What is the Sandbox?

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

To post to the Sandbox, scroll to the bottom of this page or click on the "Add Proposal" link below, 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. 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, replace the post here with a link to the challenge and delete it.

See the Sandbox FAQ for more information on how to use the Sandbox.

The Sandbox works best if you sort posts by "active".

Add Proposal

Search the Sandbox

Browse your pending proposals

Get the Sandbox Viewer to view the sandbox more easily

To add an inline tag to a proposal use shortcut link syntax with a prefix: [tag:king-of-the-hill]

| |

2872 Answers 2872

45 46
48 49

Decorate the Christmas Tree



It's a week before Christmas and the your family just bought a Christmas tree. Your little sister has a box of ornaments which she wants you to decorate the tree with.


Write a function or program to randomly decorate the tree with all the ornaments from the box. All possible arrangements of ornaments on the tree must have a non-zero probability of occurrence.


  1. height of the tree n
  2. 'box' of ornaments (a string of printable ascii characters where each character represents an ornament) in which order doesn't matter. Also, you can assume that '#' will not show up as an ornament since it is used to draw bare branches.

Given the number of branches on a tree of size n is (n-1)^2 you can assume that the number of ornaments in the box will be less than or equal to the number of branches on the tree.

Example of a box of ornaments: o = '**@@***@@$$**OOOO....'. Since this string is 21 characters long it can only be valid input for n > 5.


Print out the tree with its decorations

The tree is the following structure of ascii characters where n specifies the height of the tree:

//Tree where n=9                         //Tree where n=6
        #                                        #
       ###                                      ###
      #####                                    #####
     #######                                  #######
    #########                                #########
   ###########                                  |||

Bare branches will always be represented with '#' and the bottom trunk is always ||| while centered with center character of each row.

Examples of output with ornaments:

//n=9, o='X@@X%%%**&&'                   //n=6, o='OO*X*X'
        #                                        O
       ##%                                      O##
      #X###                                    #X#*#
     ###@##%                                  *######
    #&#######                                #####X###
   #@######X##                                  |||



  • Is this too close to a duplicate?
  • Are there any improvements or clarifications required?
| |
  • 2
    \$\begingroup\$ By random, do you mean that every possible distribution of decorations should be equally likely? Or just that the distribution changes with each run? \$\endgroup\$ – frank Dec 17 '19 at 20:10
  • 1
    \$\begingroup\$ What does the tree look like if n is even? That is, how does the trunk center? \$\endgroup\$ – AdmBorkBork Dec 17 '19 at 20:14
  • \$\begingroup\$ Every level of the tree will have an odd number of characters, no matter what the value of n is, so the trunk should remain centered with the top level of the tree. \$\endgroup\$ – Grumpy_Boy Dec 17 '19 at 21:09
  • \$\begingroup\$ What can you put in the box of ornaments? Printable ascii characters? Unicode? Also, what if there's more ornaments than there is area of tree to put them on? \$\endgroup\$ – KeizerHarm Dec 17 '19 at 22:41
  • 1
    \$\begingroup\$ To add to frank's comment: codegolf.meta.stackexchange.com/a/10909/36398 \$\endgroup\$ – Luis Mendo Dec 17 '19 at 23:14
  • 1
    \$\begingroup\$ Regarding "a string of printable ascii and/or unicode characters" You should know that handling unicode (millions of possible characters, from Urdu اُردُو‎ to cuneiform 𒀣) is a lot harder in many golfing languages than printable ascii (which is mostly just the Latin alphabet and some punctuation marks) - to the point where using the tag unicode is advisable if you want the solution to handle unicode characters. If you want the possibility to handle every conceivable tree decoration, you should be more explicit about it than use an and/or construction. \$\endgroup\$ – KeizerHarm Dec 18 '19 at 9:20
  • 1
    \$\begingroup\$ Fair enough. Sticking to printable ascii should be fine then. \$\endgroup\$ – Grumpy_Boy Dec 18 '19 at 22:44
  • \$\begingroup\$ This feels very familiar; I think we may have had it before but, maybe, without the inputs. \$\endgroup\$ – Shaggy Dec 18 '19 at 23:54

For any integer a and any positive odd integer n the Jacobi symbol is defined as follows: $$ \left(\frac{a}{n}\right) = \left(\frac{a}{p_1}\right)^{\alpha_1} \left(\frac{a}{p_2}\right)^{\alpha_2} \cdots \left(\frac{a}{p_k}\right)^{\alpha_k} $$ where $$n = p_1^{\alpha_1}p_2^{\alpha_2}\cdots{p_k^{\alpha_k}}$$ is the prime factorization of n. The Legendere symbol is defined for all integers a and odd primes p as $$\left(\frac{a}{p}\right) = \begin{cases} 0 & \text{if $a \equiv 0 \pmod{p}$},\\ 1 & \text{if $a \not\equiv 0 \pmod{p}$ and for some integer $x$: $a \equiv x^2 \pmod{p}$},\\ -1 & \text{if $a \not\equiv 0 \pmod{p}$ and there is no such $x$}. \end{cases} $$

Your task is to write a function that will take two parameters: a and v, where a is a positive integer, and v is a list of n values. The function should return the index of the first value in v for which the Jacobi symbol is -1. For this challenge, you may assume that all values in the v array are odd primes, and are greater than or equal to a. The length of v will be between 4 and 100 values. If no value in v produces a -1 result for the Jacobi symbol, simply return the length of v.

Because each invocation of the function will execute very quickly, your function will be tested against a wide variety of inputs and the execution time will be summed. This process will be repeated 50 times and the best execution time will serve as your score. My machine is a 2018 Macbook Pro, with the following specs:

CPU: Intel(R) Core(TM) i9-8950HK CPU @ 2.90GHz
GPU: Intel UHD Graphics 630 1536MB
RAM: 32GB 2400MHz DDR4

| |

Even-Odd chunks

(Inspired by the Keg utility of this challenge)

Given an input string, e.g. s c 1= e(a"E"), split the input into even-odd chunks.


This input string, when mapped to its code points, yields the list [115, 32, 99, 32, 49, 61, 32, 101, 40, 97, 34, 69, 34, 41]. When applied modulo-2 for every item, this returns [1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1].

In this list let's find the longest possible chunk that is consistent with even and odd code points:

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

For the first chunk, this yields [1, 0, 1, 0, 1] because this is the longest chunk that follows the pattern

Odd Even Odd Even Odd Even ...


Even Odd Even Odd Even Odd ...

. Adding another codepoint into [1, 0, 1, 0, 1, 1] breaks the pattern, therefore it is the longest possible even-odd chunk that starts from the beginning of the string.

Using this method, we should split the input into chunks so that this rule applies. Therefore the input becomes (the ; here is simply a separator; this can be any separator that is not an empty string):

s c 1;= e(a"E")


  • This is so the shortest solution wins. Let it be known that flags don't count towards being in the pattern. They also don't count towards byte count in this challenge.
| |
  • \$\begingroup\$ Right now, the output specified in the introduction (Longest even-odd chunk) doesn't agree with the example, where you have instead the string split into even-odd chunks. Both would work as challenge, but you have to choose which. \$\endgroup\$ – AlienAtSystem Dec 21 '19 at 12:24
  • \$\begingroup\$ I chose the latter because it is easier to specify. \$\endgroup\$ – user85052 Dec 21 '19 at 12:30

Convert Character Substrings to Numeric

I often have to take character data and categorize it numerically. A common thing I do is to take character type variables and convert them to numeric type characters, keeping same categories according to the level of work I'm doing. (The longer the substring, the more in depth, shorter substrings for broad level). Enough backstory...

The challenge: In as few bytes as possible, convert the input part A, a vector/list of unique strings, into the output, a vector/list of numbers, keeping unique categories within the length of substrings the same length, which is input part B.


