443
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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.

To post to the Sandbox, scroll to the bottom of this page or click on the "Add Proposal" link below, and click "Answer This Question". Click "OK" when it asks if you really want to add another answer. Write your challenge just as you would when actually posting it. You may also add some notes about specific things you would like to clarify before posting it. Other users will help you improve your challenge by rating and discussing it. When you think your challenge is ready for the public, go ahead and post it, replace the post here with a link to the challenge and delete it.

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

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

Add Proposal

Search the Sandbox

Browse your pending proposals

Get the Sandbox Viewer to view the sandbox more easily

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

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  • 2
    \$\begingroup\$ Suggestion: instead of having a notice on the top answer ("note: if you are..."), you'd better just put a moderator notice below the question \$\endgroup\$ – nicael Mar 19 '18 at 19:35
  • 3
    \$\begingroup\$ @nicael We can only choose from three post notices: citation needed, current event, and insufficient explanation. \$\endgroup\$ – Dennis Apr 7 '18 at 14:43
  • \$\begingroup\$ If you remove a post but didn't post it you can replace the text body with [](lots of text here to reach the min chars) to make it much smaller when removed \$\endgroup\$ – Christopher Apr 13 '18 at 17:54
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    \$\begingroup\$ @Christopher Please don't do that for old proposals. It clutters the first page with an answer nobody cares about anymore, instead of staying hidden on page 10 where it will bother nobody. \$\endgroup\$ – Dennis Apr 13 '18 at 18:17
  • \$\begingroup\$ @Dennis ? what are you talking about. As if if you didn't post it like you just removed you own sandbox because dupe or something \$\endgroup\$ – Christopher Apr 13 '18 at 18:18
  • 4
    \$\begingroup\$ @Christopher If your proposal is still on the first few pages, you can replace the proposal with a stub to save vertical space on these pages. However, if your proposal is already on page 10, editing your proposal will bump it to page 1, where space is more precious than on page 10. \$\endgroup\$ – Dennis Apr 13 '18 at 18:21
  • \$\begingroup\$ @Dennis ohh that makes sense \$\endgroup\$ – Christopher Apr 13 '18 at 18:25
  • \$\begingroup\$ codegolf.meta.stackexchange.com/questions/12599/… \$\endgroup\$ – Redwolf Programs Apr 17 '18 at 17:38
  • 1
    \$\begingroup\$ Maybe it's time to consider cleaning some of this up a bit. There's just too much to go through and some of these proposals are years old and obviously not going anywhere (even some of the good ones). Perhaps cull anything that is two years old and has likewise been inactive for as long? \$\endgroup\$ – ouflak Aug 6 '18 at 9:07
  • 1
    \$\begingroup\$ @ouflak You can sort posts by "active". That seems to resolve all of the problems you describe. \$\endgroup\$ – FryAmTheEggman Sep 27 '18 at 19:04
  • \$\begingroup\$ I already posted this, but codegolf.stackexchange.com/questions/176599/… \$\endgroup\$ – 2br-2b Nov 27 '18 at 2:38
  • \$\begingroup\$ How are tags added to questions? \$\endgroup\$ – guest271314 Jan 9 at 7:51
  • \$\begingroup\$ It seems like there is a rollback war with moderators and the Community user to add and remove the featured tag. \$\endgroup\$ – smileycreations15 Mar 21 at 21:13
  • 3
    \$\begingroup\$ @smileycreations15 That's unfortunately unavoidable. Community is an automatic script, and, since most featured questions are only temporarily so, it assumes that we don't want this question to be featured forever. However, we do, so a mod has to edit the tag in every now and then. \$\endgroup\$ – Erik the Outgolfer Mar 24 at 15:22
  • 3
    \$\begingroup\$ @EriktheOutgolfer Yeah. Maybe they can create a special [featured-pin] tag which will both feature it and pin it from removal by the Community user. \$\endgroup\$ – smileycreations15 Mar 24 at 17:20

2449 Answers 2449

4
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Comment Java Code

Little Billy is in an AP computer science class and his teacher requires him to comment his code (in Java) even though Little Billy thinks it is redundant. Little Billy is a lazy person and so he wants a program to comment for him.

The Challenge:

Make a program in any language that gets a text file as input and saves the commented version to a new text file (or to STDOUT) changing it in the following ways:

  • Add comments before all contructors:

/**
* Constructs a new instance of (class whatever).
* if applicable: @param nameOfVariable is the name of variable
*/

  • Add comments before any functions:

/**
* if applicable: @param nameOfVariable is the name of variable.
* if applicable: @return nameOfReturnVariable is the name of return variable.
*/

  • Add a comment before static void main(String[] args) if present:

/**
* This is where the program starts.
* @param args is the command line arguments passed to the program.
*/

  • Add comments before the class declaration (not applicable for inner classes)

/**
* class NameOfClass is another great example of Object Oriented Programming!
*/

For example, given this:

public class Car {
    private double gas;
    private double mpg;

    public static void main(String[] args) {
        new Car(20,20);
    }

    public Car(){}

    public Car(double gasInVehicle, double milesPerGallon){
        gas = gasInVehicle;
        mpg = milesPerGallon;
    }

    public void drive(double milesToDrive){
        gas-=(milesToDrive/mpg);
    }

    public double getGas(){
        double gasolineInTank = gas;
        return gasolineInTank;
    }

}

Your program should output this:

/**
 * class Car is another great example of Object Oriented Programming!
 */
public class Car {
    private double gas;
    private double mpg;
    /**
     * @param args is the command line arguments passed to the program.
     */
    public static void main(String[] args) {
        new Car(20,20);
    }
    /**
     * Constructs a new instance of Car.
     */
    public Car(){}
    /**
     * Constucts a new instance of Car.
     * @param gasInVehicle is the gas in vehicle.
     * @param milesPerGallon is the miles per gallon.
     */
    public Car(double gasInVehicle, double milesPerGallon){
        gas = gasInVehicle;
        mpg = milesPerGallon;
    }
    /**
     * @param milesToDrive is the miles to drive
     */
    public void drive(double milesToDrive){
        gas-=(milesToDrive/mpg);
    }
    /**
     * @return gasolineInTank is the gasoline in tank.
     */
    public double getGas(){
        double gasolineInTank = gas;
        return gasolineInTank;
    }
}

Extended Info

  • Because Little Billy's class is only on week 3, Little Billy's knowledge of complex Java statements is pretty limited. In other words, only common functions, class declarations, constructors and instance variables will be included; objects/variables/resources are limited to Java primitives and their wrapper classes and Math.
  • Each variable name is separated for its definition according to standard camel case rules, with an exception of the first letter, which may be capitalized. For example,

milesToRun = miles to run.
milestoRun = milesto run.
MilesToRun = miles to run.
parseXML = parse xml - each successive capital letter (with a minimum of 3 in a row) is considered an acronym, so GetABuffer would become get a buffer. parseXM is parse x m.

If it is a parameter -

@param milesToRun is the miles to run.

  • Your program should be able to handle multiple arguments to functions
  • There should be no lines with only " * ".
  • Assume all variables and parameters are formatted correctly (e.g. all camel case, only letters, etc.)
  • Indentation is not critical. However, all newlines should remain.
  • You may also assume that statements ending in a semicolon are 1 line each

Scoring

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

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  • 1
    \$\begingroup\$ when you say "assume all camel case"... what about the unlikely method name GetABuffer? If that is valid and should produce "get a buffer", then what about ParseXML? \$\endgroup\$ – Martin Ender Sep 11 '14 at 10:12
  • 1
    \$\begingroup\$ Similar to codegolf.stackexchange.com/q/3241/194 . The thing that would potentially make this more difficult is parsing Java, but then you need to provide some details on the grammar. I also notice that your spec and examples avoid all of the tricky stuff around generics. \$\endgroup\$ – Peter Taylor Sep 11 '14 at 11:32
  • \$\begingroup\$ I think it should be parseXml. \$\endgroup\$ – Soham Chowdhury Sep 13 '14 at 15:07
  • \$\begingroup\$ Hello! This looks like a good but abandoned meta post, would you be willing to offer it for adoption? (If you want to, you can still post to main.) Due to community guidelines, if you don't respond to this comment in 7 days I have permission to adopt this. \$\endgroup\$ – programmer5000 Jun 9 '17 at 18:39
4
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Coloring Book

Given a black-and-white raster line drawing (no anti-aliasing, with fully enclosed regions), write a program that will color it in. Something maybe like below, though I'd clean up messy JPGs into clean B&W PNGs to start with:

http://wallalay.com/wp-content/uploads/2014/07/Coloring-Books-27.jpg

Could either be a , to give freedom in create more interesting output (gradients, patterns, how to select "better" colors, other images). Along the lines of patterns and gradients, I'd opt for simpler stock "prompts", preferably animals, to show off solutions:

http://www.frontiernet.net/~goofis1/Images/Dinosaurs/SlateBack.jpg

...or perhaps if each region has a specified color given by a swatch inside it: so the program would basically need to find-the-color then flood-fill. Sounds boring, I prefer more creativity.

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  • \$\begingroup\$ The flood fill would indeed be quite trivial, but as a popularity contest this could become too much of an art contest and less of a programming contest. \$\endgroup\$ – Martin Ender Oct 10 '14 at 9:45
  • \$\begingroup\$ In my mind's eye I kinda pictured an elementary school classroom wall with the same line drawing jazzed up every which way. \$\endgroup\$ – Nick T Oct 10 '14 at 10:18
  • \$\begingroup\$ I do think there is reasonable technical skill involved in creating something to dynamically create said art. Gradients, patterns, masking other images into regions, coloring outside the lines, and something even more interesting than what I can come up with all seem to take reasonable skill. Especially for a few arbitrary input drawings. \$\endgroup\$ – Nick T Oct 10 '14 at 10:26
  • 2
    \$\begingroup\$ If you can come up with a testable criterion, you might be able to get an interesting code-golf which asks for cel shading rather than just floodfill. \$\endgroup\$ – Peter Taylor Oct 10 '14 at 10:42
  • \$\begingroup\$ You could specify a rule for which colour should be used. For example, you could colour based on area or perimeter of the region being filled (or some combination of both). This would colour similar regions in similar colours and give the impression of being chosen for the image, but would still be rigorously defined enough to make it a code golf. \$\endgroup\$ – trichoplax Aug 26 '15 at 14:50
  • \$\begingroup\$ Hello! This looks like a good but abandoned meta post, would you be willing to offer it for adoption? (If you want to, you can still post to main.) Due to community guidelines, if you don't respond to this comment in 7 days I have permission to adopt this. \$\endgroup\$ – programmer5000 Jun 9 '17 at 18:43
  • \$\begingroup\$ @programmer5000 feel free to take this one. Are you thinking about doing code-golf somehow or a popularity contest? \$\endgroup\$ – Nick T Jun 9 '17 at 23:09
4
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Score a Mahjong Hand

So you've built a solver for which tiles you need to complete a Mahjong hand, but you've heard that certain hands are worth more than others and sometimes a hand that scores too low can't even win.

So, you decide to build a program that will score hands for you.

This time, however, you will be making it for the full mahjong set. There are a total of 136 tiles in this variant, four copies of each of the following:

  • 1 coin to 9 coins, represented from 1p to 9p.
  • 1 stalk to 9 stalks, represented from 1s to 9s.
  • 1 myriad to 9 myriads, represented from 1m to 9m.
  • four winds, represented as EE, SS, WW, and NN.
  • three dragon tiles, represented as ZZ, FF, and BB.

The winds and dragons are known as honour tiles, and they are not part of any suit.

A standard mahjong hand consists of four sets and a pair. A set can be any of:

  • A chow, which is a sequence of three tiles in the same suit. Winds and dragon tiles cannot be part of a chow.
  • A pong, which is a group of three of the same tile.
  • A kong, which is a group of four of the same tile. Every kong increases the total number of tiles in a player's hand by 1.

Every chow and pong has three tiles in it, while a kong has four. So a winning hand can have anywhere from 14 to 18 tiles, depending on the number of kongs.

Now, the score that a hand has is based on fan. A valid winning hand has no fan by default. However, the following patterns count for fan:

Sets

  • A pong or kong of dragons (e.g. ZZ ZZ ZZ): 1 fan
  • A pong or kong of winds (e.g. EE EE EE or SS SS SS SS): 1 fan

Patterns

  • A straight, 1-9 in a suit as three sets (e.g. 1s 2s 3s 4s 5s 6s 7s 8s 9s + 1 more set and a pair): 3 fan
  • A broken straight, 1-9 in a suit plus an additional 3, 5, and 7 (1s 2s 3s 3s 4s 5s 5s 6s 7s 7s 8s 9s): 5 fan
  • A pong or kong of all three dragons (e.g. ZZ ZZ ZZ FF FF FF BB BB BB BB): 7 fan
  • Four winds (EE EE EE SS SS SS WW WW WW NN NN NN + a pair): 10 fan
  • Nine Gates (1 1 1 2 3 4 5 6 7 8 9 9 9 of any single suit, plus one more of any tile in the suit): 14 fan (effectively 20 fan because of one-suit)

Entire hand patterns

  • Entire hand is chows (e.g. 1s 2s 3s 6s 7s 8s 2p 3p 4p 7m 8m 9m NN NN): 1 fan
  • Entire hand is pongs or kongs (but not all kongs) (e.g. 1m 1m 1m 3s 3s 3s 7p 7p 7p NN NN NN EE EE): 3 fan
  • Entire hand is kongs only (e.g. 1m 1m 1m 1m 3s 3s 3s 3s 7p 7p 7p 7p NN NN NN NN EE EE): 10 fan
  • All tiles from one suit + honours (e.g 1p 1p 1p 4p 4p 4p 7p 8p 9p SS SS SS EE EE): 3 fan
  • All tiles from one suit (e.g. 1p 1p 1p 4p 4p 4p 7p 8p 9p 6p 6p 6p 5p 5p): 6 fan
  • All tiles honours (e.g NN NN NN ZZ ZZ ZZ BB BB BB SS SS SS EE EE): 9 fan

Special hands

The hands below are scored specially and do not follow the four-sets-and-a-pair rule.

  • Seven Pairs (seven pairs of any tiles): 2 fan
  • Seven pairs in the same suit: 10 fan
  • Seven honour pairs: 20 fan
  • Thirteen Orphans (1p 9p 1s 9s 1m 9m EE SS WW NN ZZ FF BB and one more of any of these 13): 10 fan

Note that you only have to account for the value of the hand itself - you don't need to care about what the prevailing wind is or whether the tile was self-drawn, by discard, or anything else.

Also note that a hand may fulfill multiple criteria. If it does, add up all the fan from each criterion it fulfills, except if the hand is a special hand.

