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  • \$\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

4730 Answers 4730

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77 78

Is a coincident point in a pair of rotated hexagonal lattices closest to the origin?

I've deleted from here because I agree it is takes too much time to understand at the moment.


This is a question where the technique must search or exclude from search all possibilities. As soon as I figure out how to pose it, there will be a separate question for mathematical techniques other than a simple search, so please don't use them here.

Below is quoted from Math SE determining if a coincident point in a pair of rotated hexagonal lattices is closest to the origin?:

A pair of hexagonal lattices with one scaled by the square root of a rational number \$r = \sqrt{\frac{m}{n}}\$ and then rotated will produce a variety of different hexagonal lattices of coincident points.

For the first lattice let

$$x, y = i+\frac{1}{2}j, \ \frac{\sqrt{3}}{2}j$$

and for the second

$$x, y = r\left(k+\frac{1}{2}l\right), \ r\left(\frac{\sqrt{3}}{2}l\right).$$

Per this and this helpful answer the squares of the distances to unit lattice points are given by Loeschian numbers (A003136) equal to \$i^2+ij+j^2\$ so in this case a point \$i, j\$ on the first lattice will coincide with a point \$k, l\$ on the second lattice once rotated by some amount if

$$n(i^2+ij+j^2) = m(k^2+kl+l^2).$$

For example if \$m, n = 13, 7\$ then both \$(i, j) = (5, 6)\$ and \$(6, 5)\$ will coincide with \$(k, l) = (5, 3)\$ at rotation angles of about 5.2 and 11.2 degrees as given by.

$$\theta = \arctan\left( \frac{\frac{\sqrt{3}}{2}l}{k+\frac{1}{2}l} \right) - \arctan\left( \frac{\frac{\sqrt{3}}{2}j}{i+\frac{1}{2}j} \right)$$

However, while the first solution is part of the hexagonal superlattice built on the much closer point \$(i, j), (k, l) = (1, 3), (1, 2)\$ the second point represents the shortest possible coincident distance and therefore a far lower density coincident lattice.

coincident hexagonal lattices

plotting script: https://pastebin.com/pZFCGXbE


Given the rational number (m, n) e.g. (13, 7) and pairs of known coincident lattice points, e.g. (5, 6), (3, 5) or (6, 5), (3, 5) (besides the origin) we want to find out if this is one of the six closest coincident lattice points, or if it is a member of a coincident lattice with points closer.

  • If it's one of the six closest: return some flag letting us know there wasn't anything closer. You may also return either the same point, or one of the other five of identical distance.

  • If it isn't: return one of the six points that was closest along with (but not only) a flag letting us know that a closer point was found and the original point wasn't one of the closest. Reminder that this will be a closer point in a coincident lattice that also contains the original input.

Do this by some combination of searching/testing all possible pairs of lattice points (one from each lattice) to see if they are coincident and closer and potentially excluding blocks of combinations that don't need searching. The problem is finite because one only searches points that aren't obviously farther from the origin.

Feel free to use tricks to exclude large fractions from search as long as they rely on simple rules, for example a point near the origin in one lattice will never be coincident with a point near a distant point in the other.

But if you find yourself considering computations like matrix division or using Eisenstein integers or Euclid's algorithm in the complex plane please save that for the follow-up question.

This is so shortest code wins.


  • Input will have six integers \$(m, n), (i, j), (k, l)\$ as described above, but can have any order or hierarchy, or additional (but uninformative) place holders (e.g. zero padding, blanks...)
  • \$(m, n)\$ will be positive, but the other four can be positive, negative or zero, excluding a \$(0, 0)\$ pair (the origin).

Recipe for making test cases

In addition to the two mentioned above (13, 7), (5, 6), (5, 3) and (13, 7), (6, 5), (5, 3) you can roll your own:

  1. Pick two Loeschian numbers \$L_1\$ and \$L_2\$ and find some integer pairs (i, j), (k, l) that can make them. (Find all integer pairs that produce a given Loeschian number)
  2. Make new pairs by choosing two nonzero integers \$a, b\$, then:

\begin{align} i' & = ai - bj\\ j' & = aj + b(i+j)\\ k' & = ak - bl\\ l' & = al + b(k+l). \end{align}

If \$i, j\$ and \$k, l\$ were coincident, then \$i', j'\$ and \$k', l'\$ will be as well.

(This just says that the coincidence lattice between two hexagonal lattices is also a hexagonal lattice.)


As a Web Frontend developer, I feel out of place most of the time coming here. And that is despite the existence of established challenges like One Div in the field of web development. But they aren't here, because objective criteria are hard to do. (Sparing more unkind reasons.)

Maybe the following would be a way to draw others. It's not a One Div challenge, but a way to set different web technologies against one another in a sort of code golf.

To give you an idea, compare these two Codepens:

(To be fair, thebabydino doesn't know at the time of writing there is a challenge, it was a spontaneous idea after seeing her code.)

It was rightly pointed out to me that capturing the idea of good coding practice will probably fail. But let me give you a vision: What if the challenges were not only about learning new coding tricks, but also about identifying the "cheats"?

There is a whole industry out there that tries to make objective rules to ultimately measure code quality. Their work is mostly to tilt against windmills. Maybe there is a way to make a playfull challenge a tool to identify new features for linters.

If good coding wins, so much better for it. If everyone thinks the winner was going squarely against the spirit of good coding practice, professionalism gains a discussion.

(Somehow this reminds me of the discussion about doping in sports.)

I hope to make the playing field understandable also to non-web people, so please bear with me if I add comments that would be obvious to everyone in the profession.

One goal here is performance. There are a lot of champions for the approach to minimize external dependencies loaded into the browser. I am trying to capture this with a "no external request" rule. This is not only about the size of files loaded but also the debt of interpreting extensive JS frameworks. Look at current discussions about web site performance, and you will understand what I am talking about.

Challenge: Recreate the image with web technologies

A real question would contain a raster image in any format. Most likely, they will be PNG or GIF (to show animations).

The code must be runnable on Codepen

You must create a pen and post its URL here. Your own authorship should be on a honors base. Anonymous pens are acceptable, but if your nick is the same here as there, you must use your account.

There are a good number of web development environments around, but they all offer different features. The criterium for selection was the ease to enforce the "no external resources" rule - and popularity, of course.

In addition, the code must be posted in its entirety here.

No external code

The pen must not include any external sources.

  • No import statements in any language
  • no statements that trigger any external requests
  • no external stylesheets or scripts imported via the pen settings

An important case here is the use of JS frameworks that offer server-side rendering, or template languages like JSX. Should their source code be made legal? This sort of code is mostly undistinguishable from code used client-side, with the difference that it must be compiled before it is delivered to the client.

No base64-encoding

No part of the code should be unreadable at first glance. So neither HTML, nor CSS, nor JS are allowed to contain any base64-encoded strings

  • No data: URI that contains the base64 option
  • The use of atob() and btoa() in JS is ok, provided there is no string literal containing a base64-encoded string
  • The use of <canvas> and/or URL.createObjectURL() is certainly ok, provided there is no string literal containing a base64-encoded string
  • The use of Blob or ArrayBuffer to construct images is certainly ok.

The use of preprocessors is allowed

Any preprocessor that is offered by CodePen can be used. What counts for the solution is the source code in that language.

Using Babel as a preprocessor inside CodePen is ok, but the "no imports" rule applies just the same.

Currently the Babel preprocessor offered by CodePen has a fixed configuration. If that changes in the future, it might be valid or not to use an URI to an external config file or npm module or not. That is in the future. The basic idea here is "if it is executed on the server side, or by native browser code, go for it"

Linting must pass

To give languages a fair chance that rely on fixed indents, and to maintain readability, the Javascript part must pass ESLint without any warning or error.

  1. Which preset does the best to enforce good code quality? It was pointed out that there will probably remain a number of loopholes, but I would like to try to get them as few and obscure as possible.
  2. I am not so fluent with linters for other languages. Proposals?

No minification

All code should be pretty.

  • Running prettier must not change the code formatting

    This might not cover all languages. Please review and propose other/additional formatters.

  • No code line can be longer than 80 columns

Matching the source image

I have still to research that. My first instinct is to use ImageMagicks magick compare and set a cutoff value to account for antialiasing and rounding differences, but that does not solve how to capture animations. Maybe capture all frames with Lighthouse?

Wining criterium: aggregated Number of lines of code

The number of lines of source code in markup, stylesheet and script are added. The least number of lines wins.

  • Numbers are counted using sloc. Only the Source output value is relevant. For example on the command line

      sloc pen.css | grep Source

    is there a sensible way to scrape the code lines directly from Codepen? Or is this available on the site and I am simply not aware of it?

Please give your answer a header that qoutes the number of lines in each language separately.

  • \$\begingroup\$ While giving 'languages a fair chance' by enforcing no base64 and counting lines of code only is admirable, I think you will find that there are too many ways to cheese it. For example, I could write every line as eval('...' + '...' + '...') to escape minification. It might be better just to use a more conventional scoring criteria (bytes) and drop the base64/formatting requirements. \$\endgroup\$
    – Sisyphus
    Commented Oct 22, 2020 at 2:56
  • \$\begingroup\$ I am not giving up on this so easily. If there is to be a challenge, it should be on the basis that is done with good coding practice. Otherwise, it just makes no sense. Linters should be able to catch a lot of the loopholes, and if not, at least this would be a way for others to learn about them. Isn't this also the goal of this community? My vision would be that if this gains popularity, Linter manufactureres look at "cheating" solutions for improvements to their product. \$\endgroup\$
    – ccprog
    Commented Oct 22, 2020 at 12:06
  • \$\begingroup\$ @ccprog Unfortunately, "good coding practice" is a subjective thing, and we strive for objective scoring criteria. Most answers on the site will happily sacrifice things such as readability and good practice in order to reduce their score. \$\endgroup\$ Commented Oct 22, 2020 at 13:00
  • \$\begingroup\$ @cairdcoinheringaahing Did you see my latest edit? Maybe this can be a productive goal in itself. \$\endgroup\$
    – ccprog
    Commented Oct 22, 2020 at 13:01
  • \$\begingroup\$ @ccprog I was mainly responding to your comment, I was still reading through the full spec as I wrote that. I do very much like the idea of this challenge, and I think it should be fully able to do while still fitting within our site rules (objective scoring criteria etc.). I'm not too familiar with the topic of web technologies, so forgive any mistakes I may make, but one thing sticks out to me as needing improvement when reading this challenge: the attempts to cover things in all languages. Usually, we tend to avoid language-specific challenges on the site, but I think this would be a (cont) \$\endgroup\$ Commented Oct 22, 2020 at 13:05
  • \$\begingroup\$ (cont) good example of where limiting the language would help rather than harm the challenge. From my understanding, it looks as though restricting answers to Javascript would allow you to make the challenge much more objective with the rules about minification and base64, as you won't have to worry about edge cases in esoteric languages such as Jelly or ><> that are commonly used on the site. \$\endgroup\$ Commented Oct 22, 2020 at 13:07
  • \$\begingroup\$ That, on the other hand is unrealistic. Most Javascript code nowadays is somehow cross-compiled, either from other languages (for example the Typescript superset) or through the use of frameworks (Angular, React). I'll have to think more about th "no imports" rule here, as one of the banes of current web development is that a lot of devs loose the sight of whether their code is executed server or client side. \$\endgroup\$
    – ccprog
    Commented Oct 22, 2020 at 13:15
  • \$\begingroup\$ As far as esoteric languages are concerned, there is an implicit stop gag: Forcing Codepen as execution environment. If the language isn't supported there,, the answer fails. And believe me, Codepen is an industry tool, not just any website. Every post there is considered to be a showcase for the developer and counts for his/her professional reputation. \$\endgroup\$
    – ccprog
    Commented Oct 22, 2020 at 13:21
  • \$\begingroup\$ @ccprog I actually have no problem with the challenge as specified. I just think that the winning solutions will cheat the tool(s) you use to score. But if that's ok with you, I don't mind this challenge at all, and will look forward to see the answers. \$\endgroup\$
    – Sisyphus
    Commented Oct 23, 2020 at 1:07

Get the Systematic Chemical Symbol


  • \$\begingroup\$ "prints the corresponding chemical symbol." → "prints the corresponding systematic chemical symbol." \$\endgroup\$
    – Adám
    Commented Oct 21, 2020 at 17:18
  • \$\begingroup\$ You should state clearly that taking a list of digits is also acceptable. \$\endgroup\$
    – Adám
    Commented Oct 21, 2020 at 17:19

C Code Compressor (WIP)

Your task is to create a lossless compression scheme that is optimized for ASCII-only C code. Whitespace and indentation should be preserved.