  1. w, Vector/list of unique strings of equal character length. n <= 10

    • These strings may be any combination of uppercase letters and numbers. Sorry if it seems my examples follow a pattern, I just created them after a similar pattern I see in the data I work with.
  2. s, where 1 <= s <= n

Output: May take input and output in the same order, or you can convert alphabetically, but output in the same order. See example 2. (I've included comments in my output to clarify, this is not required)

Example Input 1:

#Already alphabetized, but this input is not always guaranteed

s = 3, w = 

Example Output 1:

[1, #ABC
 2, #ABD
 3, #ABE
 4, #ACA
 5, #ACB

Example Input 2: s = 4, w =


Example 2 Output:

#Categorize by order
[1, #X1Z1
 2, #X1Y1
 3, #X1X1
 4, #X1X2
 5, #X2Z1
 6, #X2Z2

OR if conversion follows alphabetical formatting,

#Categorize alphabetically, but output in same order as input.
[6, #X1Z1
 5, #X1Y1
 3, #X1X1
 4, #X1X2
 1, #X2Z1
 2, #X2Z2
| |
  • \$\begingroup\$ I'm a little confused; the input matches the regex [A-Z]+[0-9]+ and the task is to replace the inputs with unique integer identifiers based on the first s letters? Something like as.integer(as.factor(substring(w,1,s)))? I think it's a good challenge but needs a little more clarification. \$\endgroup\$ – Giuseppe Dec 23 '19 at 19:58
  • \$\begingroup\$ Sorry, I suppose I didn't specify how the input will appear. The input is not supposed to follow any specific regex pattern, but some inputs will be similar enough. It could be all uppercase letters or all numbers. \$\endgroup\$ – Sumner18 Dec 23 '19 at 20:24

Partial tq interpreter

In this task you are expected to provide a list output given an input tq program. The tq programs will not contain whitespace inside them. (I find tq extremely difficult to implement within a short time, therefore I consider it to be a nice challenge.)

What is tq, in the first place?

tq is a lazy-evaluated language that is designed with the idea that array items should be accessable during the definition of them. In tq, there is no explicit separator of an array, only special arrangements of monadic/dyadic functions and nilads.

The following program is a program printing [123] (Pretend that tq doesn't support strings because we aren't dealing with them in this case):


This defines a list with the first item being the number 123, after which all items in the list will be outputted inside a list.

In tq, numbers are supported to allow multiple-digits. So this defines a list with 2 items:


In this test case, you are expected to output the list [24,5]. Let's explain it step by step.

12+12   # This evaluates 12 + 12 in the current item in the list, returning 24
     ,  # A separator. This separates two items when they could be potentially
        # ambiguous when they are applied without a separator.
      5 # This evaluates 5 in the current item in the list, returning 5
        # The comma is simply a no-op that doesn't require parsing.

So you think that tq is not hard at all to implement? Well, remember that tq also has a special feature of accessing the items in an array before the array is defined!


We introduce two new atoms:

  • t (tail) means access the last item in the list
  • h (head) means access the first item in the list

Therefore our list is going to yield:


Now take a look at this program:


We introduce 2 more atoms:

  • p Yield the next item before (previous) the current position
  • s Yield the next item after (succeeding)the current position

This yields the list:


A quick reference of the tq language

Just assume that you only have two operands for the operators.

  • [0-9] starts a number. Numbers will only be positive integers, i.e. no decimals and negative numbers.
  • , This is a separator of different items when it is given that two consecutive indexes will be ambiguous with each other without a separator. In tq all of the remaining characters can act as a separator, but in this case it is a good idea to implement only , for the ease of your implementation.
  • + and * These are arithmetic operators. Usually in tq, they may be applied multiple times, e.g. 1+2+3, but in your implementation, input will be provided so that this will not happen and only 1+2 will happen (there will not be applications multiple times).
  • t return the last item in the list. If the code is t1,2,3 it shall return [3,1,2,3].
  • h return the first item in the list. If the code is 1,2,3h it shall return [1,2,3,1].
  • p returns the item before the current item. If the code is 0,1,p,3,4 the code shall return [0,1,1,3,4].
  • s returns the item after the current item. If the code is 0,1,s,3,4 the code shall return [0,1,3,3,4].

More test cases

  • 4p*p will yield [4,16]
  • 1p+s2 will yield [1,3,2]
  • 1,2,3h+t4,5,6 will yield [1,2,3,7,4,5,6]
  • 3ppss6 will yield [3,3,3,6,6,6]
  • You also have to implement multiple hops. E.g. 1th should yield [1,1,1]
  • If you know that something is going to form a loop, e.g. 1sp2, the cells that form the loop should be removed. Therefore the previous example will yield [1,2].
  • Out of bounds indexing will yield the closest index of the indexed item that is a number. E.g. 1,2s should yield [1,2,2]
| |

Internal Truth Machine

It's a normal truth machine but instead of taking input, it uses the first character of the program. Thus, internal.

Example: 0abcd prints 0 and halts, and 1abcd prints 1 infinitely.

| |

Balanced interval tree

You will write a balanced, online interval tree.

  • A function, method, or procedure that performs insertion into a query of intervals that overlap with an interval over a, b in O(log n) time, given that a is less than b, including:
    • intervals that fully enclose a, b
    • intervals fully within a, b
    • intervals that start to the left of a but end before b
    • intervals that start to the right of a but end after b
  • A function, method, or procedure that performs insertion ov an arbitrary interval a,b into a data structure in O(log n) time
  • Both O-bounds must hold after arbitrary series of insertions and queries


If you write non-method functions, your submission may take the data structure as a global variable or receive the data structure as the first parameter.

You must informally prove your submission falls within the required O-bounds or name the data structure your program implements. If the name of the data structure you are implementing is obscure, you may name the paper of the data structure.

This is , so the submission with the shortest length in bytes wins.

| |

Posted: Find the Inverse Neighbor Pairs

| |

Recolour my Table (WIP)

I have an oak table. It's a very nice oak table. But recently, I've noticed that my table has been getting some sort of white patches on it:

Insert picture of my table here.

Now, I like my table, and I don't want my table getting all dirty. I'm asking y'all to fix this for me, so that my table will once again be clean.

The Challenge

Given a picture of the table in question, replace all pixels that are white with the correct table colour. I know I'll need to specify this area better, but I don't know how to. Suggestions are welcome.




This is , so the answer with the fewest amount of bytes wins.

| |
  • \$\begingroup\$ Maybe try limiting the table format. Is it top down or sideways? Also can we assume that the original table color (or slight variation to a certain HSL difference, maybe) is the majority of the image? \$\endgroup\$ – Roman Gräf Feb 6 at 19:50

floating-point error matters

Write a expression of floating-point numbers in any languages. When calculating the expression without floating-point errors (as what a human do), it should be 0. But with floating-point errors (as what happened in your language), it yield 1 instead.

  • Floating-point numbers are some numbers which store a finite number digits (binary or decimal or in any other bases) of fraction, plus an exponent in computer. It may be IEEE 754, but not must be.
  • Loss of precision due to integer types (which do not has a exponent) are not allowed in the expression. You are still allowed to include integers (or even other types) in your expression as long as operations such as rounding into an integer are not the root of errors.

Shortest codes win as code-golf.


  • Is there any duplicates here?
  • Is asking for an expression instead of full program allowed?
| |
  • \$\begingroup\$ Yes, you can allow expressions. Good fit for this one, imho. \$\endgroup\$ – Adám Feb 6 at 11:33

No title yet.

By Zekendorf's theorem every non-negative integer has a unique representation as the sum of Fibonacci numbers where no two numbers coincide or are adjacent.

In The minimum fibonacci challenge! the challenge was to output the list of Fibonacci numbers. However, you can instead consider the list of coefficients of the sum \$ \small x_0F(2) + x_1F(3) + x_2F(4) + \ldots \$ (since \$ \small F(0) = 0 \$ and \$ \small F(1) = F(2) \$ never appear in the Zekendorf representation) and represent that as a binary number \$ \ldots x_2 x_1 x_0 \$, e.g. \$ \small 67 = \small 1F(2) + 0F(3) + 1F(4) + 0F(5) + 1F(6) + 0F(7) + 0F(8) + 0F(9) + 1F(10) \$ which we can represent using the binary number \$ \small 100010101 \$ or \$ \small 277 \$ in decimal.

We can readily convert the binary representation back into the original integer by calculating the sum of the relevant Fibonacci numbers. However, I would like you to, given an input integer \$ \small n \$, output the decimal integer whose binary representation encodes the Zekendorf representation of \$ \small n \$ in this way.

