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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.

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Igpā Atinlā (Pig Latin)

This is .

In this challenge, we will be translating strings of words to Pig Latin.

Input: A string of words (a "word" is a continuous sequence of the characters A-Za-z) (ASCII only).

Output: The translated version of the input. Translations described below.

Some Definitions:

First, the following are vowels: "A,a, E,e, I, i, O, o, U, u". Any alphabet character that is not a vowel is a consonant. A consonant cluster is any continuous sequence of consonants surrounded on both sides by non consonant characters (or beginning/end of input). An example, the clusters are in bold:

"I am two hundred years young, you child-mother."

A word is a set of alphabet characters surrounded on both sides by non-alphabet characters.

Translation:

  1. If the string "'s" or "'d" or "'t" appears ("apostrophe s/d/t"), remove the apostrophe.

For each word in the string, do the following:

  1. If a word only contains capital letters (A-Z), ignore the next step.

  2. If the beginning of the word was a capital letter (A-Z) AND a consonant, change it to its lower-case equivalent (a-z). Then capitalize the first vowel in the word. If no vowel exists, then recapitalize the letter. e.g. Stretch --> strEtch, "Twxx" --> "Twxx"

  3. If the word begins with a consonant cluster, move that consonant cluster to the end of the word. e.g. stretch --> etchstr

  4. Append the "hard a" character to the word. If every letter is capital, append 'Ā'. Otherwise append 'ā' e.g. etchstr --> etchstrā, "ATC" --> "ATCĀ"

    • If your language is unable to output 'ā' and 'Ā', you may use "ay" and "AY" respectively.
  5. If the word is "A" or "a", ignore previous instructions. Transform the word to "Anā" or "anā", respectively. (because in Pig Latin, everything begins with a vowel, so we use the article "an" instead of "a")

Test Cases: (I think these are right)

  • "I want to be a cat." --> "IĀ antwā otā ebā anā atcā"

  • "That's a really nice... ass-car?" --> "Atsthā anā eallyrā icenā... assā-arcā?"

  • "[CR][NL]'ssssssssssssssssssTRUExxxxxxxxIAMSOCOOL" --> "[CRĀ][NLĀ]UExxxxxxxxIAMSOCOOLssssssssssssssssssTRā"

  • "THIS CHALLENGE IS PROBABLY A DUPE" --> "ISTHĀ ALLENGECHĀ ISĀ OBABLYPRĀ Anā UPEDĀ"

  • "I SAT ON an APPLE" --> "IĀ ATSĀ ONĀ anā APPLEĀ"

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  • \$\begingroup\$ Is the string: bc a consonant cluster? \$\endgroup\$ – Downgoat Mar 6 '16 at 5:30
  • \$\begingroup\$ Yeah. "bc" --> "bcā". Is that not what the spec says? \$\endgroup\$ – Liam Mar 6 '16 at 5:32
  • \$\begingroup\$ ok, I wanted to clarify if the start/end of a string counts as a "non-vowel" character. Perhaps a better description is: "a sequence of consonants"? \$\endgroup\$ – Downgoat Mar 6 '16 at 5:34
  • \$\begingroup\$ I think it's clearer now? \$\endgroup\$ – Liam Mar 6 '16 at 5:52
  • \$\begingroup\$ Very closely related. \$\endgroup\$ – FryAmTheEggman Mar 7 '16 at 15:46
  • \$\begingroup\$ @FryAmTheEggman I think mine is sufficiently more complicated that it wouldn't be a dupe. However I don't know that the complexities are interesting enough to merit posting. \$\endgroup\$ – Liam Mar 7 '16 at 19:22
  • \$\begingroup\$ Some languages might not support the special a and A. Maybe allow regular A's? \$\endgroup\$ – CalculatorFeline Mar 9 '16 at 5:43
  • \$\begingroup\$ @CatsAreFluffy I know that some languages can't but I think that makes the challenge more interesting. At least a bit. If people disagree, I might change it to "ay" and "AY" \$\endgroup\$ – Liam Mar 9 '16 at 5:49
  • \$\begingroup\$ In the THIS CHALLENGE IS PROBABLY A DUPE test case, the A should become Anā instead of ANĀ. \$\endgroup\$ – user48538 Mar 9 '16 at 7:07
  • \$\begingroup\$ I'd call this a dup. \$\endgroup\$ – mbomb007 Mar 11 '16 at 15:29
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Maximum of Two Roman Numerals

You should write a program or function which returns the maximum of two Roman numerals.

Input

  • Two positive integers between 1 and 3999 (inclusive) with their Roman numeral representation string.
  • The two strings can be separated by a space or inputted in the standard list representation of your language.
  • 4, 9, 40, ... are written as IV, IX, XL, ...
  • Trailing newline is optional.

Output

  • The larger Roman numeral as string.
  • Trailing newline is optional.
  • If the two inputs are equal you should still only return one of them.

Examples

Format is input => output (explanation)

XXIX DI => DI (29 < 501)
V X => X    

TODO more

Built-in functions involving Roman numerals are prohibited.

This is code golf so the shortest entry wins.

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  • \$\begingroup\$ Can I adopt this abandoned challenge? \$\endgroup\$ – programmer5000 Jun 9 '17 at 12:12
  • \$\begingroup\$ @programmer5000 Yep! \$\endgroup\$ – randomra Jun 10 '17 at 17:02
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Entropy Golf

This is a scoring system without a challenge.

My idea is to score entries based on the total Shannon entropy contained within them. This provides an incentive to both use fewer unique characters and to have a shorter program overall.

Given a string of characters, the score is calculated as follows, where C(x) is the number of occurrences of the given letter. To help provide a correction for multi-file programs or languages in which the program length encodes information, the EOF character at the end of every file is to be counted for the purposes of this scoring mechanism. Lowest score wins.

$$\mathrm{score} = -\sum_x{C(x)\log_2\frac {C(x)}{\mathrm{Length}}}$$

enter image description here

Anybody who knows better notation/MathJax is free to edit.

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  • \$\begingroup\$ As it is you would still get many Lenguage-like answers in the form of unary string, base convert, exec. Edit: Actually that wouldn't work, never mind. \$\endgroup\$ – feersum Jun 6 '15 at 15:51
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    \$\begingroup\$ The line between lenguage and slashes and brainfuck and real languages would be hard to define. Languages with fewer symbols tend to require more uses of those symbols to accomplish anything. \$\endgroup\$ – Sparr Jun 6 '15 at 17:01
  • \$\begingroup\$ @Sparr It's mainly about languages that encode information in the length of the program itself. It's not about number of symbols, although I could probably put a requirement that the program uses at least two symbols. \$\endgroup\$ – PhiNotPi Jun 6 '15 at 18:05
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    \$\begingroup\$ Alternatively you could consider each program to be terminated by an EOF character, which is included in the calculations. Then a $n$-character program in lenguage would score $-n \lg n/(n+1) - \lg 1/(n+1)$, which does grow indefinitely. PS I've just realised that my edits come down to the same thing as you were doing by flipping the quotient in the log upside down. Sorry about that. Revert if you want, but be aware that I might not be the only person who has $p \lg p$ so branded in their mind that anything else looks wrong. \$\endgroup\$ – Peter Taylor Jun 6 '15 at 18:46
  • \$\begingroup\$ @PeterTaylor p log (1/p) should be much more intuitive for people who see this the first time. \$\endgroup\$ – jimmy23013 Jun 7 '15 at 1:20
  • \$\begingroup\$ If you meant unique characters, Sclipting has an advantage for that selecting a character from the whole Unicode contains a lot of information (compared to the length of a short program). \$\endgroup\$ – jimmy23013 Jun 7 '15 at 1:23
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Wargame Tank Simulation

Some of you may have heard of the Wargame series of computer based real time strategy game. These games pit teams of players with primarily cold war era units to see how a hot cold war would have played out. The goal of this challenge is to simulate a tank battle in these games.

Input

  • Two "tanks" (one red and one blue) will be entered into your program or function. Each tank is classified by a rate of fire, accuracy, armor value, and attack power value.

Challenge

From the inputs above, simulate the two tanks fighting. You will do this by having each tank fire according to its rate of fire. If it hits (randomly determined by accuracy), a tank will do damage according to the armor value of its target and its own attack power. The formula for damage is floor[(attackpower - armor)/2]. Therefore a tank with 10 attack power against a tank with 5 armor would do 2 damage.

Tank crews also have morale, which follows the following rule

  • There are four possible morale values; calm, worried, scared, and panicked. Tanks always start calm. These names do not need to be in your code, in the sample below I've used calm = 1, worried = 2, etc.
  • Each morale value reduces the accuracy as follows: Calm -> 100% (no change), worried -> 75%, scared -> 50%, panicked -> 25%. Therefore a panicked tank which normally has 60% accuracy now has 0.25 * 0.6 = 15% accuracy
  • Each hit by the opposing tank degrades morale by one level, each miss upgrades the morale by one level.

For example:

morale: calm  |  worried  |  calm  |  worried  |  scared
hit:         hit         miss     hit         hit

Rules:

  • Input should be two parameters to repesent each tank (I've used two tuples in the example below). Inputs may be provided in any order, just be sure to state which one is which. Input may be provided by user input via STDIN, read from a file, or parameters passed to a function call.
  • Each tank starts with 10 health.
  • Rate of fire will either be 8.5, 7.5, 6.5, 6, or 5 seconds between shots.
  • Tanks start loaded, so each fires at time = 0. Because Communists are sneaky, red fires first.
  • Accuracy must be randomly rolled.
  • Ineffective hits (hits which do no damage) have an effect on morale! (because it probably sounds terrifying)
  • Naturally since we want to see the action, output will be an update after each shot. The update will contain the time of the shot, whom it was made by (red or blue), health of both tanks, and the morale of both tank crews. Output maybe be presented in any format so long as it contains all of the required information in a human readable fashion (items must be delimited in some way). Similar to input, please describe your output format in the answer.
  • Engagements are limited to 100 seconds. If you play the game you know this is because a plane has swooped in by then. For our purposes if both tanks are alive at this point, it is a draw.
  • After one tank reaches 0 health (or after 100 seconds), print which tank is victorious ("Red" or "Blue") or "Draw" if appropriate. I don't care about trailing whitespace or newlines.
  • Printint output may be printing to STDOUT or writing to a file
  • Shortest answer in bytes wins

Sample Python Implementation

import random, math

def Shot_Result(acc, morale, power, armor):
    Morale = {1: 1., 2: 0.75, 3: 0.5, 4: 0.25}
    actual_acc = acc * Morale[morale]
    roll = random.random()
    if roll < actual_acc:
        dmg = max(0,math.floor((power-armor)/2))
    else:
        dmg = -1
        
    return dmg
    
    
def main(Red, Blue):
    red_rate, red_acc, red_armor, red_pow = Red
    blue_rate, blue_acc, blue_armor, blue_pow = Blue
    
    red_health = blue_health = 10
    red_morale = blue_morale = 1
    
    red_shots = [("Red", shot/100.) for shot in range(0,10000,int(red_rate*100))]
    blue_shots = [("Blue", shot/100.) for shot in range(0,10000,int(blue_rate*100))]
    
    Shots = sorted(red_shots + blue_shots, key=lambda x: x[1])
    
    print "{:^6}|{:^6}|{:^12}|{:^12}|{:^12}|{:^12}|".format("Shot","Time","Red Health","Blue Health","Red Morale","Blue Morale")
    
    for shot in Shots:
        if shot[0] == "Red":
            dmg = Shot_Result(red_acc, red_morale, red_pow, blue_armor)
            if dmg >= 0:
                blue_health -= dmg
                blue_morale = min(blue_morale+1,4)
            else:
                blue_morale = max(blue_morale-1,1)
        else:
            dmg = Shot_Result(blue_acc, blue_morale, blue_pow, red_armor)
            if dmg >= 0:
                red_health -= dmg
                red_morale = min(red_morale+1,4)
            else:
                red_morale = max(red_morale-1,1)
                
        print "{:^6}|{:^6}|{:^12}|{:^12}|{:^12}|{:^12}|".format(shot[0], shot[1], red_health, blue_health, red_morale, blue_morale)
        
        if red_health <= 0:
            print "Blue tank is victorious!"
            break
        if blue_health <= 0:
            print "Red tank is victorious!"
            break
    else:        
        print "It's a draw!"

