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

Sandbox FAQ


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

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

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


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

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

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

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

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


Search the sandbox / Browse your pending proposals

The sandbox works best if you sort posts by active.

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

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

4706 Answers 4706

98 99
101 102

Fizzbuzz for tensorflow

Since I was a new and naive user, I posted this question. In one day it got over 25 upvotes but was shut down for being too broad. Clearly there is community interest -- so by the suggestion of @DrGreenEggsandHamDJ I'll try it again in the sandbox. I'd really like to see some of the answers here, so I appreciate any help you can give turning this question into a proper submission.

Original text copied below:

Inspired by the job-interview with Joel Grus, the goal of this challenge is to write a tensorflow (or other deep/machine learning) program that learns Fizzbuzz and correctly prints out the answers to the positive integers less than 1000.

You can assume there are files named train.csv and test.csv and each contain a sorted list of sequential integers and the fizzbuzz answer:

 100, buzz
 101, 101
 102, fizz
 103, 103
 104, 104
 105, buzz
 150000, fizzbuzz

test.csv spans 1-1000 and train.csv spans 1001-150000.


  1. You must not hard-code the rules to Fizzbuzz anywhere in your program. The output must be from a machine learned representation that is learned while running the code.
  2. You must utilize train.csv in the training set and check your output against test.csv. You cannot use test.csv during training.
  3. You must get all outputs correct from test.csv (but as is case with deep-learning, we'll allow your code to fail this rule no more than 5% of the time).
  4. You may use any language and any external module (eg. python/tensorflow) as long they explicitly perform some kind of learning. Reference both the language and the module in the title of your post.
  5. This is a popularity contest, so the submission with the most votes after one week wins.
  • \$\begingroup\$ What are the odds this will work well without an obscene number of iterations? A week doesn't seem like much time to test more than a single method or two. \$\endgroup\$
    – Geobits
    Commented May 25, 2016 at 2:58
  • \$\begingroup\$ @geobits I'm not sure, but I can train a decent random forest model on Netflix predictions or a CNN to recognize basic images in about in hour. I figured the community would enjoy the challenge of a basic programming task in an unconventional manner. Surely fizzbuzz can't be that hard... \$\endgroup\$
    – Hooked
    Commented May 25, 2016 at 3:05
  • \$\begingroup\$ Well, one problem that comes to mind immediately is that you need it to output valid code at all. So you're going to need to compile/run/whatever for every slight variation in the code. So then you need to figure out what building blocks you can give it to start with. A list of keywords to try, just random ascii, somewhere in between? That part in particular seems underspecified, but could make or break the odds of it working imo. \$\endgroup\$
    – Geobits
    Commented May 25, 2016 at 3:08
  • 1
    \$\begingroup\$ I foresee arguments over where exactly the line falls for "hard-cod[ing] the rules to Fizzbuzz". \$\endgroup\$ Commented May 25, 2016 at 11:45
  • \$\begingroup\$ Hmmm....if you come up with arbitrary rules of fizzbuzz (like different numbers, different amounts of numbers), that might work to prevent hardcoding. \$\endgroup\$ Commented May 25, 2016 at 14:24

What can I build?

The Rules

Today I have decided to make geometric shapes out of toothpicks and gumdrops! However, I have a limited supply, so you have to figure out what I can build. I will give you an input in the format m n, where m is toothpicks (edges) and n is gumdrops (vertices.) Your output should be, in any output format of your choice, all 3D geometric shapes such that the amount of edges=m and the amount of vertices=n. The list of 3D geometric shapes you will use is this: Gist

  • You may NOT access the Internet.
  • No builtins relating to geometry or solids
  • This is code golf, so shortest code wins.
  • For no solids and invalid input, output nothing


In: 3 2 Out: <empty>
In: CodeGolf123 Out: <empty>
In: 12 6 Out (Bonus): {regular tetrahedron,unit equilateral square pyramid,unit equilateral triangular dipyramid,unit equilateral triangular prism,unit equilateral pentagonal pyramid,regular octahedron} Out (Regular): {regular octahedron} `

  • \$\begingroup\$ The first 3 rules can be deleted, as they're ppcg defaults \$\endgroup\$
    – Bálint
    Commented May 22, 2016 at 20:03
  • \$\begingroup\$ Not the first... \$\endgroup\$
    – user46167
    Commented May 22, 2016 at 20:04
  • \$\begingroup\$ I think this would work better with 1 solid with exactly n and m edges and sides. \$\endgroup\$
    – Riker
    Commented May 22, 2016 at 20:29
  • \$\begingroup\$ How about doing that, but having this as a bonus? \$\endgroup\$
    – user46167
    Commented May 22, 2016 at 20:37
  • 4
    \$\begingroup\$ 1. KISS. Ditch the stuff about the external file: the overhead to load it and the overhead generated by requiring it to be UTF-8 mean that no-one would want to use it anyway. 2. "No builtins" literally bans people from using any language. Specify what built-ins are banned. 3. Make the data available in a usable format: i.e. a text file hosted on gist.github.com or pastebin. 4. The bonus is a no-brainer: a 10% saving for changing two == to <=. Either make it compulsory or remove it entirely, because as it stands it's just complication. \$\endgroup\$ Commented May 22, 2016 at 22:38
  • \$\begingroup\$ @petertaylor: I am currently on a phone so I can't put a gist... \$\endgroup\$
    – user46167
    Commented May 23, 2016 at 10:42
  • 2
    \$\begingroup\$ There are a lot of shapes in that gist. Some of them are specified in different formats. If they're all to be supported it would be nice to have a standard format to represent them. Also some of these shapes specify different edge lengths. Will the challenge assume all edges are length 1 or will you potentially need to break toothpicks and keep track of the remaining pieces? \$\endgroup\$
    – Poke
    Commented May 26, 2016 at 21:02
  • 1
    \$\begingroup\$ @Poke 1. Will fix 2. Irrelevant-- you have all sorts of toothpicks, some miniscule and others huge. \$\endgroup\$
    – user46167
    Commented May 26, 2016 at 23:43

Code me music


Write a program that will play music based on input.


When your program is run it will be given a small song. Each note in the song will have 3 components:


where octave is the octave for the note to be played in, pitch is the key of the note (a b c d e f g), and duration is the length of the note in milliseconds. For example, 4c1000 would be middle c played for one second. Notes in a song are separated by spaces. Flats and sharps are possible, and they go after the note like this: 5gb1000 (5th octave g flat) for flats and 5g#1000 (5th octave g sharp) for sharps.


Your program must produce sound based on the input. If one of the notes in the input is 3f500, your program must play f in the third octave for a half of a second.

Other notes

  • This is code golf, so shortest program in (insert period of time) wins.

  • No functions, only full programs.

  • The sound can be whatever you please.

  • Here are the frequencies of notes in the 4th octave in hertz:

    • 4c - 261.63
    • 4c#/4db - 277.18
    • 4d - 293.66
    • 4d#/4eb - 311.13
    • 4e - 329.63
    • 4f - 349.23
    • 4f#/4gb - 369.99
    • 4g - 392.00
    • 4g#/4ab - 415.30
    • 4a - 440.00
    • 4a#/4bb - 466.16
    • 4b - 493.88
  • A list of all frequencies is here.


  • Is the challenge objective clear?
  • This challenge may be hard for some languages, is that a problem?
  • Is this already a challenge?
  • Any positive feedback is welcome.
  • \$\begingroup\$ I think this could be improved by explaining what the frequencies are for the pitches at a given octave, and then explaining how an octave relates to that. That information is necessary to answer in any language which doesn't handle that itself, so I think it warrants being in the post rather than being behind a (potentially stale) link. In addition, you probably need to have some kind of leniency about frequency and duration, machines are not perfect after all. \$\endgroup\$ Commented May 27, 2016 at 14:35
  • \$\begingroup\$ @FryAmTheEggman Thank you for your feedback, updated challenge. \$\endgroup\$
    – lapras
    Commented May 27, 2016 at 14:54
  • \$\begingroup\$ Yes, this is already a challenge. (And, curiously, the second sandbox proposal which is a variant on that challenge in just a week). \$\endgroup\$ Commented May 28, 2016 at 10:55

Plan a special tournament

tags: [more tags required]

I host a special tournament with any number n >= 2 of participants.

