituted with (S(s1)(s2)), where s1 and s2 are the result of performing any necessary substitutions on (x1>e1) and (x1>e2).

Naturally function application associates left-to-right, so in the case of (x1>c1(c2)(e3)(c4)) this needs to be substituted using the last rule as if it was (x1>(c1(c2)(e2))(c4)), which requires generating substitutions for (x1>c1(c2)(e2)) and (x1>c4), but then (x1>c1(c2)(e2)) can use the penultimate rule, turning into c1(c2)(x1>e2), with (x1>e2) still needing to be substituted.

Given an input expression of lambda calculus, please convert it to SKI-notation. You must use the above substitutions at a minimum, but you can perform additional reductions if you wish, so for instance if you use a string-based notation then you may simplify ((S)(K)) to (SK) or even SK depending on context.

This is , so the shortest program or function that breaks no standard loopholes wins.

  • \$\begingroup\$ I'm a little unclear as to whether we can output any SKI expression so long as it is equivalent to the input and you are just laying out an example algo or whether we have to output the result of that algo (plus optional substitutions). \$\endgroup\$
    – Wheat Wizard Mod
    Commented May 29, 2021 at 11:23
  • \$\begingroup\$ I think also a definition of what it means to be equivalent between the two systems and within the systems themselves would be helpful. \$\endgroup\$
    – Wheat Wizard Mod
    Commented May 29, 2021 at 11:24
  • \$\begingroup\$ @WheatWizard You have to show that you do no worse than the example aglo. In theory the post will have enough examples so that you just need to be able to match or beat all of the examples. \$\endgroup\$
    – Neil
    Commented May 29, 2021 at 12:44
  • \$\begingroup\$ I'll have to work on that definition because I don't want to accidentally be too restrictive or indeed lenient. \$\endgroup\$
    – Neil
    Commented May 29, 2021 at 12:48
  • \$\begingroup\$ Closely related: unrestricted conversion challenge, metagolf version \$\endgroup\$
    – Bubbler
    Commented Jun 7, 2021 at 7:01

Conic Sections (simplified)

Given the equation of a non-parabolic conic section, output its characteristics.


This assumes prior knowledge of hyperbolas and ellipses, as well as their characteristics. (This includes circles; they are a special case of the ellipse with eccentricity of zero.)


A non-parabolic (to simplify things) conic section given in the standard equation form. To simplify things further (because the main point is not to perform linear algebra magic) there will be no xy term. This is an example of a valid equation:

x^2+6x+y^2-8y+15=0 // string form
[1,6,1,-8,15]      // array form

These are not:

x^3+5x^2+7x+4=0 // because the degree of the equation is 2
x^2+5xy+y^2-4=0 // because there should be no `xy` term
x^2+3x+7=0      // because there should be `x` and `y` terms.

Note that the conic section can also be taken as an array as shown above. If so, please specify the order of the array; I am flexible when it comes to this format, As long as there are 5 elements with nonexistent terms represented by zero (like no y term in x^2+5x+y^2+14=0) and that the terms they represent are x^2 x y^2 y c where c is a constant. The equation will always be <expression> = 0.


Output should be the type of section, center, horizontal radius, vertical radius, foci and eccentricity (in whatever desired order). This can be output as a string or an array as long as it is clear. A valid output for x^2+6x+y^2+8y+16=0 (or its array equivalent) would be:

["ellipse", [-3, -4], 3, 3, [[-3, -4]], 0]


-3 -4
-3 -4

or similar.

(no need to output "circle" because it is a special case of the ellipse)


  • duplicate?
  • should I edit in info on conic sections?
  • should I create the characteristics tag?
  • \$\begingroup\$ One issue with the x^2+5xy+y^2-4=0: It is a valid conic section (hyperbola). \$\endgroup\$
    – WarpPrime
    Commented May 30, 2021 at 21:01
  • \$\begingroup\$ @fasterthanlight I know, but we are not considering xy terms in this challenge. \$\endgroup\$
    – user100690
    Commented May 31, 2021 at 8:01

Is this a trivial brainfuck NOP?

Brainfuck is Turing-complete*, which means determining the behaviour of an arbitrary program is undecidable. However, we can get arbitrarily close to this if we limit ourselves to detecting a limited subset of programs.

For this challenge, we will detect a limited subset of NOPs - that is, snippets of programs which have no effect including on the brainfuck interpreter's state - which I have declared "trivial NOPs".

Here is a recursive definition of a trivial NOP:

  • the empty string is a trivial NOP
  • snippets consisting only of characters outside the set []<>+-,. (that is, characters which are ignored in Brainfuck), are trivial NOPs
  • the concatenation of two trivial NOPs is also a trivial NOP
  • snippets of any of the following forms, where X is another trivial NOP, are also trivial NOPs:
    • +X-
    • -X+
    • <X>
    • >X<

We will assume brainfuck has unlimited tape length. Overflow also does not need to be considered, because if a cell rolls over due to a +, a - can then underflow it back and it will still be a NOP.

Any snippet containing , or . is not a NOP because it performs impure I/O, and any snippet containing [ or ] is undecidable, so must be assumed to be not a NOP.

Your task is to, given a brainfuck snippet as a string or list of characters, determine whether it is a trivial NOP, and output two distinct values corresponding to true and false.



(empty string)


<+>+<->-             (note: even though this is a NOP, it's not a trivial NOP)



  • Can you think of any other classes of brainfuck snippet that are simple to detect as doing nothing?
  • Is this a duplicate?
  • Is this clear enough?
  • Any other feedback?

Back to the sandbox.

Challenge will be initially done on TBD, although I may do it again after that sometimes. Also, programs will be launched with a launcher shell script.


You are a programmer who has been requested to create a program that wipes out other similar programs. However, you need to have your program fight others to see how it does in the field.


You are to make a program that can run on a Debian 10 VM. If the language you use is not available by default on Debian 10, you must provide instructions or a link to the compiler.


Submissions are to be formatted as follows:

# Name, Language



More description and explanation

Link to compiler download and/or download instructions for Debian 10 GNU/Linux (also add compiler options to use when compiling.)

Requirements for being included

Your answer must have at least one upvote. You also need to include an emergency off switch and test the switch to be included, and provide the source code and have tested it on a Debian 10 VM. Also, its growth rate must be hypothetically sub-exponential (this is to stop excessively laggy answers. If you don't know what growth rate to go with, go with linear (\$O(x)\$) or Polynomial (\$O(x^2)\$,\$O(x^3)\$,etc.).