Your score is the total compressed size of various samples from well known open source C projects. A sample of the largest few files will be taken from one specific commit. Both header files and source files will be included from the following projects:

  • The CPython Interpreter
  • The Linux Kernel
  • Git

Comments will be stripped from the source files for the set of scoring, therefore you do not need to optimize for comments, however you still need to handle them.

Three additional files will also be included in the corpus, which must be handled correctly, but do not contribute to your score:

  • A C source file with comments
  • A Python source file
  • A non-programming plain text file (probably a short story or poem from the public domain. Jabberwocky?)


$$ Score = {C + D + L^2 \over O} $$


  • \$C\$ is the total size, in bytes, of the compressed text
  • \$O\$ is the total size, in bytes, of the original text
  • \$D\$ is the total length, in bytes, of all strings in any predefined dictionary
  • \$L\$ is the length, in bytes, of the longest string in any predefined dictionary

Lowest score wins

Predefined Dictionaries

(Needs refinement)

A predefined dictionary, for the purposes of this challenge, is a collection of predefined output strings of length 2 or greater which are baked into your encoding format.

  • An entry in a predefined dictionary must have at least two different characters to be counted in scoring, so repeating a single character \$n\$ times does not count as a dictionary entry.
  • Leading and trailing whitespace on a predefined string does not count as a distinct entry from one without matching leading or trailing whitespace

A predefined dictionary may, for instance, be helpful for keywords and common identifiers, but it is ultimately up to you what approach you take.

For instance, if you map \x80 to int and \x81 to float, your dictionary score would be \$8\$ for the total length, plus \$25\$ for the longest string (float), for a total of \$33\$. However, mapping runs of tabs and spaces to \xC0-\xFF would not count toward dictionary size in this case.


  • Standard rules and loopholes apply.
  • Although scoring only depends on how well you can compress C, your algorithm needs to work even when the input text is not valid C; it must work for all ASCII input text.
  • Your algorithm must be deterministic, meaning it should always produce the same compressed text every time for any given input, regardless of external factors such as time.
  • \$\begingroup\$ Once you reveal the test cases, couldn’t I just create an “encoding” where 1 represents the Linux kernel, 2 represents the Git source, etc? And if the test cases aren’t fixed but can change if an answer overfits some, then is there really an objective winning criterion? \$\endgroup\$ Commented Oct 10, 2020 at 7:26
  • \$\begingroup\$ @water_ghosts Optimizing exclusively for the test cases given is one of the standard loopholes \$\endgroup\$
    – Sara J
    Commented Oct 10, 2020 at 13:02
  • \$\begingroup\$ Are you planning on scoring based entirely on compression amount, or also on code size? Seems like it will be a fun challenge, especially is code size is part of it. \$\endgroup\$ Commented Oct 10, 2020 at 22:05
  • \$\begingroup\$ @RedwolfPrograms I've done a compression challenge before and I found that making code size a part of the score got in the way of making things interesting since it typically encouraged calling out to gzip since that takes way fewer bytes than making an actual algorithm. I might consider a code size limitation or work in dictionary size into score though. \$\endgroup\$
    – Beefster
    Commented Oct 28, 2020 at 15:56

Is It A Rainbow Color

Posted in main

  • \$\begingroup\$ "obviously not an index" isn't so clear. Is 7 "obviously not an index" if we use 0-based indexing? \$\endgroup\$
    – Adám
    Commented Oct 27, 2020 at 23:09
  • \$\begingroup\$ Adding to that: In some languages, -1 and false are valid indices. \$\endgroup\$
    – Adám
    Commented Oct 27, 2020 at 23:10
  • \$\begingroup\$ Ok updated it to make it a bit more definitive. \$\endgroup\$
    – Scott
    Commented Oct 27, 2020 at 23:35
  • \$\begingroup\$ Doesn't this give an unfair advantage to languages that use -1 for "not found" when asking for an index? Other languages give the next index after the last valid index. \$\endgroup\$
    – Adám
    Commented Oct 27, 2020 at 23:37
  • \$\begingroup\$ What would you suggest then? I feel no matter what, there's always going to be some languages that are better suited to a particular challenge. I tried to offer as many ways as possible for handling invalid input to be more language-inclusive. \$\endgroup\$
    – Scott
    Commented Oct 28, 2020 at 0:40
  • 1
    \$\begingroup\$ Well, let's take a step back and look at the essence of the challenge. There are two things: a look-up, and the compression of the words. I'd focus on one, i.e. either let "not found" have undefined behavior, or simply be a cover for look-up (given a list and a value, find the index, or … if not found). Since the latter is probably too boring, go with the simple lookup. \$\endgroup\$
    – Adám
    Commented Oct 28, 2020 at 0:44
  • \$\begingroup\$ I strongly recommend removing the invalid-input case and just guaranteeing that the input is valid. It means answers don't have to store all valid inputs in some form to check against them, which gives rooms for more interesting methods of fingerprinting. \$\endgroup\$
    – xnor
    Commented Oct 29, 2020 at 1:35
  • \$\begingroup\$ I feel like we've already have some challenge(s) about mapping a set of n strings to the numbers 1 through n, though I don't quite remember what they were or how to find them. The concept is a good one, but it may have been done already. \$\endgroup\$
    – xnor
    Commented Oct 29, 2020 at 1:38
  • \$\begingroup\$ This is some valuable feedback. I agree that we've had some very similar challenges, how would it work if I rework the challenge to be something along the lines of this: output whether a given color is in the rainbow. The possible inputs would be pre-defined and output can be truthy/falsy \$\endgroup\$
    – Scott
    Commented Oct 29, 2020 at 2:12
  • \$\begingroup\$ I can't help thinking this will boil down to a "let xnor find some magic trick that simplifies it significantly and then adapt it to every other programming language" challenge \$\endgroup\$
    – pxeger
    Commented Oct 30, 2020 at 8:13
  • \$\begingroup\$ I think this actually should have good variety across languages. It's string-based, and different languages have various abilities to use regexes, hashes, conversion to code points, etc, that differ in relative length. This is in contrast to purely arithmetic ones where a single formula can be ported easily. \$\endgroup\$
    – xnor
    Commented Nov 3, 2020 at 4:44

Fix the message on my calculator

My calculator has lots of mathematical buttons, but they're boring; it's much more fun to write messages using letters. You can access the letters of the alphabet with the ALPHA button which turns each button press into a corresponding letter which is noted above each key (or space, in the case of .).

My calculator looks like this*:

 A    B    C    D    E    F
log   ln   ^2  sin  cos  tan
 G    H    I    J    K    L
 /    ->   (    )    ,   Esc
 M     N     O
 7     8     9 
 P     Q     R     S     T
 4     5     6     *     %
 U     V     W     X     Y
 1     2     3     +     -
 Z    (Space)
 0       .  

For example, to type HELLO I would press Esc cos -> -> 9 ..

However, recently my ALPHA button was broken, so I can't get to the letters any more. When I try to write a message, only the mathematical symbols get input.

Without ALPHA, the keys do the following:

  • The Esc key does nothing
  • The digit keys, ., (, ), ,, ->, +, -, *, and % enter those strings verbatim
  • The log, ln, sin, cos, and tan keys enter function_name() and place the cursor before the closing bracket
  • The / and ^2 keys enter a / or ^2 respectively, and then move the cursor to before the new symbol, iff the symbol preceding it is one of + - * % ( , ->

For the example above, instead of getting the message HELLO, I would get ->cos(9). Note that since the key for L (Esc) does nothing, it is unfortunately impossible to know that the letter L was pressed, and this information is lost.

Your task is to convert a string of maths operations into the correct text.


  • Some combinations are ambiguous; for example CGI, CIG, ICG all could have come from (/^2. In this case you can output any or all of them.
  • You should assume the input will always be a valid combination that could have been produced using the rules above.
  • You should never output Ls
  • You may output the string in any case, with an optional trailing newline, but no other extra whitespace.
  • This is . The shortest code in bytes wins.
  • Standard rules and loopholes apply.


More coming soon.

      Input                 Output
789                    MNO
log(ln(^2))            ABC
cos((cos((9)           EIEIO
-->-                   YHY
3.14159                W UPUQO
(/^2                   CGI or CIG or ICG
<empty string>         <empty string>

* The % and / keys are both for division on the original calculator, but labelled something like ÷ and ☐/☐. Here I've just replaced them with simple ASCII equivalents. (did you know that both % and ÷ represent a fraction?) Also, there are other keys not shown (including the mentioned ALPHA), but they don't matter for this challenge.


  • Any feedback or suggestions?
  • Is this clear enough?
  • I considered adding some regular expressions to show more clearly what the rules mean, but decided not to because I though that should be part of the challenge. Was this the right decision?
  • I will probably make another challenge that is the reverse of this - given a list of buttons that were pressed, produce the correct mathematical symbols. It might have made sense to post that challenge first as context for this one, but I thought this one is more interesting
  • Regarding ambiguity: I could change some of the syntax rules to reduce the ambiguity but it would make things a lot more complicated and it couldn't be removed completely without straying significantly from the behaviour of the original calculator

Set of subsets without subsubsets

You are given two integers \$n\$ and \$k\$, and are to output a random collection of \$k\$ sets \$\mathcal A_1,\ldots,\mathcal A_k\$ such that:

  • each \$\mathcal A_i\$ is a subset of \$\{1, 2, \ldots, n\}\$;
  • no \$\mathcal A_i\$ is a subset of another \$\mathcal A_j\$.

Any collection which satisfies these constraints should have positive probability of being output.

Input/output is flexible.

You may assume that there exists at least one collection which satisfies the constraint; by Sperner's theorem, this is equivalent to the condition \$k\leq \tbinom{n}{\lfloor n/2\rfloor}\$.

Example outputs

The output is random, so these example collections should merely have positive probability of being output.

| n | k | possible output
| 2 | 2 | { {1}, {2} }
| 4 | 3 | { {1}, {2,3}, {3,4} }
| 4 | 6 | { {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4} }
| 5 | 3 | { {1,2,3,4}, {2,5}, {3,5} }

This is , so shortest code in each language wins.

  • \$\begingroup\$ What does random output do to benefit this challenge? Randomness just adds boilerplate and prevents some languages from competing. That doesn't make randomness bad universally, but it does suggest that there should be a reason you involve randomness. \$\endgroup\$
    – Wheat Wizard Mod
    Commented Nov 28, 2020 at 21:07
  • \$\begingroup\$ @WheatWizard One alternative would be "output any collection of subsets which satisfies the constraints". But that challenge is trivial: it suffices to output k different subsets, all of size \$n\choose {n/2}\$; many challenges have a built-in for this. I included the randomness to sidestep this. \$\endgroup\$ Commented Nov 28, 2020 at 22:35
  • 1
    \$\begingroup\$ @WheatWizard Thinking about it, I now realize there is another alternative: list all acceptable collections of subsets, rather than a random collection. That seemed more tedious to me when I wrote the proposal, but it might actually be better; I'll think about it. Thanks for the comment! \$\endgroup\$ Commented Nov 28, 2020 at 22:37

(Pan)consummate Vs

An integer \$v\$ is said to be consummate if there is an integer \$n\$ and a base \$b\$ such that \$n\$ divided by the sum of its base \$b\$ digits is equal to \$v\$.