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  • \$\begingroup\$ I won't be around for long. But sure, if you can make a chatroom and link to it here, I'll check in on it periodically. \$\endgroup\$ – Joe Z. Dec 19 '14 at 3:40
  • \$\begingroup\$ Chat link: here \$\endgroup\$ – Sp3000 Dec 19 '14 at 3:47
  • \$\begingroup\$ Oh, just using that one? It might get muddled up in the rest of the messages, though. \$\endgroup\$ – Joe Z. Dec 19 '14 at 5:25
  • \$\begingroup\$ Some ambiguous case that never happens in a real game: 111122233345555, which can be either 1111 222 33 345 555 or 11 123 123 234 5555. Both are 6 fan, though. Btw: psm are Japanese and ZFB are Chinese. Shouldn't that be at least consistent? (Or is that the standard somewhere?) \$\endgroup\$ – jimmy23013 Dec 19 '14 at 11:01
  • \$\begingroup\$ @user23013 In my question I was using Japanese, just so I could plug tests into Tenhou's solver. They use 1z2z3z... for honours. ZFB is indeed Chinese, but unfortunately taking the initials for Japanese gives CHH. I'm not sure if there's a standard anywhere. \$\endgroup\$ – Sp3000 Dec 19 '14 at 14:21
  • \$\begingroup\$ @Sp3000 In fact it would be HHC in the Japanese order... I think most Japanese people don't use letters for honors. And Chinese people don't use letters at all. I did some search and found some English-speaking people are just using something worse: Red, Green, ... \$\endgroup\$ – jimmy23013 Dec 19 '14 at 15:48
  • \$\begingroup\$ ZFB is Chinese, and psm was Japanese/Cantonese but there I was just following the convention from the previous question. I can change it to BSW if you want, though. \$\endgroup\$ – Joe Z. Dec 19 '14 at 20:30
  • \$\begingroup\$ I was just nitpicking... never mind. \$\endgroup\$ – jimmy23013 Dec 20 '14 at 9:56
  • \$\begingroup\$ The multiple maching criteria should be clarified: For example seven pairs in the same suit, scores just 10 total - not 12. In otherwords, it isn't considered to match both "Seven Pairs" and "Seven Pairs in the same suit". \$\endgroup\$ – MtnViewMark Jan 2 '15 at 15:42
4
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Invert the matrix

The name says it all. I am surprised that there is no preexisting question on calculating matrix inversion

Given an nXn invertible matrix, your task is to output its inverse.

Rules

  • Floating point precision of at least 3 significant decimal places is required
  • You can assume that the input matrix will always have an inverse
  • You cannot use any inbuilt methods (or external libraries) to perform any of the following tasks:
    • Calculate the inverse
    • Solve system of equations
    • Calculate determinant
    • Multiply two matrices or calculate dot product.
  • In other words, your code has to calculate the inverse using any of the mathematical methods itself. (If I missed out any inbuilt methods in the above rule, this rule should make it clear that it cannot be used)
  • You cannot calculate inverse by randomly generate random matrix and then multiply with the input to check for unit matrix.

Input

Input can be in a format of your choice. For example:

[[ 1, 3, 4],
 [ 5, 7, 8],
 [ 3, 4, 5]]

or

1 3 4
5 7 8
3 4 5

etc

Output

The output should be in the same format as the input, except for 1 requirement that each row of the matrix should be in a separate line with elements of the same row on the same line.

This is code-golf so shortest code in bytes win.

Sandbox notes

What do you think of the problem ? Too mundane ? Too trivial ? Need any extra rule to make it more interesting ?

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  • \$\begingroup\$ There is a pre-existing question, although it only covers 2x2. \$\endgroup\$ – Peter Taylor Jan 3 '15 at 8:28
  • \$\begingroup\$ @PeterTaylor Yeah, also, it is based on system of equations, which is related but not exactly same as Inverse (Inverse is a part of solving equations). So size plus this fact should make this question different enough ? \$\endgroup\$ – Optimizer Jan 3 '15 at 8:32
  • \$\begingroup\$ "Floating point precision upto 3 decimal places is required" - did you mean "Floating point precision of at least 3 decimal places is required"? \$\endgroup\$ – John Dvorak Jan 3 '15 at 10:11
  • \$\begingroup\$ @JanDvorak err. \$\endgroup\$ – Optimizer Jan 3 '15 at 10:13
  • 4
    \$\begingroup\$ The whole business of numerical accuracy is a minefield for this because there are many ill-conditioned instances. You can't talk in terms of decimal places unless you either require people to work in arbitrary precision or guarantee some very conservative bounds on the input and output values. (Unless you want to merely go for conservative bounds and include some test cases which will disqualify naïve Gaussian elimination and force people to do a Bareiss-style approach, but that will lead to having not more than one or two correct answers and probably a dozen incorrect ones). \$\endgroup\$ – Peter Taylor Jan 3 '15 at 16:34
  • \$\begingroup\$ To get around thorny questions of ill-conditioned matrices and the possibility of divide-by-zero, how about only requiring that the algorithm work with high probability on random Gaussian matrices of size at most 100? \$\endgroup\$ – xnor Jan 3 '15 at 18:00
  • \$\begingroup\$ @PeterTaylor I might not have understood even 10% of that comment. Can you please tell me what can I put in place of that rule then ? \$\endgroup\$ – Optimizer Jan 3 '15 at 19:14
  • \$\begingroup\$ @xnor what do you mean by ill-conditioned matrices ? \$\endgroup\$ – Optimizer Jan 3 '15 at 19:15
  • \$\begingroup\$ No, I can't. I don't consider myself qualified to write a spec for inverting matrices. I know what the problem is, but I don't understand it well enough to formulate validity constraints. I think it would be best to leave the idea for someone who has a firm grasp of the numerical analysis of linear algebra to pick up and finish. \$\endgroup\$ – Peter Taylor Jan 3 '15 at 19:41
  • \$\begingroup\$ @PeterTaylor I disagree with you. The purpose of this challenge is not to get an mathematically perfect matrix inversion code, but instead, a code that calculates the inverse in the shortest and smartest way. Martin suggested in chat that the floating point precision upto certain significant digits can be made a compulsion till a matrix size of say 10. Simply hoping to get a mathematically experienced person who has mastery in matrix inversion to pick up the challenge is a very pessimistic approach. \$\endgroup\$ – Optimizer Jan 3 '15 at 20:38
  • \$\begingroup\$ @Optimizer Here's the issue: say I write a row reduction algorithm. I want to subtract a multiple of the first row from the second in order to zero out the first entry of the second row. Do I have to consider the case where the first entry of the first row is zero, so I get a divide-by-zero error? Ok, say I have a separate check if equals 0. What if that entry is supposed to be 0, but as a result of a previous calculation is 0.00000001, failing the check? Now my second row is ginormous and it shouldn't be. \$\endgroup\$ – xnor Jan 4 '15 at 2:47
  • \$\begingroup\$ @Optimizer Alternatively, finding the inverse mod 2 would avoid accuracy issues. \$\endgroup\$ – xnor Jan 7 '15 at 4:21
  • \$\begingroup\$ @xnor mod 2 as in ? \$\endgroup\$ – Optimizer Jan 7 '15 at 4:27
4
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Responsible governance

Background

For years, the Federal government of the United States has been in chaos. The Presidency and Congress are under different parties, and there are no compromises in sight. Instead of passing a budget or improving the tax code, the politicians are squabbling as the country teeters on the brink of economic collapse. It seems that a few more years of political dysfunction will soon reduce the United States to rubble.

Yet, some have noticed that many state governments seem to be functioning. Taxes are being reformed, education is being improved, and even compromise can be found on occasion!

The solution is clear. The federal government must be dissolved. Each state will control its own eduction, infrastructure, and healthcare, but the budgets will be pooled. As the governor of your state, you must use the country's funds to improve your constituents' lives, but also be frugal enough to let your neighbors prosper as well.

Rules

This system of government will be in place for 10 years (120 months), after which the results of this experiment will be analyzed. Each month, every governor will be given the opportunity to use the country's funds to pay for one project of their choice.

The government starts with $1,000,000 in the bank, times the number of states. In every state, life expectancy is 75 years, education is at 95 IQ, and the infrastructure gets a rating of 65 points.

Every year, the following happens:

  • First, taxes are collected and centralized expenses are paid.
  • For each of the next twelve months:
    • Each governor, in order, is allowed to finance a project.
  • Finally, information about each states' education, healthcare, and infrastructure is updated to reflect the projects built.

Scoring

The final score is the government's total funds, times average state's life expectancy, times the average state's IQ, times the average state's infrastructure rating. (These averages are not weighted by population.)

This entire experiment will be performed many times. Each time, a different governor will be excluded. The winner will be the governor such that the final score is the worst when they are excluded, since they must have had the biggest impact on the government's success.

Details

  • N is the number of governors (not counting the excluded one)
  • L is the life expectancy of a particular state.
  • La is the average life expectancy.
  • E is the IQ of a particular state.
  • Ea is the average IQ.
  • T is the infrastructure rating of a particular state.
  • Ta is the average infrastructure rating.

    Revenues and expenses

  • Annual revenue: $ N Ea2 Ta

  • Annual expenses: $10,000,000,000,000 N / La3 / Ta

    Projects

  • Each of the three factors (health, education, and infrastructure) is calculated on a state-by-state basis. L starts at 75, E starts at 95, and T starts at 65.

  • For each factor, the bonuses are cumulative.
  • For example, the bonuses L +2, L +5 mean that L = ( 75 + 2 + 5 ) = 82.
  • Each project can be done only once by each state.
  • Each project has an associated six-letter code (for I/O).

    List of projects

  • [ENOUGH] Enough medicine: L +5, cost: $300,000

  • [CANSCR] Early cancer screenings: L + 4, cost $200,000
  • [FREECC] Free community college: E +10, cost: $400,000
  • [LIBRAR] Library system: E + 3, cost: $150,000
  • [MUSEUM] Museums E + 1 in each state, cost: $500,000

many more in the actual challenge...

Your program

Write a function in Java, Python 2, or Ruby, which accepts an integer (the government's funds) and returns an optional six-letter code, corresponding to the project chosen. If nothing is returned, or the returned value is invalid, or the chosen project was already built by you, or costs more than the government's funds, no project will be built for this month.

You may not use any I/O, except for one file called "<program-name>.txt, which you may use however you wish, provided it's kept under 2 megabytes at all times.


Some questions:

  1. Are the rules completely clear?
  2. Are there any perverse incentives? The goal should be to improve one's own state while being frugal with the government's funds.
  3. In the first section too political?
  4. Is the challenge too complicated?
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  • \$\begingroup\$ I wonder about the formulas used for score/revenue/expenses. Have you run this a few times with dummy states to see if it works in practice? I'd hate to see a simple, constant choice resulting in a clear winner with no good way to stop it. \$\endgroup\$ – Geobits Jan 26 '15 at 2:23
  • \$\begingroup\$ That's a good point; thanks for the feedback. The existing formulas are sort of placeholders for now. Once the other details are worked out, then I'll focus on getting the formulas balanced. \$\endgroup\$ – Ypnypn Jan 26 '15 at 2:48
4
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Real-Time Hovercraft Battle

Hovercrafts are cool. Hovercrafts with rail guns are even cooler. In this challenge, you will write a program to controller a battling hovercraft.

The hovercrafts in this game obey the laws of physics. A hovercraft with no acceleration will continue moving in a straight line. Also, a spinning hovercraft will continue to spin.

At each moment in time, the hovercraft's motion is defined by a few variables: the location, angle, linear velocity, and angular velocity of the hovercraft.

Hovercrafts are circles of radius 4, with a clear "front" and "back" end. On the left and right there are two forward-facing fans which serve as propulsion. Each fan has 8 power settings, from -3 to +4. These fans control additional aspects of the hovercraft's motion, the linear and angular acceleration.

Todo: determine the mass and moment of inertia for the hovercraft, which will then allow me to determine which power settings have which effects.

The real fun begins with the rail gun. The rail gun fires from the front of the hovercraft and deals damage when it strikes the side of the opposing hovercraft. The ammunition travels at a fast speed but finite, and the angle of the shot depends on the motion of the hovercraft. Firing the railgun also produces a recoil effect.

Note: I need to prevent the "rotating turret" strategy: spin really fast and fire when you find yourself pointed at something. An a truly instantaneous shot would never miss in this circumstance.

In order to power the rail gun, you must first charge a capacitor. Capacitor charging can be turned on and off. When the capacitor is charging, the fans will have reduced thrust. Once the capacitor stops charging (either it is full or charging is turned off) the thrusters will act normally again. You can still fire the rail gun when the capacitor is charging: the shot will have reduced damage. By default, the capacitor starts empty with charging on.

The Tournament

All of the participants will participate in a series of 1v1v1v1 battles. The goal is to survive the longest.

The battle arena is a radius 400 circle. Hovercraft can collide with the boundary, bounce,and take a little damage. At the start of the match, one hovercraft will be randomly located (1 per quadrant).

The finish order for a round will be determined by the order of death. After a certain period of time, all remaining hovercrafts will be deemed co-champions of that round.

After a decent number of battles (a number not currently determined, but something less than an exhaustive search but enough to determine the overall winner with non-negative confidence), the finish orders for each battle will be treated as ballots in a form of ranked voting scheme. By using a mathematically sound voting scheme, the final order should represent the results of the battles pretty well.

I have not started the controller yet.

This will be a real-time game. Each hovercraft program can request updates and send commands in real-time. This means that game ticks can happen while your program is running, and each game tick is a very small increment of time (as fast as feasible).

\$\endgroup\$
  • \$\begingroup\$ Check out Robocode. It is a long-running contest with comparable rules. What vision do the players have? Can you see the orientation of other participants? \$\endgroup\$ – Franky Feb 15 '15 at 10:49
4
\$\begingroup\$

Best Approximating Polynomial

(?)

Let f(x) and g(x) be two continuous real-valued functions over the interval [a, b], where a < b.  The mean squared error (or simply "error" hereafter) between f(x) and g(x) is defined as

Equation 1

Roughly speaking, the smaller the error is, the closer f(x) and g(x) are.

In this challenge, f(x) will be some user-supplied target function and g(x) will be a polynomial of some maximal degree n.  The goal is to find the polynomial g(x) that minimizes the error w.r.t. f(x) (such a polynomial exists and is unique;) we say that g(x) is the best approximation of f(x) over the interval [a, b] using a polynomial of degree at most n.

For example, let's take the function f(x) = cos x over the interval [0, ½π] (which contains all the "interesting" information about this function), and look for a polynomial of degree at most 2. As you might know, the Taylor expansion of cos x up to degree 2 is t(x) = 1 - ½x2.  The error between t(x) and f(x) is approximately 6.25×10-3.  We can do better than that! If I got my math right, the best approximation of cos x over the interval [0, ½π] using a polynomial of degree at most 2 is (brace yourselves):

Equation 2

or approximately g(x) ≈ -0.34 x2 - 0.13 x + 1.02.  The error between g(x) and f(x) is approximately 7.03×10-5—two orders of magnitude better! The difference is very notable, as the following plot shows:

Plot 1

Challenge

Write a program or a function that takes a function f(x), an interval [a, b], where a < b, and a nonnegative integer n, and returns the polynomial of degree at most n that best approximates f(x) over the given interval.