This is , so the smallest function or program that breaks no standard loopholes wins!

| |
  • 1
    \$\begingroup\$ I would clarify that your binary representation is "flipped", i.e. 0-extends infinitely to the left. \$\endgroup\$ – Jonathan Frech Feb 8 at 18:32
  • \$\begingroup\$ @JonathanFrech flipped in comparison to what? \$\endgroup\$ – RGS Feb 10 at 20:18
  • \$\begingroup\$ @RGS I wrote the Fibonacci coefficients from left to right but the bits in the binary number from right to left and it wasn't so clear before. \$\endgroup\$ – Neil Feb 10 at 21:15
  • \$\begingroup\$ Oh ok, but that still matches what we don in binary. I can also write 11 = 1 + 2 + 8 and \$11 = 1011_2\$ :) but sure, going for perfect clarity is better than trusting other people's common sense \$\endgroup\$ – RGS Feb 10 at 21:54
  • 2
    \$\begingroup\$ @RGS Having to guess the specifications from a single example is not really common sense as it is annoying. \$\endgroup\$ – Jonathan Frech Feb 11 at 21:41

Calculate Mahjong points

Introduction to Mahjong tiles

Mahjong (麻雀) is a board game that originates from China. Mahjong tiles used in this challenge are in Unicode points U+1F000 – U+1F021:


They are categorized as:

  • Winds (風牌): 🀀(East Wind), 🀁(South Wind), 🀂(West Wind), 🀃(North Wind)

  • Dragons (三元牌): 🀄(Red Dragon), 🀅(Green Dragon), 🀆(White Dragon)

  • Ten Thousands (萬子): 🀇(One of the Ten Thousands) through 🀏(Nine of the Ten Thousands)

  • Bamboos (索子): 🀐(One of the Bamboos; note that it depicts a bird) through 🀘(Nine of the Bamboos)

  • Circles (筒子): 🀙(One of the Circles) through 🀡(Nine of the Circles)

  • Winds and Dragons are together called Honors (字牌).

  • Ten Thousands, Bamboos and Circles are each called a suit and together called Numbers (数牌). They have ranks from 1 to 9.

  • Among Numbers, the Ones and the Nines are called Terminals (老頭牌). The Twos through the Eights are called Simples (中張牌).

  • Honors and Terminals are together called Orphans (幺九牌).

Every Mahjong tiles have 4 copies of themselves.

Mahjong Hand

A legal Hand will be given as the input. It consists of:

  • A Head, which is 2 identical tiles.

  • 4 Bodies, each which is either:

    • A Sequence, which is 3 consecutive Numbers of the same suit.

    • A Triplet, which is 3 identical tiles.

Hence, a Hand consists of 14 tiles.

There are exceptions. See "Seven Heads" and "Thirteen Orphans" below.


The objective is to calculate how many points a Hand has. Points are gained by having Yakus (役), which are cumulative.

Also I apologize that I named these Yakus to be more intuitive for the code golf, making them differ from the 'usual' names. Sorry!

1 Point Yakus

  • Menzen Tsumo (門前清自摸和), which is worth 1 point, will always be assumed.

  • Pinfu (平和): Get a non-Dragon Head and 4 Sequences.

    • Example: 🀃🀃🀇🀈🀉🀑🀒🀓🀔🀕🀖🀟🀠🀡
  • One Pair (一盃口): Get 2 identical Bodies.

    • Example: 🀂🀂🀂🀇🀇🀇🀈🀉🀓🀓🀔🀔🀕🀕
  • Dragon Triplet (役牌): Get a Triplet of Dragons. Multiple Dragon Triplets are cumulative.

    • Example: 🀆🀆🀆🀇🀈🀉🀊🀊🀒🀓🀔🀛🀜🀝
  • All Simples (断幺九): Let the Hand consist only of Simples.

    • Example: 🀈🀉🀊🀓🀓🀓🀚🀛🀛🀜🀜🀝🀠🀠

2 Points Yakus

  • Seven Heads (七対子): Get 7 Heads. It cannot contain 2 identical Heads.

    • Example: 🀀🀀🀂🀂🀆🀆🀈🀈🀊🀊🀐🀐🀛🀛
  • Colorful Sequences (三色同順): Get a sequence for each suit, consisting of the same set of ranks.

    • Example: 🀀🀀🀁🀁🀁🀉🀊🀋🀒🀓🀔🀛🀜🀝
  • Full Sequence (一気通貫): Get 3 sequences of the same suit, consisting of ranks of 123, 456, and 789.

    • Example: 🀀🀀🀁🀁🀁🀇🀈🀉🀊🀋🀌🀍🀎🀏
  • Semi-Orphans (チャンタ): Let all Heads and Bodies have at least 1 Orphan.

    • Example: 🀀🀀🀁🀁🀁🀇🀈🀉🀐🀑🀒🀡🀡🀡
  • Concealed Three (三暗刻): Get 3 Triplets.

    • Example: 🀀🀀🀁🀁🀁🀊🀊🀊🀋🀋🀋🀑🀒🀓
  • Colorful Triplets (三色同刻): Get a Triplet for each suit, consisting of the same rank.

    • Example: 🀀🀀🀉🀊🀊🀊🀊🀋🀓🀓🀓🀜🀜🀜
  • Dragons Minor (小三元): Let the Head and 2 Bodies consist of Dragons. Cumulative with the 2 Dragon Triplets.

    • Example: 🀄🀄🀅🀅🀅🀆🀆🀆🀊🀋🀌🀒🀓🀔
  • All Orphans (混老頭): Let the Hand consist only of Orphans. Supersedes Semi-Orphans. Cumulative with Seven Heads.

    • Example: 🀁🀁🀂🀂🀃🀃🀆🀆🀇🀇🀏🀏🀘🀘
  • Three of a Kind (一色三順): Get 3 identical Bodies. Supersedes One Pair.

    • Example: 🀅🀅🀇🀇🀇🀈🀈🀈🀉🀉🀉🀙🀚🀛 (Note that this example also presents Semi-Orphans, which makes it supersede Concealed Three.)

3 Points Yakus

  • Two Pairs (兩盃口): Get 2 pairs of identical Bodies. Supersedes One Pair and Seven Heads.

    • Example: 🀃🀃🀇🀇🀈🀈🀉🀉🀑🀑🀒🀒🀓🀓
  • Semi-Terminals (純チャンタ): Let all Heads and Bodies have at least 1 Terminal. Supersedes Semi-Orphans.

    • Example: 🀇🀇🀇🀈🀉🀍🀎🀏🀏🀏🀏🀙🀙🀙
  • Semi-Flush (混一色): Let the Numbers consist of a single suit.

    • Example: 🀂🀂🀇🀈🀉🀉🀉🀉🀊🀋🀌🀌🀍🀎

6 Points Yaku

  • Flush (清一色): Let the Hand consist of a single suit of Numbers. Supersudes Semi-Flush.

    • Example: 🀇🀇🀇🀈🀉🀉🀉🀉🀊🀋🀌🀌🀍🀎


Yakumans (役満) worth 13 points, and supersede all Yakus above. Multiple Yakumans are cumulative.

Without Yakumans, the points are capped at 13.

  • Concealed Four (四暗刻): Get 4 Triplets.

    • Example: 🀀🀀🀉🀉🀉🀌🀌🀌🀒🀒🀒🀞🀞🀞
  • Thirteen Orphans (国士無双): Collect all 13 Orphans, plus an additional Orphan.

    • Example: 🀀🀁🀂🀃🀄🀅🀆🀇🀇🀏🀐🀘🀙🀡
  • Nine Gates (九蓮宝燈): Get Flush of ranks of 1112345678999, plus an additional Number of the same suit.

    • Example: 🀇🀇🀇🀈🀉🀊🀋🀋🀌🀍🀎🀏🀏🀏
  • All Greens (緑一色): Let the Hand consist only of Green Dragons and the Twos, Threes, Fours, Sixes, and Eights of the Bamboos.

    • Example: 🀅🀅🀅🀑🀑🀒🀒🀓🀓🀕🀕🀕🀗🀗
  • All Honors (字一色): Let the Hand consist only of Honors. Cumulative with Concealed Four.

    • Example: 🀀🀀🀀🀁🀁🀂🀂🀂🀅🀅🀅🀆🀆🀆
  • All Terminals (清老頭): Let the Hand consist only of Terminals. Cumulative with Concealed Four.

    • Example: 🀇🀇🀇🀏🀏🀏🀐🀐🀐🀘🀘🀙🀙🀙
  • Dragons Major (大三元): Get a Triplet for each of Dragons.

    • Example: 🀄🀄🀄🀅🀅🀅🀆🀆🀆🀇🀈🀉🀑🀑
  • Winds Minor (小四喜): Let the Head and 3 Bodies consist of Winds.

    • Example: 🀀🀀🀁🀁🀁🀂🀂🀂🀃🀃🀃🀖🀗🀘
  • Straight Flush (連七対): Get Seven Heads with Numbers with consecutive ranks of the same suit.

    • Example: 🀚🀚🀛🀛🀜🀜🀝🀝🀞🀞🀟🀟🀠🀠
  • Four of a Kind (一色四順): Get 4 identical Bodies.