Here is a list of plans of tournaments for n = 2 to 20:

 2: 1) DE-2c-1w
 3: 1) RR-3c-1w
 4: 1) DE-4c-1w
 5: 1) RR-5c-2w 2) DE-2c-1w
 6: 1) RR-6c-3w 2) RR-3c-1w
 7: 1) RR-7c-3w 2) RR-3c-1w
 8: 1) DE-8c-1w
 9: 1) RR-9c-4w 2) DE-4c-1w
10: 1) RR-10c-5w 2) RR-5c-2w 3) DE-2c-1w
11: 1) RR-11c-5w 2) RR-5c-2w 3) DE-2c-1w
12: 1) RR-12c-6w 2) RR-6c-3w 3) RR-3c-1w
13: 1) RR-13c-6w 2) RR-6c-3w 3) RR-3c-1w
14: 1) RR-14c-7w 2) RR-7c-3w 3) RR-3c-1w
15: 1) RR-15c-7w 2) RR-7c-3w 3) RR-3c-1w
16: 1) DE-16c-1w
17: 1) RR-17c-8w 2) DE-8c-1w
18: 1) RR-18c-9w 2) RR-9c-4w 3) DE-4c-1w
19: 1) RR-19c-9w 2) RR-9c-4w 3) DE-4c-1w
20: 1) RR-20c-10w 2) RR-10c-5w 3) RR-5c-2w 4) DE-2c-1w

Explanation of the plan

  • The entire tournament enters the first round, which has c = n participants.
  • For each round:
    • If c is a power of 2, then this round will be a double elimination round, with 1 winner. After this round, the tournament ends.
    • Else, this round will be a round robin round, with floor(c/2) winners that continue to the next round.
      • If only one winner continues, the tournament ends.
      • Else, let c be the number of winners, and start again from "For each round".

The Challenge

Given n, return a plan of the special tournament with n participants.

This is a , so shortest code wins.

TODO: Reword the explanation clearly, write more content.

  • \$\begingroup\$ How does a round robin with five people work, where you have two winners? Most round robins would have a bye, so you'd end up with three winners continuing on. (same for any odd number) \$\endgroup\$
    – Geobits
    Commented Jun 4, 2016 at 19:02

Golf a 2d maze (Yes, a whole game)

Nowadays, I am interested in making games. especially mazes. In this golf round of code, You should make a maze game. You can do it with just preset mazes, but, if you make it randomly generate, I will be a-maze-d and will give you bonus points.(Huh. is that a pun?) So, Go on, Why don't you try right now?


  • Input W(Up), A(Left), S(Down), D(Right) until the player gets to the finish.
  • Display the maze and the player each input.
  • Move the player Up if the input is 'W', Down if 'S', Left if 'A', and Right if 'D'.
  • The Character for the wall and the player is undefined. you choose.
  • The Character for the wall, the player and the end square should be all different.


  • You should get the Width and height in the input.
  • Not Necessary, but you can use Prim or Kruskal.
  • Also you can use the method mentioned in here.
  • 1
    \$\begingroup\$ The spec as is probably has too many ambiguities, e.g. 1) What counts as a "maze"? 2) If I was to golf this question right now, I'd put the exit next to the player and make it so that the only valid move is into the exit. That would save a ton of bytes since I only have to check one input. 3) If random generation is optional, it's almost certain that it won't be done (but if you do make it mandatory, then you would need to specify what random generation means) 4) Do we have to handle invalid input from the user? \$\endgroup\$
    – Sp3000
    Commented Jun 5, 2016 at 6:50
  • \$\begingroup\$ In order for this to fit as a challenge on this site, it needs to have a winning criterion. For example, code-golf (shortest code wins), or fastest-code. You can also use code-challenge if you define a score based on something else, but there must be some way of assigning a score to each solution so they can be put in order and encourage competition. \$\endgroup\$ Commented Jun 6, 2016 at 13:52
  • \$\begingroup\$ I think this could be an interesting challenge, but it needs to be well defined before it will be ready. \$\endgroup\$ Commented Jun 6, 2016 at 13:57
  • \$\begingroup\$ It is code-golf. \$\endgroup\$
    – user54200
    Commented Jun 6, 2016 at 14:01

Light black box: generate mirror processors



The list of black box output values for all black box input values from 0 to 255 (inclusive), in any reasonable format. Each value is also in the range 0 to 255.


A text representation of a rectangular grid showing the locations and orientations of mirrors that will generate the appropriate black box output for each black box input.


  • The input bits arrive from the left of the grid, in the top 8 squares, initially travelling right.
  • The output bits depart from the right of the grid, in the top 8 squares, travelling right.
  • For both black box input and black box output, the most significant bit is at the top.
  • Bits move through the grid horizontally or vertically until they encounter a mirror or the edge of the grid.
  • Bits that leave the edge of the grid are lost.
  • A bit that would move onto a mirror at the next step instead changes direction by 90 degrees clockwise or anticlockwise according to the mirror type, and takes a step in that direction (a bit never stays on the same square).
  • If two or more bits arrive on the same square at the same timestep, all of them annihilate. Two bits can still pass through each other if this does not involve sharing a square (traveling in opposite directions, being adjacent facing towards each other on one step, and adjacent facing away from each other on the next).

Note that because a bit changes direction just before reaching a mirror, there cannot generally be a mirror in an input or output square, as this would prevent the bit entering/leaving by that square. However, a grid taking advantage of input/output that has no requirement for a given bit by placing a mirror that blocks a given input or output square is still valid provided it gives the correct behaviour.




The score for each input is the area of the resulting grid. The total score will be the sum of the scores for each of the test inputs.

Sandbox questions

  • Is this too similar to domino circuits? The photons leave no trail so they can cross over their own and each other's paths arbitrarily many times. Also a given mirror can affect arbitrarily many photons, and each an arbitrary number of times.
  • Is this a duplicate of anything else?
  • Should the score be just the area of the grid, or also include the time between input and output? The time will have to be the worst case over all inputs, as reading the input before then would give a false result in some cases.
  • Should output bits have to arrive at the same time?
  • Should the time from input to output be required to be constant over all inputs?
  • There cannot be more output bits than input bits. Should test cases reflect this, or should a mechanism be introduced to make all patterns of output possible?
  • Should the particles be referred to as "bits", "photons", or something else?
  • Currently tie break is first posted. Should the number of mirrors be taken into account as a tie break first?
  • Should the grid wrap? I initially thought not, but then I realised there would need to be an extra row above the I/O rows to allow mirrors to redirect onto the top output bit. It might be simpler to keep to having the I/O rows at the very top, and simply put redirecting mirrors in the bottom row. This would make it possible to redirect onto rows 0 and 7 with only 8 rows in total.
  • I've chosen mirrors that change the bit's direction just before impact, rather than on impact. I liked the fact that this gives an asymmetry - reversing the direction will not reverse the route taken. This introduces the potential for sending a bit back along part of the same path without trivially sending it back to its origin. Is there any reason to stick to the symmetrical case instead?
  • The other thought that occurred to me was to have the mirrors change direction too, flipping between the two possibilities at each impact.
  • There are 256**256 possible ways to assign a value from 0 to 255 to each of 256 different inputs. This is far too large a space for hardcoding all solutions. However, this doesn't necessarily mean that the worst case grids will be huge. A lower bound on the worst case grid area is log(256**256,3), which is just over 1292 (since there are 3 possible states for each square). The maximal lower bound is likely to be far higher than that, but I have no idea how much higher. I'm likely to settle on a small enough space that code can be expected to deal with any input, but still a large enough space that hardcoding is impossible (I don't want to explicitly rule it out). (Here ** indicates exponentiation, as used in python.)
  • Mirrors that split bits, and bits annihilate on collision with each other. This will allow for arbitrary input and output rather than being restricted by the initial number of bits. I'm considering a number of possible approaches:
    1. One mirror type only: always splits a single bit into two bits in the two directions perpendicular to the current direction.
    2. 4 mirror types - each having a dead direction. A bit arriving from that direction is destroyed. A bit arriving from any other direction is split into two bits, one for each remaining direction.
    3. 4 mirror types - each having a dead direction. A bit is only split into two bits if the two perpendicular directions are not dead. Otherwise only one bit is produced, in the available perpendicular direction (or no bits if the bit approaches from the dead direction).
  • Can bits be left in the grid that do not terminate? Can a solution specify that the output should be measured at a set time, even if some bits will later reach the output squares altering the result? Should the output only be measured once all activity has ceased - excluding the possibility of using a repeating cycle?

Implement an HTTP Tunnel

I'm bored at work and stuck behind a draconian proxy.

They can take my other ports but they can never take my port 80 freedom!

Help me get all the internets!

This is a challenge.