Sample Output (Yours does not need to be this fancy)

 Shot | Time | Red Health |Blue Health | Red Morale |Blue Morale |
 Red  | 0.0  |     10     |    5.0     |     1      |     2      |
 Blue | 0.0  |     10     |    5.0     |     1      |     2      |
 Blue | 6.5  |     10     |    5.0     |     1      |     2      |
 Red  | 8.5  |     10     |    5.0     |     1      |     1      |
 Blue | 13.0 |    5.0     |    5.0     |     2      |     1      |
 Red  | 17.0 |    5.0     |    5.0     |     2      |     1      |
 Blue | 19.5 |    5.0     |    5.0     |     1      |     1      |
 Red  | 25.5 |    5.0     |    5.0     |     1      |     1      |
 Blue | 26.0 |    0.0     |    5.0     |     2      |     1      |
Blue tank is victorious!
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  • \$\begingroup\$ Its a good challenge, but you need to define your input/outputs a little more. I'd recommend indicating that tanks are passed in as a tuple/list, which tank is passed in first, and then define what the functions should return if red wins/blue wins/tie. \$\endgroup\$ – Nathan Merrill Mar 28 '16 at 15:32
  • \$\begingroup\$ @NathanMerrill, I've made some slight updates to the rules. I want to keep things open as I often find myself frustrated with arbitrary I/O restrictions. \$\endgroup\$ – wnnmaw Mar 28 '16 at 15:58
  • \$\begingroup\$ Flexible I/O is the way to go :) Looks good from my end. \$\endgroup\$ – Nathan Merrill Mar 28 '16 at 16:05
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Diagonals of an Array

The k-th diagonal of a two-dimensional array is a list of all elements in positions (a,a+k). Your task is to output all of the diagonals.

For example, the diagonals of

[[3,1,4,1],[5,9,2,6],[5,3,5,8]]

are:

[[1],[4,6],[1,2,8],[3,9,5],[5,3],[5]]

Which can be visualized thusly:

k -2-1 0 1 2 3
    \ \ \ \ \ \
     \ \ 3 1 4 1
      \ \ \ \ \
       \ 5 9 2 6
        \ \ \ \
         5 3 5 8

Input

A rectangular nested array or 2D array of positive integers, which will be nonempty.

Output

Its diagonals in any consistent order.

Test cases

[[3,1,4,1],[5,9,2,6],[5,3,5,8]]
[[1],[4,6],[1,2,8],[3,9,5],[5,3],[5]]

[[42]]
[[42]]

[[57,72,15,66,49,01,53,28,60,60,65,12,09,00,82]]
[[57],[72],[15],[66],[49],[1],[53],[28],[60],[60],[65],[12],[9],[0],[82]]

[[57],[72],[15],[66],[49],[1],[53],[28],[60],[60],[65],[12],[9],[0],[82]]
[[57],[72],[15],[66],[49],[1],[53],[28],[60],[60],[65],[12],[9],[0],[82]]

Maybe a second question about n-dimensional arrays?

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  • \$\begingroup\$ I suggest a requirement that the order of the output diagonals be consistent (the same irrespective of input); and add a row matrix and column matrix as test cases \$\endgroup\$ – Luis Mendo Mar 26 '16 at 18:21
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    \$\begingroup\$ Now almost a duplicate. Also, Jelly has a 2-character built-in for this. (Just to let you know.) \$\endgroup\$ – Martin Ender Apr 6 '16 at 8:51
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Help! Everything is in Wingdings!

Your task is to write a program or function that converts wingdings to a readable font. The input to the program will be the path to a png file with between 1 and 10 wingdings characters. Your output should be the message contained in the wingdings characters.

Specifications

Each test case will be a screenshot of the message as rendered in Microsoft Word (Black text on a white background, size 36, and 100% zoom). The image will be cropped to the bounding box of the message.

[insert 20 test cases, 2 of each size between 1 and 10]

[insert link to test cases + reference images]

Scoring:

(% of characters right in test cases)*10 - code length

(Highest score wins)

Rules

  • Standard loopholes apply.
  • No catering to the test cases.

(Note that this is just a concept for a challenge. I will expand the challenge if this receives a positive response).

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  • \$\begingroup\$ @EasterlyIrk The input is an image containing wingdings characters. \$\endgroup\$ – Daniel M. Apr 10 '16 at 3:34
  • \$\begingroup\$ Related \$\endgroup\$ – Leaky Nun Apr 10 '16 at 3:35
  • \$\begingroup\$ @KennyLau This is different because the test cases will have up to 10 characters in the same image (and catering to the test cases is a standard loophole). I'll work on this more tomorrow. \$\endgroup\$ – Daniel M. Apr 10 '16 at 3:44
  • \$\begingroup\$ The description of the image seems underspecified to me. Will the characters all be the same font size? How will they be laid out? Colours? Anti-aliasing? \$\endgroup\$ – Peter Taylor Apr 10 '16 at 21:36
  • \$\begingroup\$ What is the subpixel rendering setting? \$\endgroup\$ – John Dvorak Apr 11 '16 at 10:22
  • \$\begingroup\$ @JanDvorak I'm not completely sure, so it's probably at whatever the default for Windows 10 is. If needed I can run the images through a program that creates a 1-bit image. \$\endgroup\$ – Daniel M. Apr 11 '16 at 10:56
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KOTH: Black Hole

I've recently seen this video (by Tom Scott), in which it featured a game called Black Hole. Let's play that for KOTH!

Overview (the original version in the video)

  • Two "players" have 10 counters, each labelled 1 to 10.
  • These two players take it in turns to place counters in increasing order on a grid of 21 "spaces" arranged in a triangle (i.e. a triangle with 6 rows).
  • Once all 21 counters have been placed, the remaining space that hasn't been used is the "black hole", and the dots surrounding the black hole are "sucked in".
  • The person with the lowest sum of the counters "sucked in" wins.

For the purpose of KOTH, the triangular grid will consist of 120 spaces (i.e. the triangle "pyramid" will be 15 rows high), and each player will have 59 counters (Yes, that means that there will be 2 black holes).

Point System

  • Win: 10 points, Draw (very rare): 3 points, Loss: no points

Instructions

  • The bot must output a number between 1 and 120 each turn, and a "counter" with the same number as the current no. turn made by the bot will be placed there.
  • After 118 turns, the black holes activate, the score is counted, and points are awarded.

NOTE: If there is no output within 5 seconds, the bot immediately loses.

"Battle" System

  • The competition will be a round-robin, except each bot battles every other bot 5 times.
  • The top three bots will then play each other 10 times each.
  • The final winner is the winner of the competition.

Info on controller stuff

The controller will be in Python, but I don't have the first clue how to make an controller. Any help?

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  • \$\begingroup\$ Well I think that you have to choose a way of inputting and outputting data first to have an idea how to write the controller. If it's going to be single language KOTH you can just ask people to write you functions/classes in some way and you can attach those to a regular program performing the game. On the other hand the challenge seems simple enough that you may just give the input as arguments to an execution of the programs and read output for them (calling an instance of the bot for each move). \$\endgroup\$ – Lause Apr 11 '16 at 10:12
  • \$\begingroup\$ Yeah... I'll probably do it as an input/output thing, regardless of language. \$\endgroup\$ – clismique Apr 11 '16 at 10:55
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    \$\begingroup\$ 1. I think you've changed the size of the board so many times that there are figures in the question from at least three different board sizes. 2. In the video, the counters must be placed in increasing order of value, but the question doesn't mention this anywhere. 3. From my experience of other koths, best-of-50 is going to be far too time-consuming. I suggest making it best-of-5 or at most best-of-7. \$\endgroup\$ – Peter Taylor Apr 11 '16 at 21:39
  • \$\begingroup\$ Ah. Thanks for the tips... sorry if it's confusing. Will edit. \$\endgroup\$ – clismique Apr 12 '16 at 1:10
  • \$\begingroup\$ Interesting game. There is discussion about the different options for input and output on meta, and also there's a general tips for writing KotHs. \$\endgroup\$ – trichoplax Apr 12 '16 at 13:14
  • \$\begingroup\$ So... if there are multiple languages, will I have to install other things that can run those languages? That's a lot of stuff to prepare for, considering it's only for a short while (maybe for longer if I keep on doing KotHs). \$\endgroup\$ – clismique Apr 13 '16 at 10:13
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Generate a waveform from audio

Given an audio file, in any common audio format (such as MP3, OGG, WMA, M4A, or WAV) of your choice, as input, output an image, in any common image format (such as PNG, JPG, PPM, BMP, SVG, or GIF) of your choice, of a waveform representing the audio input.

Here is an example image, generated with Audacity from this audio:

waveform example

For reference, here is what an amplitude 1 sine wave at 440 Hz looks like:

sine 440

And this is what it sounds like (warning: it doesn't sound good).

Restrictions

  • The background of the image must be white (#FFFFFF in HTML notation)
  • The waveform may be any color that is sufficiently distinguishable from the background
  • The input audio must be sampled at 44.1 kHz with 32-bit floats (in the inclusive range [-1, 1])
  • For raster graphics:
    • Sample points must be plotted every 5 pixels
    • The height must be an odd number of pixels, no less than 101
  • Because vector graphics can scale indefinitely, the above restrictions do not apply
  • The horizontal line in the middle of the image represents 0 amplitude, and the top and bottom of the image represent 1 and -1 amplitude, respectively
  • The vertical scale must be linear, not logarithmic (the amplitude of the wave, not the relative loudness)
  • The color of the wave, height of the image, input format, and output format must be the same every time your submission is run (for example, you cannot output as a PNG for one input, and a JPG for another)
  • For the sake of testing/verification, submissions must run in under 1 minute on my machine (Core i3-3240 quad core, 3.4 GHz, 8 GB RAM) for any input.
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  • \$\begingroup\$ "common format"? \$\endgroup\$ – Addison Crump Mar 2 '16 at 21:37
  • \$\begingroup\$ @CoolestVeto Examples for audio include MP3, WAV, WMA, M4A, and OGG (basically any format Audacity can export to). Examples for graphics include PNG, JPG, BMP, SVG, PPM, GIF, etc. \$\endgroup\$ – user45941 Mar 2 '16 at 21:40
  • \$\begingroup\$ So we're permitted to use any of those formats, not that we must allow for any of those formats? \$\endgroup\$ – Addison Crump Mar 2 '16 at 21:42
  • \$\begingroup\$ @CoolestVeto Correct. \$\endgroup\$ – user45941 Mar 2 '16 at 21:43
  • \$\begingroup\$ Horizontal scale is specified, but (minimum) vertical scale is not. \$\endgroup\$ – Peter Taylor Mar 2 '16 at 22:31
  • \$\begingroup\$ @PeterTaylor What do you mean? The horizontal center of the image should represent 0 (silence), and the top/bottom should represent ±1 (the maximum/minimum measureable sample amplitude). The vertical scale should be linear, not logarithmic (measuring amplitude of the wave, not the loudness in decibels). \$\endgroup\$ – user45941 Mar 2 '16 at 22:40
  • \$\begingroup\$ A 3-pixel high image could meet that spec. \$\endgroup\$ – Peter Taylor Mar 3 '16 at 7:06
  • \$\begingroup\$ @PeterTaylor Oh, I see the issue now, thanks. \$\endgroup\$ – user45941 Mar 3 '16 at 19:51
  • \$\begingroup\$ Re The input audio must be sampled at 44.1 kHz with 32-bit floats (in the inclusive range [0, 1]), did you maybe mean the inclusive range [-1, 1]? \$\endgroup\$ – zwol Apr 21 '16 at 23:13
  • \$\begingroup\$ @zwol I did indeed, thanks. \$\endgroup\$ – user45941 Apr 21 '16 at 23:23
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The free monoid on two letters 'x' and 'y' is the set of all finite strings you can make up from them, including the empty string, with their concatenation a binary operation. So the elements of this free monoid look like this: "", "x", "y", "xy", "yx", "xx", "yy" and so on. Strings like "xxxx" are often written as powers and we don't generally use the quotes: x^4.

This challenge is about a certain binary relation on this free monoid. Suppose we decide to treat x^2 abd y^2 as "the same". Which other strings (or "words", as they're called) will have to become "the same" (or congruent under the relation x^2 = y^2) as a result? In xxy, we can substitute yy for xx, which means that xxy is congruent to yyy. But then, the last yy in that string can be turned into xx as well, so yxx is congruent to the previous ones as well. A nice thing about this relation is that any congruent words will have to have the same length, so the number of words congruent to a given one is always finite. (To read up on this, you can look up things like "string rewriting systems", "transitive closure", "congruence relation".)

Take a string as input. The string is assumed to be of any length (including the empty string) and to contain at most two different characters (chosen by the user from the printable ASCII set or equivalent), say x and y. The program will interpret the string as a word in the free monoid on x and y and output all the other words of that monoid, also as strings, using the characters chosen by the user, congruent to the input word under the relation x^2 = y^2, in any order. Every congruent word must occur exactly once, including the input.

The words in the output have to be consistently separated by any non-empty string of printable characters not chosen by the user. If the input word contains less than two distinct characters, your program will assume the missing one(s) according to your choice, but that must not overlap with the separator. How exactly you want to take the input and output the words is up to you, but it must be possible to enter any valid string as input.

This has to be a complete program, not a function. No loopholes please.

Examples of input and output:

in: empty string
out: empty string

in: xx
out: xx yy

in: xy
out: xy

in: xyx
out: xyx

in: xyy
out: xyy xxx yyx

in: xxxx
out: xxxx yyxx yyyy yxxy xyyx xxyy

in: xyyxy
out: xyyxy xxxxy yyxxy yyyyy xxyyy xxyxx yyyxx yxxxx yxyyx yxxyy

(Note that the whitespace is a valid input character -- your program must choose a different separator if the user chooses to use it as a letter.)

Don't worry about things like the maximal length of the input or the output. The logic of your program has to work for all input word lengths, but the program only needs to work for reasonably long inputs.