    • Example: 🀃🀃🀓🀓🀓🀓🀔🀔🀔🀔🀕🀕🀕🀕
  • Winds Major (大四喜): Worths 2 Yakumans. Get a Triplet for each of Winds. Cumulative with Concealed Four.

    • Example: 🀀🀀🀀🀁🀁🀁🀂🀂🀂🀃🀃🀃🀖🀖
  • The Septentrions (大七星): Worths 2 Yakumans. Get a Head with each of Honors. Supersedes All Honors.

    • The only example: 🀀🀀🀁🀁🀂🀂🀃🀃🀄🀄🀅🀅🀆🀆

Other rules about Mahjong

When a Hand can be interpreted as different combinations of Heads and Bodies, the combination with the most points will be chosen. (See Three of a Kind and Two Pairs above)


🀅🀅🀋🀋🀌🀌🀍🀍🀑🀒🀓🀟🀠🀡 : Menzen Tsumo + One Pair = 2 points.

🀉🀊🀋🀌🀍🀎🀑🀒🀓🀔🀔🀞🀟🀠 : Menzen Tsumo + Pinfu + All Simples = 3 points.

🀀🀀🀀🀁🀁🀁🀂🀂🀂🀃🀃🀃🀆🀆 : Winds Major + Concealed Four + All Honors = 52 points. (Most points possible)

Rules about code golf

  • Input type and format doesn't matter, but it must consist of the Unicode characters above. In C++, valid examples include std::u8string (sorted or not) and std::multiset<u32char_t>.

  • Output type and format doesn't matter either.

  • Invalid Hands (not exactly 14 tiles, contains 5 copies of the same tile, etc) fall into don't care situation.

  • If and only if your language doesn't support Unicode, use the following table to parse Mahjong tiles:

| |
  • \$\begingroup\$ A related, more golfable challenge posted here: codegolf.stackexchange.com/q/199202/89459 \$\endgroup\$ – Dannyu NDos Feb 10 at 6:30
  • \$\begingroup\$ Maybe you need more testcase. At least one per yaku. \$\endgroup\$ – tsh Feb 10 at 14:40
  • \$\begingroup\$ Should output for 11112233778899m be 13 or 14? \$\endgroup\$ – tsh Feb 16 at 7:02
  • \$\begingroup\$ @tsh Since it has no Yakuman, its point is capped at 13. \$\endgroup\$ – Dannyu NDos Feb 16 at 7:06
  • \$\begingroup\$ Is Straight Flush (連七対) required to be All Simples (断幺九)? \$\endgroup\$ – tsh Feb 16 at 7:15
  • \$\begingroup\$ @tsh No. 小車輪, 大車輪, and their correspondent to other suits are all Straight Flush. \$\endgroup\$ – Dannyu NDos Feb 16 at 7:17

Be second to last

This is a challenge with a game with a clear winning strategy, but the winner is not the one who comes in first, but the one who comes in second to last! While losing is easy (just throw every game), just barely not losing is quite a task.

The Game

The bots will be made to play games of normal Nim against each other. The rules are as follows:

  • The board consists of a list of unsigned integers
  • In alternating turns, the players reduce one non-zero element by an amount of their choosing, as long as the element is not negative afterwards
  • The player who reduces the list to all entries being 0 wins

The length of the Nim list and its entries will be randomly determined for each game.

The Tournament

All bots will compete against each other in a Danish-Style tournament:

  • Initially, players are assigned positions in a list randomly
  • After each round, the list is sorted by the amount of wins with an order-preserving method
  • In each round, the players with the odd numbers play against those with the following even numbers (so 1st against 2nd, 3rd against 4th, and so forth)
  • If the number of players is odd, the one in the middle position has a bye and is given 1 win without playing.
  • The tournament ends after \$\lceil \log_{2}(N_{Players}) \rceil\$ (Binary logarithm of the number of players, rounded up) rounds.

The player in the second to last position of the list counts as the total winner of the tournament.

King of the Hill

Each time the contest is run, 100 tournaments are played. The bots with the highest number of being second to last is the overall winner.

Challenge Rules

  • Each bot is a Python 3 class implementing the following functions:
    • __init__(ID, n) passes the bot its randomly assigned ID for this tournament, and the total number of players.
    • nim(self,list) which takes in the Nim board state and returns a tuple (index, amount), specifying from which list index to subtract what number. This function is repeatedly called during each game of Nim played, until a winner has been determined.
    • rank(self,IDs,scores) which takes in the current order of ranking in the tournament, and the list of scores of each bot, ordered by ID. It returns nothing. This will be called for each bot after each round of tournament, as well as before the first round, to provide the ranking information if the bot requires it.
  • Bots are explicitly allowed to implement further functions and store data for private use. Bots will only be deleted and re-initialized after each full tournament.
  • Programming meta-effects are forbidden, meaning any attempts to directly access other bots' code, the Controller's code, causing Exceptions or similar. Any bot doing so is disqualified until fixed.
  • The following will also be set up to cause Exceptions:
    • nim returning an index of which the element is already 0
    • nim return an amount larger than the element at that index
  • Other languages are allowed only in case they can be easily converted to Python 3.
  • Class names have to be unique
  • Multiple bots per person are allowed, but only the latest version will be taken of iteratively updated bots.
  • As per Standard Loopholes, copies of bots are not allowed. This includes bots who differ from other bots only by a trivial change in strategy (e.g. a change in the pseudorandom seed).

Controller and Examples

Watch this space.

| |
  • \$\begingroup\$ this sounds fun! I have no idea how it fits this community, but I can't wait to see this in the main site! \$\endgroup\$ – RGS Feb 11 at 16:04
  • \$\begingroup\$ @RGS king-of-the-hill challenges are common -- not all challenges have to be code-golf. \$\endgroup\$ – S.S. Anne Feb 11 at 23:55
  • \$\begingroup\$ @S.S.Anne I know not everything has to be code golf, I'm just saying that in the little time I've been active, I've never seen a KotH challenge \$\endgroup\$ – RGS Feb 12 at 7:01
  • \$\begingroup\$ Me neither. This sounds great! \$\endgroup\$ – PkmnQ Mar 11 at 5:21

Find the most important character

The challenge is to write a program that outputs the most occurrent non-trivial* character in the input string, excluding these bracketing characters ()[]{}. A solution is scored by feeding the program's source into its input. Bracketing characters are allowed in the program, but may not be selected by the program as the most occurrent character. In cases where letters have the same percentile score, any of the characters can be returned, or all of them.

Write a program that has a maximal percentage of a single non-trivial* character, excluding these bracketing characters ()[]{}. Bracketing characters are allowed, but may not be selected as the highest-percentage character. Answer is scored as by running the program, with the program's source as its own input. In cases where letters have the same percentile score, any of the characters can be returned, or all of them.

*Non-trivial is defined in this case to be a character that contributes to the functionality of the program. If the character can be removed without influencing how the program runs, it is a trivial character.


Score is determined as follows:

$$(\frac{n_{char}}{n_{total}}) * 100\% $$

With nchar being the number of occurrences of that character in the input, and ntotal being the total number of non-trivial characters in the input. With this scoring criteria, all scores should be within the range (0,100]

Highest scoring solution per language wins.


Solutions should take in any valid string as input. Scoring is done using the program's source code as input(with all non-trivial charcters removed).


A single character with the highest percentage occurrence. Each solution should be capable of outputting standard ASCII, as well as the language's codepage (if applicable).