An integer \$v\$ is said to be panconsummate if it is consummate in all bases \$b\geq 2\$. Panconsummate numbers are A058226 in the OEIS.

Your task:

Write a full program or function that takes a positive integer \$v\$ and returns two distinct, consistent values, one if \$v\$ is panconsummate, and the other if \$v\$ is not. However, the sum of your code's bytes must be panconsummate as well. Your code must work theoretically for any integer.

Truthy values:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 18, 20, 21, 23, 24, 31, 34, 36, 37, 39, 40, 43, 45, 53, 54, 57, 59, 61, 69, 72, 73, 77, 78, 81, 85, 89, 91, 121, 127, 144, 166, 169, 211, 219, 231, 239, 257, 267, 271, 331, 337, 353, 361, 413, 481, 523, 571, 661, 721, 1093, 1291, 3097

Falsey values:

13, 16, 17, 19, 22, 25, 26, 27, 28, 29, 30, 32, 33, 35, 38, 41, 42, 44, 46, 47, 48, 49, 50, 51, 52, 55, 56, 58, 60, 62, 63, 64, 65, 66, 67, 68, 70, 71, 74, 75, 76, 79, 80, 82, 83, 84, 86, 87, 88, 90, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 122, 123, 124, 125, 126, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 167, 168, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200

Note that panconsummate numbers \$v>3097\$ must be at least \$10^6\$, and the OEIS speculates that the truthy values above are all panconsummate numbers.



The Slices Puzzle Puzzle


Slices is a puzzle game. Placed on a 2d map is a set of islands; possibly just a single island. Placed on each island is a set of points. The task is to draw a number of straight lines across the islands such that no point can be reached from another without crossing one of the drawn lines, or leaving an island. In other words, you partition the points with lines such that each partition contains at most one point on each island. For each level, you are given a maximum number of lines that can be drawn.

In the game illustrated below, drawing a straight line between the first and second row of points, and then a line between the second and third row of points, would create a valid partition of points.

Game illustration


Your task is to write a solver for this game. The solver takes as input a set of islands, the set of points for each island plus the maximum number of lines which can be drawn. The input and output may be represented in any way, and the input may be hard-coded. Hard-coded input must show that your solution is sufficiently general. Any format will be accepted. For instance, you might implement this as a linear programme in the CPLEX format -- I haven't verified that this is possible -- or you might write some sort of brute force solution in in an imperative language. Maybe some other declarative solution works even more nicely.


Solutions will be judged by popularity. Things such as number of bytes in the solution, computational complexity (justify this if you want) and elegance could be considered.

  • \$\begingroup\$ 1) Should the lines (separating the points) be straight lines? 2) Allowing hardcoded input is different from our I/O defaults, and may interfere with code golf scoring. 3) Complexity is often hard to measure, and elegance is not an objective winning criterion. My suggestion is pure code-golf, pure fastest-code, or maybe popularity-contest. (For fastest code, you'll need to prepare some test cases, and get ready to run all submissions on your own machine. This is how we achieve objective scoring for speed.) \$\endgroup\$
    – Bubbler
    Commented Dec 10, 2020 at 8:13
  • \$\begingroup\$ @Bubbler The lines should be straight. I would like to allow hard-coded inputs, as this makes it possible to submit a nice linear programme if that is possible. I think I will go for popularity given that there are so many interesting dimensions. I will update the question with this in mind. \$\endgroup\$
    – justinpc
    Commented Dec 10, 2020 at 11:01
  • \$\begingroup\$ @Bubbler What do you think about the task now? \$\endgroup\$
    – justinpc
    Commented Dec 10, 2020 at 11:06
  • 1
    \$\begingroup\$ Regardless of the challenge type, you need to provide some test cases for verifying submissions - at least a couple of easy levels and a couple of harder ones, maybe some that requires drawing lines at unexpected angles. \$\endgroup\$
    – Bubbler
    Commented Dec 11, 2020 at 2:44
  • \$\begingroup\$ @Bubbler Will do. \$\endgroup\$
    – justinpc
    Commented Dec 11, 2020 at 6:51
  • \$\begingroup\$ Computational geometry problems have a lot of nuance. e.g are the points in general position, or can they be collinear? Can a cut just touch a vertex, or must it actually cross an edge? Is an island just the convex hull of its points, or is it given as a polygon? Are the convex hulls (resp. polygons) of the islands guaranteed to be mutually disjoint? \$\endgroup\$ Commented Dec 11, 2020 at 22:49
  • \$\begingroup\$ @user1502040 I will clarify these questions. I would like to make the representation of the solution as general as possible. Can you clarify what you mean by "Can a cut just touch a vertex, or must it actually cross an edge". By edge, do you mean edge of an island? Lines should not intersect with vertices. I should clarify that lines are not line segments, and that they should extend to infinity. \$\endgroup\$
    – justinpc
    Commented Dec 12, 2020 at 5:59
  • \$\begingroup\$ @justinpc Sorry, I can see how that was confusing. I meant the conceptual edge joining two points in the same island. \$\endgroup\$ Commented Dec 13, 2020 at 0:56
  • \$\begingroup\$ @justinpc I know most of your feedback has been discouraging, but I think this would be a fun challenge. Don't be nervous about posting it! \$\endgroup\$ Commented Jan 3, 2021 at 22:25
  • \$\begingroup\$ @user1502040 You're right. It would be a fun challenge. I have been busy recently, and would like to put time into this when I have a spare minute. \$\endgroup\$
    – justinpc
    Commented Jan 4, 2021 at 11:27

Given a list of strings like this (input):

["some test", "{", "some", "_subst", "ring}", "some", "other text"]

Need to combine all the elements in to one for the given substring (input).

For example:

substring = "{some_substring}"

The output list of strings will be

["some text", "{some_substring}", "some other text" ]

Here are few more possible lists of strings:

input_list = ['some test', '{', 'sub', 'string}', 'som other text']
input_list2 = ['some test', '{sub', 'string}', 'som other text']
input_list3 = ['some test', '{', 'sub', 'string}', 'som other text']
input_list4 = ['some test', '{', 'sub', 'string', '}', 'som other text']
input_list5 = ['some test {', 'sub', 'string', '}', 'som other text']
input_list6 = ['some test {', 'sub', 'string', '} som other text']

substring can be any text and it does not need to be wrapped in to round braces. If substring appears multiple times in the input string than it should be replaced as well.

Note: Elements of the string that do not contain part of the input string should be the same instances. So creating a new list from joned string will not work for this problem.

Let say these strings carry additional information like style. So removing or replacing them will remove style as well

Signature of expected function to better understand problem.

def join_variable(input_list, substring):
    # implementation
    return output_list
  • \$\begingroup\$ You need to specify a winning criterion (usually code-golf) \$\endgroup\$ Commented Dec 22, 2020 at 17:00
  • \$\begingroup\$ Not exactly sure what do you mean. What kind of criterion? Can share some examples? \$\endgroup\$
    – Pol
    Commented Dec 22, 2020 at 18:57
  • 3
    \$\begingroup\$ We don't host challenges where the goal is solve the problem. Rather, there must be a way to "score" solutions so they compete with each other. See why. \$\endgroup\$
    – Adám
    Commented Dec 23, 2020 at 0:27
  • \$\begingroup\$ Got it. Alright then. Should I remove it? \$\endgroup\$
    – Pol
    Commented Dec 23, 2020 at 13:34
  • 1
    \$\begingroup\$ You can use code-golf as a scoring criterion, I think - shortest code wins. It's the most commonly used tag here. \$\endgroup\$
    – user
    Commented Dec 23, 2020 at 18:05

Map the alternating group A5 to the rotations of a dodecahedron

8 years ago, the Math Stack Exchange was able to prove, in 523 characters of arcane incantation, that the alternating group A5 (ie all rearrangements of five objects that can be created by swapping two objects an even number of times) is isomorphic to the group of rotations of a dodecahedron that map vertices to vertices, edges to edges, and faces to faces. In my mind, 523 characters is far too many -- we can do better!


To solve this challenge, take as input a member a of A5, represented as an even permutation of the first 5 integers, and output a 3x3 rotation matrix that maps a dodecahedron to itself. Your program must be an isomorphism: This means that your program must output a unique rotation matrix for each element a, and for any elements of A5 a, b, c such that a composed with b yields c, then YourProgram(a) * YourProgram(b) = YourProgram(c) (to within floating point precision, of course). Shortest program wins!

Example Input and Output

We present as an example one valid isomorphism, but any valid isomorphism is a permitted answer.

Example Dodecahedron

We assosciate the colored cubes [red, green, yellow, blue, black] with the numbers [1, 2, 3, 4, 5]. Then, for each permutation a in A5, our example outputs a matrix that rotates the figure shown so as to permute the colored cubes according to a.

[1, 2, 3, 4, 5] -> [[1, 0, 0][0, 1, 0],[0, 0, 1]]

[2, 3, 1, 4, 5] -> [[0, 0, 1], [1, 0, 0], [0, 1, 0]] (a 120 degree rotation along the axis (1, 1, 1)

[3, 4, 1, 2, 5] -> [[-1, 0, 0], [0, 1, 0], [0, 0, -1]] (a 180 degree rotation along the y axis ie coming towards the viewer)

[4, 2, 3, 5, 1] -> [[-0.3090170, 0.5, 0.8090170] [0.5, 0.8090170, -0.3090170] [-0.8090170, 0.3090170, -0.5]]

(a 60 degree rotation along the vector (1 / phi, phi, 0)


Battleships Board Validation

Do you know the game Battleship? Well, I want to play with my little brother, but before we can begin we need to set up our ships on the board. This is your input.

Now, we need to check that the ships that we set up are valid. This is where you come in to help. Your task is to write a program or function which checks whether the given 2d array (your board/input) is a valid board or not.

The input will be a 2d array, where 1 represents part of a ship and 0 represents part of the ocean.

The rules:

  • There must be:
    • One battleship (size 4)
    • Two cruisers (size 3)
    • Three destroyers (size 2)
    • Four submarines (size 1)
  • Any additional ships are not allowed, and neither are missing ships
  • Each ship must be either vertical or horizontal (aside from submarines, which are a single grid space)
  • The ships cannot overlap, but may be adjacent

Solutions may not use streams.

in addition, you must solve this with a BF function (normally the class will also be called BF) which receives the 2d array(input) and uses a validate function. Afterward, you are free to manipulate the 2d array however you want and add any functions that you want.

here are some more examples that you can use to check your code: is valid:


wrong ships, result is false-


missing ships, is false->


check contact, is true->


check another one with contact->


check invalid, is false->


random board which is true->


Who's code will be the shortest byte solution that passes all these examples successfully?! Good luck. ;)

Any language is allowed, but I'd like to see an answer in Java especially.


link to the original post- Validating a Battleship board

this also my first post that I am doing. Whats a good way to decide who is the winner and in what time frame? help would be appreciated

(any sidenotes on how to improve?anything needing clarity?)