Please give a short explanation of why your program works.

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  • \$\begingroup\$ You might want to include some more background about monoids and congruent words. Also, test cases should include the expected output as well. The test cases would also be more readable in a code block without quotes. And I get that you want to show with the test cases that the choice of characters is arbitrary, but that's fairly clear from the spec. It would be more useful if you stuck to one set of characters in the test cases to make it easier for people to transform all of them at once into their own input format (instead of having to rewrite each one independently by hand). \$\endgroup\$ – Martin Ender Feb 16 '16 at 8:17
  • \$\begingroup\$ @MartinBüttner I guess I'm not attached to the test cases. I got rid of them and added a short introduction. \$\endgroup\$ – ymar Feb 17 '16 at 0:47
  • \$\begingroup\$ I didn't mean you should get rid of the test cases. They're important. I just meant, use the same pair of characters throughout to make them more usable. \$\endgroup\$ – Martin Ender Feb 17 '16 at 8:15
  • \$\begingroup\$ @MartinBüttner I've added new test cases. \$\endgroup\$ – ymar Mar 8 '16 at 14:35
  • \$\begingroup\$ Why are functions banned? \$\endgroup\$ – CalculatorFeline Mar 9 '16 at 5:38
  • \$\begingroup\$ @CatsAreFluffy This is personal preference - it's hard for me to understand why a function should be OK. It doesn't do the job (by itself) and it seems unfair to the people who write runnable programs. I feel that they do more. I'm new here and new to code golf so I don't know. Is there a reason functions should be allowed other than to make the top answer this much shorter? \$\endgroup\$ – ymar Mar 9 '16 at 7:29
  • \$\begingroup\$ Well, in powerful golfing languages, making a function can actually be longer (eg Cjam, Golfscript) But in Java, you have to include class Q{public static void main(String[]a){...}} in all your programs. Even in other (normal) languages, fewer bytes for input/output could help beat a golfing language. tl;dr It helps languages like Java have a chance against Python or other less-boilerplatey languages. \$\endgroup\$ – CalculatorFeline Mar 9 '16 at 15:20
  • \$\begingroup\$ Why are you demanding that the program allow its user to input a separator? What's the benefit in that as opposed to allowing the program to use whatever separator (not one of the letters, obviously) the programmer wants? \$\endgroup\$ – msh210 Apr 27 '16 at 19:25
  • \$\begingroup\$ @msh210 I don't think I'm saying that. My intention was to say exactly what you're saying. The separator is chosen by the programmer/program, but it can't be one of the letters. The letters are chosen by the user from a certain set and the separator is chosen from the same set. \$\endgroup\$ – ymar Apr 27 '16 at 19:52
  • \$\begingroup\$ Oh, okay, thanks for the edit. \$\endgroup\$ – msh210 Apr 27 '16 at 20:04
  • \$\begingroup\$ Allowing the user to choose any letters and forcing the program to come up with a different separator seems like it only adds arbitrary complexity that’s unrelated to the interesting part of the challenge. You should either guarantee that the valid input letters are x and y, or allow the programmer to choose the valid input letters. \$\endgroup\$ – Anders Kaseorg May 4 '16 at 22:13
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Determine whether one graph is a subgraph of the other

Given two unlabelled graphs as adjecency matrices with the same number of vertices, the goal is determining whether the first graph is a subgraph of the second one.

Definitions

A graph G=(V,E) comprises a set of vertices V={1,2,3,...,n} and a set of edges E ⊆ V x V = {(u,v) | u,v ∈ V}. The adjecency matrix A={a(i,j)}of a graph G is defined entry wise:a(i,j) = 1 if (i,j) ∈ E, 0 otherwise.

A graph H=(W,F) with adjecency matrix B={b(i,j)} is a subgraph of G iff all following statements hold:

  • V=W
  • There is a permutation p:V→W=V such that b(p(i),p(j)) ≤ a(i,j) for all i,j ∈ V=W

Examples

First the trivial ones:

  • Obviously, every graph is a subgraph of itself.
  • If H has more edges than G then it cannot be a subgraph.
  • If you remove edges from a valid subgraph, the result will again be a subgraph.
  • If every of the n nodes in G has edges to every other node, then every graph H with n nodes is a subgraph.

6 vertices, isomorphic (one big cycle) (remove some ones from Graph 1 in order to generate more valid subgraphs)

Permutation: [5 6 4 1 3 2]
Graph 1:     [0 1 0 0 0 0;0 0 1 0 0 0;0 0 0 1 0 0;0 0 0 0 1 0;0 0 0 0 0 1;1 0 0 0 0 0]
Graph 2:     [0 0 1 0 0 0;0 0 0 0 1 0;0 1 0 0 0 0;1 0 0 0 0 0;0 0 0 0 0 1;0 0 0 1 0 0]

Visualization of the above example by @KennyLau:

10 vertices, same number of edges, non isomorphic (g1 has minimal cycles of length 3,4,5, g2 has minimal cycles of length 3,4,4)

Graph 1:    [0 1 0 1 0 0 0 0 0 0;0 0 1 0 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0;0 0 0 0 0 1 1 0 0 0;1 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 0 1 0 0;0 0 0 0 0 0 0 0 1 0;0 0 0 0 0 0 0 0 0 1;0 0 0 0 1 0 0 0 0 0]
Graph 2:    [0 0 0 0 0 0 0 1 0 1;0 0 0 0 0 0 1 0 0 0;0 0 0 0 0 1 0 0 0 0;1 0 0 0 0 0 0 0 0 0;0 1 0 1 0 0 0 0 0 0;1 0 0 0 0 0 0 0 0 0;0 0 0 0 1 0 0 0 0 0;0 0 1 0 0 0 0 0 0 0;0 0 0 0 0 1 0 0 0 0;0 0 0 0 1 0 0 0 0 0]
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  • \$\begingroup\$ Sample input output? \$\endgroup\$ – Leaky Nun Apr 30 '16 at 11:22
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    \$\begingroup\$ I added some now. \$\endgroup\$ – flawr Apr 30 '16 at 14:06
  • \$\begingroup\$ @KennyLau How did create those visualizations? \$\endgroup\$ – flawr Apr 30 '16 at 15:21
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    \$\begingroup\$ Microsoft Paint. \$\endgroup\$ – Leaky Nun Apr 30 '16 at 15:27
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Strip Iterated Prisoner's Dilemma

Inspired by https://xkcd.com/696/, of course.

The Prisoner's Dilemma is a classic game (more in the game theory sense than the family fun sense) where two agents - two accomplices to a crime, in the original formulation - must choose whether to sell out the other.

If each player chooses not to betray the other, both win. If one player chooses treachery and the other does not, it wins, but if both betray each other, neither wins. The "iterated" variation is where the game is played multiple times with the same players, and both players know all the decisions each player has made in the past.

Of course, that's all a bit dull, and has been done before besides. We're going to tart it up a bit.

(Unfortunately due to the nature of the test all solutions must be in the same language, and one that supports executing strings. I've chosen Python 3, since it's my favorite and I have to write the runner.)

The Game:

Submit a program - specifically a Python 3 function - that plays Iterated Prisoner's Dilemma against another such program. If both functions betray each other, each one gets one character deleted from the end of its source code. If one function betrays the other, the betrayed function gets two characters deleted from the end of it, and the traitor gets a # appended to it, immunizing it against one future loss. If neither turns traitor, both go unharmed. Functions will be restored to their original text between each contest with a new opponent.

Submissions are scored based on failure rate and length; specifically, score = round( (number_of_trials_failed / total_number_of_trials) * ( length / 2 ) ). The submission with the lowest score wins.

The length of a submission is the number of non-whitespace characters in the submission. Comments are not counted, but are also removed before the contest begins, so a commented out "guard" at the end will not protect your code from deletions. Length also does not count the function specifier (the def submission(y, o, t): which should not be included in your submission text anyway.)

Your function will be called with three parameters, y (your history), o(opponent's history), and t (text of the opponent itself). t is the very same text that the game runner will execute as your function's opponent, which you may analyze or otherwise use to run simulations. It will then return True if it wishes to betray its opponent or False if it does not.

Every possible 2-combination of submissions will be contested against each other - including each submission against itself. Each contest consists of 2,500 trials.

Other Rules:

  • All submissions must be in Python 3. (Specifically, they should run under Python 3.4.4)

  • The submission text must be the only code that is run to produce the result; you may not import any libraries, ask for user input, read from /dev/urandom/or equivalent, and of course you can't pull results from some webserver (which is already a violation of the standard loophole rules.) You MAY execute the given opponent text, and of course you're allowed to call all the builtins.

  • The submission must terminate and return an answer within 1 second (this will be run on a 12GB i7 gaming computer, so this should be plenty of time).

  • A submission that emits an uncaught exception or returns an invalid value loses that round.

  • Submissions that do not return in one second or less are immediately disqualified.

  • Submissions will be closed on [date posted + 8 days] and programmatically judged, results will be appended to the challenge. There will be a "trial run" in the evening of [date posted + 4 days].

API:

The "submission text" is valid, 4-space-indented Python 3 source code that will have the function specifier line def submission(y, o, t):\n appended to the beginning and a single 4-space-indent added to the beginning of each line. So:

if len(y) == 0: return False
else: return (not y[-1])

...is run as...

def submission(y, o, t):
   if len(y) == 0: return False
   else: return (not y[-1])

The function should return True to betray its opponent or False to trust its opponent. Returning any other value (such as if your return statement has been deleted and you return None`) is an error and will result in an automatic loss of that round.

o is a list of True-es or False-es, representing the opponent's previous moves when playing against you; o[n] equals the decision your opponent made in round n, and of course this list will be empty in the first trial. y is similar except it's your previous moves. t is the text of the opponent submission, formatted as specified above.

This is the program that will perform the contest:

[said code here]

Example Submissions: These will also be included in the actual contest.

Chronic Villain Syndrome:

#always choose 'betray'
return True
11111111 #padding

The Patsy:

#always choose 'trust'
return False
11111111 #padding

Do Onto Others...:

#always choose what opponent chose last round
if o:
   return o[-1]
return True

Professor X:

me="""
#Read opponent's mind and choose optimally.
exec("def e(y, o, t):\n"+t)
return False e(o, y, me)
#Opponent's choice this turn if we betray...
tb = e(o,y+[True], me)
#if we trust...
tt = e(o,y+[False], me)

#Opponents choice NEXT turn if we betray then betray...
tbb = e(o+[tb], y+[True], me)
#... betray then trust...
tbt = e(o+[tb], y+[False], me)
#... trust then betray...
ttb = e(o+[tt], y+[True], me)
#... trust then trust...
ttt = e(o+[tt], y+[False], me)

#Maps (your_choice,opponent_choice) to desirability
#Betrayed = -2, mutual betrayal = -1,
#mutual cooperation = 1, betray opp. = 2
v = {(False, True): -2, (True, True): -1,
     (False, False): 1, (True, False): 2}
#Best outcome next turn of trusting now
ftv = max(v[ttt], v[ttb])
#... of betraying now
fbv = max(v[tbt], v[tbb])
#value of betraying now
bv = v[tb] + fbv
#... of trusting now
tb = b[tt] + ftv

#Tuple comparison is done l to r, so
#this returns True if tv >= bv,
#False if bv > tv.
return max( (bv, True), (tv, False) )
"""
exec(me)

(p.s. this one gets disqualified every time - why is left as an exercise to the reader)

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  • \$\begingroup\$ So this is a KOTH? \$\endgroup\$ – Rɪᴋᴇʀ May 13 '16 at 22:07
  • \$\begingroup\$ And why is the professor X bot disqualified? \$\endgroup\$ – Rɪᴋᴇʀ May 13 '16 at 22:10
  • \$\begingroup\$ Finally, should there be a rule against running the opponents code to pick the best output for yourself? \$\endgroup\$ – Rɪᴋᴇʀ May 13 '16 at 22:12
  • \$\begingroup\$ It's a KOTH, yes, with a little bit of golf mixed in. "Professor X" gets disqualified because it will eventually play against itself, which means unbounded recursion, which means never halting and thus getting kicked out by the "must return in one second" rule. So any bot that tries to run its opponent is going to have to at least have some way of stopping its opponent from running it without corrupting the answer. All in all that kind of "mind reading" is cool enough and easy enough to defeat that I think it should be left in. \$\endgroup\$ – Schilcote May 14 '16 at 1:26
  • \$\begingroup\$ Modified quines aren't hard, and neither are narcissist programs. I think you should add a rule against that. I'll see if I can create a bot like that. \$\endgroup\$ – Rɪᴋᴇʀ May 14 '16 at 2:03
  • \$\begingroup\$ @EᴀsᴛᴇʀʟʏIʀᴋ Right, but it should be pretty easy to pull stuff like having side effects (store something in a global for example) or unwinding the stack to see who your caller is. \$\endgroup\$ – Schilcote May 14 '16 at 2:41
  • \$\begingroup\$ Can't you do something like this? Wouldn't that almost always win? \$\endgroup\$ – Rɪᴋᴇʀ May 14 '16 at 3:23
  • \$\begingroup\$ @No, for a few reasons; one, there'd probably be other "mind reader" variations (that's the whole reason Professor X is in there) which means you'd have to check for ALL of them, which would mean a massive character count. For another, I can think of at least two ways (stack unwinding & using globals to check for two calls that are given same sized y) to check if your opponent has done that and alter your decision accordingly. I don't want to make "mind readers" impossible, I just want to put interesting challenges in their way, and I think this ruleset does that. \$\endgroup\$ – Schilcote May 14 '16 at 17:33
  • \$\begingroup\$ @Shilcote okay, forgot that more mind readers could exist. \$\endgroup\$ – Rɪᴋᴇʀ May 14 '16 at 18:14
  • \$\begingroup\$ What does number_of_trials_failed mean? The number where you were disqualified? And why 4-space-indented? That guarantees that when you lose a space from the indentation you will lose the next three rounds due to parse errors. \$\endgroup\$ – Peter Taylor May 14 '16 at 19:14
  • \$\begingroup\$ @PeterTaylor No, when you're disqualified you're totally out, you don't even get a score. number_of_trials_failed is the number of times you got betrayed and lost. Good thought on the indentation; maybe it should delete one non-whitespace character? Or better yet, one statement? \$\endgroup\$ – Schilcote May 15 '16 at 0:29
  • \$\begingroup\$ This is likely going to be a one-up challenge. Given a set of programs, you'd always be able to write another program that beats all previously existing programs. \$\endgroup\$ – Nathan Merrill May 16 '16 at 18:37
  • \$\begingroup\$ I'm sure there'll be a lot of that, but I don't think it'll be exclusively that... \$\endgroup\$ – Schilcote May 17 '16 at 1:10
2
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Nested brackets in source code


Introduction

This challenge is somewhat unusual. Basically, your inputs are a correctly matched string of brackets S and a number n ≥ 0, and your output is the number of matched pairs at nesting level n in S. The outermost pairs are at level 0, those inside them are at level 1, and so on. The twist is that the string S is part of your source code, and incorrectly matched strings must result in compilation or runtime errors.