Creative solutions are encouraged. Boring solutions with long padded strings substituted in for numbers are discouraged. Standard loopholes are disallowed.









a or b or ab

| |
  • \$\begingroup\$ Since I don't see it mentioned, you may want to add we should aim for an as high as possible percentage as score. \$\endgroup\$ – Kevin Cruijssen Feb 12 at 8:25
  • \$\begingroup\$ And if I understand it correctly the input is the only possible input for the program? So it won't have to work for other inputs as well (outputting the 'most important' character of that input). \$\endgroup\$ – Kevin Cruijssen Feb 12 at 8:36
  • 1
    \$\begingroup\$ So something like '?? in 05AB1E (push "?" and output it, ignoring the input) would score 66.7%? \$\endgroup\$ – Kevin Cruijssen Feb 12 at 9:01
  • 1
    \$\begingroup\$ You are correct on aiming for the highest score. My intention was that the code should work on any inputs, however scoring is only done on the program source itself. \$\endgroup\$ – JPeroutek Feb 12 at 11:14
  • \$\begingroup\$ Ah ok, that indeed seems better. You may want to clarify that a bit, since at the Input section it now only mentions "The program's source code", hence my confusion. \$\endgroup\$ – Kevin Cruijssen Feb 12 at 11:44
  • \$\begingroup\$ You’re right, that bit is a tad vague. I will update when I get to work. \$\endgroup\$ – JPeroutek Feb 12 at 11:46
  • 1
    \$\begingroup\$ @KevinCruijssen I've edited the challenge to reflect these changes. Does this better reflect the challenge? \$\endgroup\$ – JPeroutek Feb 12 at 13:39
  • 1
    \$\begingroup\$ Yep, that's indeed a lot clearer! :) One more question: can the input potentially contain any unicode character, or is it limited to just printable ASCII and/or the language's used codepage? \$\endgroup\$ – Kevin Cruijssen Feb 12 at 13:41
  • \$\begingroup\$ Hmm, well if the character isn't on the languages codepage then it likely wouldnt be used in a solution right? I'd think that any character that the program can successfully output would be valid. \$\endgroup\$ – JPeroutek Feb 12 at 13:43
  • 1
    \$\begingroup\$ "I'd think that any character that the program can successfully output would be valid." This is different than having to support the entire codepage or extended ASCII, though. I.e. an answer in Unary/Lenguage using a single character and printing that single used character would score 100%, since the only possible input is any amount of that chosen character. And an answer in Whitespace would only have to support spaces, tabs, and newlines as potential input in that case. If that's your intention than it's fine by me, but it's different than allowing the entire codepage as input. :) \$\endgroup\$ – Kevin Cruijssen Feb 12 at 14:06
  • 1
    \$\begingroup\$ Ahh I see the distinction you are making. I'd say that the program should at least be able to printout standard ASCII, as well as the languages codepage. Does that sound reasonable? \$\endgroup\$ – JPeroutek Feb 12 at 14:10
  • \$\begingroup\$ Yep, that indeed sounds as a good solution. \$\endgroup\$ – Kevin Cruijssen Feb 12 at 14:27

Shift the digits

Here, x (supplied as input) and n (the result of your computation) are both positive integers. n * x = n shifted. Find n.

Here's an example of shifting:

123456789 -> 912345678
abcdefghi -> iabcdefgh (letters = any 0~9 digit)
123       -> 312


  • Preceding zeros count after shifting. If the number is 10 and is multiplied by 0.1 (0.1 isn't a valid input), the result is 1, which isn't equal to 01 (10 after shifting).
  • Your code has to run on Try It Online without timing out.
  • If your number only has one digit, the shifted result is your number:
1 -> 1
4 -> 4
9 -> 9

Test cases

Just to show that it's possible ...

9 -> 10112359550561797752808988764044943820224719
(In this test case, x = 9 and n = 10112359550561797752808988764044943820224719.
n shifted = n * x =              91011235955056179775280898876404494382022471)

Don't believe it? Try it online.

| |
  • \$\begingroup\$ Is this the point of the challenge: "you hardcode any number you want in your code, I give you a number as input, you multiply your number by mine and right shift it once"? \$\endgroup\$ – Lyxal Feb 16 at 7:29
  • \$\begingroup\$ @Lyxal The challenge should be self-explanatory. Can you understand it now? \$\endgroup\$ – Third-party 'Chef' Feb 16 at 10:01
  • \$\begingroup\$ Aha! So it's "x * n = x shifted, find n". I get it now! \$\endgroup\$ – Lyxal Feb 16 at 10:26
  • \$\begingroup\$ I feel there is something I'm missing. Are you going to give as input only numbers that have a multiple that corresponds to said shifting? \$\endgroup\$ – RGS Feb 16 at 18:06
  • \$\begingroup\$ I've been doing some thinking and I believe that asking to find the smallest possible integer is impossible for numbers which aren't all the same digit. Some test cases would be helpful. \$\endgroup\$ – Lyxal Feb 16 at 20:33
  • \$\begingroup\$ I've seen a really large integer (in a book) that shifts itself right once after being multiplied by 9. If that's really neccecary I'd try to type it here. \$\endgroup\$ – Third-party 'Chef' Feb 17 at 2:46
  • \$\begingroup\$ @Lyxal It's possible, see my test case. \$\endgroup\$ – Third-party 'Chef' Feb 17 at 3:40
  • \$\begingroup\$ @RGS I'm going to give any number that has a multiple; after division of that multiple by that number, the result will be the reverse of the shifting of the number. (I've changed my challenge since Lyxal "got" the formula.) \$\endgroup\$ – Third-party 'Chef' Feb 17 at 3:43

Posted here

| |

Approximate Alternating Triangles

| |
  • \$\begingroup\$ I think your added explanation is good, though I admit I still don't see how the resulting image is a triangle. It may just be me, but maybe it would be easier to see if provided a drawing? \$\endgroup\$ – FryAmTheEggman Mar 2 at 0:34
  • \$\begingroup\$ Ah, I see, but why \` (sorry comment markdown is messing that up, just the one backslash...) on the left and not |? You could also try naming it "approximate alternating triangles" or something. \$\endgroup\$ – FryAmTheEggman Mar 2 at 0:41
  • \$\begingroup\$ look less like a triangle Why do you want to make it look _less than a triangle? Why not | to make it look more like a triangle? \$\endgroup\$ – Luis Mendo Mar 4 at 11:45

Japanese Encoding Conversion

| |
  • \$\begingroup\$ I think switching between hex and decimal in the definitions is more likely to cause confusion than sticking to just one of them. I think you also don't indicate that the tuples you define are bytes that are concatenated to the final result. I'm not sure if I'm misunderstanding, but doesn't the clipping mean converting from shift JIS is ambiguous? Separately, you can use \left and \right to make your brackets the right height. \$\endgroup\$ – FryAmTheEggman Mar 1 at 18:40
  • \$\begingroup\$ @FryAmTheEggman For the Shift_JIS case, the range used is [0x81-0x9F,0xE0-0xEF] for the first byte and [0x40-0x7E, 0x80-0xFC] for the second byte, which is of size precisely 8,836. Since 区 won't exceed 128, and the starting points of 点 in both cases are separated by 95, the first case won't overlap with the second. \$\endgroup\$ – Shieru Asakoto Mar 2 at 3:24
  • \$\begingroup\$ @FryAmTheEggman I wrote kuten in decimal and bytes in hexadecimal because they were so defined; edited to only show in decimal, but keep accepting both decimal and hexadecimal input/output form. \$\endgroup\$ – Shieru Asakoto Mar 2 at 3:36
  • \$\begingroup\$ Ah yes, I see where I made an error now. I think the rewrite you did makes it easier to read. I don't see anything else, though of course I don't speak for everyone. Good luck! \$\endgroup\$ – FryAmTheEggman Mar 2 at 20:06
  • \$\begingroup\$ Typo: monas should be moras. \$\endgroup\$ – Grimmy Mar 6 at 14:24