  • a single web page that processes GET and POST data and produces output
  • if GET m=start
    • creates any files you feel are necessary (named pipes, scripts, what-have-you)
    • forks a process which will create a tcp connection
    • connects to host and port based on GET variables h and p
    • these variables should be cleansed so as not to allow command injection
  • if GET m=in
    • write the raw POST data to your forked process's tcp connection
  • if GET m=out
    • get all data available from the forked process's tcp connection and write as response
    • should always return right away (let's say, in less than 1 second)
  • if GET m=stop
    • kill your forked process
    • clean up any files it has created

Test data

my first instinct was to have this challenge be three separate pieces of code, a client which listens on a port locally and interacts with the web page, the web page itself, and a script which will be forked by the web page. your score would be the sum of all their byte counts. i decided to remove the last as the start process would likely have to create other files so why not have it create the script to run as well, and decided to remove the first option as well to make it nice and even.

is this a feasible challenge? i will add more explanations and test data soon i think. adding my own client would probably be helpful

  • \$\begingroup\$ HTTP is stateless, so this is a fundamentally broken way of designing a tunnel over it. I can't remember the exact headers, but there are ways in HTTP/1.1 of reusing a single connection for multiple bidirectional data transfers, and that would be the correct way of doing it. \$\endgroup\$ Commented Jun 7, 2016 at 11:02
  • \$\begingroup\$ That's why the "start" mode is going to have to fork another process to handle the actual connection. I have this working with my php script forking a nc process which reads and writes from named pipes. I'll post it later. Sure you could use Connection: keep-alive but that connection usually times out pretty quickly, you'd have to implement your own pings to keep it alive, and there's no guarantee how long it would stay alive for. \$\endgroup\$
    – Nacht
    Commented Jun 7, 2016 at 23:28

Code Golfed Rosetta Code Code Golfer

(any others? maybe , and/or )

Browsing examples on the Rosetta Code site, I can't help but think that all the code is just so long-winded, inefficient and ...well, readable. Something needs to be done.


Choose a language, then write a function that takes the example source code, in the same language, for specific tasks on Rosetta Code and returns a golfed version of that code.

Winner is the person who can golf the example tasks the most. However, this is a code golf challenge so the length of your own source matters too.


  1. Write a function in your chosen language that takes a string as input and returns a string as output (or equivalent - reading/writing to stdio or file, etc is also ok)
  2. Input is the source code implementation, in your chosen language, of the following three specific tasks as shown on Rosetta Code (your function runs three times, once for each):
  3. Output is a code golfed version of the input with identical functionality (again in the same language)
  4. If there is more than one implementation of a task for a specific language, you must use the first listed
  5. If your chosen language doesn't have an implementation for one of the tasks, then you need to add it yourself (following the Rosetta Code rules - don't go messing up their site just to get a better score here)
  6. With the exception of the rule above, you may not, in any way, modify the content or order of examples on Rosetta Code
  7. You must leave the logical flow of the algorithm mostly intact (eg. you can't simply replace the J quick sort code with /:~)


Score for each individual task is calculated as the output character count as a percentage of the input character count. Implementation score for your own code is simply its character count. Total score for a submission is the sum of the three task scores, plus the implementation score.

Submission with the lowest score wins.

So, assuming your function is 100 chars long and running it against the test tasks gives you the output counts shown, your overall score would be calculated as follows:

 Task      | Input char count | Output char count | Score
 Quicksort |            600   |            400    |  67%
 Happy Nos |            400   |            200    |  50%
 GCD       |            200   |            150    |  75%
 Implementation score:                            |  100
 Overall score:                                   |  292

Things I'm not sure about...

Before I post this as a challenge, it would be good to get input on a few things:

  • Will the "if your language doesn't have an implementation" rule cause problems, or can people be trusted to provide sensible implementations that follow the intent of Rosetta Code and don't simply artificially improve their score on the challenge? Is it better just to deny entries from languages which don't already have implementations?
  • With scoring, obviously it's a balancing act, a really terse language will likely get a solid "implementation" score, but should be less able to improve the length of the examples, whereas a verbose language will be the opposite. So, the having too few "tasks" included in the challenge will benefit terse languages, and too many will benefit verbose languages. I want to find a middle ground, so does three tasks seem reasonable?
  • Will someone just find an edge case language which has a really easily golf-able Rosetta example, that will make it unbeatable?
  • \$\begingroup\$ In the given example score, if an empty program echos then it would be better. More worryingly, this seems to allow coding to the test cases. Are the programs required to do something sensible with inputs other than the three test cases? \$\endgroup\$ Commented Jun 9, 2016 at 9:42
  • \$\begingroup\$ @PeterTaylor - re: an empty program, I agree, this absolutely creates a minimum bar to beat, but even the most rudimentary whitespace stripping javascript function: (s) => {return s.replace(/[\s]{2,}/g,"");}; results in a score of 263, so my example score sheet is more the problem. \$\endgroup\$
    – Alconja
    Commented Jun 9, 2016 at 11:09
  • \$\begingroup\$ @PeterTaylor - Re: coding to the tests, yes this is a bigger problem... The obvious solution is to simply include more tasks, since that would force people to target more generic things, rather than each individual task, but as stated the more you add the more you'll reward verbose languages (I think?)... One possible work around could be just to double things (i.e. have six tasks and make your implementation score, your code length x 2). \$\endgroup\$
    – Alconja
    Commented Jun 9, 2016 at 11:11

Golf a golf-scoring program!

Given a space delimited array of integers, find the smallest number. It's that simple.

[Meta] This may in fact be a duplicate. Please tell me if so.

  • 1
    \$\begingroup\$ This is basically a dupe of any regular sorting question, isn't it? Particularly something like Sign that word. \$\endgroup\$ Commented Jun 9, 2016 at 12:52

Distance between two words

You are given an input of two strings consisting entirely of letter characters. The "distance" between two such strings is the number of operations from the following list that it takes in order to transform one word into the other:

  1. Adding a letter anywhere
  2. Removing a letter anywhere
  3. Changing a letter's case

Since your boss wants to avoid wear on the office keyboards as much as possible, you have to write a very short program to determine the distance between words so you can fix the typos.

  • 2
    \$\begingroup\$ Closely related. \$\endgroup\$
    – Leaky Nun
    Commented Jun 10, 2016 at 13:56
  • \$\begingroup\$ Almost a duplicate of Leaky Nun's link. The only difference is that the linked challenge allows for straight substitution, whereas here it's two operations (a deletion and an insertion). \$\endgroup\$ Commented Jun 10, 2016 at 13:59
  • \$\begingroup\$ Thanks guys, I had a brief look for dupes but couldn't find anything. If I come up with a good twist I'll edit the OP otherwise I suppose this is dead. \$\endgroup\$
    – A Simmons
    Commented Jun 10, 2016 at 13:59

Cross Validated

(This challenge is almost complete, but something doesn't feel quite right)

Continuing the theme of using site names as challenges...

Your task

Write a program or function that accepts a string, and prints out the location, size, and validity of each cross in the string.


  • Valid Cross- A cross where all four spokes of the cross are equal in length. The size of the cross is the number of segments of each cross.

Valid, size 0:


Valid, size 3:

  • Invalid cross- A cross where one or more spoke is a different length. The size is the number of segments of the longest spoke.

Invalid, size 2:


Invalid, size 6:



  • Each input will have one or more crosses.
  • The program should print out the location of the center of each cross. The location is zero-indexed and measured in characters/lines from the top left.
  • The size and validity, as defined above, of each cross should be printed out.
  • Each test case must pass without printing to STDERR.
  • Crosses will not overlap
  • Your program can take input via a string (with line breaks), an array of strings (each representing a line), or a 2d array of characters.
  • This is so shortest program, in bytes, wins

Test Cases

+ +-

(0,0) size 0, valid

(2,0) size 1, invalid

   |     |
   |     |
---+---  |
   |     |
   |  ---+

(3,3) size 3, valid

(9,5) size 4, invalid

(empty test case)

(must not crash or print to STDERR)


This question isn't fair.

I want you to tell me the chances of flipping an coin n times and ending up on tails.

Naturally this isn't a fair coin. In fact, it isn't even a standard unfair coin, where the chance of flipping tails is always p. This is a sticky coin, where the chance of the coin staying the same is p; the coin is fair at other times.

The coin starts off heads, so when n is 0 or p is 1 then the answer is always 0.

Your program or function should be capable of calculating the result to at least six significant digits.

This is , so the shortest program wins!

  • \$\begingroup\$ What does "the coin is fair at other times" mean? \$\endgroup\$
    – Leaky Nun
    Commented Jun 16, 2016 at 13:12
  • \$\begingroup\$ @LeakyNun Sorry, I'm not sure what you're trying to ask there. A fair coin is one which is equally likely to produce heads or tails but there is no way of predicting which. This coin isn't always fair; a proportion of the time p it repeats the last result. \$\endgroup\$
    – Neil
    Commented Jun 16, 2016 at 14:22
  • \$\begingroup\$ You said that the chance of the coin staying the same is p. Then, isn't the chance of it being different 1-p? So, do you mean that 1-p = 50%? \$\endgroup\$
    – Leaky Nun
    Commented Jun 16, 2016 at 14:25
  • 1
    \$\begingroup\$ I think it means that the chance of the coin staying the same is p + (1-p)/2 and the chance of the coin changing is (1-p)/2. \$\endgroup\$
    – Emigna
    Commented Jun 16, 2016 at 14:37
  • \$\begingroup\$ @LeakyNun Ah, sorry, the chance of it being sticky is p, and of being fair is ¬p. \$\endgroup\$
    – Neil
    Commented Jun 16, 2016 at 15:24
  • \$\begingroup\$ Meaning, that the chance of it being the same is p+(1-p)/2 and the chance of it being different is (1-p)/2. What a meaningful obfuscation. \$\endgroup\$
    – Leaky Nun
    Commented Jun 16, 2016 at 18:42
  • 2
    \$\begingroup\$ That means that the chance of it not changing is p + (1-p)/2 = (1+p)/2, and Sp3000's closed form needs changing to (1 - p^n)/2. It's still not exactly an interesting function to golf. \$\endgroup\$ Commented Jun 16, 2016 at 18:43
  • \$\begingroup\$ I can see I should have described the coin of having a chance q of flipping, and you needed to calculate the probability of an odd number of flips in n trials. \$\endgroup\$
    – Neil
    Commented Jun 16, 2016 at 18:53
  • \$\begingroup\$ (Removed my previous comment because I misunderstood and thought the probabilities were p and 1-p for same/change, but ditto Peter's comment) \$\endgroup\$
    – Sp3000
    Commented Jun 16, 2016 at 22:59

"Optimize" a RegEx


Inspired by atrociously large regex's like


I propose a challenge to create a string representing an optimized and sorted regex from an input string of characters expected to be matched.


