Sandbox for Proposed Challenges

Question

This "sandbox" is a place where Code Golf users can get feedback on prospective challenges they wish to post to main. 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.

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Answer 1

Pattern in Prime

Given a positive integer less than 100 with no preceding zeros we have to find all 12 digit primes that contains the maximum number of time the given pattern.

Note: Pattern's occurance can overlap. For example 11 occur 6 times in 1111111

Example:

pattern = 1 , count = 13 the maximum occurance is in 111111110111, eleven, and those 12 digit primes are {101111111111,111011111111,111111011111,111111110111,111111111101,111111111511,111111111611,111111211111,111113111111,111211111111,111511111111,116111111111,311111111111}

pattern = 22 , count = 1 {522222222229}

pattern = 69, count = 162

pattern = 37, count = 151

Output format: first line should be count of such primes, and then print the primes in sorted order

If possible while answering please write a short explanation of what you did. This is code-golf, so the shortest answer in bytes per language wins.

code-golf number prime pattern-matching

Answer 2

Best Rotation

code-challenge If we rotate a raster image by some angle that is not a multiple of \$90°\$, we will have to use some kind of interpolation. Depending on what kind of interpolation method we use, we get a better or a worse quality. This is especially apparent if we repeat the rotation multiple times. In the following image we se the original on the top left, and then the result of different interpolation methods when used to rotate the original \$360\$ times by \$1°\$.

\$\renewcommand\phi\varphi\$

Challenge

Given some angle \$\phi\$ Your task is to find a method to find a (deterministic) function \$f_\phi\$ that takes some image \$I\$ and rotates it by \$\phi\$.

Scoring (not solved yet)

META this is the issue. Ideally we'd like to have some objective criterion. But so far I have not managed to find one that cannot be gamed:

1. If we just let the participants rotate an image by some angle \$\phi\$ how can we then compare it to the "exact" solution? Maybe their method is better than the best standard method, even if they have to apply it repeatedly. So it seems we can only truly check against a total rotation that is a multiple of \$90°\$.

2. Let's say we say we use an angle of \$\phi = 2\pi/n\$ and let them apply their "rotation" \$n\$ times to measure e.g. the \$l^2\$-error \$E = \Vert I - f_\phi(\ldots f_\phi(I)\ldots)\Vert_2\$ . This means they could just use \$f_\phi = id\$.

3. If we instead use a total rotation of e.g. \$\pi/2\$ (that is \$\phi = \frac{\pi}{2n}\$, with \$n\$ rotations). Then the participants could just use some permutation of pixels that just happens to rotate the image after \$n\$ rotations exactly (or maybe exactly except for some pixels).

So does anyone have another suggestion of how to objectively measure it without introducing any hand-wavey rules?

Answer 3

code-golf

Halve Code Regen

TODO: This is a horrible title.

Your challenge is to write a program, and when I half the program, the output must stay the same. Then, I will halve it and add the last two characters, and the output should be the same with the program's last two characters at the end.

Sometimes the code can't be halved evenly, and when that happens, I will do something similar to floor division:

blahy => bl
horse => ho
meddle => med (regular halving)
oof => o

Example for "the output should be the same with the program's last two characters at the end"

Let's say we have this code:

q|w_a2e(o+2ei2ere

and the output is

q|w_a2e(o+2ei2ere

when I change it to

q|w_a2e(o+2ei2erere

by adding the last two characters (re), the output should look like

q|w_a2e(o+2ei2erere

(Note the extra re at the)

Rules

• No padding with comments
• The program must have at least 2 characters.
• Standard loopholes apply
• The output cannot be empty.

Scoring

This is code-golf, so the answer with the least bytes wins.

How I came up with the random code in the example two sections ago

q means quine. | is a separator. w_a2e means when the last two characters are added to the end. (o+2e means add the last two characters to the end. i2ere means ignore a repeated 2e at the end.