Input

You may read the input through STDIN, the command line, as function arguments or an equivalent method. Note that you don't have to accept a, b and n in any specific format; particularly, you don't have to read a and b as [a, b].

You should accept the function f(x) in one of the following forms:

  • As a language-level function, function-like object, polymorphic object or any other equivalent method your language uses to communicate functions.
  • As a string containing an expression. The actual format of the string is flexible, but it should be expressive enough to allow basic arithmetic operations, exponentiation, and preferably also logarithms and basic trigonometric functions. The intention is for it to be used with something like eval. Note that you're allowed to require the string to be formatted in some convenient way (within reason), for example, pre-wrapped in a function
    (as in "function (x) { return x + Math.sin(x); }").
  • If, and only if, both of the above methods are impractical in your language, you may not take the function as input and assume that the function is already defined by the user. In this case, add to your score a +16byte penalty.

You may assume that the input is valid.

Please specify how your program accepts its input in your post.

Output

You may write the output to STDOUT, return it as the function's result or use an equivalent method. You may express the polynomial as a list of coefficients (e.g., [1, -2, 1]), using some pretty-printed form (e.g., x^2 - 2x + 1) or some other equivalent method. Either way, it should be easy to determine the coefficients of the different terms from your output; so, for example, you may not return the polynomial as an opaque function.

Note that unlike the example given at the beginning of the post, you don't have to produce symbolic output (unless you want to); a numeric output is fine.

Please specify how your program returns its output in your post. If the order of the coefficients may be ambiguous, make sure to clarify it.

Accuracy and Run Time

The numerical accuracy of your program is most likely going to be dependent on the target function, the degree of the polynomial and the amount of time you allow it to run. As a result, giving a general accuracy requirement is impractical. However, your program should process each of the below test cases in less than a minute and produce a polynomial whose error w.r.t. the target function is no worse than the error specified in the test case.

Additional Rules

  • You may not use any function that performs this specific task.

Scoring

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

Test Cases

Recall that your program should complete each of the following test cases in less than a minute, with an error that is less than or equal to the specified error. That being said, this is not a hard and fast rule; if your program struggles with a couple of test cases, that's fine. Note that the approximating polynomials listed in the test cases are just an approximation and are only given for illustration. Although theoretically each case has a unique optimal polynomial, your program might produce a notably different one; this is especially true for the higher degree polynomials. However, both polynomials should behave similarly over the given interval, which is why the correctness of your program is determined by the error. Likewise, the errors listed in the test cases are only an approximation and are not necessarily the errors of the corresponding listed polynomials—they're just a lower-bound for accuracy.

Test 1
f(x) = sin(x)
From: -pi
To: pi
Max. Deg.: 0
g(x) = 0
e^2 = 5.0001 * 10^-1

Test 2
f(x) = sin(x)
From: -pi
To: pi
Max. Deg.: 1
g(x) = 0.304 x
e^2 = 1.9604 * 10^-1

Test 3
f(x) = cos(x)
From: -pi
To: pi
Max. Deg.: 2
g(x) = -0.231 x^2 + 0.76
e^2 = 3.8032 * 10^-2

Test 4
f(x) = sin(x)
From: -pi
To: pi
Max. Deg.: 5
g(x) = 0.005643 x^5 - 0.155 x^3 + 0.988 x
e^2 = 1.8490 * 10^-5

Test 5
f(x) = cos(x)
From: 0
To: 3pi
Max. Deg.: 8
g(x) = 6.69465*10^-9 x^8 - 1.01005*10^-4 x^7 + 0.00332742 x^6 - 0.0400508 x^5 + 0.205029 x^4 - 0.351657 x^3 - 0.120773 x^2 - 0.169562 x + 1.01830
e^2 = 1.9764 * 10^-5

Test 6
f(x) = e^(-x^2)
From: -2
To: 2
Max. Deg.: 10
g(x) = -0.00143879 x^10 + 0.0211686 x^8 - 0.133468 x^6 + 0.473915 x^4 - 0.992265 x^2 + 0.999624
e^2 = 8.4405 * 10^-8

Test 7
f(x) = 1/(x^2 + 1)
From: -3
To: 3
Max. Deg.: 16
g(x) = 3.16504*10^-6 x^16 - 1.23974*10^-4 x^14 + 0.00203230 x^12 - 0.0181511 x^10 + 0.0966544 x^8 - 0.318123 x^6 + 0.662508 x^4 - 0.923653 x^2 + 0.996846
e^2 = 2.6 * 10^-6

Test 8
f(x) = e^((sin(x))^3)
From: 0
To: 2pi
Max. Deg.: 1
g(x) = -0.224 x + 1.87
e^2 = 2.6845 * 10^-1

Test 9
f(x) = (cos(x))^5
From: pi
To: 2pi
Max. Deg.: 17
g(x) = 6.31854*10^-12 x^17 - 1.63320*10^-11 x^16 - 2.78192*10^-10 x^15 - 8.73964*10^-10 x^14 - 2.74564*10^-9 x^13 + 5.59537*10^-8 x^12 - 2.70984*10^-8 x^11 - 2.61779*10^-5 x^10 - 2.67450*10^-7 x^9 - 8.40220*10^-7 x^8 + 0.3247516 x^7 - 7.3046333 x^6 + 72.908485 x^5 - 408.43197 x^4 + 1362.34294 x^3 - 2657.0269 x^2 + 2737.319113682029 x - 1100.1514
e^2 = 5.8 * 10^-5

Test 10
f(x) = sqrt(1 - x^2)
From: -1
To: 1
Max. Deg.: 20
g(x) = -627.601 x^20 + 2853.92 x^18 - 5499.47 x^16 + 5845.71 x^14 - 3738.80 x^12 + 1472.42 x^10 - 350.970 x^8 + 47.6426 x^6 - 3.39128 x^4 - 0.414096 x^2 + 0.999632
e^2 = 1.2 * 10^-6

Test 11
f(x) = sqrt(1 - x^2) + 0.1*sin(4*pi*x)
From: -1
To: 1
Max. Deg.: 18
g(x) = -203.563 x^18 + 141.013 x^17 + 821.952 x^16 - 757.063 x^15 - 1378.59 x^14 + 1771.07 x^13 + 1241.44 x^12 - 2345.58 x^11 - 648.127 x^10 + 1895.10 x^9 + 197.425 x^8 - 928.025 x^7 - 33.5970 x^6 + 254.982 x^5 + 2.70132 x^4 - 32.7510 x^3 - 0.590680 x^2 + 1.25167 x + 1.00047
e^2 = 1.9 * 10^-6

Test 12
f(x) = abs(x)
From: -1
To: 1
Max. Deg.: 6
g(x) = 1.46667 x^6 - 2.82024 x^4 + 2.30724 x^2 + 0.0854450
e^2 = 5.08635 * 10^-4

Test 13
f(x) = x*sin(x^1.2) + 10*sqrt(x)
From: 0
To: 2.5pi
Max. Deg.: 3
g(x) = -0.129 x^3 + 1.046 x^2 + 1.392 x + 6.98
e^2 = 7.3389 * 10^0

Test 14
f(x) = sin(log(1 + e^cos(x)))
From: 2
To: 10
Max. Deg.: 10
g(x) = 9.52190*10^-10 x^10 - 6.30342*10^-6 x^9 + 3.3746525*10^-4 x^8 - 0.0078575036 x^7 + 0.103708803 x^6 - 0.84926923 x^5 + 4.4421097 x^4 - 14.766225 x^3 + 30.153933 x^2 - 34.72483 x + 17.8389
e^2 = 5.4 * 10^-5

Test 15
f(x) = abs(log(x))
From: 1/e
To: e
Max. Deg.: 15
g(x) = 6.07034*10^-8 x^15 + 9.85444*10^-5 x^14 + 3.35349*10^-7 x^13 + 7.88206*10^-7 x^12 + 1.85260*10^-6 x^11 + 4.35436*10^-6 x^10 - 3.4245574 x^9 + 42.093925 x^8 - 228.714688 x^7 + 711.8547 x^6 - 1383.06 x^5 + 1720.8859 x^4 - 1357.596 x^3 + 651.1596 x^2 - 173.8175 x + 20.680
e^2 = 1.587 * 10^-4

Test Program

The following snippet can be used to compute the best approximating polynomial for a given function, as well as to calculate the error of your own approximations. Note that it has limited accuracy, and becomes numerically unstable for polynomials of degree over 20–30.

<style>#main {display: none;}#status_container {padding: 4px;}#status {padding: 5px;background-color: #fffdce;box-shadow: 1.5px 1.5px 3.5px #aaaaaa;font-size: 10pt;word-wrap: break-word;display: none;}#status[loading] {display: inline;}</style><span id="main"><table id="main_table"><tr><td><div id="plot_area"><div id="plot_float"><div id="plot_container"><div id="plot"></div></div></div></div></td><td><table class="field_table" id="field_table"><tr><td class="field_name">𝑓(𝑥)&nbsp;=</td><td><table class="padding_table"><tr><td><div class="input_container"><div class="input_underlay" id="expression0_underlay"></div><div class="input_error_underlay" id="expression0_error_underlay"></div><textarea class="input" id="expression0" spellcheck="false" oninput="update(0)">e^-(x/6) cos x</textarea></div></td></tr></table></td></tr><tr><td colspan="2"><table class="horz_field_table"><tr><td><table class="field_table"><tr><td class="field_name small_field_name">From</td><td><table><tr><td><div class="input_container"><div class="input_underlay" id="from_underlay"></div><div class="input_error_underlay" id="from_error_underlay"></div><textarea class="input" id="from" spellcheck="false" oninput="update()">0</textarea></div></td></tr></table></td></tr></table></td><td><table class="field_table"><tr><td class="field_name small_field_name">To</td><td><table><tr><td><div class="input_container"><div class="input_underlay" id="to_underlay"></div><div class="input_error_underlay" id="to_error_underlay"></div><textarea class="input" id="to" spellcheck="false" oninput="update()">2.5pi</textarea></div></td></tr></table></td></tr></table></td><td><table class="field_table"><tr><td class="field_name small_field_name">Max.&nbsp;Deg.</td><td><table><tr><td><div class="input_container"><div class="input_underlay" id="max_deg_underlay"></div><div class="input_error_underlay" id="max_deg_error_underlay"></div><textarea class="input" id="max_deg" spellcheck="false" oninput="max_deg_add(0)" onkeydown="return spinner_keydown(event, 'max_deg', max_deg_add)">3</textarea></div></td></tr></table></td><td><div class="button_group small_button_group"><table><tr><td><button class="increment pos_button fixed_button" id="max_deg_inc" title="Increment" onclick="max_deg_add(+1)">▲</button></td></tr><tr><td><button class="decrement neg_button fixed_button" id="max_deg_dec" title="Decrement" onclick="max_deg_add(-1)">▼</button></td></tr></table></div></td></tr></table></td><td class="separator"></td><td><div class="button_group"><button class="check_button" id="hold" title="Supress update" onclick="hold()">Hold</button></div></td></tr></table></td></tr><tr class="separated_row" id="expression0_0"><td class="field_name">𝑔(𝑥)&nbsp;=</td><td><table class="padding_table"><tr><td><div class="output_container"><div class="output polynomial" id="poly"></div></div></td></tr></table></td></tr><tr id="expression0_1"><td class="field_name"><span class="variable">ϵ</span><sup>2</sup>&nbsp;=</td><td><table class="horz_field_table"><tr><td><div class="output_container"><div class="output" id="poly_error"><table class="horz_field_table"><tr><td class="error_value" id="poly_error_value"></td><td class="error_change" id="poly_error_change"></td></tr></table></div></div></td><td><div class="button_group"><button class="add_expr pos_button fixed_button" id="add_expr0" title="Add function" onclick="add_expr(0, event.ctrlKey + 2 * event.shiftKey)">+</button><button class="remove_expr neg_button fixed_button" style="display: none;" id="remove_expr0" title="Remove function" onclick="remove_expr(0, event.ctrlKey)" disabled>-</button></div></td></tr></table></td></tr><tr class="options separated_row" id="flag_row"><td colspan="2"><table class="horz_flag_table"><tr><td><label class="flag_container" title="Plot legend"><table><tr><td><input type="checkbox" class="flag" id="legend" onchange="update(null, true)" checked></td><td>Key</td></tr></table></label></td><td><label class="flag_container" title="Extra accuracy"><table><tr><td><input type="checkbox" class="flag" id="extra_accuracy" onchange="update(undefined, true)"></td><td>Acc.</td></tr></table></label></td><td><label class="flag_container" title="Scientific notation"><table><tr><td><input type="checkbox" class="flag" id="scientific_notation" onchange="render_output()"></td><td>Sci.</td></tr></table></label></td><td><label class="flag_container" title="Decimal exponent style"><table><tr><td><input type="checkbox" class="flag" id="decimal_exponent" onchange="render_output()"></td><td>Dec.</td></tr></table></label></td><td><label class="flag_container" title="Ascending order"><table><tr><td><input type="checkbox" class="flag" id="poly_ascending" onchange="render_output()"></td><td>Asc.</td></tr></table></label></td><td><select id="poly_mode" title="Polynomial style" onchange="render_output()"><option value="disp">Display</option><option value="text">Text</option><option value="prog">Program</option><option value="list">List</option></select></td></tr></table></td></tr></table></td></tr></table></span><div id="status_container"><span id="status" loading>Loading...</span></div><link rel="stylesheet" type="text/css" href="https://gist.githack.com/anonymous/6059e31443745ba122dd/raw/9ca2667d0027fc4c4764533098885a8c0cc31674/poly.css"><!--[if lte IE 8]><script language="javascript" type="text/javascript" src="http://www.flotcharts.org/flot/excanvas.min.js"></script><![endif]--><script language="javascript" type="text/javascript" src="http://www.flotcharts.org/flot/jquery.js"></script><script language="javascript" type="text/javascript" src="http://www.flotcharts.org/flot/jquery.flot.js"></script><script async type="text/javascript" src="https://gist.githack.com/anonymous/6059e31443745ba122dd/raw/aa24abd74de60be00eb3d5b4cbb4e39ee8cbf3d6/poly.js"></script>

Sandbox Notes

The "no functions that perform this specific task" rule seems to be too ambiguous after all. I'm returning this to the sandbox for now.