| |
  • 1
    \$\begingroup\$ What happens if all integers are equally probable? Is that answer disqualified? Or does it score infinity? I also take it that there is a factor of randomness here, so perhaps you could mention that. \$\endgroup\$ – Lyxal Feb 14 at 6:21
  • 1
    \$\begingroup\$ Of course, I may have missed the mark completely. Are we required to create our own scale of randomness (i.e. our own deviation from uniform randomness if that makes sense)? \$\endgroup\$ – Lyxal Feb 14 at 6:22
  • 1
    \$\begingroup\$ If all integers are equally probably, wouldn't that give a score of effectively 0? (probability of 1/Integer.Max for the "most frequently occuring", which could be any of the integers)? \$\endgroup\$ – simonalexander2005 Feb 14 at 12:08
  • 3
    \$\begingroup\$ @Lyxal It is mathematically impossible for all integers to have equal probability. \$\endgroup\$ – Ad Hoc Garf Hunter Feb 14 at 14:07
  • 3
    \$\begingroup\$ This seems easy to score arbitrarily close to zero. For example, keep generating a random number from 1 to N until you get a 1, and count how many tries it takes before you succeed. To include negative outputs, pick a random sign. Make N large and the score tends to 0. \$\endgroup\$ – xnor Feb 14 at 14:45
  • \$\begingroup\$ @xnor There is a byte limit which stops "arbitrarily close" formulations. Did you miss this or am I missing what you are suggesting? \$\endgroup\$ – Ad Hoc Garf Hunter Feb 14 at 15:28
  • 4
    \$\begingroup\$ You're right, the byte limit stops you from actually getting arbitrarily close to 0. But, I expect there to be plenty of room in 100 bytes to stuff in some ginormous number bust-beaver style, at least in fairly compact languages. So, I think this challenge will mostly come down to "what's the biggest number you can express in 100-X bytes" where X is however many bytes the random part takes. \$\endgroup\$ – xnor Feb 14 at 15:33
  • \$\begingroup\$ My thought is that there are a number of possible things that you would be able to use as the random part each with different drawbacks and strengths. In my mind the challenge is sort of coming up with the random part that best fits the language. There is definitely going to be a big number component to this that is important, but since that has mostly been explored in other challenges I would expect borrowing there. \$\endgroup\$ – Ad Hoc Garf Hunter Feb 14 at 15:36
  • 1
    \$\begingroup\$ What do you think then about the challenge being just to generate a random integer so that each one has nonzero probability? I think that as is, the big number generation is so important to the score that it makes this a chameleon challenge. \$\endgroup\$ – xnor Feb 14 at 15:42
  • \$\begingroup\$ Alternatively, the "big number" N could be an input, and you stipulate that as N grows, the maximum chance of any integer to be chosen must approach zero. \$\endgroup\$ – xnor Feb 14 at 15:43
  • \$\begingroup\$ @xnor I think that I am pretty happy with the challenge as is. I prefer the current challenge over the suggested challenges since this one I feel has an incentive to loosen the distribution whereas I do not feel the others do. The big number generation is important, but I consider that as a more general technique. \$\endgroup\$ – Ad Hoc Garf Hunter Feb 14 at 15:48
  • \$\begingroup\$ Maybe I'm missing something about loosening the distribution. I'm envisioning solutions in a golfing language to be about 7 bytes of "take a number N and make a flat random distribution based on it" and 93 bytes of "generate a huge N". For regular languages with a random() function, maybe split it 25 bytes / 75 bytes. Maybe I'm missing something, but I can't really see a viable strategy that isn't effectively the N-input challenge I suggested but with big number generation code copy-pasted in for N. Do you think there is? \$\endgroup\$ – xnor Feb 14 at 15:59
  • \$\begingroup\$ Let us continue this discussion in chat. \$\endgroup\$ – xnor Feb 14 at 16:07
  • 1
    \$\begingroup\$ I would allow for some entropy input to give pure languages a chance. \$\endgroup\$ – Jonathan Frech Feb 25 at 1:16
  • \$\begingroup\$ @JonathanFrech I would like to but it is a little difficult for entropy to make it work. If you have a concrete idea for how I might fairly introduce entropy, I am open to ideas. \$\endgroup\$ – Ad Hoc Garf Hunter Feb 25 at 1:35

Making Minimial Memory Masterpieces

In this challenge you will be asked to write a small computer program to paint an approximation of the painting Fine Wind, Clear Morning.

Your submission

In this challenge we are going to start with a blank canvas. We are going to add an "ant" in the top left corner. You will be telling the ant how to paint the picture by writing its brain.

An ant is a simple creature, at a given time it knows two things.

  • Some memories that it has

  • The color of the pixel it is currently standing on

And at every step the ant consults these two things it knows and then

  1. Draws one pixel of any color where it is standing

  2. Moves one pixel in a cardinal direction

  3. Replaces its all of its memory

The ant can thus be thought of as a function which takes a color and a value and spits out a color, a value and a cardinal direction.

For example here is the brain of an ant that draws a zigzag pattern in red

\$ f(c,m) = \left\{\begin{matrix}(\mathrm{Red}, & \mathrm{South}, & 1) & \mathrm{if} & m = 0\\ (\mathrm{Red}, & \mathrm{East}, & 0) & \mathrm{if} & m = 1\end{matrix}\right. \$

However our canvas is not infinite, so this ant would run into a border eventually. We will stop the ant if it tries to move off, by canceling its move and leaving it on the square it is on. You are free to use this behavior to your advantage.

The one issue here is that currently the ant will never stop, it will just keep painting forever. Which is why ants come with a builtin kill switch. When an ant's memories are equal to an exact value the ant explodes or something, ending the drawing.

Your submission will thus consist of 3 things

  • The starting memory for your ant

  • A description of the ant's function (more on this later)

  • The ending memory for your ant


The goal of this challenge is to have the ant that requires the least memory to operate. We will count this by the number of different states your ant's memory can have.

Thus to score your answer you should run it on the canvas provided. Once the ant has finished your score will be the total number of distinct memories used by your ant through the process. We will include the initial memory even if it does not appear again, but you should not count the ending memory (the one that kills the ant).

The lowest score will be the winner.

The painting

The painting will be a low resolution version of the one from the wikipedium. It will use 3 bit color resolution.

I will decide on the exact specific sizes and make the image in a bit.

A valid answer must produce this image exactly when run on the canvas.

Verification Tool

I will make a tool for running and verifying programs, using a standardized format. It will be runnable in browser.

| |
  • 1
    \$\begingroup\$ I assume based on the scoring the ant is invalid if it fails to paint exactly the required picture? \$\endgroup\$ – FryAmTheEggman Mar 9 at 20:37
  • \$\begingroup\$ @FryAmTheEggman Yes thank you. That also reminded me of another thing I forgot to say in the challenge. \$\endgroup\$ – Ad Hoc Garf Hunter Mar 9 at 20:47
| |
  • 2
    \$\begingroup\$ Interesting challenge, looks solid. Will there always be at least one empty column between lines or lane dividers, as seems to be in all examples? \$\endgroup\$ – xnor Mar 10 at 11:08
  • 1
    \$\begingroup\$ With your new clarification, can there be two adjacent lines? Also, can there be an input with just one line, or just empty space? Are trailing empty spaces a possibility? \$\endgroup\$ – xnor Mar 10 at 11:59
  • \$\begingroup\$ I meant in the input, but I guess whether the output can add or omit them could also be a question. \$\endgroup\$ – xnor Mar 10 at 12:12
  • 1
    \$\begingroup\$ Yes, I think you answered everything. So, if I got this right, I think this is a uniquely-matching regex of all valid inputs as sequences of columns, with e,l,d indicating empty, lane, and divider: e*(le(e|d)*e|le|l)*le*|e*. \$\endgroup\$ – xnor Mar 10 at 12:26
  • \$\begingroup\$ @xnor Took a little work to analyze but I think it is. \$\endgroup\$ – Ad Hoc Garf Hunter Mar 10 at 12:39

How many balloons do I need in each corner?


| |
  • \$\begingroup\$ Does it have to be the minimum number of balloons? For example, doubling the four numbers should still keep the board balanced. \$\endgroup\$ – wastl Feb 17 at 14:15

Bot Factory

Main Post

| |

Golf yourself a real calculator [draft]

We will only cover the characters(0123456789+-*/=%ñ) in our tutorial.

The = operations

Unlike most desktop calculators, our household calculator is a tacit language. Therefore it is able to do a lot more than other infix calculators.

Take a simple calculation as an example. The non-scientific calculator does not have the exponentiation operator. What do you do to calculate 2^5?


However, there's a shortcut for doing that. Since 2 is already in the expression buffer, you can simply do


The calculator automatically fills in the current expression during the inputting.

Here is a demonstration of how this works:

(A template for easy copy&paste.

Pressed Key      : 
Expression buffer: 
Output buffer    : 

Pressed Key      : 2
Expression buffer: 2
Output buffer    : 2

Pressed Key      : *
Expression buffer: 2 *
Output buffer    : 2

Pressed Key      : =
Expression buffer: * 2
Output buffer    : 4

Pressed Key      : =
Expression buffer: * 2
Output buffer    : 8

Pressed Key      : =
Expression buffer: * 2
Output buffer    : 16

Pressed Key      : =
Expression buffer: * 2
Output buffer    : 32

Implicit 0 before calculation

Suppose you enterede the following expression:


Now, don't get me wrong, the household calculator of course doesn't have pointers. So, why doesn't it raise a syntax error though? (The output is 0 by the way.) Here's why.

The calculator initially has the expression starting at 0, therefore it prepends a 0 to the expression. Therefore the full form of our expression is:


What we've learned so far

  • The output buffer is a part of the calculator storing the latest-evaluated integer. All entered numbers get appended to the output buffer as well as the expression buffer.
  • The expression buffer is a part of the calculator storing the latest instruction. After a = operator, it stores the latest applied expression for later application.
  • The = operator tries to evaluate the instruction buffer. If that's a syntax error, it tries to evaluate that concatenated the output buffer. If that still fails, it tries to evaluate the output buffer concatenated with the instruction buffer. After that operation, the expression starting from the newest-entered dyadic operator is saved in the expression buffer.
| |

Mimic the X11 Primary selection


In X11 (a windowing system used by a lot of Unix-like OS), what you would call the clipboard behave a bit differently than on other OSes like MacOS or Windows. While the "traditional" clipboard using ctrl+v/ctrl+c works, there is also another clipboard, called PRIMARY selection, that behave as following:

  • when you select a piece of text, this selection is added to the clipboard
  • when you use the middle mouse button, the content of that selection is pasted where your mouse is.