  • only printable ASCII characters with code points 32 - 127 need to be supported:

The RegEx output string should:

  • Group 3 or more consecutive code points in the input string together like begin-end
  • Sort the characters and groups in order of ascending code point
  • Escape the literal character - to \- to differentiate it from a range

The RegEx output string should not:

  • Escape the characters !$()*+./=?[\]^{|}
  • Support any RegEx escape sequences like \w, \d or \s





Just so people can see what the last one's pattern is, here are the ASCII indices:


Spoiler Alert

Example implementation in JavaScript ES6:

f=s=>[...s] // spread string into array
  .sort() // sort array by ASCII indices
    c=>c.charCodeAt() // convert each one to ASCII index
  ).reduce( // reduce sorted indices
      ~p[0][0]+c? // if last index is not one less than current index
        p.unshift([c]): // then start new run with this index
        p[0].unshift(c) // else continue existing run
    [[]] // start reduce with empty run
  ).map( // map array of runs
        a=a.map( // map each run
          n=>( // convert index back to ASCII
            c=='-'? // if '-'
              '\-': // then escape it
        a.length>2? // if run has more than 2 indices
          [a[0],'-',a.pop()]: // keep only the begin and end
          a // else keep whole run
      ).reverse().join`` // reverse and join run
  ).reverse().join`` // reverse and join array

  • 2
    \$\begingroup\$ This question seems to be about optimising regex character classes, not full regexes. Is that correct? \$\endgroup\$ Commented Jun 17, 2016 at 19:46
  • \$\begingroup\$ @FryAmTheEggman I suppose you could say that. That's a more wordy title though so I just kept it simple. \$\endgroup\$ Commented Jun 17, 2016 at 19:47
  • \$\begingroup\$ I could understand not requiring any escaping on the grounds that it's a minor detail. I could understand requiring enough escaping to make this actually a useful tool. But it seems really odd to require escaping - but not ]. \$\endgroup\$ Commented Jun 17, 2016 at 22:07
  • \$\begingroup\$ @PeterTaylor I decided to escape only the characters that are necessary to determine whether an execution is correct or not. If you have an alternate suggestion, please feel free. \$\endgroup\$ Commented Jun 18, 2016 at 1:53

Exercise your kids

I want you to output the nth verse of this kid's exercise song:

Head, and shoulders, knees and toes, knees and toes,
Head, and shoulders, knees and toes, knees and toes
And eyes and ears and mouth and nose,
Head, and shoulders, knees and toes, knees and toes.

Each verse is the same as the previous except that all occurrences of one word is masked by replacing it with a dash (you can use a two-byte dash of your choice) wherever it appears in the verse. The words are replaced in order however the word "and" is never replaced, therefore the next verse should look like this:

—, and shoulders, knees and toes, knees and toes,
—, and shoulders, knees and toes, knees and toes
And eyes and ears and mouth and nose,
—, and shoulders, knees and toes, knees and toes.

Also, the last verse should look like this:

—, and —, — and —, — and —,
—, and —, — and —, — and —
And — and — and — and —,
—, and —, — and —, — and —.

Your answer should specify whether n will range from 0 to 8 or 1 to 9.

  • 1
    \$\begingroup\$ Could you show us what the second verse would look like? \$\endgroup\$
    – Leaky Nun
    Commented Jun 21, 2016 at 10:26
  • \$\begingroup\$ Although it can be deduced from the numbers, I'd recommend changing "one word is masked" to "all occurrences of one word are masked" to be explicit. \$\endgroup\$ Commented Jun 21, 2016 at 10:32
  • \$\begingroup\$ Is this a challenge involving Unicode as the main topic? If not, can we replace (U+2014) by a simple hyphen - (U+002D)? \$\endgroup\$
    – Leaky Nun
    Commented Jun 21, 2016 at 10:34
  • \$\begingroup\$ @LeakyNun No, but you may replace it with a double hyphen, as that's still 2 bytes. \$\endgroup\$
    – Neil
    Commented Jun 21, 2016 at 10:38
  • \$\begingroup\$ Actually, it is 3 bytes. \$\endgroup\$
    – Leaky Nun
    Commented Jun 21, 2016 at 11:05
  • \$\begingroup\$ Something to note is that "and" is the only word in the verse that starts with the letter "a". That means something like \b(?!a)\w+ will have its first match be the word to replace each time. Also, I don't believe any of the words are prefixes/postfixes of each other, so once you have them you can blindly replace all occurrences of them. This isn't a problem or anything, I just wanted to make sure you knew in case you wanted it to be harder. \$\endgroup\$ Commented Jun 21, 2016 at 12:50
  • \$\begingroup\$ codegolf.stackexchange.com/q/100153/34718 \$\endgroup\$
    – mbomb007
    Commented Nov 17, 2016 at 15:27

Calculate the Average Squared Error

Given a line y = mx + b and a set of n points (xi, yi), find the average square error between the given line and each set of points.


Your goal is to create a function or program that given the values m, b, and the set of points (xi, yi), outputs the average square error according to the formula above.


  • This is so the shortest solution wins.
  • Builtins that compute this value are not allowed. This includes builtins that compute a result which is a scaled value of this.
  • Floating-point inaccuracies will not be counted against you.

Test Cases



Index sum and strip my matrix

Given a matrix/2d array in your preferable language


  • The matrix will always have an odd length
  • The matrix will always be perfectly square
  • The matrix values can be any integer in your language (positive or negative)


1  2  3  4  5  6  7
2  3  4  5  6  7  8
3  4  50 6  7  8  9
4  5  6 100 8  9  10
5  6  7  8 -9  10 11
6  7  8  9  10 11 12
7  8 900 10 11 12 0


  • The "central number" is defined as the number that has the same amount of numbers to the left,right,up and down

In this case its middlemost 1000

  • The "outer shell" is the collection of numbers which their x and y index is or 0 or the matrix size

1  2  3  4  5  6  7
2                 8
3                 9
4                 10
5                 11
6                 12
7  8 900 10 11 12 0

Your task:

Add to the central number the sum of each row and column after multiplying the values in each by their 1-based index

A single row for example

4  5  6  7  8

for each number

number * index + number * index.....

4*1 + 5*2 + 6*3 + 7*4 + 8*5 => 100


 2 -3 -9  4  7  1  5  => 61
-2  0 -2 -7 -7 -7 -4  => -141
 6 -3 -2 -2 -3  2  1  => -10
 8 -8  4  1 -8  2  0  => -20
-5  6  7 -1  8  4  8  => 144
 1  5  7  8  7 -9 -5  => 10
 7  7 -2  2 -7 -8  0  => -60
78 65 60 45 -15 -89 10   => 154
                     => -16
  • For all rows and columns you combine these values..
  • Now you sum these too => 154-16 = 138
  • You add that number to the "central number" and remove the "outer shell" of the matrix

 0 -2 -7 -7 -7     => -88
-3 -2 -2 -3  2     => -15
-8  4 1+138 -8  2  => 395
 6  7 -1  8  4     => 69
 5  7  8  7 -9     => 26

19 69 442 30 -26

do this untill you end up with a single number

-2 -2 -3     => -15
 4  1060 -8  => 2100
 7 -1  8     => 29

27 2115 5
  • Add 2114+2147 to 1060
  • Remove the "outer shell" and get 5321
  • Now we have a single number left

this is the output!

test cases:



-7 -1  8
-4 -6  7
-3 -6  6


 6  7 -2  5  1
-2  6 -4 -2  3
-1 -4  0 -2 -7
 0  1  4 -4  8
-8 -6 -5  0  2


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


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

  • \$\begingroup\$ yes, that is much better. hard when English is your 3rd language \$\endgroup\$ Commented Jun 21, 2016 at 18:02
  • \$\begingroup\$ No worries, English is a pretty incomprehensible language no matter who is speaking it ;) Anyway, you identify the central number as "the zero", but the example you gave actually has two zeros. Perhaps change the array or change the wording to the "middlemost zero"? \$\endgroup\$ Commented Jun 21, 2016 at 18:07
  • \$\begingroup\$ For the process to work I think you need the matrix to be square, but I don't see a statement of that anywhere. \$\endgroup\$ Commented Jun 21, 2016 at 20:26
  • \$\begingroup\$ added that @FryAmTheEggman also changed \$\endgroup\$ Commented Jun 22, 2016 at 16:52

White Water Rafting

This problem is about finding the best path through a bunch of rocks on a 5-column wide river, without crashing your raft. A river looks like this (* reprsents a rock):

. . . R .
. R . . .
R . . . .
. . R . .
R . . . .
. . . . R

Each row will contain exactly 1 rock, no more, no less. Your raft can start on any space without a rock.

