# 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 your first try can be difficult, and there is a much better chance of your challenge being well received if you post it in the Sandbox first.

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

# Background

Slices is a puzzle game. Placed on a 2d map is a set of islands; possibly just a single island. Placed on each island is a set of points. The task is to draw a number of straight lines across the islands such that no point can be reached from another without crossing one of the drawn lines, or leaving an island. In other words, you partition the points with lines such that each partition contains at most one point on each island. For each level, you are given a maximum number of lines that can be drawn.

In the game illustrated below, drawing a straight line between the first and second row of points, and then a line between the second and third row of points, would create a valid partition of points.

Your task is to write a solver for this game. The solver takes as input a set of islands, the set of points for each island plus the maximum number of lines which can be drawn. The input and output may be represented in any way, and the input may be hard-coded. Hard-coded input must show that your solution is sufficiently general. Any format will be accepted. For instance, you might implement this as a linear programme in the CPLEX format -- I haven't verified that this is possible -- or you might write some sort of brute force solution in in an imperative language. Maybe some other declarative solution works even more nicely.

Solutions will be judged by popularity. Things such as number of bytes in the solution, computational complexity (justify this if you want) and elegance could be considered.

• 1) Should the lines (separating the points) be straight lines? 2) Allowing hardcoded input is different from our I/O defaults, and may interfere with code golf scoring. 3) Complexity is often hard to measure, and elegance is not an objective winning criterion. My suggestion is pure code-golf, pure fastest-code, or maybe popularity-contest. (For fastest code, you'll need to prepare some test cases, and get ready to run all submissions on your own machine. This is how we achieve objective scoring for speed.) – Bubbler Dec 10 '20 at 8:13
• @Bubbler The lines should be straight. I would like to allow hard-coded inputs, as this makes it possible to submit a nice linear programme if that is possible. I think I will go for popularity given that there are so many interesting dimensions. I will update the question with this in mind. – justinpc Dec 10 '20 at 11:01
• @Bubbler What do you think about the task now? – justinpc Dec 10 '20 at 11:06
• Regardless of the challenge type, you need to provide some test cases for verifying submissions - at least a couple of easy levels and a couple of harder ones, maybe some that requires drawing lines at unexpected angles. – Bubbler Dec 11 '20 at 2:44
• @Bubbler Will do. – justinpc Dec 11 '20 at 6:51
• Computational geometry problems have a lot of nuance. e.g are the points in general position, or can they be collinear? Can a cut just touch a vertex, or must it actually cross an edge? Is an island just the convex hull of its points, or is it given as a polygon? Are the convex hulls (resp. polygons) of the islands guaranteed to be mutually disjoint? – user1502040 Dec 11 '20 at 22:49
• @user1502040 I will clarify these questions. I would like to make the representation of the solution as general as possible. Can you clarify what you mean by "Can a cut just touch a vertex, or must it actually cross an edge". By edge, do you mean edge of an island? Lines should not intersect with vertices. I should clarify that lines are not line segments, and that they should extend to infinity. – justinpc Dec 12 '20 at 5:59
• @justinpc Sorry, I can see how that was confusing. I meant the conceptual edge joining two points in the same island. – user1502040 Dec 13 '20 at 0:56
• @justinpc I know most of your feedback has been discouraging, but I think this would be a fun challenge. Don't be nervous about posting it! – user1502040 Jan 3 at 22:25
• @user1502040 You're right. It would be a fun challenge. I have been busy recently, and would like to put time into this when I have a spare minute. – justinpc Jan 4 at 11:27

Given a list of strings like this (input):

["some test", "{", "some", "_subst", "ring}", "some", "other text"]

Need to combine all the elements in to one for the given substring (input).

For example:

substring = "{some_substring}"

The output list of strings will be

["some text", "{some_substring}", "some other text" ]

Here are few more possible lists of strings:

input_list = ['some test', '{', 'sub', 'string}', 'som other text']
input_list2 = ['some test', '{sub', 'string}', 'som other text']
input_list3 = ['some test', '{', 'sub', 'string}', 'som other text']
input_list4 = ['some test', '{', 'sub', 'string', '}', 'som other text']
input_list5 = ['some test {', 'sub', 'string', '}', 'som other text']
input_list6 = ['some test {', 'sub', 'string', '} som other text']


substring can be any text and it does not need to be wrapped in to round braces. If substring appears multiple times in the input string than it should be replaced as well.

Note: Elements of the string that do not contain part of the input string should be the same instances. So creating a new list from joned string will not work for this problem.

Let say these strings carry additional information like style. So removing or replacing them will remove style as well

Signature of expected function to better understand problem.

def join_variable(input_list, substring):
# implementation
return output_list

• You need to specify a winning criterion (usually code-golf) – pppery Dec 22 '20 at 17:00
• Not exactly sure what do you mean. What kind of criterion? Can share some examples? – Pol Dec 22 '20 at 18:57
• We don't host challenges where the goal is solve the problem. Rather, there must be a way to "score" solutions so they compete with each other. See why. – Adám Dec 23 '20 at 0:27
• Got it. Alright then. Should I remove it? – Pol Dec 23 '20 at 13:34
• You can use code-golf as a scoring criterion, I think - shortest code wins. It's the most commonly used tag here. – user Dec 23 '20 at 18:05

# Map the alternating group A5 to the rotations of a dodecahedron

8 years ago, the Math Stack Exchange was able to prove, in 523 characters of arcane incantation, that the alternating group A5 (ie all rearrangements of five objects that can be created by swapping two objects an even number of times) is isomorphic to the group of rotations of a dodecahedron that map vertices to vertices, edges to edges, and faces to faces. In my mind, 523 characters is far too many -- we can do better!

# Challenge

To solve this challenge, take as input a member a of A5, represented as an even permutation of the first 5 integers, and output a 3x3 rotation matrix that maps a dodecahedron to itself. Your program must be an isomorphism: This means that your program must output a unique rotation matrix for each element a, and for any elements of A5 a, b, c such that a composed with b yields c, then YourProgram(a) * YourProgram(b) = YourProgram(c) (to within floating point precision, of course). Shortest program wins!

# Example Input and Output

We present as an example one valid isomorphism, but any valid isomorphism is a permitted answer.

We assosciate the colored cubes [red, green, yellow, blue, black] with the numbers [1, 2, 3, 4, 5]. Then, for each permutation a in A5, our example outputs a matrix that rotates the figure shown so as to permute the colored cubes according to a.

[1, 2, 3, 4, 5] -> [[1, 0, 0][0, 1, 0],[0, 0, 1]]

[2, 3, 1, 4, 5] -> [[0, 0, 1], [1, 0, 0], [0, 1, 0]] (a 120 degree rotation along the axis (1, 1, 1)

[3, 4, 1, 2, 5] -> [[-1, 0, 0], [0, 1, 0], [0, 0, -1]] (a 180 degree rotation along the y axis ie coming towards the viewer)

[4, 2, 3, 5, 1] -> [[-0.3090170, 0.5, 0.8090170] [0.5, 0.8090170, -0.3090170] [-0.8090170, 0.3090170, -0.5]]

(a 60 degree rotation along the vector (1 / phi, phi, 0)

Battleships Board Validation

Do you know the game Battleship? Well, I want to play with my little brother, but before we can begin we need to set up our ships on the board. This is your input.

Now, we need to check that the ships that we set up are valid. This is where you come in to help. Your task is to write a program or function which checks whether the given 2d array (your board/input) is a valid board or not.

The input will be a 2d array, where 1 represents part of a ship and 0 represents part of the ocean.

The rules:

• There must be:
• One battleship (size 4)
• Two cruisers (size 3)
• Three destroyers (size 2)
• Four submarines (size 1)
• Any additional ships are not allowed, and neither are missing ships
• Each ship must be either vertical or horizontal (aside from submarines, which are a single grid space)
• The ships cannot overlap, but may be adjacent

Solutions may not use streams.

in addition, you must solve this with a BF function (normally the class will also be called BF) which receives the 2d array(input) and uses a validate function. Afterward, you are free to manipulate the 2d array however you want and add any functions that you want.

here are some more examples that you can use to check your code: is valid:

 {1,1,1,1,0,0,0,0,0,0},
{1,1,1,1,0,0,0,0,0,0},
{1,1,1,1,0,0,0,0,0,0},
{1,1,1,1,0,0,0,0,0,0},
{1,1,1,1,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0},


wrong ships, result is false-

 {1,0,0,0,0,1,1,0,0,0},
{1,0,1,0,0,0,0,0,1,0},
{1,0,1,0,1,1,1,0,1,0},
{1,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0},
{0,0,0,0,1,1,1,0,0,0},
{0,1,0,0,0,0,0,0,1,0},
{0,0,0,1,0,0,0,0,0,0},
{0,0,0,0,0,0,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0},


missing ships, is false->

 {0,0,0,0,0,1,1,0,0,0},
{0,0,1,0,0,0,0,0,1,0},
{0,0,1,0,1,1,1,0,1,0},
{0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0},
{0,0,0,0,1,1,1,0,0,0},
{0,0,0,0,0,0,0,0,1,0},
{0,0,0,1,0,0,0,0,0,0},
{0,0,0,0,0,0,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0},


check contact, is true->

 {1,1,1,0,0,0,0,0,0,0},
{1,1,0,0,0,0,0,0,1,0},
{1,1,0,0,1,1,1,0,1,0},
{1,0,0,0,0,0,0,0,0,0},
{1,0,0,0,0,0,0,0,1,0},
{0,0,0,0,1,1,1,0,0,0},
{0,0,0,0,0,0,0,0,1,0},
{0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0},


check another one with contact->

 {1,1,1,0,0,0,0,0,0,0},
{1,1,0,0,0,0,0,0,1,0},
{1,1,0,0,1,1,1,0,1,0},
{1,0,0,0,0,0,0,0,0,0},
{1,0,0,0,0,0,0,0,1,0},
{0,0,0,0,1,1,1,0,0,0},
{0,0,0,0,0,0,0,0,1,0},
{0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,1,0,0},
{0,0,0,0,0,0,0,0,0,0},


check invalid, is false->

 {0,1,1,1,0,0,0,0,0,0},
{0,0,0,1,1,1,0,0,0,0},
{0,1,1,1,0,0,0,0,0,0},
{0,0,0,1,1,1,0,0,0,0},
{0,1,1,1,0,1,1,0,0,0},
{0,1,0,0,0,0,1,1,0,0},
{0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0},


random board which is true->

{0,1,0,0,0,0,0,0,0,0},
{1,1,1,1,0,0,0,0,0,0},
{1,0,1,1,1,0,0,0,0,0},
{1,1,0,0,0,0,0,0,0,0},
{1,1,0,0,0,0,0,0,0,0},
{0,1,0,0,0,0,0,0,1,0},
{0,0,1,0,0,0,0,0,1,0},
{0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,1,0,0,0,0,0},
{0,0,0,0,1,0,0,0,0,0},


Who's code will be the shortest byte solution that passes all these examples successfully?! Good luck. ;)

Any language is allowed, but I'd like to see an answer in Java especially.

link to the original post- Validating a Battleship board

this also my first post that I am doing. Whats a good way to decide who is the winner and in what time frame? help would be appreciated

(any sidenotes on how to improve?anything needing clarity?)