The task

Your task is to write four strings, A, B, L, and R, that satisfy the following conditions.

  • The strings L and R are non-empty and distinct. They represent a left and right bracket.
  • If S is a concatenation of Ls and Rs that is correctly matched, then the concatenation ASB is a valid program (full runnable program or function definition) is your programming language of choice. It takes an integer n ≥ 0 and outputs the number of L-R pairs in S at nesting level n.
  • If S is a concatenation of Ls and Rs that is not correctly matched, then ASB either fails to compile, or throws an error on every possible input.

Example

Suppose that the strings A, B, L, and R are BEGIN;, END;, DO(i++; and );, respectively, in an imaginary programming language. Then the string DO(i++;);DO(i++;DO(i++;);); is a correctly matched concatenation of Ls and Rs. On input 0, the program

BEGIN;DO(i++;);DO(i++;DO(i++;););END;

should output 2, because there are two matched pairs at level 0. However, the program

BEGIN;DO(i++;DO(i++;);END;

should result in an error, because the brackets are not correctly matched.

More examples

Here is a table of some programs, inputs and expected outputs.

Program          Input  Output
------------------------------
AB               <any>  0
ALRB             0      1
ALRB             1      0
ALLRRB           1      1
ALRLRB           0      2
ALRLLRRB         0      2
ALRLLRRB         1      1
ALLRLLLRRRLRRLRB 0      2
ALLRLLLRRRLRRLRB 1      3
ALLRLLLRRRLRRLRB 2      1
ALLRLLLRRRLRRLRB 3      1
ALLRLLLRRRLRRLRB 4      0
ARB              <any>  error
ALB              <any>  error
ARLB             <any>  error
ALLRB            <any>  error
ALLRRRLRB        <any>  error
ALLRRLLRB        <any>  error
ALLRRRLLLRRB     <any>  error

Rules and scoring

Your score is the sum of the lengths of the four strings, lower score being better. You must identify the strings in your answer. You are not allowed to read your source code directly or indirectly. Standard loopholes are also disallowed.


Sandbox notes

  • Is the task of counting pairs on a nesting level too easy? I don't want it to be trivial, but not so hard either that it shadows the source layout aspect. A slightly more difficult variant would be to count the number of peaks (substrings LR) at nesting level n.
  • Is the restriction of erroring on mismatched brackets interesting? I could also require the program to always return -1 in this case, or something similar.
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  • \$\begingroup\$ Shouldn't DO(i++;);DO(i++;DO(i++;)) be DO(i++;);DO(i++;DO(i++;););? \$\endgroup\$ – Peter Taylor May 14 '16 at 19:23
  • \$\begingroup\$ @PeterTaylor Yes it should, good catch. \$\endgroup\$ – Zgarb May 14 '16 at 19:25
  • \$\begingroup\$ Thought about this more. By making A end in ", L be (, R be ), and B start with ", I can reduce it to matching brackets in a string. If it's not a duplicate of an existing question, it has at least neutered the twist. \$\endgroup\$ – Peter Taylor May 16 '16 at 14:30
  • \$\begingroup\$ @PeterTaylor That's a valid approach, but I'm hoping that more quine-like solutions will be shorter, at least in "normal" languages. The error rule is intended to help with that too. \$\endgroup\$ – Zgarb May 16 '16 at 17:15
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Shortest program with unknown halting status

Rules

  1. Post a term on the binary λ-calculus (BLC) whose termination is unknown.

  2. If someone proves that your term does or doesn't terminate, you entry is disqualified.

  3. The term with the smallest number of bits on the BLC wins.

Don't forget to also post a quick description of what you did and the original source code, otherwise we will just have to trust your random string meets the specs!

Example submission

Size: 579 bits

Program: 01001001000100010001000101100111101111001110010101000001110011101000000111001110
10010000011100111010000001110011101000000111001110100000000111000011100111110100
00101011000000000010111011100101011111000000111001011111101101011010000000100000
10000001011100000000001110010101010101010111100000011100101010110000000001110000
00000111100000000011110000000001100001010101100000001110000000110000000100000001
00000000010010111110111100000010101111110000001100000011100111110000101101101110
00110000101100010111001011111011110000001110010111111000011110011110011110101000
0010110101000011010

Explanation: this term, if it terminates, reduces to a list with all church-encoded natural numbers of the sequence of Collatz (A006577) from 0 to 2^256. It is not known if collats(n) halts for all n; we only know up to about 2^64, so my submission satisfies the specification. For a longer explanation, I've set this repository. The original code was written on Caramel and also on the repository. Here is a brief:

-- Receives fix and a natural, returns the number of
-- recursive calls until the collatz function halts.
collatz fix n = (fix go n)

    -- The recursive search
    go go n = (succ (even_odd_or_leq_one n even odd leq1))
        even = (go go (half n))
        odd  = (go go (succ (mul n 3)))
        leq1 = 0
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    \$\begingroup\$ The only thing I'd like to comment on is that the challenge shouldn't be limited to BLC, and that other languages should be allowed so long as they are valid. For example, there is no reason to disallow a python solution that does the same thing as the above BLC code. \$\endgroup\$ – Zwei Jun 7 '16 at 3:39
  • \$\begingroup\$ I agree with @Zwei, but also ban IO and other nondeterministic code. \$\endgroup\$ – HEGX64 Jun 7 '16 at 9:56
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    \$\begingroup\$ As commented on main, you need to define the system of axioms which are permitted for the proof. \$\endgroup\$ – Peter Taylor Jun 7 '16 at 11:03
  • \$\begingroup\$ Fair enough. I feel like this will never get enough momentum for any relevant answer, though. I might just not ask it. \$\endgroup\$ – MaiaVictor Jun 7 '16 at 11:10
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    \$\begingroup\$ @Dokkat This challenge is interesting imo, there are just a few things that need to be worked out such as how does one prove that a program does or does not terminate. Remove the BLC only rule and define the axiom system and this challenge should be good for main. \$\endgroup\$ – Zwei Jun 7 '16 at 14:59
  • \$\begingroup\$ I think it's better to keep the language restriction. \$\endgroup\$ – feersum Jun 7 '16 at 20:18
  • \$\begingroup\$ @feersum for what reason? \$\endgroup\$ – Zwei Jun 8 '16 at 0:15
  • \$\begingroup\$ @Zwei the whole point of the challenge is that I want the program with the lowest Kolmogorov Complexity. It is hard to talk about this metric when you don't have a fixed language. There is a very clear and trivial measure of the complexity of a term on the binary lambda calculus. In fact, I'd say the only way for a code golf challenge to be an objective competition (not just a popularity contest) is by using a fixed language with a clear complexity metric, and the BLC is about as good as we know. \$\endgroup\$ – MaiaVictor Jun 8 '16 at 13:50
  • \$\begingroup\$ @Dokkat fair enough. \$\endgroup\$ – Zwei Jun 8 '16 at 15:23
  • \$\begingroup\$ That link is to binary combinatory logic, not binary lambda calculus. \$\endgroup\$ – CalculatorFeline May 31 '17 at 17:55
2
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Determine the winner of Beggar my Neighbour

The card game Beggar my Neighbour is boring in that the final outcome is entirely determined by the initial arrangement of the deck, so long as certain rules are followed for the order in which cards are picked up from the playing field and moved to decks.

The Game

  1. Both players are dealt 26 cards.

  2. Players play their top card alternately, starting with the player who won the previous stack or Player 1 at the beginning of the game.

  3. Play is interrupted when either player plays a picture card. In that case their opponent must play a number of cards equal to the value of the picture card above 10, i.e. Jack = 1, Queen = 2, King = 3, Ace = 4. The player then wins all the played cards which are returned to the bottom of their hand, unless the opponent themselves plays a picture card, in which case this rule interrupts their play.

  4. If at any point one of the players needs to draw a card from their deck, but their deck is empty, they immediately lose the game.

Example play

Player 1 starts with 7; Player 2 plays 3; subsequent plays are 9; 9; T; A; 6, J; A; 2, 3, 7, 6: Player 2 adds the cards 7399TA6JA2376 to his deck.
Player 2 starts: J; K; 9, 4, A; 5, 2, J; 2: Player 1 adds the cards JK94A52J2 to his deck.
Player 1 starts: 6; T; T; 5; 9; A; A; 3, K; K; 7, Q; 5, T: Player 1 adds the cards 6TT59AA3KK7Q5T to his deck.

The Challenge

Given two lists of cards in the players' decks, in any convenient format, output a truthy value if Player 1 wins, and a falsey value if Player 2 wins.

For convenience, a 10 card will be represented with a T, and face cards will be abbreviated (Ace -> A, King -> K, Queen -> Q, Jack -> J), so that all cards are one character long. Alternatively, ranks may be represented with decimal integers 2-14 (Jack -> 11, Queen -> 12, King -> 13, Ace -> 14) or hex digits 2-E (10 -> A, Jack -> B, Queen -> C, King -> D, Ace -> E). Since suits don't matter, suit information will not be given.

You may assume that all games will terminate at some point (though it may take a very long time), and one player will always run out of cards before the other.

There are variations for more than two players but they will not be considered here.

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Pareto frontier

A point (x1,y1) dominates another point (x2,y2) if both x1≥x2 and y1≥y2. A set of points is a Pareto frontier if no point dominates another point. In other words, any increase in one coordinate must be met with a decrease in the other coordinate.

Your task is to decide whether a given set of points is a Pareto frontier.

Input: A collection of two or more points, which are pairs of positive integers. No two points will have the same x-value or the same y-value. You may not assume the points are given in a particular order.

Output: A consistent Truthy value if it is, and a consistent Falsey value if it's not.

True:

[(12, 1), (6, 4)]
[(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)]
[(4, 4), (3, 8), (12, 3), (20, 1)]
[(106, 106), (107, 102), (104, 127)]

False:

[(5, 9), (4, 8)]
[(1, 1), (11, 11)]
[(5, 3), (2, 4), (7, 7), (1, 2)]
[(15, 2), (7, 8), (4, 14), (6, 6)]

Sandbox:

  • Is it better to do the decision problem, or the filtering problem of finding the upper Pareto frontier of non-dominated points?

  • Should the input be allowed to be taken pre-zipped, as two lists of n numbers? What about a 2D 2-by-n array versus an n-by-2 array?

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  • \$\begingroup\$ I'd recommend only doing the filtering problem if there is a nicer solution than subsets -> filter on decision -> take longest that you can enforce with restraints of some kind. Otherwise I feel like it might be kind of chameleon-y. \$\endgroup\$ – FryAmTheEggman Jul 13 '16 at 1:49
  • \$\begingroup\$ The decision problem can be done as filter and check equality; or sort on one axis and check that the other axis is in reverse order. The filter problem can be done by filtering the power set with the decision problem and taking the longest; or by folding a removal of dominated points. IMO the filter problem is more interesting, because sort builtins are very common and typically cheap. \$\endgroup\$ – Peter Taylor Jul 13 '16 at 10:37
  • \$\begingroup\$ @PeterTaylor The largest Pareto-incomparable subset might contain points dominated by the rest of the set. I expect filtering would be done by folding removal. Does that affect your opinion? \$\endgroup\$ – xnor Jul 20 '16 at 5:16
  • \$\begingroup\$ Ah, true. Folding removal still seems more interesting to me than sorting and checking reverse order. \$\endgroup\$ – Peter Taylor Jul 20 '16 at 7:39
2
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Lets play Stratego!