  • \$\begingroup\$ Hi there, new person! I'm happy to see you using the sandbox first! Generally it is good to avoid language-specific challenges. Also, is there a particular reason you're making this a fastest-code challenge rather than code-golf? \$\endgroup\$
    – Beefster
    Commented Jan 7, 2021 at 23:28
  • \$\begingroup\$ not really im just used to using java \$\endgroup\$
    – Rocky cohn
    Commented Jan 8, 2021 at 6:15
  • \$\begingroup\$ not to sure overall what the challenge itself would be so any suggestions would be great \$\endgroup\$
    – Rocky cohn
    Commented Jan 8, 2021 at 6:17
  • 1
    \$\begingroup\$ Hi there. Great first challenge! As others have mentioned, I would drop the Java code to allow all languages, and just give test cases (since I usually answer in Java and 05AB1E, I might still answer in Java when the challenge goes live though ;) ). As for the win-condition, I would suggest to use [code-golf], since it's the most used and popular, and probably most appropriate for this challenge as well. (Code-golf means to create a function/program as short as possible in terms of used bytes.) \$\endgroup\$ Commented Jan 8, 2021 at 8:48
  • \$\begingroup\$ @KevinCruijssen would it still be good for me to leave the condition that the input is a 2d array into the BF function and must have the validate function? I think that it also makes it more challenging \$\endgroup\$
    – Rocky cohn
    Commented Jan 8, 2021 at 8:58
  • 1
    \$\begingroup\$ @Rockycohn Input as a 2d array is indeed good. But I would remove the Java function and validate code completely and just allow all languages. You can still mention you prefer to see a Java answer especially, though. Here a pastebin what I would personally change it to as example. \$\endgroup\$ Commented Jan 8, 2021 at 9:04
  • \$\begingroup\$ gotcha. did it thanks:) \$\endgroup\$
    – Rocky cohn
    Commented Jan 8, 2021 at 9:19
  • 1
    \$\begingroup\$ That looks a lot better. :) I've upvoted and voted to re-open your challenge, since I saw you made the same edit there. \$\endgroup\$ Commented Jan 8, 2021 at 9:52
  • \$\begingroup\$ would it be better to delete the main post repost it since the title has Java and I changed it to all languages? \$\endgroup\$
    – Rocky cohn
    Commented Jan 8, 2021 at 10:18
  • \$\begingroup\$ Unless there's some rep constraint that I'm not aware of, you should be able to edit the title of the main post. There's no need to repost it. (Or just ping me so that I do the update myself if you're not allowed to for some reason.) \$\endgroup\$
    – Arnauld
    Commented Jan 8, 2021 at 10:44
  • \$\begingroup\$ it doesn't seem to let me @Arnauld if you could remove "Java," from the post that would be great \$\endgroup\$
    – Rocky cohn
    Commented Jan 8, 2021 at 10:46
  • \$\begingroup\$ Also, please note that you should not make assumptions about language features. You should remove things like Solutions may not use streams or you must solve this with a BF function. \$\endgroup\$
    – Arnauld
    Commented Jan 8, 2021 at 10:46
  • 1
    \$\begingroup\$ Title updated. :-) \$\endgroup\$
    – Arnauld
    Commented Jan 8, 2021 at 10:47

Solve a Cubic Equation


  • Your program will take in the integer coefficients of the equation \$ax^3+bx^2+cx+d=0\$ as inputs (a, b, c, and d)
  • All solutions are between 1000 and -1000.
  • a is nonzero


  • All real solutions of the input equation, with an accuracy of at least the thousandths place.


  • Built-in equation solver are not allowed
  • Native math libraries that do not solve equations may be used
  • If you have any questions, please ask in comment



1 2 3 4



Another Valid Output:

-1.651 -1.651 -1.651


1 3 0 -1


-2.879 -0.653 0.532

Worked Example (C++) doesn't work with two roots

Try it online!

#include <iostream>
#include <cstdio>
using namespace std;
int main()
   double a,b,c,d;
   for(double i=-100;i<=100;i+=0.0001)
      double j=i+0.0001;
      double y1=a*i*i*i+b*i*i+c*i+d;
      double y2=a*j*j*j+b*j*j+c*j+d;
         double x=(i+j)/2;
         printf("%.3lf ",x);
  • 1
    \$\begingroup\$ I don't know why is the section with your code named "Worked Example", but it doesn't seem to solve the equation x^3 + x^2 = 0. \$\endgroup\$ Commented Jan 10, 2021 at 11:36
  • \$\begingroup\$ I'd suggesting guaranteeing that a is nonzero so that quadratic or lower cases don't have to be handled. Or, possibly just guarantee a monic polynomial with a=1. \$\endgroup\$
    – xnor
    Commented Jan 11, 2021 at 3:43

Win a K vs KQ endgame


The goal of this challenge is to create a program or function which will win a Chess game with a King and Queen against a lone King. The user will specify three squares, representing the locations of the computer's King, the computer's Queen, and the user's King. The computer will then a output a move which will eventually lead to checkmate.


The program or function will first take as input three squares, representing the locations of the computer's King, the computer's Queen, and the user's King (not necessarily in that order). It can be assumed that the input is a legal position.

Parsing input is not the point of this challenge, so all reasonable forms of input/output are allowed, including but not limited to

  • Strings with algebraic chess notation such as "Ke4" or "Qf6"

  • Triples representing pieces and coordinates such as ('K', 0, 2)

After three squares are taken as input, the computer outputs a single legal move. Behaviour on invalid input is undefined.


This procedure must terminate using your program or function:

  • User sets up a legal KQ vs K position on a physical chessboard.

  • User inputs the board position. The computer outputs a legal move. If the move is a checkmate, STOP.

  • User makes the computer's move on the physical board.

  • User makes a legal move for the lone king on the physical board.

  • User goes to step 2 and repeats.

In other words, the computer must eventually win by checkmate, through repeatedly using your program or function.

Furthermore, from any legal starting position the checkmate must occur in 50 or fewer moves by the computer.

(Of course, a physical chessboard is in no way necessary to test the code; I only mentioned it to help visualize the procedure. The chessboard could just as well be visualized in the user's head.)

Possible test cases

The squares are given in the order: computer's Queen, computer's King, user's King

  • c2, h8, a1 (must avoid stalemate)
  • a1, a2, a8
  • a8, a1, e5


  • The checkmate must occur in 50 or fewer moves by the computer, but it does not need to be as fast as possible.
  • Chess libraries are not permitted.
  • Shortest program in each language (in bytes) wins.

Questions for sandbox

How can I word the rules better if they are not currently sufficiently clear?

  • \$\begingroup\$ 'It can be assumed that the input is a legal position from which it is possible to force checkmate in 50 or fewer moves.' It's always possible to force checkmate in a KQ vs K endgame (when the side with the queen has the move) in 10 moves or fewer. I wonder if you should consider lowering the 50-move limit? \$\endgroup\$
    – Dingus
    Commented Jan 5, 2021 at 8:39
  • \$\begingroup\$ I don't want to lower the limit too much because it is conceivable that some good solutions might require more than 10 moves. \$\endgroup\$
    – 79037662
    Commented Jan 5, 2021 at 14:20
  • \$\begingroup\$ Fair enough. I still think you should mention that mating in 10 moves is always possible, even if you allow for more. Right now the question reads as if the possibility of forcing checkmate in 50 moves is an assumption (which it is not). \$\endgroup\$
    – Dingus
    Commented Jan 5, 2021 at 22:41
  • \$\begingroup\$ In a pedantic reading, 'The user will specify three squares' could be understood to mean that the same starting position (chosen by the user) is always used. Can you make it clearer that the program should work for any legal starting position (I assume this is what you intend)? \$\endgroup\$
    – Dingus
    Commented Jan 5, 2021 at 22:42
  • \$\begingroup\$ @Dingus The fact that any position can be won in 10 or fewer moves isn't really relevant to the specification of the challenge, but if you think it improves the prompt I'll go ahead and add it. \$\endgroup\$
    – 79037662
    Commented Jan 6, 2021 at 18:05
  • \$\begingroup\$ @Dingus I explain what I mean in the next section, and I'll be surprised if anyone reads this and thinks I meant to use the same starting position every time. \$\endgroup\$
    – 79037662
    Commented Jan 6, 2021 at 18:08
  • \$\begingroup\$ I agree that the 10 moves doesn't affect the spec. Where I see the possible relevance is as an answerer, where I might keep that knowledge in the back of my mind as a reference point to focus my golfing strategy. To your Sandbox questions: 1. I won't offer a strong opinion either way, just an observation: banning libraries will likely lead to fewer answers. Library-based answers might be boring but voters tend to vote accordingly. 2. I agree that there's no reason to specify which side is which. \$\endgroup\$
    – Dingus
    Commented Jan 6, 2021 at 22:26
  • \$\begingroup\$ @Dingus Thanks for your input. \$\endgroup\$
    – 79037662
    Commented Jan 7, 2021 at 15:47
  • \$\begingroup\$ Disallowing chess libraries sounds alright, since there's only 3 pieces to account for. More complex problems have been answered without libraries, but many tend to use libraries since it's simply golfier and simpler. I think this question would benefit from fastest-code scoring, or something similar, rather than code-golf. \$\endgroup\$
    – Razetime
    Commented Jan 8, 2021 at 11:13
  • 1
    \$\begingroup\$ I think fastest code would be tricky to test due to the potentially numerous user inputs needed per code run, and testing how short of a mate the programs can find is not really what I had in mind for this challenge. \$\endgroup\$
    – 79037662
    Commented Jan 8, 2021 at 21:48
  • \$\begingroup\$ Just commenting since you requested, I think this is good. My only real disagreement was the interactivity and since you have changed it and the spec remains clear, it has my thumbs up for as much as that is worth. \$\endgroup\$
    – Wheat Wizard Mod
    Commented Jan 12, 2021 at 2:27



Is this a narrow Thumb instruction?

ARM Thumb was originally a 16-bit only subset of the 32-bit ARM instruction set.

However, later versions added 32-bit "wide" instructions which were more flexible, and called the original, more restrictive 16-bit instructions "narrow" instructions.

The assembler now chooses between narrow and wide instructions automatically, depending on how the instruction was written. However, this meant that the syntax had to be changed to have specific rules.

Your job is to be this assembler.

However, the programs you parse are not that interesting; they will only ever consist of add and adds.

More specifically:

Your task is to write a function or program that will take an add/adds instruction, and return a truthy value if it is a valid narrow instruction, or a falsey value if it is not.

Syntax rules

  • ARM has 16 registers, r0-r12, r13 (aka sp), r14 (aka lr), and r15 (aka pc). For ease of parsing, we are going to refer to all registers by their number, instead of using the special register names.
  • Do note the names when reading the official docs, as there are a lot of special cases for sp and pc.
  • In Thumb mode, these are split into "Lo registers", which are r0-r7, and "Hi registers" which are r8-r15. Many instructions can only use Lo registers.
  • Many instructions use the same source register as the destination register, even if they are written with three operands.
  • add and adds are distinct instructions. adds affects the condition flags, while add does not. That is the difference, if you were wondering.

The following 6 forms are valid for narrow instructions (adapted from here):

  1. adds x, y, #imm: x and y must both be Lo registers, and imm is a 3-bit constant from 0-7.
  2. adds x, y, z: x, y, and z must all be Lo registers.
  3. add x, x, y: x and/or y must be Hi registers. Note that x is repeated twice.
  • We are ignoring the fact that ARMv6 relaxed this rule to keep it interesting.
  1. adds x, x, #imm: x must be a Lo register. imm is an 8-bit constant from 0-255. Again, note that x is repeated twice.
  2. add r13, r13, #imm: imm is a constant multiple of 4 in the range 0-508.
  3. add x, y, #imm: x must be a Lo register, and y must either be r13 or r15. imm is a constant multiple of 4 in the range 0-1020.

Everything else is either a wide instruction or not valid.

Other notes

Standard loopholes, everything must be self-contained, and you are only allowed to treat it as text. You can't feed it to an assembler (unless you include the assembler source code in the result, but.. why).

The input can either be a string argument or text from stdin.

You can assume the format will match the following format (all lowercase, separators being a single space):

{add or adds} reg, reg, {#imm or reg}

Where imm is a non-negative number in base 10 (yes, including zero).

As a regex pattern:

^adds? r([0-9]|1[0-5]), r([0-9]|1[0-5]), (#[0-9]+|r([0-9]|1[0-5]))$

Reference implementation

In case the rules are difficult to follow, here is a reference implementation I made in C. Yes, I deliberately overabstracted it to make you do all the work.