You can't maneuver your raft too much, so as you travel down the river, there are only 3 next places you can go (x is the next space):

. . @ . .
. x x x .

If there is a rock in your way, you can't go there. Your raft can't fit through rocks that are diagonally adjacent to each other. You can't beach your raft either. Finally, you can't go anywhere that would result in you crashing your raft.

@ . . . .
R x . . .   <- Can't go on rock, can't go oob

R . . . .
. R @ . .
. . R X .   <- Can't go through diagonal rocks

. @ . R .
. R x . .   <- Can go through non-adjacent diagonals,
R . . . .      Can't go to dead end.

Because you don't have much space on the raft to write this code, shortest code wins.

Test cases:


. R . . .
. R . . .
. . . R .
. . . R .
R . . . .


R x . . .
. R x . .
. x . R .
x . . R .
R x . . .


R . . . .
R . . . .
R . . . .
R . . . .
R . . . .


R x . . .
R x . . .
R x . . .
R x . . .
R x . . .


. R . . .
. R . . .
. R . . .
. R . . .
R . . . .


. R x . .
. R x . .
. R x . .
. R x . .
R x . . .


. . . . R
. . . R .
. . R . .
. R . . .
. . R . .


x . . . R
x . . R .
x . R . .
x R . . .
x . R . .


. . . . R
. . . R .
. . R . .
. R . . .
R . . . .


[nothing] or [empty/blank array/matrix]


  • Input can be in array of indexes, array of truthy/falsey values, or any other input format most comfortable to your language.
  • Output should indicate the left-most valid path.
  • Output can be in array of indexes, array of truthy/falsey values, or any other input format most comfortable to your language.
    • Output does not have to be in the same format as input.
    • Output nothing/(empty/blank) (array/matrix) if there is no valid path
  • Standard loopholes are forbidden by default.

This is my first challenge, so please let me know if I have left anything out or something is unclear.

Related problems

I couldn't find any dupe targets looking through , so I'll look again in and later.

  • \$\begingroup\$ I don't have 10 minutes now to sort through all of the dupe targets to work out which one is the most similar, but I guarantee that there is something similar enough that this is a dupe. \$\endgroup\$ Commented Jun 23, 2016 at 9:55


The year is 19XX.

You are a spy of some distant country, and your job is to send messages across the globe.

Unfortunately, because, frankly, you suck at being a spy, you need a way to obfuscate your information, so that when the opposition catches you (which they will), they won't know what the heck is written.

How are you going to do this?

Your task is, using two inputs (the first input the encoding "cypher" and the second the message), encode the message.

This is how the encoding works:

  • The message only consists of the lowercase letters and the numerals.
  • Because there are 36 different characters in total, we will convert each individual letter of the message to "base 9" (a = 00, b = 01, c = 02... 8 = 37, 9 = 38). This will be called the FSO, or First-Stage Obfuscation.
    • For example, the message hello1 would then be translated to 07 04 12 12 15 28.
  • Each individual "bit" of the FSO is then stripped of its first part. This will be known as the SSO, or Second-Stage Obfuscation.
    • The example 07 04 12 12 15 28 is then translated to 7 4 2 2 5 8.
  • This is where the encoder comes in handy! The encoder will consist of a string of numbers 0 to 3 (e.g. 1212003).
  • You then add to the start of each digit of the encoder to the corresponding digit of the SSO. This is now the TSO, or Third-Stage Obfuscation.
    • The example 7 4 2 2 5 8 with the encoder 1212003 is then converted to 17 24 12 22 05 08.
    • With a shorter encoder (say 121), this step will "wrap around", so 7 4 2 2 5 8 with the encoder 121 will end up with 17 24 12 12 25 18.
  • We then change the TSO back into readable characters, using the same "base 9" method.
    • The two examples 17 24 12 22 05 08 and 17 24 12 12 25 18 will be converted to QWLUEI and QWLLXR respectively.

So, in summary:

Encoder: 121

Message: hello1

hello1 => 07 04 12 12 15 28 => 7 4 2 2 5 8 => 17 24 12 12 25 18 => QWLLXR

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


  • This is really confusing, and I don't really know how to phrase the "how the encoding works" bit better. Can anyone help me phrase this better? I can offer clarification on certain things if needed.
  • \$\begingroup\$ This is basically a mildly astandard and irreversible Vigenère cipher, so it's virtually a dupe of this question. (I would also say that it violates one of the key criteria for being a good question, which is to have a motivation. The backstory IMO isn't a motivation because it doesn't explain why anyone would want to implement or use what's really a supremely bad hash function). \$\endgroup\$ Commented Jun 23, 2016 at 13:36
  • \$\begingroup\$ @PeterTaylor Yeah, IK. I need a lot of things to fix with this question (I was extremely tired at the time, couldn't think of a nice backstory). Also, a few things about the dupe: ONE, it's 5 years ago, so there's bound to be some new answers out there (if it's even a dupe in the first place), and TWO, how is it a dupe? \$\endgroup\$
    – clismique
    Commented Jun 24, 2016 at 10:55
  • \$\begingroup\$ It's the difference between for (i=0 to n-1) s[i] = handleWrapping(s[i] + k[i % klen]) and for (i=0 to n-1) s[i] = handleWrapping(s[i] % 9 + k[i % klen] * 9). As far as I'm concerned that's a very minor transformation. \$\endgroup\$ Commented Jun 24, 2016 at 11:03
  • \$\begingroup\$ @PeterTaylor Yeah, I kinda understand now. Still some questions, though. ONE: What language is that? TWO: Is there any way I could improve on my explaining? \$\endgroup\$
    – clismique
    Commented Jun 24, 2016 at 11:22
  • \$\begingroup\$ 1. It's pseudocode, because I didn't want to faff around working out what escaping a less-than sign needs in comments. 2. I'm not quite sure what you're asking, but I would ditch this idea completely and try to find something more original. See e.g. meta.codegolf.stackexchange.com/q/1475/194 \$\endgroup\$ Commented Jun 24, 2016 at 11:49
  • \$\begingroup\$ @PeterTaylor Yeah, I can see those, but most of those ideas have been taken already, and I kinda want to move away from numbers for a bit and play around a bit with strings and whatnot. Any suggestions? \$\endgroup\$
    – clismique
    Commented Jun 24, 2016 at 12:36

Regex golf: Match the Thu'um-s


Skyrim is a game made by Bethesda and came out in 2011. One of the objectives of the game is to collect every 27 shouts (or Thu'um-s). In this challenge, you need to match every shout and nothing else.


This is the list of the available shouts in the game:

Raan Mir Tah
Laas Yah Nir
Mid Vur Shaan
Feim Zii Gron
Gol Hah Dov
Od Ah Viing
Hun Kaal Zoor
Lok Vah Koor
Ven Gaar Nos
Zun Haal Viik
Faas Ru Maar
Mul Qah Diiv
Joor Zah Frul
Gaan Lah Haas
Su Grah Dun
Yol Toor Shul
Fo Krah Diin
Liz Slen Nus
Kaan Drem Ov
Krii Lun Aus
Rii Vaaz Zol
Tiid Klo Ul
Strun Bah Qo
Dur Neh Viir
Zul Mey Gut
Fus Ro Dah
Wuld Nah Kest

Additionally, every shout can have 3 levels depending on what the player collected so far, each level adds a new word to each shout, so you need to be able t match the separate words without the full shout.

The words need to be uppercase. A shout should only be matched, if it is a separate word, for example: Golf shouldn't be matched.

The separate words for the same shout in the order as in the list appear next to each other, then they need to be in the same match.

The input strings will only contain ASCII letters and spaces as word separators.

Test cases

The matched text is bold

Hydrogen Sulphur Krah Coal Gaan Lah Haas

one two three four Kaal six seven Joor Zah Frul eleven

red green blue Gol Hah orange purple Qo violet

Ran Miir Taah Raan Mir Tah raan mir tah

Golf Gol Middleage Mid


  • The answer should be a .NET type RegEx and should not contain any other langauge.