Stratego is a game of imperfect information which centers around strategy foresight and deception.

https://en.wikipedia.org/wiki/Stratego

Your mission should you choose to accept it. is to write a bot which either uses polymorphism in a java class file, and/or communicate with the arena via an stdin/stdout wrapper.

The arena code and/or stdin out wrapper will be made available if there is interest. Here is the summary.


Piece Summary

? = Unkown enemy piece
lowercase characters = your pieces
Upper case or special characters (!@#$%^&*()) or F B is the enemy pieces
f or F represents an immobile flag if the enemy moves on to this square it is game over.
b or B represents an immobile if any hostile piece steps on to it then it dies. If an enemy miner (8) steps on it will succesfully defuse or capture the bomb.
s or S gets killed when attacked by any piece, but can kill the 10 by stepping on it.
9 or ) can move any number of squares orthagonally.
Otherwise all pieces can capture pieces with a higher number than themselves. I.E 1 captures two which captures three which captures a 4 which captures a five etc...
When you attack a piece, the values of both pieces are revealed (thus the enemy will no longer see a "?") If the attacking piece is stronger it takes the spot, if it is weaker, it dies and the defender remains unharmed, and if they are equal both disappear. 

Spaces empty are indicated with a space. Spaces with an impassable lake are indicated with an L. The wikipedia article has some great sample strategies and explanations. The protocol for communication will involve Outputting four zero indexed coordinates for the start and end locations of a piece. Invalid moves will simply result in no action taken.

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  • \$\begingroup\$ do you have a bot API? \$\endgroup\$ – noɥʇʎԀʎzɐɹƆ Jul 23 '16 at 18:56
  • \$\begingroup\$ Not yet. It was something I was pondering in my head. If there is any interest in this I can slap together an api. @AgentCrazyPython do you think this is a doable challenge \$\endgroup\$ – Rohan Jhunjhunwala Jul 23 '16 at 19:27
  • \$\begingroup\$ I think it's doable. You'd get a whole ton of rep for it. \$\endgroup\$ – noɥʇʎԀʎzɐɹƆ Jul 23 '16 at 21:17
  • \$\begingroup\$ @AgentCrazyPython lol I need the rep. I will write up an arena controller. Should I do it through an stdin/stdout interface or an abstract class? I think I will first write an interface that allows you to inherit from an abstract class. Than I would write a wrapper. The only thing is that I have no clue how to go about implementing a sample bot. \$\endgroup\$ – Rohan Jhunjhunwala Jul 23 '16 at 21:23
  • \$\begingroup\$ depends on if you want to let one language or all languages. I think you should go just one langauge: JS. JS can be demoed easily in the browser and it's great for writing bots. the spacewar! challenge is a good example \$\endgroup\$ – noɥʇʎԀʎzɐɹƆ Jul 23 '16 at 21:25
  • \$\begingroup\$ I need to learn JS, but I am slightly discouraged after learning about jsf*** lol. It seems like an <s>odd</s> unique language lol. \$\endgroup\$ – Rohan Jhunjhunwala Jul 23 '16 at 21:27
  • \$\begingroup\$ I will first prototype a Java arena, and then I will first write a wrapper (probably by modifying @Moogie 's) wrapper. Then I will stare at a wall idly considering porting it to Java script. \$\endgroup\$ – Rohan Jhunjhunwala Jul 23 '16 at 21:28
  • \$\begingroup\$ @AgentCrazyPython ^ Forgot to ping you on the above comments \$\endgroup\$ – Rohan Jhunjhunwala Jul 23 '16 at 21:29
  • \$\begingroup\$ js is super easy to learn. JS has really similar syntax \$\endgroup\$ – noɥʇʎԀʎzɐɹƆ Jul 23 '16 at 22:10
  • \$\begingroup\$ Yeah, but the dynamic typing stuff really gets me. I do have some python background which should help, but I am I java dev mainly. @Agent CrazyPython That being said, I will try to learn java as stack snippets do appeal to me. \$\endgroup\$ – Rohan Jhunjhunwala Jul 23 '16 at 23:00
  • \$\begingroup\$ java ≠ javascript!!!!!!! why is dynamic typing hard - it's looser than static typing \$\endgroup\$ – noɥʇʎԀʎzɐɹƆ Jul 23 '16 at 23:48
  • \$\begingroup\$ @AgentCrazyPython I know that JS!=Java lol. THats why I am scared of it. For someone reason I do prefer being sure of type safety as I code rather than at run time \$\endgroup\$ – Rohan Jhunjhunwala Jul 24 '16 at 3:39
  • \$\begingroup\$ Let us continue this discussion in chat. \$\endgroup\$ – Rohan Jhunjhunwala Jul 24 '16 at 3:39
2
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Reversing the Game of Life

Though computing successive generations of Conway's Game of Life may seem simple enough, reversing the process is not.

Given an initial board configuration in Conway's Game of Life, either return a predecessor, or some value indicating that it is a Garden of Eden (and therefore contains an orphan), meaning it has no predecessor. A predecessor is a previous generation of a board configuration, meaning that after running the game for a some number of generations, you will arrive at the successor generation, the board configuration that you were initially given.

Any given configuration may have zero or more predecessors. Some have infinitely many, such as still life patterns.

It will probably be simplest to find a parent, a predecessor configuration from the preceding generation, such that reaching the successor takes one generation.

Two parents of the Block

If given a state bounded by a 6x6 rectangle, for example, a parent (if any) will be found by searching the 8x8 rectangle around it. This is a fastest-code challenge, since brute force solutions would take a long time, even for relatively small inputs.

All Game of Life patterns bounded by a 6x6 rectangle have a predecessor. Source


Example Input

Input can be how you like. If you prefer a list of active points, or a matrix of Booleans, that's fine. Just make your output be the same format.

This input represents a block:

Block still life

[(1, 1), (1, 2), (2, 1), (2, 2)]

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

The original Garden of Eden:

Garden of Eden

All cells outside the image are dead (white).


Example Output:

This is a parent to the block.

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

Garden of Eden:

No predecessor

Resources:

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  • \$\begingroup\$ Not exactly a dupe, but searching for a parent was how I approached an earlier fastest-code game-of-life question. \$\endgroup\$ – Peter Taylor Jul 22 '16 at 21:34
  • \$\begingroup\$ Why is it undecidable? Theoretically, you could just enumerate over every possible parent state until one generates the pattern (or you run out of possible parents in the case of a Garden of Eden). \$\endgroup\$ – LegionMammal978 Jul 22 '16 at 21:41
  • \$\begingroup\$ @LegionMammal978 I don't understand either, but from the linked Wikipedia article: For one-dimensional cellular automata, orphans and Gardens of Eden can be found by an efficient algorithm, but for higher dimensions this is an undecidable problem. Nevertheless, computer searches have succeeded in finding these patterns in Conway's Game of Life. \$\endgroup\$ – mbomb007 Jul 22 '16 at 21:43
  • \$\begingroup\$ I think it's saying that there is no way that a decidable AND efficient algorithm exists? Not sure. \$\endgroup\$ – mbomb007 Jul 22 '16 at 21:44
  • \$\begingroup\$ @mbomb007 Maybe it's talking about polynomial-time algorithms or something, last time I checked, bruteforce wasn't undecidable \$\endgroup\$ – LegionMammal978 Jul 22 '16 at 21:53
  • \$\begingroup\$ @LegionMammal978 K, I'll remove that part. Is there enough explanation, do you think? There is probably required reading from the links I provided in order for a decent non-brute-force solution to be developed. \$\endgroup\$ – mbomb007 Jul 22 '16 at 21:55
  • \$\begingroup\$ @mbomb007 Looks fine to me \$\endgroup\$ – LegionMammal978 Jul 22 '16 at 21:56
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    \$\begingroup\$ I feel like the performance of an algorithm will heavily depend on how test cases are generated. You should specify this. \$\endgroup\$ – xnor Jul 22 '16 at 23:17
  • \$\begingroup\$ @LegionMammal978: The "undecidable" part is probably due to how you can have spaceships crash together from a distance to make new patterns. \$\endgroup\$ – El'endia Starman Jul 23 '16 at 12:39
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    \$\begingroup\$ @El'endiaStarman However, all that matters for finding the previous generation is the part of the spaceship within the n+2 X m+2 box that can affect the pattern \$\endgroup\$ – LegionMammal978 Jul 23 '16 at 14:34
  • \$\begingroup\$ @LegionMammal978: Hmm, true. I don't know either. \$\endgroup\$ – El'endia Starman Jul 23 '16 at 14:47
  • \$\begingroup\$ @LegionMammal978, the undecidable problem is "Given the rules for a 2-dimensional cellular automaton, does any Garden of Eden exist with those rules?" \$\endgroup\$ – Peter Taylor Jul 25 '16 at 20:38
  • \$\begingroup\$ @PeterTaylor See the Garden of Eden theorem. It says that any CA with multiple patterns that evolve to the same pattern must contain some Garden of Eden (presumably by a variant of the pigeonhole principle). \$\endgroup\$ – LegionMammal978 Jul 25 '16 at 21:04
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    \$\begingroup\$ @mbomb007 One way to do the test cases would be random grids. Most of these would likely have many predecessors and be doable with a heuristic search. If you want hard examples, maybe it's possible to generate ones with few predecessors and mix in gardens of eden. I don't know how doable that is. For returning a g.o.e. in an mxn grid, can't one just return a known minimal one if it fits and falsey otherwise? Generating optimal gardens of eden seems really hard even if you have good heuristics. \$\endgroup\$ – xnor Jul 25 '16 at 21:48
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    \$\begingroup\$ @LegionMammal978, so the existence of twins is logically also undecidable, but I'm not sure why you think that's important to point out. \$\endgroup\$ – Peter Taylor Jul 26 '16 at 8:42
2
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Do X without Y​ again!

Here is an X done with Y:

YyYy        YyYy
 YyYy      YyYy 
  YyYy    YyYy  
   YyYy  YyYy   
    YyYyYyYy    
     YyYyYy     
     yYyYyY     
    yYyYyYyY    
   yYyY  yYyY   
  yYyY    yYyY  
 yYyY      yYyY 
yYyY        yYyY

As you can see, there are 92 Y/ys in this X. Conveniently, there are also exactly 92 printable ASCII characters in the range ! to ~ if you exclude Y and y. Your challenge is to write a program to output or a function to return the above X with all the Y/ys replaced by any permutation of those 92 characters. Leading and trailing white space is permitted. Shortest code wins!

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  • \$\begingroup\$ Should that say "each of those 92 characters" (in the range ! to ~, excluding Yy)? \$\endgroup\$ – nanofarad Jul 30 '16 at 13:15
  • \$\begingroup\$ Just drop the exclamation mark already. \$\endgroup\$ – Leaky Nun Jul 30 '16 at 13:15
  • \$\begingroup\$ Maybe I'm misunderstanding the question, but you seem to be requiring all answers to be exactly 92 characters long and within that constraint aiming to be the shortest... \$\endgroup\$ – Peter Taylor Jul 30 '16 at 17:12
  • \$\begingroup\$ @PeterTaylor: He doesn't specify a length of an answer... He says that in the above ascii X each Y is replaced by another character of the 92 printable characters that are not y or Y; each char is only used once. \$\endgroup\$ – KarlKastor Jul 30 '16 at 19:41
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    \$\begingroup\$ I'd make it obvious that a permutation of the 92 characters are to be used (no repeating a character) \$\endgroup\$ – Nathan Merrill Jul 31 '16 at 1:13
  • \$\begingroup\$ Have you abandoned this altogether already? \$\endgroup\$ – Leaky Nun Aug 18 '16 at 0:16
  • \$\begingroup\$ @LeakyNun It only got 1 upvote, unless you know a better way of telling when it's ready. \$\endgroup\$ – Neil Aug 18 '16 at 0:17
2
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The Enemy's Gate is Down! (Ender's Game)

Please note this is a work in progress

Your challenge should you choose to accept it is to play ender's game to win style, in honor of Martin Ender first receiving 100k rep! You will split up into teams. You are red if you are an even post and blue if you are odd.

Here is a simplified ascii representation of the map.

WWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW
W       1                        F                         2    W
W                  WwW                                          W
W        WWW                                                    W
W                                            WWWWWWWW           W
W                WWWWWWWWWWWWWWWWW                              W  
W                                                               W  
W                                                               W  
W             WWWWWWWWW                    WWWWWWWWWWW          W  
W                                                               W  
W                              WWWWW                            W  
W                                                               W
W       3                        G                         4    W
WWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW

Your gate is represented by the "F" and the enemies gate is "G". W's represent walls which will be arranged in an unbiased distribution. The goal is to get to the enemies gate first. Each turn you will output a number from 0-15.

 0-7 will allow you to "push" off of a block assuming there is a block in the 8 blocks closest to you (orthagonally or diagonally")
8-17 allows you to shoot a bullet in a given direction which will continue to travel in a given direction. 
It will break upon hitting a wall or another player. The bullet represented by "P" on the ascii map is iterated after each step. Any player that drifts into it will also be killed. 

Directions

567
4U0
321

The winning team is the team which either captures the flag first the last man (bot/woman) standing!

All submission are deterministic java programs. A psuedorandom generator will be provided. The winning team is the team which wins the first game, (if there are ties).

Games will be halted after a significant number (1000) turns. I may allow multiple copies of the same bot on each team to make it more interesting.

Collisions with players are treated like collisions with walls.

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  • \$\begingroup\$ What about collisions between players? \$\endgroup\$ – Destructible Lemon Aug 3 '16 at 2:44
  • \$\begingroup\$ Both die (i guess) \$\endgroup\$ – Rohan Jhunjhunwala Aug 3 '16 at 3:07
  • \$\begingroup\$ @DestructibleWatermelon a \$\endgroup\$ – Rohan Jhunjhunwala Aug 3 '16 at 3:08
  • \$\begingroup\$ Would it be possible to add an Stdin/out wrapper? \$\endgroup\$ – Destructible Lemon Aug 3 '16 at 3:18
  • \$\begingroup\$ What if friendly bots can be in the same square, and push off each other to confuse the enemy? That seems cool. Probably make it so it doesn't affect the non-pushing bot just to reduce troublemaker bots ;) (Also, I guess each player would need an indication that they have a teammate in the same square or something) \$\endgroup\$ – Destructible Lemon Aug 3 '16 at 3:32
  • \$\begingroup\$ @DestructibleWatermelon ok I can make it so all players atreated like walls u can push off of. \$\endgroup\$ – Rohan Jhunjhunwala Aug 3 '16 at 11:46
  • \$\begingroup\$ It would be nice for the programs to be able to be in any language (using subprocess pipe) \$\endgroup\$ – Blue Aug 3 '16 at 15:11
  • \$\begingroup\$ @Blue I can write an stdin/stdout wrapper once I get an arena. The onl issue is that by starting subproceses the performance may be significantly harmed \$\endgroup\$ – Rohan Jhunjhunwala Aug 3 '16 at 16:56
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Golf an InterpretMe interpreter (in any language other than InterpretMe)

This is a very simple challenge.

The joke language InterpretMe consists of one command; *, which causes the program to take input of an InterpretMe program and execute it. An InterpretMe program will interpret as many InterpretMe programs as there are * in input. Your goal is to create a program that interprets InterpretMe in as few bytes as possible.

Test cases consist not of input and output, but input and termination. A newline denotes a new input to be interpreted as InterpretMe.

1. *   (executes input as an interpret me program, finishes)
2. *   (^same)
3. **  (executes input as an interpret me program, then does this again after the first program is done, finishes)
4. hi  (nothing, first star of previous line finishes)
5. **  (same as the other two star line)
6. hi  (nothing, first star of previous line finishes)
7. hi  (nothing, second star of line 5 finishes, so second star of line 3 finishes, so line 2 finishes, so line one finishes)
[termination]

hi  (does nothing and finishes)
[termination]

*hi  (executes inputted program, finishes)
*yo  (executes inputted program, finishes)
hey  (nothing)
[termination]

Sandbox

How can I make this challenge more clear? I understand that this challenge is simple, but that is part of the point; it sees how small a program can be for this purpose

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  • 2
    \$\begingroup\$ To make the challenge more clear, you could supply the specification for the language which has to be interpreted. It shouldn't be necessary for each answerer to search for the basic definition of the problem. \$\endgroup\$ – Peter Taylor Aug 3 '16 at 14:46
  • \$\begingroup\$ @PeterTaylor, Excuse me, it's right there. An InterpretMe interpreter takes input, and runs an InterpretMe interpreter (which takes new input) as many times as "*" occurs in input. (dumb blockquotes) \$\endgroup\$ – Destructible Lemon Aug 3 '16 at 23:09
  • \$\begingroup\$ Maybe if you state up front that it's a joke language that would prime the reader to not expect it to do anything useful. \$\endgroup\$ – Peter Taylor Aug 4 '16 at 7:19
  • \$\begingroup\$ @PeterTaylor Thanks \$\endgroup\$ – Destructible Lemon Aug 4 '16 at 7:31
2
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I wanna be the very best...

Wow, there's so many Pokemon in my Pokedex! Well, there seems too many... I wish there was some good way of sorting all of them!

Can you help?

Given a list of Pokemon with their stats - level and HP - and a sorting criterion, output the sorted list.

An example input would be:

([["Squirtle", 2, 10], ["Charizard", 58, 140], ["Mew", 75, 160], ["Pichu", 10, 25]], "alphabetical")

As you can see, each Pokemon is shown like so:

[pokemon_name, level, hp]

There are three sorting options:

  • "alphabetical": The Pokemon are sorted alphabetically.
    • If two or more Pokemon have identical names, then they are sorted by level, then HP.
  • "level": The Pokemon are sorted by level.
    • If two or more Pokemon have the same level, then they are sorted alphabetically, then by HP.
  • "hp": The Pokemon are sorted by HP.
    • If two or more Pokemon have the same HP, then they are sorted alphabetically, then by level.

Specs:

  • You are guaranteed that:
    • The hp and level of every Pokemon are integers.
    • The level of a Pokemon will not exceed 100.
    • The hp will not exceed 200.

If two or more Pokemon share all stats, then they can be arranged however you like.

The output will be the sorted list.

This is , so shortest code in bytes wins.

Meta:

  • Is the challenge too easy/hard?
  • Any improvements to my explanation?
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  • 1
    \$\begingroup\$ Could you perhaps add some test cases? Also, since you haven't mentioned it I'm assuming the input is unrestricted? (I.e. we're allowed to use an array as parameter, a comma-separated string, a list, etc. Everyone's own choice?) And one very important question, does the list potentially include all 721 Pokémon (with more incoming in generation 7..) \$\endgroup\$ – Kevin Cruijssen Aug 4 '16 at 12:38
  • \$\begingroup\$ I think this challenge is pretty chameleon-y. I'm not sure if there is a dupe target floating around, but even if it isn't a dupe I'm not convinced this adds very much to the basic sorting challenges. \$\endgroup\$ – FryAmTheEggman Aug 4 '16 at 18:26
  • \$\begingroup\$ @KevinCruijssen It can contain anything, really - it could contain all 721 Pokemon available, but it doesn't have to contain Pokemon at all. \$\endgroup\$ – clismique Aug 5 '16 at 7:58
  • \$\begingroup\$ @DerpfacePython I know, it's just a string. My last question was kinda a sarcastic one.. :) \$\endgroup\$ – Kevin Cruijssen Aug 5 '16 at 8:25
2
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Make a Four Color Map

The Four Color Theorem states that it is possible to color any map separated into contiguous regions using only four colors such that no two adjacent regions are the same color. While the Five Color Theorem has been proven, no proof exists for only using four colors (though there have also been no counterexamples). Given an image containing white regions separated by black borders, generate a four color map. You may assume that the borders of the image are also region borders and that shared corners do not count as adjacencies.

Related: Four Color Theorem

Input

An image in a standard format containing white regions separated by black borders

Output

Displaying or writing an image file in any standard format that contains the original image colored according to the Four Color Theorem

Examples

Your colorings do not need to match mine, they just need to be valid solutions ====> ====>

Note that the map of the US is not colored by state, it is colored by contiguous borders

This is so shortest code wins!

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  • \$\begingroup\$ Isn't the first example invalid? I see two adjacent red regions in the bottom left. \$\endgroup\$ – Business Cat Aug 5 '16 at 17:11
  • \$\begingroup\$ Whoops! Thanks, I'll fix that. \$\endgroup\$ – theLambGoat Aug 5 '16 at 17:14
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    \$\begingroup\$ I thought the four-color one was definitively proven a couple years ago? \$\endgroup\$ – AdmBorkBork Aug 5 '16 at 18:19
  • \$\begingroup\$ It might have been, I was just going off of wikipedia. I just read the Wolfram article on it and it looks like it was proven for maps that are flat (and maybe all maps?) but I don't 100% follow it so not positive. \$\endgroup\$ – theLambGoat Aug 5 '16 at 18:28
  • 1
    \$\begingroup\$ 1. The four colour theorem was proven in the 70s. 2. This is essentially a dupe of this question but with a really bad input format. \$\endgroup\$ – Peter Taylor Aug 5 '16 at 21:09
  • \$\begingroup\$ Here's the relevant Wikipedia section if you're looking for more info on the proof \$\endgroup\$ – trichoplax Aug 6 '16 at 14:42
  • \$\begingroup\$ The theorem was "disproved" by counterexample on some mailing list I believe. It was a joke. Someone produced a pretty complex map and claimed that it couldn't be colored with only 4 colors. I think he was just seeing if he could get people to waste time looking at it. \$\endgroup\$ – Liam Aug 10 '16 at 23:48
2
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Your job is to take an input like below:

Hello, world!

π is cool!

and output it as a series of <kbd>.

Every character, including enter and space get their own <kbd> HTML element, like so:

Hello,SPACEworld!NEWLINE

OPT + PSPACEisSPACEcool!

Other characters like space and newline that are invisible and typeable (so no option, command, control, function, escape, delete, arrow keys, etc.) are all uppercase, like SPACE and NEWLINE. OPT is an exception because it is used to create new characters. The code for these characters are:

<kbd>SPACE</kbd> <kbd>NEWLINE</kbd>

Special characters needing to be typed using option use the following format OPT + P instead of π. The following code is used for these:

<kbd><kbd>OPT</kbd> + <kbd>P</kbd></kbd>

The same applies to characters that are typed using OPT + SHIFT + K instead of .

We're assuming this is a QWERTY layout computer.

Here are the available option keys:

Option keys

Any of the orange keys are not allowed in this.

Here are the OPT + SHIFT keys:

Option shift keys

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  • 5
    \$\begingroup\$ +1 for well laid out challenge -1 for being mac specific \$\endgroup\$ – Rohan Jhunjhunwala Aug 9 '16 at 19:33
  • \$\begingroup\$ @RohanJhunjhunwala It's not technically Mac-specific, but rather uses the same shortcut keys to type special characters as a Mac. \$\endgroup\$ – haykam Aug 9 '16 at 19:35
  • 1
    \$\begingroup\$ Yeah, Ina all seriousness here have an upvote, It does seem like targetting the opt key may be somewhat confusing for some S̶a̶n̶e̶ pc users. \$\endgroup\$ – Rohan Jhunjhunwala Aug 9 '16 at 19:39
  • 1
    \$\begingroup\$ For each letter you check which of the three character sets it is in then add some fixed strings to it. It doesn't seem to me like there is much interesting stuff to do in there? I'm not sure this adds much to the other keyboard focused challenges that we have. \$\endgroup\$ – FryAmTheEggman Aug 9 '16 at 20:23
  • \$\begingroup\$ Do all of those characters even have Unicode codepoints? \$\endgroup\$ – Peter Taylor Aug 10 '16 at 5:31
2
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How synchronized are my clocks?

I have two clocks, A and B, and A always shows the exact time, however B is off by a certain amount of time, that I'd like to find out. (I know A and B run at the exact same speed.) I cannot read the exact time of both simultaneously, that means I can only switch back and forth and read the time sequentially. (For the sake of simplicity, both show their time as a real number (lets say hours), and each real number encodes an unique point in time.)

So here is an example: B: 1, A: 3, A: 5, B: 4.5. We see because of the first two entries B: 1, A: 3 that B is behind A by at most 2 hours, and that because of the last two entries A: 5, B: 4.5 we see that B is behind A by at least half an hour. So the possible interval of how much B is off is [-2,-0.5].

Challenge

Given a list of timestamps with their labels, return the possible interval by how much B can be off.

Details

  • The list can be in any convenient format like [(timestamp, label),...] or as two lists [timestamp,...],[label,...] e.t.c
  • You can assume that all the readings from A are in ascending. (and the same for B)
  • If there is no such time interval, output something falsy.

Testcases:

(more to be added)

label  A  B  A
time   1  4  5
output [-1,3]

label  A B B A
time   1 2 5 2.5
output false

Meta:

  • Should only valid cases be considered, or should there be a check for invalid ones (remove point 2 or 3 of the details?)

  • Should the challenge be restricted to alternating readings (ABABABABABA) or should random ones be allowed (ABBAAABABAABBBABBBBB)? Restricting to alternating readings would make it pretty much trivial.

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  • \$\begingroup\$ IMO go with alternating readings \$\endgroup\$ – Downgoat Aug 11 '16 at 23:26
  • \$\begingroup\$ @Downgoat But that makes it almost trivial I think... \$\endgroup\$ – flawr Aug 11 '16 at 23:27
  • 1
    \$\begingroup\$ We see that B is at least two hours behind, but at most half an hour behind. I find that wording confusing. How about this? We see (B: 1, A: 3) that B is behind A by at most 2 hours, and that (A: 5, B: 4.5) B is behind A by at least half an hour. \$\endgroup\$ – Luis Mendo Aug 11 '16 at 23:42
  • \$\begingroup\$ @LuisMendo Thank you, I'm going to correct that. \$\endgroup\$ – flawr Aug 12 '16 at 8:41
2
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Print a booklet

I want to take a PDF document and put four pages of that document onto the front and four pages onto the back of a sheet of paper. Then I'll fold and cut that page so that I end up with a 1/4─size booklet that holds 8 pages. This would produce two folios comprising one signature.

           Fold here
            ↓
       ┌────┬────┐
       │    │    │
       │    │    │ ← Outer folio
Cut    │    │    │
here → ╞════╪════╡
       │    │    │
       │    │    │ ← Inner folio
       │    │    │
       └────┴────┘

LaTeX's pdfpages package allows me to place 2x2 pages of a PDF per sheet of paper with a given page ordering.

If I were to specify the option pages={2,7,4,5,8,1,6,3} to pdfpages, I would get:

       Front of sheet

        (1)     (8)     Back of sheet
          \     /
        ┌────┬────┐      (7)     (2)
        │    │    │        \     /
Outer   │  2 │  7 │      ┌────┬────┐
folio → │    │    │      │    │    │
        ╞════╪════╡      │  8 │  1 │ ← Outer folio
Inner   │    │    │      │    │    │
folio → │  4 │  5 │      ╞════╪════╡
        │    │    │      │    │    │
        └────┴────┘      │  6 │  3 │ ← Inner folio
        /          \     │    │    │
     (3)            (6)  └────┴────┘
                        /           \
                      (5)            (4)

(Numbers put in parenthesis are referring to the back of the page.)

Why do we need that strange order of numbers? So that the fronts and backs of pages line up when put into the signature. This is how you read the book:

    Start here at (1) on the back. Continue to 2 on the front.
      │
      │      End here at (8)
      ↓     /
    ┌────┬────┐
    │    │    │
    │  2 │  7 │ ← Outer folio
    │    │    │
    └────┴────┘
      │
On to (3) in
inner folio
      │
      ↓     ↑
          On to (6)
          in outer folio
            │
    ┌────┬────┐
    │    │    │
    │  4 │  5 │ ← Inner folio
    │    │    │
    └────┴────┘

But that's just for two folios from one sheet of paper. What if I want to use two sheets of paper, make four folios, and still combine all of them in one signature?

            Start at (1)
             │
             │      End here at (16)
             ↓       │
            (1)    (16)  (5)    (12)
              \     /      \     /
            ┌────┬────┐  ┌────┬────┐
            │    │    │  │    │    │
Outermost → │  2 │ 15 │  │  6 │ 11 │ ← Folio #3
 folio (#1) │    │    │  │    │    │
            ╞════╪════╡  ╞════╪════╡
            │    │    │  │    │    │
 Folio #2 → │  4 │ 13 │  │  8 │  9 │ ← Innermost
            │    │    │  │    │    │    folio (#4)      
            └────┴────┘  └────┴────┘
              /     \      /     \
            (3)    (14)  (7)    (10)

            Front of      Front of
             sheet 1       sheet 2

And your LaTeX option would be:

pages={2,15,4,13,16,1,14,3,6,11,8,9,12,5,10,7}

Objective

Write a function taking an integer n of the number of pages in the final booklet (8 and 16 in the examples above) and returning a list integers (of length n and ranging from 1 to n) for the page numbers in the right order.

Example:

> f(8)
=> [2,7,4,5,8,1,6,3]

> f(16)
=> [2,15,4,13,16,1,14,3,6,11,8,9,12,5,10,7]

Since we're dividing a sheet of paper into 4 pieces and using front and back, the input is always a multiple of 8. If the input is not a multiple of 8, the output is not defined but would prefer that it's rounded up to the next multiple of 8.

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  • 1
    \$\begingroup\$ Is the code required to work for integers that don't fit and leave unused pages? Should such integers be rounded up to the next integer that fits? \$\endgroup\$ – trichoplax Aug 12 '16 at 11:40
  • 2
    \$\begingroup\$ 1. I don't understand the repeated mention of US Letter. Surely this is completely independent of the paper size? 2. I don't understand the first two diagrams. It seems to me that the instructions in the first diagram give a booklet with page order 2,3,4,1, and the folds in the two diagrams can only be consistent with each other if one of the 1s is turned upside down. 3. The meaning of folio doesn't seem to be consistent between the 4-page example (which has two folios, implying that a folio is a half-sheet of paper) and the 8-page example (which has just an outer and an inner folio). \$\endgroup\$ – Peter Taylor Aug 12 '16 at 12:09
  • \$\begingroup\$ I specified what to do with integers that don't fit, removed mention of US letter, and simplified the diagrams. \$\endgroup\$ – Caleb Paul Aug 12 '16 at 16:52
  • \$\begingroup\$ @PeterTaylor, the first example is 8 pages (1 physical sheet of paper divided into 2 folios), the second example is 16 pages (2 physical sheets of paper divided into 4 folios). A folio is just a half sheet of paper. I cleaned up the terminology of sheet vs page in the question. \$\endgroup\$ – Caleb Paul Aug 12 '16 at 16:56
  • \$\begingroup\$ Ok, I think it's clearer now, but to check that I've understood what the task is. Would the following serve as a reference implementation? Given n which is a multiple of 8, form the array [1 2 ... n-1 n]. Then while the array is non-empty, remove the first four elements and the last four elements, apply permutation [2 7 4 5 8 1 6 3] to them, and recurse on the remaining elements. \$\endgroup\$ – Peter Taylor Aug 12 '16 at 17:40
  • \$\begingroup\$ Might turn into something interesting, but if you just print, cut, fold, then sort and glue, there is nothing keeping you from switching upper with lower halfs, turning half-pages around or swapping half-pages between pages. You either need to specify the exact way the paper will be combined or you should add information which solutions are acceptable. \$\endgroup\$ – MarLinn Aug 13 '16 at 1:20
2
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Count Langford pairings

A Langford pairing is a permutation of the numbers 1, 1, 2, 2, ..., n, n such that there is one number between the 1s, two numbers between the 2s, etc. E.g. (with the pairs marked)

+-----------+ +---------------+
| +-----+   | | +---------+   |
| |     |   | | |         |   |
5 2 8 6 2 3 5 7 4 3 6 8 1 4 1 7
    | |   |       | | | |   |
    | |   +-------+ | | +---+
    | +-------------+ |
    +-----------------+

If we reverse a Langford pairing then obviously we get another Langford pairing. The number of distinct Langford pairings (i.e. modulo this symmetry) for given n is OEIS sequence A014552.

Write a program which takes n as either a command-line argument or on stdin and prints the number of distinct Langford pairings for that n. You may assume that the input given will be a positive integer no greater than 32.

To avoid hard-coding, your program must be capable of calculating the number of Langford pairings for n=32, optionally modulo a number of your choice which is at least 230; and the only case splitting permitted for valid input is to split on the value of n % 4.

To avoid brute-forcing, your program must be capable of calculating the number of Langford pairings for n=16 in less than 15 minutes on my reference machine. (TODO). The standard approach is an algebraic technique due to Mike Godfrey and works by evaluating a generating function at {-1,1}^2n, but variants such as evaluating Godfrey's generating function at {0,1}^2n and using inclusion-exclusion are also possible.


The time limit is about twice the time required by my (partially optimised) reference solution, an algebraic approach in Java; a fully optimised approach in C should have a lot of slack. That gives people a trade-off in the symmetries they use, and should allow slow scripting languages to submit valid answers but at a penalty of having to spend more code on handling symmetries than faster languages.

However, I'm worried that it might allow trivial modification of answers to Langford strings , so I probably need a reference implementation which works by enumeration for comparison.

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  • 4
    \$\begingroup\$ Wow, that's really unhelpful. That page for A014552 has a forumla of a(n) = A176127(n)/2, and guess what the page for A176127 has? Yep, a(n) = 2 * A014552(n). :P \$\endgroup\$ – Doorknob Oct 8 '14 at 11:38
  • \$\begingroup\$ @PeterTaylor could you give us some hints on optimisation? As in, the basic hacks that nearly everyone will do? \$\endgroup\$ – Soham Chowdhury Oct 8 '14 at 12:39
  • \$\begingroup\$ There's this now. \$\endgroup\$ – Martin Ender Aug 14 '16 at 14:18
  • 1
    \$\begingroup\$ @MartinEnder, I ran across the linked page the other day when I was tidying my bookmarks (it now 404s), but I didn't remember that this was in the sandbox. IMO the performance requirement and ability to avoid enumeration mean that it wouldn't be a dupe. May try to write a reference implementation tomorrow, since it's a public holiday here. \$\endgroup\$ – Peter Taylor Aug 14 '16 at 16:35
  • \$\begingroup\$ Yeah I wasn't sure whether it would be a dupe (or whether there would be ways to avoid enumeration), but I figured you'd be the best judge of that, so I just thought I'd let you know. \$\endgroup\$ – Martin Ender Aug 14 '16 at 17:40
2
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The Coin Flop KoTH

In this challenge, you start out with a stack of 100 coins, alternating between gold and silver:

1.   Gold
2.   Silver
3.   Gold
4.   Silver
     ...
99.  Gold
100. Silver

Now, Gold, Inc. will pay you for any gold coins you can give them. Furthermore, they really like bulk shipment. If a shipment contains N coins, they will pay you N^2 for that shipment.

However, you've only got 1 truck to share between you and your opponent, who has a similar contract with Silver, Inc.

Game play

Each turn, the following steps occur:

  1. Flip a section of coins (like pancakes)
  2. Collect the top coins (of the same type)
  3. Ship off the collected coins (worth N^2)

1. Flip:

You or your opponent can each select a range to reverse. For example, if you selected the range [1,3], the following would occur:

Gold                        Gold
Silver                      Gold
Gold          ->            Gold
Gold                        Silver
Silver                      Silver

Notice how the top stack now has 3 golds in a row, which would be worth 9.

However, you only get to flip every other turn, and your opponent gets to flip on the other turns.

2. Collect

We collect all similar coins from the top of the stack, and put them in the truck:

Silver
Silver
Gold
Silver
Gold

Would result in 2 Silvers being put in the truck.

3. Ship

If the truck contains gold coins, then we send it off to Gold, Inc and we get paid N^2, where N is the number of coins in the truck. (Our opponent gets paid if the truck contains silver coins)

After the stack is empty, the player with the most money wins!

Sandbox questions:

  1. Is this clear?
  2. Does this seem interesting at all (if it isn't, please say so)? What are some interesting strategies you can come up with?
  3. Is there a never-lose strategy?
  4. I'm debating adding a "Freeze" as an alternative to "Flip", which would cause 2 collect/ship actions to occur. (Your opponent's turn would then be next)
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  • \$\begingroup\$ Could you add tags so it's clear this is a KOTH? \$\endgroup\$ – Zgarb Aug 15 '16 at 15:31
  • \$\begingroup\$ "If the truck contains gold coins, then we send it off to Gold, Inc (otherwise Silver, Inc), and we get paid" Is this correct? Earlier you said that we had a contract with Gold, Inc and the opponent with Silver, Inc, so it seems that we should be paid for gold and the opponent for silver. \$\endgroup\$ – Peter Taylor Aug 15 '16 at 21:02
  • \$\begingroup\$ @PeterTaylor does that make it clearer? \$\endgroup\$ – Nathan Merrill Aug 15 '16 at 21:09
  • \$\begingroup\$ Yes, thanks. Follow-up: I presume the controller will show different things to the two players so we both think we're the one collecting gold? \$\endgroup\$ – Peter Taylor Aug 16 '16 at 5:38
  • \$\begingroup\$ @PeterTaylor correct. The only difference players will notice is whether or not Gold is the first coin. \$\endgroup\$ – Nathan Merrill Aug 16 '16 at 12:20
2
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Still to-do:

  1. Add more test cases
  2. Make input requirements a little looser
  3. Make output requirements looser
  4. Specify that a full program is not required
  5. Change the name of gem-elements maybe?
  6. Come up with a title
  7. Come up with a cool story

I saw this as a problem over on CodeReview, here is the original question. I thought it would be fun to golf.


You have discovered various rocks. Each rock is composed of various elements, and each element is represented by a lowercase letter from 'a' to 'z'. The same element can be present multiple times in a rock. An element is called a 'common-element' if it occurs at least once in each of the rocks. Given the list of rocks you have to determine how many different kinds of common-elements you have.

Input Format

  • Each rock is a string which consists of lowercase letters from 'a' to 'z' representing elements
  • You can take the strings in whatever way is easy for you (lines from STDIN, pipe delimited STDIN, array of strings as a method param, etc.)

Output Format

  • The number of different types of common-elements for the given list of rocks. This can be as an integer or string

Constraints

  • There will always be at least one rock
  • Each rock will always have at least one element
  • You do not need to make a full program (functions & methods are allowed)

Sample Input

abcdde
baccd
eeabg 

Sample Output

2

Explanation

Only 'a' and 'b' are common-elements since these are the only characters that occur in each of the rocks' composition.