• Hi there, new person! I'm happy to see you using the sandbox first! Generally it is good to avoid language-specific challenges. Also, is there a particular reason you're making this a fastest-code challenge rather than code-golf? – Beefster Jan 7 at 23:28
• not really im just used to using java – Rocky cohn Jan 8 at 6:15
• not to sure overall what the challenge itself would be so any suggestions would be great – Rocky cohn Jan 8 at 6:17
• Hi there. Great first challenge! As others have mentioned, I would drop the Java code to allow all languages, and just give test cases (since I usually answer in Java and 05AB1E, I might still answer in Java when the challenge goes live though ;) ). As for the win-condition, I would suggest to use [code-golf], since it's the most used and popular, and probably most appropriate for this challenge as well. (Code-golf means to create a function/program as short as possible in terms of used bytes.) – Kevin Cruijssen Jan 8 at 8:48
• @KevinCruijssen would it still be good for me to leave the condition that the input is a 2d array into the BF function and must have the validate function? I think that it also makes it more challenging – Rocky cohn Jan 8 at 8:58
• @Rockycohn Input as a 2d array is indeed good. But I would remove the Java function and validate code completely and just allow all languages. You can still mention you prefer to see a Java answer especially, though. Here a pastebin what I would personally change it to as example. – Kevin Cruijssen Jan 8 at 9:04
• gotcha. did it thanks:) – Rocky cohn Jan 8 at 9:19
• That looks a lot better. :) I've upvoted and voted to re-open your challenge, since I saw you made the same edit there. – Kevin Cruijssen Jan 8 at 9:52
• would it be better to delete the main post repost it since the title has Java and I changed it to all languages? – Rocky cohn Jan 8 at 10:18
• Unless there's some rep constraint that I'm not aware of, you should be able to edit the title of the main post. There's no need to repost it. (Or just ping me so that I do the update myself if you're not allowed to for some reason.) – Arnauld Jan 8 at 10:44
• it doesn't seem to let me @Arnauld if you could remove "Java," from the post that would be great – Rocky cohn Jan 8 at 10:46
• Also, please note that you should not make assumptions about language features. You should remove things like Solutions may not use streams or you must solve this with a BF function. – Arnauld Jan 8 at 10:46
• Title updated. :-) – Arnauld Jan 8 at 10:47

# Solve a Cubic Equation

## Input

• Your program will take in the integer coefficients of the equation $$\ax^3+bx^2+cx+d=0\$$ as inputs (a, b, c, and d)
• All solutions are between 1000 and -1000.
• a is nonzero

## Output

• All real solutions of the input equation, with an accuracy of at least the thousandths place.

## Rules

• Built-in equation solver are not allowed
• Native math libraries that do not solve equations may be used

## Examples

Input:

1 2 3 4


Output:

-1.651


Another Valid Output:

-1.651 -1.651 -1.651


Input:

1 3 0 -1


Output:

-2.879 -0.653 0.532


## Worked Example (C++) doesn't work with two roots

Try it online!

#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
double a,b,c,d;
scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
for(double i=-100;i<=100;i+=0.0001)
{
double j=i+0.0001;
double y1=a*i*i*i+b*i*i+c*i+d;
double y2=a*j*j*j+b*j*j+c*j+d;
if(y1>=0&&y2<=0||y1<=0&&y2>=0)
{
double x=(i+j)/2;
printf("%.3lf ",x);
}
}
}

• I don't know why is the section with your code named "Worked Example", but it doesn't seem to solve the equation x^3 + x^2 = 0. – the default. Jan 10 at 11:36
• I'd suggesting guaranteeing that a is nonzero so that quadratic or lower cases don't have to be handled. Or, possibly just guarantee a monic polynomial with a=1. – xnor Jan 11 at 3:43

# Win a K vs KQ endgame

## Summary

The goal of this challenge is to create a program or function which will win a Chess game with a King and Queen against a lone King. The user will specify three squares, representing the locations of the computer's King, the computer's Queen, and the user's King. The computer will then a output a move which will eventually lead to checkmate.

## Input/Output

The program or function will first take as input three squares, representing the locations of the computer's King, the computer's Queen, and the user's King (not necessarily in that order). It can be assumed that the input is a legal position.

Parsing input is not the point of this challenge, so all reasonable forms of input/output are allowed, including but not limited to

• Strings with algebraic chess notation such as "Ke4" or "Qf6"

• Triples representing pieces and coordinates such as ('K', 0, 2)

After three squares are taken as input, the computer outputs a single legal move. Behaviour on invalid input is undefined.

## Requirements

This procedure must terminate using your program or function:

• User sets up a legal KQ vs K position on a physical chessboard.

• User inputs the board position. The computer outputs a legal move. If the move is a checkmate, STOP.

• User makes the computer's move on the physical board.

• User makes a legal move for the lone king on the physical board.

• User goes to step 2 and repeats.

In other words, the computer must eventually win by checkmate, through repeatedly using your program or function.

Furthermore, from any legal starting position the checkmate must occur in 50 or fewer moves by the computer.

(Of course, a physical chessboard is in no way necessary to test the code; I only mentioned it to help visualize the procedure. The chessboard could just as well be visualized in the user's head.)

## Possible test cases

The squares are given in the order: computer's Queen, computer's King, user's King

• c2, h8, a1 (must avoid stalemate)
• a1, a2, a8
• a8, a1, e5

## Rules

• The checkmate must occur in 50 or fewer moves by the computer, but it does not need to be as fast as possible.
• Chess libraries are not permitted.
• Shortest program in each language (in bytes) wins.

## Questions for sandbox

How can I word the rules better if they are not currently sufficiently clear?

• 'It can be assumed that the input is a legal position from which it is possible to force checkmate in 50 or fewer moves.' It's always possible to force checkmate in a KQ vs K endgame (when the side with the queen has the move) in 10 moves or fewer. I wonder if you should consider lowering the 50-move limit? – Dingus Jan 5 at 8:39
• I don't want to lower the limit too much because it is conceivable that some good solutions might require more than 10 moves. – 79037662 Jan 5 at 14:20
• Fair enough. I still think you should mention that mating in 10 moves is always possible, even if you allow for more. Right now the question reads as if the possibility of forcing checkmate in 50 moves is an assumption (which it is not). – Dingus Jan 5 at 22:41
• In a pedantic reading, 'The user will specify three squares' could be understood to mean that the same starting position (chosen by the user) is always used. Can you make it clearer that the program should work for any legal starting position (I assume this is what you intend)? – Dingus Jan 5 at 22:42
• @Dingus The fact that any position can be won in 10 or fewer moves isn't really relevant to the specification of the challenge, but if you think it improves the prompt I'll go ahead and add it. – 79037662 Jan 6 at 18:05
• @Dingus I explain what I mean in the next section, and I'll be surprised if anyone reads this and thinks I meant to use the same starting position every time. – 79037662 Jan 6 at 18:08
• I agree that the 10 moves doesn't affect the spec. Where I see the possible relevance is as an answerer, where I might keep that knowledge in the back of my mind as a reference point to focus my golfing strategy. To your Sandbox questions: 1. I won't offer a strong opinion either way, just an observation: banning libraries will likely lead to fewer answers. Library-based answers might be boring but voters tend to vote accordingly. 2. I agree that there's no reason to specify which side is which. – Dingus Jan 6 at 22:26
• @Dingus Thanks for your input. – 79037662 Jan 7 at 15:47
• Disallowing chess libraries sounds alright, since there's only 3 pieces to account for. More complex problems have been answered without libraries, but many tend to use libraries since it's simply golfier and simpler. I think this question would benefit from fastest-code scoring, or something similar, rather than code-golf. – Razetime Jan 8 at 11:13
• I think fastest code would be tricky to test due to the potentially numerous user inputs needed per code run, and testing how short of a mate the programs can find is not really what I had in mind for this challenge. – 79037662 Jan 8 at 21:48
• Just commenting since you requested, I think this is good. My only real disagreement was the interactivity and since you have changed it and the spec remains clear, it has my thumbs up for as much as that is worth. – Wheat Wizard Jan 12 at 2:27

# Is this a narrow Thumb instruction?

ARM Thumb was originally a 16-bit only subset of the 32-bit ARM instruction set.

However, later versions added 32-bit "wide" instructions which were more flexible, and called the original, more restrictive 16-bit instructions "narrow" instructions.

The assembler now chooses between narrow and wide instructions automatically, depending on how the instruction was written. However, this meant that the syntax had to be changed to have specific rules.

Your job is to be this assembler.

However, the programs you parse are not that interesting; they will only ever consist of add and adds.

More specifically:

Your task is to write a function or program that will take an add/adds instruction, and return a truthy value if it is a valid narrow instruction, or a falsey value if it is not.

### Syntax rules

• ARM has 16 registers, r0-r12, r13 (aka sp), r14 (aka lr), and r15 (aka pc). For ease of parsing, we are going to refer to all registers by their number, instead of using the special register names.
• Do note the names when reading the official docs, as there are a lot of special cases for sp and pc.
• In Thumb mode, these are split into "Lo registers", which are r0-r7, and "Hi registers" which are r8-r15. Many instructions can only use Lo registers.
• Many instructions use the same source register as the destination register, even if they are written with three operands.
• add and adds are distinct instructions. adds affects the condition flags, while add does not. That is the difference, if you were wondering.

The following 6 forms are valid for narrow instructions (adapted from here):

1. adds x, y, #imm: x and y must both be Lo registers, and imm is a 3-bit constant from 0-7.
2. adds x, y, z: x, y, and z must all be Lo registers.
3. add x, x, y: x and/or y must be Hi registers. Note that x is repeated twice.
• We are ignoring the fact that ARMv6 relaxed this rule to keep it interesting.
1. adds x, x, #imm: x must be a Lo register. imm is an 8-bit constant from 0-255. Again, note that x is repeated twice.
2. add r13, r13, #imm: imm is a constant multiple of 4 in the range 0-508.
3. add x, y, #imm: x must be a Lo register, and y must either be r13 or r15. imm is a constant multiple of 4 in the range 0-1020.

Everything else is either a wide instruction or not valid.

### Other notes

Standard loopholes, everything must be self-contained, and you are only allowed to treat it as text. You can't feed it to an assembler (unless you include the assembler source code in the result, but.. ｗｈｙ).

The input can either be a string argument or text from stdin.

You can assume the format will match the following format (all lowercase, separators being a single space):

{add or adds} reg, reg, {#imm or reg}


Where imm is a non-negative number in base 10 (yes, including zero).

As a regex pattern:

^adds? r([0-9]|1[0-5]), r([0-9]|1[0-5]), (#[0-9]+|r([0-9]|1[0-5]))\$


### Reference implementation

In case the rules are difficult to follow, here is a reference implementation I made in C. Yes, I deliberately overabstracted it to make you do all the work.

I resisted the urge to post the reference implementation in ARM Thumb assembly, as that would be genuinely evil. 😏

You will not need to do the same error checking I did here. You can always assume the string itself is valid. The error checks in the main function are mostly to show what CAN'T happen.