\$\endgroup\$
  • 1
    \$\begingroup\$ Challenge or not, your code snippet is a mighty useful tool for math impaired people like me :). \$\endgroup\$ – user16991 Feb 7 '15 at 18:01
  • \$\begingroup\$ "You may not use any function that performs this specific task." But functions which perform symbolic integration, enumerate families of orthogonal polynomials, or solve systems of linear equations are ok? \$\endgroup\$ – Peter Taylor Feb 7 '15 at 18:46
  • \$\begingroup\$ @PeterTaylor I'm still mulling it over. I tried coming up with a list of no-no functions but it ended up too long. I'm aware that with the current spec Mathematica folks are going to have a field day :) \$\endgroup\$ – Ell Feb 7 '15 at 19:03
  • \$\begingroup\$ Is it allowed to examine the function in ways other than numerically evaluating it? \$\endgroup\$ – feersum Feb 8 '15 at 0:38
  • \$\begingroup\$ @feersum Assuming you mean something like inspecting the string or doing some symbolic analysis, then yes, that's ok. Like I said to Peter, though, I might disallow some builtin functions, like integration (symbolic or otherwise). \$\endgroup\$ – Ell Feb 8 '15 at 13:06
4
\$\begingroup\$

Programming Puzzle or Code Golf?

(A judging books by covers question)

^Might need a better title.

This question is based off the "Let's Judge Some Books By Their Covers" question.

Browsing the site, I see that 1859 of our 2692 questions (69%) are tagged . My question is: what's the difference?

Your goal is to write a program (or function) to predict, given only the title of a question, whether or not that question is tagged code-golf. Your program will receive a title as input and should output either a truthy (if it's code golf) or falsey (it it's not) value.

Additionally, your program should contain no more than 1000 bytes.

Scoring

The test data will be all of the questions on this website, excluding closed/migrated/deleted questions. Your score will be the Phi coefficient calculated by comparing the results of your program with the actual data. Higher values (closer to 1) are considered better.

The Phi coefficient is calculated via the following formula:

                actual
guess      puzzle  golf  total
  puzzle   A       B     Y
  golf     C       D     Z
  total    W       X

Phi = (AD - CB) / sqrt(WXYZ)

The benefit of this scoring method is that any form of random guessing (output not affected by input) results in an average score of zero.

The exact data set is yet to be generated.

Notes

I believe this challenge is an improvement over the previous challenge due to a few reasons:

  • The question title probably has a much stronger relationship to its tags than to votes, so there's hopefully more room for improvement and competition.
  • Although there will be special-casing (for words like "short") it won't be for single questions. There's no massive outliers in the data.
\$\endgroup\$
  • \$\begingroup\$ You might want to force the program to be deterministic to prevent return rand<.69 \$\endgroup\$ – FryAmTheEggman Nov 11 '14 at 15:59
  • 2
    \$\begingroup\$ Please don't change the title, it's perfect :P \$\endgroup\$ – undergroundmonorail Nov 11 '14 at 16:36
  • \$\begingroup\$ This has the issue of hardcoding. A near-perfect program can probably be written by compressing the 2692 bits needed into a 337-bytes magic string and using a hash table with enough expansion to make collisions rare. Ideally, you'd have a secret test set that's separate from the training set, but I don't know how to restrict that here except for "honor system" given that the data is public. \$\endgroup\$ – xnor Nov 14 '14 at 10:22
4
\$\begingroup\$

Find a Diagonal

Given a (possibly concave) polygon of n ≥ 4 sides, output a valid diagonal, a line segment joining two distinct vertices which, aside from the endpoints, is completely contained within the interior of the polygon.

For example, for the polygon

[(0, 0), (3, 0), (1, 1), (0, 4)]

a valid diagonal (in fact, the only possible diagonal) is:

[(0, 0), (1, 1)]

Diagonal

Some invalid diagonals are:

[(3, 0), (0, 4)]     Lies outside the polygon
[(0, 0), (3, 0)]     Is an edge of the polygon - the interior of the line is not inside
[(0, 0), (0, 0)]     Two identical vertices

Input/output

Input will be n pairs of integers representing the vertices of the polygon in order. There is no fixed orientation for the input — it could be clockwise or anticlockwise. You may write either a function or a full program for this challenge, and assume any clear (all integers distinguishable) and convenient list/string format for the input.

You may assume that no three consecutive vertices of the polygon are collinear, i.e. there are no 180 degree angles. You may also assume that all coordinates are between 0 and 255 inclusive.

Output will be 2 pairs of integers representing a diagonal, which may also be in any clear and convenient list/string format.

Rules

  • You must work in the integers or rationals. In particular, you cannot use floating point integers, due to imprecision.

  • You may not use any polygon-related builtins.

  • This is code-golf, so the program in the fewest bytes wins.

Test cases

For each case, the first line is the input polygon, and the second line is all possible edges which are a valid diagonal. You only need to output one valid diagonal, and the vertices may be in either order.

Vertical diagonal
[(0, 0), (1, 1), (2, 0), (1, 3)]
[(1, 1), (1, 3)]

Horizontal diagonal
[(5, 0), (3, 4), (8, 8), (6, 4)]
[(3, 4), (6, 4)]

The relevant images are given below, in test case order (click the thumbnails to view).

Test 1 Test 2

(More cases to be added)

\$\endgroup\$
  • 2
    \$\begingroup\$ Your first example of a valid diagonal has one end-point which isn't actually a vertex of the polygon. The example would benefit from an image. Since you're restricting people to exact arithmetic, you should specify a bound on the vertex coordinate values so that people can work out whether they're at risk of overflow. \$\endgroup\$ – Peter Taylor Mar 17 '15 at 8:00
4
\$\begingroup\$

The Predator of my Predator is my Prey

(Three Team KotH)

Three teams: Red, Green, Blue

  • Red kills Green
  • Green kills Blue
  • Blue kills Red

As in Red vs Blue, each entrant is assigned a colour based on their userid. Your objective is to ensure your team has the most surviving members at the end of the game.

The rules are simple but the dynamics may not be obvious. For example, wiping out your prey colour early on seems like success, but it leaves your predator colour with no predators of their own, and free to wipe you out. This means early on it may be better to herd your prey rather than kill them, but this could back-fire if your team leaves it too late...

Possible game styles

Whatever the style, when a Red bot touches a Green bot, the Green bot becomes a Red bot (the bot's code is replaced by its attacker's code). The total number of bots is therefore constant throughout the game. There are a number of settings in which such a game could be played:

  • pixels in an open arena (like Red vs Blue)

  • pixels in an arena with obstacles/walls/mazes

  • bots in a continuous arena (no grid), free to turn smoothly through 360 degrees

I like the idea of a continuous arena, and bots only seeing a small radius semi-disc ahead of them. With no vision behind them they would have to either turn regularly, or coordinate with their team mates to get more information on their surroundings. Bots would be able to write messages and read the messages of other bots on the same team.


Sandbox questions

  • Stack Snippet / full multi-language KotH?

  • which of the game styles suggested would be most interesting?

\$\endgroup\$
4
\$\begingroup\$

ASCII Art of the Day Series

My new found love for ASCII art has lead me to a lot of good (trivial and non-trivial) ideas for ASCII ART challenges. Here are the ideas :

1. Double Knot

2. Flow Snakes

3. Chinese Shrine

4. Zodiac Signs


6. Snow Flakes

Its time for another ASCII Art of the Day. This time, we are going back to the winters and drawing Snow Flakes (not to be mixed up with Flow Snakes ;) ). The snow flakes are generative based on random walk so each run should give a different pretty snow flake ASCII.

Challenge

Given an input integer N, draw an ASCII snow flake of radius N using the construction instructions provided below.

Construction

The Snow Flake will have 6-fold rotational symmetry and 3-fold reflection symmetry. You will ideally only generate 1 out of the 12 wedges in a snow flake and then rotate/mirror them to get the other wedges.

Lets consider the following snow flake for N = 5:

     \__    __/
     /_/ /\ \_\
    __ \ \/ / __
    \_\_\/\/_/_/
__/\___\_\/_/___/\__
  \/ __/_/\_\__ \/
    /_/ /\/\ \_\
     __/ /\ \__
     \_\ \/ /_/
     /        \

Lets name its wedges as:

       4      3
     5\__    __/2
      /_/ /\ \_\
     __ \ \/ / __
     \_\_\/\/_/_/
6__/\___\_\/_/___/\__1
7  \/ __/_/\_\__ \/  12
     /_/ /\/\ \_\
      __/ /\ \__
      \_\ \/ /_/
     8/        \11
       9     10

For creating the above snow flake, all we need is to first construct a single wedge (say 1) and fit others in place based on wedge 1.

Wedge Construction

Lets consider the wedges 1 and 2 from the above example and shade all the blocks which belong to wedge 1 with x.

    /2       x
   _\     xxx
  / __ xxxxx
 /_/xxxxxxx
/xxxxxxxxx1

We can see that exactly 25 x belong to wedge 1 in a 50 block trapezium of wedge 1 and 2. The height of the trapezium corresponds to the input integer N and the base length of the trapezium is 2N.

To construct the wedge 1, follow these instructions:

  • Use only the area marked for your wedge (For example, the x area in the above image for wedge 1)
  • Use uni-directional random walk to fill up the x using the characters /, _ and \
  • There should be at least 1 path in the random walk which connects the left most x with any one of the right most x of each row.

A few examples of valid random walks for wedge 1 are:

[TBC]


Random Ants

This is still a bit hazy but the challenge would involve making random ants out of the following ants:

 \_/
'-0-'
--0--
.-0-.

 \_/
'-0-'
--0--
.-0-.

 \_/
'-0-'
--0--
.-0-.


 \O/
'-O-'
 /o\
  ^

 \O/
'-O-'
 /o\
  ^

 \O/
'-O-'
 /o\

Fish Aquarium

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
         .               `         /
                          .    ,../...       .
          .                .  /       `\  /  .
     \    .        o         < '  )     =<
     /\  .                    \ \      /  \   .  __
   >=)'>                       `'\'"'"'         /o \/
     \/ .    /         o              /,        \__/\    .:/
     /   .  /--\ /         /         <')=<     .      ,,///;,   ,;/
           <o)  =<      . / \         \`         .   o:::::::;;///
            \__/ \       <')_=<                     >::::::::;;\\\
             \            \_/            .            ''\\\\\'' ';\
    (                      \              .   __
     )                                       <'_><          (
    (          (                ,/..          `              )
     )     (    )             <')   `=<                )    (
    (       )  (               ``\```                 (      )
_____)_____(____)______________________________________)____(___________
\$\endgroup\$
  • 1
    \$\begingroup\$ I like (1) and (2) more than (3) and (4) because they are more structured, so more can be done with algorithms rather than compression. \$\endgroup\$ – xnor May 21 '15 at 5:20
  • \$\begingroup\$ @xnor 3 is not finalized yet, so I may do 4 before 3. But 3 would have more structure to it once done. I agree that 4 does not have much similarity, but there still is scope for reuse in 4. \$\endgroup\$ – Optimizer May 21 '15 at 5:49
4
\$\begingroup\$

Sandbox note : Big change in specs, mainly for balancing issues (thanks to @trichoplax for your sugeestions :))

Major changes : Starting, Turn deroulement (2 phases from now), tie-break, Cloning. Merging may be removed due to the new specs (rendered pretty useless).

With the current specs, 1000 turn migh be WAAAAAAY to much, as you could easily go up to the 17 range... well, you could reach it in 3 turns, and with 3 turns more you could wipe everything with an explode.... Will fix that

Summoner war

tags :

You're a Summoner controlling Demons. Regulary, some tournament are done to dertermine who will be the king. And the king is always the strongest !

As it is a turn-based game, each Summoner will play twice against each opponent. So, for each pair of submission, they will all fight while playing first AND playing last.

A fight is limited to 1000 turns. A turn is the cumulated actions of both Masters.

The winner of a fight is the one who totally destroyed all the demons of the opponent. The number of Demons you killed will be used as tie-breaker

The winner of the tournament is the Summoner who won the most fights. The same tie-breaker as for fights is used, except it is the cumulated amount of kill :).

The Grid

The grid is 9*17 and the cells are placed as above :

.................
.................
.6..1.......1..6.
.................
.5..2.......2..5.
.................
.4..3.......3..4.
.................
.................

The numbers indicates the position of the starting Demons. Each Summoner will see itself as the left Summoner. Demons can move and attack on horizontal lines, vertical lines and diagonals.

Starting

You will chose up to 6 Demons to summon, determining their stats and their capacities. You can give a maximum of 12 stat point distributed among your Demons Each Demon you summon (at this moment) will cost you 2 points.

Summoning 6 demons will cost 6*2=12 points

Summoning 3 demons will cost you 3*2=6 points, leaving you 6 points to increase their statistics.

Statistics

Demons have 4 statistics :

 Life 
    Hp = 2+Life
    If the hp of your Demon reach 0, it dies.
 Attack
    Determine the damages you deal/heal
 Defense
    Reduct all incoming damages by defense/2 rounded to superior. 
    You can't take less than 1 (except if you're attacked for 0 hit points)
 Range 
    Determine the range of your attacks

Basic Moves

There's 3 moves that every Demons can do :

 Move
    The Demons will move by one unit.
 Attack
    Deal damages to the Demon in an adjacent cell.
    Damages: attack+1
 Explode
    Destroy the Demon and deal damages to it's surrounding.
    It only can be used if your Demon have more than 40% hp.
    It does raw damages (defense doesn't influence it).
    Damages: attack/2 (raw)
    Radius : 1+sqrt(range)  (square shaped, truncated)

Capacities

Each Demon will have one capacity in the following list:

 Merge
    Allows a Demon to absorb an other adding their stats 
    During the merge process, you chose if you want to keep Merge as a capacity, 
    or if you want to take the other Demon's capacity.
    After merging, a Demon won't be able to clone again.
 Heal
    Heal nearby Demons but not self.
    Heal: attack/2  rounded to superior
    Radius : range+1(square)
 Cross
    Deal damages on all lines or diagonals around your Demon
    It will damages on lines if the direction you specify is
    North/South/East/West/Self
    Otherwise, it will damages on the diagonals.
    Damages : attack
    Radius : range+1(lines or diagonals)
 Eat
    Steal some stats if it kills the target
    The target must be on an adjacent cell.
    Damages : attack
    Stat stolen : range (maximum)
                The stat stolen will be distributed randomly between the stats
                of your Demon. It cannot steal more point that the opponent 
                Demon have.

Misc

You may give an optional name for your Demons. This name will only be known by you, and could be useful if you want to remember the role you gave to your Demons. This name could be changed at anytime.

Turn deroulement

Each player's turn has 2 steps : the enhance phase and the battle phase.

A turn is derouling as following :

player 1's enhancing phase
player 2's enhancing phase
player 1's battle phase
player 2's battle phaes

Enhance phase

Each turn, you will be given 6+nbTurn/50 stats points.

On turn 1, you will have 6 points.
On turn 50, 7 points. 
On turn 1000, 26 points. 

Those points will stack upon time if you don't use them.

On turn 1, I had 6 points, I used 4 remain :2
Turn 2, I gain 6 points more, I can use 8.
etc ...