  • This is a code-golf, so the shortest answer in bytes wins

  • 2
    \$\begingroup\$ Very closely related. What if a word appears in lower case in the input? What if it appears inside another word like Golf? If the latter should not be matched please clarify what characters can be in the input and which of those are valid word delimiters. All that said, I don't see anyone coming up with a good way to compress the words within the limit framework of regex. \$\endgroup\$ Commented Jun 23, 2016 at 12:18
  • \$\begingroup\$ @MartinEnder I clarified it a but \$\endgroup\$
    – Bálint
    Commented Jun 23, 2016 at 13:16
  • 1
    \$\begingroup\$ Still doesn't say what characters can appear in the input and what characters count as word separators (I'm assuming only letters and spaces, and spaces are separators, but if that's your intention you should say so explicitly, and if not, you should add further test cases). Looks good otherwise. To prevent confusion you might want to say explicitly that people should submit only a regex and which flavours are allowed (and whether flavours like Perl are allowed to make use of their eval features to execute code in the hosting language). \$\endgroup\$ Commented Jun 23, 2016 at 13:21
  • \$\begingroup\$ If only one language is allowed this would limit the participation in your question. I don't understand why it should be only one language. \$\endgroup\$
    – george
    Commented Jun 29, 2016 at 19:38
  • \$\begingroup\$ @george I don't think you know what regex-golf is \$\endgroup\$
    – Bálint
    Commented Jun 29, 2016 at 19:44
  • \$\begingroup\$ @Bálint Whoops I didn't see your title, my mistake. \$\endgroup\$
    – george
    Commented Jun 29, 2016 at 19:47

Find all the Vampire numbers

Shamelessly stolen from https://stackoverflow.com/q/17352108

A "Vampire" number is defined as the product of two numbers of equal length (known as the "fangs") that uses the same digits as the two numbers being multiplied. Examples:

21 * 60 = 1260
15 * 93 = 1395
30 * 51 = 1530

Your task is to find all the Vampire numbers whose fangs have n digits.

This is , so the shortest solution wins.

  • \$\begingroup\$ What about double fangs, triple fangs etc? \$\endgroup\$
    – clismique
    Commented Jun 24, 2016 at 11:30
  • \$\begingroup\$ @DerpfacePython Double fangs? But I suppose I could generalise it to m fangs of n digits if there's enough support (and if there are actually solutions!) \$\endgroup\$
    – Neil
    Commented Jun 24, 2016 at 12:19
  • \$\begingroup\$ Related \$\endgroup\$ Commented Jun 24, 2016 at 13:23
  • \$\begingroup\$ @FryAmTheEggman My bad for not searching first. I might as well delete this. \$\endgroup\$
    – Neil
    Commented Jun 24, 2016 at 14:14
  • \$\begingroup\$ I think all you have to do is add the double/triple fang thing and it probably is different enough? Not sure, but there is probably a way to get it to work. \$\endgroup\$ Commented Jun 24, 2016 at 14:20

Title TBD - Generate Emoticons :)

Your task is to create a program that generates the most emoticons using the least amount of code.


  • The only valid emoticons are listed on Wikipedia, under the heading Western/horizontal emoticons: List of emoticons page (version 722951221)
  • The output can be any format, as long as a delimiter exists between emoticons.


  • Scores will reflect the character (not byte) count.
  • For every emoticon after 10 that is output by your program, subtract 1 from your score. Emoticons over 20, subtract 2, over 30 subtract 3, and so on.


Is the scoring fair/reasonable? Should I edit down the list/create a new list of acceptable emoticons?


Forecast Romantic Dates

Sort of inspired by this.

A Romantic date is a date that, when the year, month and day are converted to Roman Numerals the individual values contain no more than two symbols. For example, in YY-MM-DD format: the Romantic date 20-04-15 would become XX-IV-XV.

For the purpose of this challenge, years will only be tracked by the two least significant digits of the year, as otherwise the last Romantic date was in the 15th century. In addition, they wouldn't add much to the challenge as the omissions of the leap year every 400 years is irrelevant, as February the 29th is not a Romantic date.

Romantic dates

For your convenience, here is a list of all of the two digit numbers that can be represented with two or fewer symbols in Roman numerals:

[1, 2, 4, 5, 6, 9, 10, 11, 15, 20, 40, 50, 51, 55, 60]

These were determined using the "standard method" that negative groups would only be used with the symbols that are powers of ten and only on the values that are five or ten times that symbol's value. So I only combines with V and X for example.

Dates which include only numbers from this list are Romantic dates. For the purpose of this challenge, assume an ideal Western calendar: no dates are ever skipped or repeated, 12 months per year, and more than 20 days per month. Assume there is no year, month or day zero (i.e. year 99 loops to 1 not to 0).


Given a date as input, output that date if it is Romantic, or output the next Romantic date.

Input and Output

You may accept input in any consistent ordering of year, month and day with any consistent separator. You may specify if the input should have the numbers padded to be two digits. If the numbers are padded, you may choose to have no separator. Your output must have the same form as your input.

Test Cases

The following test cases are all in the format YY MM DD, with no padding.

1 1 1 => 1 1 1
20 4 3 => 20 4 4
15 7 1 => 15 9 1
51 9 7 => 51 9 9
70 9 7 => 1 1 1
20 6 24 => 20 9 1
47 12 1 => 50 1 1
60 11 24 => 1 1 1

Here is the script that I used to generate these.


Did I miss any Romantic numbers? I just did that by hand.

Allow unary? I'm unsure about this because it sort of violates the reasoning behind Romantic dates for the values to have >2 symbols...

Should I explain more about parts of dates that are not useful? For example, the length of the months is entirely irrelevant as the later days are all skipped. My concern is that the current one feels clunky already

Should I allow both outputting the input if the date is already Romantic or the strictly next Romantic date (as long as it is consistent)? There doesn't seem to be much different, but I don't know if that'd be too broad? Personally leading towards allowing it.

I'm also somewhat tempted to make use of the silly title a bit more, but I'm not sure if that'd be going overboard.

Too boring / compression based? I've particularly been trying to think of a way for fewer results to wrap back around to 1.

  • \$\begingroup\$ That rule should be spelt out explicitly in the question, because although some people insist on it it's a modern innovation. There are actual Roman inscriptions which do use e.g. IC. \$\endgroup\$ Commented Jun 24, 2016 at 21:05
  • \$\begingroup\$ @PeterTaylor You're right, I originally left out the reasoning because I thought it might clutter up the spec, but I realise now I just left that comment undeleted to prevent people from asking the same question. I don't have time right now but I'll edit it in once I get a chance. Also, I figured it was better with this rule because I thought say VL was rather unintuitive, does that make sense or should I be more laissez-faire about it? \$\endgroup\$ Commented Jun 24, 2016 at 22:04

Stacks and Stacks and Stacks...

Write a program that, with the input as n, finds the first n-gonal and n-gonal pyramid number that is NOT 1.

n is guaranteed to be larger than or equal to 3.


  • n = 3: 10
  • n = 4: 4900
  • n = 2: The output can be nothing, False, or anything that you want, just as long as it can be distinguished from an actual output.

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


  • If your code output both the name of the n-gon and the number: You get a big fat -50% off of your byte count (see below for examples).
    • n = 4: Square 4900
    • n = 3: Triangle 10
    • n = 5: Pentagon ??? (the ??? is a placeholder because I have no idea what the number is)


  • Is the bonus a good idea?
  • \$\begingroup\$ I don't think the bonus is a good idea because 1) the name compression takes away from the original challenge and likely isn't worth it and 2) you haven't defined the naming scheme (e.g. 12-gon vs dodecagon) \$\endgroup\$
    – Sp3000
    Commented Jun 26, 2016 at 7:10
  • \$\begingroup\$ @Sp3000 Ah, OK. I really want to incorporate the use of strings in the challenge, but if the idea sucks, then I'll scrap it. Any further suggestions? \$\endgroup\$
    – clismique
    Commented Jun 26, 2016 at 7:12
  • 2
    \$\begingroup\$ Adding strings just makes it feel like squeezing two challenges into one, unfortunately. I'd recommend posting the shape names as a separate challenge, but I see we have this challenge. As for suggestions, maybe 1) remove the part about n = 2, since you already say n is guaranteed to be at least 3 and 2) maybe make it explicit that the output should be both n-gonal and n-gonal pyramidal (and maybe explain, say, that 10 is the 4th triangular and 3rd triangular pyramid number) \$\endgroup\$
    – Sp3000
    Commented Jun 26, 2016 at 7:16
  • \$\begingroup\$ Hmm my other problem is - what happens if there's no solution for a given n? Note the only reason I'm asking is because the number of solutions for any n could be finite, e.g. A027568. (6 is 946, 8 is 1045, 10 is 175 and 11 is 23725 I believe) \$\endgroup\$
    – Sp3000
    Commented Jun 26, 2016 at 9:28
  • \$\begingroup\$ Yeah, that could be a problem... maybe a time limit? Or maybe check numbers up to a given range. \$\endgroup\$
    – clismique
    Commented Jun 28, 2016 at 6:50
  • \$\begingroup\$ One alternative could be to allow solutions to potentially infinite loop/hit memory or recursion errors in the case of no solution/large solution. Numerical limit to check up to could work too, that'd probably be better than a time limit (since it reduces a dependency) \$\endgroup\$
    – Sp3000
    Commented Jun 28, 2016 at 9:02
  • \$\begingroup\$ @Sp3000 I would probably go with the numerical limit/memory limit thing, whichever comes first. But what about golfing/esoteric languages? There might not necessarily be a memory/numerical bound for those. \$\endgroup\$
    – clismique
    Commented Jun 29, 2016 at 9:04

Pyth Meta-Golf Golf Battery

(now that's a name, isn't it?)