Meta

• Is this even possible?
• If so, any other suggestions?

Answer 4

Integer points on Hyperspheres

Given two non-negative integers \$r\$ and \$d\$, write a program that will output the number of integer points that lie on all hyperspheres of dimension \$d\$ or lower with radius \$r\$ or lower.

For example, given \$r = 5, d = 2\$, the program would do something like this:

The hypersphere of dimension 2 is a circle. The radius is 5. There are 12 integer points on this circle, as follows: $$(5,0),(4,3),(3,4),(0,5),(-3,4),(-4,3),(-5,0),(-4,-3),(-3,-4),(0,-5),(3,-4),(4,-3)$$

If the radius is 4, there are 4 integer points: $$(4,0),(0,4),(-4,0),(0,-4)$$

If the radius is 3, again it's 4 integer points. If the radius is 2, again 4. Radius 1 - again, 4. Radius 0 has only 1 integer point. If we add all these up it's 12 + 4 + 4 + 4 + 4 + 1 , or 29 points total for the 2 dimensional hyperspheres with radius \$<=5\$.

Now if we take the dimension 1 hypersphere, it is interpreted as 2 points in a one dimensional space, distance r from the origin. For example the 1 dimensional hypersphere with radius 5 has only two points, $$(5),(-5)$$ and their coordinates are only single dimensional. So, the dimension 1 hypersphere with radius 5 has two integer points. The dimension 1 hypersphere with radius 4 also has 2 integer points, and so on. The total integer points on all 1 dimensional hypspheres with radius 0 to 5 is 2 + 2 + 2 + 2 + 2 + 1, or 11.

The dimension 0 hypersphere is defined for this challenge to have 0 points.

So the final output of a program with input \$r = 5, d = 2\$ will be 29 + 11 + 0, or 40.

• You may assume that r and d will be low enough so that the integer type of your language will be large enough so that it can represent the answer. In other words if your language has only 16 bit integers, then you can write a program that assumes r and d will be low enough so that the final answer is between 0 and 65535.

• This is Code Golf, lowest # of bytes wins

Answer 5

Same String Regardless of Repetition

Challenge

Write a full program that prints a string that has a length of at least 1. When the source code is repeated any amount of times, that string should still be outputted. For example, if my source code is ABC and it prints Hello, World!, then ABCABC will still print Hello, World! and ABCABCABC will also print Hello, World! and etc.

Rules

• It has to be a full program that takes no input (or have an unused input if this is impossible) and prints the string to STDOUT.
• The program and the string have to be at least 1 byte long.
• The strings that are outputted with each repetition have to be exactly the same. Trailing or leading spaces make the strings different.
• There must be only one output.
• There is no code between repetitions.
• Comments in your code are not allowed.
• This is code-golf, so programs are scored in bytes, with less bytes being better.

Answer 6

In 1960 Andrey Kolmogorov conjectured that any algorithm to multiply two integers would require \$\Omega(n^2)\$^* steps.

Within a week of presenting this conjecture it was proven false by graduate student Anatoly Karatsuba. Karatsuba's algorithm is an algorithm to multiply two numbers which takes in the worst case \$O(n^{\log_2 3})\$ time.

Your task will be to implement this algorithm. You will receive as input two binary strings. These can be lists of ints, arrays of bools etc.

You should then output the binary representation of their product in the same format.

It's hard to say in precise terms what it means to implement a particular algorithm. So here we will require only that your algorithm have asymptotic time complexity of \$O(n^{\log_2 3})\$ and \$\Omega(n^{\log_2 3})\$ where \$n\$ is the total number of bits in the input. Which is to say that it is neither better nor worse than Karatsuba's algorithm up to a constant factor.

This is code-golf so answers will be scored in bytes with lower bytes being the goal.

---

^{* In this challenge we will use \$f\in \Omega(g)\$ to mean that \$\displaystyle\limsup_{x\to \infty}\dfrac{f(x)}{g(x)}\$ does not converge to \$0\$. Also stated as \$f \in \Omega(g) \iff \exists k.\forall x.\exists x'.f(x')\geq k\cdot g(x')\$. This is the Hardy-Littlewood definition.}

Answer 7

Solve linear equations over the integers

All variables in this question are integer valued.

Input

4 integers w, x, y, z. They can be positive or negative and will be less than 1048576 in absolute value.

Output

The general solution to the equation.

\$ aw+bx+cy+dz = 0 \$.

The variables \$a, b, c, d\$ must all be integer values.

Output format

Your output should consist of three tuples each with four parts, one for each of the values a, b, c, d. Let me explain by example:

Input: -118, 989, 918, -512

Answer: b = 2 n_0 
        c = 9 n_0 + 256 n_1 + 81 a
        d = 20 n_0 + 459 n_1 + 145 a

Explanation: n_0 and n_1 are integers that you can set to anything you like. The solution says: a can also be set to any integer value, b must be twice whatever you set n_0 to. This means that a can be set to any integer, c can now be calculated in terms of three variables we have already set and so can d.

The format of your output should be 3 tuples (#,#,#,#), (#,#,#,#), (#,#,#,#). We can assume three free integer variables n0, n1 and n2 and so (a,b,c,d) = (#,#,#,#)n0 + (#,#,#,#)n1 + (#,#,#,#)n2. In the example above the output would therefore be:

Output: (0, 2, 9, 20), (0, 0, 256, 459), (1, 0, 81, 145)

Examples

Example one:

 Input: -6, 3, 7, 8

 Answer:  c = 2a + 3b + 8n
          d = -a - 3b - 7n 
          n is any integer

Output: (1, 0, 2, -1), (0, 1, 3, -3), (0, 0, 8, -7)

Example two:

Input: -116, 60, 897, 578

Answer: c = 578 n + 158 a + 576 b
        d = -897 n - 245 a - 894 b 
        n is any integer

Output: (1, 0, 158, -245), (0, 1, 576, -894), (0, 0, 578, -897)

Example three:

Input: 159, -736, -845, -96

Output: (1, 0, 27, -236), (0, 1, 64, -571), (0, 0, 96, -845)

Discussion

To understand this challenge further it is worth looking at this possible general solution which does not work [(z, 0, 0, -w), (0, z, 0, -x), (0, 0, z, -y)]. The problem with this is that there are solutions to the problem instances above which are not the sum of any integer multiples of those tuples. For example: take input -6, 3, 7, 8 from Example 1. The proposed solution would give the tuples:

(8, 0, 0, 6), (0, 8, 0, -3), (0, 0, 8, -7)

Why doesn't this work?

There is a solution for this instance with a = 1, b = 1, c = 13, d = -11 because -6+3+7*13-11*8 = 0. However there are no integers n_0, n_1, n_2 to make n_0 * (8, 0, 0, 6) + n_1 * (0, 8, 0, -3) + n_2 * (0, 0, 8, -7) = (1, 1, 13, -11) .

Answer 8

Is someone eavesdropping? (WIP)

Alice and Bob, who are quantum physicists, are being watched by Eve, a quantum FBI agent. Eve has quantum tunneled underneath Bob's house and is tapping all his quantum channels. Luckily, Alice and Bob are using the BB84 protocol to exchange quantum keys to encode their quantum messages. Your job is to write a program/function to help Alice and Bob determine if Eve is evesdropping.

Answer 9

Posted here

Answer 10

posted here

Answer 11

Implement every dirname (1p)

code-golf string

Implement the dirname utility from scratch. It can be either a program or a fucntion. Assume input string satisfies these constraints:

• it contains portable filename character set only, and
• its length is up to 256 bytes, including trailing a NULL character (as in _POSIX_PATH_MAX from limits.h).

The dirname utility, however, has two kinds of implementations. This is because some POSIX systems treat //foo/bar differently from /foo/bar. So in this challenge, you must output every possible outputs, in any order. They can be duplicated.

Here is the algorithm to implement the utility, provided string to be input:

1. If string is //, skip steps 2 to 5.
2. If string consists entirely of <slash> characters, string shall be set to a single <slash> character. In this case, skip steps 3 to 8.
3. If there are any trailing <slash> characters in string, they shall be removed.
4. If there are no <slash> characters remaining in string, string shall be set to a single <period> character. In this case, skip steps 5 to 8.
5. If there are any trailing non- <slash> characters in string, they shall be removed.
6. If the remaining string is //, it is implementation-defined whether steps 7 and 8 are skipped or processed.
7. If there are any trailing <slash> characters in string, they shall be removed.
8. If the remaining string is empty, string shall be set to a single <slash> character.

The final string is the output.

Standard I/O rules apply. Standard Loopholes apply. No builtins or libraries that does exactly same functionality. Shortest code wins.

Examples

Some examples are taken from POSIX explaination of basename().

* means an empty string. 1st column is input and 2nd and 3rd are possible outputs.

usr               .
usr/              .
*                 .
..                .
../               .
/                 /
//                /          //
///               /
/usr/             /
//usr/            /          //
///usr/           /
/usr/lib          /usr
//usr//lib//      //usr
/home//dwc//test  /home//dwc

Meta

• ?

Answer 12

Write a Stack Exchange compliant brainfuck explainer

Answer 13

Escape the maze

code-golf maze

Introduction

Here is a random maze:

#####
#M.#E
#.##.
#.##.
#....
#####

Here M is the starting point and E is the endpoint. # is a maze wall and . is a path.

Now we can get out of this maze by following the sequence sssdddwww. (s is dow, d is right, w is up, a is left.)

Your challenge

Given a maze, output the shortest possible route to the endpoint. (E) You may assume the maze is solvable.

Test cases

#####
#M.#E
#.##.
#.##.
#....
#####

outputs

#####
#..#E
#M##.
#.##.
#....
#####

#####
#..#E
#.##.
#M##.
#....
#####

#####
#..#E
#.##.
#.##.
#M...
#####

#####
#..#E
#.##.
#.##.
#.M..
#####

#####
#..#E
#.##.
#.##.
#..M.
#####

#####
#..#E
#.##.
#.##.
#...M
#####

#####
#..#E
#.##.
#.##M
#....
#####

#####
#..#E
#.##M
#.##.
#....
#####

#####
#..#M
#.##.
#.##.
#....
#####

(Notice the newlines between the steps.)

Scoring

This is code-golf, so shortest code wins.

---

Todo

• Add more test cases
• Maybe some custom scoring or atomic-code-golf

Answer 14

Solve the halting problem for Minyrinth code-golf decision-problem interpreter halting-problem

Introduction

Minyrinth is the stripped-down version of Labyrinth. It has the same routing semantics as Labyrinth but only four non-wall commands, and only one register (that can hold unbounded signed integers) instead of two stacks. The register is initialized to zero.

Commands

• " is a no-op path.
• @ halts the program.
• ) increments the register.
• ( decrements the register.

You may assume that the input only contains the four command characters plus spaces (non-path) and newlines (used for 2D layout).

Execution semantics

Copied directly from the README, with some parts edited for Minyrinth:

The source code consists of single-character instructions and is interpreted as a 2D grid. The instruction pointer starts at the first non-whitespace character in the file (in reading order) going right.

Labyrinth is interpreted in a simple loop. At each step, the command under the instruction pointer is executed, then the new movement direction is determined, and then the instruction pointer moves one cell in that direction. The edges of the grid are not connected.

The instruction pointer will generally follow "corridors" of instructions. Junctions can be used for non-trivial control flow. How the new movement direction is determined depends on the number of available steps (i.e. number of direct neighbours with known commands):

• 4 neighbours: The top of the main stack is examined. If it's 0, keep moving straight ahead. If it's negative, turn left. If it's positive, turn right.

• 3 neighbours: Do the same as for four neighbours, but if you hit the wall, reverse the direction. Hence, a T-junction hit from the side differentiates between 0 and non-zero. A T-junction hit from the bottom on the other hand sends negative/positive to the left/right whereas a 0 value reverses the direction.

• 2 neighbours: The first rule here is, don't turn around. So if you came from one of the two directions, continue in the other direction. If this is not the case, but one of the two directions is straight ahead, follow that one (this can happen, for instance, at the start of the program in a corner).

• 1 neighbour: Go towards the only available direction. Usually, this means you have hit a dead end and turn around on the spot (executing the command you turn around on only once).

• 0 neighbours: Remain where you are without changing your direction. This can occur at the very start of the program.

Challenge

Solve the halting problem for Minyrinth. Unlike Labyrinth which is Turing-complete, Minyrinth simulates a specific case of a pushdown automaton whose halting problem is decidable.

For output, you can choose to

• output truthy/falsy using your language's convention (swapping is allowed), or
• use two distinct, fixed values to represent true (affirmative) or false (negative) respectively.

Standard code-golf rules apply. The shortest code in bytes wins.

Some hints can be found in CS Stack Exchange: Decidability of halting problem for DPDAs with \$\epsilon\$-transitions?, Counter Machine (Halting Problem)

Answer 15

Plan my factory

"Factory games", such as Factorio, Mindustry, and Satisfactory are my favourite genre of video games, and the way I play them involve a lot of ratio calculations to make sure the factory is running as efficiently as possible.

Given a list of recipe specifications, and a desired production rate of an end product, output the necessary input rates of raw materials, where "raw material" is any ingredient that does not have a provided recipe.

For example:

Recipes:

[
    {
        name: "Iron Plate",
        time: 3.2,
        quantity: 1,
        input: [{
            name: "Iron Ore",
            quantity: 1
        }]
    },
    {
        name: "Copper Plate",
        time: 3.2,
        quantity: 1,
        input: [{
            name: "Copper Ore",
            quantity: 1
        }]
    },
    {
        name: "Iron Gear",
        time: 0.5,
        quantity: 1,
        input: [{
            name: "Iron Plate",
            quantity: 2
        }]
    },
    {
        name: "Automation Science Pack",
        time: 5,
        quantity: 1,
        input: [
            {
                name: "Iron Gear",
                quantity: 1
            },
            {
                name: "Copper Plate",
                quantity: 1
            }
        ]
    },
]

Desired production rate:

{
    name: "Automation Science Pack",
    rate: 60
}

Expected output:

[
    {
        name: "Iron Ore",
        rate: 120
    },
    {
        name: "Copper Ore",
        rate: 60
    }
]

Both input and output may be any standard IO format that represents item names as strings and quantities/rates as integers.

Testcases

TBA

Sandbox

Is this a chameleon challenge?

Answer 16

Golf a Friedman's tree sequence for 3 colors

code-golf math busy-beaver

Friedman proved that given a sequence of 3-labelled rooted trees with following conditions, TREE sequence, must be finite and defined maximum length of such sequence as TREE(3).

• \$i\$th tree, \$T_i\$, should have no more than \$i\$ vertices. (index is 1-based)
• No tree \$T_i\$ can be homeomorphically embeddable into any of the following tree \$T_j\$ if \$i < j\$.
  • A tree \$A\$ is said to be homeomorphically embeddable to tree \$B\$ if and only if there exists a function \$f\$ from vertices of \$A\$ to vertices of \$B\$ which satisfies following conditions:
    1. \$f\$ preserves label. That is, \$v\$ and \$f(v)\$ has same label.
    2. If \$u\$ is ancestor of \$v\$, then \$f(u)\$ is ancestor of \$f(v)\$.
    3. If \$u_1\$ and \$u_2\$ are children of \$v\$, then the path from \$f(u_1)\$ to \$f(u_2)\$ contains \$f(v)\$.
  • There is a guide with examples in this youtube video.

It is well known that TREE(3) is far larger than \$2^{64}\$.

Challenge

Find a TREE sequence with length of \$2^{64}\$ or longer. Your task is to implement a program or a function or a subroutine that takes one 64-bit integer \$i\$ and outputs a tree \$T_i\$ of your TREE sequence.

Rules

• Standard loopholes apply.
• Standard I/O rules apply.
• Shortest code wins.
• Output can be any format as long as it describes a 3-labelled tree.
• Describe how your outputs form valid TREE sequence.

Example of valid sequence, length 16

1 - {}              # 1 vertex tree with label {}. Note that any tree after this cannot contain node with label {}.
2 - [[]]            # 2 vertex tree with label both []. Any tree after this cannot contain two ancestor-child node with both label [].
3 - [(())]
4 - ([][])          # Root with label (), 2 children with label [].
5 - ([]((())))
6 - ([](()))
7 - [()]
8 - []
9 - ((((((((()))))))))
10 - (((((((())))))))
11 - ((((((()))))))
12 - (((((())))))
13 - ((((()))))
14 - (((())))
15 - ((()))
16 - (())

Meta

This is my first challenge. Suggestions?

I'm sure that normal computers will have serious difficulty to execute code for this problem, but also I don't want answers to just brute force TREE sequence and print ith term. How can I balance the question?

Answer 17

OEIS A049190

your task is to print all numbers of this sequence.

is separated by newline, is separated by space or separated by comma.

the first few numbers are:

1, 3, 5, 59, 245, 2491, 235253, 127756731, 330567489269, 258479716298484155, 36823182192123209878050549, 25576412117054296344209353299113896379, 10994511204169842163496446583221775727830456269734123253

How to get numbers of this sequence:

this is look and say but in binary and convert to decimal

Meta

• How to clarify this?

Answer 18

Convert codepoint to UTF-1

Answer 19

Infinite ordinals from a well-ordering

Answer 20

Baloney Sandwich KOTH

Tags: [king-of-the-hill], [python]

Introduction

In this King Of The Hill Challenge, your bots will be playing Baloney Sandwich.

How Baloney Sandwich is played

1. First, the entire deck is dealt equally to all players (EVERY BOT PLAYS; if there is more than 26 bots, it will be 2 decks mixed together).
2. Then, the player with the ace of spades goes first.
3. They play all of their aces
   • However! They could opt to "cheat" and instead play cards that are not aces, or mix and match aces with other cards.
4. Every other player has their call_bs (bs = Baloney Sandwich) method called after this play, in turn until a player returns True; if a player returns True then the cards played are "revealed".
   • Every player has their bs_call_outcome method to "reveal" the cards
   • If no player returns True but the player was lying, the pb method is called (pb = Peanut Butter)
5. Play continues with twos, threes, etc.; after kings, start over at aces

Creating A Player

Your player should be in this format:

class ExamplePlayer(AbstractBSPlayer):
    def __init__(self, hand: list[int]):
        super().__init__(hand)
        # Any necessary initialization goes here

    def call_bs(self, player_name: str, card_rank: int, number_of_cards: int) -> bool:
        """
        player_name is the name of the bot who just played
        The number of cards they played is number_of_cards
        The claimed rank is card_rank (1 is Ace; 11-13 are face cards)
        """
        # Magic decision maker
        return False # never calls "Baloney Sandwich"
        # return True would always call "Baloney Sandwich"
    
    def bs_call_outcome(self, player_name: str, caller_name: str, card_rank: int, cards: list[int], cheated: bool):
        """
        player_name is the name of the bot who just played
        caller_name is the name of the bot who called "Baloney Sandwich"
        The claimed rank is card_rank
        The cards they played is cards
        cheated is True if the player cheated
        """
        pass  # Couldn't care less
    
    def pb(self, player_name: str, card_rank: int, number_of_cards: int):
        """
        player_name is the name of the bot who just played
        The number of cards they played is number_of_cards
        The claimed rank is card_rank (1 is Ace; 11-13 are face cards)
        """
        pass # Couldn't care less

    def play_cards(self, card_rank: int) -> list[int]:
        """
        The rank you will claim to play is card_rank
        """
        # Magic decision maker
        return self.hand[0] # Always plays the first thing in hand

A player has access to the following instance variables:

• self.store: A dictionary that is empty by default. The only instance variable allowed to be written to.
• self.hand: A list of ints, each representing a card (1 is Ace, 11-13 are face cards)

Testing Your Player

The controller can be found at https://github.com/sethpeace/bs-koth. I apologize in advance that it is janky beyond comprehension.

Rules

Runtime Disqualifications

• You must play at least one card every turn
• You can't play cards you dont have in your hand
• You can't take longer than one second to return from a function

Pre-runtime Disqualifications

• No reading or writing to controller, runtime, or other submissions (bots can't read them; you can if you feel like it)
• Only write to self.store instance variable
  • Other variables inside function scope are of course OK
• Don't design a bot to defend or support specific bots
• Bots can't use the same strategy as another bot
• Standard loopholes apply

Answer 21

Implement xorshift128+

I don't have time to finish writing this draft at the moment.

Might also consider something like xoroshiro if it's more interesting to implement, but since xorshift128+ is so common it seems like a clearer choice.

(Possible idea: Given initial seed, determine number of iterations it'll take to get a certain random number with xorshift128+. It's not a CSPRNG, so there might be something more interesting than brute force.)

Answer 22

Radiate a prime checker into a sum program

Write a program or function that takes a number n and returns/print truthy if it is prime, and falsy otherwise. Additionally, if you remove 1 character of the source code, the program must now take a list of one or more numbers and return/print the sum. This must work for at least one character, and your score is however many characters that this does /not/ work for. Lowest score wins.

meta:

let me know if i need to clarify anything etc also: would it make sense to allow the programs to be the other way around (i.e. a sum checker where removing 1 character makes it a prime checker instead)? Just in case it could lead to a more interesting golf

Answer 23

Quickly pour liquid nitrogen

I have a container of liquid nitrogen, but nothing to measure it with. I do have some containers, whose volumes I know, and I know how much I start with. My goal is to fill any container with a given amount of liquid nitrogen, with the fewest possible pours.

Importantly, liquid nitrogen boils away as I pour it. Every time I pour liquid nitrogen into a container, a milliliter is lost.

Pouring mechanics:

If I have two containers, a and b, and I pour a into b:

• If both are empty, nothing changes
• If b's unused volume is greater than or equal to the volume of liquid nitrogen in a, a is emptied into b, and a milliliter boils away from b
• Otherwise, as much as possible is poured from a into b, and an additional milliliter is taken from a (so b will still be full)

If I have three containers, x, y, and z, they all have volumes of 25, and they contain 25, 10, and 18 milliliters of liquid nitrogen respectively, here are some example pours (one after another, not indiviudual):

1. x into y: y is filled, and x is left with 9 milliliters
2. y into z: z is filled, and y is left with 1 milliliter
3. z into x: x remains full, z is left with 24 milliliters
4. y into z: z remains at 24 milliliters, y is emptied

(Note that in your input, not all containers will have the same volume)

Task:

Given a list of container sizes, where the first container is initiallyfull and the amount of liquid nitrogen you must measure out, return the minimum number of pours that would be needed to leave that much in any container.

You may assume volumes are non-negative integers, and the initial amount of liquid nitrogen will never be less than the amount you should end with. The actual sequence of pours you use doesn't matter, as long as you output the minimum possible count.

Test cases:

Containers          Target      Output

7, 4                3           5
12, 3, 3            4           2
100, 10, 8          84          6
8, 8                5           3
10, 0               8           2
100, 20, 20, 17     73          7

Other:

I don't think this is a good challenge after doing the test cases

Answer 24

Chunk Sort a Sequence

Answer 25

Tinla Resqua (Latin Square of rotated words)!

Based on this (currently closed) question on Puzzling, your task is to take a single word generate a valid 'latin square', not by permuting the letters in a word, but by rotating the word.

E.g. if one row was CENTURY then ENTURYC is a valid row, but ENTCTURY is not.

Also unlike some Latin Squares, no diagonal may contain repeated letters either.

1	2	3	4	5	6	7
C	E	N	T	U	R	Y
R	Y	C	E	N	T	U
T	U	R	Y	C	E	N
E	N	T	U	R	Y	C
Y	C	E	N	T	U	R
U	R	Y	C	E	N	T
N	T	U	R	Y	C	E

Is a valid grid.

latin-square code-golf

Answer 26

This was heavily inspired by the donut.c program

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Challenge:

You have to come up with a program that prints (in console) a shape that looks exactly like the shape of your code.

For example, if your code's shape looks like this:

    2343
3kf5
     3o*/f
   3%kt^7

then the output's shape should look like this:

    9I8b
&yt^
     )ph67
   {}fgh]

(the byte don't need to match, just the shape)

Rules

• Empty programs/whitespace-only programs are not allowed.

• Your program should have more than 2 lines/statements of code.

• Each line should not be identical to every other line.

• All whitespace characters (space, tab,...) don't have to be identical.

• Whitespaces characters don't count towards the minimum bytes count.

• Your program's source code and the output should not be identical. It should be at least has one different character.

• The output should have at least two different bytes on one line.

• You should not use comments.

Scoring criteria

• Longest code (in bytes) wins!

• If two same programs have the same byte count, the one with the most unique characters in the output wins!

• If two same programs have the same unique characters, the one with the most different characters in output compared to source code wins!

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Meta

• Does this question have any duplicates?

• My wording is bad, can you improve it?

• Is this question interesting? Fit within the rules of the site?

Answer 27

Get all binary between a binary number and its reverse

Answer 28

Challenge

You should write a program that has one in (your code in bytes) chance to output a 4 (could be a function that returns a 4). So if your code has 10 bytes, your code should have 1 in an 10 chance to print out a 4. Otherwise, output "finding my dice".

Rules

• You can use whitespaces, comments or any unnecessary code to get to the number of bytes that you want.
• You shouldn't use any loopholes.

Winning criteria

This is code-golf, so shortest code in bytes wín!

Answer 29

Find duplicates, quickly

In this challenge, you'll be given a list of positive integers or non-empty alphabetical strings (your choice), and must return a list containing all duplicates. You must do this with the minimum time complexity you can, and as few bytes as possible.

Task:

Given a list of either positive integers or non-empty alphabetical (lower or uppercase, your choice) strings, return a list containing all duplicates. You may contain a duplicate value any number of times, so for the input [a, b, b, c, d, d, d, d], any of [b, d], [b, d, d, d], [b, b, d, d, d, d], or even [b, b, b, b, d] would be fine.

Scoring:

Scoring is by both average time complexity and bytes. A better time complexity will always beat an answer with a lower one, and byte count is a tie breaker. E.g., if there are two \$O(n^2)\$ answers, at 10 and 20 bytes, and an \$O(n\log{n})\$ answer at 140 bytes, it would rank higher than either of the others (until either a shorter \$n\log{n}\$ answer or an answer faster than \$n\log{n}\$ showed up).

As in code-golf, this is per-language, so Vyxal and Vim, Charcoal and C, or Japt and Java are not competing.

Answer 30

Check if the string is reshaped.