I resisted the urge to post the reference implementation in ARM Thumb assembly, as that would be genuinely evil. 😏

You will not need to do the same error checking I did here. You can always assume the string itself is valid. The error checks in the main function are mostly to show what CAN'T happen.

#include <stdlib.h>
#include <stdint.h>
#include <stdio.h>
#include <stdbool.h>
#include <errno.h>
#include <inttypes.h>

struct thumb_add_insn {
    char opcode[5];
    uint32_t op1;
    uint32_t op2;
    char op3_prefix;
    uint32_t op3;

// Returns whether the opcode ID is adds.
static inline bool is_adds(const char *opcode)
    return strcmp(opcode, "adds") == 0;

// Returns whether this register ID belongs to a Lo register,
// specifically r0-r7.
static inline bool is_lo_reg(uint32_t reg_id)
    return reg_id <= 7;

// Returns whether this register ID belongs to a Hi register,
// specifically r8-r15.
static inline bool is_hi_reg(uint32_t reg_id)
    return reg_id >= 8;

// Returns whether the operand prefix is for an immediate
// value, specifically, '#'.
static inline bool is_imm(char c)
    return c == '#';

// adds x, y, #imm3
static bool is_form_1(const struct thumb_add_insn *insn)
    return is_adds(insn->opcode)
        && is_lo_reg(insn->op1)
        && is_lo_reg(insn->op2)
        && is_imm(insn->op3_prefix)
        && insn->op3 <= 7;

// adds x, y, z
static bool is_form_2(const struct thumb_add_insn *insn)
    return is_adds(insn->opcode)
        && is_lo_reg(insn->op1)
        && is_lo_reg(insn->op2)
        && !is_imm(insn->op3_prefix)
        && is_lo_reg(insn->op3);

// adds x, x, #imm8
static bool is_form_3(const struct thumb_add_insn *insn)
    return is_adds(insn->opcode)
        && is_lo_reg(insn->op1)
        && insn->op1 == insn->op2
        && is_imm(insn->op3_prefix)
        && insn->op3 < 256;

// add x, x, y
static bool is_form_4(const struct thumb_add_insn *insn)
    return !is_adds(insn->opcode)
        && !is_imm(insn->op3_prefix)
        && (is_hi_reg(insn->op1) || is_hi_reg(insn->op3))
        && insn->op1 == insn->op2;

// add r13, r13, #imm
static bool is_form_5(const struct thumb_add_insn *insn)
    return !is_adds(insn->opcode)
        && insn->op1 == 13
        && insn->op1 == insn->op2
        && is_imm(insn->op3_prefix)
        && insn->op3 <= 508
        && insn->op3 % 4 == 0;

// add x, y, #imm, y == r13 or r15
static bool is_form_6(const struct thumb_add_insn *insn)
    return !is_adds(insn->opcode)
        && is_lo_reg(insn->op1)
        && (insn->op2 == 13 || insn->op2 == 15)
        && is_imm(insn->op3_prefix)
        && insn->op3 <= 1020
        && insn->op3 % 4 == 0;

// Parses a Thumb add/adds instruction.
// Returns 1 if it is a narrow instruction, 0 if it is not,
// and -1 on an error.
int is_narrow_add(const char *str)
    // Note that you do not have to do error checking for the
    // competition.

    if (str == NULL) {
        errno = EINVAL;
        return -1;

    // Allocate a 24 byte struct on the heap for good measure
    struct thumb_add_insn *insn = calloc(1, sizeof(*insn));
    if (insn == NULL) {
        return -1;

    // Parse the instruction with sscanf.
    // {adds} r{0}, r{3}, {#}{3}
    if (sscanf(str, "%4s r%"SCNu32", r%"SCNu32", %c%"SCNu32,
               &insn->op3) != 5
       || (strcmp(insn->opcode, "add") != 0
          && strcmp(insn->opcode, "adds") != 0)
       || insn->op1 > 15
       || insn->op2 > 15
       || (insn->op3_prefix != 'r' && insn->op3_prefix != '#')
       || (insn->op3_prefix == 'r' && insn->op3 > 15)
    ) {
        errno = EINVAL;
        return -1;

    int ret;
    // Test against each of the forms
    if (is_form_1(insn)) {
        ret = 1;
    } else if (is_form_2(insn)) {
        ret = 1;
    } else if (is_form_3(insn)) {
        ret = 1;
    } else if (is_form_4(insn)) {
        ret = 1;
    } else if (is_form_5(insn)) {
        ret = 1;
    } else if (is_form_6(insn)) {
        ret = 1;
    } else { // not a match
        ret = 0;

    return ret;

Test cases

adds r6, r3, #0    // true, form 1
adds r0, r1, #7    // true, form 1
add r0, r1, #3     // false, must be "adds"
adds r0, r9, #1    // false, r9 is a Hi register
adds r0, r1, #9    // false, must be 0-7

adds r0, r0, r0    // true, form 2
adds r7, r1, r2    // true, form 2
adds r4, r4, r1    // true, form 2
add r7, r1, r2     // false, must be "adds"
adds r13, r14, r6  // false, r13 and r14 are Hi registers (this isn't even valid as a wide instruction)

adds r0, r0, #0    // true, form 3
adds r5, r5, #249  // true, form 3
add r6, r6, #31    // false, must be "adds"
adds r3, r3, #256  // false, must be 0-255
adds r8, r8, #72   // false, r8 is a Hi register

add r4, r4, r11    // true, form 4
add r8, r8, r5     // true, form 4
add r9, r9, r9     // true, form 4
add r14, r14, r12  // true, form 4
add r8, r9, r10    // false, Rd must be the same
add r1, r1, r0     // false, one must be a Hi register (we are ignoring the ARMv6 change)

add r13, r13, #0   // true, form 5
add r13, r13, #48  // true, form 5
adds r13, r13, #64 // false, must be "add"
add r13, r13, #17  // false, not a multiple of 4
add r13, r13, #512 // false, must be 0-508

add r0, r15, #0    // true, form 6
add r4, r13, #1000 // true, form 6
add r11, r13, #32  // false, r11 is a Hi register
add r2, r13, #4000 // false, must be 0-1020
adds r7, r15, #384 // false, must be "add"
add r3, r15, #127  // false, not a multiple of 4

Things you can safely ignore:

// String will never be empty
adds r1, r2 // don't worry about implicit middle operand
adds R4, #12 // same
adds r3, r3, #-3 // adding a negative is not even a thing
add r0, r0, r99 // the only registers are r0 - r15
add r13, r13, #0x32 // it is base 10
subs r1, r1, r2 // only add and adds need to be handled
add r2, r2, lr // you don't need to handle the special names
add r0, sp, #0 // same
add #3, r1, r1 // only the last one will be an immediate
adds r3, r3, 32 // all immediates are prefixed with #
ADDS R0, R0, R1 // everything is lowercase
adds        r2,     r3 , r4 // only one space
adds r2,r3,r4 // there will always be spaces
addeq r0, r0, r1 // no IT blocks
adds.n r0, r0, r1 // no manual width specifiers
add r1, r2, r3, lsl #8 // no barrel shifting

This is , so the shortest answer in bytes per language wins.

Proposed tags: and maybe but I think that is for things you must write in assembly, not parsing assembly itself.


Hanabi playing bot


Hanabi is a cooperative card game with limited communication. It won the German "Spiel des Jahres" award in 2013.

The game can be played by 2-5 players, each of which has a hand of 5 cards (4 cards for 4-5 players), which they can't see themselves, but the other players can. In each turn, you can either play or discard a card, or give one other player a hint about their cards.

Common goal is to play out the cards in each color in ascending order.

As this is a cooperative game, each answer needs to cooperate with other instances of itself.

I/O format

Option 1: stateless, for all languages

Your program is called once for each turn (and is supposed to terminate afterwards). It will receive the current game state via standard input, and reply with an action via standard output. Invalid output means this game is counted as forfeited (lost, 0 points).


Input is a line-based ASCII-format, with a prefix indicating what kind of data it is. The last line of the input is the current turn number (see below). Input will be in this order:

  1. Meta-information: n: n = number of players (2 .. 5) y: you = own player id (1 .. n)

  2. visible cards of the other players:

    • c 1: cards for player 1
    • ...
    • c n: cards for player n

    The cards for you are omitted in the list (you don't see your own cards).

    Each card has a color (one of r, y, g, b, w) and a number (1 to 5), written like r1 or y5, comma-separated.

    Example: c 2: w4,y1,g2,r3 means that player 2 has a white 4, a yellow 1, a green 2 and a red 3, in this order, on her hand.

  3. Hints given about each player's cards, in a similar list:

    • h 1: hints for player 1
    • ...
    • h n: hints for player n

    For each card (in the same order as before), all hints are noted, in a comma-separated list. We also note negative information, i.e. when a card was present when a hint was given, but was not selected, by prefixing with !. For example 5 is a card which is known as 5 but of unknown color, w1 is a card which is known as a white 1, 3!y is a 3 which is known as not yellow, g!15 is a green card which is known as neither being a 1 nor 5.

    For example, h 2: ,y!2,2!yr,r means that for player 2, the first card is completely unknown (was taken after the last hint), the second card is known as yellow and as not a 2, the third card is known as a 2 which is neither red nor yellow, and the fourth card is known as red.

  4. p color: number – already (successfully) played out cards (one per line). Here we just note the highest card of each color.

    • p r: 2 means for red, the 1 and two were played out (i.e. red 3 is the next one to play).
    • p w: 0 means no white card was played out yet (i.e. the white 1 is the next one to play).
  5. d: – Discarded cards A list of all cards which were either intentionally discarded or unsuccessfully played out, comma-separated.

    For example, d: w3,b5,y2,y2 means that white 3, blue 5 and two yellow 2s were already discarded.

  6. game status information (each on one line, in this order): lh: number of hints left, ld: number of discards left (those two always sum to 8), lb: number of "bad plays" left lc: number of cards left in the deck t: current turn number

(The input will be closed here.)


The action to take. One of

  • h player-id color or number – give a hint to another player. E.g. h 4 y will give player 4 a hint which of his cards are yellow. h 3 1 will give player 3 a hint which of her cards are a 1. This action is only possible if the number of hints left is positive. (The number of hints left will be reduced by 1, the number of discards will be increased by 1.)
  • p card# – play one of your own cards (identified by its number (1..5)). (If this card fits into the cards already played, it's added there. (If this is the last 5, the game ends immediately with full score (25)). Otherwise it is discarded and the number of bad plays is reduced by 1. If it reaches 0, the game ends (unsuccessfully).)
  • d card# – discard one of your own cards (idenfified by its number (1..5)). This is only possible if the number of discards is positive. (The number of hints left will be increased by 1, the number of discards will reduced by 1).

After playing or discarding a card, this card is removed from the hand, and a new card is drawn from the deck and is added to the list of cards of this player (on the left). (The hints are automatically updated.)

Option 2 (stateful)

Your program receives a live transcript of everything what happens (including the actions of the other players, and the results thereof). The controller will read one line of output from the program when it's its turn.


Most input lines have the same format as before.

Initial input (as for the stateless version):

  1. Meta information (n: , y: )
  2. other player's cards (as before)
  3. hints for cards (as before)
  4. played cards (initially just p r: 0, p w: 0, etc.)
  5. discarded cards (initially just d: )
  6. game status, ending with t: 1.

After each player's turn:

  1. a line is given with that player's action: a player id: the action as defined in the output, e.g. a 1: h 4 y means that player 1 gave a hint to player 4 about yellow cards
  2. Those parts of the card situation which changed, e.g. c and d lines if a player discarded a card (or unsuccessfully played a card) and drew a new one), c and p lines if a player played out a card successfully, a h line if a player gave a hint.
  3. Updated status information, e.g. lh + ld when hint was given or a card discarded, lc when a new card was drawn, lb when a card was played unsuccessfully.
  4. t: indicating the next turn number.