#include <stdlib.h>
#include <stdint.h>
#include <stdio.h>
#include <stdbool.h>
#include <errno.h>
#include <inttypes.h>

char opcode[5];
uint32_t op1;
uint32_t op2;
char op3_prefix;
uint32_t op3;
};

// Returns whether the opcode ID is adds.
static inline bool is_adds(const char *opcode)
{
}

// Returns whether this register ID belongs to a Lo register,
// specifically r0-r7.
static inline bool is_lo_reg(uint32_t reg_id)
{
return reg_id <= 7;
}

// Returns whether this register ID belongs to a Hi register,
// specifically r8-r15.
static inline bool is_hi_reg(uint32_t reg_id)
{
return reg_id >= 8;
}

// Returns whether the operand prefix is for an immediate
// value, specifically, '#'.
static inline bool is_imm(char c)
{
return c == '#';
}

static bool is_form_1(const struct thumb_add_insn *insn)
{
&& is_lo_reg(insn->op1)
&& is_lo_reg(insn->op2)
&& is_imm(insn->op3_prefix)
&& insn->op3 <= 7;
}

static bool is_form_2(const struct thumb_add_insn *insn)
{
&& is_lo_reg(insn->op1)
&& is_lo_reg(insn->op2)
&& !is_imm(insn->op3_prefix)
&& is_lo_reg(insn->op3);
}

static bool is_form_3(const struct thumb_add_insn *insn)
{
&& is_lo_reg(insn->op1)
&& insn->op1 == insn->op2
&& is_imm(insn->op3_prefix)
&& insn->op3 < 256;
}

static bool is_form_4(const struct thumb_add_insn *insn)
{
&& !is_imm(insn->op3_prefix)
&& (is_hi_reg(insn->op1) || is_hi_reg(insn->op3))
&& insn->op1 == insn->op2;
}

static bool is_form_5(const struct thumb_add_insn *insn)
{
&& insn->op1 == 13
&& insn->op1 == insn->op2
&& is_imm(insn->op3_prefix)
&& insn->op3 <= 508
&& insn->op3 % 4 == 0;
}

// add x, y, #imm, y == r13 or r15
static bool is_form_6(const struct thumb_add_insn *insn)
{
&& is_lo_reg(insn->op1)
&& (insn->op2 == 13 || insn->op2 == 15)
&& is_imm(insn->op3_prefix)
&& insn->op3 <= 1020
&& insn->op3 % 4 == 0;
}

// Returns 1 if it is a narrow instruction, 0 if it is not,
// and -1 on an error.
{
// Note that you do not have to do error checking for the
// competition.

if (str == NULL) {
errno = EINVAL;
return -1;
}

// Allocate a 24 byte struct on the heap for good measure
struct thumb_add_insn *insn = calloc(1, sizeof(*insn));
if (insn == NULL) {
return -1;
}

// Parse the instruction with sscanf.
if (sscanf(str, "%4s r%"SCNu32", r%"SCNu32", %c%"SCNu32,
insn->opcode,
&insn->op1,
&insn->op2,
&insn->op3_prefix,
&insn->op3) != 5
|| insn->op1 > 15
|| insn->op2 > 15
|| (insn->op3_prefix != 'r' && insn->op3_prefix != '#')
|| (insn->op3_prefix == 'r' && insn->op3 > 15)
) {
errno = EINVAL;
free(insn);
return -1;
}

int ret;
// Test against each of the forms
if (is_form_1(insn)) {
ret = 1;
} else if (is_form_2(insn)) {
ret = 1;
} else if (is_form_3(insn)) {
ret = 1;
} else if (is_form_4(insn)) {
ret = 1;
} else if (is_form_5(insn)) {
ret = 1;
} else if (is_form_6(insn)) {
ret = 1;
} else { // not a match
ret = 0;
}

free(insn);
return ret;
}


### Test cases

adds r6, r3, #0    // true, form 1
adds r0, r1, #7    // true, form 1
adds r0, r9, #1    // false, r9 is a Hi register
adds r0, r1, #9    // false, must be 0-7

adds r0, r0, r0    // true, form 2
adds r7, r1, r2    // true, form 2
adds r4, r4, r1    // true, form 2
adds r13, r14, r6  // false, r13 and r14 are Hi registers (this isn't even valid as a wide instruction)

adds r0, r0, #0    // true, form 3
adds r5, r5, #249  // true, form 3
adds r3, r3, #256  // false, must be 0-255
adds r8, r8, #72   // false, r8 is a Hi register

add r4, r4, r11    // true, form 4
add r8, r8, r5     // true, form 4
add r9, r9, r9     // true, form 4
add r14, r14, r12  // true, form 4
add r8, r9, r10    // false, Rd must be the same
add r1, r1, r0     // false, one must be a Hi register (we are ignoring the ARMv6 change)

add r13, r13, #0   // true, form 5
add r13, r13, #48  // true, form 5
add r13, r13, #17  // false, not a multiple of 4
add r13, r13, #512 // false, must be 0-508

add r0, r15, #0    // true, form 6
add r4, r13, #1000 // true, form 6
add r11, r13, #32  // false, r11 is a Hi register
add r2, r13, #4000 // false, must be 0-1020
add r3, r15, #127  // false, not a multiple of 4


Things you can safely ignore:

// String will never be empty
adds r3, r3, #-3 // adding a negative is not even a thing
add r0, r0, r99 // the only registers are r0 - r15
add r13, r13, #0x32 // it is base 10
subs r1, r1, r2 // only add and adds need to be handled
add r2, r2, lr // you don't need to handle the special names
add r0, sp, #0 // same
add #3, r1, r1 // only the last one will be an immediate
adds r3, r3, 32 // all immediates are prefixed with #
ADDS R0, R0, R1 // everything is lowercase
adds        r2,     r3 , r4 // only one space
adds r2,r3,r4 // there will always be spaces
addeq r0, r0, r1 // no IT blocks
adds.n r0, r0, r1 // no manual width specifiers
add r1, r2, r3, lsl #8 // no barrel shifting


This is , so the shortest answer in bytes per language wins.

Proposed tags: and maybe but I think that is for things you must write in assembly, not parsing assembly itself.

# Hanabi playing bot

## Overview

Hanabi is a cooperative card game with limited communication. It won the German "Spiel des Jahres" award in 2013.

The game can be played by 2-5 players, each of which has a hand of 5 cards (4 cards for 4-5 players), which they can't see themselves, but the other players can. In each turn, you can either play or discard a card, or give one other player a hint about their cards.

Common goal is to play out the cards in each color in ascending order.

As this is a cooperative game, each answer needs to cooperate with other instances of itself.

## I/O format

### Option 1: stateless, for all languages

Your program is called once for each turn (and is supposed to terminate afterwards). It will receive the current game state via standard input, and reply with an action via standard output. Invalid output means this game is counted as forfeited (lost, 0 points).

Input

Input is a line-based ASCII-format, with a prefix indicating what kind of data it is. The last line of the input is the current turn number (see below). Input will be in this order:

1. Meta-information: n:  n = number of players (2 .. 5) y:  you = own player id (1 .. n)

2. visible cards of the other players:

• c 1: cards for player 1
• ...
• c n: cards for player n

The cards for you are omitted in the list (you don't see your own cards).

Each card has a color (one of r, y, g, b, w) and a number (1 to 5), written like r1 or y5, comma-separated.

Example: c 2: w4,y1,g2,r3 means that player 2 has a white 4, a yellow 1, a green 2 and a red 3, in this order, on her hand.

3. Hints given about each player's cards, in a similar list:

• h 1: hints for player 1
• ...
• h n: hints for player n

For each card (in the same order as before), all hints are noted, in a comma-separated list. We also note negative information, i.e. when a card was present when a hint was given, but was not selected, by prefixing with !. For example 5 is a card which is known as 5 but of unknown color, w1 is a card which is known as a white 1, 3!y is a 3 which is known as not yellow, g!15 is a green card which is known as neither being a 1 nor 5.

For example, h 2: ,y!2,2!yr,r means that for player 2, the first card is completely unknown (was taken after the last hint), the second card is known as yellow and as not a 2, the third card is known as a 2 which is neither red nor yellow, and the fourth card is known as red.

4. p color: number – already (successfully) played out cards (one per line). Here we just note the highest card of each color.

• p r: 2 means for red, the 1 and two were played out (i.e. red 3 is the next one to play).
• p w: 0 means no white card was played out yet (i.e. the white 1 is the next one to play).
5. d:  – Discarded cards A list of all cards which were either intentionally discarded or unsuccessfully played out, comma-separated.

For example, d: w3,b5,y2,y2 means that white 3, blue 5 and two yellow 2s were already discarded.

6. game status information (each on one line, in this order): lh:  number of hints left, ld:  number of discards left (those two always sum to 8), lb:  number of "bad plays" left lc:  number of cards left in the deck t:  current turn number

(The input will be closed here.)

Output

The action to take. One of

• h  player-id   color or number – give a hint to another player. E.g. h 4 y will give player 4 a hint which of his cards are yellow. h 3 1 will give player 3 a hint which of her cards are a 1. This action is only possible if the number of hints left is positive. (The number of hints left will be reduced by 1, the number of discards will be increased by 1.)
• p  card# – play one of your own cards (identified by its number (1..5)). (If this card fits into the cards already played, it's added there. (If this is the last 5, the game ends immediately with full score (25)). Otherwise it is discarded and the number of bad plays is reduced by 1. If it reaches 0, the game ends (unsuccessfully).)
• d  card# – discard one of your own cards (idenfified by its number (1..5)). This is only possible if the number of discards is positive. (The number of hints left will be increased by 1, the number of discards will reduced by 1).

After playing or discarding a card, this card is removed from the hand, and a new card is drawn from the deck and is added to the list of cards of this player (on the left). (The hints are automatically updated.)

### Option 2 (stateful)

Your program receives a live transcript of everything what happens (including the actions of the other players, and the results thereof). The controller will read one line of output from the program when it's its turn.

Input

Most input lines have the same format as before.

Initial input (as for the stateless version):

1. Meta information (n: , y: )
2. other player's cards (as before)
3. hints for cards (as before)
4. played cards (initially just p r: 0, p w: 0, etc.)
5. discarded cards (initially just d: )
6. game status, ending with t: 1.

After each player's turn:

1. a line is given with that player's action: a player id:  the action as defined in the output, e.g. a 1: h 4 y means that player 1 gave a hint to player 4 about yellow cards
2. Those parts of the card situation which changed, e.g. c and d lines if a player discarded a card (or unsuccessfully played a card) and drew a new one), c and p lines if a player played out a card successfully, a h line if a player gave a hint.
3. Updated status information, e.g. lh + ld when hint was given or a card discarded, lc when a new card was drawn, lb when a card was played unsuccessfully.
4. t:  indicating the next turn number.

*(TODO: Do we need an indication that's now your turn? That can be calculated by y == t mod n, but an explicit prompt might be easier to handle.

When the game ends, the input will be closed. (Your bot should terminate then.)

Output

As in the stateless version, one line indicating the player's action.

### Option 3/4 (JVM only, stateless or statefull)

To be defined. As I will be writing the controller in a JVM language, it should be possible/easy to provide a Java API to be implemented by the bots.

## Other game rules:

I tried to give most of the details above, but here are some which might be missing/unclear:

• There are three × 1, two × 2 to 4 and one 5 in each of the five colors (50 cards in total, 10 per color).

• When playing with 2 or 3 players, each player has 5 cards, when playing with 4 or 5 players, each player has 4 cards in their hand.

• If you have three bad plays (i.e. the lb counter reaches 0), you lose immediately. This is counted as score 0.

• If you succeed to play all 25 cards (i.e. all 5s are played successfully), you win immediately, with a score of 25.

• When the drawing deck is exhausted, one more round is played (i.e. each player has one more turn), then the game ends and the final score is the number of cards successfully played out.

## Competition Rules

• While for human play, the amount of extra communication is "subject to negotiation", here I want to explore what is possible with just what the rules provide. Any communication between your bot instances (except as provided by the defined interface, i.e. via game actions) is strictly forbidden. Strategy needs to be encoded in the source code, not discussed during the game.

• There is also no communication between your individual games (i.e. no persistence).

• I will nominate an overall winner, and one for the stateless category. (The stateful ones have a bit more information, so they could emulate the stateless ones.)

• I will run contestants 1000 times with random decks of cards, once for each number of players from 2 to 5. If your bot only works with a specific number of players, state this in your answer. The competition score is the average score for all the runs. [I'll need to experiment to see how this varies, maybe I'll increase or decrease the count.]

• A bot needs to provide output in a reasonable time (to be defined). The stateless version needs to terminate after providing output, the stateful one keeps running, but should terminate after end of input (i.e. after the game ended).

• I will provide the controller on Github, feel free to test your bot with it (and compare it with other competition entries).

• The programming language needs to have an interpreter or compiler which is available free of cost for Ubuntu 20.4 (otherwise I can't run your bot to score it).

• I reserve the right to not run a bot when I suspect malicious code in it.

• This still needs to be refined, and I actually need to write (and test) the controller. I won't post it before that is done. – Paŭlo Ebermann Jan 17 at 23:22
• What tags should this get? code-competition? – Paŭlo Ebermann Jan 17 at 23:23

# Write a Length interpreter

Length is a simple stack-based esolang where instructions are encoded as line lengths The instruction set is as follows:

Line Length Name Description
9 inp Pushes the ascii value of the first byte of stdin to the stack.
10 add Adds the top two values on the stack and pushes the result onto the stack.
11 sub Subtracts the top two values on the stack and pushes the result onto the stack.
12 dup Duplicates the top value of the stack.
13 cond If the top value of the stack is 0, skip the next instruction. Then pop it.
14 gotou Sets the program counter to the value of the line under the instruction.
15 outn Pops the top of the stack, and outputs it as a number.
16 outa Pops the top of the stack, and outputs its ascii value.
20 mul Multiplies the top two values on the stack and pushes the result onto the stack.
21 div Divides the top two values on the stack and pushes the result onto the stack.
24 gotos Sets the program counter to the value at the top of the stack

In case the table doesn't work, here is the esolangs page: https://esolangs.org/wiki/Length
Test inputs are too long to put here, they can be found here
helloworld.len - Outputs Hello, World!
truth.len - A truth machine
bottles.len - Outputs the lyrics to 99 bottles of beer
This is a code golf, so shortest program wins!

• table is broken, it looks fine in preview – Nailuj29 Jan 18 at 20:49
• Fixed. But this should be reported on SE Meta. – Adám Jan 18 at 20:56

Your task is to create a program which takes the other program's source as input and output the index of the byte that should be deleted from the other program. (zero-indexed)

Each program will be radiated at the same time. The first program to fail to return a valid index after being radiated loses (whether by compiler error, runtime exception, out-of-bounds output, or some other means), or it is considered a draw if both fail at the same time.

Each program will compete against each other program. The program receives 1 point per round survived. The overall winner is the one with the most points.

Programs are limited to a length of 1024 bytes.

Alternate possibilities:

(Inspired by @Dingus) A hash of the opponent's original source code and a list of the indexes of bytes deleted so far is passed in instead of the current source code, making it a bit more of a blind guess as to what you radiate. If at any time, a program makes a guess it has already made, it loses. This turns it into a sort of "Radiation Battleship"

• Interesting challenge. How will it be tested, if different languages are allowed? – user Jan 19 at 19:12
• @user it shouldn't be terribly difficult to make a shell-based controller. Submissions would probably need to include compiler flags or shebangs separately so that the controller knows exactly how to run the program. – Beefster Jan 19 at 19:19
• That would work for "practical" languages, but I'm a bit worried about esolangs. I guess TIO can probably deal with that, though. – user Jan 19 at 19:20
• If Program A has all its bytes deleted before Program B does, does B win? I'm imagining a pathological scenario in which A is empty and outputs by exit code. – Dingus Jan 19 at 23:46
• Also, wouldn't something like this be impossible to beat? – Dingus Jan 20 at 0:12
• @Dingus I suppose that would be impossible to beat. I wonder how I could make things different with resisting arbitrary insertions... probably not viable though. – Beefster Jan 20 at 16:14
• To prevent fixed-output programs, could you perhaps tie the indexing to the unmodified code? In other words, force the output to be different every round? – Dingus Jan 20 at 22:22
• @Dingus I think that might work (implying that both the original program and guesses are passed to the program), but there's probably another similar edge case I'm missing. Kinda turns it into radiation battleship. If a program returns a guess it has already made, it instantly loses. – Beefster Jan 20 at 23:58
• Following that thought, it might be kind of interesting if some opaque id (e.g. a hash of the source code) is passed in instead of the current state of the source code. – Beefster Jan 21 at 0:02
• @Beefster what's the point of the hash? (what can submissions do with it?) – the default. Jan 21 at 4:37
• @thedefault. the hash is only used to assign a distinct id to each program and it could be used to seed a random number generator. – Beefster Jan 21 at 16:16
• @Beefster I thought submissions aren't supposed to be designed to beat specific opponents (and the only way to use a hash is to optimize to beat specific opponents) (and providing a hash of an unknown string as input just in case somebody wants to seed a RNG with it is a weird decision) – the default. Jan 22 at 16:47
• Also, if the length of the source code is not known, it's effectively impossible to make guesses that are guaranteed not to be out-of-bounds. (also, I expect 90% answers to this challenge to output something like 0,1,2,3,4,5,...) – the default. Jan 22 at 16:58

# Print a 3D shape

## Posted

• Suggestion: use characters other than • and ¶ as they are multi-byte characters which not all languages will handle easily. ASCII characters instead would be better – caird coinheringaahing Jan 20 at 21:12
• I modified them, thank you. – Sheik Yerbouti Jan 20 at 21:35
• @caridCoinheringaahing I was just looking to try some other character taking as reference this ASCII table. The two characters that I changed are numbers 166 and 167 of the table. I am confused. – Sheik Yerbouti Jan 21 at 23:45
• Most languages can handle code points between 32 and 127 inclusive. Some use custom code pages (e.g. Jelly) that won't necessarily have extended ASCII characters, but will have regular ASCII – caird coinheringaahing Jan 21 at 23:48
• Oh all right thank you for the clarification! – Sheik Yerbouti Jan 21 at 23:55
• @Davide the table you linked is not actually an ASCII table. ASCII only goes from 0 to 127; there are more code pages that extend it to 256 and the creator of that table has clearly picked one. This confusion arises because almost all code pages include ASCII as the first 128 and then extend with some symbols like ¶ on top of that. Nowadays most things use UTF-8 which encodes some characters with more than one byte and can use millions of different characters from Unicode – pxeger Jan 22 at 7:58
• @pxeger Oh thank you for this definitive explanation! – Sheik Yerbouti Jan 22 at 13:02

# Quine Countdown! quinecode-golf

Write a program that accepts a single parameter n and outputs another program that outputs another program etc until the nth call outputs the original input n again.

Scoring is a modified version of codegolf: For input 100, add together the code length of each program in the chain, excluding the final 100. This is your score. Lowest score wins!

• Actually it's not much harder than a typical quine program. Because 100 is so large, most programs will do it the general quine way. – user202729 Jan 29 at 7:18
• Besides, the score would be easier to read if the score is the average size instead of the maximum. – user202729 Jan 29 at 7:18
• "Because 100 is so large, most programs will do it the general quine way." That's the intention. It's a quine with something extra. Without the scoring rule, everyone would just build nested multi-escaped prints, which isn't as interesting. – Sinthorion Feb 9 at 11:07

# Counting set bits in a byte

Here is a bite-sized problem I ran in to when trying to implement Conway's game of life on a microcontroller, and was trying to count the amount of neighbours: How can you check how many bits are set in a byte?

## Challenge

Given one byte of input data and an integer N between 0 and 8, check if there are exactly N bits set in the byte.

## Test cases

N = 0, input = 0b00000001 -> False
N = 0, input = 0b00000000 -> True
N = 3, input = 0b01001001 -> True
N = 3, input = 0b11100000 -> True
N = 3, input = 0b00001111 -> False
N = 7, input = 0b11101111 -> True
N = 7, input = 0b11111110 -> True
N = 8, input = 0b11111111 -> False


# Sandbox question

• My first intuition to solving this problem was to shift the bits out one by one, AND with 0x01 and count them. I feel however it must be possible to do something more efficient in terms of CPU cycles used. How can I make a challenge that is about optimizing instruction count and memory usage rather than on program-size? I have seen the tag, but I don't know what scoring method best to use.

# Edit: Closing after some good comments and possible solutions

Arnauld and CristoLosoph gave some great comments and led me to conclude that my issue is maybe to hardware/language specific to fit in a nice coding challenge. CristoLosoph showed me this interesting code snippet which I think is quite efficient:

uint8_t count_bits(uint8_t x){
x = ((x & 0b10101010) >> 1) + (x & 0b01010101);
x = ((x & 0b11001100) >> 2) + (x & 0b00110011);
x = ((x & 0b11110000) >> 4) + (x & 0b00001111);
return x;
}


Two other things I learned:

• Some hardware have a builtin POPCNT instruction in their instruction set, and gcc has a __builtin_popcount() method that does exactly what I was looking for
• I found in this question that it's an interesting trade-off (as with most embedded functions probably) to just make a lookup table containing all 256 possible return values. It takes some memory but not too much.

I also learned that probably would have been a good fit for this type of challenge!

Once again thanks for the interesting comments!

• Hello RvdV. Hm, seems like your motivation does not really fit. Fastest-code and smallest code is not the same, rather opposite things. When you program microcontrollers you are very restricted in the language but code golfing is the opposite (if you do not specifically restrict languages to one single one). Of course, there is a very performant solution for bitcount in the language which you might be looking for which however is quite uninteresting for code golf. I'll post you a fast snippet in C below. – ChrisoLosoph Jan 29 at 10:20
• – Arnauld Jan 29 at 10:25
• You may consider using atomic-code-golf, but be aware that it's pretty hard to specify correctly. – Arnauld Jan 29 at 10:26
• x = ((x & 0xAAAAAAAA) >> 1) + (x & 0x55555555); x = ((x & 0xCCCCCCCC) >> 2) + (x & 0x33333333); x = ((x & 0xF0F0F0F0) >> 4) + (x & 0x0F0F0F0F); x = ((x & 0xFF00FF00) >> 8) + (x & 0x00FF00FF); x = ((x & 0xFFFF0000) >> 16) + (x & 0x0000FFFF); You only need the first 3 lines if you only need it for 8-bit integers and then you can cut the hexadecimal literals down to two digits. – ChrisoLosoph Jan 29 at 10:29
• But if you work with GCC, please use __builtin_popcount() since that is the function that the compiler people tried to optimize exactly for your purpose and for your selected architecture. – ChrisoLosoph Jan 29 at 10:31
• Thanks a lot for your comments Arnauld and ChrisoLosoph! You are right that it's hard to define, and I didn't realize even how much language but also hardware-bound my problem was, maybe it's too difficult to define indeed (and if you define it language-agnostic it becomes the related post Arnauld linked). I will update the post with you suggestions! – RvdV Jan 29 at 17:51

# Mix my colors

This challenge is inspired by the Color Alchemy Patch on NetHack, notably incorporated by UnNetHack 3.5.2.

## Objective

Given two strings indicating colors, mix them according to the rules below, then output it.

## Colors

There are 16 colors in total that are valid inputs. They are categorized to 8 chromaticities and 2 brightnesses, like below:

        Light    Dark
Hueless white    black
Red     pink     ruby
Blue    sky-blue indigo
Yellow  yellow   golden
Orange  orange   amber
Green   emerald  dark-green
Purple  puce     magenta
Brown   ochre    brown


(I'm omitting gray, for it will make this challenge just cumbersome in this regard.)

Note that all color names are in lowercase.

## Color-mixing rules

• Above all, mixing colors is idempotent.

• Mixing colors is commutative unless noted below.

• When mixing two primary colors (Red, Blue, or Yellow):

• If they have same chromaticities, the result will also be in the same chromaticity.

• Mixing Red and Blue results in Purple.

• Mixing Red and Yellow results in Orange.

• Mixing Blue and Yellow results in Green.

• When mixing two secondary colors (Orange, Green, or Purple):

• If they have same chromaticities, the result will also be in the same chromaticity.

• All other combinations result in Brown.

• For all cases covered by above, mixing two Light (resp. Dark) colors will result in corresponding Light (resp. Dark) color.

• For all cases covered by above, mixing a Light color and a Dark color shall result in either Light or Dark. This is the only rule that may break the commutativity.

• Mixing a Light color with black results in corresponding Dark.

• Mixing a Dark color with white results in corresponding Light.

• All combinations not covered by above fall in don't care situation.

## Examples

### Valid outputs

• Mixing white and white results in white.

• Mixing pink and pink results in pink.

• Mixing pink and ruby results in either pink or ruby.

• Mixing ruby and golden results in amber.

• Mixing sky-blue and ruby results in either puce or magenta.

• Mixing emerald and dark-green results in either emerald or dark-green.

• Mixing emerald and orange results in ochre.

• Mixing puce and amber results in either ochre or brown.

• Mixing white and ruby results in pink.

• Mixing ochre and black results in brown.

### Don't care situations

• Mixing pink and white falls in don't care situation. There is no rule that covers this case.

• Mixing ochre and brown falls in don't care situation. There is no analogous rule for tertiary colors.

• Mixing indigo and magenta falls in don't care situation. There is no rule for mixing a primary color and a secondary color.

• Mixing indigo and orange falls in don't care situation.

• Mixing pink and ochre falls in don't care situation.

• Mixing white and black falls in don't care situation. (In the game, it results in gray, but I'll ignore this, for sake of simplicity of this challenge.)

## Rules for code golf

• Input format is flexible. In particular, it can be two strings, or one string separating colors by whitespaces. It's implementation-defined whether to accept leading or trailing whitespaces.

• Output format is also flexible. Outputting leading or trailing whitespaces is okay.

• Invalid inputs fall in don't care situation.

• The rule looks clear enough, but you can provide a table of 256 possible outputs just to be extra sure. – user202729 Jan 31 at 1:57

# Longest Common Suffix

Given arbitrarily many (more than 4) words, your goal is to find the longest common suffix of all of them.

## Rules

1. The suffix shouldn't be longer than one third of the length of any word, and should be longer than 2 characters.
2. The words can have at most one "exception" among them, that is, it doesn't have the same suffix others have. You should ignore these "exceptions".
3. If no suffix can satisfy all rules above, do not make any output.
4. All given words are separated by spaces; they only contain lower case English alphabets.
5. You can use any way to accept input, but output should only be in STDOUT.

## Example

Given: television operation delegation repetition
Output: ion

Given: vision decision subtraction observation
Output:


Why? The suffix, ion, is longer than 1/3 the length of vision and decision. There're two exceptions.

Given: interested congratulated excited overjoyed
Output: ted


Why? overjoyed is an exception, because others have the suffix ted, but it doens't. So we ignore the word overjoyed.

Given: abcdefghijkl bcdefghijkl cdefghijkl defghijkl
Output: jkl


Why? defghijkl is the longest, but it is longer than 1/3 the length of defghijkl.

This is code golf, so the shortest code win.

It's not recommended to use built-in functions which directly returns the result.

• Most answerers prefer "any (standard) way to get output" as well, but it's up to you. Practically, it's mostly not a problem. – user202729 Feb 1 at 9:30
• Regarding the output "No common...", people don't like hard coding long error messages, it would be better received if it's changed to "any value that signifies that there isn't...". – user202729 Feb 1 at 9:31

# Compile Roman Numerals to <some language>

Your code will, given an integer i, output code code(i) in any programming language that will evaluate to (print, push, return) i by itself. code(i) ++ code(i') should evaluate to i + i' for an i' <= i and i' - i for an i' > i. You do not have to handle cases like code(1) ++ code(2) ++ code(3) where there are more than one number less than 3 before 3, but you should handle code(5) ++ code(3) ++ code(4) => 6 with more than two numbers.

Clarification: the generating code takes a number, not a roman numeral, but the generated code is expected to have the behaviour of a roman numeral when concatenated with other outputs.

## Examples

if  yourcode(10) -> 'X'
and yourcode(5)  -> 'V'
then eval('XV')  -> 15.

if  yourcode(1)  -> 'I'
and yourcode(10) -> 'X'
and yourcode(5)  -> 'V'
then eval('XIV') -> 14

if  yourcode(7)  -> 'a'
and yourcode(3)  -> 'b'
then eval('ab')  -> 10
and  eval('ba')  -> 4


## Scoring

Your answer is scored by the total bytes of the generating code.

• Sandbox: how do I explain this? what tags does this get as a challenge that requires an answer that generates code? – Wezl Feb 1 at 14:48
• Any upper limit? – Adám Feb 1 at 15:21
• @Adám no, besides any restrictions on integers built-in to the language. – Wezl Feb 1 at 15:25
• This sounds very difficult. The code snippets would need to carry a state. – Adám Feb 1 at 15:36
• Can the code take something like 7 (not representable as a single character in Roman numeral) as input? – user202729 Feb 1 at 15:43
• @user202729 yes, they would need to handle other numbers – Wezl Feb 1 at 15:44
• What should something like eval(code(7) + code(5)) evaluate to? – user202729 Feb 1 at 15:46
• @user202729 13 because 7 > 5. – Wezl Feb 1 at 15:47
• How do you define "Roman numeral" then? (when digits can take values other than the standard values) – user202729 Feb 1 at 15:48
• @user202729 OK, I've now defined how any integers should be handled when next to eachother. – Wezl Feb 1 at 16:07
• So, given a sequence $p_1,\ldots p_n$ where there's no $p_i < p_{i+1} < p_{i+2}$, the result of $\operatorname{exec}(\operatorname{generate}(p_1) + \ldots + \operatorname{generate}(p_n))$ (where $+$ denotes concatenation) should be $p_n + \sum _{i=1} ^{n-1} p_i \times (-1)^{[p_i < p_{i+1}]}$, right? – user202729 Feb 1 at 16:30
• @user202729 I guess so, although I'm not good at reading math :P. – Wezl Feb 1 at 16:35

# Posted

• Clearly defined enough (not very interesting) -- except that is it guaranteed that the value exists? – user202729 Feb 1 at 3:14
• @user202729 Oh thank you, I forgot to cover this. I think that in case it doesn't exist, we can output a something like the smallest integer in the array. I would like to not restrict the input at all. – Sheik Yerbouti Feb 1 at 12:03
• Possible test case: empty array. (by the way don't patch "edit" into the sandbox post, include it in the text itself) – user202729 Feb 2 at 2:58
• @user202729 thank you, I edited the challenge. I will add test cases, including the empty array. – Sheik Yerbouti Feb 2 at 10:26

# Golf this Thumb-2 constant!

• Not very related to the question, but have you tried using some size-optimization option of some existing compiler? (-Os) – user202729 Feb 1 at 9:32
• Otherwise, clear enough. – user202729 Feb 1 at 9:33
• I explained the pattern better, and the last pattern does cover the rest, but it isn't the shortest. – EasyasPi Feb 4 at 0:43
• Okay. (just to check, there are $4862 > 2^{12} = 4096$ distinct values represented with the Imm12 format, right?) – user202729 Feb 4 at 0:53
• Yes. There are 4096 possible encodings for the Imm12 format. – EasyasPi Feb 4 at 1:48
• Then I must understand something wrong, because (as I've said above) there are 4862 different values that can be encoded.) – user202729 Feb 4 at 1:51
• Oh yeah I was wrong. I think I am going to go back to it just being a byte rotated, even if it isn't accurate. As the true encoding is too complex to be fun to calculate. 🤷‍♂️ – EasyasPi Feb 4 at 3:02

(this is the core of the BF memory optimizer challenge.)

(At the moment I still need to make some test cases; however you can still review the rest of the challenge.)

note: This problem is reducible from Simple Max Cut, therefore it's NP-complete. (https://doi.org/10.1016/0304-3975(76)90059-1)

note: While I did get a bunch of test cases from this site, I'm not sure how can I write a reasonable algorithm to compete with...

### Proof

(actually this is not part of the sandbox challenge, but I'll post it here because it's related)

First, for convenience, assume that the problem is represented by a undirected graph, where the number of rows/columns of the matrix is equal to the number of nodes, and the corresponding weight is the sum of the value of the edges connecting the corresponding two nodes.

With that representation, the value to be minimized is the sum of the product of the edge lengths and the edge weights, with the graph nodes embedded into the point $$\ 1, 2,\ldots, |V| \$$.

From a Simple Max Cut problem of the form:

Given $$\ n \$$ variables $$\ x_1, x_2,\ldots, x_n \$$, maximize the value of $$\ \sum_{i=1}^m [a_i \ne b_i] \$$, where each of $$\ a_i, b_i \$$ represents either a variable or its negation.

It can be transformed to an instance of this problem:

First, construct $$\ 2n \$$ nodes on a graph, denoted $$\ p_1, p_2,\ldots, p_n, q_1, q_2,\ldots, q_n \$$. Let $$\ a \$$ be some positive integer. Connect those vertices:

• $$\ p_1 \$$ and $$\ q_1 \$$, with cost $$\ 2n^2 a \$$,
• the 4 pairs of vertices $$\(p_1, p_i),(p_1, q_i),(q_1, p_i),(q_1, q_i)\$$ with cost $$\ (n+1-i) a \$$, for each $$\ i=2, 3,\ldots, n \$$,
• and some other edges with small weights (the sum of their weights should be less than $$\a\$$ -- (1)) that mainly does not affect the optimal configuration.

The sum of the edge weights (except the first one) is $$\ 4 ((n-1)+(n-2)+\ldots+1)a=2n(n-1)a \$$.

The edge between $$\ p_1 \$$ and $$\ q_1 \$$ has a weight larger than the sum of all the others, it's obvious to see that in the optimal (minimum cost) configuration, these two must be adjacent.

Then, regardless where those 2 vertices are placed, the $$\ i \$$'th smallest distance-pair to those are at least the $$\ i \$$'th value in the sequence are

$$(1, 2),(1, 2),(2, 3),(2, 3),(3, 4),(3, 4),\ldots,(n-1, n),(n-1, n)$$

.

And by the rearrangement inequality, it's optimal to place the weight so that the vertices with the smaller edge-weight to $$\ p_1, q_1 \$$ are placed further from the vertices. Therefore the only optimal placement is

$$z_n, z_{n-1}, \ldots, z_2, z_1, z_1, z_2, \ldots, z_{n-1}, z_n$$

where each $$\ z_i \$$ is either $$\ p_i \$$ or $$\ q_i \$$ ($$\1 \le i \le n\$$).

Encode the condition "vertex $$\ i \$$ is on the left side of the cut" by "$$\ p_i \$$ is to the left of $$\ q_i \$$ in the permutation". (2)

Assuming that the edge weights are allowed to be fractional.

For each condition (in the Simple Max Cut problem) that "there's an edge between vertices $$\u\$$ and $$\v\$$ ($$\ u\le v \$$)", add an edge between $$\ p_u \$$ and $$\ q_v \$$ with weight $$\\frac 1 {2u-1}\$$ to this problem.

The weight of this edge can either be $$\ u-v \$$ or $$\ u-v+(2u-1)\$$ in the configuration that minimizes the total weight of the edges constructed in the previous section.

Therefore, if the vertices $$\ u \$$ and $$\ v \$$ are on different sides of the cut (according to the encoding (2)), the total weight is decreased by $$\ 1 \$$ if $$\ u \$$ and $$\ v \$$ are on different sides; and the configuration with the minimum weight is exactly the one with maximum number of edges cut.

However, in the actual problem edge weights must be an integer. We replace each edge weight $$\\frac 1 {2u-1} \$$ by $$\\lceil\frac c {2u-1}\rceil \$$, where $$\ c=2nm\$$.

Because there are $$\ m \$$ edges in total, if the sum of any $$\ k-1 \$$ increment values $$\\lceil\frac c {2u-1}\rceil (2u-1) \$$ is strictly less than the sum of any $$\ k \$$ increment values, for $$\1\le k\le m \$$, then the optimal sum is also the maximum cut.

Observe that because $$\ c=2nm \$$ and $$\ x\le 2n-1 \$$, each increment value must be between $$\ 2nm \$$ (inclusive) and $$\ 2nm+2n-1 \$$ (exclusive). Therefore the maximum sum of $$\ k-1 \$$ values is $$\ (2nm+2n-2)(k-1) \$$, which is less than the minimum sum of $$\ k \$$ values $$\ 2nmk \$$ when $$\ 1\le k\le m \$$.

The sum of all those does not exceed $$\ \lceil \frac {2nm}{1} \rceil m \$$. Therefore if $$\ a \$$ is chosen to be $$\ 2nm^2+1 \$$, then the condition (1) is satisfied.

### Minimum cost matrix permutation

(or ? Obviously the latter would be more useful in practice code golf)

Given a matrix $$\w\$$ in $$\{\mathbb N_0}^{n\times n}\$$, define the symmetric matrix $$\d\$$ in $$\{\mathbb N_0}^{n\times n}\$$ by the formula $$\d_{i,j}=\left| i-j \right|\$$, find a permutation matrix $$\P\$$ such that the sum of elements in the matrix $$\(P^{\mathsf T} \cdot w \cdot P ) \,\odot\,d\$$ (where $$\\odot\$$ denotes the Hadamard product/element-wise product) is smallest.

The result will be the mean (TODO: median? mean of the 50% maximum? mean result/naive ratio?) of the score over these test cases, for as long as you can run your program.

# (WIP) Settling the Lands of Codegolfia

The challenge controller will randomly generate a 200x200 map representing the terrain of the land

Your task is to write an AI whose goal is to have the largest population after 500 turns.

## Start

Each player begins with 1 cell claimed and a population of 100.

## Turns

On each turn, you have the opportunity to claim land cells. You can claim 1 cell per turn, plus 1 per 1000 population. Claimed cells must be orthogonally adjacent to a cell you already own. You cannot claim cells belonging to other players.

Turns happen simultaneously. In the event that two players attempt to claim the same cell, both players will lose 10 people from their population and neither player will claim the cell.

At the end of your turn, your population grows by 10% (rounded up), up to the maximum size your colony can support.

## Population Support

Without any land claimed, your colony can support up to 150 people.

Supporting larger populations requires claiming land. Each terrain type increases the amount by some amount. Combinations of terrain expand this further.

## Terrain

There are 4 terrain types:

• Plains
• Supports 20 people by itself
• Supports 5 per adjacent owned plains
• Forest
• Supports 10 people by itself
• Supports 5 per adjacent owned forest
• Supports 50 additional people if there are at least 10 owned plains cells within 10 cells (Manhattan distance)
• Mountains
• Supports 5 by itself
• Supports 5 per owned plains within 15 cells
• Supports 10 per owned forest within 5 cells
• Water
• Each plains cell can support 500 additional people if it is within 5 cells of an owned water cell.
• Each forest cell can support 200 additional people if it is within 10 cells of an owned water cell.

### Alternative ideas for terrain

• Plains support raw population. Same as above
• Forest cells increase population support for all plains within 5 cells by a factor of 5% (stacks multiplicatively)
• For every water cell and 10 plains cells, reproduction rate increases by 1%
• Each mountain cell increases the range of each owned mountain and forest within 3 cells by 1. (stacks additively)
• I was just thinking of something like this! (but different enough I can still do mine :p). Currently I don't see too many options for strategy other than comparing each orthagonal square each turn and picking the best. – Redwolf Programs Feb 11 at 19:38
• @RedwolfPrograms yeah. Maybe I should make each of the terrains more mechanically different. Maybe one increases reproduction rate, one expands synergy range, and another enhances the others. – Beefster Feb 11 at 20:24
• That sounds like a good idea, depending on how it's implemented. Maybe being close to water would let you claim extra land per turn near the body of water, sort of like exploration? – Redwolf Programs Feb 11 at 20:31

# Heads I win, tails you booze: coin tossing, pub trivia style code-golfprobability-theorygame

At my old pub trivia night, a free jug of beer or bottle of wine was awarded to the team that won Heads or Tails. This game requires players to correctly guess the outcome of successive coin tosses. The game proceeds as follows:

1. Before each toss the remaining players, who are standing, place both hands on their head (for heads) or their bottom (for tails). If all remaining players choose the same outcome (which would lead to a draw) the host refuses to toss the coin until at least one player changes his/her guess.
2. The host then tosses a coin (assumed fair) and announces the outcome. All players who guessed incorrectly are eliminated and have to sit down.
3. Play continues until there is only one person left standing to claim the prize for their team.

In perfect play, the remaining players on each team always divide as evenly as possible between heads and tails:

• If the number of remaining players on a team is $$\2k\$$ for some positive integer $$\k\$$, then exactly $$\k\$$ of them guess heads and $$\k\$$ of them guess tails.
• If the number of remaining players on a team is $$\2k+1\$$, then with equal probability
• exactly $$\k\$$ of them guess heads and $$\k+1\$$ of them guess tails, or
• exactly $$\k+1\$$ of them guess heads and $$\k\$$ of them guess tails.

An interesting consequence of players not acting independently is that a team's probability of winning is not, in general, equal to $$\\dfrac{\text{number of players on the team}}{\text{number of players on all teams}}\$$, as the examples below illustrate.

### Challenge

Assume that there are $$\n_0\$$ players on our team and $$\n_1, n_2,\ldots, n_N\$$ players on the opposing $$\N\$$ teams ($$\N\$$ and the $$\n_i\$$ are all positive integers). Given the $$\n_i\$$ as input, write the shortest program or function that calculates the probability that our team wins the booze, assuming perfect play by all teams.

Input format is flexible: you may (for example) take

• an integer $$\n_0\$$ and a list $$\[n_1, n_2, \ldots, n_N]\$$ (this format is used in the test cases),
• a flat list $$\[n_0, n_1, \ldots, n_N]\$$,
• a nested list $$\\left[n_0, [n_1, n_2, \ldots, n_N]\right]\$$.

The $$\n_i\$$ may be taken in any order. Output should be in a suitable numeric format. Floating-point approximations are acceptable provided that the underlying algorithm is theoretically exact. You may not return a list of winning probabilities for all teams (I don't care how likely it is that another team wins).

### Examples

Example 1: $$\n_0=1\$$, $$\n_1=2\$$

Suppose there are only two teams: our team has one player and the opposing team has two players. After the first toss, exactly one player from the opposing team remains in the game (because one player guesses heads and the other guesses tails). For our team:

• Half of the time, our player has guessed incorrectly and is eliminated. The other team wins.
• Half of the time, our player has guessed correctly. A second toss occurs—a head-to-head battle with the remaining player on the other team—which our player wins half of the time.

The probability that our team wins is therefore $$\\dfrac{1}{2}\cdot0 + \dfrac{1}{2}\cdot\dfrac{1}{2} = \dfrac{1}{4}\$$.

Example 2: $$\n_0=2, n_1=3\$$

In this case, both teams survive the first toss. Exactly one player remains on our team. For the other team:

• Half of the time, one player remains. A second toss decides the winner.
• Half of the time, two players remain. From this point, the remaining tosses play out exactly as in Example 1.

The probability that our team wins is therefore $$\\dfrac{1}{2}\cdot\dfrac{1}{2} + \dfrac{1}{2}\cdot\dfrac{1}{4} = \dfrac{3}{8}\$$.

### Test cases

$$\n_0\$$, $$\[n_1, n_2, \ldots, n_N]\$$ -> Output

1, [2] -> 1/4
2, [3] -> 3/8
5, [5, 5, 5] -> 1/4
5, [3, 5, 7] -> 5929/24576
8, [1, 2, 4] -> 485/768
6, [4, 5, 5, 6, 8] -> 45/256

• Even then, that "perfect strategy" strategy is ambiguous -- what if there's an odd number of players? Why isn't there a (us=H, them=H) case in the "round 2" of example 1? [please review other sandbox posts] – user202729 Feb 13 at 5:41
• @user202729 Yes to your first question. For your second question, there's no (H, H) because one of the players on the other team was eliminated in Round 1. Do the edits help? – Dingus Feb 13 at 5:46
• @user202729 Just saw your edit. (us=H, them=H) is a draw (by which I mean the round would just be repeated until they guess differently from each other). It's supposed to be covered by 'the pathological scenario in which all players repeatedly guess the same outcome never occurs in practice'. – Dingus Feb 13 at 5:47
• So if there are two ways to divide evenly (odd number), assume that each team choose one of them independently with the same probability? – user202729 Feb 13 at 5:49
• By the way, unless there's some clever formula to calculate the result, I expect that all the submissions will be too inefficient to calculate the last test case. – user202729 Feb 13 at 5:52
• @user202729 I'll add some text to clarify what happens with odd-sized teams and a small test case with all odd-sized teams of different sizes. Fair point about the inefficiency. – Dingus Feb 13 at 5:57
• Okay that part is good now, but you might want to add "if all players guessed wrong, then no players is eliminated" – user202729 Feb 13 at 11:24
• @user202729 I think that's covered by 'If all remaining players choose the same outcome (which would lead to a draw) the host will refuse to toss the coin until at least one player changes his/her guess', but it should be clearer now that I've moved that sentence earlier in the description. – Dingus Feb 13 at 11:36
• By the way, do you have a proof that the mentioned algorithm is optimal? – user202729 Feb 13 at 12:47
• "until at least one player changes..." can be ambiguous (what is their strategy to change the choice?) I think "skips the round and all player choose [...] again" is better. – user202729 Feb 13 at 12:52
• By the way, do you have a polynomial time solution? – user202729 Feb 15 at 2:07
• (this is a (probably polynomial time, or at least can be made polynomial time) implementation . tio.run/… ) – user202729 Feb 16 at 6:43
• @user202729 I appreciate all the comments, thank you. Clearly there's a bit to think about but I won't have time to look at this again in detail for a few days, probably. (It may be that this challenge has more potential in a category other than code golf, but I'm not sure yet.) – Dingus Feb 16 at 7:02

# Score an approximation code challenge code-golf

Given a list of inputs (strings) and their expected outputs (integers or floats)[1], and a black-box program[2], calculate the score using the following scoring system:

Let $$\R_n\$$ be the expected output of the $$\n^{th}\$$ input, let $$\A_n\$$ be the actual output given by the black-box program, and let $$\ j \$$ be the total number of input/output pairs. (All values are positive and non-zero)

Then the score is defined as: $$S=\left\lceil L\times\max_{1\le i \le j}\left({\max\left(\frac{A_i}{R_i},\frac{R_i}{A_i}\right)^2}\right)\right\rceil$$ where $$\L\$$ is the length of the black-box program in bytes.

Example:

If the size of the program is $$\100\$$ bytes and the worst approximation is on the input "moon", where the program outputs $$\1000\$$ instead of the expected $$\1737\$$, then the score would be:

$$S=\left\lceil 100\times{\left(\frac{1737}{1000}\right)^2}\right\rceil=302$$

This system is taken shamelessly from this challenge by @Arnauld. Here is his reference implementation.

[1]: You may take the inputs and outputs either zipped ([(in1, out1), (in2, out2), ...]), or not zipped (([in1, in2, ...], [out1, out2, ...]), with both lists of the same length), at your option.

[2]: You may take the black-box program as either:

• a black-box function, and its length in bytes, as two separate inputs
• a string of code to be evaluated, as one input (the length will not be given separately unless necessary)

# Rules

• (1) is it a problem if the program only support integers? (2) is it guaranteed that both values (actual and submission output) are strictly positive? The formula is not well defined when either value is zero. – user202729 Feb 15 at 13:17
• @user202729 all values will be positive; the idea was that you can support integers or floats, whichever is easier – pxeger Feb 15 at 13:23
• The "whichever is easier" part is currently not mentioned in the challenge. – user202729 Feb 17 at 10:36

# Stack half full or half empty? (worldview of a programming language)

Is the glass half full or half empty? It's a common rethoric question to determine a person worldview, which can be optimistic or pessimistic based on the answer given. If i were dealing with a machine i would probably ask "Is the stack half full or half empty?" Let's try and see if programming languages have a worldview too.

Write a full program or a function which prints or returns one of "The stack is half full." or "The stack is half empty." when given as input the question "Is the stack half full or half empty?" but wait.. you have to provide a stack to be observed.. which stack you provide is up to you: input one or check an already existing stack. Your solution will then output one of the two sentences mentioned above for a non empty / non full stack, your choice which one to print, just make sure your answer is as short as possible because this is . For an empty stack it must produce the "empty" output while for a full stack "full".

# Survey

I will make a chart of the pessimistic Vs optimistic languages based on the half outputs. If you find the same byte count for both choices you can provide both, in which case they will count on both sides of the chart, but you just can select one if you like.

## input

• A sentence in any reasonable method.
• The behavior is defined only for the question "Is the stack half full or half empty?" and exactly this.
• Indeed your solution can also completely ignore the input as long as it works with the specified input.
• Question mark is mandatory.

## output

• One of "The stack is half full." or "The stack is half empty.", for a non full / non empty stack.
• "full" or "empty" for a full stack or an empty stack respectively
• upper / lower case or a mix are fine.
• ending dot not mandatory.

# Rules

• Only the mentioned input allowed.
• Loopholes allowed.
• This is and the answer with the fewer bytes of source code wins.
• This would be a really cool challenge, but the tasks are too similar. I doubt anything other than print "The glass is half full" will be competitive. Maybe add an additional thing they have to do, in addition to printing the text? As an example, for half empty, take a number of inputs and discard every second one. For half full, print every other fibonacci number. – Redwolf Programs Feb 18 at 20:41
• So undefined behaviour is fine for all inputs that are not one of the two specific questions? – Adám Feb 18 at 20:41
• Your initial description only mentions one possible input, while your "input" section mentions both. – Adám Feb 18 at 20:42
• Thanks @Redwolf Programs ! To be honest I had this crazy idea but don't really knew how to make it interesting.. I ended developing it in the simplest and most obvious way.. Maybe I 'll change many things during the sandbox – AZTECCO Feb 18 at 20:48
• @Adám yes for UB, for the input I'm sorry, my fault.. Initially I was using only one question but then decided it was reasonable to handle both combinations.. Gonna edit thanks! – AZTECCO Feb 18 at 20:51
• @Redwolf Programs I changed a bit, I hope this stack thing will add some flavor without exiting the context, please let me know if you want to share some opinions – AZTECCO Feb 21 at 18:43
• @Adám some changes.. If you have any opinions please let me know – AZTECCO Feb 21 at 18:45
• @AZTECCO I still think half empty and half full should require doing different things. Currently I doubt anyone will pick half empty, because half full is shorter and there's no other difference. – Redwolf Programs Feb 22 at 0:24
• @Redwolf Programs yeah that's true. I was wondering if languages with built in string compression, string methods or regex would compete the plain print"..answer.." method which should be around 25~30 of score but.. I don't know.. I just feel like a do this or that based on input is a different challenge but it seems to be the only way – AZTECCO Feb 22 at 9:21
• @AZTECCO The concept of the answers choosing between two options is really cool and unique; the current options are just too similar. Maybe one thing that tests control flow, like a truth machine sort of thing, and one that tests input/output, like discarding every other output? The stack validating thing (where you have to print empty or full) feels kind of irrelevant to the cool part of the challenge. Having the input be the question isn't really necessary IMO; you could replace the input with something like a list of numbers (cont.) – Redwolf Programs Feb 22 at 16:19
• (cont.) and have the answers do one of two things with that list (totally up to you, although I'd recommend tasks that use different capabilities of the language, like one control flow and one math), then print The glass is half (empty|full). – Redwolf Programs Feb 22 at 16:20
• @Redwolf Programs thanks for explaining your opinion, your insight are really important and I totally agree, I was losing the scope and your words got me back on track! – AZTECCO Feb 22 at 19:27

# Bytewise look-and-say sequence code-golf

The look-and-say sequence is a sequence which begins with 1, 11, 21, 1211, 111221, 312211. To get a term of the sequence from the previous, read the previous out literally:

312211 => one three, one one, two twos,two ones => 13112221

So the next term is 13112221.

Given this javascript code (Node.js), which outputs the look-and-say sequence indefinitely, delimited by newlines:

function nextTerm(x){
return x.replace(/(.)\1*/g,z=>z.length+z[0])
}
function printUpTo(x){
var str = '1';
for(var i = 0; i < x; i++){
console.log(str);str = nextTerm(str);
}
}
printUpTo(Infinity)


It will output a string that begins 1\n11\n21\n1211\n111221 etcetera. Your job is to write a function that takes a number as input and outputs that byte of the output string.

## Rules

You may output a different character for newline instead of \n, e.g. # or something.

## Scoring

Your program should be able to handle 100 million.

The first answer to handle 1e+18 correctly will receive a 100-rep bounty. This cannot be hard-coded.

Standard loopholes apply.

Test cases:

1 => \n
5 => 2
10 => 1
20 => 3
50 => 3
100 => 1
1000 => 1
100000 => 3
10000000 => 1


This is , so shortest bytes wins.

• Note that restricted-complexity means "solutions' time complexity must be polynomial in the input size" or something similar. "In a minute" means restricted-time, but then you have to specify the specification of your machine and test all solutions (although in most cases it will be obvious and you don't need to) – user202729 Feb 20 at 9:39
• And an implementation of the naive algorithm in a low-overhead language (C++) would be able to do that as well. – user202729 Feb 20 at 9:40
• @user202729 I'm not actually sure how to write this myself - notice that there isn't a testcase for 1 billion - that's because I couldn't figure out how to do an actually efficient version. Also, by 'In a minute', I mean on TIO. I'll add that. – A username Feb 20 at 9:44
• And TIO isn't suitable for timing submissions (personally I think it's really suitable, but people don't agree.) – user202729 Feb 20 at 9:45
• @user202729 What do you mean? Is it something to do with how busy the server is or similar? – A username Feb 20 at 9:46
• codegolf.meta.stackexchange.com/questions/12707/… – user202729 Feb 20 at 9:48
• @user202729 Ok, redoing the whole thing, since I don't want to install loads of different compilers/interpreters. – A username Feb 20 at 9:50
• Ok, publishing. – A username Feb 22 at 4:30
• Wait, I did warn you that a naive implementation in C++ would be able to do it. – user202729 Feb 22 at 4:30
• Do you want to allow that? – user202729 Feb 22 at 4:31
• Ok, 1e+12 should be fine. – A username Feb 22 at 4:31
• Wait are you sure that there actually is a polynomial time solution? – user202729 Feb 22 at 4:32
• I feel like there should be one, but I don't know. Never mind. – A username Feb 22 at 4:33
• Actually there probably isn't since there's no way to calculate the nth term or length of nth term without calculating the previous. So never mind. – A username Feb 22 at 4:36
• There probably is. (I didn't come up with one.)Although people could let their programs run for a month. 1e+18 is definitely safe, but might be a little too large for some polynomial approaches. – user202729 Feb 22 at 4:36

# Self improving program

In this challenge you will write a program or function which when run will output a faster solution to this challenge.

Formally speaking: you will write a program or function, $$\T_0\$$, which takes an integer $$\x\$$ as input and outputs a program or function (in the same language), $$\T_1\$$, which is itself a valid solution to this challenge, such that there exists a function $$\f\$$ where $$\T_1\$$ is in the time complexity class $$\O\left(f\right)\$$, but $$\T_0\$$ is not. That is to say the output program must have a strictly faster asymptotic time complexity.

Note that since $$\T_1\$$ must be a valid solution to the challenge it must output a program or function $$\T_2\$$ which is even faster, and so on and so forth, creating an infinite chain each faster than the last.

Answers will be scored by their length in bytes with fewer being better.

# Precision

In keeping with the tradition of this site, and to remain inclusive: The order notation of a function will be assumed to be measured by the algorithm implemented by your answer. Thus we will pretend not to notice the behavior of your actual program for inputs out of the range of your language's numeric type.

However that being said, in order to ensure that you are actually changing the algorithm at every step, for constants initialized in your program this leniency is not given. e.g. If you use a double precision floating point 1.0000000000000001 is equal to 1.0. So if your method relies on an increasing or decreasing constant embedded in the program you should be careful that the algorithm actually changes.

Simply put for $$\T_n(x)\$$ you are forgiven for errors arising from very large $$\x\$$, but not from very large $$\n\$$.

• (which is the same thing is $T_0\in o(T_1)$.) – user202729 Feb 20 at 9:54
• How is this possible? You'd need to get faster and faster, and even if you do it exponentially, you can't change the speed by tiny fractions of a second each time. Please explain, very confused. – A username Feb 20 at 9:54
• And what prevents programs from doing for i in range(int(n**1.00001)): pass? – user202729 Feb 20 at 9:55
• Thinking about it again, since programs can simply run a loop to have any asymptotic complexity (>= complexity to parse input), it's not that hard. And you could just ask programs to print out an always-decreasing floating point value plus the new program. – user202729 Feb 20 at 9:57
• @user202729 But the always-decreasing float is the problem here, because eventually it will get too small to be distinguished by the language from 0. – A username Feb 20 at 10:01
• @Ausername An easier way to have it constantly decreasing is to have O(n^2) -> O(n log(n)) -> O(n log(log(n))) -> O(n log(log(log(n)))) etc. This approaches O(n) from above and remains distinguishable for really large inputs, without requiring anything finicky about precision. – Wheat Wizard Feb 20 at 10:02
• It means that programs doing things like (Python) n=1000000000000000000000000000000; for i in range(int(input())**(1+1/n): pass and (1+1/n) evaluates to 1 (because of IEEE754, even though the value of n is stored exactly) will fail? – user202729 Feb 20 at 11:45
• @user202729 Yes. Precision rules are always pretty non-concrete and non-objective. I think it is best to not try to push at the boundaries. – Wheat Wizard Feb 20 at 12:49

Posted at Convert Gemtext to HTML but moved here to discuss

Gemtext is a very simple markup format used by the alternative web protocol Gemini. Write a Gemtext to HTML converter.

From the Wiki:

A line of text is a paragraph, to be wrapped by the client. It is is independent from the lines coming before or after it.

A list item starts with an asterisk and a space. Again, the rest of the line is the line item, to be wrapped by the client.

A heading starts with one, two, or three number signs and a space. The rest of the line is the heading.

A link is never an inline link like it is for HTML: it’s simply a line starting with an equal-sign and a greater-than sign: “⇒”, a space, an URI, and some text. It could be formatted like a list item, or like a paragraph. Relative URIs are explicitly allowed.

Example:

# This is a heading
This is the first paragraph.
* a list item
* another list item
This is the second paragraph.
=> http://example.org/ Absolute URI
=> //example.org/ No scheme URI
=> /robots.txt Just a path URI


This should produce this HTML tree (just the equivalent tree, not the exact formatting):

<h1>This is a heading</h1>
<p>This is the first paragraph.</p>
<ul>
<li>a list item</li>
<li>another list item</li>
</ul>
<p>This is the second paragraph.</p>
<a href="http://example.org/">Absolute URI</a>
<a href="//example.org/">No scheme URI</a>
<a href="/robots.txt">Just a path URI</a>


HTML in the text must be escaped, e.g paragraph <p> is cool becomes <p>paragraph &ltp&gt is cool</p>.

This is so the shortest code in bytes wins

• Can you clarify "whitespace ... doesn't matter"? – Wezl Mar 1 at 19:30
• @Wezl edited question – R Harrington Mar 1 at 19:36
• What parts of HTML do we need to handle? Can we convert every character (even unreserved ones like A) to its entity number? Is the input guaranteed to be ASCII only or do we need to handle other characters? I don't know Gemtext well enough to ask about it, but are there escape characters or something that would allow mapping, say, to HTML: <p># A paragraph</p>? – FryAmTheEggman Mar 1 at 19:53

# Dedekind, cut!

## Objective

Given a Dedekind cut and a nonnegative integer $$\n\$$, print the real number represented by the Dedekind cut up to $$\n\$$ precisions, rounded.

## Dedekind cut

A Dedekind cut representing a real number $$\x\$$ is a boolean-valued function on $$\\mathbb{Q}\$$ that gives a falsy value for rational numbers under $$\x\$$, and a truthy value otherwise.

Note that Dedekind cuts can represent arbitrary real numbers. Irrational, nonconstructible, transcendental, or even uncomputable numbers.

## Rules

• Direction of rounding is implementation-defined.

• For the inputted Dedekind cut, it shall accept arbitrary rational numbers. That is, the numerator and the denominator must be arbitrary-length integers. (Though the denominator is nonzero, of course) Accepting the rational number as a pair of integers is acceptable. In this case, the denominator is expected to be positive, and the fraction is expected to be irreducible.

• Invalid inputs fall in don't care situation. This also applies to the rational numbers the Dedekind cut accepts.

• Note that negative numbers also must be handled. For nonnegative numbers, however, it is implementation-defined whether to output a leading plus sign.

• Outputting leading or trailing whitespaces is permitted.

• All digits after the decimal point shall be considered significant in this regard. The amount of significant digits after the decimal point must be exactly $$\n\$$. Trailing zeros cannot be discarded.

• If $$\n = 0\$$, the decimal point may be omitted.

• If the decimal representation has no significant digits before the decimal point, the leading zero may be omitted.

• The omission from the previous two rules cannot be present simultaneously.

## Examples

• For $$\x = 0\$$ and $$\n = 0\$$, output one of 0, 0., +0, and +0..

• For $$\x = 0\$$ and $$\n = 2\$$, output one of .00, 0.00, +.00, and +0.00.

• For $$\x = ½\$$ and $$\n = 0\$$, output one of 0, +0, 1, +1, 0., +0., 1., and +1..

• For $$\x = ½\$$ and $$\n = 3\$$, output one of .050, 0.050, +.050, or +0.050.

• For $$\x = \sqrt{2}\$$ and $$\n = 3\$$, output either 1.414 or +1.414.

• For $$\x = -\pi\$$ and $$\n = 3\$$, output -3.142. Note the rounding.

## Ungolfed solution

An implementation with exponential time complexity.

import Text.Printf

showDedekind :: (Rational -> Bool) -> Int -> String
showDedekind x n = go (0.5 * 10 ^^ negate n)
where
go p = let
epsilon = 10 ^^ negate n
in if x p
then if x (p - epsilon)
then go (p - 2 * epsilon)
else let
str = printf "%0*d" n (floor (p / epsilon) :: Integer)
(str1, str2) = splitAt (length str - n) str
in str1 ++ '.' : str2
else go (p + epsilon)

• So to be precise, the input value x is actually given as a black-box function, right? For output format, I recommend to keep it simple. You can even omit it entirely so that we can assume a convenient format for each language in use (under default I/O rules). Should the solutions theoretically support arbitrarily large x (with both signs)? – Bubbler Mar 2 at 9:08

# Let's rise higher!

(I am posting this challenge in sandbox because this kind of exceptional challenges are often closed, and then downvoted to oblivion, so I want to get your feedback and with kind help you fix the specifications in this challenge, and make it interesting, Thanks!)

# Challenge

The main objective of the challenge is pretty simple, this is an answer chaining contest, where you have to serially print numbers from 1. That means User 1's answer will print 1, then User 2's answer will print 2 and so on. But there are some rules to make the contest tougher.

# Rules

• You can't use the characters of the source code used in the previous answer.

• Each answer cannot use more than 12 distinct bytes.

• You cannot post 2 answers in a row, let me explain, suppose you have written answer no. 5, now you cannot write answer no. 6, you have to wait someone to post a valid answer no. 6 (That prints 6 without using characters in answer no. 5), and after that you can write a valid answer no. 7 following the rules.

• Program cannot take input or access internet.

• Standard loopholes apply.

# Scoring criterion

This is not challenge, so there is a custom objective scoring criteria. Try to make your score higher.

• Each answer's initial score is 1. (Only chain beginning answer has score 0).

• For each distinct byte you use, your score is increased by the number of distinct bytes. That means if your number of distinct bytes is 6, then your score increases by 1+2+3+4+5+6, and if your number of distinct bytes is 7, then your score increases by 1+2+3+4+5+6+7 and so on.

• You have to calculate distinct byte difference with the previous answer, and add/subtract score using the 2nd rule. So if byte difference is 3 then you will get a +1+2+3 score, or if -3 (Previous answer has 3 more distinct bytes than yours), then -1-2-3.

• For first 40 bytes you will get a 0.15 score, but from then you have to minus 0.5 score. An example if your score has total 45 bytes then your score bonus will be (40*0.15)-(5*0.5)=6-2.5=3.5, so you have to add +3.5 to your score. (The score criteria is to encourage medium length smart answers.)

A total example:

Suppose,

• Your previous answer has 4 distinct bytes, and a total of 20 bytes
• Your answer has 7 distinct bytes, and a total of 56 bytes

Distinct byte difference: (7-4)=3
Total score: 1+(1+2+3+4+5+6+7)+(1+2+3)+(40*0.15)-(16*0.5)=33


So you get a score of 33 for this condition.

# Answer Number. Language Name, x Bytes, y Distinct bytes, Score: Z

source code

(TIO/Any other interpreter link) -> Not required, but preferred


example

1. Powershell, x Bytes, y Distinct Bytes, Score: Z

......


# Winning criterion

Winner will be determined after 30 days, winner is determined by total score. But users are welcome to continue the chain after that!

Suppose you have three answers with score 33,16,78 then your total score will be 33+16+78=127.

And there is two special prizes from me after 30 days:

• +150 bounty for the winner.
• +50 bounty for the single answer with highest score! (Such as man of the match in Cricket)

# 1. PowerShell, 1 byte, 1 distinct byte, Score: 0

1


Try it online!

---

# Meta

• Are all rules clear?

• Please improve the scoring criterion and rules. (Especially please improve the second rule, I am not very sure about that.)

• Please make me a javascript snippet for leaderboard, to sort users by their total score.

• There are a bunch of small things that could be clarified (particularly the scoring), but I think there is a larger problem you should try to fix first. Answer-chaining challenges rely on the task becoming more difficult as more answers are posted, but since you are only restricted by the previous response getting an optimal score will probably not be difficult until the number of answers is very large (much larger than 100). If you fix this, then the problem becomes that if finding a good answer is difficult, posting for more score becomes a race, which usually isn't fun. – FryAmTheEggman Mar 3 at 18:32
• @FryAmTheEggman if the number of answers are large, then you would sort the question by active. What are the other problems please say. – Wasif Mar 4 at 6:58
• The number of answers being large isn't the problem. Please re-read my comment - it is talking about how the challenge isn't a good answer-chaining challenge because the approach for each link in the chain is too similar. Try writing what you think the next solution would be, and the one after that, and I think you will start to see the problem. – FryAmTheEggman Mar 4 at 14:42

# Finish what John McCarthy started (WIP) code-golfinterpreterlisp

This does not have anything to do with Joseph McCarthy or communism. Some background from Wikipedia (you can skip ahead if you like):

John McCarthy published the first paper on Lisp in 1960 while a research fellow at the Massachusetts Institute of Technology. In it he described a language of symbolic expressions (S-expressions) that could represent complex structures as lists. Then he defined a set of primitive operations on the S-expressions, and a language of meta-expressions (M-expressions) that could be used to define more complex operations. Finally, he showed how the meta-language itself could be represented with S-expressions, resulting in a system that was potentially self-hosting.[3] The draft version of this paper is known as "AI Memo 8".[4]

Example M-expressions

McCarthy had planned to develop an automatic Lisp compiler (LISP 2) using M-expressions as the language syntax and S-expressions to describe the compiler's internal processes. Stephen B. Russell read the paper and suggested to him that S-expressions were a more convenient syntax. Although McCarthy disapproved of the idea, Russell and colleague Daniel J. Edwards hand-coded an interpreter program that could execute S-expressions.2 This program was adopted by McCarthy's research group, establishing S-expressions as the dominant form of Lisp.

Interpret a subset of M-expressions, where 9 primitive functions have already been defined.

## Syntax

Atoms: An atom is a series of any characters excluding [, ], ;. While an atom name can include whitespace, leading and trailing whitespace is ignored. Some examples include foo, lambda (also a function name), and nil (also a list). Atoms act as both variables and data (they're like strings).

Booleans: The atom t is truthy, whereas the atom nil is falsy.

quote: A series of expressions between square brackets, delimited by semicolons. [a; [b; c]; de] is equivalent to the S-expression (QUOTE (A (B C) DE)). It is somewhat like this JS list: ["a", ["b", "c"], "de"].

cond: A series of if-then pairs between square brackets, delimited by semicolons, with an arrow -> (you can use any other character(s)) separating the if-then pairs. e.g. [atom[x] -> x; t -> car[x]] yields x if it's an atom, or the first element of x if it's a list. It is guaranteed that at least one of the cases will be true.

Function application: Functions can be called using functionname[arg1; arg2; ...; argN].

• car - Return the first element of its argument.
• cdr - Return its argument without the first element.
• cons - Prepend its first argument to its second argument.
• eq - Check if its two arguments are equal.
• atom - Check if its argument is an atom.
• lambda - Define an anonymous function using the syntax lambda[[param1; param2; ...; paramN]; bodythatusesparams].
• label - Store a function using a name. For example, label[drop2; lambda[[xs]; cdr[cdr[xs]]]] defines a function drop2 that drops the first two elements of its first argument.

# Rules

• Functions may use lexical or dynamic scope, or some crazy mixture of both.

Much of this question is copied from this challenge and the Wikipedia article on M-expressions.

# Questions for Meta

• For cond, should I use the [condition -> res; condition2 -> res2; ...] syntax, or should I keep it like the linked challenge?
• Should answers support higher-order functions?
• Does anyone have any good examples using Mexprs?
• I did a double-take at the title, which sounds like the challenge is to root out communists still hiding in the US state department to finish what Joseph McCarthy started... – xnor Jan 20 at 10:23
• @xnor Uh...that certainly was not what I intended. I could probably say "Finish what John McCarthy started" or just "Interpret M-expressions", although the latter sounds more boring. There's also a Eugene McCarthy, apparently, so I guess just McCarthy is too ambiguous. – user Jan 20 at 13:59