To win, you need to be the last process of the bunch, by killing other processes through a system call. If you kill important processes that result in the VM breaking, that round is ignored for scoring. A total of 100 rounds will be run, and you get a point if you are the last "malicious" process around. If there's a tie, the tiers will be put in a tiebreaker.

Pseudocode for submission example

I don't know much about coding yet, so here's some pseudocode (feel free to convert it to a real programming language and mark it as such:

Example, Pseudocode

This program simply kills processes with the word "MalicousProcess", unless it's "MalicousProcess-Example"

def Example () {
  while (system(ps aux | grep "MalicousProcess") != null) {
    for each process != "MalicousProcess-Example" {
      kill process
  return 0

fictitious compiler

Please provide more recommendations.

  • \$\begingroup\$ "This is to stop joke answers, suicidal answers (sorry EmoWolf), and malware if it shows up." I doubt it will stop joke answers (including EmoWolf). \$\endgroup\$
    – Wezl
    Commented May 25, 2021 at 21:17
  • \$\begingroup\$ Well, then it may stop bad joke answers and malware (like :(){ :|:& };: ). Also, I sure hope stack exchange doesn't run all inputs through bash. \$\endgroup\$
    – 4D4850
    Commented May 25, 2021 at 21:25
  • 2
    \$\begingroup\$ @4D4850 Don't worry about malware; we follow riules here :p \$\endgroup\$ Commented May 25, 2021 at 22:30
  • \$\begingroup\$ Ok Redwolf. I'll remove the justification part, but still keep the policy. \$\endgroup\$
    – 4D4850
    Commented May 26, 2021 at 11:50
  • \$\begingroup\$ Also, this is a little off topic, but why's your comment the only one that appears when you don't expand the comments? \$\endgroup\$
    – 4D4850
    Commented May 26, 2021 at 12:14
  • \$\begingroup\$ @4D4850 Redwolf's comment has been upvoted, so it is viewed by stack exchange as the most important/helpful/funny comment, and the others are less important \$\endgroup\$
    – Wezl
    Commented May 26, 2021 at 14:16
  • \$\begingroup\$ Anyway, can any of you provide any more recommendations, or do you think it's ready to be posted? \$\endgroup\$
    – 4D4850
    Commented May 26, 2021 at 18:01
  • \$\begingroup\$ Disallowing a rate of O(e^x) doesn't disallow all exponential solutions. Is that intended? If not you should probably require it tl be subexponential \$\endgroup\$ Commented May 28, 2021 at 6:05
  • \$\begingroup\$ Thanks for pointing that out. I'll make that change now. \$\endgroup\$
    – 4D4850
    Commented May 28, 2021 at 14:44
  • \$\begingroup\$ Provided no recommendations are made by Sunday 9:00 Central time, I'll post this question on the main site. \$\endgroup\$
    – 4D4850
    Commented May 28, 2021 at 23:37
  • \$\begingroup\$ Ok, I'm posting the question. \$\endgroup\$
    – 4D4850
    Commented May 31, 2021 at 13:11
  • \$\begingroup\$ Nevermind. Back to the Sandbox. \$\endgroup\$
    – 4D4850
    Commented May 31, 2021 at 15:51

Pickleball Doubles Scoring

  • \$\begingroup\$ Welcome to Code Golf, and thanks for using the Sandbox! I've fixed some formatting for you - on our site, you have to use \$ instead of $ for MathJax. At the moment, this challenge is missing an objective scoring criterion - I'm guessing you want code-golf? \$\endgroup\$
    – pxeger
    Commented May 31, 2021 at 7:17
  • \$\begingroup\$ @pxeger: Thank you. Yes, I was thinking code golf. Would it be better to take the END state as [0,0,0] so that the format of all inputs is the same? Is there a place to get the standard verbiage for code golf? \$\endgroup\$ Commented May 31, 2021 at 13:43
  • \$\begingroup\$ Yes, I think [0, 0, 0] is a good idea - but I think the best option is to allow answers to choose a value to represent the END state depending on their language (in some it may be easier to use null, or take a string or something...). Re verbiage, there's Standard definitions of terms within specifications, but you can also just look at the various policy posts for more information. If you have specific questions, you can always ask in our chatroom, The Nineteenth Byte. \$\endgroup\$
    – pxeger
    Commented May 31, 2021 at 14:51
  • \$\begingroup\$ I've edited this down to a stub now that it's been posted to save space \$\endgroup\$ Commented Jun 1, 2021 at 15:15

What sequence?

Stack Exchange distinguishes users by a unique id that can be found in their their profile URL.

You can find your id by navigating to https://codegolf.stackexchange.com/users/current and finding the number after the /users/ bit in the resulting URL. For example, my profile URL is https://codegolf.stackexchange.com/users/99744/alex-bries, so my user ID is 99744

You can use this id to look up the number sequence of the last submitter on OEIS:

The On-Line Encyclopedia of Integer Sequences® (OEIS) is an online database containing integer sequences. These sequences are identified via an id starting with an A followed by 6 numbers. You can search for an id with the "id:" key on the search bar of OEIS.
id:A99744 will display sequence A099744 followed by all other sequences that mention it.

Your job is to write a program that, when given a number n, produces the nth number in sequence of the submitter above.


  • Each language may only be used once.
  • You may not post 2 answers in a row
  • Every next submision may only be max 10 bytes higher than the previous one
  • The champion is the last valid submission on june 14th

golfing is not a requirement but it is in your favour

  • \$\begingroup\$ Different OEIS sequences have vastly different complexity. I'm pretty sure my user id is solvable in 6 bytes of Jelly, which can break the whole game depending on the next user's id. \$\endgroup\$
    – Bubbler
    Commented Jun 1, 2021 at 11:32
  • \$\begingroup\$ @Bubbler I like the idea of different complexities, but not sure how to prevent the colossal differences. Personally i like it, but if you have ny suggestions for how to improve, please let me know. \$\endgroup\$
    – Alex bries
    Commented Jun 1, 2021 at 11:47
  • \$\begingroup\$ I don't think this works as a [answer-chaining] challenge, but it could work as just a regular code-golf challenge (i.e. write a program which takes an integer n and outputs the nth term of the sequence with your id). It's unlikely to be fair, but its better than [answer-chaining]. I've also done a bit of editing, feel free to change anything I did \$\endgroup\$ Commented Jun 1, 2021 at 15:08
  • \$\begingroup\$ Yes, that's probably the better desicion, but after thinking about it more, its unlikely to be very fun, so ill leave it (for now). Thanks for the the constructive feedback guys. \$\endgroup\$
    – Alex bries
    Commented Jun 1, 2021 at 20:04
  • \$\begingroup\$ Using user IDs is inherently unfair. \$\endgroup\$
    – qwr
    Commented Jun 2, 2021 at 14:47

Write a program that mimics the output of the command env (with no arguments).