Some more details for those who are interested : X11: How does “the” clipboard work?


The input in this challenge is any representation of a binary input. In the following I will use 'S' for select and 'P' for paste.

Given the input, you must output the input after making the following changes :

  • put the content of the current output in the primary selection when you receive a select instruction
  • paste the content of the primary selection in the middle of the current output when you receive a paste character. If the current output is odd numbered, the middle is the length divided by 2 and truncated.


Input is SPSP :

selection = ""
output = "SPSP"
selection = "SPSP"
output = "SPSP"
selection = "SPSP"
output = "SPSPSPSP"
selection = "SPSPSPSP"
output = "SPSPSPSP"
selection = "SPSPSPSP"


Test Cases

"" -> ""
"S" -> "S"
"P" -> "P"
"PS" -> "PS"
"SP" -> "SSPP"

Reference Implementation

In Python 3 :

def clipboard(inp, select="S", paste="P"):
    out = inp
    primary_selection = ""
    for instruction in inp:
        if instruction is select:
            primary_selection = out
        if instruction is paste:
            out = out[:len(out)//2] + primary_selection + out[len(out)//2:]
    return out

Try it online!


  • You may assume that the input contains only the "paste" or "select" characters (you don't need to handle error cases)
  • You may use any kind of data representation (e.g binary data is allowed.)
  • Standard code-golf rules , shortest submission by bytes wins!
| |
  • \$\begingroup\$ Can we take the input newline-separated? (so like S\nP\nP\nS\nP) \$\endgroup\$ – user41805 Mar 15 at 8:23
  • \$\begingroup\$ @user41805 I would say that it is allowed. I would also allow using \n (or any whitespace) as either a "paste" or "select" character. \$\endgroup\$ – Lescurel Mar 16 at 18:42

Four Divisors

Given an integer array nums, return the sum of divisors of the integers in that array that have exactly four divisors.

If there is no such integer in the array, return 0.

Test Case 1:

Input: nums = [21,4,7]
Output: 32
21 has 4 divisors: 1, 3, 7, 21
4 has 3 divisors: 1, 2, 4
7 has 2 divisors: 1, 7

The answer is the sum of divisors of 21 only.

Please feel free to add more test cases.

This is code-golf so shortest submission in bytes wins! If you liked this challenge, consider upvoting it... And happy golfing!

| |
  • 1
    \$\begingroup\$ When posting, please change your post to only contain a link to the main post, then delete. \$\endgroup\$ – Adám Mar 24 at 14:23
  • \$\begingroup\$ Then delete, post in meta??? \$\endgroup\$ – Pluviophile Mar 24 at 14:56
  • \$\begingroup\$ No, post, then delete from meta. \$\endgroup\$ – Adám Mar 24 at 15:10
  • \$\begingroup\$ @Adam Please let me know, how long to wait for the post before posting it in main site. For better results. \$\endgroup\$ – Pluviophile Mar 24 at 15:20
  • \$\begingroup\$ I usually wait for a week, to catch those that stop by on a weekly basis, e.g. because they only visit while (not) at work. When I see a bunch of people in TNB, I also ask them to review. \$\endgroup\$ – Adám Mar 24 at 15:22
  • \$\begingroup\$ What is TNB? Sorry for asking, I'm new to Code-golf \$\endgroup\$ – Pluviophile Mar 24 at 15:25
  • \$\begingroup\$ The Nineteenth Byte chat room. Also, check out this list. \$\endgroup\$ – Adám Mar 24 at 15:26
  • \$\begingroup\$ What else can I do to improve the challenge?? \$\endgroup\$ – Pluviophile Mar 24 at 15:26
  • \$\begingroup\$ Too late now that you've posted to main already. Expect downvotes :-( \$\endgroup\$ – Adám Mar 24 at 15:26
  • \$\begingroup\$ If U don't mind. Please do check "Check if There is a Valid Path in a Grid" and provide any feedback/suggestions that I can use to improve the challenge \$\endgroup\$ – Pluviophile Mar 24 at 15:28
  • \$\begingroup\$ OK, but I'm a bit busy (working) atm. Hence the advice to leave for a while in the sandbox. \$\endgroup\$ – Adám Mar 24 at 15:28
  • \$\begingroup\$ Yeah sure.. Your feedback and suggestions are very valuable. Thank you. For guiding a newbie... \$\endgroup\$ – Pluviophile Mar 24 at 15:38

RRE numbers

Given a single floating-point number (which can potentially be taken as a string), output whether this decimal is an RRE number.

RRE complement for the input number

Say your input is 3.14.

  1. Replace the decimal point by the fraction bar. 3/14
  2. Reciprocal the fraction. 14/3
  3. Evaluate the fraction. 4.666...
  4. Round the decimal point to the same length as the input. 4.67
  5. If the absolute difference between the input and the output is below 2, it's an RRE number. Otherwise, it isn't an RRE number.


  1. Programs are allowed to give wrong output if the absolute difference is very close to 2.
  2. The integral part of the decimal is always nonzero.
  3. Truthy/Falsy outputs follows the language's convention, or exactly one value for truthy and another for falsy.

Test cases

Here is a sample program I use to generate the test cases.

1.0     -> True
2.9     -> True
3.14    -> True
50.2501 -> True
2.14    -> False
2.11111 -> False
3.1     -> False
51.123  -> False
51.51   -> False
24.12   -> False
| |
  • \$\begingroup\$ 1) Rounding 4.666... to two decimal places gives 4.67. 2) Floating-point numbers are always subject to precision issues. I think it's best to let the programs take a string representation as input, and state that "programs are allowed to give wrong output if the absolute difference is very close to 2". 3) If the integral part is 0, the algorithm will invoke division by 0. Is handling it part of the challenge? \$\endgroup\$ – Bubbler Mar 27 at 6:07
  • \$\begingroup\$ 4) For truthy/falsy, it currently reads like I can output the input as-is and say "the program's output format is a RRE number for truthy, a non-RRE number for falsy". What I use is "truthy/falsy following the language's convention, or exactly one value for truthy and another for falsy". \$\endgroup\$ – Bubbler Mar 27 at 6:13
  • \$\begingroup\$ I don't see how restricted source will fit here. Btw, a big truthy test case: 50.2501 \$\endgroup\$ – Bubbler Mar 27 at 6:39
  • \$\begingroup\$ @Bubbler Ahh, the challenge is just about checking whether the square of the integral part is close to the the decimal part. Should I post it? \$\endgroup\$ – Third-party 'Chef' Mar 27 at 6:43

Mom-rounding the time


Growing up with my mother, whenever she looked at a clock to check the time, she would always say "shoot, it's already X!" and then I would look at the clock and realize she was just rounding the time in a really weird way.


Given a time with hours and minutes, round it like my mom would. Rounding always occurs upwards. Say it is currently H hours and M minutes.

  • if M is 0, no rounding occurs; my mom isn't that crazy;
  • if M is 9 or less, my mom rounds to H:15;
  • if M is 19 or less, my mom rounds to H:30;
  • if M is 34 or less, my mom rounds to H:45;
  • for any other value of M, my mom rounds to H+1:00.


You will take a time that needs rounding, in any sensible format. ISO strings for date/time, two integers representing hours/numbers, a string with two integers and a separator; these are all fair game.


The string "shoot, it's already X" with X replaced with the mom rounding time.


This is so shortest answer wins. However, if your source code contains the substring shoot, it's already then you may subtract 19 from your score.

Test cases

Here is the program I use to generate the test cases.