*(TODO: Do we need an indication that's now your turn? That can be calculated by y == t mod n, but an explicit prompt might be easier to handle.

When the game ends, the input will be closed. (Your bot should terminate then.)


As in the stateless version, one line indicating the player's action.

Option 3/4 (JVM only, stateless or statefull)

To be defined. As I will be writing the controller in a JVM language, it should be possible/easy to provide a Java API to be implemented by the bots.

Other game rules:

I tried to give most of the details above, but here are some which might be missing/unclear:

  • There are three × 1, two × 2 to 4 and one 5 in each of the five colors (50 cards in total, 10 per color).

  • When playing with 2 or 3 players, each player has 5 cards, when playing with 4 or 5 players, each player has 4 cards in their hand.

  • If you have three bad plays (i.e. the lb counter reaches 0), you lose immediately. This is counted as score 0.

  • If you succeed to play all 25 cards (i.e. all 5s are played successfully), you win immediately, with a score of 25.

  • When the drawing deck is exhausted, one more round is played (i.e. each player has one more turn), then the game ends and the final score is the number of cards successfully played out.

Competition Rules

  • While for human play, the amount of extra communication is "subject to negotiation", here I want to explore what is possible with just what the rules provide. Any communication between your bot instances (except as provided by the defined interface, i.e. via game actions) is strictly forbidden. Strategy needs to be encoded in the source code, not discussed during the game.

  • There is also no communication between your individual games (i.e. no persistence).

  • I will nominate an overall winner, and one for the stateless category. (The stateful ones have a bit more information, so they could emulate the stateless ones.)

  • I will run contestants 1000 times with random decks of cards, once for each number of players from 2 to 5. If your bot only works with a specific number of players, state this in your answer. The competition score is the average score for all the runs. [I'll need to experiment to see how this varies, maybe I'll increase or decrease the count.]

  • A bot needs to provide output in a reasonable time (to be defined). The stateless version needs to terminate after providing output, the stateful one keeps running, but should terminate after end of input (i.e. after the game ended).

  • I will provide the controller on Github, feel free to test your bot with it (and compare it with other competition entries).

  • The programming language needs to have an interpreter or compiler which is available free of cost for Ubuntu 20.4 (otherwise I can't run your bot to score it).

  • I reserve the right to not run a bot when I suspect malicious code in it.

  • This is not , please keep your code readable.

  • \$\begingroup\$ This still needs to be refined, and I actually need to write (and test) the controller. I won't post it before that is done. \$\endgroup\$ Commented Jan 17, 2021 at 23:22
  • \$\begingroup\$ What tags should this get? code-competition? \$\endgroup\$ Commented Jan 17, 2021 at 23:23

Write a Length interpreter

Length is a simple stack-based esolang where instructions are encoded as line lengths The instruction set is as follows:

Line Length Name Description
9 inp Pushes the ascii value of the first byte of stdin to the stack.
10 add Adds the top two values on the stack and pushes the result onto the stack.
11 sub Subtracts the top two values on the stack and pushes the result onto the stack.
12 dup Duplicates the top value of the stack.
13 cond If the top value of the stack is 0, skip the next instruction. Then pop it.
14 gotou Sets the program counter to the value of the line under the instruction.
15 outn Pops the top of the stack, and outputs it as a number.
16 outa Pops the top of the stack, and outputs its ascii value.
20 mul Multiplies the top two values on the stack and pushes the result onto the stack.
21 div Divides the top two values on the stack and pushes the result onto the stack.
24 gotos Sets the program counter to the value at the top of the stack

In case the table doesn't work, here is the esolangs page: https://esolangs.org/wiki/Length
Test inputs are too long to put here, they can be found here
helloworld.len - Outputs Hello, World!
truth.len - A truth machine
bottles.len - Outputs the lyrics to 99 bottles of beer
This is a code golf, so shortest program wins!

  • \$\begingroup\$ table is broken, it looks fine in preview \$\endgroup\$
    – Nailuj29
    Commented Jan 18, 2021 at 20:49
  • \$\begingroup\$ Fixed. But this should be reported on SE Meta. \$\endgroup\$
    – Adám
    Commented Jan 18, 2021 at 20:56

Radiation Showdown (WIP)

Two radiation-hardened programs will go head-to-head to destroy each other.

Your task is to create a program which takes the other program's source as input and output the index of the byte that should be deleted from the other program. (zero-indexed)

Each program will be radiated at the same time. The first program to fail to return a valid index after being radiated loses (whether by compiler error, runtime exception, out-of-bounds output, or some other means), or it is considered a draw if both fail at the same time.

Each program will compete against each other program. The program receives 1 point per round survived. The overall winner is the one with the most points.

Programs are limited to a length of 1024 bytes.

Alternate possibilities:

(Inspired by @Dingus) A hash of the opponent's original source code and a list of the indexes of bytes deleted so far is passed in instead of the current source code, making it a bit more of a blind guess as to what you radiate. If at any time, a program makes a guess it has already made, it loses. This turns it into a sort of "Radiation Battleship"

  • \$\begingroup\$ Interesting challenge. How will it be tested, if different languages are allowed? \$\endgroup\$
    – user
    Commented Jan 19, 2021 at 19:12
  • \$\begingroup\$ @user it shouldn't be terribly difficult to make a shell-based controller. Submissions would probably need to include compiler flags or shebangs separately so that the controller knows exactly how to run the program. \$\endgroup\$
    – Beefster
    Commented Jan 19, 2021 at 19:19
  • \$\begingroup\$ That would work for "practical" languages, but I'm a bit worried about esolangs. I guess TIO can probably deal with that, though. \$\endgroup\$
    – user
    Commented Jan 19, 2021 at 19:20
  • \$\begingroup\$ If Program A has all its bytes deleted before Program B does, does B win? I'm imagining a pathological scenario in which A is empty and outputs by exit code. \$\endgroup\$
    – Dingus
    Commented Jan 19, 2021 at 23:46
  • 1
    \$\begingroup\$ Also, wouldn't something like this be impossible to beat? \$\endgroup\$
    – Dingus
    Commented Jan 20, 2021 at 0:12
  • \$\begingroup\$ @Dingus I suppose that would be impossible to beat. I wonder how I could make things different with resisting arbitrary insertions... probably not viable though. \$\endgroup\$
    – Beefster
    Commented Jan 20, 2021 at 16:14
  • \$\begingroup\$ To prevent fixed-output programs, could you perhaps tie the indexing to the unmodified code? In other words, force the output to be different every round? \$\endgroup\$
    – Dingus
    Commented Jan 20, 2021 at 22:22
  • \$\begingroup\$ @Dingus I think that might work (implying that both the original program and guesses are passed to the program), but there's probably another similar edge case I'm missing. Kinda turns it into radiation battleship. If a program returns a guess it has already made, it instantly loses. \$\endgroup\$
    – Beefster
    Commented Jan 20, 2021 at 23:58
  • \$\begingroup\$ Following that thought, it might be kind of interesting if some opaque id (e.g. a hash of the source code) is passed in instead of the current state of the source code. \$\endgroup\$
    – Beefster
    Commented Jan 21, 2021 at 0:02
  • 1
    \$\begingroup\$ @Beefster what's the point of the hash? (what can submissions do with it?) \$\endgroup\$ Commented Jan 21, 2021 at 4:37
  • \$\begingroup\$ @thedefault. the hash is only used to assign a distinct id to each program and it could be used to seed a random number generator. \$\endgroup\$
    – Beefster
    Commented Jan 21, 2021 at 16:16
  • 1
    \$\begingroup\$ @Beefster I thought submissions aren't supposed to be designed to beat specific opponents (and the only way to use a hash is to optimize to beat specific opponents) (and providing a hash of an unknown string as input just in case somebody wants to seed a RNG with it is a weird decision) \$\endgroup\$ Commented Jan 22, 2021 at 16:47
  • 1
    \$\begingroup\$ Also, if the length of the source code is not known, it's effectively impossible to make guesses that are guaranteed not to be out-of-bounds. (also, I expect 90% answers to this challenge to output something like 0,1,2,3,4,5,...) \$\endgroup\$ Commented Jan 22, 2021 at 16:58

Print a 3D shape


  • 4
    \$\begingroup\$ Suggestion: use characters other than and as they are multi-byte characters which not all languages will handle easily. ASCII characters instead would be better \$\endgroup\$ Commented Jan 20, 2021 at 21:12
  • \$\begingroup\$ I modified them, thank you. \$\endgroup\$
    – anotherOne
    Commented Jan 20, 2021 at 21:35
  • \$\begingroup\$ @caridCoinheringaahing I was just looking to try some other character taking as reference this ASCII table. The two characters that I changed are numbers 166 and 167 of the table. I am confused. \$\endgroup\$
    – anotherOne
    Commented Jan 21, 2021 at 23:45
  • 1
    \$\begingroup\$ Most languages can handle code points between 32 and 127 inclusive. Some use custom code pages (e.g. Jelly) that won't necessarily have extended ASCII characters, but will have regular ASCII \$\endgroup\$ Commented Jan 21, 2021 at 23:48
  • \$\begingroup\$ Oh all right thank you for the clarification! \$\endgroup\$
    – anotherOne
    Commented Jan 21, 2021 at 23:55
  • 2
    \$\begingroup\$ @Davide the table you linked is not actually an ASCII table. ASCII only goes from 0 to 127; there are more code pages that extend it to 256 and the creator of that table has clearly picked one. This confusion arises because almost all code pages include ASCII as the first 128 and then extend with some symbols like on top of that. Nowadays most things use UTF-8 which encodes some characters with more than one byte and can use millions of different characters from Unicode \$\endgroup\$
    – pxeger
    Commented Jan 22, 2021 at 7:58
  • \$\begingroup\$ @pxeger Oh thank you for this definitive explanation! \$\endgroup\$
    – anotherOne
    Commented Jan 22, 2021 at 13:02

Quine Countdown!

Write a program that accepts a single parameter n and outputs another program that outputs another program etc until the nth call outputs the original input n again.

Scoring is a modified version of codegolf: For input 100, add together the code length of each program in the chain, excluding the final 100. This is your score. Lowest score wins!

  • \$\begingroup\$ Actually it's not much harder than a typical quine program. Because 100 is so large, most programs will do it the general quine way. \$\endgroup\$
    Commented Jan 29, 2021 at 7:18
  • \$\begingroup\$ Besides, the score would be easier to read if the score is the average size instead of the maximum. \$\endgroup\$
    Commented Jan 29, 2021 at 7:18
  • \$\begingroup\$ "Because 100 is so large, most programs will do it the general quine way." That's the intention. It's a quine with something extra. Without the scoring rule, everyone would just build nested multi-escaped prints, which isn't as interesting. \$\endgroup\$
    – Sinthorion
    Commented Feb 9, 2021 at 11:07

Counting set bits in a byte

Here is a bite-sized problem I ran in to when trying to implement Conway's game of life on a microcontroller, and was trying to count the amount of neighbours: How can you check how many bits are set in a byte?


Given one byte of input data and an integer N between 0 and 8, check if there are exactly N bits set in the byte.

Test cases

N = 0, input = 0b00000001 -> False 
N = 0, input = 0b00000000 -> True 
N = 3, input = 0b01001001 -> True 
N = 3, input = 0b11100000 -> True 
N = 3, input = 0b00001111 -> False 
N = 7, input = 0b11101111 -> True 
N = 7, input = 0b11111110 -> True 
N = 8, input = 0b11111111 -> False 

Sandbox question

  • My first intuition to solving this problem was to shift the bits out one by one, AND with 0x01 and count them. I feel however it must be possible to do something more efficient in terms of CPU cycles used. How can I make a challenge that is about optimizing instruction count and memory usage rather than on program-size? I have seen the tag, but I don't know what scoring method best to use.