The program should print (to standard output, or closest equivalent) a list of all current environment variables, with each variable on it's own line. The name and content of a variable should be separated by an equal sign.

Here's a grammar:

line ::= name "=" contents
output ::= (line "\n")*

No using the env command, or any 1 command that completes the challenge on its own.

shortest code wins.

  • \$\begingroup\$ Can we use built-ins the get environment variables? \$\endgroup\$
    – Adám
    Commented Nov 15, 2020 at 15:50
  • \$\begingroup\$ @Adám If you mean "Can we use built-ins to get the environment variables", yes, I don't see a reason you couldn't, and otherwise it might be impossible to get them in some languages. \$\endgroup\$
    – binarycat
    Commented Nov 15, 2020 at 15:54
  • 1
    \$\begingroup\$ And I assume (ba|k|c|…)sh cannot participate? \$\endgroup\$
    – Adám
    Commented Nov 15, 2020 at 15:56
  • \$\begingroup\$ @Adám If they can find a way to get a list of environment variables without using env, I guess they could, unless I'm missing some other command that produces the same output. There's a lot of languages that can't complete the challenge (mostly esoteric ones), but they aren't technically banned. \$\endgroup\$
    – binarycat
    Commented Nov 15, 2020 at 23:28
  • \$\begingroup\$ How about set on Windows/DOS? \$\endgroup\$
    – Adám
    Commented Nov 16, 2020 at 8:07
  • \$\begingroup\$ @Adám I have no experience with windows, does it have similar functionality to env? if so then no. \$\endgroup\$
    – binarycat
    Commented Jun 3, 2021 at 16:46
  • \$\begingroup\$ It is indeed. Note: As a beginner challenge writer (you've never posted a challenge before), you should avoid do X without Y challenges. \$\endgroup\$
    – Adám
    Commented Jun 3, 2021 at 17:28
  • \$\begingroup\$ @Adám isn't "don't use a builtin that does exactly what you need" a standard loophole? \$\endgroup\$
    – binarycat
    Commented Jun 3, 2021 at 19:32
  • \$\begingroup\$ No, not at all. \$\endgroup\$
    – Adám
    Commented Jun 3, 2021 at 19:32
  • \$\begingroup\$ @Adám Huh, I guess you're right. I just don't want to have the challenge solved before it's even posted, with the 3 character bash program env. Does trying to prevent 1 cheeky answer really put you in the same category as forbidding something random? \$\endgroup\$
    – binarycat
    Commented Jun 3, 2021 at 20:50
  • \$\begingroup\$ Does it really bother you that much that someone will post env or set as a solution? You're under no obligation to upvote or accept such an answer. \$\endgroup\$
    – Adám
    Commented Jun 3, 2021 at 20:54
  • \$\begingroup\$ @Adám here's something you linked me: "A commonly used non-observable requirement is to avoid using a library function that solves the entire challenge. I believe these rules are OK. Like other non-observable requirements, one can find borderline cases of whether a built-in function solves all of the task or not, but the benefits of higher-quality answers outweigh the costs." this certainly seems to apply here. \$\endgroup\$
    – binarycat
    Commented Jun 3, 2021 at 20:59
  • \$\begingroup\$ @Adám I've never posted anything because whenever I post a question to the sandbox all I get is reasons why my question is bad. And putting any additional rules whatsoever gets the same "no X without Y" response. Then people don't elaborate on anything. \$\endgroup\$
    – binarycat
    Commented Jun 3, 2021 at 21:10
  • \$\begingroup\$ Don't be dismayed. We've all been there. Writing challenges is notoriously hard, especially for beginners. If you look at my account, I have lots of well-received challenges, but dig in, and you'll find that many of my early ones were flawed. I highly recommend answering a lot more existing challenges, before you attempt at writing your own. And when you do, decide on a simple unique task (this is hard, as a lot of obvious tasks have already been posted), with no special rules; just vanilla code golf. \$\endgroup\$
    – Adám
    Commented Jun 3, 2021 at 21:14

Optimize imnotdeadfish

  • \$\begingroup\$ Deadfish (at least, the original one) used i and d intead of + and -. Is there a reason you chose to use them? \$\endgroup\$
    – pxeger
    Commented Jun 3, 2021 at 10:45
  • \$\begingroup\$ @pxeger originally for the language I was going to have an input command so I swapped i and d for + and -, and I guess I never added them back. \$\endgroup\$
    – Underslash
    Commented Jun 3, 2021 at 16:02

branch golfing: sort a list by number of prime factors

Specifically, first sorted by total number of prime factors, with any ties broken by number of unique prime factors (remove this?), then broken by size. The list will never contain 1.

Test cases:

[4,5,6,9] -> [5,4,9,6]
[10,11,12] -> [11,10,12]
[360 200 12] -> [12 200 360]
[63 64 65] -> [65 63 64]
  • \$\begingroup\$ Please avoid challenges involving the prime numbers. \$\endgroup\$
    – Adám
    Commented Jun 3, 2021 at 20:42
  • \$\begingroup\$ Please avoid challenges where the scoring criteria relies on non-observable program requirements. \$\endgroup\$
    – Adám
    Commented Jun 3, 2021 at 20:46
  • 1
    \$\begingroup\$ Please avoid making assumptions about language features. \$\endgroup\$
    – Adám
    Commented Jun 3, 2021 at 20:47
  • 1
    \$\begingroup\$ Also, thank you for using the sandbox! :-) \$\endgroup\$
    – Adám
    Commented Jun 3, 2021 at 20:47
  • \$\begingroup\$ @Adám Do you have any example of something where it isn't clear whether or not it is a conditional branch? Or any advice that isn't essentially just "this is bad"? \$\endgroup\$
    – binarycat
    Commented Jun 3, 2021 at 20:56
  • \$\begingroup\$ Maybe. Are there any branches in this APL solution? The code means: the input ⊇⍨ permuted ∘⍋ according to a grade using prime factors for each , followed by ⍥≢ when counted, ∘∪ the unique ¨ for each, prime factors for each. Or in other words: Reorder according to a grading of prime factor count and then unique prime factor count. \$\endgroup\$
    – Adám
    Commented Jun 3, 2021 at 21:10
  • \$\begingroup\$ From your description, sounds like a loop construct, which I intended to include, but I could clarify. \$\endgroup\$
    – binarycat
    Commented Jun 3, 2021 at 21:21
  • \$\begingroup\$ If you count loop constructs, would you also count APL's + as a loop? It automatically loops over its arguments: Try it online! Besides, you state that "Builtin functions are presumed to be branchless." \$\endgroup\$
    – Adám
    Commented Jun 3, 2021 at 21:24
  • \$\begingroup\$ No. It loops its own code, but can't be used to perform arbitrary code in a loop. \$\endgroup\$
    – binarycat
    Commented Jun 3, 2021 at 21:28
  • \$\begingroup\$ My usage of was exactly parallel to any usage of +. They both auto-loop. However, ¨ is in fact an explicit loop. I could probably code around that too, though. An example would be extracting the odd numbers from a list. In many languages, you'd use a loop with a branch, but in APL you'd write List/⍨2|List which means L where 2 divides the list ['s elements; that's implicit]. Looping? Branching? Anyway, you get the idea. It'll be next to impossible to determine for every language (there are lots!) what exactly constitutes an explicit loop, or for that sake a branch. \$\endgroup\$
    – Adám
    Commented Jun 3, 2021 at 21:32
  • \$\begingroup\$ @Adám so they are analogous to functions that take 2 arrays as input and provide one as output? that doesn't seem like a branch to me. "I could probably code around that too, though." Yeah that's kinda the point of the challenge. \$\endgroup\$
    – binarycat
    Commented Jun 3, 2021 at 21:41
  • \$\begingroup\$ @Adám "but in APL you'd write List/⍨2|List which means L where 2 divides the list ['s elements; that's implicit]." this explanation isn't very clear. this seems like this construct (or pair of constructs, which you are confusingly explaining as one) is some kind of predicate filter/map? \$\endgroup\$
    – binarycat
    Commented Jun 3, 2021 at 22:01
  • \$\begingroup\$ Let us continue this discussion in chat. \$\endgroup\$
    – Adám
    Commented Jun 3, 2021 at 22:01

Random Prisoner's Trilemma - Python 3 KOTH


Yet another digit insertion problem


How much faster than qsort can you achieve?


Sort 1 million 32 bit integers as quickly as possible. Your code must create 1 million random non-negative integers in the range 0 to 999999 and sort them. Only the sorting should be timed.


I will run your code and compare the timings to my default C code using qsort. That is:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <stdint.h>

int cmpfunc (const void * a, const void * b) {
   return ( *(int*)a - *(int*)b );

int main(void) {
    int sz;
    printf("Enter the size of array::");
    uint32_t *arr = malloc (sz * sizeof(uint32_t));
    int i;
    clock_t begin = clock();
    qsort(arr, sz, sizeof(int), cmpfunc);
    clock_t end = clock();
    double time_spent = (double)(end - begin) / CLOCKS_PER_SEC;
    printf("%f seconds\n", time_spent);
    printf("%d\n", arr[10]);
    return 0;

Your score will be the timing for my C code divided by the timing for your code. I will run both on my PC. This means I will need clear instructions for how to compile and run your code on linux.

This means you will have to provide timing code in your solution so I am not timing the creation of the data.

Current timings

On my PC:

  • The sample C code takes 0.11 seconds.
  • Python sorted takes 0.24 seconds.


My code prints out one of the elements of the sorted array. This is to stop the opimizing compiler from removing the sorting code completely. You will have to do the same.

Builins are allowed. As ever this is a challenge per language so if your Python answer is faster than any other Python answer you have won that mini-competition.

  • 1
    \$\begingroup\$ Obligatory question - builtins, yea or nay? \$\endgroup\$ Commented Jun 7, 2021 at 14:31
  • \$\begingroup\$ That's probably the wisest choice - probably want to edit the post to say that. \$\endgroup\$ Commented Jun 7, 2021 at 14:35
  • \$\begingroup\$ I strongly disagree with banning builtins. If any builtin is faster, either the builtin is interesting enough to be posted, or the challenge is boring. Banning builtins won't improve the latter, but it will hurt the former \$\endgroup\$ Commented Jun 7, 2021 at 14:37
  • \$\begingroup\$ @cairdcoinheringaahing I can go with that. \$\endgroup\$
    – user7467
    Commented Jun 7, 2021 at 14:37
  • \$\begingroup\$ @StackMeter I deleted my comment as bultins are going to be allowed. \$\endgroup\$
    – user7467
    Commented Jun 7, 2021 at 14:39
  • \$\begingroup\$ I would argue that "only time sorting should be timed" is a bit unclear. Consider my code is: int counts[MAX],i,j,k,sort(*a,size){for(i=0;i<size;++i)++counts[a[i]];for(i=k=0;i<MAX;counts[i++]=0)for(j=counts[i];j--;)a[k++]=i;} (maybe some typos here, not tested) the initialize of counts array is required by sort. But C compiler will help me do so even begin main() entry and would probably not timed by my code. \$\endgroup\$
    – tsh
    Commented Jun 8, 2021 at 6:12
  • \$\begingroup\$ @tsh what would you suggest? \$\endgroup\$
    – user7467
    Commented Jun 8, 2021 at 7:08
  • \$\begingroup\$ Things to consider: 1) 32-bit signed or unsigned integers? Do I have a choice? 2) 32-bit (or any finite-sized) integers can be sorted using non-comparison sort algorithms, such as radix sort (which is asymptotically faster than any kind of comparison-based sort). Would you allow the answers to exploit this or not? \$\endgroup\$
    – Bubbler
    Commented Jun 14, 2021 at 1:24

Smallest number of actions to change reputation by [x] points

8: Upvote + downvote

75: Bounty + upvote + accepted answer


  • \$\begingroup\$ Potentially a dupe of this. Stoopid. \$\endgroup\$
    – lyxal
    Commented Jun 9, 2021 at 4:55
  • \$\begingroup\$ @lyxal This one's about positive or negative rep gains though, so it's more general \$\endgroup\$ Commented Jun 9, 2021 at 4:57

Finite Composition of Univariate Functions


Given $$\mathbb{F}:=\{f_i \mid f_i : \mathbb{R} \to \mathbb{R}\}$$ the set of all the real maps, and the n-ary function composition

\begin{gather} c_o(f_i) = f_1∘f_2∘f_3⋯∘f_n\\[8pt] ∀f_i ∈ \mathbb{F},~~~1≤i≤n∈\mathbb{N} \end{gather}

write a program which receives n functions of a real variable as the input and computes the in-order function composition, product of all the entered functions as a string.