11:00 -> shoot, it's already 11:00!
 3:08 -> shoot, it's already 03:15!
 1:09 -> shoot, it's already 01:15!
13:13 -> shoot, it's already 13:30!
 2:35 -> shoot, it's already 03:00!

| |
  • 1
    \$\begingroup\$ +1 because my mom's rounding is similarly weird too. One thing though: shoot, it's already has a ' which must be escaped in many languages, and the substring condition is unfairly penalizing langauges without string literal support. \$\endgroup\$ – Bubbler Apr 1 at 0:16
  • 5
    \$\begingroup\$ I suggest removing the unnecessary fluff about adding text and only keep the conversion. Bonuses in code golf are bad in general. Here it seems you try to even the playing field by explicitly disadvantaging languages with string compression, but end up making the false assumption that the substring will occur if used literally, even though some languages will need to escape the quote. \$\endgroup\$ – Adám Apr 1 at 6:17
  • 2
    \$\begingroup\$ Assuming hours wrap around in 24-hour time, some test cases showing this would be good. \$\endgroup\$ – xnor Apr 1 at 9:25
  • \$\begingroup\$ Those bullet-points should either be else-ifs instead of ifs or you should just define ranges. Currently a minute of 3 would first be rounded to 15 for being <=9, then 15 is rounded again to 30 for being <=19, and then again to 45 for being <=34. So basically: 0 remains 0; <=34 becomes 45, and >34 becomes 0 with the hour increasing, and the other bullet-points could be ignored. I think something like M=0→H:00; M=[1,9]→H:15; M=[10,19]→H:30; M=[20,34]→H:45; M=[35,59]→H+1:00 (perhaps in text form) would be clearer imo. \$\endgroup\$ – Kevin Cruijssen Apr 1 at 10:10
  • \$\begingroup\$ @petStorm thanks for your edit but I would prefer if you did not edit any reference programs into my sandboxed posts (you may comment with a TIO link) nor edited the challenge to cope with the feedback I get from commenters, nor to include a whole "test cases" section (but you can include it in your TIO link). In particular, usually when I don't include test cases right from the start is because I want to polish the spec a bit before trying to understand what test cases are really relevant and needed. \$\endgroup\$ – RGS Apr 1 at 11:50
  • 1
    \$\begingroup\$ And this is especially true because you create a whole section and then write "Here is the program I used to generate the test cases" as if you were me, which is not ok. \$\endgroup\$ – RGS Apr 1 at 11:52

Bit delivery in Bin City


Similar to a newspaper route you have been tasked with delivering bits to the residents of Bin City. Just as certain people like certain newspapers, the citizens of Bin City prefer either \$1\$s or \$0\$s delivered to their doorstep. Unfortunately, your new boss has arranged the deliveries in an awkward manner.

The residential streets of Bin City all have multiples of \$8\$ houses. So naturally, your delivery routes are organised as a series of one or more bytes. Your boss has laid out the deliveries as follows: first off, start at the house at the beginning the street on the left-side, followed by the house opposite on the right-side of the street, then the next house after the first one on the left-side of the street, then the one opposite it on the right-side, &c. The \$1\$s and \$0\$s to be delivered along this route are stored, in order, in the bits of the bytes.


You've quite rightly deduced it would be far easier an more time effective to deliver the bits starting from the first house on the left-side of the street and continue on delivering bits up the left-side to the end. Then cross the street to the house opposite on the right-side and continue on down the right-side back to what would have originally been your second delivery at the end.

Illustration for eight houses (one byte)

The original input route is on the far left next to the byte containing the bits to be delivered along that route. Beside it, to the right, is the output byte with the bits rearranged for the new improved delivery route on the far right.

Input to Output for 1 byte

Input and output

The input will be between \$1\$ and \$8\$ bytes long, organised by your boss in a back-and-forth left-side to right-side manner as above. This can be in any convenient byte-wise way (eg not a list of bits, but a list of bytes, words, or double-words &c is ok). A 64-bit integer along with the bit-length is ok too. A bit-length will be needed for any structure using elements larger than a byte. The order can be in any convenient bit-wise way (i.e. least-significant to most-significant bit or vice versa) and any convenient byte-wise way (eg first element in a byte list to last or vice versa, least--significant to most-significant byte or vice versa in a list or words or double-words &c traversed first to last or vice versa) so long as the output follows the same convention. The output must use the same structure and ordering convention as the input but with the bits rearranged to be up the left-side then back down the right-side in the manner described above.

Please state your bit-wise and byte-wise ordering as well as your i/o structure.

Scoring and winning criterion

Standard rules apply. Shortest code in bytes wins.

Test cases

As lists of bytes ordered least-significant bit to most-significant bit and first byte to last byte in the list.

[85] -> [15]
[85, 85] -> [255, 0]
[85, 85, 85] -> [255, 15, 0]
[85, 85, 85, 85] -> [255, 255, 0, 0]
[85, 85, 85, 85, 85] -> [255, 255, 15, 0, 0]
[85, 85, 85, 85, 85, 85] -> [255, 255, 255, 0, 0, 0]
[85, 85, 85, 85, 85, 85, 85] -> [255, 255, 255, 15, 0, 0, 0]
[85, 85, 85, 85, 85, 85, 85, 85] -> [255, 255, 255, 255, 0, 0, 0, 0]
[170] -> [240]
[170, 170] -> [0, 255]
[170, 170, 170] -> [0, 240, 255]
[170, 170, 170, 170] -> [0, 0, 255, 255]
[170, 170, 170, 170, 170] -> [0, 0, 240, 255, 255]
[170, 170, 170, 170, 170, 170] -> [0, 0, 0, 255, 255, 255]
[170, 170, 170, 170, 170, 170, 170] -> [0, 0, 0, 240, 255, 255, 255]
[170, 170, 170, 170, 170, 170, 170, 170] -> [0, 0, 0, 0, 255, 255, 255, 255]
[208] -> [28]
[96] -> [40]
[155, 36] -> [37, 210]
[232, 33] -> [24, 114]
[174, 18, 247] -> [66, 191, 248]
[130, 143, 125] -> [48, 111, 157]
[76, 181, 117, 107] -> [122, 159, 46, 67]
[158, 238, 106, 124] -> [166, 232, 230, 223]
[233, 87, 232, 152, 182] -> [249, 72, 182, 117, 120]
[142, 61, 195, 199, 218] -> [114, 185, 220, 153, 214]
[107, 131, 170, 25, 103, 171] -> [25, 80, 27, 175, 244, 233]
[71, 41, 113, 118, 202, 26] -> [27, 237, 72, 220, 42, 134]
[30, 226, 236, 110, 111, 211, 202] -> [134, 170, 219, 216, 233, 126, 203]
[162, 53, 89, 29, 128, 172, 134] -> [112, 125, 32, 146, 23, 68, 178]
[112, 71, 252, 192, 100, 176, 108, 71] -> [188, 142, 74, 186, 104, 35, 113, 40]
[111, 58, 224, 222, 231, 246, 214, 200] -> [75, 232, 235, 142, 149, 187, 61, 238]

Reference implementation in Python.

Sandbox: Is this a duplicate? Is it clear enough? Are the i/o restrictions reasonable?

| |
  • \$\begingroup\$ It could help a lot to add an illustration for small test cases (how to move each bit in the input to the output), and some test cases with more bytes (both even and odd). Other than that, looks good to me. \$\endgroup\$ – Bubbler Mar 31 at 1:57
  • \$\begingroup\$ @Bubbler Thanks for the feedback. How's it now? \$\endgroup\$ – Noodle9 Mar 31 at 14:32
  • \$\begingroup\$ Perfect, and +1 for reference implementation. \$\endgroup\$ – Bubbler Mar 31 at 23:00

| |
  • \$\begingroup\$ I like this idea. Do we have a default for taking "infinite lists" as input? I'd mildly suggest limiting the numbers to positive integers on general principle. \$\endgroup\$ – xnor Apr 2 at 4:26
  • \$\begingroup\$ How do you propose taking an infinite list as input? I think the natural approach would use as "input" a program (or function with no arguments) that runs forever; the list would consist of the numbers that it outputs when run. (This would be sort of like a plug-in that your answer could use.) The problem is that that might make the challenge trivial. But I'm not sure how else to take the input list. \$\endgroup\$ – Mitchell Spector Apr 2 at 6:14
  • \$\begingroup\$ In some languages the input could be a stream or iterator \$\endgroup\$ – simonalexander2005 Apr 2 at 8:36
  • \$\begingroup\$ I don't know what you want to allow, but some other possible implementations of infinite lists could be as a generator function that produces a new value on each call, or a black-box function taking a natural n and giving the n'th value. \$\endgroup\$ – xnor Apr 2 at 12:35
  • \$\begingroup\$ @xnor I don't think I will limit it to positive integers because positive integers are not bounded below, meaning sequences could only increase. This pretty radically changes the content of the challenge. \$\endgroup\$ – Ad Hoc Garf Hunter Apr 2 at 13:14
  • \$\begingroup\$ Let the first and second items of the list be x and y respectively. This checks whether input-x is divisible by y-x. (Doesn't work sometimes, I'll take a look.) \$\endgroup\$ – Third-party 'Chef' Apr 3 at 7:27
45 46
48 49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .