# What is the Sandbox?

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

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

## Get the Sandbox Viewer to view the sandbox more easily

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

### How fair are my dice?

(Inspired by a dream I recently had regarding Mario Party. This is not fully fleshed-out yet.)

Given a pair of dice, e.g., [[1,1,2,2,3,3], [1,2,3,4,5,6]], output how "closely" they resemble a pair of standard 1-6 dice (i.e., [[1,2,3,4,5,6], [1,2,3,4,5,6]]) in terms of distribution of numbers when rolled.

• Is the input always a pair of six-sided dice or can they have a variety of face counts? Sorted? How would [[2,3,4,5,6,7],[2,3,4,5,6,7]] compare considering it'd have the same distribution curve? – Veskah Dec 2 '19 at 16:20
• You need to flesh out details on how this should be measured. Do I go by simulation? Do I base the distribution on the total of each pair of rolls? What is "closely"? Sum of squared error? – Beefster Dec 6 '19 at 19:56

# A game of putting lines through dots.

This challenge is to produce a script that will accept a string as input, and generate a score for it using the rules defined below. Your score is the return value of your program given its own source as input.

The first scoring system

The first unique character is worth -1
The second unique character is worth -2
The third unique character is worth -3
etc...

The second scoring system

Each repetition of a given character incurs a further escalating penalty to your score.

The first usage of a given character incurs no additional penalty
The second usage of a given character incurs an additional penalty of -1
The third usage of a given character incurs an additional penalty of -2
The fourth usage of a given character incurs an additional penalty of -3
etc...

e.g.

• The code abc would have a score of -6
• The code aabc would have a score of -7
• The code aaaabc would have a score of -12

Rules

• Your score starts at 0
• The highest score for a given language wins.

Test

The below Stack snippet will parse a string according to the rules defined above and generate the score.

function testScore() {
var score = 0,
counter = 0,
input = document.getElementById('code').value,
x,
y,
instances = {},
keys;

input = input.split('');

for(x = 0; x < input.length; x++) {
if(typeof instances[input[x]] == 'undefined') {
instances[input[x]] = 0;
score -= ++counter;
}

instances[input[x]]++;
}

keys = Object.keys(instances);
for(x = 0; x < keys.length; x++) {
for(y = 0; y < instances[keys[x]]; y++) {
score -= y;
}
}

return false;
}
<form onsubmit="return testScore(this);">
<div>
<textarea id="code" style="width: 100%; height: 150px;"></textarea>
</div>
<div>
<button type="submit">Generate score</button>
</div>
</form>

• I'd appreciate some pointers on what tags to use for this one - it's similar to code golf but isn't quite code golf. – Scoots Dec 11 '19 at 11:48
• Similar but a slightly different scoring system – Veskah Dec 11 '19 at 13:02
• This is two challenges: First, reading your own source code. Second, creating the score for an input string. In addition, the condition of "Parse yourself" is a non-observable requirement. In this case, it's definetly bad because it forces the two separate challenges approach onto people. I understand why you want to exclude print -45, but right now, the whole quine-except-not-really thing doesn't seem helpful. – AlienAtSystem Dec 13 '19 at 14:09
• @AlienAtSystem Respectfully I disagree with your assertion that reading the source code is a challenge; I'm very explicit that any means of getting your source code in is acceptable, even copy/pasting it as a command line argument. The parsing of your own code - I thought it might encourage people to golf in a way they may not be used to due to the escalating penalties involved in the scoring system. – Scoots Dec 13 '19 at 15:36
• If you don't want the source code reading to be part of the challenge, then don't write things like it is. Really just say Takes a string as input and outputs the uniqueness score and Your score is the return value of your program when given its own source as input. – AlienAtSystem Dec 14 '19 at 9:13
• @AlienAtSystem I take your point, your phrasing is definitley better than what I originally had. – Scoots Dec 14 '19 at 11:32

# Nash equilibrium of 2-player 3-choice zero-sum game code-golfgame-theory

## Background

In a 2-player single finite game, a Nash equilibrium (NE) is a pair of strategies chosen by both players where neither has an incentive to deviate from their own strategy. For example, here is a payoff matrix of two players $$\ A,B \$$ where each player has two possible pure strategies $$\ X,Y \$$:

$$\matrix{ A \backslash B & X & Y \\ X & 3 \backslash 3 & 0 \backslash 0 \\ Y & 0 \backslash 0 & 2 \backslash 2 }$$

In this game, if $$\ A \$$ chooses $$\ X \$$, $$\ B \$$ should also choose $$\ X \$$ because it will give $$\ B \$$ the best outcome. Likewise, if $$\ B \$$ chooses $$\ X \$$, $$\ A \$$ should also choose $$\ X \$$. Therefore, $$\ (X,X) \$$ is a NE. We can observe $$\ (Y,Y) \$$ is also a NE; although it gives less outcome than $$\ (X,X) \$$ for both players, there is no reason to select $$\ X \$$ when the opponent selects $$\ Y \$$.

But a strategy can also be a mixed strategy, i.e. choosing one of the pure strategies with some chance. Let's take a mixed strategy $$\ S = \frac25X+\frac35Y \$$. If $$\ A \$$ takes $$\ S \$$, $$\ B \$$'s expected outcome is always $$\ \frac65 \$$, regardless of $$\ B \$$'s strategy (either pure or mixed). In this case, $$\ B \$$ has zero incentive to deviate from whatever strategy $$\ B \$$ is already taking. The same can be said for $$\ B \$$ taking $$\ S \$$, so $$\ (S, S) \$$ is also a NE.

Nash's existence theorem states that every game has at least one Nash equilibrium, either pure or mixed. Note that a game may have many (possibly infinitely many) NEs.

## Challenge

In this challenge, we consider 2-player 3-choice zero-sum games, where each player has three possible pure strategies, and the sum of the two players' outcomes is always zero. The payoff matrix of such a game might look like this:

$$\matrix{ A \backslash B & R & P & S \\ R & 0 \backslash 0 & -1 \backslash 1 & 1 \backslash -1 \\ P & 1 \backslash -1 & 0 \backslash 0 & -1 \backslash 1 \\ S & -1 \backslash 1 & 1 \backslash -1 & 0 \backslash 0 }$$

Since it is a zero-sum game, we can omit the payoffs of $$\ B \$$:

$$\matrix{ A & R & P & S \\ R & 0 & -1 & 1 \\ P & 1 & 0 & -1 \\ S & -1 & 1 & 0 }$$

This game is equivalent to Rock-Paper-Scissors, and the only NE is, as we all know, mixed $$\ (\frac13R + \frac13P + \frac13S,\frac13R + \frac13P + \frac13S) \$$.

The challenge is to find at least one of the Nash equilibria, given the payoff matrix for player $$\ A \$$. Remember that a Nash equilibrium is a pair of $$\ (\text{strategy of } A,\text{strategy of } B) \$$.

For an outline of finding a mixed-strategy NE, refer to this Math.SE question. Also related: why some games don't have mixed-strategy NE.

## Input and output

The input is a 3-by-3 matrix of integers which represent the payoff matrix for player $$\ A \$$. You can assume the magnitude of each integer won't exceed 100, and you can take the numbers in any order of your choice.

The output is a pair of strategies in the form of $$\ (a_XX+a_YY+a_ZZ, b_XX+b_YY+b_ZZ) \$$ where $$\ a_i \$$ and $$\ b_i \$$ are the chances of selecting one of the strategies. You don't need to format the output; outputting the 6 numbers $$\ a_X,a_Y,a_Z,b_X,b_Y,b_Z \$$ is fine. If you want to output a pure strategy, you can do something like $$\ (1, 0, 0) \$$.

## Scoring and winning criterion

Standard rules apply. The shortest code in bytes wins.

## Test cases

Matrix: (classic RPS)
0  -1   1
1   0  -1
-1   1   0
NE: A's strategy = (1/3, 1/3, 1/3), B's strategy = (1/3, 1/3, 1/3)
--------------------------------------
Matrix: (scored RPS)
0  -5   1
5   0  -2
-1   2   0
NE: A's strategy = (2/8, 1/8, 5/8), B's strategy = (2/8, 1/8, 5/8)
--------------------------------------
Matrix: (dumb game, pure strategy NE example)
0   1   2
-1   0   0
-2   0   0
NE: A's strategy = (1, 0, 0), B's strategy = (1, 0, 0)
--------------------------------------
Matrix: (asymmetric game example, infinitely many NEs)
2  -1   3
0   3  -1
-2  -3   0
NE: A's strategy = (1/2, 1/2, 0),
B's strategy = any triple (x, y, z) that satisfies
x + 2z = 2y, x + y + z = 1, 0 <= x,y,z <= 1
--------------------------------------
Matrix: (asymmetric game example, one mixed NE)
2  -1   4
0   3  -1
-2  -3   0
NE: A's strategy = (1/2, 1/2, 0), B's strategy = (2/3, 1/3, 0)

## Meta

This is a repost of the underspecified challenge.

Remove all duplicates from one list of integers. A list is simply a sequence of connected values that allows the same values to be stored at different positions in this sequence.

If an item was found to be the same value as another item in this list, keep the first occurence of the item and remove the second (third and so on...) occurence of the item.

## Input/Output

You may write a function or full program that takes a list as an input and returns a remove-duplicated list as an output.

## Rules:

• This is a challenge, so shortest answer wins!

# Join two file paths

A file path consists of letters a-z separated by slashes /. Your task is to join two paths as follows:

• If the second path doesn't start with /, concatenate them with exactly one / in between.
• If the second path starts with /, just output it.

For example,

/User/Desktop   temp/file   ->  /User/Desktop/temp/file
/User/Desktop/  temp/file   ->  /User/Desktop/temp/file
/User/Desktop   /temp/file  ->  /temp/file
/User/Desktop/  /temp/file  ->  /temp/file
relative/path   more/path/  ->  relative/path/more/path

You may assume the inputs don't contain two consecutive slashes, but they may start or end with a slash. Each input with be nonempty and not just a slash. You may use backslash \ in places of slash /.