As standard, x is the name of the parameter variable in each function. Any other literal is taken as a constant.


Example 1


number of functions: 3
f_1 = x + 1
f_2 = x - 10
f_3 = x^2 + 2


c_o(f_1,f_2,f_3) = ((x^2 + 2) - 10) + 1

Example 2


number of functions: 4
f_1 = 27x - sin(e)
f_2 = -7(x^2 + 1)^(1/2) 4y^3 + 21
f_3 = x^(3/2) + 4
f_4 = ln(z + 1)
(z, y, e are assumed as constants in the reals)


c_o(f_1,f_2,f_3,f_4) = 27(-7(((ln(z + 1))^(3/2) + 4)^2 + 1)^(1/2) 4y^3 + 21) - sin(e)
  • \$\begingroup\$ Any suggestion, improvement? please let me know. Thanks. \$\endgroup\$ Commented Jun 8, 2021 at 20:08
  • 1
    \$\begingroup\$ Thanks for using the sandbox! This seems like an interesting idea, but currently it is too underspecified to be clear. For example, the test cases imply that the input to each function is represented by x, but you haven't stated anywhere that we can assume that. Consider also a slight change to one of your cases, where one function has -y(x^2 + 1). Is y a function or a constant? There are too many specifics, so hopefully the general idea gets across that you must specify how you are representing the functions very precisely for this to work. \$\endgroup\$ Commented Jun 9, 2021 at 17:41
  • \$\begingroup\$ @FryAmTheEggman Thank you. I have edited it so that there's no ambiguity. \$\endgroup\$ Commented Jun 9, 2021 at 18:19
  • \$\begingroup\$ That is a good change, but doesn't address the ambiguity in separating functions like sin from the product of constants. It also isn't clear to me whether x^2 - 7 is a valid solution for the first example. These have some overlap in that they (probably?) won't affect a solution that just substitutes the functions appropriately, but will affect any symbolic parser based approach. \$\endgroup\$ Commented Jun 9, 2021 at 19:53
  • \$\begingroup\$ @FryAmTheEggman oh, yes, it's valid but it isn't required to go to such depths as to simplify symbolic expressions, that would be a problem on its own. I didn't specify so in the challenge and as for sin or those functions, since the problem to be addressed is the composition of functions any expression different from x should be "disregarded". I don't intend to make further computations with the output string. \$\endgroup\$ Commented Jun 9, 2021 at 21:45
  • \$\begingroup\$ What I'm trying to communicate is that your "point" for writing the challenge doesn't really matter - in some language it probably would be better to use a parser. If you just want to say "no other assumptions about the form of the function can be made" that is fine - but you have to actually say it, otherwise your challenge is unclear because I won't be able to tell if a solution is valid or not. \$\endgroup\$ Commented Jun 9, 2021 at 23:13

Interpret Unreadable


Unreadable is a programming language designed to be – as the name states – unreadable (in most fonts, anyway). Instructions are ' followed by a run of "s. So while in a code block, '""'""'""" looks just fine, it looks like '""'""'""" outside of one - it is extremely difficult to tell what is going on.

The instructions are prefix, and are as so:

Instruction Arity Behavior
'" Unary Print; output the character with codepoint x, and return x.
'"" Unary Increment; return x+1.
'""" Nilary Unit; return 1.
'"""" Binary Both; evaluate and disregard x, and return the result of evaluating y.
'""""" Binary While; keep evaluating y until x is 0. Return the last result of evaluating y.
'"""""" Binary Set; set the xth variable to y, and return y.
'""""""" Unary Get; return the value of the xth value. Undefined is 0.
'"""""""" Unary Decrement; return x-1.
'""""""""" Ternary If-else; if x is 0, return z. Otherwise, return y.
'"""""""""" Nilary Input; return the codepoint of the next character of input. EOF is -1.

For example, a cat program would be:

while(++ setvar(1 = input())) {print(getvar(1))}

Or, in its unreadable glory: '"""""'""'""""""'"""'""""""""""'"'"""""""'"""


Given a valid Unreadable program and its input separated by any non-' or " character, interpret the program as Unreadable code.

Test cases (uses ! as the separation character, I/O is represented in CP-437)

Input -> Output
'"'"""!this input does nothing lol -> ☺
'"""""'""'""""""'"""'""""""""""'"'"""""""'"""!This is a cat so I can output whatever I want. blah blah blah... -> This is a cat so I can output whatever I want. blah blah blah...
'""""'""""'""""""'"""'""'""'""'""'""'""'""'""'""'"""'"""""'"""""""'"""'""""'""""""'""'"""'""'""'""'""'""'""'""'"""""""'""'"""'""""'""""""'""'""'"""'""'""'""'""'""'""'""'""'""'""'"""""""'""'""'"""'""""'""""""'""'""'""'"""'""'""'""'""'""'""'""'""'""'""'""'"""""""'""'""'""'"""'""""'""""""'""'""'""'""'"""'""'""'""'""'"""""""'""'""'""'""'"""'""""'""""""'""'""'""'""'""'"""'""'""'""'"""""""'""'""'""'""'""'"""'""""'""""""'""'""'""'""'""'""'"""'""'""'""'""'""'""'""'""'""'""'""'""'"""""""'""'""'""'""'""'""'"""'""""'""""""'""'""'""'""'""'""'""'"""'""'"""""""'""'""'""'""'""'""'""'"""'""""""'"""'""""""""'"""""""'"""'""""'"'""'""'"""""""'""'"""'""""'"'""'"""""""'""'""'"""'""""'"'""""""""'""""""""'"""""""'""'""'""'"""'""""'"'""""""""'""""""""'"""""""'""'""'""'"""'""""'"'""'"""""""'""'""'""'"""'""""'"'""'""'""'""'"""""""'""'""'""'""'"""'""""'"'""'""'"""""""'""'""'""'""'""'"""'""""'"'""""""""'"""""""'""'""'""'""'""'""'"""'""""'"'""'"""""""'""'""'""'"""'""""'"'""'""'""'""'"""""""'""'""'""'"""'""""'"'""""""""'""""""""'"""""""'""'""'""'"""'""""'"'"""""""'""'""'"""'""""'"'""'""'""'"""""""'""'""'""'""'""'"""'"'"""""""'""'""'""'""'""'""'""'""" -> Hello, world!
'""""""'"""'""'""""""""""'""""""'""'"""'""""""""""'"'"""""'""""""'"""'""""""""'"""""""'"""'""""""'""'"""'""'"""""""'""'"""!A! -> b


  • Standard loopholes are forbidden.
  • A single trailing newline is allowed in the output, and no other trailing whitespace.
  • If possible, please link to an online interpreter (e.g. TIO) to run your program on.
  • Please explain your answer. This is not necessary, but it makes it easier for others to understand.
  • Languages newer than the question are allowed. This means you could create your own language where it would be trivial to do this, but don't expect any upvotes.
  • This is , so shortest code in bytes wins!
  • A +100 rep bounty will be given to the first Unreadable answer.
  • \$\begingroup\$ This doesn't seem like a particularly interesting challenge to me, to be honest. Decoding the program into commands is fairly trivial (split by ' and find lengths of each), in both golfing and practical languages, and the actual interpreting part of the challenge is basically the same as any other language. \$\endgroup\$ Commented Jun 10, 2021 at 1:20
  • \$\begingroup\$ @RedwolfPrograms The interesting part is that commands are prefix, so it might take some thought to make an interpreter. \$\endgroup\$
    – Makonede
    Commented Jun 10, 2021 at 1:55

Can we measure time?

You have 2 hourglasses with sand lying at bottom in both of hourglass, one that is \$1\$ sec hourglass(call it \$A\$) and other is \$\sqrt2\$ sec hourglass (call it \$B\$)

(A hourglass is said to be n second hour glass if time taken by all of the sand of hourglass to fall from top to bottom is n second)

Now we are given 2 integers \$a\$ and \$b\$ such that $$T=a+b\sqrt2 > 0$$ assume:$$\ 1000000\geq a \geq -1000000$$

$$\ 1000000\geq b \geq -1000000$$

we need to respond that can we measure this time \$T\$ with our hourglasses? ("YES" or "NO")

Some clarifications and definitions:

  • We say we can measure time \$t\$ if at that instance in at least one of hourglass sand have just stopped falling.

  • We can make a hourglass upside down(any number of time) if and only if at least one of the hour glass has empty top.

  • Assume time taken to invert a hourglass is 0 sec.(by invert means upside down,i.e the sand that was in bottom is now at top and the sand at top is now at bottom).

  • If at least one of hourglass sand have just stopped falling we can report this instant as the amount of time measured.

  • If sand on both the hour glass is at bottom and we decided to do no any operation, then it is assumed that we are reporting the time measured at this instance and the process stops.

Example 1:

\$a=0\$ and \$b=1\$

Our answer is true, because initially both the hourglass is given with sand at bottom and at the start we can invert B, so when the sand from B has fallen all to bottom at that instance a total of \$\sqrt2\$ time has passed and we can report it,hence we have measured \$\sqrt2\$ time.

Example 2:

\$ a=-1\$ and \$b=2\$

\$T = -1 +\sqrt2\$ Our answer is true, we can obtain it as follows: At time 0, turn both A and B upside down, At 1 sec A will be emptied to bottom, turn A upside down again (B remain falling as it was), now when B fall completely,turn A upside down, now when A empties the total duration is \$-1 +\sqrt2\$.

1(All the A falls, at this instance B is left with \$-1+\sqrt2\$ ) $$+$$ \$\sqrt2\$-1 (at this duration after the previous one B have fallen completely and A is left with \$1-(-1+\sqrt2)\$ at top, but we have inverted it so that means A is actually left with \$-1+\sqrt2\$ at top) $$+$$

\$-1+\sqrt2\$ (at this duration after the previous one A have fallen completely)

we can report this instant, the total time is:

\$(1)+(-1+\sqrt2)+(-1+\sqrt2)=-1 +2\sqrt2 \$

And for following input it is not possible to measure and you have to output "NO"

\$a=11\$ , \$b=-5\$

\$a=35\$ , \$b=-21\$

\$a=-4\$ , \$b=4\$

\$a=-8815\$ , \$b=6261\$

\$a=-1\$ , \$b=1\$

. . .

The idea of this problem is not my original, it appeared in a coding competition some time back.

  • \$\begingroup\$ Thanks @pxeger for editing. \$\endgroup\$
    – cheems
    Commented May 31, 2021 at 9:10
  • \$\begingroup\$ You should provide some impossible cases? \$\endgroup\$
    – l4m2
    Commented Jun 3, 2021 at 14:09
  • \$\begingroup\$ Thanks, I have added. \$\endgroup\$
    – cheems
    Commented Jun 3, 2021 at 14:53

Convert permutations to integers and back again

Convert a permutation of 0,...,n-1 to a number in the range 0,...,n!-1, and convert back again.


Write a program containing two functions (or two programs).

The first function should accept a permutation (i.e. an ordering) of the numbers 0,1,2,...,n-1 (where n is any positive integer) and output an integer in the range 0,1,...,n!-1 (where n! is n factorial). Distinct permutations of the same length n must produce distinct outputs, but there are no further restrictions.

The second function should implement the inverse of the first function (except that it also needs to know n). It should accept an integer k in the range 0,1,...,n!-1 AND a positive integer n, and should output the permutation of 0,1,2,...,n-1 that yields k when sent to the first function.

The time and space complexity of both functions MUST be at most polynomial in n, i.e. O(n**c) for some fixed c. In particular, solutions that involving computing all n! permutations are NOT acceptable. (Unofficial bonus points for O(n)).

Shortest solution (including both functions and any additional code) in each language wins.


Assuming the two functions are f and g:

Input Output
f([0]) 0
g(0,1) [0]
f([0,1]), f([1,0]) 0 and 1 in either order
g(0,2), g(1,2) [0,1] and [1,0] in either order
f([3,4,1,0,2]) some integer in range 0,...,119
g(42,5) some permutation of 0,1,2,3,4
f(g(42,5)) 42
g(f([3,4,1,0,2]),5) [3,4,1,0,2]
f([1,0,1]) not valid input (not a permutation)
f([1,3,4]) not valid input (not a permutation of 0,1,2)
g(42,4) not valid input (42 >= 4!)


Permutations can be represented in any reasonable form.

You can assume that input(s) are valid (as described).

You can assume that n! fits in the standard integer type in your language, but your algorithm should theoretically work for any n.

Standard loopholes apply.


Boxes in boxes in boxes in boxes.....

ASCII boxes are fun, but the box drawing questions seem to be limited insofar as that they seem to take a specific set of boxes. Today, I hope to fix that. The challenge here is to write a program that takes a list of sides of boxes and characters to draw them with, and outputs the boxes.

To clarify some format issues:


Input is flexible, and will look like 1) this


30 20 #$5

21 15 $%^

19 13 311


or 2) this

[17, [30, 20, "#$5], [21, 15, "$%^], [19, 13, "311"], ...]

You may support either or both of these types, but you must be consistent with this choice.

Input is given in this order:

The first number, a single number on one line, is the number of boxes to place.

Then, for as many times as said first number, you will receive 2 numbers and three characters - the two numbers represent length and width, in any order. The three characters represent, in this order:

  1. The character to use for the corners.
  2. The character to use for the edges.
  3. The character to fill the inside of the boxes.

The boxes should not overlap, but you have the freedom to place the boxes wherever you wish. The allowed symbols are !$%^&*()_-+=#~/?\@s, where the s represents space and in order to aid those who want to split on whitespace.

Input will be space separated within boxes and newline separated between boxes.


Output should be a set of boxes, conforming to the regulations and conditions above - for example,

#***# *((*
#***# ())(
#***# ())(
%###% *((*

for the input


5 5 %#*

4 4 *()


[2, [5, 5, "%#*"], [4, 4, "*()"]


This is , so shortest solution in bytes wins. Share your favourite drawings with your answer.

Additional tags:


Any issues?

Should the scoring be ?

Is the challenge specified enough?

  • \$\begingroup\$ Your scoring shouldn't be pop-con. Also, you realise people are just going to draw all the boxes newline-separated? Be clearer that the input format is flexible. \$\endgroup\$
    – emanresu A
    Commented Jun 11, 2021 at 9:23
  • \$\begingroup\$ @Ausername let me go fix that \$\endgroup\$ Commented Jun 11, 2021 at 10:29
  • \$\begingroup\$ @Ausername Should I add a bounding box to stop people just seperating the boxes by newlines \$\endgroup\$ Commented Jun 11, 2021 at 10:33
  • \$\begingroup\$ I'm not saying anything's wrong with that, just that's how people are going to do it because it's the simplest way. \$\endgroup\$
    – emanresu A
    Commented Jun 11, 2021 at 10:35

Given matrix \$A\$ and \$A^n\$, work out any possible positive \$n\$.


  • \$\begin{pmatrix}1&2&3\\3&2&1\\1&0&1\end{pmatrix},\begin{pmatrix}36&32&44\\52&40&52\\12&8&12\end{pmatrix}\rightarrow 3\$
  • \$\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}\rightarrow \text{any integer }\ge 3\$
  • \$\begin{pmatrix}0&1\\1&0\end{pmatrix},\begin{pmatrix}0&1\\1&0\end{pmatrix}\rightarrow \text{any odd positive integer }\$

Sandbox Notes

  • Since this question likely fall into pure matmul as , will this be ?
  • If so, what size and \$n\$ is reasonable?
  • Should it multiply in a modulo-\$p\$ ring?
  • If someone provide enough extra info, they have right to post this question(actually that's why I posted in sandbox)
  • \$\begingroup\$ Can you give some examples? \$\endgroup\$
    – math scat
    Commented Jun 11, 2021 at 17:01
  • \$\begingroup\$ Sometimes lots of different n can work. What should we output then? \$\endgroup\$
    – xnor
    Commented Jun 12, 2021 at 23:06
  • \$\begingroup\$ @xnor "Any possible positive n". But I agree that it should be specified better. \$\endgroup\$
    – user100690
    Commented Jun 14, 2021 at 11:22
  • \$\begingroup\$ @ophact It's edited later than comment \$\endgroup\$
    – l4m2
    Commented Jun 14, 2021 at 12:56

I am not a dad!


Given a non-empty string. Check if it starts with "I am"/"I'm" (case-insensitive) and if it is, output the dad joke. Otherwise, terminate
P.S. There must be nothing in the STDERR.


"I am an idiot" -> "Hi an idiot, I am Dad"
"I am rewuytheruty" -> "Hi rewuytheruty, I am Dad"
"i Am case insensitive" -> "Hi case insensitive, I am Dad"
"I'm A" -> "Hi A, I am Dad"
"i'M B" -> "Hi B, I am Dad"
"Iam together" -> terminated
"Nothing here!" -> terminated


  • This is , so the answer with shortest bytes wins.
  • These loopholes are, obviously, forbidden.
  • Standard code-golf rules apply.
  • Please specify the language you are using and the amount of bytes.
  • It would be great if you would put a link to a sandbox where your code can be ran in action, such as TIO.
  • Explaining your code is very welcomed.
  • \$\begingroup\$ I don't think the exceptions add anything to this challenge to be honest. And if you do keep them, they need to be better specified. \$\endgroup\$
    – pxeger
    Commented Jun 13, 2021 at 19:38
  • \$\begingroup\$ Input validation is frowned upon (validation to check if any of the exceptions appear). As well as that, I observed that there are virtually no patterns or magic bitwise operations that can handle them, just hard-coding a few extra strings and some ternary operators. Doesn't really add anything to the other dad joke challenge to be honest. \$\endgroup\$
    – user100690
    Commented Jun 14, 2021 at 12:19
  • \$\begingroup\$ And if I would remove the exceptions, will the challenge become better? Will it be worthy publishing? \$\endgroup\$ Commented Jun 14, 2021 at 12:52

Write a shortest bijective program from String to Object.

You can specify two sets of chars \$C_1\$ and \$C_2\$, both of which consist of at least 4 different chars[1], and define:

  • String: string consisting of only chars in \$C_1\$ [C₁]*
  • Name: non-empty string consisting of only chars in \$C_2\$ [C₂]+
  • Object: a finite map from Name to Object, such that every way to go down finally go to empty Object(empty map)

Reasonable I/O allowed. Outputting in Object in languages like JavaScript, Array/String with proper encoding, folder, shuld be fine.


  • [1] unary or binary is too easy.
  • I meant to map to folder but folder charset is bad, then decided to give you choice.

Sandbox Notes

  • What does proper encoding mean? Or do I just remove the allowance?

Largest number with no repeating substrings of length \$l\$

Inspired by this question, in turn inspired by this one.

When I read the above questions, I thought: "Cool". And then I thought "What if we go further?"

Therefore, in this challenge, you must find the largest number that has no repeating substrings of length n. This is : fewest bytes wins.

Meta Stuff

This seems insanely hard - is it well specified enough?

Any changes that seem reasonable to add?

  • \$\begingroup\$ I would recommend adding testcases and instead of linking to posts for people to read, you could clarify in the description. \$\endgroup\$
    – user100690
    Commented Jun 22, 2021 at 9:34
  • \$\begingroup\$ Is this the same as this? \$\endgroup\$
    – Shaggy
    Commented Jul 2, 2021 at 14:22

Iterated ultimatum

Each round, players are paired. One player is the Proposer and the other is the Receiver.

The Proposer must choose a number x such that 0 < x < 100. The Receiver must then Accept or Reject this offer. If accepted, the Proposer gains x points and the Receiver gains 100 - x points. If rejected, neither player gains anything. This is repeated several times, then the Proposer and Receiver roles are reversed, then new pairings are made.


The idea is that the Proposer must try to maximise his own gain, while the Receiver can try to coerce the Proposer to give a larger share of the 100 points by rejecting unfair offers.



  • Too similar to Iterated Prisoner's Dilemma?
  • Any other feedback?

Solve Encode a Lenguage with score 0. Shortest code wins. Quine rule applies. I'll unlikely post it.

  • \$\begingroup\$ @pxeger Qunie rule. \$\endgroup\$
    – l4m2
    Commented Jun 20, 2021 at 11:03
  • \$\begingroup\$ Zsh, 120 bytes: Try it online!. Can probably be quite a lot shorter \$\endgroup\$
    – pxeger
    Commented Jun 20, 2021 at 11:08
  • 1
    \$\begingroup\$ Undeleted as potential reference \$\endgroup\$
    – l4m2
    Commented Jul 1, 2021 at 17:53

Smallest number of panels to represent \$1\$ through \$n\$ in base \$b\$

Inspired by this question. (I have taken a lot of inspiration from Puzzling recently).

Because not all of you spend all your (non-code-golf) time looking at puzzles, let me explain the relevant details.

The challenge involved figuring out how to express the numbers 1 through 8 in only 8 panels in bases 2, 3 and 4. Today, I want to do the converse.

Your task is to, given a number \$n\$ and a base, \$b\$, find the fewest number of panels to represent 1 - n in base b. By that, I mean taking a single continuous slice of the set of panels, we can represent a number, and we can do that for 1 through n.

To clarify, each panel can take a single base-\$b\$ number, and the goal is to find a way to represent each number from 1 through n as a base b number. A continuous slice is a set of panels such that all the panels are consecutive and in the same order as in the original number - for example, 221001120 - in which every two-digit and one-digit number is represented in reading order, left-to-right, as well as 100. Note that 101 and 111 are disallowed because they are not consecutive, with 0s in between, 110, 122 and 2110 are disallowed because they read from left to right, and 121 and 212 are disallowed because they reorder the panels.

For a bigger example, let us take the base as 3 and the number as 18: a valid solution is 122121110221010011200, as every number between 1 and 18 can be represented in base 3

1: 122121110221010011200

2: 122121110221010011200

3: 122121110221010011200

4: 122121110221010011200

5: 122121110221010011200

6: 122121110221010011200

7: 122121110221010011200

8: 122121110221010011200

9: 122121110221010011200

10: 122121110221010011200

11: 122121110221010011200

12: 122121110221010011200

13: 122121110221010011200

14: 122121110221010011200

15: 122121110221010011200

16: 122121110221010011200

17: 122121110221010011200

18: 122121110221010011200

In order to verify your solution does work, provide an optimal answer by the number such that it can be checked.

Input looks like this: n b, and output like this: number solution.

Sample inputs and outputs to aid you (output is represented as o - yes I did work all of these out, and yes, I do believe it is the shortest to just find a place to append the number in):

\$b = 2\$:

n <= 8; o = n (11101000)
n = 9, 10; o = 11 (10011101000)
n = 11; o = 13 (1001011101000)
n = 12, 13, 14; o = 14 (11001011101000)
n = 15, 16; o = n (110010111010000)

\$b = 3\$:

n <= 9; o = n (221001120)
n = 10; o = 11 (22101001120)
n = 11; o = 13 (1022101001120)
n = 12; o = 14 (11022101001120)
n = 13, 14, 15; o = 15 (111022101001120)
n = 16; o = 17 (12111022101001120)
n = 17; o = 19 (12212111022101001120)
n = 18; o = 20 (122121110221010011200)

\$b = 4\$:

n <= 10; o = n (1101220213)
n = 11; o = 12 (231101220213)
n = 12, 13; o = 13 (231101220213)
n = 14, 15; o = n (33231101220213)
n = 16, 17; o = 18 (10033231101220213)
n = 18; o = 20 (1003323110122010213)
n = 19, 20; o = 22 (100103323110122010213)

(i'm not touching 5+ this took me an hour)


Are these strings within N bit flips of each other?

Given two strings a and b, and a positive integer n, determine if b can be reached by performing exactly n bit-flips on a.

A bit-flip is defined as changing one of the bits in the character encoding of a string from a 0 to a 1 or from a 1 to a 0

The same bit may be flipped multiple times, so the 0110 can be reached in exactly 2 bit-flips of itself, as well as in 0 (e.g. 0110 -> 0100 -> 0110).



n   a         b
1   hello     hullo
3   hello     hullo
5   hello     hullo
69  hello     hullo
4   hello     hello
1   x         h
1   x         p
1   x         y
2   x         q
2   x         l
2   x         x
4   x         l
4   x         x
4   x         w
4   x         e


n   a         b
2   hello     hullo
5   hello     hello
1   x         a
1   x         n
1   x         f
1   x         x


  • You may assume/require:
    • n > 0
    • a and b are the same length
    • neither string is empty
    • both strings consist only of lowercase letters a to z (or uppercase A to Z instead, if you wish)
  • You should encode the strings in ASCII (or other ASCII-compatible encoding), but if your language cannot easily handle that, ask in the comments and I'll decide on special rules
  • Your code does not need to handle high n or long a or b, but it must work in theory
  • You may use any reasonable I/O method
  • Standard loopholes are forbidden
  • This is , so the shortest code in bytes wins


  • Better title suggestions?
  • I limited to ASCII to avoid people finding loopholes that trivialise the challenge. Do you have any other suggestions that would work better?
  • Test-case suggestions?
  • Is this a duplicate?
  • Is this clear enough?
  • Any other feedback?
  • \$\begingroup\$ This question is equals to: is n - (hamming distance of a and b) a non-negative even number? \$\endgroup\$
    – tsh
    Commented Jun 30, 2021 at 10:08
  • \$\begingroup\$ @tsh yes, I've basically abandoned it \$\endgroup\$
    – pxeger
    Commented Jun 30, 2021 at 12:18
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