Those points can either be used to increase the stats of your Demons (1 point = 1 life||attack||defense||range) or to clone them.

To clone a Demon, you must use 2+life+attack+defense+range points. In result, the origin demon will have his points halved, and a clone with the other half will spawn.

Let's say I want to clone the following demon :
 life=2
 attack=2
 defense=5
 range=0
 capacity=Cross
He has 2+2+5+0 = 9 statistics points, so I would need 2+9=11 points.
The resulting Demons would both be :
 life=1
 attack=1
 defense=3 (yes, stats are rounded to superior :))
 range=0
 capacity=Cross

Battle phase At the start of the battle phase, you will be given :

  • The current tableboard
  • A list containing your Demons and all the information about them
  • A list containing the stats and position of your opponent's Demons

You will return the actions of all your Demons.

you have up to x ms to return this list (not decided yet)

Once you've chosen your actions, the following will happen :

  • Every Demons which had to move, will move. If two Demons try to move to the same location, they will both stay still.
  • All the actions which change the hp(heal/damages) will happens at the same time.
  • Finally, your turn end, and your opponent's will decide his actions.

Submissions

Each submission will be written in Java. You can't interact with other players nor the controller in any way you could imagine.

You must extend the following class :

Class not yet developped

A github link will be provided, for the controller etc.

Sandbox notes

I'm aware that there's plenty typos and errors, I will correct them before the final version. Some text and clarification might be added. As it can be complicated (lot of options), 4 demonstration Summoner will be provided (and will be in the pool), each of them using only some functionnality (prooving that you don't have to use every single thing).

I'm not done yet with all the sources for the controller/field etc. An API will be provided for extracting/using informations easily. Once it will be complete, I will put a time limit for the execution of one turn. If I have time, I will provide an interface to see the fights with colors etc.

Questions :

  • Is this too complex? I'd like to know if to much options are given, and if it's confusing (ie, if you don't know where to start).
  • Is there some points of the ruleset that are obviously dangerous for the good health of this contest.
  • Any suggestion?

Thanks for reading :)

\$\endgroup\$
  • \$\begingroup\$ I've removed all <code></code> tags and instead selected the code and pressed ctrl-K (which adds 4 spaces before the code). This formats as a code block, instead of a series of code lines with white gaps between. I find this much easier to read so I've edited it in for you to see, but feel free to rollback to the previous edit (I thought it would be easier to edit and show, rather than try to describe). \$\endgroup\$ – trichoplax Jun 5 '15 at 11:02
  • \$\begingroup\$ Ho, thank you, did not know the K^ thing ! I didn't think it was hard to read with it, that's why I did put <code> instead of spaces. Anyway, thanks :). \$\endgroup\$ – Katenkyo Jun 5 '15 at 11:11
  • \$\begingroup\$ You're welcome :) \$\endgroup\$ – trichoplax Jun 5 '15 at 11:37
  • \$\begingroup\$ @Katenkyo I'm not certain but wouldn't it be possible just to fill the whole grid by cloning? Which would make it quite hard to develop an effective strategy (because clone leaves two demon with potentially full health- it might work if it shares the health between the clones)- or is that the idea (make a massive army of demons and then fight? \$\endgroup\$ – euanjt Jun 5 '15 at 14:14
  • \$\begingroup\$ @TheE I think the AoE could compensate this point. One Demon with a good attack stat (via merging/eating) could one-shot a big number of your Demon/turn. I want the "fill with clones" strategy to be possible, but with a little more thinking to be competitive. I don't know yet if it is owerpowered. Anyway, it have to be tested ! Thanks for the case, didn't think of it. Will test it as soon as the controller is up. \$\endgroup\$ – Katenkyo Jun 5 '15 at 14:29
  • \$\begingroup\$ Minor point: the list of standard moves still refers to 4 instead of 3 moves. \$\endgroup\$ – trichoplax Jun 22 '15 at 14:09
  • \$\begingroup\$ Explode affects a radius of 1 + sqrt(range). Is this a square range or some other shape (Euclidean to give an approximate circle?). Is the sqrt rounded up or down? \$\endgroup\$ – trichoplax Jun 22 '15 at 14:11
  • \$\begingroup\$ square range, round down (so 16 range will effectively add 4 points) I did that because you'd only need 17 range to wipe out the entire map with raw damages... No you still can have a good range, but it's harder \$\endgroup\$ – Katenkyo Jun 22 '15 at 14:13
  • \$\begingroup\$ What if the cross affected both players (so you need to get into the right position or else accept that some of your own demons will be hit too)? Would that be interesting or just awkward? How would it affect the balance of the different capacities? \$\endgroup\$ – trichoplax Jun 22 '15 at 18:35
  • \$\begingroup\$ What if explode and heal also affected both players (all demons in range)? \$\endgroup\$ – trichoplax Jun 22 '15 at 18:36
  • \$\begingroup\$ @trichoplax I'm thinking about it since the beginning, but I am not able to make my mind about it... I mean, sure it is interesting, and all but... Isn't it too much? You already have to manage lots of demons, to buff/clone them, be aware of your opponents. So I don't know, I can't know, if that would be too difficult to handle... \$\endgroup\$ – Katenkyo Jun 22 '15 at 18:48
  • \$\begingroup\$ I was thinking less about how complicated it is and more about whether it would make the capacities more balanced. I think the main problem would be if one of the capacities is too strong and everyone chooses nothing else. \$\endgroup\$ – trichoplax Jun 22 '15 at 21:55
  • \$\begingroup\$ @trichoplax sounds logic. I'll look at a way to implement it :) \$\endgroup\$ – Katenkyo Jun 23 '15 at 5:32
4
\$\begingroup\$

Ironclad Tactics KoTH

This KoTH is inspired by the Ironclad Tactics paper version from Zachtronics.

The game is played on a 9x4 grid split into three areas: the North, the South, and No-Man's-Land (the center). To make things easier for you, it will always appear that you are the North (the left).

Steelclad

Each battle, each player starts with 10 Action Points (AP). In a battle, there are 4 phases: Selection, Placement, Upgrading, then Attacking.

Selection

During selection, players start by each simultaneously choosing an upgrade to ban, and then choosing 5 of the 8 or 9 remaining upgrades they wish to use. The chosen upgrades are not revealed to their opponent.

Placement

During placement, players simultaneously place an Ironclad until they decide to drop out of the phase or they have used all of their AP. Each placement costs 1AP, and Ironclads can only be placed on the player's respective side and No-Man's-Land.

Upgrading

During upgrading, players simultaneously upgrade one of their Ironclads using their remaining AP until they decide to drop out of the phase or they have used all of their AP. Each upgrade costs 2+(# of times you previously used the upgrade)-(# of times your opponent previously used the upgrade), with a minimum cost of 0.

There are 10 possible upgrades:

Upgrades

The stars represent Ironclads, and each circle represents the squares the Ironclad will attack. The arrow represents the Saboteur upgrade, which will give the player 2 additional victory points, and the diamond represents the Heavy Chassis upgrade, which requires 2 hits to be destroyed.

Upgrades can be given to any of your unupgraded Ironclads, even if the squares it would attack fall off of the board. Upgrades do not need to be given to all Ironclads.

Attacking

During attacking, all Ironclads simultaneously attack. Any Ironclad in the square of attack (even if the attack is friendly) is destroyed (unless the tank has the Heavy Chassis upgrade, in which case it must be attacked by two different tanks.

Any remaining tanks give 1 (or 3, if the tank has the *Saboteur upgrade) victory point(s) if they have a horizontal line of sight to the opposite end unobstructed by undestroyed enemy tanks.


API

You will implement a Java class. more info

Additional Info

  • You will have 10 battles against each opponent.
  • The price for upgrades resets after each opponent.
  • You beat an opponent by having more victory points than them after the 10 battles. You win this challenge by beating the most opponents.
  • If both players attempt to place in the same spot at the same time, that location becomes unplaceable for the rest of the battle, and players must place again.
  • -
\$\endgroup\$
4
\$\begingroup\$

Counting the Pattern Unlocks

Meta

Upon thinking about this challenge harder, I've realized that it is fairly trivial. Unless I am really bad at counting (which I am liable to be), there are at least 2 ways and at most 4 ways to draw any given pattern.

I think that this makes the challenge considerably less attractive. For this reason, I am thinking of ways to spice it up. For this, I would like your feedback. Currently I am thinking of:

-->Changing the rules so that dots can be used multiple times.

Let me know what you think of this change to the challenge, or suggest other changes.

The Premise

Many modern smartphones have options that allow you to choose how you would like to unlock your phone. For example, my phone allows

  1. PIN input: a 4 digit number (low security)
  2. Pattern Unlock: connect the dots in a pattern (medium security)
  3. Password entry: enter a password (high security)
  4. Fingerprint.

When you elect this second option, you are shown a 3x3 grid of dots (9 dots total) that looks something like this

.    .    .

.    .    .

.    .    .

Instead of using dots, we will use numbers in the following fashion for ease of reference.

1    2    3

4    5    6

7    8    9

You are then prompted to connect dots to form a pattern. This process is subject to the following rules:

  1. The pattern must use at least 3 dots.
  2. The pattern must be one continuous connection, that is once you lift your finger, the pattern ends.
  3. Each dot may be used at most once (there is a quasi exception to this in rule 4)
  4. If you take three collinear dots, in connecting the outer two dots, you must connect the inner dot if it is not already used. For example, if you make the connection 1-3, if 2 is not already used, you are really making the connection 1-2-3. If 2 is already used, then you are just making the connection 1-3. This is where the quasi exception to rule 3 comes in. You are passing over 2 again, even though you have already used it.

I believe that these are all the rules. Note that some phones may use different rules, but if you think I am missing something important, please let me know.

The Task

Our goal is to count how many ways there are to draw a given pattern.

For example if you see the following pattern:

1----2----3

4    5    6

7    8    8

It could have been drawn in one of four ways:

1 --> 2 --> 3    or
3 --> 2 --> 1    or
2 --> 3 --> 1    or
2 --> 1 --> 3

Input

You will be given a series of integers between 1 and 9 to stdin. They will represent connected dots on a grid.

The above example could be input via any of the following sequences:

1 2 3    or
3 2 1    or
2 3 1    or
2 1 3    or
1 3      or
3 1

If input is invalid, your program may do whatever it wants.

Output

Your program is to output the number of ways there are to draw the pattern, as an ingeter, to stdout or equivalent. In the above example, the output would be

4

Wining Criterion

This is code golf, shortest code (bytes) wins.

Bonus

Take off 20% if your answer also outputs all ways to draw the pattern (any understandable syntax acceptable).

TEST CASES

(1) grid layout

1    2    3
        /
4    5    6
|  /
7    8    9

possible inputs:

input>> 4 7 3
     >> 4 7 5 3
     >> 3 7 4
     >> 3 5 7 4
     >> 5 3 7 4
output<< 3

the ways are (for the bonus):

4 --> 7 --> 5 --> 3    or
3 --> 5 --> 7 --> 4    or
5 --> 3 --> 7 --> 4

(2)

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

input>> 3 2 1 4 7 8 9 6 5    or
     >> 3 1 7 9 6 5          or
     >> 5 6 9 7 1 3          or
     a couple others
output<< 3

example directions:

 3 --> 1 --> 7 --> 9 --> 6 --> 5   or
 5 --> 6 --> 9 --> 7 --> 1 --> 3   or
 2 --> 3 --> 1 --> 7 --> 9 --> 6 --> 5
 (or could include extra collinear points)

\$\endgroup\$
  • \$\begingroup\$ you are correct. Counting is just really hard apparently. Fixed (I think). \$\endgroup\$ – Liam Nov 4 '15 at 9:03
  • \$\begingroup\$ I've done worse... This is what the sandbox is for :) \$\endgroup\$ – trichoplax Nov 4 '15 at 9:06
  • \$\begingroup\$ It's generally good advice to write an ungolfed reference implementation while sandboxing. This will help ensure test cases are consistent, and also will sometimes highlight potential problems/changes that none of us could spot without an implementation. \$\endgroup\$ – trichoplax Nov 4 '15 at 9:08
  • \$\begingroup\$ Nice challenge, by the way. \$\endgroup\$ – trichoplax Nov 4 '15 at 9:13
  • \$\begingroup\$ Okay, thanks. I'll work on an implementation \$\endgroup\$ – Liam Nov 4 '15 at 9:24
  • \$\begingroup\$ Related and related. Might give you some ideas (or at least ensure that your claims about the pattern are consistent with theirs!) \$\endgroup\$ – Peter Taylor Nov 5 '15 at 21:51
4
\$\begingroup\$

BrainCubed

You are the proud maintainer of one of the smartest robots in the world. Well, it used to be. Now its speakers and microphones have been broken and the darn thing only seems to read Brainfuck. To make matters worse, it would appear most of its RAM... disappeared? This will be tough to explain to the boss. No matter. It seems to have figured out how to draw on a whiteboard to supplement its now-shoddy memory, and you have bigger problems on your hands.

You need to know how to fit things in cubes.

The Task

Lately, your biggest problems (aside from the embarrassing conversation with your boss later) have to do with volume. Cubing things is hard. That's why you'll get the robot to do it for you! Your goal in this challenge is to write a Brainfuck program that computes the cube of a given number. However, the robot's whiteboard isn't very big. The less memory your Brainfuck program requires, the better.

Input

You will receive a single integer as input, x.

  • You may choose to accept input in any integer base greater than 0 and less than or equal to 36, so long as this base does not vary from input to input. (e.g. binary, hexadecimal, decimal)

  • You may assume that x is in the range 0 <= x <= 2^16 - 1

  • You should take input as a string of characters, not bytes. For example, if x = 33 and my program accepts input in binary, I should receive the string "100001" (bytes: 49 48 48 48 48 49) not simply bytes containing 100001.

Output

  • Your Brainfuck program must output the value of x^3 in the same base that input was received in.

  • As it is for input, your program should output a string of ASCII characters, not a sequence of bytes.

  • The program must terminate, and should not print anything except for its numerical output.

Rules and Scoring

Your score in this challenge is defined in the following manner: Let the tape of the Brainfuck memory be described as having a 1st element at the left-most position, and then with potentially infinite cells to the right, indexed by increasing integers n.

Let N(x) denote the right-most (highest n) cell that the program ever sends the tape pointer to (not necessarily modified) for a given input x. Your score for this challenge is then sum (x = 0, 1, 2, ... 100) N(x) (modification pending)

In order to verify score, you may use [this] (soon) modified interpreter.

  • Your program must be written entirely in Brainfuck.
  • Assume the highest value a cell can hold is 255 before wrapping to 0, and that moving left off of the tape will cause the robot to suddenly and violently crash.
  • Your program should not exceed 10k bytes, nor should it take more than 10 minutes to compute x^3 for any x <= 2^16 - 1 on a relatively modern machine.
  • Standard loopholes are disallowed.

SANDBOX NOTES

What do you think of BF memory-optimization as a basis for a challenge?

I chose cubing x as a challenge that is not so trivial as to allow for different approaches, but still within the grasp of BF (if different bases are allowed). Any other ideas?

\$\endgroup\$
  • 1
    \$\begingroup\$ How is the base determined? For example, if the input is 2, how is it determined if that's base-10 (and so output 8) or base-8 (and so output 10)? \$\endgroup\$ – AdmBorkBork Nov 5 '15 at 20:50
  • \$\begingroup\$ My thought was along the same lines as TimmyD: the obvious cheat is to just read the digit, write it, and then write 00 and claim that the input was in base x. \$\endgroup\$ – Peter Taylor Nov 5 '15 at 21:48
  • \$\begingroup\$ @TimmyD Ack! My post is unclear. I meant that you, as the programmer, may choose what base you expect your input/output to be in. e.g. I can write a BF program that accepts input in base 16 and outputs in base 16, or use ternary if I think it'll help me use less memory. I didn't mean that the base changes from input to input. Thank you for the feedback, I'll certainly try to clear that up! \$\endgroup\$ – BrainSteel Nov 6 '15 at 1:48
  • \$\begingroup\$ What is the largest allowed input base? Also, the maximum input needs to be bigger: the current rules probably allow hardcoding of all the outputs. \$\endgroup\$ – feersum Nov 6 '15 at 2:08
  • \$\begingroup\$ @feersum Good points. I'll cap it at 36 to avoid the general silliness of trying to figure out what symbols to use. I understand your concern, but I want to keep the upper limit fairly small. I think 2^16 - 1 should do the trick? \$\endgroup\$ – BrainSteel Nov 6 '15 at 2:27
  • \$\begingroup\$ @BrainSteel That should be good enough, as long as there are some test cases throughout the range. \$\endgroup\$ – feersum Nov 6 '15 at 2:31
4
\$\begingroup\$

Who is the Star Abuser?

In The Nineteenth Byte some people often star chat messages without any reason. You decided to go on a quest to find out who is the star abuser.

You logged the chat actions as a string, each action represented by a character (as you are a code golfer after all).

  • Every user is marked with a letter of the alphabet.
  • An uppercase version marks the user entering chat and a lowercase one marks his/her leaving the chat.
  • At the start there is no one at the chat room.
  • A star marks a starring. We assume that each person in the chat could have starred with the same probability when a starring happened. (E.g. if there are 2 users in the chat we assume both of them had 50% chance to star.)

An example log:

MOA***a**oS**s*Cc

You should output one uppercase letter showing the letter of the user who has the highest number of expected starring. If there are multiple users, output exactly one of them.

Input:

TODO

Output:

TODO

Examples:

TODO

This is code golf so the shortest entry wins.

\$\endgroup\$
  • \$\begingroup\$ As soon as you add IO and ping me. I will upvote this \$\endgroup\$ – Akangka Nov 17 '15 at 12:06
  • \$\begingroup\$ Example : MOA***a**oS**s*Cc > M \$\endgroup\$ – Akangka Nov 18 '15 at 9:50
4
\$\begingroup\$

Pi, continued

The task: generate an arbitrarily precise rational approximation to pi by using a continued fraction.

One way to calculate pi is by using this continued fraction (from Wikipedia):

enter image description here

The first few approximations can be calculated as follows:

enter image description here

enter image description here

enter image description here

enter image description here

You may not use this particular continued fraction. The reason is that in order to calculate this sequence to arbitrary precision, you have to already know pi to arbitrary precision. The series [3,7,15,1,292,...] does not repeat, like pi's own digits.

However, there are continued fractions that have a regular structure. You may use any of these you wish (like these on Wikipedia) in your program.

The rules

  • Input: a single, non-negative integer n.
  • Output: the nth (improper) fraction in your chosen continued fraction series, in lowest terms.
  • Output may be in any form, provided that these conditions are met: 1) the same base is used for numerator and denominator (and for all fractions), 2) the numerator and denominator are clearly distinguishable, and 3) the numerator comes first.
  • Your program must use a continued fraction. It may not use any summation series like the approximation formulae here on Wikipedia.

Meta

  • I feel like some of this may be confusing. What can I clear up, and how?
  • Is it misleading to introduce continued fractions by using an unpredictable series when I want users to use ones that have a regular structure?
\$\endgroup\$
  • \$\begingroup\$ It looks good to me. As long as you don't mind the example being valid for use in an answer, yes it does seem to make sense to use one of the regular structured ones for the example. That way someone who doesn't read it properly can just use the example without realising there are other options, and you get fewer annoying questions... \$\endgroup\$ – trichoplax Nov 18 '15 at 2:20
  • 2
    \$\begingroup\$ Yes, it is misleading to start with the unpredictable series. Are you ruling out approximating pi with a built-in or other limit (that is not a summation), then computing the continued fraction from that? The examples you link are generalized continued fractions, which can have numerators other than 1 and summands that are not whole -- I take it those are allowed? \$\endgroup\$ – xnor Nov 18 '15 at 9:24
  • 1
    \$\begingroup\$ The nth term in a continued fraction series is well defined modulo possible out-by-one disagreements in the indexing. But for generalised continued fractions, there are three competing definitions: if the gcf is x = b_0 + a_1 / (b_1 + a_2 / (b_2 + ... )) is the sequence b_0, b_0 + a_1, b_0 + a_1 / b_1, b_0 + a_1 / (b_1 + a_2), b_0 + a_1 / (b_1 + a_2 / b_2), ...; or is it b_0, b_0 + a_1 / b_1, b_0 + a_1 / (b_1 + a_2 / b_2), ...; or is it b_0 + a_1, b_0 + a_1 / (b_1 + a_2), b_0 + a_1 / (b_1 + a_2 / (b_2 + a_3)), ...? \$\endgroup\$ – Peter Taylor Nov 18 '15 at 19:44
  • \$\begingroup\$ Regarding the competing definitions, I think the cleanest resolution would be to allow any pattern that adds one layer at a time in a consistent fashion, which allows any of the examples as well as off-by-one disagreements. \$\endgroup\$ – xnor Nov 18 '15 at 22:09
4
\$\begingroup\$

Rotate / Flip a Unicode Box Drawing

Given a Unicode box drawing, followed by a series of rotate and/or flip commands, output the result of those operations on the drawing. For clarification, a box drawing can be made from the following characters:

─ │ ┌ ┐ └ ┘ ├ ┤ ┬ ┴ ┼
═ ║ ╔ ╗ ╚ ╝ ╠ ╣ ╦ ╩ ╬
    ╒ ╕ ╘ ╛ ╞ ╡ ╤ ╧ ╪
    ╓ ╖ ╙ ╜ ╟ ╢ ╥ ╨ ╫

The rotate and flip commands are also presented using Unicode symbols:

↔  Flip the drawing horizontally
↕  Flip the drawing vertically
↷ Rotate the drawing 90° clockwise
↶ Rotate the drawing 90° counter-clockwise
↯  Convert all single lines to double and vice versa (Optional - 10% bonus)

All other characters should remain unchanged, but moved to fit where they would be in the modified drawing. For example, if given the inputs:

Input   Output          Input   Output
 ┌┴╖     ╓┴┐             ┌┴╖     ╔═╕
 │A║     ║A│             │E║     ╣E├
 ╘╦╝     ╚╦╛             ╘╦╝     ╙─┘
 ↔                       ↔↷

 ┌┴╖     ╒╩╗             ┌┴╖     ┌─╖
 │B║     │B║             │F║     ┤F╠
 ╘╦╝     └┬╜             ╘╦╝     ╘═╝
 ↕                       ↷↔

 ┌┴╖     ╓─┐             ┌┴╖     ╔╩╕
 │C║     ╣C├             │G║     ║G│
 ╘╦╝     ╚═╛             ╘╦╝     ╙┬┘
 ↷                      ↶↶

 ┌┴╖     ╒═╗             ┌┴╖     ╔╩╕
 │D║     ┤D╠             │H║     ║H│
 ╘╦╝     └─╜             ╘╦╝     ╙┬┘
 ↶                      ↔↕

The drawing may be of any size (let's say anywhere from 1×1 to 50×50), and not necessarily square.

The flip and rotate commands will always occur after any drawing to be flipped, and by themselves on one single line. They should be executed from left to right. (Note that order matters - the E and F examples use the same two commands but in reverse order, and produce different results.)

There may be an arbitrary number of flips and rotates, but you'll note that there are only 8 possible end-states for the final drawing. The G and H examples show two different sets of commands that produce the same result (other than the letter in the middle). It is possible that a series of commands will result in simply returning to the original drawing.

Input may be supplied via command line, user input, file I/O, or any other means you see fit. (Though it should obviously support multiple lines of input. You may use \n to represent line breaks if your input mode only supports a single line.) Likewise, output may be to the screen or a file at your discretion.

This is code-golf, so shortest code wins.

Some more complicated examples:

Input    Output        Input    Output
          ┌─┐                    ┌─┐
          │A│                    │B│
┌─╥─┐     ╞═╡          ┌─╥─┐     ╞═╡
│A║B│     │B│          │A║B│     │A│
└─╨─┘     └─┘          └─╨─┘     └─┘
↷                     ↶

   Input           Output
┌─┐╔═╗╒══╓──    ──╖══╕╔═╗┌─┐
├─┘╠═╝│  ║ ╖    ╓ ║  │╚═╣└─┤
│  ║  ╘══╙─╜    ╙─╜══╛  ║  │
↔

┌─┐╔═╗╒══╓──    ╔═╗┌─┐╓──╒══
├─┘╠═╝│  ║ ╖    ╠═╝├─┘║  │ ╕
│  ║  ╘══╙─╜    ║  │  ╙──╘═╛
↯

┌─┐╔═╗╒══╓──    ┌─┐╔═╗╒══╓──
├─┘╠═╝│  ║ ╖    ├─┘╠═╝│  ║ ╖
│  ║  ╘══╙─╜    │  ║  ╘══╙─╜
↔↯↷↷↕↶↯↶↕↔
\$\endgroup\$
  • \$\begingroup\$ You should have examples where characters other than the box move, the box is a different size and with more than one box, if that is allowed. This looks like a good challenge, but one that is likely to be rife with ambiguity. \$\endgroup\$ – FryAmTheEggman Dec 9 '15 at 14:46
  • \$\begingroup\$ @FryAmTheEggman - I chose one starting box which is asymmetrical just as a simple example. I could add a couple more complicated shapes - I was thinking this could be applied to any shape using the box-drawing symbols, not just a simple box. \$\endgroup\$ – Darrel Hoffman Dec 9 '15 at 18:55
4
\$\begingroup\$

Create a spiraling image

Introduction

WIP

Task

Given an image, output the image with a spiraling effect.

Scoring

This is a , so the submission with the highest number of votes wins!

Test cases

Note that your program should work for images with any size. The output should be a 512×512px image. Also, note that the test cases are examples, you may use any algorithm to produce these images and do not have to match the images below. Use your creativity!

Test case 1

enter image description here

Becomes:

enter image description here

Test case 2

enter image description here

Becomes:

enter image description here

Test case 3

enter image description here

Becomes:

enter image description here

Test case 4

enter image description here

Becomes:

enter image description here

Test case 5

enter image description here

Becomes:

enter image description here

Important

The full size images:

\$\endgroup\$
  • \$\begingroup\$ What counts as the Droste effect does not appear to be clearly defined here, and some of the example outputs are hard to justify as exhibiting the effect. The chat image in particular doesn't seem to have any repetition of the image within itself, only distortion of the scale. \$\endgroup\$ – trichoplax Feb 13 '16 at 23:23
  • \$\begingroup\$ @trichoplax Oh, you're right. That isn't even a Droste image. I should probably remove the "Droste" thing and make something else out of it. Maybe define it as a "spiral" image. \$\endgroup\$ – Adnan Feb 13 '16 at 23:30
  • \$\begingroup\$ Either way it sounds like the too-broad end of the popularity contest spectrum. For a Droste image challenge you could probably make it well defined enough to not need to be a pop con if you also have a template as input, defining one or more places where the image needs to reappear recursively. So for each hole in the template, the image will reappear, including the holes in the template for repeatedly smaller images. \$\endgroup\$ – trichoplax Feb 13 '16 at 23:33
  • 2
    \$\begingroup\$ @trichoplax Hmmm, okay. I should give this a lot of thought then. I do think that with the amount of possible ways to approach this problem, this should be a popularity contest. But it's definitely too broad at this stage. \$\endgroup\$ – Adnan Feb 13 '16 at 23:41
4
\$\begingroup\$

Random Physics Golf #1: Net Gravitational Force

Introduction to the Series

Every week or so I will be posting a physics challenge. My goal here is to design challenges that in the end, teach some people some physics. Overall, the challenges will be very basic with little information. All of these challenges will have the minimal information necessary to solve them, and the goal is for users like you to do some research, watch some videos, and understand how these concepts work to teach you how to approach these types of physics problems and explain how they work. Of course, I will also give two optional hints per challenge, which are there if you do not have the time or determination to do the research, or you cannot figure out how to do the problem after researching. The two hints will be "necessary equations for this challenge" and "process to solve the problem". The hints are completely optional to use and it is encourages to not use them, but as stated above to learn the information for yourself. The series will have one main leaderboard. Whoever has the least combined byte count for all of the challenges gets a to be determined prize. Each challenge will range in difficulty, with an upwards trend of difficulty. I wish you all luck and I hope you learn a thing or two!

Challenge #1: Net Gravitational Force

Note 1: this challenge highly requires knowledge of the mathematical vector quantity, if you do not know what that is, I suggest reading this and this before attempting this challenge.

Note 2: this challenge considers gravity in CLASSICAL MECHANICS. Disregard general relativity for this challenge.

Lets start with the definition of a force. A force is a vector quantity, a number with both direction and magnitude. Simply put, force is mass times acceleration. Many mathematicians will know the name of this formula as Newton's Second Law. Now, that is a well known formula, but here is something less known: all forces are classified in one of four categories: weak nuclear, strong nuclear, electromagnetic, and gravitational. These four are called the four fundamental forces of the universe. We will be focusing on the gravitational force in this called.

The gravitational force is then classified as a field force. This means that the force acts on all objects in a certain radius around another object. In this case, gravity pulls down on objects from anywhere in a radius around them. However, I still have not defined where gravity comes from. Well, the simple answer is from mass. An object with more mass has a bigger gravitational pull on objects around it. In case you are wondering, the earth has a gravitational pull of -9.81 m/s^2 (an object will gain 9.8 m/s of downward velocity every second). But here is where it gets fun: because gravity comes from mass, every object with mass has a gravitational pull. This is where you come in. I want you to calculate the net gravitational pull of all the objects surrounding another object. Here is a better explanation:

You will receive co-ordinates and masses of objects in space for input. So an example input could be visualized as this: enter image description here You can easily see each force being applied on the target object by the three larger objects. Objects will smaller masses have a smaller gravitational field (as shown by the orange force arrow). Your job in this challenge is to find the net gravitational force being acted on the target object. To do this, it is a simple vector addition problem with the forces from the other objects. So, the resultant vector (net force), may look something like this: enter image description here This is all of the information that I will give you. It is now your job to find the equations, and research how this all works.

Challenge Specs

  • Input will be several lists of numbers consisting of the x co-ordinate, y co-ordinate, and mass of each object. The numbers could be either integers or decimals. You may take this in any convienent format ([[1 2 3][4 5 6][7 8 9], [(1,2,3),(4,5,6),(7,8,9)] and 1,2,3|4,5,6|7,8,9 are all acceptable. The first list of inputs will always be the target object (red in the pictures above), and the other lists will be the other objects. All inputs will have at least two objects. No objects will have the same co-ordinates.
  • Output will be two numbers, in any human-readable format, in any order. One will be the magnitude of the net force, using your desired unit system (SI, Planck units, Imperial units, etc.) and the other will be the direction in radians OR degrees of the net force. Output must be precise to the least number of significant figures in the input (Input: 2, 3.5, 6.1 -> Output: 200 (232.34 before rounding), note this is not an actual test case). Output may or may not be in scientific notation, its up to you.
  • You may assume input will not cause any error during execution, and you may assume all inputs will be valid.

META NOTE: Help me decide the precision of the output: http://strawpoll.me/6825341

Test Cases

Meta Note: WIP

Hints

These hints are for those who do not want to put in the time and effort of research, or those who could not find a solution. So, here are the two hints:

Hint 1: Equations:

You need the following equations for this challenge:

enter image description here

You will also need the standard vector equations for this challenge:

enter image description here

Hint 2: Sample Solution Process:

It would take up too much space to fit it all here, so I made this to aid you for this hint.

Leaderboard

Meta Note: blah blah blah, working leaderboard will eventually go here! This leaderboard will contain and combine scores for all of the weekly challenges. It will only be visible on this question, though.

\$\endgroup\$
  • 1
    \$\begingroup\$ The input format currently is cumbersome, requiring string parsing. I would recommend loosening it to allow input as a list of 3-tuples (or similar representations). You should also specify whether the inputs will be integers or decimals. \$\endgroup\$ – Mego Feb 5 '16 at 3:45
  • \$\begingroup\$ @Mego I took your recommendations. Does that look good? \$\endgroup\$ – GamrCorps Feb 5 '16 at 21:55
  • 1
    \$\begingroup\$ "Output must be precise to at least three decimal places" would get a rebuke from every physics teacher I ever had. Surely it should be to a number of significant figures which depends on the s.f. of the input? \$\endgroup\$ – Peter Taylor Feb 6 '16 at 13:45
  • 1
    \$\begingroup\$ @PeterTaylor Well, this is a programming challenge, and sig-fig calculations would (IMO) overly complicate the answers. (If your comment was a joke/sarcasm sorry ;) \$\endgroup\$ – J Atkin Feb 13 '16 at 19:56
4
\$\begingroup\$

Temperature in a line of rooms

You have a line of rooms that are different temperatures.

      1       2       3  
 1.2  |  3.5  |  4.0  |  3.7

The doors between adjacent rooms start out closed. When you open a door, the now-connected rooms average out their temperatures. For example, opening door 2 gives

      1       2       3   
 1.2  |  3.75 _  3.75  |  3.7

Then, opening door 1 equalizes the first three rooms to their average (1.2+3.75+3.75)/3 = 2.9

      1       2       3   
 2.9  _  2.9 _  2.9  |  3.7

Finally, if we close door 2 and then open door 3, the last two rooms will average out without affecting the other rooms.

      1       2       3   
 2.9  _  2.9 |  3.3  _  3.3

You can think of the instructions to open and close doors as a sequence of toggles that switch between open and closed, here 2, 1, 2, 3, with the doors starting closed. Given the initial temperatures of the rooms and the sequence of door toggles, output the final temperatures. Fewest bytes wins.

Input:

  • A list of initial room temperatures, which are positive reals. There will be at least two rooms.
  • (Optional) The number of rooms n.
  • A list of doors to toggle in order, which range from 1 to n-1. Optionally, these may be zero-indexed.

Output: The list of final room temperatures to some reasonable precision.

TODO: Test cases.

\$\endgroup\$
4
\$\begingroup\$

Senior Prank

We're graduating to a full site soon, and there's only one thing left to do before graduation: pull a senior prank! I think we should do a variation on the classic "fill a hallway with cups of water" gag.

Challenge

Your program will read in text and output that text, covered in upside-down cups of water. An upside-down cup of water looks like this: /~\
These cups can only be placed in whitespace in the input, and can only be placed so that all three characters of the cup are directly above a non-whitespace character (otherwise the water would spill out!). Cups cannot be stacked on top of other cups. Cups must be placed in every available opening, and it is assumed that every input is surrounded by an infinite field of whitespace.
We need to pull the prank off quickly and without anyone noticing, so fewest bytes in each language wins.

Test Cases

Input:

     ____________________________________________
    /   ___    /   ___    /   ______/   ________/
   /   /__/   /   /__/   /   /     /   /_______
  /   _______/   _______/   /     /   //__    /
 /   /      /   /      /   /_____/   /___/   /
/___/      /___/      /_________/___________/

Output:

     /~\/~\/~\/~\/~\/~\/~\/~\/~\/~\/~\/~\/~\/~\
     ____________________________________________
    /   ___    /   ___    /   ______/   ________/
   /   /__//~\/   /__//~\/   /     /   /_______
  /   _______/   _______/   //~\  /   //__    /
 //~\/      //~\/      //~\/_____//~\/___//~\/
/___/      /___/      /_________/___________/

Input:

 L
LOL  ROFL:ROFL:LOL:ROFL:ROFL
 L\\        ____I____
    ========    |  |[\
            \___O==___)
            ___I_I__/

Output:

 L   /~\/~\/~\/~\/~\/~\/~\
LOL  ROFL:ROFL:LOL:ROFL:ROFL
 L\\/~\/~\  ____I____
    ========/~\ |  |[\
            \___O==___)
            ___I_I__/
\$\endgroup\$
  • \$\begingroup\$ @TimmyD Initially that gap was two wide. I must have added the space after making the output. \$\endgroup\$ – Mike Bufardeci Mar 3 '16 at 18:19
  • \$\begingroup\$ @FryAmTheEggman Yeah, that's a typo. I should really double check when I edit on my phone. \$\endgroup\$ – Mike Bufardeci Mar 4 '16 at 4:39
4
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City Life

A cellular automation war game.

In this game, each player will control group of cites on a grid. Each city takes up one cell, and all cells with no city are "wilderness", and have no owner. The game will consist of a series of rounds, called "generations". Play continues until a player gets 1000 points, or 200 rounds, whichever happens first.

Setup

The board will start with one city controlled by each player. It will be square with sides length ceil(sqrt(25*n)) for an n player game. Cities will be placed randomly in such a way that no two cities will see each other the first round.

Phase 1: give orders

At the start of each generation, each city gets n actions where n = # of adjacent wilderness spaces + 2. So a city surrounded by wilderness gets 10 actions while a city surrounded by cites gets only 2. The Actions will be divided into these three categories:

Attack/Spread : used to Attack Cites or spread into the wilderness.

  • Takes a direction as a parameter. Will add one "Attacker" to that cell, even if the cell is a city with the same owner. (See resolve attacks)

Defend : used to protect your city.

  • Will add one to the defender count of the city performing this action. (If no defend actions are used, the city will become wilderness.)

Score : used to win.

  • Adds one to the score the cites owner.

Phase 2: resolve attacks

After all cites have put in orders, all cells are checked for takeover.

  • A wilderness cell will become a city if at least three attackers are there. The new city's owner will be determined randomly from among the attackers.(for example, if player A sent 2 attackers and player B sent 3, than A has a 2/5 chance of owning the new city)

  • A city will become wilderness if the number of attackers is equal to the number of defenders (even if both are 0).

  • A city is taken over if their are more attackers than defenders. The city's new owner is determined randomly from among the attackers, as above.

After the round is complete, all attackers and defenders are reset.


To enter the competition, you must create a bot to perform the "give orders" step. All bots will be written in JavaScript.

I/O

You will provide a character to represent your city and a function that takes as parameters:

  1. Your vision. each city can see a 5x5 square with the city as its center. It will be represented as an array of arrays of characters, " " representing wilderness, and each players character to represent their cites.

Example:

if you have a map like This ("Y" represents you)

+-----+   
|AA  B|    N
|A    |    ▲
|  YY | W< O >E
|C   C|    V
|   C |    S
+-----+    

Your sight parameter will be:

[["A","A"," "," ","B"],["A"," "," "," "," "],[" "," ","Y","Y"," "],["C"," "," "," ","C"],[" "," "," ","C"," "]]`
  1. The number of actions you can perform this turn. (which can be calculated, but I will give it to you as most bots will need it.)

You must return an array of strings, each string representing an order.

"N" - Attack North

"NE"- Attack North east

"W" - Attack West

...

"D" - Defend

"$" - Score

If you return more moves than you have actions allotted, the moves at the end of the array will be ignored. If you have less, extra moves will be set to "D".


I have made a controller that is reasonable, although I would like add to it and finalize the rules before publishing. If you have any advice or criticism, please comment below.


Example Answer:


Random Bot [?]

function(map, moveCount){
    var allMoves = ["N","S","E","W","NE","NW","SE","SW","D","$"];
    var orders = [];
    while(orders.length < moveCount){
        orders.push(allMoves[Math.floor(10*Math.random())]);
    }
    return orders;
}

This bot will just assign a random move for each action. Its cites are represented by ?.

\$\endgroup\$
  • \$\begingroup\$ I think you left off a row in the map. You should also clarify whether attackers and defenders are persistent or must be re-assigned each turn. \$\endgroup\$ – ballesta25 Jan 18 '16 at 20:15
  • \$\begingroup\$ "A city can become wilderness if the number of attackers is equal to the number of defenders" implies to me that there's a random element. Is that so? If not, I suggest replacing can with will. \$\endgroup\$ – Peter Taylor Jan 23 '16 at 19:08
  • \$\begingroup\$ @PeterTaylor Thank you, fixed. \$\endgroup\$ – MegaTom Jan 23 '16 at 20:52
  • \$\begingroup\$ I've been thinking more. It would be good to state the initial density (or, in other words, how the size of the world varies with the number of players). In the interests of fairness, probably also worth guaranteeing that the initial cities will have a certain buffer region (and in particular, that two of them won't start next to each other and a third in a nice open space); and that the topology will be toroidal (so the board wraps both horizontally and vertically). Finally, I would make it explicit that you can attack your own city, but not "support" it (i.e. lend it defenders). \$\endgroup\$ – Peter Taylor Jan 23 '16 at 22:25
4
\$\begingroup\$

Super Smash Bots

This is an idea for a KOTH based off of the Super Smash Bros video game series by Nintendo.

The basic mechanics of this KOTH would be an every-man-for-himself battle between a large number of players simultaneously. Players can execute a variety of moves, like short/medium/long-ranged attacks or blocking. As opposed to most combat-based games, players do not have a health bar, but rather die when knocked off of the stage (the arena). When an attack hits a player, it deals damage to that player but also causes knockback. The amount of knockback a player experiences is proportional to the total amount of damage he's received so far in the game.

The Arena

The game takes place on a vertical stage with gravity. There will be several fixed platforms surrounded by empty space. A player too far from a platform is killed. More ideas about the design of the stage are covered in the Stage Design section.

Character Selection?

The actual game offers players the selection of different characters, each of which have different abilities, strengths, and weaknesses. If I am to include character selection, that will be a way for players to pick the strategy they think works best and choose a bot that has those strengths. Another KOTH which had this feature was the Pokemon-themed KOTH (citation needed).

On the other hand, balancing stuff can be hard.

Actions

  • Move. Simple as that. Well, not exactly. I think this could make a very good contender for a continuous-surface (not a grid) area. On a surface, the character would walk, while airborne, the character's movement doesn't respond as quickly.
  • Jump. Is like a move, but vertical. More specifically, this gives the player an upward velocity. Double jumping might be possible.
  • Short Range. This deals damage to an immediately adjacent player.
  • Medium Range. This deals damage to players within a certain range. It would likely involve your player physically moving as well.
  • Long Range. This creates a projectile, which can deal damage to a player in the specified direction, no matter the distance.
  • Area of Effect. All players within a given range take a little damage.

Respawning?

Is is often typical that players have three lives, and thus must be killed three times to be eliminated. By respawning players, the variance of each match outcome should be reduced.

Game Ticks

If I'm doing a continuous-field, then I would want something the emulates continuous movement. In order to give fair processing time, I can't really have all of the entrant programs running at once in an asynchronous fashion. Some possible solutions are as follows.

  1. Priority Queue
    • Each action taken creates a certain time delay before the player can move again. Standing still is shortest delay, moving is second-shortest, and long-ranged attacks have the longest delay.
    • Turn order is determined by a priority queue. From the list of players, the player with the least delay is selected. Then, the game's physics are simulated for that amount of time ("virtual" time, not "true" time) and that much time is subtracted from everyone's delay. The selected player picks an action, and he is put back into the queue with that move's delay.
    • This creates a period of vulnerability after an attack which may be game-mechanically interesting.
  2. Random Time-Steps
    • Play occurs in a semi-random order, with random, non-uniform time-steps between moves. This makes it so that the player cannot predict exactly what the world will be like in the future, or guarantee the exact timings of any moves.
    • After each player has taken a single move, the order of players would be scrambled.
    • Time-step duration would be approximately constant with a random variation of +/-20% or something like that.

Animation

This would be really cool to watch, but I don't think I could possibly animate this by myself.

Stage Design

Stage design will be important, because a large chunk of the entrant programs will be tailored the stage. I don't really have any clue what I'm doing here.

One idea I've had is to crowd-source the stage design.

Maybe something with a few ledges, like this? (scaled down for ASCII-artability)

..........................
..........................
...XXXX...................
...............XXXXXXX....
..........................
....XXXXXXXXXXXXX.........
..........................

I could add some more features like so:

..........................
....XXXX..................
........XXXXX.....XXXX...
......................\...
..XXXXXXXXXXXXX....XXXXX..
................../.......
............XXXXXX........
..........................

The slants represent ramps that the player could walk up.

Player Navigation

I foresee one of the most difficult things for entrants to do is to navigate the stage. Each player would definitely receive a copy of the current stage and players as input every turn (or as arguments/parameters, more details on that below).

I may choose to offload a bunch of pathfinding stuff onto the controller so that entrants, if they so desire, can give the destination and have their character move there. Given that the stage would be constant, this should not be difficult for the controller to do. On the other hand, the continuous-field design can make pathfinding more complicated.

One thing to consider during stage design is the ease of pathfinding.

Vertical or Horizontal?

Pretty much the whole proposal has been assuming a vertical map. I could change this to horizontal to allow a larger number of people to fight in one match.

Classes or Full Programs, and Language?

Personally, I think this would be easiest to do as a Java KOTH with classes. It will run quick(-er than several other methods) and I could give entrants access to a variety of methods that give information about the stage.

Controller

Literally no work has been done yet.

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4
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Simultaneous variable updates

Suppose we have the following code snippet:

a = 2*b + c
b = a - b + c
c = a + 3*c

In most programming languages, the above three lines would execute one after the other. This means that the a on the second and third lines would refer to the updated value of a, rather than the value of a before this code block ran.

If we wanted to see how a, b, c actually changes as the result of running the above three lines, we'd have to substitute earlier variable updates into later expressions, like so:

a = 2*b + c
b = a - b + c = (2*b + c) - b + c = b + 2*c
c = a + 3*c   = (2*b + c) + 3*c   = 2*b + 4*c

Hence, the first code block is equivalent to the following in languages like Python, which support simultaneous variable updates:

a, b, c = 2*b + c, b + 2*c, 2*b + 4*c

We can rewrite the original code block and the updated code block in matrix form, like so:

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

For example, the second rows of the input/output matrices above refer to b = a - b + c and b = b + 2*c respectively. Expressing the update step in the latter, simultaneous form is useful since we can perform multiple updates very easily, using matrix exponentiation (which we will not do in this challenge).

Here's one more example, for clarity:

                                              Input matrix:
a = a - b + d                                 [  1 -1  0  1 ]
b = 2*b + c                                   [  0  2  1  0 ]
c = -2a - b + d                               [ -2 -1  0  1 ]
d = b + c                                     [  0  1  1  0 ]

                                              Output matrix:
a = a - b + d                                 [  1 -1  0  1 ]
b = 2*b + c                                   [  0  2  1  0 ]
c = -2*(a-b+d)-(2*b+c)+d = -2*a - c - d       [ -2  0 -1 -1 ]
d = (2*b+c)+(-2*a-c-d) = -2*a + 2*b - d       [ -2  2  0 -1 ]

The task

Given a square input matrix representing a sequence of variable updates one after the other, output the corresponding matrix which represents the variable updates being applied simultaneously.

You may write either a function or a full program. The exact input/output format is flexible, as long as rows are separated from other rows and individual matrix entries are distinguishable. You may assume that entries in the input/output are integers in the range [-127, 127], and that the matrix will be at least 2x2.

This is , so the goal is to reduce the number of bytes in your program as much as possible.

Test cases

TODO. Will contain:

  • Several simple 2x2 examples
  • The above examples, plus a few more 3x3 or 4x4s
  • An example with input rows all zero
  • An example with input columns all zero
  • Large examples: dense 10x10, sparse 10x10

Questions for the sandbox

  • Any potential of being a dupe?
  • Any interesting test cases?
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  • \$\begingroup\$ @PeterTaylor I guess the main difference is the update step, which would be a product with previous rows for elements below the main diagonal plus the elements on or above the main diagonal kept intact. Would you say that that's too trivial of a change? \$\endgroup\$ – Sp3000 May 22 '16 at 15:45
  • \$\begingroup\$ No, ok, the Gaussian elimination adaption does too many updates and gets out of kilter. \$\endgroup\$ – Peter Taylor May 23 '16 at 18:11
4
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Golf all the 16 logic gates with 2 inputs and 1 output!

This question asked for 16 independent functions. I would like the opposite: a single function that takes an additional parameter that specifies which of the 16 logic gates is required using an integer from 0 to 15. If you don't want to use a 0-based index of the list in the linked question then you should specify which integers map to which logic gate (but they should still be 0 to 15).

Examples:

 0,0,0  falsey
 1,0,1  falsey
 2,1,0  truthy
 3,1,1  truthy
 4,1,1  falsey
 5,0,1  truthy
 6,1,0  truthy
 7,0,0  falsey
 8,0,0  truthy
 9,0,1  falsey
10,1,0  truthy
11,1,1  truthy
12,1,1  falsey
13,0,1  truthy
14,1,0  truthy
15,0,0  truthy

This is , so the shortest answer in bytes wins.

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  • \$\begingroup\$ Here you go \$\endgroup\$ – Peter Taylor Jun 15 '16 at 10:37
  • 2
    \$\begingroup\$ @PeterTaylor That one is a kolmogorov-complexity. \$\endgroup\$ – Leaky Nun Jun 15 '16 at 10:47
  • \$\begingroup\$ @Mego How is generating 0101010101010101\n0011001100110011\n0000111100001111\n0000000011111111 the same as golfing 16 logic gates? \$\endgroup\$ – Leaky Nun Jun 21 '16 at 9:48
  • \$\begingroup\$ You may want to ban builtins. J has a two-byte solution, b. \$\endgroup\$ – Conor O'Brien Jun 25 '16 at 22:18
  • \$\begingroup\$ @CᴏɴᴏʀO'Bʀɪᴇɴ How do I go about looking up what b. means? \$\endgroup\$ – Neil Jun 25 '16 at 23:53
  • \$\begingroup\$ @Neil Voila \$\endgroup\$ – Conor O'Brien Jun 25 '16 at 23:54
  • \$\begingroup\$ @CᴏɴᴏʀO'Bʀɪᴇɴ Thanks, I can see that excluding builtins will be necessary. (Strictly speaking, I asked how to look it up, rather than for a direct link, which wouldn't help me look up any other J code should I need to for any reason.) \$\endgroup\$ – Neil Jun 26 '16 at 0:03
  • 1
    \$\begingroup\$ @Neil J is a hard language reverse. I would postix your search on google with site:jsoftware.com \$\endgroup\$ – Conor O'Brien Jun 26 '16 at 0:05
4
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Fireworks


Make me some fireworks !
And since we are super-late for the 4th make them as quick (read short) as possible !

Input
2 integer, the fuse (any value equal or bigger than 5) and the radius (1 - 2 - 3).

The fuse define the lenght of the tail, the last character of the tail is the center of the explosion. The tail must be centered with the explosion.
The radius define the explosion.

Rule
No need for exception handling, the input will be a valid one.
You may or may not padd your firework, the choice is up to you.
Input, Output and the choice beetwen full program or function is, once again, up To you and your lenguage of choice.
Standard loophole rules apply.
Hopefully no built-in (i'm looking at you mathematica) exist.

Example

fuse 5, radius 1  
     * *    
    * * *
     *|*
      |
      |
      |


fuse 10, radius 3  

   *     *
    *   *
     * *    
* * * * * * *
     *|*
    * | *
   *  |  *
      |
      |
      |
      |
      |
      |


while stretching the fuse is not a big deal, ence no limit to it, I found interesting see if it's gonna be cheaper to have hardcoded the strings for the part where the firework cross the fuse or is gonna be cheaper some fancy algorithm, ence I' ve set a limited number of alternative for the radius.

I'm really sorry if my english is bad, I usually can get my idea trought but more than often i stumble with some verbs.

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  • 1
    \$\begingroup\$ Welcome to PPCG and thanks for using the sandbox! "Number" isn't very descriptive, I assume that the fuse and radius should be positive integers (i.e. not including zero)? What happens if the radius is larger than the fuse? \$\endgroup\$ – FryAmTheEggman Jul 13 '16 at 17:00
  • \$\begingroup\$ thanks for the welcome, i'm actually lurking since a bit, but never had something new \ better golfed to add. anyway 2 time my bad :/ . First with the number (i'm used to deal alot with database where number make sense) and then i must have forgot about min and max value while rewriting my question. edited ! \$\endgroup\$ – Jackyz Jul 13 '16 at 17:29
4
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Peter Piper and the Peck of Pickled Peppers

tags:


Without an introduction, output the following tongue twister:

Peter Piper picked a peck of pickled peppers.
A peck of pickled peppers Peter Piper picked.
If Peter Piper picked a peck of pickled peppers,
Where's the peck of pickled peppers Peter Piper picked?

with or without a trailing newline.

This is , so shortest code wins.

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  • \$\begingroup\$ I did some searching for duplicates, I came up with: this and this. There were a couple others that were also rather similar (slim shady and old macdonald), but they had a source restriction or some input as well. I'm not sure if any of them are duplicates, but these seem awfully close. \$\endgroup\$ – FryAmTheEggman Jul 26 '16 at 21:05
  • 3
    \$\begingroup\$ I wish the words didn't come in the same chunks. The substrings "Peter Piper picked" and "peck of pickled peppers" are most of the text, and the rest has little structure: X a Y. A Y X. If X a Y, Where's the Y X? Is there another tongue-twister where the words are permuted more? \$\endgroup\$ – xnor Jul 27 '16 at 1:21
4
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Connect the Dots

Given a black and white image with a white background and a set of black dots, paint a set of white pixels in red such that there is a path between each pair of black pixels. A path is a set of connected pixels (8-neighbourhood connectivity). Black pixels can be used as part of the paths. The goal is trying to minimize the set of red pixels under the above conditions, and outputing a corresponding image. You don't have to find the optimal solution. A trivial and at the same time worst solution is just painting all the white pixels red.

Example: (Pixels are enlarged for visibility.)

Details

Given a pixel image (in any suitable format) return another image with the dots connected as specified above, as well as a integer indicating how many red pixel were used.

The Score is the product of (1+the number of red pixels) for each testcase.

The goal is having the lowest score.

Testcases

enter image description hereenter image description hereenter image description here

enter image description hereenter image description hereenter image description here

Meta

I am aware that it is conceptually the same challenge as this one: https://codegolf.meta.stackexchange.com/a/7342/24877 However I think it does call for a totally different appraoch, due to the topology on the one hand, and on the other hand that we are operating on a discrete grid.

** I think the sum is not an ideal scoring system, as different test cases requre a imensely different minimal number of pixels. Would a weighted sum be suitable? (My idea is finding almost optimal solutions (by hand), and then letting the participans divide their number of pixels by the "almost optimal" one, such that all entries are more or less normed.)

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  • \$\begingroup\$ This does seem different from your Steiner Tree post. This one doesn't use Euclidean distance (8-connected pixels seems equivalent to Chebyshev distance). The fact that it is discrete does seem a significant difference too. \$\endgroup\$ – trichoplax Aug 13 '16 at 12:02
  • \$\begingroup\$ I haven't tried solving this but I suspect it will be significantly easier than finding a Steiner Tree. It's worth investigating carefully what size problems (image size and number of black pixels) can be solved optimally. Then you can decide whether to make it golf/fastest-code insisting on optimal, or ensure that the test cases are definitely too large to be solved optimally, if that turns out to be practical. \$\endgroup\$ – trichoplax Aug 13 '16 at 12:06
  • \$\begingroup\$ Both challenges sound interesting. My personal bias makes me more interested in this one (the discrete version). \$\endgroup\$ – trichoplax Aug 13 '16 at 12:07
  • \$\begingroup\$ Presumably the score is derived from the number of pixels used for each test case? So a "bad" solution just won't have a good score? \$\endgroup\$ – trichoplax Aug 13 '16 at 12:12
  • \$\begingroup\$ @trichoplax I forgot to mention that, yes, they can be as bad as they want, but the are going to have a big score, which is not desirable. \$\endgroup\$ – flawr Aug 13 '16 at 12:23
  • \$\begingroup\$ Is the first test case one that requires zero red pixels? \$\endgroup\$ – trichoplax Aug 15 '16 at 18:56
  • \$\begingroup\$ No, it needs just one pixel. \$\endgroup\$ – flawr Aug 15 '16 at 19:06
  • \$\begingroup\$ I suspect this will need some bigger test cases (both bigger images and more black pixels) \$\endgroup\$ – trichoplax Aug 15 '16 at 19:40
  • \$\begingroup\$ @trichoplax Do you think the current ones are too bruteforceable, or what is the reason? \$\endgroup\$ – flawr Aug 18 '16 at 19:43
  • \$\begingroup\$ I haven't tried it, but I suspect optimal solutions will be found (yes I'd think a brute force plus intelligent starting points approach would solve most of these). I don't have a good grasp of what strategies are possible, so I'd be inclined to make the test cases too difficult rather than try to guess and find that someone comes up with an approach much better than we expected. \$\endgroup\$ – trichoplax Aug 19 '16 at 9:16
4
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Transpose a Ragged Array

Given an array of arrays of integers where the rows may not be of equal length, pad those rows with nulls, and transpose the array.

Rules

  • Use any sane input for the array.
  • Specify which null(s) you are using for this function.
  • The output should be a transposed array, printed in whatever way is sane for your language.
  • This is code golf. Aim for the shortest code possible.

Test cases

I: [[1, 2], [3], [4, 5]]          # Padding with nil here
O: [[1, 3, 4], [2, nil, 5]]

I: [[1], [2, 3], [4, 5]]
O: [[1, 2, 4], [nil, 3, 5]]

I: [[1, 4, 5], [8, 3, 2], [1, 7, 9, 6]]
O: [[1, 8, 1], [4, 3, 7], [5, 2, 9], [nil, nil, 6]]

I: [[1], [2]]
O: [[1, 2]]

I: [[1, 2]]
O: [[1], [2]]

I: [[4, 5, 6, 7], [8, 9]]                  # Padding with spaces here
O: [[4, 8], [5, 9], [6, ' '], [7, ' ']]    # as an example of a different null

As always, if the problem is unclear, please let me know. Good luck and good golfing!

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  • \$\begingroup\$ Having arrays that contain both integers and strings seems odd (and unrelated to the challenge) to me. \$\endgroup\$ – Nathan Merrill Aug 19 '16 at 6:29
  • \$\begingroup\$ @NathanMerrill In my head, I was trying to allow as many nulls as possible by restricting what data the arrays would have. You're right, though, and I have removed the reference to numbers and strings. \$\endgroup\$ – Sherlock9 Aug 19 '16 at 6:32
  • \$\begingroup\$ There's still a test case with it :) \$\endgroup\$ – Nathan Merrill Aug 19 '16 at 13:54
  • \$\begingroup\$ @NathanMerrill That's because I still want arrays that can contain any data. \$\endgroup\$ – Sherlock9 Aug 19 '16 at 13:59
  • \$\begingroup\$ @NathanMerrill Alright I rewrote the test case \$\endgroup\$ – Sherlock9 Aug 19 '16 at 17:00
  • \$\begingroup\$ Perhaps "ragged" is a better term for "uneven" \$\endgroup\$ – Conor O'Brien Sep 10 '16 at 15:54
  • \$\begingroup\$ Does the "null" have to be constant or can it depend on the input? \$\endgroup\$ – Dennis Sep 16 '16 at 6:20
  • \$\begingroup\$ @Dennis, I think the nulls should be constant for all inputs. \$\endgroup\$ – Sherlock9 Sep 16 '16 at 10:01

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