Edit: Closing after some good comments and possible solutions

Arnauld and CristoLosoph gave some great comments and led me to conclude that my issue is maybe to hardware/language specific to fit in a nice coding challenge. CristoLosoph showed me this interesting code snippet which I think is quite efficient:

uint8_t count_bits(uint8_t x){
     x = ((x & 0b10101010) >> 1) + (x & 0b01010101); 
     x = ((x & 0b11001100) >> 2) + (x & 0b00110011); 
     x = ((x & 0b11110000) >> 4) + (x & 0b00001111); 
     return x;

Two other things I learned:

  • Some hardware have a builtin POPCNT instruction in their instruction set, and gcc has a __builtin_popcount() method that does exactly what I was looking for
  • I found in this question that it's an interesting trade-off (as with most embedded functions probably) to just make a lookup table containing all 256 possible return values. It takes some memory but not too much.

I also learned that probably would have been a good fit for this type of challenge!

Once again thanks for the interesting comments!

  • 1
    \$\begingroup\$ Hello RvdV. Hm, seems like your motivation does not really fit. Fastest-code and smallest code is not the same, rather opposite things. When you program microcontrollers you are very restricted in the language but code golfing is the opposite (if you do not specifically restrict languages to one single one). Of course, there is a very performant solution for bitcount in the language which you might be looking for which however is quite uninteresting for code golf. I'll post you a fast snippet in C below. \$\endgroup\$ Commented Jan 29, 2021 at 10:20
  • 1
    \$\begingroup\$ Closely related. \$\endgroup\$
    – Arnauld
    Commented Jan 29, 2021 at 10:25
  • 1
    \$\begingroup\$ You may consider using atomic-code-golf, but be aware that it's pretty hard to specify correctly. \$\endgroup\$
    – Arnauld
    Commented Jan 29, 2021 at 10:26
  • 1
    \$\begingroup\$ x = ((x & 0xAAAAAAAA) >> 1) + (x & 0x55555555); x = ((x & 0xCCCCCCCC) >> 2) + (x & 0x33333333); x = ((x & 0xF0F0F0F0) >> 4) + (x & 0x0F0F0F0F); x = ((x & 0xFF00FF00) >> 8) + (x & 0x00FF00FF); x = ((x & 0xFFFF0000) >> 16) + (x & 0x0000FFFF); You only need the first 3 lines if you only need it for 8-bit integers and then you can cut the hexadecimal literals down to two digits. \$\endgroup\$ Commented Jan 29, 2021 at 10:29
  • 1
    \$\begingroup\$ But if you work with GCC, please use __builtin_popcount() since that is the function that the compiler people tried to optimize exactly for your purpose and for your selected architecture. \$\endgroup\$ Commented Jan 29, 2021 at 10:31
  • \$\begingroup\$ Thanks a lot for your comments Arnauld and ChrisoLosoph! You are right that it's hard to define, and I didn't realize even how much language but also hardware-bound my problem was, maybe it's too difficult to define indeed (and if you define it language-agnostic it becomes the related post Arnauld linked). I will update the post with you suggestions! \$\endgroup\$
    – RvdV
    Commented Jan 29, 2021 at 17:51

Mix my colors

This challenge is inspired by the Color Alchemy Patch on NetHack, notably incorporated by UnNetHack 3.5.2.


Given two strings indicating colors, mix them according to the rules below, then output it.


There are 16 colors in total that are valid inputs. They are categorized to 8 chromaticities and 2 brightnesses, like below:

        Light    Dark
Hueless white    black
Red     pink     ruby
Blue    sky-blue indigo
Yellow  yellow   golden
Orange  orange   amber
Green   emerald  dark-green
Purple  puce     magenta
Brown   ochre    brown

(I'm omitting gray, for it will make this challenge just cumbersome in this regard.)

Note that all color names are in lowercase.

Color-mixing rules

  • Above all, mixing colors is idempotent.

  • Mixing colors is commutative unless noted below.

  • When mixing two primary colors (Red, Blue, or Yellow):

    • If they have same chromaticities, the result will also be in the same chromaticity.

    • Mixing Red and Blue results in Purple.

    • Mixing Red and Yellow results in Orange.

    • Mixing Blue and Yellow results in Green.

  • When mixing two secondary colors (Orange, Green, or Purple):

    • If they have same chromaticities, the result will also be in the same chromaticity.

    • All other combinations result in Brown.

  • For all cases covered by above, mixing two Light (resp. Dark) colors will result in corresponding Light (resp. Dark) color.

  • For all cases covered by above, mixing a Light color and a Dark color shall result in either Light or Dark. This is the only rule that may break the commutativity.

  • Mixing a Light color with black results in corresponding Dark.

  • Mixing a Dark color with white results in corresponding Light.

  • All combinations not covered by above fall in don't care situation.


Valid outputs

  • Mixing white and white results in white.

  • Mixing pink and pink results in pink.

  • Mixing pink and ruby results in either pink or ruby.

  • Mixing ruby and golden results in amber.

  • Mixing sky-blue and ruby results in either puce or magenta.

  • Mixing emerald and dark-green results in either emerald or dark-green.

  • Mixing emerald and orange results in ochre.

  • Mixing puce and amber results in either ochre or brown.

  • Mixing white and ruby results in pink.

  • Mixing ochre and black results in brown.

Don't care situations

  • Mixing pink and white falls in don't care situation. There is no rule that covers this case.

  • Mixing ochre and brown falls in don't care situation. There is no analogous rule for tertiary colors.

  • Mixing indigo and magenta falls in don't care situation. There is no rule for mixing a primary color and a secondary color.

  • Mixing indigo and orange falls in don't care situation.

  • Mixing pink and ochre falls in don't care situation.

  • Mixing white and black falls in don't care situation. (In the game, it results in gray, but I'll ignore this, for sake of simplicity of this challenge.)

Rules for code golf

  • Input format is flexible. In particular, it can be two strings, or one string separating colors by whitespaces. It's implementation-defined whether to accept leading or trailing whitespaces.

  • Output format is also flexible. Outputting leading or trailing whitespaces is okay.

  • Invalid inputs fall in don't care situation.

  • \$\begingroup\$ The rule looks clear enough, but you can provide a table of 256 possible outputs just to be extra sure. \$\endgroup\$
    Commented Jan 31, 2021 at 1:57

Longest Common Suffix

Given arbitrarily many (more than 4) words, your goal is to find the longest common suffix of all of them.


  1. The suffix shouldn't be longer than one third of the length of any word, and should be longer than 2 characters.
  2. The words can have at most one "exception" among them, that is, it doesn't have the same suffix others have. You should ignore these "exceptions".
  3. If no suffix can satisfy all rules above, do not make any output.
  4. All given words are separated by spaces; they only contain lower case English alphabets.
  5. You can use any way to accept input, but output should only be in STDOUT.


Given: television operation delegation repetition
Output: ion
Given: vision decision subtraction observation

Why? The suffix, ion, is longer than 1/3 the length of vision and decision. There're two exceptions.

Given: interested congratulated excited overjoyed
Output: ted

Why? overjoyed is an exception, because others have the suffix ted, but it doens't. So we ignore the word overjoyed.

Given: abcdefghijkl bcdefghijkl cdefghijkl defghijkl
Output: jkl

Why? defghijkl is the longest, but it is longer than 1/3 the length of defghijkl.

This is code golf, so the shortest code win.

It's not recommended to use built-in functions which directly returns the result.

  • \$\begingroup\$ Most answerers prefer "any (standard) way to get output" as well, but it's up to you. Practically, it's mostly not a problem. \$\endgroup\$
    Commented Feb 1, 2021 at 9:30
  • \$\begingroup\$ Regarding the output "No common...", people don't like hard coding long error messages, it would be better received if it's changed to "any value that signifies that there isn't...". \$\endgroup\$
    Commented Feb 1, 2021 at 9:31

Compile Roman Numerals to <some language>

Your code will, given an integer i, output code code(i) in any programming language that will evaluate to (print, push, return) i by itself. code(i) ++ code(i') should evaluate to i + i' for an i' <= i and i' - i for an i' > i. You do not have to handle cases like code(1) ++ code(2) ++ code(3) where there are more than one number less than 3 before 3, but you should handle code(5) ++ code(3) ++ code(4) => 6 with more than two numbers.

Clarification: the generating code takes a number, not a roman numeral, but the generated code is expected to have the behaviour of a roman numeral when concatenated with other outputs.


if  yourcode(10) -> 'X'
and yourcode(5)  -> 'V'
then eval('XV')  -> 15.

if  yourcode(1)  -> 'I'
and yourcode(10) -> 'X'
and yourcode(5)  -> 'V'
then eval('XIV') -> 14

if  yourcode(7)  -> 'a'
and yourcode(3)  -> 'b'
then eval('ab')  -> 10
and  eval('ba')  -> 4


Your answer is scored by the total bytes of the generating code.

  • \$\begingroup\$ Sandbox: how do I explain this? what tags does this get as a challenge that requires an answer that generates code? \$\endgroup\$
    – Wezl
    Commented Feb 1, 2021 at 14:48
  • \$\begingroup\$ Any upper limit? \$\endgroup\$
    – Adám
    Commented Feb 1, 2021 at 15:21
  • \$\begingroup\$ @Adám no, besides any restrictions on integers built-in to the language. \$\endgroup\$
    – Wezl
    Commented Feb 1, 2021 at 15:25
  • 1
    \$\begingroup\$ This sounds very difficult. The code snippets would need to carry a state. \$\endgroup\$
    – Adám
    Commented Feb 1, 2021 at 15:36
  • \$\begingroup\$ Can the code take something like 7 (not representable as a single character in Roman numeral) as input? \$\endgroup\$
    Commented Feb 1, 2021 at 15:43
  • \$\begingroup\$ @user202729 yes, they would need to handle other numbers \$\endgroup\$
    – Wezl
    Commented Feb 1, 2021 at 15:44
  • \$\begingroup\$ What should something like eval(code(7) + code(5)) evaluate to? \$\endgroup\$
    Commented Feb 1, 2021 at 15:46
  • \$\begingroup\$ @user202729 13 because 7 > 5. \$\endgroup\$
    – Wezl
    Commented Feb 1, 2021 at 15:47
  • \$\begingroup\$ How do you define "Roman numeral" then? (when digits can take values other than the standard values) \$\endgroup\$
    Commented Feb 1, 2021 at 15:48
  • \$\begingroup\$ @user202729 OK, I've now defined how any integers should be handled when next to eachother. \$\endgroup\$
    – Wezl
    Commented Feb 1, 2021 at 16:07
  • 1
    \$\begingroup\$ So, given a sequence \$p_1,\ldots p_n\$ where there's no \$p_i < p_{i+1} < p_{i+2}\$, the result of \$\operatorname{exec}(\operatorname{generate}(p_1) + \ldots + \operatorname{generate}(p_n))\$ (where \$+\$ denotes concatenation) should be \$p_n + \sum _{i=1} ^{n-1} p_i \times (-1)^{[p_i < p_{i+1}]}\$, right? \$\endgroup\$
    Commented Feb 1, 2021 at 16:30
  • \$\begingroup\$ @user202729 I guess so, although I'm not good at reading math :P. \$\endgroup\$
    – Wezl
    Commented Feb 1, 2021 at 16:35

Next to the middle


  • 1
    \$\begingroup\$ Clearly defined enough (not very interesting) -- except that is it guaranteed that the value exists? \$\endgroup\$
    Commented Feb 1, 2021 at 3:14
  • \$\begingroup\$ @user202729 Oh thank you, I forgot to cover this. I think that in case it doesn't exist, we can output a something like the smallest integer in the array. I would like to not restrict the input at all. \$\endgroup\$
    – anotherOne
    Commented Feb 1, 2021 at 12:03
  • 1
    \$\begingroup\$ Possible test case: empty array. (by the way don't patch "edit" into the sandbox post, include it in the text itself) \$\endgroup\$
    Commented Feb 2, 2021 at 2:58
  • \$\begingroup\$ @user202729 thank you, I edited the challenge. I will add test cases, including the empty array. \$\endgroup\$
    – anotherOne
    Commented Feb 2, 2021 at 10:26

Golf this Thumb-2 constant!