The Challenge

Write a program in a language of your choice that takes an input of code in the same language and outputs a code that does the same thing in Pyth (though not necessarily the shortest code).


Given the test battery set below, you should try to find the best way to minimize your score with the following rules:

  • You must minimize the source code of the submission.
    • You may not used compressed versions of the input source for each test; the Pyth code must be output procedurally.
    • The code must theoretically do the same conversion for any input code, not necessarily just these test cases.
  • You must minimize the input code of each test.
    • If an input is shown to be shortenable by without non-standard libraries of the submitted language, you must change it immediately to the shorter answer. If your answer does not support this change, then you must change your answer to accommodate.
  • You may not submit answers written in Pyth.

Scoring is done with the following equation:

(source code)*((Pyth src out, test 1)/(input src in, test 1)+(same for test 2)+...)

Objectives of each test

Test 1:

Output the string "Hello, world!".

Test 2:

Given an integer input, multiply that input by three and output it.

Test 3:

Quine. (must be a valid quine in the submitted language and in Pyth)

Test 4:

Produce infinite output.

Test 5:

Cat program.

  • 2
    \$\begingroup\$ I'm not a Pyth expert, but I strongly suspect that what this asks is impossible in many languages. E.g. I understand that Python's multi-threading support is extremely limited. Even if true impossibility is not an option, answers in many languages won't fit inside the 30000-character limit. E.g. I don't think the grammar for Java fits inside the limit, let alone a lexer and parser. \$\endgroup\$ Commented Jun 27, 2016 at 9:56
  • 1
    \$\begingroup\$ I agree with Peter, that this seems like unless you use only trivial languages it probably won't be possible to answer. The scoring would also make this confusing: to make sure a solution is correct one would have to not only test 10 programs, you would also have to golf 5 programs. Further, there is a bit of a problem in Pyth's $ operator, which runs literal Python, which means you should probably ban Python as well. I'm not really sure of how to turn this into a good question, too heavily restricting the type of program seems to be the only way, but it also seems to defeat the purpose. \$\endgroup\$ Commented Jun 27, 2016 at 13:09

Don't even think about non-42-related numbers!

Introduction and Credit

We all love our Answer to the Ultimate Question of Life, the Universe, and Everything which, of course is 42. So let's take this unworthy Fibonacci-Sequence thingy and adapt it to be worthy of 42!



The input will be a positive integer.


The output will be either true or false.

What to do?

Your task is to implement the predicate that the number under consideration is an element of the generalized Fibonacci-Sequence given by:

a_1 = 14
a_2 = 42
a_n = 41 * a_{n-1} + 43 * a_{n-2}

Where 42 is the ultimate answer, 41 and 43 are the primes next to it and 14 is the preceding catalan number.

Potential corner cases

The number will always be greater than zero. Your program must pass all (32-bit) test vectors below.

Who wins?

This is code-golf so the shortest answer in bytes wins!
Standard rules apply of course.

Test Vectors

14 -> true
42 -> true
2324 -> true
4080622 -> true
97090 -> true
171480372 -> true
7 -> false
1 -> false
41 -> false
43 -> false
  • \$\begingroup\$ a_3 is already outside the range of 8-bit unsigned integers, a_6 is outside the range of 32-bit signed integers, and a_11 is outside the range of 64-bit signed integers. It would be worth addressing this issue and at least ensuring that languages which are inherently 8-bit can't just special-case the two values which they can handle. \$\endgroup\$ Commented Jun 25, 2016 at 21:58
  • \$\begingroup\$ @PeterTaylor I'm unsure how to formulate this without disqualifying legitimate approaches. Would saying "your program must pass all test vectors" (with 32-bit test vectors) also be considered OK for most people? \$\endgroup\$
    – SEJPM
    Commented Jun 25, 2016 at 22:16
  • \$\begingroup\$ I feel like this is too much two separate challenges: one to test for membership in the sequence, and the other to takeWhile on a condition. I also suspect the sequence has a direct arithmetic membership test like the one where n is a Fibonacci number exactly if either 5*n*n+4 or 5*n*n-4 is a square. \$\endgroup\$
    – xnor
    Commented Jun 25, 2016 at 23:27
  • \$\begingroup\$ What is takewhile? \$\endgroup\$
    – clismique
    Commented Jun 26, 2016 at 7:02
  • \$\begingroup\$ @xnor, I originally only wanted to do the takeWhile, but needed a (mediumly) complex, interesting predicate, so I came up with this one. Of course I'm open to suggestion for more suitable interesting predicates. \$\endgroup\$
    – SEJPM
    Commented Jun 26, 2016 at 10:18
  • \$\begingroup\$ @DerpfacePython, the higher-order functionality described in the second paragraph of "what to do?" is also called takewhile, I've clarified this though. \$\endgroup\$
    – SEJPM
    Commented Jun 26, 2016 at 10:21
  • \$\begingroup\$ @SEJPM I think the other way -- to do a challenge about takeWhile, make the predicate as simple possible. Beware chameleon challenges and needless fluff. One clean condition would be integers being positive. \$\endgroup\$
    – xnor
    Commented Jun 26, 2016 at 10:56

Compute the mincut of a graph

Given a graph, compute a division of the graph such that the edges stranded between the cut.

hayo mouseover readers; leave a comment if you see this!

Red line: a cut

Green line: a mincut


The first line will contain the number of nodes. The rest of the lines will contain pairs of positive integer IDs separated by spaces showing connectedness between the nodes with those IDs. Here's an example; for a graph where 1 is connected to 2 and 2 is connected to 3:

1 2
2 3
  • You may assume that the nodes are numbered consecutively from one to the number of nodes.
  • However, you may not assume that the list of pairs of nodes is in any specific order.


Simply output a comma-separated list of the IDs of the nodes of one of subgraphs created by the cut.

Additional Rules!

  • You cannot implement brute-force search. Other than that, feel free to use Karger's Algorithm* or another algorithm. Remember that Karger's algorithm is likely the easiest to implement.
  • Notice: you must run Karger's algorithm at least this many times to ensure a low chance of failure and a low chance of failure

*Karger's algorithm

For your convenience, I've included a simple description of Karger's algorithm.

  1. find two adjacent nodes and merge them into one node (so that all nodes that where connected to the original two nodes are connected to the new node), concatenating the labels
  2. repeat step one until there are only two nodes
  3. the result is any label of one of the nodes
  4. repeat steps 1-3 at least this many times to ensure a low probability of failure, and choose the result that occurs the most often
  • 1
    \$\begingroup\$ 1. Wouldn't it be better to take the graph as an adjacency matrix or list? 2. Your description of the minimum cut is somewhat confusing. 3. Karger's algorithm is probabilistic, which isn't allowed by our defaults (I don't think). Allowing probabilistic algorithms opens up a whole can of worms (for instance I could write a program that just returns a random cut) -- you should probably make it so that the algorithm must return the minimum cut two-thirds of the time or something similar if you want to allow them. \$\endgroup\$ Commented Jul 5, 2016 at 0:07
  • \$\begingroup\$ @quartata 1. it's an adjacency list 2. yeah I need help with that 3. I made sure you had to repeat it insert some math equation here amount of times \$\endgroup\$ Commented Jul 5, 2016 at 0:13
  • \$\begingroup\$ Sorry I misunderstood the input. \$\endgroup\$ Commented Jul 5, 2016 at 0:14
  • \$\begingroup\$ Generally adjacency lists are done like [node1, connected_node1, connected_node2, ...] and not in pairs like you have it; this is more flexible and you don't have to specify the number of nodes (it is just the length of the input list) \$\endgroup\$ Commented Jul 5, 2016 at 0:35
  • 2
    \$\begingroup\$ "You cannot implement brute-force search" is too vague. What about a basically brute force search that shortcuts some obviously wrong possibilities? I think what you want is a running time bound. \$\endgroup\$
    – xnor
    Commented Jul 5, 2016 at 1:55
  • 1
    \$\begingroup\$ 1. The I/O description seems to assume that all answers will be programs taking input on stdin and writing output to stdout, but our defaults are more flexible. In particular, by default we allow answers to be functions which take arrays and return arrays. Comma-separating is also IMO unnecessarily constrained, especially as the input isn't comma-separated. \$\endgroup\$ Commented Jul 5, 2016 at 7:51
  • 1
    \$\begingroup\$ 2. "Feel free to use Karger's algorithm or another algorithm". There's an implicit licence here to use another non-deterministic algorithm, but although you give an explicit iteration count for Karger's algorithm you don't for e.g. randomised Kruskal's algorithm, which it's based on. 3. Besides which, in general I don't think that questions should tell people which algorithm to use. Specify the task and constraints (e.g. "Randomised algorithms are allowed, but must find the correct answer with probability at least foo. All answers must be polynomial-time"). \$\endgroup\$ Commented Jul 5, 2016 at 7:54
  • \$\begingroup\$ 4. But if you're going to include an algorithm description, be careful to get it right. Karger's algorithm is randomised, but in the description given there's no mention of where the random selection occurs or of what uniformity constraints are required to get the desired behaviour. \$\endgroup\$ Commented Jul 5, 2016 at 7:55
  • 1
    \$\begingroup\$ Food for thought: outputting the value of the min cut instead might lend to more approaches. Also, any rules on min cut/max flow/possibly other optimisation builtins? \$\endgroup\$
    – Sp3000
    Commented Jul 5, 2016 at 10:34
  • \$\begingroup\$ I'm going to add a story to this soon. \$\endgroup\$ Commented Jul 5, 2016 at 13:55