Sandbox: Is this too boring? Too likely a built-in?

# String Blockify™ a Hexagon

an obvious rip-off of Hexagonify™ a String Block

### What is Hexagonification?

Hexagonification is a transformation that creates a hexagon with 3 copies of a rectangle block™, each skewed to the right by 30 degrees and then rotated by 0, 120 and 240 degrees respectively, as shown in the following image. A triangle hole may appear in the middle, but that isn't a great deal.

### Challenge

Write a program or function that receives a hexagonified block of string™ as an input and outputs the original string block™. The input and output formats are flexible. You may receive a single string, a list of lines or a 2D array as input, and output a single string or a list of lines. You may have leading/trailing spaces on each line and leading/trailing newlines, provided that the block is properly formed. See Sample IO for how the blocks™ should be positioned.

### Sample IO

Input:

3 x 3
e s q u
u r a r e
3 q a a s 3
x s r q x
3 e u 3

Output:

3x3
squ
are

Input:

l o n g e r . . . . .
. r e c t a n g u l a r
r . b l o c k . . . . . .
. a .                 b r l
. l .               l e o
. u .             o c n
. g .           c t g
. n k         k a e
r a c       . n r
e t o     . g .
g c l   . u .
n e b . l .
o r . a .
l . r .

Output:

longer.....
rectangular
block......

Input:

v b
8 e l
x l r o
2 a   t c
k c     i k
c i       c 2
o t         a x
l r           l 8
b e l a c i t r e v
v 8 x 2 k c o l b

Output:

vb
el
ro
tc
ik
c2
ax
l8

Input:

1 . l i n e
e           1
n         .
i       l
l     i
.   n
1 e

Output:

1.line

Input:

1
e .
n   l
i     i
l       n
.         e
1 e n i l . 1

Output:

1
.
l
i
n
e

### Scoring

Your score will be the area of the source code as a String Block™. Assuming your code can just fit inside a rectangle of width $$\w\$$ and height $$\h\$$, the score will be:

$$w \cdot h$$

### Winning Criteria

The submission with the lowest score in each language wins.

• The obvious problem here is that the reverse challenge is "cluttered" with the information being available three times. The solutions won't profit from the structure being a full hexagon, one of the three skewed rectangles is enough. – AlienAtSystem Dec 22 '19 at 16:47

# Joyous Kwanzaa!

Kwanzaa is an annual week-long celebration of African culture and history, and takes place on December 26 to January 1. During Kwanzaa it is traditional to light the seven candles of a kinara, one on each day. The middle candle is lit first, followed by the rest of the candles from left to right.

## Challenge

Your challenge will be to write a program or function that when given ASCII art (in any convenient format) of candles on a kinara (as specified below), some of which may be lit, will return or output a ASCII art of what the kinara would look like on the next day.

## Specification

A kinara will contain of seven even columns of equal length with equal spacing between them, representing the candles. The candles will be constructed with some consistent printable, non-whitespace ASCII character.

Some of the candles might have a flame above them, which will be represented by a single printable, non-whitespace ASCII character, possibly the same as the candle's character. If no candles are lit, there will be an empty line above the candles.

Here is an example of a kinara ASCII art:

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

## Input/Output

The input will be a kinara with seven candles, 0-6 of them lit according to the rule (middle candle first, then the rest from left to right).

The output will be a kinara with candles and flames identical to the input kinara, but with one more lit candle than the input. The extra candle will be lit according to the rule. If no candles are lit in the input, you may use any character for the flame.

You may assume the input is valid according to the specification given. You may assume the input has no trailing or leading whitespace in any of the lines, but trailing or leading whitespace is acceptable in the output as long as the kinara has the proper shape.

## Examples

In/out:

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

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

In/out:

** *
|||||||

****
|||||||

In/out:

5 5 5 5
6 6 6 6 6 6 6
6 6 6 6 6 6 6

5 5 5 5 5
6 6 6 6 6 6 6
6 6 6 6 6 6 6

In/out:

# pretend this is an empty line
q q q q q q q

0
q q q q q q q

## Rules

• No standard loopholes

• Shortest code in bytes wins

## Questions

• Is this too complicated? If so how could I simplify it?

• Is the wording unclear?

• Is this too similar to an existing challenge? I found another challenge which involves creating a menorah, but this one takes candles as input instead of a date.

# Hit the most Balloons with one Arrow

You got 1 arrow to hit as much balloons as possible. A common way to solve this, is to mark the tangent points of every balloon and project the angles on the source's position. As seen in the following image this has been down already. The task is now to find the optimal and angle-centered solution to hit the most Balloons.

The Input:

A json array of length 360 (for every 1° one entry) representating the number of Balloons hit at this angle.

Find an algorithm that solves the task with the following requirements:

• return the number of most hits
• return the max range of angles with the most hits
• return the the centered angle of this range
• there are no format restrictions how you are returning the values, it could be 3 prints or an array... whatever, as long as it is clear to see whats the hits, range and center

Solution of given JSON:

• Most Hits: 6
• Range: 358 - 0
• Center: 359

Notes:

Remember its a circle, so you need to check if the range crosses the 360°/0° mark!!! A range can be like 300° - 10° for example.

For simplicity the angles are 0° - 359° as 360° would be 0°.

• That's a lot of words and a picture wasted because your challenge is really just "Find the longest max-value sequence in the array", which is probably a duplicate. I think the challenge would be a lot better if you actually had to calculate this angle distribution from an input of balloon coordinates and a size. – AlienAtSystem Dec 23 '19 at 6:29

# Bunda-Gerth parser for abstract APL-family language code-golfparsingapl

## Background

APL has four kinds of tokens: arrays, functions, monadic operators (mops) and dyadic operators (dops).1) Let's denote them A (arrays), F (functions), M (mops), and D (dops) respectively. Usually when programming in APL, we use a mental model like this:

• Strand notation: A A ... A -> A
• Monadic function application: F A -> A
• Dyadic function application: A F A -> A
• Mop with left operand: (A|F) M -> F
• Dop with both operands: (A|F) D (A|F) -> F
• Mops and dops bind first from left to right (producing functions), and then function applications are evaluated from right to left

But J.D.Bunda and J.A.Gerth proposed another way to parse APL expressions. In this method, each (ordered) pair of tokens is assigned a binding strength and a resulting token. Below is the translation of the above mental model. Note that some token pairs do not bind at all, and we have a new token type AF, which stands for a function with bound left argument.

Binding strength | Token pair(s) -> result
--------------------------------------------
5 | A A -> A        # Strand notation
4 | D (A|F) -> M    # Dop + Right operand -> Mop
3 | (A|F) M -> F    # Left operand + Mop  -> Function
2 | A F -> AF       # Array + Function    -> Left-bound function
1 | (F|AF) A -> A   # Function + Array    -> Array

Then the actual parsing proceeds as follows: given a stream of tokens,

1. Evaluate the binding strengths between adjacent tokens.
2. Select the token pair whose binding strength is the rightmost local "peak", i.e. the x y pair in w x y z satisfying wx < xy >= yz. w and/or z can be empty. Note that the leftmost pair is bound first in A A ... A.
3. Group the pair and produce a new token as specified in the transition table.
4. Repeat from 1 until single token is left (successful parse) or no more reduction is possible (syntax error).

## Challenge

Given a list of tokens, output the order of binding (refer to the I/O section for specification) using the transition table and algorithm described above.

## Input and output

The input is a non-empty list of A/F/M/D tokens. You may use any number or single character to represent a token. You may assume that the input will always parse successfully, and the input won't contain any parentheses (since it isn't specified in the table above).

For the output, the order of binding represents in which order each gap between the tokens is closed:

Given the tokens:       A A F M A
Assign each gap an ID:   w x y z
Binding strength:        5 2 3 0
Rightmost local peak:       F M
Bind them first:        A A (F M) A     -> y = 1
Bind A-A next:         (A A) (F M) A    -> w = 2
Bind A-F next:        ((A A) (F M)) A   -> x = 3
Bind AF-A last:      (((A A) (F M)) A)  -> z = 4

Given the tokens:       A A F M A
The order of binding:    2 3 1 4  (the answer)

Note that, if the input contains L tokens, the answer is always a permutation of 1..L-1 (or 0..L-2 if you choose zero-based numbering).

## Scoring and winning criterion

Standard rules apply. The shortest code in bytes wins.

## Test cases

Input:  A A A A A
Output:  1 2 3 4
-----------------------------
Input:  A F F A A F F A
Output:  6 7 5 2 3 4 1
-----------------------------
Input:  A  F D A A  M  A A D F M F A
Output:  11 9 8 7 10 12 3 4 2 5 6 1

## Footnote

1. An actual implementation of APL has some special tokens, e.g. dot ., jot , and index notation [x]. We ignore them here for simplicity.
• Maybe an easy answer format is simply inserting the order of binding? A F F A A F F A((A F) (F (((A A) F) (F A))))((A5F)7(F6(((A1A)2F)4(F3A)))) and maybe even returning the permutation vector 5 7 6 1 2 4 3? – Adám Dec 17 '19 at 8:42
• @Adám That could work, but I don't think it's any easier to produce than a fully structured output. Also, alternative algorithms can produce the same tree with different binding order. – Bubbler Dec 17 '19 at 23:27
• It will be really hard to verify with the broad output allowance. Maybe just require fully parenthesising? – Adám Dec 17 '19 at 23:55
• @Adám IMO it's not a problem, as long as the answers specify which output format they're using. – Bubbler Dec 18 '19 at 0:16
• On second thought, I decided to emphasize the algorithm itself (over the resulting parse tree). – Bubbler Dec 26 '19 at 2:04

# Potentially Prime Punch-Card Patterns code-golfprimesnumber-theory

## Punch-cards

Let us define a punch-card to be a set of 'open' and 'closed' holes which can slide over the positive integer number line:

_________________________
|   ___     ___     ___   | -->
|  |   |   |   |   |   |  |
1  |  | 3 |   | 5 |   | 7 |  |  9   10  11  12  13  14
|  |___|   |___|   |___|  |
|_________________________| -->

When the first open hole is on the integer $$\ p \$$, then this punchcard (as it slides across the number line) yields the integers $$\p, p+2,\$$ and $$\p+4\$$. So, we can represent this card as the set $$\\{0, 2, 4\}\$$.

Given a punch-card we may wonder whether it is possible to slide the card to such a position that only prime numbers are under the 'open' holes.

Clearly, for $$\\{0, 2, 4\}\$$, we can position the punch-card to make 3, 5, and 7 visible. However, this is the only solution. Taking each element modulo 3 yields $$\\{0, 2, 1\}\$$, which contains every possible remainder under division by 3, so one visible integer must be divisible by 3. But we only want primes, and so 3 itself must be visible; this leaves a finite number of possible positions, of which only one is a valid solution.

More generally, if there exists a prime $$\q\$$ such that every integer $$\0 \le n < q\$$ appears in the punch-card set modulo $$\q\$$, then the punch-card is inadmissible and has a finite number of positions where all visible integers are prime: as we are bounded by the condition that $$\q\$$ must be visible. We can place $$\q\$$ in each hole in turn, and perform primality tests to count the valid solutions.

However, if there exists no such $$\q\$$, then the punch-card is admissible and we cannot assume there are finite solutions; the K-Tuple Conjecture in fact hypothesizes that every admissible punch-card can assume infinitely many positions where all visible integers are prime.

## The Challenge

Your task is to write a program or function which, given a list of ordered positive integers representing a punch-card set, outputs the number of positions the punch-card can assume where all visible integers are prime. If the set is admissible, then give a distinct output such as -1, null, Inf - anything that is not a non-negative integer.

Coming soon.

## Meta

There is a related challenge - testing for admissible sequences. However, I believe this is distinct enough to be a duplicate as rather than being a simple , if a set is admissible this program will have to then try different positions of the punch-card and count the valid solutions; whereas the previous challenge considers admissible sets in isolation, this challenge applies it to prime numbers.

• Given that the admissibility test is already a challenge, how about you split off the two-challenge and special case part: The input can be assumed to be an admissible sequence, the task is simply finding the positions resulting in valid solutions. – AlienAtSystem Dec 28 '19 at 5:54
• @AlienAtSystem i just feel as though that it is much less interesting to golf, and returning the distinct different output would open more interesting golfing opportunities. – FlipTack Dec 28 '19 at 6:09
• Handling special cases is very rarely interesting to golf, for a quite simple reason: Checking if the special case is present requires bytes. Being told the simple case is the case saves those bytes. – AlienAtSystem Dec 29 '19 at 20:29
• @AlienAtSystem but for there to be more golfing opportunities, don't there have to be more bytes to golf? – FlipTack Dec 29 '19 at 20:37
• Not really. The issue is that for multiple task challenges, the byte count is 90% of the time optimal_bytes(task1)+optimal_bytes(task2). Often, one of the tasks is considerably longer to golf, to the point where optimizing the other task is almost irrelevant because it's so small compared to the other. Therefore, the site consensus is to split challenges into their individual components as much as possible, and especially avoid input validation, because it's boring and requires lots of bytes. – AlienAtSystem Jan 1 at 8:29
• But it's not input validation in the sense of there being erroneous or invalid inputs: it's just that the answer to the question "how many prime positions does this have?" may be 0,1,2..., and also (probably) infinity. – FlipTack Jan 1 at 11:04

# How many ways can I count on n?

By using addition of natural numbers {1, 2, 3...} and multiplication of natural numbers larger than 1, we can reach the same outcome in several ways. For example 4 = 2 x 2 but also: 4 = 2 + 1 + 1. Using normal mathematical operator precedence, there are actually 6 ways to express 4 and 9 ways to express 5.

Since addition is commutative, a + b is counted as the same solution as b + a. The same holds for multiplication. So 7 = 1 + 2 x 3 is the same solution as 7 = 3 x 2 + 1.

The (trivial) solution n (using no addition or multiplication) is also counted as one of the solutions of n.

Multiplication with 1 is forbidden, because you can do this infinitely.

For a given input n, output c(n), which is defined as the number of ways n can be uniquely expressed using zero, one or more additions and multiplications of natural numbers, using normal mathematical operator precedence and without multiplication with 1.

## Rules

• Input and output are integers (your program should at least support input and output in the range of 1 up to 32767)
• Invalid input (0, floats, strings, negative values, etc.) may lead to unpredictable output, errors or (un)defined behaviour.
• Default I/O rules apply.
• Default loopholes are forbidden.
• This is , so the shortest answers in bytes wins

## Final note

I can think of several ways to approach this problem. I think it's interesting to see how the different approaches impact the length of the solution.

## Sandbox questions

• Please let me know if this task/problem is stated clear enough.
• Please let me know if there are any loopholes that I should cover in the question.
• I intentionally omitted more examples (because I hope contesters will think about solutions rather than just reproduce an OEIS sequence; although I'm not sure if this question had an OEIS sequence).
• I will tag this question with .

## Morse with Binary code-golfmorsebinarykolmogorov-complexity

For every input string (in ASCII, containing only lowercase alphabetic characters), output the alphabetic string converted from Morse code to Binary code.

In this case we are assuming that . is 0 and - is 1.

## How to do the conversion

For reference, here are a few Morse code libraries:

a .-
b -...
c -.-.
d -..
e .
f ..-.
g --.
h ....
i ..
j .---
k -.-
l .-..
m --
n -.
o ---
p .--.
q --.-
r .-.
s ...
t -
u ..-
v ...-
w .--
x -..-
y -.--
z --..

What's amazing about those codes is that the code length never exceeds 4 codes.

And now we are trying to cofuse binary with Morse code. First let's convert sample to morse code:

... .- -- .--. .-.. .

And consider the whole thing as a binary:

.... ---. --.. -...

Here is an alternative table for mapping the characters:

a ....
b ...-
c ..-.
d ..--
e .-..
f .-.-
g .--.
h .---
i -...
j -..-
k -.-.
l -.--
m --..
n --.-
o ---.
p ----

The result of this operation, after conversion, is aomi. You are never going to end up with a letter past p.

• What if there is a group of characters remaining? Input: e, Output: ? – Element118 Jan 1 at 5:23

# Quine Generator Generator... of any* length! code-golfmeta-golf

A Quine Generator Generator... is a quine when the input is empty or 0. Otherwise, for any other sufficiently large positive integer input, it should print a Quine Generator Generator... of that specified length (in the same language, same options).

Let S be the (most likely theoretically infinite) set of Quine Generator Generator... you can generate from your initial Quine Generator Generator....

Your score is the smallest N such that any Quine Generator Generator... in S can generate a Quine Generator Generator... of length N or longer (for reasonably sized N).

Input can be from standard input or as an argument of a function.

Lowest score wins.

# Sandbox Meta:

Typically, code is scored by bytes, but now the length would depend on how that is interpreted, since it is included in the description of the question. Hence would it be a problem to put: "You may choose to define characters in terms of bytes or characters (should they differ) for purposes of the program length and scoring purposes."

# Integer Keys and Duplicates

Given a list/vector of positive integers, write a function to check the following conditions in as few bytes as possible.

1. Take the first integer (the key, or k1) and check that the next k1 values have no duplicate values, excluding k1.
2. Take the last integer (the second key, or k2) and check that the k2 values before k2 have no duplicate values, excluding k2.

Note that both keys, k1 and k2, are elements of the list/vector as either key could contain the other.

Also, k1 and/or k2 can be greater than the number of integers within the list, which means you should check every element of the list besides the given key for duplicates, excluding the key.

If both steps return True, return True, else, return False.

### Test Cases

[5,1,2,5,3,4,3] is TRUE because [k1=5][1,2,5,3,4] has no duplicates, nor does [5,3,4] have any duplicates, excluding 3

[6,9,12,15,18,19,8,8,3] is FALSE because [k1=6][9,12,15,18,19,8] has no duplicates while [19,8,8][k2=3] has a duplicate.

[100,100,100,100,101,102,3] is TRUE because [k1=100][100,100,100,101,102,3] has no duplicates, and [100,101,102][k2=3] has no duplicates.

[100,100,100,100,101,102,4] is FALSE. [k1=100][100,100,100,101,102,4] has no duplicates, but [100,100,101,102][k2=4] has duplicates.

[6,6,6,6,6,6,6,3,3,3,3] is TRUE. [k1=6][6,6,6,6,6,6] has no duplicates and [3,3,3][k2=3] has no duplicates.

[1,2] is TRUE (clearly)

[1] is TRUE (clearly)

[] the empty list is also TRUE (if you can make a valid argument why it should be FALSE then I might give it to you)

# Drum fill generator

Create a program that generates a drum fill. Your program will output a pattern of L (left hand hits), 'R' (right hand hits), and K for kick drum hits.

## Rules

• The pattern must never have more than 2 of the same hits consecutively.
• The pattern must be loopable, so it mustn't have more than 2 of the same hits when it loops.
• Your program accepts 1 argument which is the length of the pattern. You can assume this will always be an integer > 0.
• Program output must be random each time it's ran.
• IO can be used with any convenient method.
• Standard loopholes are forbidden.
• This is code-golf, so smallest program wins!

Example valid output:

RLRRKKLL
LRLLKKLR
LKKRLRKLRRKL
LRLR

Example invalid output:

LRLL // 3 characters when looped
LRLRRRLLR // 3 characters in a row
RRLRLLRR // 4 characters when looped
• What's a paradiddle? You should include a description in your question – Jo King Jan 4 at 5:18
• How's that? I added more detail on what the rules/patterns are. – TMH Jan 4 at 21:18
• No it must be random each time the program is ran, I'll add that to the rules now, thanks :). – TMH Jan 5 at 21:16

# Theorem proving code golf: a+(b+c)=b+(c+a) code-golf

## Notes

This challenge is restricted to the languages with dependent type support. Some examples include Idris, Agda, Coq, Lean (which are designed for theorem proving), and Haskell with certain extensions enabled.

In these languages, a theorem is considered equivalent to a type, and writing a proof of a theorem is equivalent to providing a value of the given type (due to Curry-Howard correspondence). For example, the "modus ponens" rule

$$P,(P \rightarrow Q) \rightarrow Q$$

could be expressed as a function type

(p : P) -> (pq : P -> Q) -> Q

and have a proof by function application like the following:

thm : (p : P) -> (pq : P -> Q) -> Q
thm p pq = pq p

But a recursive definition requires an additional constraint that it is terminating, i.e. the recursive calls are done with smaller arguments so that it eventually reaches the base case. This is to prevent erratic proofs like a : False := a, which evokes an infinite loop to fake a value of empty type False.

A recursive proof is usually analogous to a proof by (weak or strong) induction.

Given the Peano natural numbers and addition

$$n : \mathbb{N} := \text{Zero } | \text{ Succ }n \\ a+ \text{Zero} := a, a+ \text{Succ } b := \text{Succ }(a+b)$$

express the following statement in code and prove it:

$$\forall a,b,c\in \mathbb{N}, a+(b+c)=b+(c+a)$$

The theorem should have the type of (or your language's equivalent of)

forall a b c : nat, a + (b + c) = b + (c + a)

or, in function notation,

(a : nat) -> (b : nat) -> (c : nat) -> a + (b + c) = b + (c + a)

## Condition

You're free to use a library provided with your language of choice. If the library uses the flipped definition of Peano addition, it's fine to use it instead. As usual for , you need to count the bytes used for imports.

For theorem-proving languages (where infinite loops are rejected by the compiler), a successful compilation usually means a correct proof. If it is not the case, you should provide an explanation that the definition is indeed terminating.

If you find a bug in some theorem-proving language that allows to prove anything, using it is not allowed, as it is not considered as a solution to the given task.

## Scoring and winning criterion

Standard rules apply. The shortest code in bytes wins.

## Meta

• Is it a good idea for a CGCC challenge? I once learned Coq and a bit of Lean, and I thought minimizing code by clever usage of tactics and/or explicit definitions could be interesting.
• Is the task appropriate for an introductory challenge in this topic? I just made up a statement that looks less likely to be in a standard library, but is easy enough to prove by commutativity and associativity.
• There is already proof-golf, which has all well received questions (though they've all been asked by one user). This is a bit different, and I think you need to try to work out some of the kinks if you want to ask it this way. Mostly, your definition of "prove" feels a tad loose. I've not used such languages, but I recommend trying to find a common element amongst them so you can be more precise. Perhaps you will want to prove something over a finite field instead, so an exhaustive proof can terminate? – FryAmTheEggman Jan 7 at 21:41
• @FryAmTheEggman Thanks for the suggestion. I tried to clarify what a proof means in this context. I initially thought about proof-golf too, but listing out the rules of entire CIC theory is probably out of the question. (Probably I could extract the needed rules if I limit to the natural number type; I'll think about it later.) – Bubbler Jan 8 at 0:44

# Interpret LCGFuck™ code-golfinterpreter

### Introduction

LCGFuck™ is a Brainfuck-like esoteric programming language invented by me. Inspired by Brainfuck and linear congruential generators (LCGs), LCGFuck™ combined both things into one.

An LCG in LCGFuck™ is defined with 5 integers, namely multiplier a, adder b, modulus c, offset d and seed e. Let x be its current state, then the corresponding output will be x + d, and the next internal state x' of the LCG can be calculated by x' = (a * x + b) mod c. All numbers should be non-negative except for d. Moreover, b and e should be in the range [0, c).

### Technical details

An LCG is defined with 5 numbers, like 12345 678 90 -123 45. The last two numbers are optional and default to 0.

LCGFuck™ has 3 storage variables, the LCG list, the number list and the number memory. The functions are as follows:

• LCG list: Cyclic list that stores all LCGs created. Every entry is in turn a list of 5 numbers in the order [a, b, c, d, e] as notated in the introduction. It has a pointer that controls which LCG is chosen at the moment, and moves in a cycle through the list.
• Number list: Ordinary list that stores the numbers for LCG definition. It can store at most 5 numbers.
• Number memory: A single variable that can be read and written with an input or an LCG output.

LCGFuck™ has 14 operators, namely:

• \n (newline): Creates an LCG with the numbers in the order of [a, b, c(, d(, e))] and clears the list if the number list contains 3 or more numbers. No-op if the number list contains no more than 2 numbers.
• <: Shifts to the previous LCG circularly (back to the last when moving from the first).
• >: Shifts to the next LCG circularly (back to the first when moving from the last).
• +: Moves the current LCG to the next state (calculates x' = (a * x + b) mod c)
• c: Prints the output of the current LCG as a character (with the output as the codepoint).
• n: Prints the output of the current LCG as a number (with a trailing space)
• o: Writes the output of the current LCG to number memory
• i: Reads a value from the input and writes it to number memory. There is two input mode, one reading a character one time, and one reading a number one time. You do not need to implement this operator in this challenge.
• s: Reads the value from number memory and seeds the current LCG with it
• m: Reads the value from number memory and pushes it to the number list
• []: Output loop. Executes the loop if the current LCG is non-zero
• {}: State loop. Executes the loop if the state of the current LCG is non-zero

An integer, optionally with a negative sign at the front, pushes the number to the number list. Any characters other than the newline \n and any of -0123456789[]{}<>+cimnos are no-ops.

The code runs linearly from the first characters, except when loop ends are encountered. Loops can be nested but must be paired accordingly from the innermost loop to the outermost loop.

### Challenge

Write an interpreter, making it as short as possible, that interprets the LCGFuck source code given as the input. You need not implement the i operator in this challenge. The output should be printed to STDOUT or returned in case of writing a function.

1. An LCG is always defined before any of the commands that manipulate the LCG or the LCG list pointer
2. At any moment the number list contains at most 5 numbers
3. All brackets are paired accordingly

Standard loopholes are forbidden by default.

Sample implementation with operator i (Primality check)

### Sample

The LCG list need not be printed out. It is just for illustration purpose.

1. The "Hello world!" program in LCGFuck™ is as follows:

1 29 30 72
1 3 9 108
1 1 2 32
c+c>cc+co>c
1 8 9 m
>+c<<c+c+c<+c>>+c

• Output: Hello world!

• LCG list after execution:

[1, 29, 30, 72, 28]
[1, 3, 9, 108, 0]
[1, 1, 2, 32, 1] < Current LCG
[1, 8, 9, 111, 8]

2. An example that outputs the values of x - 16, where x is the states of the LCG x' = (19x + 17) mod 32, starting with seed x = 4, until the output is 0:

19 17 32 -16 4
[n+]n

• Output: -12 13 8 9 -4 5 -16 1 4 -3 -8 -7 12 -11 0

• LCG list after execution:

[19, 17, 32, 16, -16] < Current LCG

3. An example that outputs all possible values of 5(x + 12), where x is the states of the LCG x' = (5x + 3) mod 8:

5 3 8 12
5 0 99
o>s+n<+{o>s+n<+}

• Output: 60 75 70 85 80 95 90 65

• LCG list after execution:

[5, 3, 8, 12, 0] < Current LCG
[5, 0, 99, 0, 65]

4. An example illustrating the 3-number restriction of \n (the 9 in the 3rd line is pushed to the number list but the LCG is defined before the 9 is pushed):

12 34
56 78
9
n

• Output: 78

• LCG list after execution:

[12, 34, 56, 78, 0] < Current LCG

5. An example that determines whether 97 is a prime:

1 -1 97 0 -1
{o+1 1 m 0 97
}>{>}o+{o+}<{<}s
1 1 1 32
1 4 12 69 4
1 1 7 78 2
{<+++++c+c+++++c+++
<<c>>>{+}}<c++c<c+c+c

• Output: PRIME

• LCG list after execution:

[1, -1, 97, 0, 0]
[1, 1, 96, 0, 1]
[1, 1, 95, 0, 2]
[1, 1, 94, 0, 3]
[1, 1, 93, 0, 4]
[1, 1, 92, 0, 5]
[1, 1, 91, 0, 6]
[1, 1, 90, 0, 7]
[1, 1, 89, 0, 8]
[1, 1, 88, 0, 9]
[1, 1, 87, 0, 10]
[1, 1, 86, 0, 11]
[1, 1, 85, 0, 12]
[1, 1, 84, 0, 13]
[1, 1, 83, 0, 14]
[1, 1, 82, 0, 15]
[1, 1, 81, 0, 16]
[1, 1, 80, 0, 17]
[1, 1, 79, 0, 18]
[1, 1, 78, 0, 19]
[1, 1, 77, 0, 20]
[1, 1, 76, 0, 21]
[1, 1, 75, 0, 22]
[1, 1, 74, 0, 23]
[1, 1, 73, 0, 24]
[1, 1, 72, 0, 25]
[1, 1, 71, 0, 26]
[1, 1, 70, 0, 27]
[1, 1, 69, 0, 28]
[1, 1, 68, 0, 29]
[1, 1, 67, 0, 30]
[1, 1, 66, 0, 31]
[1, 1, 65, 0, 32]
[1, 1, 64, 0, 33]
[1, 1, 63, 0, 34]
[1, 1, 62, 0, 35]
[1, 1, 61, 0, 36]
[1, 1, 60, 0, 37]
[1, 1, 59, 0, 38]
[1, 1, 58, 0, 39]
[1, 1, 57, 0, 40]
[1, 1, 56, 0, 41]
[1, 1, 55, 0, 42]
[1, 1, 54, 0, 43]
[1, 1, 53, 0, 44]
[1, 1, 52, 0, 45]
[1, 1, 51, 0, 46]
[1, 1, 50, 0, 47]
[1, 1, 49, 0, 48]
[1, 1, 48, 0, 1]
[1, 1, 47, 0, 3]
[1, 1, 46, 0, 5]
[1, 1, 45, 0, 7]
[1, 1, 44, 0, 9]
[1, 1, 43, 0, 11]
[1, 1, 42, 0, 13]
[1, 1, 41, 0, 15]
[1, 1, 40, 0, 17]
[1, 1, 39, 0, 19]
[1, 1, 38, 0, 21]
[1, 1, 37, 0, 23]
[1, 1, 36, 0, 25]
[1, 1, 35, 0, 27]
[1, 1, 34, 0, 29]
[1, 1, 33, 0, 31]
[1, 1, 32, 0, 1]
[1, 1, 31, 0, 4]
[1, 1, 30, 0, 7]
[1, 1, 29, 0, 10]
[1, 1, 28, 0, 13]
[1, 1, 27, 0, 16]
[1, 1, 26, 0, 19]
[1, 1, 25, 0, 22]
[1, 1, 24, 0, 1]
[1, 1, 23, 0, 5]
[1, 1, 22, 0, 9]
[1, 1, 21, 0, 13]
[1, 1, 20, 0, 17]
[1, 1, 19, 0, 2]
[1, 1, 18, 0, 7]
[1, 1, 17, 0, 12]
[1, 1, 16, 0, 1]
[1, 1, 15, 0, 7]
[1, 1, 14, 0, 13]
[1, 1, 13, 0, 6]
[1, 1, 12, 0, 1]
[1, 1, 11, 0, 9]
[1, 1, 10, 0, 7]
[1, 1, 9, 0, 7]
[1, 1, 8, 0, 1]
[1, 1, 7, 0, 6]
[1, 1, 6, 0, 1]
[1, 1, 5, 0, 2]
[1, 1, 4, 0, 1]
[1, 1, 3, 0, 1]
[1, 1, 2, 0, 1]
[1, 1, 1, 0, 0]
[1, 1, 1, 32, 0]
[1, 4, 12, 69, 0] < Current LCG
[1, 1, 7, 78, 4]

### Winning criteria

Since this is a code-golf challenge, the shortest solution for every language wins.

## I'm stepping down code-golfarithmeticinteger

You are given 4 positive integers: volume of the first container (v1), volume of the second container (v2), volume of the liquid in the first container (l1), and volume of the liquid in the second container (l2). Your task is to move (or "step down", if appropriate) some of the liquid from container 1 to container 2, making the amount of empty space in the containers equal to each other, outputting how much liquid should be moved.

## An example

Here is an example of some possible input (formatted for the ease of test cases. The input isn't formatted.):

8 11
6  5

Here, you need to make sure that the differences between the volumes and the containers are equal. Currently the difference is not equal:

8 11
- 6  5
======
6 2

So we need to try to make them equal, by taking some of the values in one container to another container:

8 11
- 4  7
======
4  4

After you have succeeded, output how much liquid you need to take from container 1 to container 2. Therefore the above example should output 2.

## Test cases

15 16
9  2

We move into this state:

15 16
- 5  6
======
10 10

Therefore 4 is the expected result.

## Test case #2

16 12
13  1

We move into this state:

16 12
- 9  5
======
7  7

Therefore 4 is the expected output.

## Test case #3

20 12
10 2

Moved:

20 12
-10 2
======
10 10

Therefore 0 is the expected output.

## Rules

• The test cases will be made so that the result is always possible as a positive integer. The result will never be a decimal.
• Input can be taken in any convenient and reasonable format.
• Output can be given in any convenient and reasonable format.
• The input string will be made so that a moving is always possible without involving decimals or negative numbers. Also there will not be more liquid than how much a container can hold.
• Of course, this is , so the answer in each language consisting of the fewest number of bytes wins. Happy golfing!
• An extra restriction for the test cases: you can assume that there is always going to be space available in the second container. So a test case like this:
10 2
2 2

is not going to be a valid testcase.

• So, it just calculate abs(v1-l1-v2+l2)/2? It seems trivial to me. And what should I do if the calculate result is not an integer? – tsh Jan 13 at 6:03
• May I assume the solution always exits? What should I do if input is v1=10,l1=2,v2=2,l2=2? – tsh Jan 13 at 9:17
• The test cases will not contain calculations that yield decimal results; also you can assume that there is always enough empty space in the second container to move the liquid. – user85052 Jan 13 at 13:26
• I find the challenge title confusing. Please change it. – Beefster Jan 20 at 17:40
• @Beefster The challenge is already posted, and after at least 35 people agreed with you, someone else changed the title. @ a'_': please edit this Sandbox post so it only contains the title and link to the posted challenge and delete it from the Sandbox please (as mentioned in bold in the "What is the Sandbox?" post at the top). – Kevin Cruijssen Jan 22 at 10:49

# He. Might. Go. All. The. Way. Touchdown!

[in-progress]

Premise:

In American Football, a team has to drive up the field to their opponent's end-zone to score points. But here's the catch, they have 4 tries (called downs) to go at least 10 yards to gain a new set of downs. They repeat this until they score a Touchdown, or until they fail to get 10 yards and (at least in this challenge) punt it away or go for a Field Goal.

Challenge:

You are to output the state of each down, i.e. which down it is, how far to the next down (or the end-zone), and where the ball is. Football location counts up from a team's end-zone (0yd line) up to the 50yd line (the middle), then back down to 0 for the opponent's end-zone.

We differentiate the sides of the field by prefixing the location with the team's name. In this challenge, you can use a 1 character label or use positive and negative values. It must go 0-49,50,49-0 and have a way to differentiate between the sides. Your choice on who owns the 50yd line.

Sample Output: (Our team is A, the opponent's team is B)

1st & 7 on A 13
2nd & 10 on A 48
3rd & 12 on 50 OR 3rd & 12 on A 50 OR 3rd & 12 on B 50
4th & 8 on B 10

Your team will start on your own 1 yard line on 1st & 10 (1st down, 10 yards to go for another first down). You will then gain a uniformly random number of yards between [-3,10] called N. If you didn't get enough for a 1st down, it will now be 2nd & (10 - N). Repeat drawing another number between [-3,10] and adding the yardage for 2nd and 3rd downs if it's still not enough for a 1st down. If you do gain enough for a 1st down, you simply go back to 1st & 10 on the next go and continue going down the field.

On 4th Down, your team is playing safe and will either punt or go for a Field Goal. If you are further away than their 40, output P. If you're within 40 yards, you will attempt a Field Goal with 100 - Yards Away% chance of success. If you succeed, output FG. If you miss, output NG ("No Good"). Afterwards, terminate.

However, there are two special situations that must be handled.

If you get within 10yd (inclusive) and gain a 1st Down, It will then be 1st & Goal and there are no more opportunities to gain 1st downs. Do or die! If you score, output TD and terminate, otherwise you'll follow the normal Field goal rules.

If you lose enough yards to go into your own end-zone, that's called a Safety. Simply output a S and terminate.

Rules:

• No usable input will be provided
• Output is flexible. Tuples and lists of lists are all fine.

Sample Runs:
To-Do

Tags: Code-golf, random, game

Feedback? Does the Field goal add enough meat to be worth including?

# What order(s) can I fire my tonics in?

Helsing's Fire is a classic mobile game about light. You fire blasts of holy retribution (from chemicals called tonics), using cover to selectively hit enemies in the right sequence.

For this challenge, we're going to ignore the line-of-sight stuff and focus on just the tonic order. Each monster is represented by a string like RBBGB—this indicates that, in order to kill it, you need to hit it with a red tonic, followed by 2 blue tonics, then one green tonic and finally one more blue tonic.

Since you're allowed to ignore cover and pick any subset of the monsters to hit, any of the following sequences will kill it:

RBBGB
GRBBGBRR
RBBBBBBBBBBBBBBBGGGGBR
RRRBGGGBBRGRBB

Your goal is to take in a list of monsters as well as how many of each tonic you have, and output every possible ordering of tonics that kills every monster, including ones where not every tonic is fired. For instance:

R; 2 red -> R, RR
R; 1 red/1 blue -> R, RB, BR
No monsters, but tonics -> every possible permutation of the tonics listed
No tonics, but monsters -> no output, empty set output, or single trailing separator
Neither tonics nor monsters -> empty string
RB,BR; 2 red/2 blue -> RBR, BRB, BRBR, RBBR, RBRB, BRRB
RR,RR,GRG,R,GBR,GR,GRRR,GGR,GB,G,R,BBR; 3 red/2 blue/2 green -> BGRRBGR, BGRRGBR, BGRBRGR, BGRBGRR, BGRGRBR, BGRGBRR, BGBRRGR, BGBRGRR, GRRBBGR, GRRBGBR, GRRGBBR, GRBRBGR, GRBRGBR, GRBBRGR, GRBBGRR, GRBGRBR, GRBGBRR, GRGRBBR, GRGBRBR, GRGBBRR, GBRRBGR, GBRRGBR, GBRBRGR, GBRBGRR, GBRGRBR, GBRGBRR, GBBRRGR, GBBRGRR
BR,R,BR,RBBR,RBR,RGB,RGBR,GB,RGR,GG,GRB,G; 3 red/2 blue/2 green -> RBGRBRG, RBGRBGR, RBGRGBR, RBGGRBR, RGRBBRG, RGRBBGR, RGRBGBR, RGRGBBR, RGBRBRG, RGBRBGR, RGBRGBR, RGBGRBR, RGGRBBR, RGGBRBR, GRRBGBR, GRRGBBR, GRBRGBR, GRBGRBR, GRBGBR, GRBGBRR, GRGRBBR, GRGBRBR, GRGBBR, GRGBBRR

Would there be a duplicate for this?

## Elect the Doge of Venice!

The Venetian election system was... complicated.

The Great Council came together and put in an urn the ballots of all the councilors who were older than 30. The youngest councilor went to St Mark's Square and chose the first boy he met who drew from the urn a ballot for each councillor and only those 30 who got the word ‘elector' remained in the room. The 30 ballots were then placed back in the box and only 9 contained a ticket, so the 30 were reduced to 9, who gathered in a sort of conclave, during which, with the favourable vote of at least seven of them, they had to indicate the name of 40 councillors.

With the system of ballots containing a ticket, the 40 were reduced to 12; these, with the favourable vote of at least 9 of them, elected 25 others, which were reduced again to 9 who would elect another 45 with at least 7 votes in favour. The 45, again at random, were reduced to 11, who with at least nine votes in favour elected another 41 that finally would be the real electors of Doge.

These 41 gathered in a special room where each one cast a piece of paper into an urn with a name. One of them was extracted at random. Voters could then make their objections, if any, and charges against the chosen one, who was then called to respond and provide any justification. After listening to him, they preceded to a new election, if the candidate obtained the favourable vote of at least 25 votes out of 41, he was proclaimed Doge, if they were unable to obtain these votes a new extraction took place until the outcome was positive.

Venice is no longer an independent republic, but if they were, they would be dying to automate this system! (because we all know electronic voting is the future!) Time for you to step in. Your program is to do the following.

• Here is the list of members of the Great Council (who are all older than 30). Take this as input, by perhaps reading it from a file, or whatever you prefer. The number of councillors varied over time, so your program should work for any list of sufficient length.
• Take the youngest member of the council. Because there are no ages given, you'll have to guess. Pick a person at random, and print: "[Name] goes to St Mark's Square".
• The boy at the square will pick thirty members from an urn. So, randomly choose 30 people from the list (not including the youngest councillor). Print "1 selected 30: " followed by the names of each of the thirty members in this round, separated by ", ".
• Of those thirty, nine are randomly selected to go to the next round. So randomly choose 9 from that group, and print "30 reduced to 9: " followed by those nine electors.
• Those nine electors have to choose forty different councillors. So, from the list of councillors, excluding the electors (but including the twenty-one people from the previous round of thirty who did not become electors), pick forty members. Print "9 selected 40: " followed by those forty people.
• The forty were reduced to twelve by lot. Pick twelve from these members at random, print "40 reduced to 12: " followed by the dozen.
• The dozen elected twenty-five councillors. You know the rules by now: pick 25 councillors excluding the 12 (but including anyone not in the 12 who partook in previous rounds), and print "12 selected 25: " followed by the twenty-five.
• The twenty-five got reduced to nine again. Pick 9 random people from the 25 and print: "25 reduced to 9: " followed by those nine.
• Those nine selected forty-five councillors. Pick 45 people not in the 9, and print: "9 elected 45: " followed by those people.
• The forty-five were reduced to eleven. Pick 11 random councillors from those 45, and print "45 reduced to 11: " followed by those 11.
• The eleven picked forty-one councillors who would elect the doge. Pick 41 people not in the 11, print "11 selected 41: " followed by those people.
• Finally, these people will elect the Doge of Venice. Pick 1 person, randomly, from the 41. Then print "The new doge is [Name]" And then you can rest and watch the sun set on a democratic universe.

This horrendously complicated system was made to reduce corruption; so however you implement it, every member of the Great Council must have an equal chance of getting elected in the end. I will run your implementation ten thousand times to ensure that's roughly what happens. Abberating programs are corrupt and therefore disqualified.

Other rules:

• Capitalisation of names must be kept. Expect only names with readable ascii characters (I had to remove a few Nicolò's from the list to make the challenge mildly easier).
• Always print a list of names separated by the string ", ". A period at the end is not necessary.
• Print every new instruction on a new line.
• The list will always have at least 54 people, so enough for this process.
• For every non-overlapping Italian word of at least 4 characters that can be found in your source code, remove 3/4ths of the length of that word (round up) from your score. A word counts as a sequence of alphanumeric characters that does not have to be consecutive, so "new random: (am-m)[a]" would count because it contains "mamma". Recent English loanwords (words that originate in English, even if they happen to occur in modern Italian, and are still spelled the same way) are not allowed. For golfing languages, modifications/variants of letters like ʀ and ʁ are allowed to count as the letters they are recognisable as. I'll be the judge if it's ambiguous.
• This is , so your score is the number of bytes in your code. Lowest score wins!

• This is rather full of unnecessary requirements that don't seem to add much of anything to the challenge. Why not just require each of these random selections to be unambiguously separated in any format? Why demand odd prefaces to each subtask? Finally the Italian bonus is very weird, and seems to require that you provide an entire lexicon of valid words to be fully specified. – FryAmTheEggman Jan 17 at 20:02
• I have to be honest (not a native Anglophone): I am not sure I can understand what your first two concerns say. The reason I ask for every step to be printed out, if that's what you mean, is because I cannot read Malbolge or Befunge and if I didn't get an indication that this repeated selection is being followed to a tee, I would not be able to tell if the solutions were doing it at all. As for the Italian, that was just a creative addition similar to the even-or-odd puzzle that requires alternating cardinality for source code characters. I'm not married to it though; it just felt right to me. – KeizerHarm Jan 17 at 20:36
• I'm sorry that I wasn't clear. I intend to ask why you require outputs like "30 reduced to 9:". Since the order of the operations is fixed, there's no reason to require this. I guarantee your challenge will be better received if you remove those requirements. The Italian addition would be fine, only it is deeply unclear. "I'll be the judge if it is ambiguous" isn't the standard we hold challenges to - you need to specify precisely what counts and what doesn't if you want to keep it. – FryAmTheEggman Jan 17 at 20:41
• @FryAmTheEggman Alright. Funny, I did not expect the printing to be the less desirable requirement, but I'll follow your judgement. And I can specify the lexicon, sure - I'll take an Italian word list, and subtract an English one (because loanwords). Just, taking into account golfing languages is going to be ambiguous - they frequently use modifications of Latin letters, and I want to allow those to be treated as their equivalent letters, but you might run into ambiguous situations like whether a Cyrillic к (don't know if any language uses it) can be a regular k. – KeizerHarm Jan 17 at 21:00

# Fun with Lasers and Prisms (WIP)

Given a rectangular grid of objects, one or more laser pointers, and a target, determine if any laser beam will hit the target.

# Objects

ASCII will be used for illustration purposes

• Laser Pointer: ^, V, <, > - a beam will shoot up, down, left, right, respectively, starting from this cell.
• Target: O - return true if a beam reaches this cell
• Mirror: /, \ - reflects a beam 90 degrees
• Prism: # - the laser will split into three beams, one for each direction
• One-way block: A, U, (, ) - a beam will pass up, down, left, right, respectively, but not other directions
• Corridor block: =, " - a beam will only pass horizontally or vertically, respectively
• Gate block: I, H - a beam will pass through horizontally if another beam touches it vertically, or vice-versa, respectively

# Rules

• Use any convenient representation
• I think it is likely that your gate blocks prevent this from being a dupe of other similar challenges, though I haven't thoroughly checked yet. – FryAmTheEggman Jan 18 at 3:19

# How low can you go?

Time to play so ascii-art limbo!

Here's the bar:

|--|
|  |
|  |

Can you fit under it?

## Goal

Write a program or function that takes an ascii string representing a some shape, and a positive integer representing a bar height.

Output the shape from the input after it has attempted to do the limbo.

## Details

In limbo you lean back to make yourself as small as possible to fit under the bar, and that is just what the input shapes will try and do.

If the input shape contains any repeating patterns in its lines, then you can remove all but the last of the repetitions that are in the pattern. In additions the pattern must start at the top line, and once the repeating pattern is broken no more sections can be removed.

If there is a repeating pattern that contains another repeating pattern, only the innermost pattern is stripped.

For example this is how the following inputs would look after "Leaning back":

1. XXX              2. xxx            3. xxx           4. xxx
YYY                 xxx       xxx     xxx              xxx      xxx
XXX   -->   XXX     yyy  -->  yyy     xxx  -->         yyy      yyy
YYY         YYY     zzz       zzz     xxx              yyy      yyy
zzz         zzz     aaa       aaa     xxx      xxx     xxx  --> xxx
xxx      xxx
yyy      yyy
yyy      yyy

Note how in example number 4 there was a repetition ox xxx insinde of another repetition of xxx, xxx, yyy, yyy. In this case only the inner repeating lines got reduced.

The bars will be drawn as shown below for the given heights:

1 -> |--|    2 -> |--|   3 -> |--|  etc...
|  |        |  |
|  |

If the given input shape in its reduced form does not fit underneath the bar then the bar will be drawn on the ground like this |__|

## Exmaples

Inputs:

a. height: 3      b.  height: 2     c.  height: 1
shape: xxx         shape: xxx        shape: (emptystring)
xxx                xxx
xxx                xxx
yyy                yyy

Outputs:

a. |--|           b. |  | xxx       c. |--|
|  | xxx          |__| yyy
|  | yyy

## Notes

1. You can assume only valid inputs will be given, handle invalid input however you want
2. The input shape will be drawn 1 space after the bar, and the bottom of the input shape will always line up with the bottom of the bar.
3. Extra whitespace after is fine as long as all the lines are properly aligned.
4. If there is whitespace in the given input then that should be included in the output
5. The shape will not necessarily always line up in a perfect rectangle as I drew it.

This is for code-golf so the answer using the fewest bytes wins.

Please let me know what you think / if anything is unclear and needs to be improved. Hope you guys like it!

EDIT: Would whoever down voted please explain whats wrong with it?

• I'm not the voter, but I'd bet they voted because what you have right now is very difficult to understand. I had to read this three times to figure out what you wanted. It may be worth revisiting the concept of "leaning over" as it is deeply unintuitive to me at the moment. On top of that, this requires a lot of "boring" golfing for the required output format when true/false seems to be basically the same. I hope this is helpful! – FryAmTheEggman Jan 20 at 20:24
• @FryAmTheEggman Hmm, I thought I'd explained it fairly clearly - its just removing any lines that form a repetitive pattern, but I will try to update it. The output is slightly more than just T/F because of wanting to see the "limbo'ed" input shape with the shaved lines, and having to draw the bar different if it fails... though maybe that is what you meant by boring golf? – Quinn Jan 20 at 20:30
• @FryAmTheEggman I tried updating the explanation, though I'm not sure if it is any clearer, please let me know if it makes more sense now – Quinn Jan 20 at 20:33
• Explaining things is often harder than one would guess. Here I think a big problem is that your definition of bending is not what I would expect, so it makes the whole idea harder to grasp - particularly the rules about which repetitions happen first. Since I don't get the why I struggle to get the what. Maybe try explaining this to someone verbally to see if you can get rapid and direct feedback. What you have now is better, but I still think I'd have a hard time following on a first read. – FryAmTheEggman Jan 20 at 20:35

# Factorise a floating point number

Given a target floating point number, $$\T\$$ and a set of $$\N\$$ floating point numbers $$\\{x_{1},..,x_{N}\}\$$ and a permissible error $$\tol\$$ , find a set of integer coefficients $$\A\$$, $$\\{m_{1},..,m_{N}\}\$$ such that: $$A\prod\limits_{i=1}^{N}x_{i}^{m_{i}} = T\pm{}tol$$

## Input

Input will be a target number, a set of real numbers and a tolerance as a decimal or percentage (note which) in any order or format required by your language.

## Output

Output should be $$\A\$$, $$\\{m\}\$$ and the corresponding error as a percentage. If multiple combinations are valid any or all sets of $$\A\$$ and $$\\{m\}\$$ are within tolerance.

## General rules

• This is , so shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
• Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
• Default Loopholes are forbidden.

## Examples

$$\3*(3.2^4*7.1^2)=15857.614848\$$

15857.6 [3.2, 7.1] 0.05 => 3, [4, 2]

$$\2*(e^2*\pi{}^2)=145.854121188\$$

145.85 [2.7182, 3.1415] 0.1 => 2, [2, 2]

$$\3*(\pi^0*0.1^1)\$$

0.3 [3.14159, 0.1] 0.05 => 3, [0, 1]

# Periodic table

The periodic table is a large tablecitation needed where we can find the chemical elements written out with their chemical symbols. For example, Helium shows up as He and Carbon shows up as C. You can read Wikipedia's article on the periodic table, if you want.

You have to write a function/program/procedure/... that, given a chemical element's name, returns its chemical symbol. For the purposes of this challenge, we will use the 118 elements listed below:

['Hydrogen', 'Helium', 'Lithium', 'Beryllium', 'Boron', 'Carbon', 'Nitrogen', 'Oxygen', 'Fluorine', 'Neon', 'Sodium', 'Magnesium', 'Aluminium', 'Silicon', 'Phosphorus', 'Sulfur', 'Chlorine', 'Argon', 'Potassium', 'Calcium', 'Scandium', 'Titanium', 'Vanadium', 'Chromium', 'Manganese', 'Iron', 'Cobalt', 'Nickel', 'Copper', 'Zinc', 'Gallium', 'Germanium', 'Arsenic', 'Selenium', 'Bromine', 'Krypton', 'Rubidium', 'Strontium', 'Yttrium', 'Zirconium', 'Niobium', 'Molybdenum', 'Technetium', 'Ruthenium', 'Rhodium', 'Palladium', 'Silver', 'Cadmium', 'Indium', 'Tin', 'Antimony', 'Tellurium', 'Iodine', 'Xenon', 'Cesium', 'Barium', 'Lanthanum', 'Cerium', 'Praseodymium', 'Neodymium', 'Promethium', 'Samarium', 'Europium', 'Gadolinium', 'Terbium', 'Dysprosium', 'Holmium', 'Erbium', 'Thulium', 'Ytterbium', 'Lutetium', 'Hafnium', 'Tantalum', 'Tungsten', 'Rhenium', 'Osmium', 'Iridium', 'Platinum', 'Gold', 'Mercury', 'Thallium', 'Lead', 'Bismuth', 'Polonium', 'Astatine', 'Radon', 'Francium', 'Radium', 'Actinium', 'Thorium', 'Protactinium', 'Uranium', 'Neptunium', 'Plutonium', 'Americium', 'Curium', 'Berkelium', 'Californium', 'Einsteinium', 'Fermium', 'Mendelevium', 'Nobelium', 'Lawrencium', 'Rutherfordium', 'Dubnium', 'Seaborgium', 'Bohrium', 'Hassium', 'Meitnerium', 'Darmstadtium', 'Roentgenium', 'Copernicium', 'Nihonium', 'Flerovium', 'Moscovium', 'Livermorium', 'Tennessine', 'Oganesson']

with the respective element symbols:

['H', 'He', 'Li', 'Be', 'B', 'C', 'N', 'O', 'F', 'Ne', 'Na', 'Mg', 'Al', 'Si', 'P', 'S', 'Cl', 'Ar', 'K', 'Ca', 'Sc', 'Ti', 'V', 'Cr', 'Mn', 'Fe', 'Co', 'Ni', 'Cu', 'Zn', 'Ga', 'Ge', 'As', 'Se', 'Br', 'Kr', 'Rb', 'Sr', 'Y', 'Zr', 'Nb', 'Mo', 'Tc', 'Ru', 'Rh', 'Pd', 'Ag', 'Cd', 'In', 'Sn', 'Sb', 'Te', 'I', 'Xe', 'Cs', 'Ba', 'La', 'Ce', 'Pr', 'Nd', 'Pm', 'Sm', 'Eu', 'Gd', 'Tb', 'Dy', 'Ho', 'Er', 'Tm', 'Yb', 'Lu', 'Hf', 'Ta', 'W', 'Re', 'Os', 'Ir', 'Pt', 'Au', 'Hg', 'Tl', 'Pb', 'Bi', 'Po', 'At', 'Rn', 'Fr', 'Ra', 'Ac', 'Th', 'Pa', 'U', 'Np', 'Pu', 'Am', 'Cm', 'Bk', 'Cf', 'Es', 'Fm', 'Md', 'No', 'Lr', 'Rf', 'Db', 'Sg', 'Bh', 'Hs', 'Mt', 'Ds', 'Rg', 'Cn', 'Nh', 'Fl', 'Mc', 'Lv', 'Ts', 'Og']

# Input

The code you write should receive an element name in any sensible format, such as a string "helium". You may assume whatever capitalization that suits your needs.

# Output

The code you write should return the element's symbol as a string, with any capitalization that suits your needs. Bonus imaginary internet points if you return the symbol with the standard capitalization.

# Scoring

This is so your answer doesn't win by being the shortest! You will be provided 118 test cases. Your score will be your code's byte count divided by the percentage of test cases your code passes correctly. Lowest score wins!

E.g. my code has 1 byte and I get 1 test case correct. My score is $$\1 / \frac{1}{118} = 118 \$$. Someone else writes some code with 110 bytes but gets all the test cases correct. The other person scores $$\ 110 / \frac{118}{118} = 110 \$$, meaning the other person has a better score than me.

# Test cases:

'Hydrogen' -> 'H'
'Helium' -> 'He'
'Lithium' -> 'Li'
'Beryllium' -> 'Be'
'Boron' -> 'B'
'Carbon' -> 'C'
'Nitrogen' -> 'N'
'Oxygen' -> 'O'
'Fluorine' -> 'F'
'Neon' -> 'Ne'
'Sodium' -> 'Na'
'Magnesium' -> 'Mg'
'Aluminium' -> 'Al'
'Silicon' -> 'Si'
'Phosphorus' -> 'P'
'Sulfur' -> 'S'
'Chlorine' -> 'Cl'
'Argon' -> 'Ar'
'Potassium' -> 'K'
'Calcium' -> 'Ca'
'Scandium' -> 'Sc'
'Titanium' -> 'Ti'
'Chromium' -> 'Cr'
'Manganese' -> 'Mn'
'Iron' -> 'Fe'
'Cobalt' -> 'Co'
'Nickel' -> 'Ni'
'Copper' -> 'Cu'
'Zinc' -> 'Zn'
'Gallium' -> 'Ga'
'Germanium' -> 'Ge'
'Arsenic' -> 'As'
'Selenium' -> 'Se'
'Bromine' -> 'Br'
'Krypton' -> 'Kr'
'Rubidium' -> 'Rb'
'Strontium' -> 'Sr'
'Yttrium' -> 'Y'
'Zirconium' -> 'Zr'
'Niobium' -> 'Nb'
'Molybdenum' -> 'Mo'
'Technetium' -> 'Tc'
'Ruthenium' -> 'Ru'
'Rhodium' -> 'Rh'
'Silver' -> 'Ag'
'Indium' -> 'In'
'Tin' -> 'Sn'
'Antimony' -> 'Sb'
'Tellurium' -> 'Te'
'Iodine' -> 'I'
'Xenon' -> 'Xe'
'Cesium' -> 'Cs'
'Barium' -> 'Ba'
'Lanthanum' -> 'La'
'Cerium' -> 'Ce'
'Praseodymium' -> 'Pr'
'Neodymium' -> 'Nd'
'Promethium' -> 'Pm'
'Samarium' -> 'Sm'
'Europium' -> 'Eu'
'Terbium' -> 'Tb'
'Dysprosium' -> 'Dy'
'Holmium' -> 'Ho'
'Erbium' -> 'Er'
'Thulium' -> 'Tm'
'Ytterbium' -> 'Yb'
'Lutetium' -> 'Lu'
'Hafnium' -> 'Hf'
'Tantalum' -> 'Ta'
'Tungsten' -> 'W'
'Rhenium' -> 'Re'
'Osmium' -> 'Os'
'Iridium' -> 'Ir'
'Platinum' -> 'Pt'
'Gold' -> 'Au'
'Mercury' -> 'Hg'
'Thallium' -> 'Tl'
'Bismuth' -> 'Bi'
'Polonium' -> 'Po'
'Astatine' -> 'At'
'Francium' -> 'Fr'
'Actinium' -> 'Ac'
'Thorium' -> 'Th'
'Protactinium' -> 'Pa'
'Uranium' -> 'U'
'Neptunium' -> 'Np'
'Plutonium' -> 'Pu'
'Americium' -> 'Am'
'Curium' -> 'Cm'
'Berkelium' -> 'Bk'
'Californium' -> 'Cf'
'Einsteinium' -> 'Es'
'Fermium' -> 'Fm'
'Mendelevium' -> 'Md'
'Nobelium' -> 'No'
'Lawrencium' -> 'Lr'
'Rutherfordium' -> 'Rf'
'Dubnium' -> 'Db'
'Seaborgium' -> 'Sg'
'Bohrium' -> 'Bh'
'Hassium' -> 'Hs'
'Meitnerium' -> 'Mt'
'Roentgenium' -> 'Rg'
'Copernicium' -> 'Cn'
'Nihonium' -> 'Nh'
'Flerovium' -> 'Fl'
'Moscovium' -> 'Mc'
'Livermorium' -> 'Lv'
'Tennessine' -> 'Ts'
'Oganesson' -> 'Og'

I used this code to shape the list into the test cases, might be useful to you.

Elements and symbols extracted from https://www.thoughtco.com/element-list-names-atomic-numbers-606529, visited at the 9th of February of 2020. At the time of writing, 118 elements were available. Source included "Aluminum" and "Aluminium" as alternatives, dropped "Aluminum" for the purposes of this challenge.

• This is just a massive lookup table. I don't like it. – S.S. Anne Feb 10 at 0:50
• "Aluminum" might be in the dictionaries of some languages. I suggest allowing either but and/or both in the solution and specifying which one was chosen. – S.S. Anne Feb 10 at 1:07
• Dupe – Jo King Feb 10 at 2:55
• @JoKing this being a code challenge doesn't make it different from the dupe you linked? – RGS Feb 10 at 7:02
• I think the problem you have is that your scoring strongly incentivises submitting H as the best answer. The other question had a shortest solution of 200 bytes - which can't get a better score than the 1 byte 1 answer program. – FryAmTheEggman Feb 10 at 19:52
• @FryAmTheEggman thanks for the feedback. Would you suggest tweaking the score or dropping this challenge? – RGS Feb 10 at 20:19
• You might consider a similar idea but with something that hasn't been done as much. I think it depends on how you feel, it certainly isn't unsalvageable. – FryAmTheEggman Feb 10 at 20:27
• @FryAmTheEggman very nice idea. I am thinking airport codes, country codes or military rank abbreviations... – RGS Feb 10 at 21:21
• I really like this idea, and I think allowing approximate answers makes it distinct enough from this that it's not a dupe. However, the current scoring encourages very inaccurate, trivial answers. The best I could find in 05AB1E is (05AB1E to output the first 2 characters of the input), which gets 45 test cases correct, scoring better than a 6 character answer that gets everything right. – Grimmy Feb 14 at 16:03
• @Grimmy thanks for your very detailed feedback! Like I mentioned above, I'm exploring the possibility of using something other than chemical symbols. In your opinion, would you tweak the scoring or would you suggest something other than chemical symbols? – RGS Feb 14 at 16:20

# Compress even permutations

## Factorial number system

Every nonnegative integer can be encoded by factorial number system. Factorial number system doesn't have a fixed radix, but uses the factorial of nonnegative integers as radices. The place of $$\0!\$$ can have $$\0\$$ as the digit, the place of $$\1!\$$ can have $$\0\$$ or $$\1\$$ as the digit, the place of $$\2!\$$ can have $$\0\$$, $$\1\$$, or $$\2\$$ as the digit, and so on. Subscript exclamation mark denotes that the integer is encoded in factorial number system. For example:

$$41010_! = 4 \times 4! + 1 \times 3! + 0 \times 2! + 1 \times 1! + 0 \times 0!= 103_{10}$$

## Permutations

Numbers encoded by factorial number system has one-to-one correspondence to permutations:

$$41010_! \cong \begin{pmatrix} 0 \\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 1 & 0 \\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix} 3 & 2 & 1 & 0 \\ 0 & 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} 4 & 3 & 2 & 1 & 0 \\ 3 & 2 & 1 & 0 & 4 \end{pmatrix} = \begin{pmatrix} 4 & 3 & 2 & 1 & 0 \\ 1 & 3 & 2 & 0 & 4 \end{pmatrix}$$

That is, each the place of $$\n!\$$ represents a right-direction rotation of $$\n\$$ through $$\0\$$, and its digit represents what $$\0\$$ becomes after the rotation.

It follows that the parity of the permutation is same as the parity of the sum of digits at the places of factorials of odd numbers. For $$\41010_!\$$, the permutation is even because $$\1 + 1 = 2\$$ is even.

## Objective

An even permutation of 5 elements, or any equivalent object will be the input. That leaves $$\5! \div 2 = 60\$$ distinct permutations. Compress it to $$\\lceil \log_2{60}\rceil=6\$$ bits.

## Rules

• Input type and format doesn't matter. Possible choices of input format include (in C++):

• std::map<int,int> scrambling the numbers from $$\4\$$ to $$\0\$$

• int[5] containing the digits of the factorial number system

• int[4] containing the digits of the factorial number system, except the place of $$\0!\$$ (it doesn't contribute to the permutation)

• int encoded by the factorial number system

• Output type and format doesn't matter either. It may be:

• bool[6]

• int8_t

• a bit field

• std::string containing ASCII digit 0s and 1s

• Invalid inputs fall in don't care situation.

# Simplify a directed graph

## Input

A connected directed graph, in any convenient format. A valid format (and probably the most convenient) would be a list of edges.

## Simplification

These two reductions are performed as often as possible. It does not suffice to apply one reduction first, then the other, as each of them can cause more of the other one.

Let $$\E\$$ be the set of all edges of the graph in the following explanations. For example $$\E = \{(0, 1), (1, 2), (2, 0)\}\$$ represents the following graph:

0 ----> 1
^       |
'-- 2 <-'

Let $$\E_a\$$, where $$\a\$$ is a vertex, indicate the set of edges involving $$\a\$$:

$$E_a := \{(u, v) \in E : u = a \vee v = a\}$$

Two graphs are isomorphic iff all vertices from one graph can be relabeled to make it equal to the other graph.

### Deduplication

If there are disjoint sets of vertices $$\M\$$ and $$\N\$$ ($$\M \cap N = \emptyset\$$) not directly connected ($$\((M, N) \cup (N, M)) \cap E = \emptyset\$$) and the graphs with edges $$\E \setminus E_M\$$ and $$\E \setminus E_N\$$ are isomorphic, then the vertices $$\M\$$ (or $$\N\$$, but not both, are removed).

This means that of any two non-overlapping, non-directly-connected subgraphs that leave isomorphic graphs behind when removed, one is removed.

Simple example:

b --> 1
^  ==>  b --> 1
|
a

### Relay removal

If there are vertices $$\a\$$ and $$\b\$$ and a non-empty set of vertices $$\N\$$, such that $$\b \notin N\$$ and $$\E_b = \{(a, b)\} \cup (b, N)\$$, then $$\b\$$ is removed from the graph and edges $$\(a, N)\$$ are added if they do not alreay exist.

This means that a vertex with exactly one edge ending there, at least one edge starting there and no self-loop can be removed, moving the starting edges to the start of the ending edge.

Simple example:

1             1
^             ^
|             |
O --> a --> b  ==>  O --> a ----.
^     |             ^     |
|     v             |     v
'---- 2             '---- 2

## Output

The output is the result of the reductions, in any convenient format. If identifiers are used for edges or vertices, as in a list of edges, these identifiers are not required to correspond to identifiers in the input.

A different format may be used for input and output.

## Scoring

This is : Lowest bytecount in each language wins. No answer will be accepted, as there is no overall winner.

## Other rules

• Standard loopholes apply
• Functions or programs
• Any input and output methods (STDIN, arguments, prompt(), ...)

not yet

## Meta:

• Everything clear?
• Better explanations?
• Better title?
• I'm not sure whether output might depend on the order in which reductions are performed.

# Let's Play Countdown!

(The Numbers round this time)

Countdown is a British TV game show composed of three different styles of rounds; the letters round, the conundrum, and the numbers round.

The conundrum could be solved with the same program you'd make for the Letters round, so let's tackle the third option that hasn't been done yet!

## Challenge

Take in a set of numbers. One of which is the "Target" number, and the rest are the building numbers.

The Countdown Number Round asks you to take the building numbers and to construct the Target number only using the four elementary operators. Every step must result in another strictly-positive integer (so non-perfect division is disallowed). Output the method to which you can construct the target number. If that's impossible, get as close as possible (above or below are scored the same). Numbers do not need to be used, but may NOT be reused.

Note - You will perform at most N-1 operations, where N is the number of building numbers. Every elementary operator takes in two inputs and provides one, so you "lose" one from your ranking every operation. That should give you an idea of the size of your output.

I/O is in any reasonable format, but target vs building numbers must be obviously distinct (either by the target being the first or the last number, or outside an array, or a different type, etc).

Output needs to explain exactly what operations are being performed on what numbers, and what the output for each operation would be, but can be done in whichever way seems reasonable.

## Example I/O

In these examples, the first element is the target.

[888 100 2 75 3 1 10]
75-1=74
10+2=12
74x12=888

[766 22 10 8 3 1]
22+10=32
32x8=256
256-1=255
255x3=765 #You cannot get this one exactly, but one-off is close.

## Sandbox Questions

I'm... like 85% sure this isn't anywhere here yet. I did a chunk of searching and couldn't find anything that fit the bill, so I think this is clear?

Any other neat examples you guys got?

• As you say, I don't think we have exactly this challenge, but there are many similar ones. I doubt you will get anything besides brute forcing all possible arrangements then sorting them by nearness to the target. Separately, if there are multiple tying solutions you don't require that e.g. the shortest be output, correct? It is probably worth mentioning that. – FryAmTheEggman Feb 21 at 21:24