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4636 Answers 4636

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73 74

Asyncronous Breakfast

I haven't seen any asyncronous challenges here, even tho it's an important part of coding, so here's my idea, based on an example (https://docs.microsoft.com/en-us/dotnet/csharp/programming-guide/concepts/async/) that explains asyncronous coding:

You are making breakfast. The things you want to serve are:
Cooked eggs, Bacon, Coffee, Orange Juice and Toast.

You start boiling the eggs, cooking the bacon, making coffee, and toasting toast. 
When everything is finished, you're serving the orange juice.

After every finished food, the console outputs "[Food] is ready."
  • 1
    \$\begingroup\$ I think you should generally only specify things about the input and the output, and not the program's internal behaviour ("do addition without the + operator"-like questions are now discredited). Can you define a required output in a way that the only way to produce it is for the program to run asynchronously? \$\endgroup\$
    – KeizerHarm
    Dec 16, 2019 at 14:29
  • 1
    \$\begingroup\$ How about checking the multi-threading tag? \$\endgroup\$
    – Adám
    Dec 16, 2019 at 22:03


This is kept fairly simple, compared to a real Blackjack-Game:

Write a program that creates a blackjack game, where the input "h" means "hit" and "s" means "stand". Everything else outputs nothing, it just waits for the right input.

- You can draw as many cards as you want, also to make it easier, there is no card limit.
  This means you could draw more than 4 Asses.
- The cards you can get are
  standard-cards (2-10, J, Q, K, A) where the Values are the default ones:

2-9 = 2-9,
10, J, Q, K = 10

- A is a special card. So to make it easy, the default value is 11.
  If a player gets over 21, the value gets reduced to 1.
  In the case that a player has two or more A-Cards, all but one A
  reduces to 1, giving a total value of 14 for 4 Asses. (11+1+1+1)

- The dealer needs to get in between 18-21. If the bank has reached a value in 
this radius,
it can't draw another card (For example: Bank reached 18.
                          It can't get another card to get closer to 21)

- The game runs forever
- hit means „draw another card“, stand means „stop with cards and wait for 
the dealer to reach his limit“


- The player gets 2 cards at the beginning, shown by "Player:" in the console 
  (Example: Player gets 2 and 5 at the beginning. 
  Console output: "Player: 2, 5").

- The cards can be shown in the console by their values
  (For example: 2, 10, 1 | No need for: 2, K, A), but you can also use 
  the card-names instead. Your choice. But you need to show the maximum
  value you have
  (for example:
  first draw console output "Player: 2, 4, 5 = 11",
  second draw console output "Player: 2, 4, 5, 7 = 18")

- The input by the player is "h" for "hit" and "s" for "stand". Also allowed 
  is 1 and 0

- In the console, the player is displayed by "Player:"
- In the console, the dealer is displayed by "Dealer:"

- After the player got two cards, the dealer gets one card.
  Then the player can hit or stand.
  After every card drawn from the player,
  the dealer gets another card until he reaches his limit. (18-21)

  If the dealer reaches 21 before the player does, the dealer wins.
  If the Player reaches 21 first, the player wins.
  If the dealer reached his limit, and the player stands on the same value, it's a draw.
  If the dealer is over 21 first, the player wins.
  If the player is over 21 first, the dealer wins.

- At the end you get the output "Win", "Lose", "Draw",
  or "Blackjack", depending on the game

This game is a challenge, even the smallest result won'tbe under 150 bytes at least

  • \$\begingroup\$ @Noodle9 1. If you want to write 'h' and mistype, then you basically ruined your win 2. Blackjack rules: win = more points than dealer or Blackjack | lose = less points than Dealer | draw = same points as dealer | Blackjack = 21 points straight | 3. As it's the easy version there is only one hand, will add that \$\endgroup\$ Dec 16, 2019 at 18:14
  • \$\begingroup\$ Do you know blackjack? You draw cards and try to get as close to 21 as possible. If you get 21 you already have won the game. if you get 20 you could still draw a card, maybe it's an Ass. But you still need to get higher points than the dealer, but not more than 21 \$\endgroup\$ Dec 16, 2019 at 18:37
  • \$\begingroup\$ if you play with 52 cards you could draw up to 7 cards (4*Ass+ 3*2 = 20) \$\endgroup\$ Dec 16, 2019 at 18:38
  • \$\begingroup\$ Yes I've played the game T_T But questions/challenges need to be self-contained with a full explanation for anyone who's doesn't know the game. Also you should edit your sandbox to include the things above. \$\endgroup\$
    – Noodle9
    Dec 16, 2019 at 19:59
  • \$\begingroup\$ If a player gets over 21, the value gets reduced to 1. this contradicts the bit below of all but one A reduces to 1, giving a total value of 14 for 4 Asses. (11+1+1+1) \$\endgroup\$
    – Corsaka
    Dec 20, 2019 at 22:36
  • \$\begingroup\$ @Corsaka It doesn't. First ace you have 11, second ace makes that 22 so it gets reduced to 12, third ace makes that 23 so it gets reduced to 13, forth ace makes that 24 so it gets reduced to 14. \$\endgroup\$
    – Noodle9
    Jan 3, 2020 at 21:48
  • \$\begingroup\$ Ah, I misread that. Apologies. \$\endgroup\$
    – Corsaka
    Jan 5, 2020 at 13:20

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, 2019 at 20:10
  • 1
    \$\begingroup\$ What does the tree look like if n is even? That is, how does the trunk center? \$\endgroup\$ Dec 17, 2019 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, 2019 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, 2019 at 22:41
  • 1
    \$\begingroup\$ To add to frank's comment: codegolf.meta.stackexchange.com/a/10909/36398 \$\endgroup\$
    – Luis Mendo
    Dec 17, 2019 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, 2019 at 9:20
  • 1
    \$\begingroup\$ Fair enough. Sticking to printable ascii should be fine then. \$\endgroup\$
    – Grumpy_Boy
    Dec 18, 2019 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, 2019 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\$ Dec 21, 2019 at 12:24
  • \$\begingroup\$ I chose the latter because it is easier to specify. \$\endgroup\$
    – user85052
    Dec 21, 2019 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, 2019 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, 2019 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


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, 2020 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\$ Feb 8, 2020 at 18:32
  • \$\begingroup\$ @JonathanFrech flipped in comparison to what? \$\endgroup\$
    – RGS
    Feb 10, 2020 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, 2020 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, 2020 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\$ Feb 11, 2020 at 21:41

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, 2020 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, 2020 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, 2020 at 7:01
  • \$\begingroup\$ Me neither. This sounds great! \$\endgroup\$
    – PkmnQ
    Mar 11, 2020 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\$ Feb 12, 2020 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\$ Feb 12, 2020 at 8:36
  • 1
    \$\begingroup\$ So something like '?? in 05AB1E (push "?" and output it, ignoring the input) would score 66.7%? \$\endgroup\$ Feb 12, 2020 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, 2020 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\$ Feb 12, 2020 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, 2020 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, 2020 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\$ Feb 12, 2020 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, 2020 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\$ Feb 12, 2020 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, 2020 at 14:10
  • \$\begingroup\$ Yep, that indeed sounds as a good solution. \$\endgroup\$ Feb 12, 2020 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, 2020 at 7:29
  • \$\begingroup\$ @Lyxal The challenge should be self-explanatory. Can you understand it now? \$\endgroup\$
    – user92069
    Feb 16, 2020 at 10:01
  • \$\begingroup\$ Aha! So it's "x * n = x shifted, find n". I get it now! \$\endgroup\$
    – lyxal
    Feb 16, 2020 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, 2020 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, 2020 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\$
    – user92069
    Feb 17, 2020 at 2:46
  • \$\begingroup\$ @Lyxal It's possible, see my test case. \$\endgroup\$
    – user92069
    Feb 17, 2020 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\$
    – user92069
    Feb 17, 2020 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\$ Mar 2, 2020 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\$ Mar 2, 2020 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, 2020 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\$ Mar 1, 2020 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\$ Mar 2, 2020 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\$ Mar 2, 2020 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\$ Mar 2, 2020 at 20:06
  • \$\begingroup\$ Typo: monas should be moras. \$\endgroup\$
    – Grimmy
    Mar 6, 2020 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, 2020 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, 2020 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\$ Feb 14, 2020 at 12:08
  • 3
    \$\begingroup\$ @Lyxal It is mathematically impossible for all integers to have equal probability. \$\endgroup\$
    – Wheat Wizard Mod
    Feb 14, 2020 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, 2020 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\$
    – Wheat Wizard Mod
    Feb 14, 2020 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, 2020 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\$
    – Wheat Wizard Mod
    Feb 14, 2020 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, 2020 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, 2020 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\$
    – Wheat Wizard Mod
    Feb 14, 2020 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, 2020 at 15:59
  • \$\begingroup\$ Let us continue this discussion in chat. \$\endgroup\$
    – xnor
    Feb 14, 2020 at 16:07
  • 1
    \$\begingroup\$ I would allow for some entropy input to give pure languages a chance. \$\endgroup\$ Feb 25, 2020 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\$
    – Wheat Wizard Mod
    Feb 25, 2020 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.

  • 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, 2020 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, 2020 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, 2020 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, 2020 at 12:26
  • \$\begingroup\$ @xnor Took a little work to analyze but I think it is. \$\endgroup\$
    – Wheat Wizard Mod
    Mar 10, 2020 at 12:39

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.

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, 2020 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, 2020 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, 2020 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\$
    – user92069
    Mar 27, 2020 at 6:43

Are these the same time?


When asked about the time (i.e. hours and minutes), people naturally reply with any one of a given set of fairly common sentences:

  • (A) it is M past H
  • (B) it is M to H
  • (C) it is H minus M

Where M above refers to some amount of minutes and H to some amount of hours. Concrete corresponding examples, all referring to the time 3:40 pm:

  • (A) it is 40 past 3
  • (B) it is 20 to 4
  • (C) it is 4 minus 20


Given two of these sentences, output a Truthy value if they represent the same time and a Falsy value if they do not.


Your input will be two sentences of the above, where references to minutes will always be rounded to the nearest multiple of 5 (i.e. the minutes will always be one of 5, 10, 15, 20, ..., 50, 55.

Because all sentences start with "it is " you may ommit that from your input sentences.


A Truthy value if the two times are the same, a Falsy value otherwise.

Test cases

Here is a sample program for checking the test cases.


Should the minutes and hours in the input com as integers instead of English words?

  • \$\begingroup\$ Interesting challenge. Yes, the minutes and hours in the input should come as integers. Otherwise, this becomes a chameleon challenge that appears to be about parsing relation words, but actually is about parsing English numbers. I think you can make the challenge more interesting by adding (D) it is H M. Please address 1) how to distinguish AM/PM or that we don't need to, 2) how to deal with roll-overs like "5 to 0", and 3) if H and M have upper and lower bounds. \$\endgroup\$
    – Adám
    Apr 1, 2020 at 6:29
  • \$\begingroup\$ I kid you not, I have never heard any one call it "H minus M". Still, I agree with @Adám that y'all need to ensure that input and output formats are what I like to call "reasonable and convenient", with extra emphasis on the "convenient" part. \$\endgroup\$
    – lyxal
    Apr 1, 2020 at 6:52
  • 1
    \$\begingroup\$ @Lyxal Me neither. But actually, that can be fixed by changing "minus" to "in", as in "4 in 20 [minutes]" \$\endgroup\$
    – Adám
    Apr 1, 2020 at 6:55
  • \$\begingroup\$ An alternative you might consider to checking if two sentences represent equal times, is to have code take just one and produce any "canonical form" of it, such that the canonical form can be anything where two inputs give the same canonical form if and only if they are equal. \$\endgroup\$
    – xnor
    Apr 1, 2020 at 9:15
  • \$\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. The feedback is very good and I will take care of it, but I prefer to do it myself so I can do the changes I see necessary: e.g. if I am accepting hours and minutes as integers, I no longer want the minutes to be in the set 5, 10, 15, ..., 55. \$\endgroup\$
    – RGS
    Apr 1, 2020 at 11:48

\$\Theta(N\cdot\sqrt N)\$ sort

The challenge is to write a program that sorts an array of distinct positive integers in ascending order. You may input the array and output the result using the default IO methods.

However, the worst-case time complexity of the algorithm used must be \$\Theta(N \cdot \sqrt N)\$, where \$N\$ is the length of the input array.

You may not assume your built-in sorting functions to have any time complexity in particular. While you can implement a fast (e.g. \$O(N \log N)\$) sort and then perform pointless operations to increase the complexity, direct algorithms exist.

This question is tagged , so the shortest code wins!

Sandbox stuff

I have noticed that a possible solution is, for example, to create a sorted multiset from the array and read it back. I would probably like to disallow that. Is there a way to achieve that without making the validity criteria subjective?

  • \$\begingroup\$ I feel like this will be answered by implementing sorting efficiently, then doing something pointless for the required number of steps. \$\endgroup\$
    – xnor
    Apr 1, 2020 at 8:55
  • \$\begingroup\$ @xnor That would involve implementing a O(n log n) sort or radix sort, which can be more complicated than a O(n*sqrt(n)) algorithm. There's, for example, a gap sequence that results in O(n * sqrt(n)) complexity for Shellsort. \$\endgroup\$ Apr 1, 2020 at 9:04

Similar Numbers



Modify The Stack


  • \$\begingroup\$ Specifying that the first element is the top of the stack could help understanding the challenge a lot. Can we use some other values instead of the stack elements a, b, c and/or the commands s, d, t? Also, the operation a b c -> b c a is often called rotate or roll. \$\endgroup\$
    – Bubbler
    Mar 30, 2020 at 2:38
  • \$\begingroup\$ For test cases: a c b is tts. a b a b c can be done with dtdtt or dtdts. \$\endgroup\$
    – Bubbler
    Mar 30, 2020 at 2:41
  • \$\begingroup\$ thanks, but dtdtt and dtdts doesn't work. \$\endgroup\$ Mar 30, 2020 at 5:29
  • \$\begingroup\$ Sorry, I had the t operation mistaken, but your tdtsdt doesn't work either. The first operation should be s. \$\endgroup\$
    – Bubbler
    Mar 30, 2020 at 5:39
  • \$\begingroup\$ right, I fixed that \$\endgroup\$ Mar 30, 2020 at 5:46
  • \$\begingroup\$ I don't think the test cases are very helpful here until they are actually proven optimal... Do you really want to require input validation? If so, it is probably a very bad idea to post this without writing a reference implementation to make sure it's possible and not too annoying. (besides: I silently downvoted before because I simply dislike the idea, and not because of some specification issues, so I had nothing to comment) \$\endgroup\$ Apr 5, 2020 at 16:05
  • 1
    \$\begingroup\$ @mypronounismonicareinstate Sandbox is all about feedback. Silent downvotes don't help the challenge writer and the community. If you don't like the whole idea, you could say so in the comments in the first place (preferably with why you think that way), and then the challenge writer could consider to rewrite it or abandon it and try out something different. \$\endgroup\$
    – Bubbler
    Apr 5, 2020 at 23:25

Compactify the input


  • \$\begingroup\$ Off topic: BASIC programmers were often recommended to name their variables this way. \$\endgroup\$
    – user92069
    Apr 8, 2020 at 14:26
  • \$\begingroup\$ I agree that it's better to compress a single word instead. \$\endgroup\$
    – user92069
    Apr 8, 2020 at 14:26
  • \$\begingroup\$ hmm ok, I'll do that \$\endgroup\$ Apr 8, 2020 at 17:42
  • \$\begingroup\$ I think the mention of compression and natural language is misleading, since it leads the reader to expect some compression based on the statistical properties of text. \$\endgroup\$
    – xnor
    Apr 8, 2020 at 18:36
  • \$\begingroup\$ hmm ok, do you have an idea for a better name? \$\endgroup\$ Apr 9, 2020 at 4:53
  • \$\begingroup\$ Do you think Compactify the name is a good idea? \$\endgroup\$
    – user92069
    Apr 9, 2020 at 8:44
  • \$\begingroup\$ Related. \$\endgroup\$
    – user92069
    Apr 9, 2020 at 8:46
  • \$\begingroup\$ yes, I like that name, maybe Compactify the input though? As it doesn't have to be a name \$\endgroup\$ Apr 9, 2020 at 9:02
  • 2
    \$\begingroup\$ I like Compactify the input \$\endgroup\$
    – xnor
    Apr 10, 2020 at 12:12
  • \$\begingroup\$ Just an FYI, the regex \B[aeiou] matches each character to be removed. You may receive a lot of answers that are basically just that. \$\endgroup\$ Apr 12, 2020 at 5:10

How Many Ways To Empty The Glove Box?



Compress Numbers

Write two programs, a compressor and a decompressor.

The compressor

  • The compressor will accept a sequence of integers of any value from 0 to 263-1, expressed in any convenient format.
  • You may specify the required format as long as any arbitrary sequence of integers in the required range may be expressed in this format.
  • Behaviour is undefined for any input that does not conform to your required format.
  • The output will be a self contained sequence of bytes.

The decompressor

  • The input will be an unmodified sequence of bytes produced by a valid input to the compressor.
  • Behaviour is undefined for any other sequences of bytes.
  • The output will be the same input to the compressor program that produced the provided sequence of bytes.


The winning entry will be the valid entry that produces the smallest intermediate sequence of bytes for a sequence of integers that will be produced by the question setter that will be revealed after some number of entries have been submitted and only entries submitted prior to that reveal will be eligible to win.

This sequence will be generated by joining these following sequences into a single sequence and then randomly shuffling that single sequence.

  • 1000 repetitions of the same randomly selected number from 0 to 9.
  • 1000 repetitions of the same randomly selected number from 262 to 263-1.
  • For each x in (8, 16, 32, 63):
    • 1000 random numbers from 0 to 2x-1.

The question setter will answer the challenge with GZIP/GUNZIP at the highest compression setting with no additional processing. If that entry wins, the glory of winning will belong to the authors of GZIP.


If two or more entries produce produce byte sequences of the same size, the following criteria will decide the winner:

  1. If one of those entries is the GZIP entry posted by the question setter, that entry will win.
  2. The entry with the highest voting score wins.
  3. The entry posted first wins.
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