More test cases

a

0
aa

0
aa
aa

1
zyxabc

0
abc
def

0
abc
a

1
abc
ab
cb

1
abc
cba

3
abcdefghijklmnopqrstuvwxyz
qwertyuiopasdfghjklzxcvbnm

26
defabc
bfgcde
chfgde
gedfhi
fgiejh

2
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  • 3
    \$\begingroup\$ N doesn't seem to actually be a useful input, so I'd consider making it optional. In addition, tight I/O requirements are usually frowned upon, I'd consider modifying it to allow any format that doesn't add any additional information. I'd also recommend removing the quote block, and adding at least one test case with N=1 and one where the result itszero. Good luck with your challenge and thanks for using the sandbox! :) \$\endgroup\$ – FryAmTheEggman Jun 23 '16 at 19:47
  • 1
    \$\begingroup\$ Also can we take the input in the default list/array format for our language? As FryAmTheEggman, good luck with the challenge! \$\endgroup\$ – Blue Jun 23 '16 at 20:04
  • \$\begingroup\$ How do I say "any format that doesn't add additional information" in a way that doesn't sound silly? Is that fine as-is? \$\endgroup\$ – Captain Man Aug 11 '16 at 18:36
  • \$\begingroup\$ Nice edits, looking much better! If you want an example of a nice way to word the preamble, you could try emulating this question. I think "any reasonable format" is a fine way to say what you want, but what you have is fine, too. \$\endgroup\$ – FryAmTheEggman Aug 11 '16 at 19:44
  • 2
    \$\begingroup\$ This might be fun to golf in Java, but in the languages which are predominantly used on this site nowadays it's two built-ins and something like 3 to 6 characters depending on the language, so it's actually a pretty boring question. \$\endgroup\$ – Peter Taylor Aug 11 '16 at 21:44
2
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Stratego

This is a KOTH challenge based on the popular board game Stratego.

Rules

Stratego is played on a 10x10 board which looks like this (starting position):

  12345678910

A **********
B **********
C **********
D **********
E ..~~..~~..
F ..~~..~~..
G xxxxxxxxxx
H xxxxxxxxxx
I xxxxxxxxxx
J xxxxxxxxxx

*: enemy piece; x: your piece, .: empty space, ~: lake

There are twelve types of pieces: the flag (F), the bomb (B), the spy (S), and pieces numbered 2-10 (ten is 0). You can see the identities of your pieces, but not your opponent's. Each player starts with:

  • 6 bombs
  • 1 flag
  • 1 ten
  • 1 nine
  • 2 eights
  • 3 sevens
  • 4 sixes
  • 4 fives
  • 4 fours
  • 5 threes (miners)
  • 8 twos (scouts)
  • 1 spy

A player may arrange their pieces however they wish within their starting area.

The players take turns moving one piece into an adjacent square.

  • The bomb and flag may not move.
  • Twos (scouts) may move any number of squares in one direction, like a chess rook.
  • You may not move into a square containing one of your pieces.
  • If you attempt to move into a piece containing an enemy piece:
    • The identities of both pieces are revealed to the other player.
    • If both pieces are equal, they both lose.
    • If both pieces are numbered, the higher piece wins.
    • If the attacked piece is a bomb…
      • and the other piece is a miner, the miner wins.
      • otherwise, the bomb wins.
    • If the attacked piece is the flag, then the player owning the flag loses.
    • If the attacking piece is the spy…
      • and the defending piece is a 10, the spy wins.
      • otherwise, the spy loses.
    • If the attacked piece is the spy, it loses.
    • If the attacking piece won, the attacked piece is removed from the board and the attacking piece moves into its place.
    • If the attacked piece won, the attacking piece is removed from the board and the attacked piece remains in place.
    • If both pieces lose, they are both removed from the board.
  • A scout's move may end on an enemy piece, but it may not go over enemy pieces during its move.

Play continues until one player loses their flag or is unable to move on their turn, at which point that player loses.

Protocol

Bots may be written in any language that I can get to run on macOS. They will communicate with the server using newline-separated JSON on stdin/stdout. When a bot is started, it should send the following message:

{
  "type": "start",
  "layout": "..." /* starting layout, as 4x10 string; last line is edge of board */
}

When it is the bot's turn, the server will send the following message:

{
  "type": "turn",
  "board": "...", /* current board state, as 10x10 string; you half of the board is always on the bottom */
  "yourPreviousTurn": { /* result of your previous turn; not present on your first turn */
    "from": "B8",
    "to": "B9", /* you moved a piece from B8 to B9 */
    "movedPiece": "5" /* you moved a 5 */
    "attackedPiece": "8" /* the piece you attacked was an 8; only present if you attacked a piece on your last turn */,
    "winner": "them" /* their piece won; only present if you attacked a piece */
  },
  "theirPreviousTurn": { /* result of the turn they just took; not present on first turn of the game */
    "from": "H8",
    "to": "G8",
    "movedPiece": "4" /* they moved a 4; only present if they attacked */
    "attackedPiece": "9" /* they attacked your 9; only present if they attacked */,
    "winner": "you" /* your piece won; only present if they attacked */
  },
  "yourGraveyard": ["B", "3", ...], /* array of your pieces that have already died */
  "theirGraveyard": ["7", "4", ...] /* array of their pieces that have already died */
}

You must respond with:

{
  "type": "move",
  "from": "A6",
  "to": "A7"
}

Answering

Submit your bot as a pull request to [github link added here]. You do not need to include your entire bot in the SE answer, but please include the main/interesting parts of the code; I would recommend not including the layouts that you are using so that they remain secret (not that secret, but better than nothing). I will run bots regularly and post the latest scores here.

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  • \$\begingroup\$ 1. Can you give references to indicate that there is no legal issue here (with respect to copyright, trademarks, etc)? 2. Why 2 to 9 and then 0? Would it be simpler to use 1 to 9? 3. The list of starting pieces could be presented much more succinctly as a single string (BBBBBBF9...), and the combat table as a grid with columns and rows headed by B, F, etc and using one of three symbols to indicate the outcome. 4. Docker images? Are you serious? You should be trying to lower the barrier of entry, not raise it. \$\endgroup\$ – Peter Taylor Aug 25 '16 at 9:57
  • 1
    \$\begingroup\$ This is a dupe of my stratego challenge.... \$\endgroup\$ – Rohan Jhunjhunwala Aug 25 '16 at 14:10
  • \$\begingroup\$ meta.codegolf.stackexchange.com/a/9653/46918 \$\endgroup\$ – Rohan Jhunjhunwala Aug 25 '16 at 14:11
  • \$\begingroup\$ 1. I feel that the copyright on the game is unlikely to be enforced for something this small. 2. The original game uses 2-10, so I am staying with that. 3. That would be more succinct, but less useful for understanding the game -- it would be nice not to have to analyze a table to understand things like "larger number wins." 4. Agreed. I will edit to remove the Docker requirement. \$\endgroup\$ – Gaelan Aug 25 '16 at 15:29
  • \$\begingroup\$ @Peter see above \$\endgroup\$ – Gaelan Aug 25 '16 at 15:31
  • \$\begingroup\$ Whether the copyright holder decides to take action or not is irrelevant: if posting this challenge would violate copyright then this site's legal terms prohibit it. (CC @RohanJhunjhunwala). \$\endgroup\$ – Peter Taylor Aug 25 '16 at 16:48
  • \$\begingroup\$ @PeterTaylor game rules can not be copyrighted: copyright.gov/fls/fl108.pdf \$\endgroup\$ – Gaelan Aug 25 '16 at 17:09
  • 1
    \$\begingroup\$ @PeterTaylor from what I understand, developing an AI to play a game isn't really breaching the copyright of the game holder. However I feel we should take this to meta. \$\endgroup\$ – Rohan Jhunjhunwala Aug 25 '16 at 17:31
  • \$\begingroup\$ @PeterTaylor feel free to continue to discuss this here meta.codegolf.stackexchange.com/q/9925/46918 \$\endgroup\$ – Rohan Jhunjhunwala Aug 25 '16 at 17:35
  • \$\begingroup\$ Gaelan, the document you link states that game rules are subject to copyright: it's the mechanics which aren't. The design elements probably extend to such matters as board layout and tile distribution: or, at least, the major Scrabble clones' owners weren't confident enough of winning against Hasbro, and changed their layouts and distributions under legal pressure. And I (and the SE legal terms) also mentioned trademarks: Stratego is a registered trademark in the US. \$\endgroup\$ – Peter Taylor Aug 25 '16 at 21:49
  • \$\begingroup\$ @RohanJhunjhunwala, the AI might not be, but a king-of-the-hill doesn't work without a controller. \$\endgroup\$ – Peter Taylor Aug 25 '16 at 21:50
  • \$\begingroup\$ @PeterTaylor I will rename the challenge to avoid the trademark issue. Nothing in the PDF says anything about rules being copyrighted. It says the specific wording of the rules that came with the game are copyrightable, but I wrote the rules in my own words from memory. Also, see the dupe link in the meta question. \$\endgroup\$ – Gaelan Aug 25 '16 at 21:57
  • \$\begingroup\$ @Gaelan you have failed to address the issue that it is a dupe of my challenge \$\endgroup\$ – Rohan Jhunjhunwala Aug 25 '16 at 22:00
  • \$\begingroup\$ @PeterTaylor sure I can edit the name of my challenge to avoid stomping on a trademark. (But from what I understand you can saw the name of a company, without infringing their trademarks) \$\endgroup\$ – Rohan Jhunjhunwala Aug 25 '16 at 22:01
  • \$\begingroup\$ @RohanJhunjhunwala sorry, I didn't realize your challenge was this recent. Are you still planning on running your challenge? \$\endgroup\$ – Gaelan Aug 25 '16 at 22:03
2
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De-Parenthesize Ruby

Reuben wants to start programming and golfing in Ruby. However, since he learned programming from his siblings Cecil and Emma, who program in C and ECMAScript respectively, he's developed a habit of adding parentheses to all sorts of functions when he doesn't need to. However, he can't just remove all the parentheses either! Help him by writing a short program that will remove some of the parentheses for him.

The Challenge

Given a snippet of Ruby code (you don't need to worry if it's valid code or not), apply the following rules to remove parentheses from function calls wherever possible. Whenever parentheses are removed, put a whitespace in between the function call and the arguments (even if in actual golfing practice it isn't needed, such as in the case of print"hello")

For the purposes of this challenge, not all parts of Ruby need to be checked:

  • Variables/functions will contain alphanumerics and underscore only, and won't start with numbers. (No functions like array.slice!(1,4).include?(6), and no variables like $a)
  • Blocks (curly braces after functions, or for declaring stabby-lambdas) are not present. No array.map{|i|i+1}
  • Function definitions like def f(x) will not be present.
  • No backslashes \ so you don't have to parse something like "\"hello\""
  • No comments. (Comments in Ruby start with #.) If a # appears in code, it's going to be part of a string.

These are the only rules you need to check for:

Functions without any parameters always have their parens removed.

array.size()+array.something().length() -> array.size+array.something.length

Functions at the end of a method chain can be removed only if they are not next to an arithmetic operator.
For the purposes of this challenge, the operators used will be +, -, *, /, %.

num.parse(3,7) -> num.parse 3,7
string.gsub("this","that").count("t") -> string.gsub("this","that").count "t"
1+string.count("T") -> 1+string.count("T") # no change
1+(string.count("T")) -> 1+(string.count "T")
puts(3,8*7) -> puts 3,8*7

If a function contains another function call as its only function argument, apply the other rules to that function as well.

print(print(x.sub(a,b.to_s(16)))) -> print print x.sub a,b.to_s(16)

If a string substitution operator #{...} is present within a string (enclosed with double quotes), remove parens from the functions in accordance to the other rules.
Assume that the contents within the string substitution don't include any literal strings, including single-character "question mark" strings like ?a. This means that things like "#{"a}#{"}" or "#{?}}" do not need to be dealt with.

"hello() #{world().string(5)}" -> "hello() #{world.string 5}"
"hello() # {world()}" -> "hello() # {world()}" # no change

Might add more rules or other things in the future if people want, before it gets published of course.

Would this be a duplicate of something like Remove unnecessary parentheses? It's a similar concept but with different rules

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\$\endgroup\$
  • \$\begingroup\$ The "only if they are not in an expression with an arithmetic operator" part seems like it might potentially be a problem. What about 1+(string.count("T"))? Or, how about: print(foo().bar())+1? I think you may want to more aggressively remove some of these edge cases. \$\endgroup\$ – FryAmTheEggman Aug 19 '16 at 12:45
  • \$\begingroup\$ @FryAmTheEggman that's a good point. I will think about how to phrase that case with parentheses to make more sense. For the second example, since the first rule says that empty parens are always removed, it'd become print(foo.bar)+1 as expected anyways. \$\endgroup\$ – Value Ink Aug 19 '16 at 18:40
  • \$\begingroup\$ I think a whitelist of language constructs would be clearer than listing a lot of things that aren't allowed. \$\endgroup\$ – feersum Aug 28 '16 at 11:59
  • \$\begingroup\$ @feersum I don't understand what you mean by that. Should I say something like "only alphanumerics, the dot operator ., these arithmetic operators, and the string substitution operator in strings"? \$\endgroup\$ – Value Ink Aug 29 '16 at 17:23
  • \$\begingroup\$ Yes, or better some sort of grammar describing the allowed input. \$\endgroup\$ – feersum Aug 29 '16 at 22:36
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