  • \$\begingroup\$ Not very related to the question, but have you tried using some size-optimization option of some existing compiler? (-Os) \$\endgroup\$
    Commented Feb 1, 2021 at 9:32
  • \$\begingroup\$ Otherwise, clear enough. \$\endgroup\$
    Commented Feb 1, 2021 at 9:33
  • \$\begingroup\$ I explained the pattern better, and the last pattern does cover the rest, but it isn't the shortest. \$\endgroup\$
    – EasyasPi
    Commented Feb 4, 2021 at 0:43
  • \$\begingroup\$ Okay. (just to check, there are \$4862 > 2^{12} = 4096\$ distinct values represented with the Imm12 format, right?) \$\endgroup\$
    Commented Feb 4, 2021 at 0:53
  • \$\begingroup\$ Yes. There are 4096 possible encodings for the Imm12 format. \$\endgroup\$
    – EasyasPi
    Commented Feb 4, 2021 at 1:48
  • \$\begingroup\$ Then I must understand something wrong, because (as I've said above) there are 4862 different values that can be encoded.) \$\endgroup\$
    Commented Feb 4, 2021 at 1:51
  • \$\begingroup\$ Oh yeah I was wrong. I think I am going to go back to it just being a byte rotated, even if it isn't accurate. As the true encoding is too complex to be fun to calculate. 🤷‍♂️ \$\endgroup\$
    – EasyasPi
    Commented Feb 4, 2021 at 3:02

(this is the core of the BF memory optimizer challenge.)

(At the moment I still need to make some test cases; however you can still review the rest of the challenge.)

note: This problem is reducible from Simple Max Cut, therefore it's NP-complete. (https://doi.org/10.1016/0304-3975(76)90059-1)

note: While I did get a bunch of test cases from this site, I'm not sure how can I write a reasonable algorithm to compete with...


(actually this is not part of the sandbox challenge, but I'll post it here because it's related)

First, for convenience, assume that the problem is represented by a undirected graph, where the number of rows/columns of the matrix is equal to the number of nodes, and the corresponding weight is the sum of the value of the edges connecting the corresponding two nodes.

With that representation, the value to be minimized is the sum of the product of the edge lengths and the edge weights, with the graph nodes embedded into the point \$ 1, 2,\ldots, |V| \$.

From a Simple Max Cut problem of the form:

Given \$ n \$ variables \$ x_1, x_2,\ldots, x_n \$, maximize the value of \$ \sum_{i=1}^m [a_i \ne b_i] \$, where each of \$ a_i, b_i \$ represents either a variable or its negation.

It can be transformed to an instance of this problem:

First, construct \$ 2n \$ nodes on a graph, denoted \$ p_1, p_2,\ldots, p_n, q_1, q_2,\ldots, q_n \$. Let \$ a \$ be some positive integer. Connect those vertices:

  • \$ p_1 \$ and \$ q_1 \$, with cost \$ 2n^2 a \$,
  • the 4 pairs of vertices \$(p_1, p_i),(p_1, q_i),(q_1, p_i),(q_1, q_i)\$ with cost \$ (n+1-i) a \$, for each \$ i=2, 3,\ldots, n \$,
  • and some other edges with small weights (the sum of their weights should be less than \$a\$ -- (1)) that mainly does not affect the optimal configuration.

The sum of the edge weights (except the first one) is \$ 4 ((n-1)+(n-2)+\ldots+1)a=2n(n-1)a \$.

The edge between \$ p_1 \$ and \$ q_1 \$ has a weight larger than the sum of all the others, it's obvious to see that in the optimal (minimum cost) configuration, these two must be adjacent.

Then, regardless where those 2 vertices are placed, the \$ i \$'th smallest distance-pair to those are at least the \$ i \$'th value in the sequence are

$$(1, 2),(1, 2),(2, 3),(2, 3),(3, 4),(3, 4),\ldots,(n-1, n),(n-1, n)$$


And by the rearrangement inequality, it's optimal to place the weight so that the vertices with the smaller edge-weight to \$ p_1, q_1 \$ are placed further from the vertices. Therefore the only optimal placement is

$$z_n, z_{n-1}, \ldots, z_2, z_1, z_1, z_2, \ldots, z_{n-1}, z_n$$

where each \$ z_i \$ is either \$ p_i \$ or \$ q_i \$ (\$1 \le i \le n\$).

Encode the condition "vertex \$ i \$ is on the left side of the cut" by "\$ p_i \$ is to the left of \$ q_i \$ in the permutation". (2)

Assuming that the edge weights are allowed to be fractional.

For each condition (in the Simple Max Cut problem) that "there's an edge between vertices \$u\$ and \$v\$ (\$ u\le v \$)", add an edge between \$ p_u \$ and \$ q_v \$ with weight \$\frac 1 {2u-1}\$ to this problem.

The weight of this edge can either be \$ u-v \$ or \$ u-v+(2u-1)\$ in the configuration that minimizes the total weight of the edges constructed in the previous section.

Therefore, if the vertices \$ u \$ and \$ v \$ are on different sides of the cut (according to the encoding (2)), the total weight is decreased by \$ 1 \$ if \$ u \$ and \$ v \$ are on different sides; and the configuration with the minimum weight is exactly the one with maximum number of edges cut.

However, in the actual problem edge weights must be an integer. We replace each edge weight \$\frac 1 {2u-1} \$ by \$\lceil\frac c {2u-1}\rceil \$, where \$ c=2nm\$.

Because there are \$ m \$ edges in total, if the sum of any \$ k-1 \$ increment values \$\lceil\frac c {2u-1}\rceil (2u-1) \$ is strictly less than the sum of any \$ k \$ increment values, for \$1\le k\le m \$, then the optimal sum is also the maximum cut.

Observe that because \$ c=2nm \$ and \$ x\le 2n-1 \$, each increment value must be between \$ 2nm \$ (inclusive) and \$ 2nm+2n-1 \$ (exclusive). Therefore the maximum sum of \$ k-1 \$ values is \$ (2nm+2n-2)(k-1) \$, which is less than the minimum sum of \$ k \$ values \$ 2nmk \$ when \$ 1\le k\le m \$.

The sum of all those does not exceed \$ \lceil \frac {2nm}{1} \rceil m \$. Therefore if \$ a \$ is chosen to be \$ 2nm^2+1 \$, then the condition (1) is satisfied.

Minimum cost matrix permutation

(or ? Obviously the latter would be more useful in practice code golf)

Given a matrix \$w\$ in \${\mathbb N_0}^{n\times n}\$, define the symmetric matrix \$d\$ in \${\mathbb N_0}^{n\times n}\$ by the formula \$d_{i,j}=\left| i-j \right|\$, find a permutation matrix \$P\$ such that the sum of elements in the matrix \$(P^{\mathsf T} \cdot w \cdot P ) \,\odot\,d\$ (where \$\odot\$ denotes the Hadamard product/element-wise product) is smallest.

The result will be the mean (TODO: median? mean of the 50% maximum? mean result/naive ratio?) of the score over these test cases, for as long as you can run your program.


(WIP) Settling the Lands of Codegolfia

The challenge controller will randomly generate a 200x200 map representing the terrain of the land

Your task is to write an AI whose goal is to have the largest population after 500 turns.


Each player begins with 1 cell claimed and a population of 100.


On each turn, you have the opportunity to claim land cells. You can claim 1 cell per turn, plus 1 per 1000 population. Claimed cells must be orthogonally adjacent to a cell you already own. You cannot claim cells belonging to other players.

Turns happen simultaneously. In the event that two players attempt to claim the same cell, both players will lose 10 people from their population and neither player will claim the cell.

At the end of your turn, your population grows by 10% (rounded up), up to the maximum size your colony can support.

Population Support

Without any land claimed, your colony can support up to 150 people.

Supporting larger populations requires claiming land. Each terrain type increases the amount by some amount. Combinations of terrain expand this further.


There are 4 terrain types:

  • Plains
    • Supports 20 people by itself
    • Supports 5 per adjacent owned plains
  • Forest
    • Supports 10 people by itself
    • Supports 5 per adjacent owned forest
    • Supports 50 additional people if there are at least 10 owned plains cells within 10 cells (Manhattan distance)
  • Mountains
    • Supports 5 by itself
    • Supports 5 per owned plains within 15 cells
    • Supports 10 per owned forest within 5 cells
  • Water
    • Each plains cell can support 500 additional people if it is within 5 cells of an owned water cell.
    • Each forest cell can support 200 additional people if it is within 10 cells of an owned water cell.

Alternative ideas for terrain

  • Plains support raw population. Same as above
  • Forest cells increase population support for all plains within 5 cells by a factor of 5% (stacks multiplicatively)
  • For every water cell and 10 plains cells, reproduction rate increases by 1%
  • Each mountain cell increases the range of each owned mountain and forest within 3 cells by 1. (stacks additively)
  • \$\begingroup\$ I was just thinking of something like this! (but different enough I can still do mine :p). Currently I don't see too many options for strategy other than comparing each orthagonal square each turn and picking the best. \$\endgroup\$ Commented Feb 11, 2021 at 19:38
  • \$\begingroup\$ @RedwolfPrograms yeah. Maybe I should make each of the terrains more mechanically different. Maybe one increases reproduction rate, one expands synergy range, and another enhances the others. \$\endgroup\$
    – Beefster
    Commented Feb 11, 2021 at 20:24
  • \$\begingroup\$ That sounds like a good idea, depending on how it's implemented. Maybe being close to water would let you claim extra land per turn near the body of water, sort of like exploration? \$\endgroup\$ Commented Feb 11, 2021 at 20:31

Score an approximation code challenge

Given a list of inputs (strings) and their expected outputs (integers or floats)[1], and a black-box program[2], calculate the score using the following scoring system:

Let \$R_n\$ be the expected output of the \$n^{th}\$ input, let \$A_n\$ be the actual output given by the black-box program, and let \$ j \$ be the total number of input/output pairs. (All values are positive and non-zero)

Then the score is defined as: $$S=\left\lceil L\times\max_{1\le i \le j}\left({\max\left(\frac{A_i}{R_i},\frac{R_i}{A_i}\right)^2}\right)\right\rceil$$ where \$L\$ is the length of the black-box program in bytes.


If the size of the program is \$100\$ bytes and the worst approximation is on the input "moon", where the program outputs \$1000\$ instead of the expected \$1737\$, then the score would be:

$$S=\left\lceil 100\times{\left(\frac{1737}{1000}\right)^2}\right\rceil=302$$

This system is taken shamelessly from this challenge by @Arnauld. Here is his reference implementation.

[1]: You may take the inputs and outputs either zipped ([(in1, out1), (in2, out2), ...]), or not zipped (([in1, in2, ...], [out1, out2, ...]), with both lists of the same length), at your option.

[2]: You may take the black-box program as either:

  • a black-box function, and its length in bytes, as two separate inputs
  • a string of code to be evaluated, as one input (the length will not be given separately unless necessary)


  • \$\begingroup\$ (1) is it a problem if the program only support integers? (2) is it guaranteed that both values (actual and submission output) are strictly positive? The formula is not well defined when either value is zero. \$\endgroup\$
    Commented Feb 15, 2021 at 13:17
  • \$\begingroup\$ @user202729 all values will be positive; the idea was that you can support integers or floats, whichever is easier \$\endgroup\$
    – pxeger
    Commented Feb 15, 2021 at 13:23
  • \$\begingroup\$ The "whichever is easier" part is currently not mentioned in the challenge. \$\endgroup\$
    Commented Feb 17, 2021 at 10:36
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