Generate a random Vietnamese syllable


The Vietnamese syllable space is interesting, because it is huge.

TODO: Describe the space and why it is interesting.

Here's how such syllables are made:

The onset matches the regex ^([bcdđghklmnprstvx]|qu|[cgkpt]h|ng|tr)$

The vowel is one of the following massive list:

a à a' ã á a.
â â` â' â~ â´ â.
a. ă` ă' ă~ ă´ ă.
e è e' e~ é e.
ê ê` ê' ê~ ê´ ê.
i ì i' ĩ í i. ia iê
o ò o' õ ó o. oa oă oe
ô ô` ô' ô~ ô´ ô.
o' ò' o" õ' ó' o'.
u ù u' ũ ú u. ua uâ uê ui uô uo' uy
u' ù' u" ũ' ú' u'. u'a u'o'
y y` y' y~ ý y. ya yê

The coda matches the regex ^([iouycptmn]|ch|ng|nh)$

(thanks Peter Taylor!)

The onset, vowel and coda are concatenated to make the result syllable.


The objective is to generate random Vietnamese syllables. Your program has to take no input and as output include only the syllable, with an optional trailing new line.


  • Each syllable must be generated with a non zero probability.

I think it's unclear. Contributions are so much welcome.

  • \$\begingroup\$ 1. I'm not sure what you mean by can be with. 2. You don't mention randomness anywhere excwpt the clarification. 3. Object should probably be Objective. \$\endgroup\$
    – Dennis
    Commented Jul 3, 2016 at 18:44
  • \$\begingroup\$ 1. c can also be with h means that h can follow c as the 2nd stage letter in the syllable. 2. Where should I also mention it? 3. Ah k :) \$\endgroup\$
    – user48538
    Commented Jul 3, 2016 at 18:46
  • \$\begingroup\$ If I'm reading this correcting then it can be vastly simplified by saying that the onset matches the regex ^([bcdđghklmnprstvx]|qu|[cgkpt]h|ng|tr)$, the vowel is one of a massive set of options (I don't see any benefit to splitting that into "stage 2" and "stage 3"), and the coda matches the regex ^([iouycptmn]|ch|ng|nh)$ \$\endgroup\$ Commented Jul 4, 2016 at 16:36

Let's play some Briscola

Briscola is an Italian game, played with a deck of 40 cards, divided in 4 suits - coins (denari - D), swords (spade - S), cups (coppe - C) and clubs (bastoni - B).

The values on the cards range numerically from one through seven, plus 3 special cards - knave (11), knight (12) and king (13).


After the deck is shuffled, each player is dealt three cards. The next card is placed face up on the playing surface, and the remaining deck is placed face down. This card is the Briscola, and represents the trump suit for the game.

First player starts by playing one card face up on the playing surface. Each player subsequently plays a card in turn, until all players have played one card.

The winner of that hand is determined as follows: If any briscola (trump) has been played, the player who played the highest valued trump wins, else the player who played the highest card of the lead suit (suit of the first card played) wins.


Briscola has a special type of ranking:

1   ace
3   three
13  king
12  knight
11  knave


Standard loopholes apply.


As an input, you must accept 5 values (cards), in a reasonable format, for example:

briscola (trump card), 1. card, 2. card, 3. card, 4.card


You must output the winning card

Example input and output:

4S 7D 12B 13B 2S -> 2S
5D 1D 5D 12S 3C -> 1D
3B 2C 4S 5S 7D -> 2C
12D 3S 11B 1B 7S -> 3S
  • \$\begingroup\$ As mentioned in chat, I think this is probably a duplicate of this challenge. Just adding so other people don't have to go looking. \$\endgroup\$ Commented Jul 6, 2016 at 21:15

nth number that multiplies k equals its reverse

Tags: ,

It's quite simple, given n and k, output the nth number such that, if the number is multiplied by k and its digits reversed, it equals the original number. Both input and output are positive numbers.

The challenge originally is from Mego, posted on my broken challenge. Firstly, I used 4 instead of k, but based on my tests, only 1 and 4 values gives output, so I decided to put 4 instead of k, finally I put k back. But the challenge would be ruin with that putting "9"*(n-1) between 2178, so no loopholes will be permitted.

I just posted here for further discussions, suggestions and improvements.

  • \$\begingroup\$ Those numbers are positive right? \$\endgroup\$
    – Fatalize
    Commented Jul 6, 2016 at 7:52
  • \$\begingroup\$ Please add some examples of expected outputs. \$\endgroup\$
    – Fatalize
    Commented Jul 6, 2016 at 7:53
  • \$\begingroup\$ Also you might want to prevent people from hardcoding 2178 in any fashion in their code so that they have to compute the numbers, because it seems they all are of the form 21X...X78 where X...X is a series of nines (except for the first one, which is 0). \$\endgroup\$
    – Fatalize
    Commented Jul 6, 2016 at 7:56
  • \$\begingroup\$ According to the community advises, I'm not allowed to prevent people use methods those work perfectly. \$\endgroup\$
    – Ehsaan
    Commented Jul 6, 2016 at 8:23
  • \$\begingroup\$ Let's wait to see what others think. I personally don't think it's very interesting if people are allowed to hardcode the "format" of those numbers. \$\endgroup\$
    – Fatalize
    Commented Jul 6, 2016 at 8:26
  • \$\begingroup\$ Me neither, I think the challenge isn't interesting at all. \$\endgroup\$
    – Ehsaan
    Commented Jul 6, 2016 at 8:43
  • 1
    \$\begingroup\$ I think there's no good way to prevent hardcoding. Maybe making "4" were an input parameter as well would make solutions actually search for an answer? \$\endgroup\$
    – xnor
    Commented Jul 6, 2016 at 9:01
  • \$\begingroup\$ @xnor You mean make 4 as k input? \$\endgroup\$
    – Ehsaan
    Commented Jul 6, 2016 at 9:33
  • \$\begingroup\$ @Ehsaan Yes, exactly. \$\endgroup\$
    – xnor
    Commented Jul 7, 2016 at 9:09
  • \$\begingroup\$ 9 works too: 1089 * 9 = 9801. \$\endgroup\$
    – Neil
    Commented Jul 10, 2016 at 17:36

Write a program that can determine the median value of a read-only (static, const, immutable) sequence of unsorted numbers (array, list, stream) but minimises storage, without completely sacrificing speed.

The basic bracket is that if we copied all the values into a sorted list and then picked the middle one (or average of the middle pair), it would require storage of the whole sequence, so the storage would be 'n', and the performance would be O(n log n).

The score is the total cost of finding the median of 1 bn values, divided by 1 bn, at a cost of 8 per value stored, 1 per comparison or numerical operation and 1 per read, for the worst case. Thus if our insertion sort costs exactly n*log2(n), the for 1 bn values the total score is 1 for the read, 29.8 for the sort + 8 for the storage, for a total of 37.8.

If instead we skimmed the whole range to get the average (costing 1 for the read and 1 for the summation), we could then only store some portion of the range to sort; but then we would need a second pass to be sure that there were an equal number of values above and below this median (at the cost of another 2).

Lowest score wins, low-level languages (C/C++/D) only so that we can count the actual operations.

  • 3
    \$\begingroup\$ 1. It's not clear to me what counts as a "value stored" or a "read", and I think there are probably gray areas with "comparison or numerical operation" too. (E.g. in C is if (foo) a comparison?) 2. "The score is the total cost ... for the worst case." For any non-trivial algorithm, the full calculation of this score risks being longer than the code. There's a reason that complexity theorists deal with Landau notation rather than exact operation counts. \$\endgroup\$ Commented Jul 7, 2016 at 13:39
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