Valid Badminton Score?

Introduction:

I saw there was only one other badminton related challenge right now. Since I play badminton myself (for the past 13 years now), I figured I'd add some badminton-related challenges. Here the first one:

Challenge:

Input: Two integers
Output: One of three distinct and unique outputs of your own choice. One indicating that the input is a valid badminton score AND the set has ended with a winner; one indicating that the input is a valid badminton score AND the set is still in play; one indicating the input is not a valid badminton score.

With badminton, both (pairs of) players start with 0 points, and you stop when one of the two (pairs of) players has reached a score of 21, with at least 2 points difference, up to a maximum of 30-29.

So these are all possible input-pairs (in either order) indicating it's a valid badminton score AND the set has ended:

[[0,21],[1,21],[2,21],[3,21],[4,21],[5,21],[6,21],[7,21],[8,21],[9,21],[10,21],[11,21],[12,21],[13,21],[14,21],[15,21],[16,21],[17,21],[18,21],[19,21],[20,22],[21,23],[22,24],[23,25],[24,26],[25,27],[26,28],[27,29],[28,30],[29,30]]

And these are all possible input-pairs (in either order) indicating it's a valid badminton score BUT the set is still in play:

[[0,0],[0,1],[0,2],[0,3],[0,4],[0,5],[0,6],[0,7],[0,8],[0,9],[0,10],[0,11],[0,12],[0,13],[0,14],[0,15],[0,16],[0,17],[0,18],[0,19],[0,20],[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[1,7],[1,8],[1,9],[1,10],[1,11],[1,12],[1,13],[1,14],[1,15],[1,16],[1,17],[1,18],[1,19],[1,20],[2,2],[2,3],[2,4],[2,5],[2,6],[2,7],[2,8],[2,9],[2,10],[2,11],[2,12],[2,13],[2,14],[2,15],[2,16],[2,17],[2,18],[2,19],[2,20],[3,3],[3,4],[3,5],[3,6],[3,7],[3,8],[3,9],[3,10],[3,11],[3,12],[3,13],[3,14],[3,15],[3,16],[3,17],[3,18],[3,19],[3,20],[4,4],[4,5],[4,6],[4,7],[4,8],[4,9],[4,10],[4,11],[4,12],[4,13],[4,14],[4,15],[4,16],[4,17],[4,18],[4,19],[4,20],[5,5],[5,6],[5,7],[5,8],[5,9],[5,10],[5,11],[5,12],[5,13],[5,14],[5,15],[5,16],[5,17],[5,18],[5,19],[5,20],[6,6],[6,7],[6,8],[6,9],[6,10],[6,11],[6,12],[6,13],[6,14],[6,15],[6,16],[6,17],[6,18],[6,19],[6,20],[7,7],[7,8],[7,9],[7,10],[7,11],[7,12],[7,13],[7,14],[7,15],[7,16],[7,17],[7,18],[7,19],[7,20],[8,8],[8,9],[8,10],[8,11],[8,12],[8,13],[8,14],[8,15],[8,16],[8,17],[8,18],[8,19],[8,20],[9,9],[9,10],[9,11],[9,12],[9,13],[9,14],[9,15],[9,16],[9,17],[9,18],[9,19],[9,20],[10,10],[10,11],[10,12],[10,13],[10,14],[10,15],[10,16],[10,17],[10,18],[10,19],[10,20],[11,11],[11,12],[11,13],[11,14],[11,15],[11,16],[11,17],[11,18],[11,19],[11,20],[12,12],[12,13],[12,14],[12,15],[12,16],[12,17],[12,18],[12,19],[12,20],[13,13],[13,14],[13,15],[13,16],[13,17],[13,18],[13,19],[13,20],[14,14],[14,15],[14,16],[14,17],[14,18],[14,19],[14,20],[15,15],[15,16],[15,17],[15,18],[15,19],[15,20],[16,16],[16,17],[16,18],[16,19],[16,20],[17,17],[17,18],[17,19],[17,20],[18,18],[18,19],[18,20],[19,19],[19,20],[20,20],[20,21],[21,21],[21,22],[22,22],[22,23],[23,23],[23,24],[24,24],[24,25],[25,25],[25,26],[26,26],[26,27],[27,27],[27,28],[28,28],[28,29],[29,29]]

Any other pair of integer would be an invalid badminton score.

Challenge rules:

I/O is flexible, so:
- You can take the input as a list of two numbers; two separated numbers through STDIN or function parameters; two strings; etc.
- Output will be three distinct and unique values of your own choice. Can be integers (i.e. [0,1,2], [1,2,3], [-1,0,1], etc.); can be Booleans (i.e. [true,false,undefined/null/empty]); can be characters/strings (i.e. ["valid & ended","valid","invalid"]); etc.
- Please specify the I/O you've used in your answer!

You are allowed to take the input-integers pre-ordered from lowest to highest or vice-versa.

The input integers can be negative, in which case they are of course invalid.

General rules:

This is code-golf, 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.

If possible, please add a link with a test for your code (i.e. TIO).

Also, adding an explanation for your answer is highly recommended.

Test cases:

These test cases are valid, and the set has ended:

These test cases are valid, but the set is still in play:

These test cases are invalid:

edited Mar 26 at 16:03

asked Mar 26 at 14:59

Kevin Cruijssen

42.3k570217

add a comment |

Introduction:

Challenge:

So these are all possible input-pairs (in either order) indicating it's a valid badminton score AND the set has ended:

[[0,21],[1,21],[2,21],[3,21],[4,21],[5,21],[6,21],[7,21],[8,21],[9,21],[10,21],[11,21],[12,21],[13,21],[14,21],[15,21],[16,21],[17,21],[18,21],[19,21],[20,22],[21,23],[22,24],[23,25],[24,26],[25,27],[26,28],[27,29],[28,30],[29,30]]

And these are all possible input-pairs (in either order) indicating it's a valid badminton score BUT the set is still in play:

[[0,0],[0,1],[0,2],[0,3],[0,4],[0,5],[0,6],[0,7],[0,8],[0,9],[0,10],[0,11],[0,12],[0,13],[0,14],[0,15],[0,16],[0,17],[0,18],[0,19],[0,20],[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[1,7],[1,8],[1,9],[1,10],[1,11],[1,12],[1,13],[1,14],[1,15],[1,16],[1,17],[1,18],[1,19],[1,20],[2,2],[2,3],[2,4],[2,5],[2,6],[2,7],[2,8],[2,9],[2,10],[2,11],[2,12],[2,13],[2,14],[2,15],[2,16],[2,17],[2,18],[2,19],[2,20],[3,3],[3,4],[3,5],[3,6],[3,7],[3,8],[3,9],[3,10],[3,11],[3,12],[3,13],[3,14],[3,15],[3,16],[3,17],[3,18],[3,19],[3,20],[4,4],[4,5],[4,6],[4,7],[4,8],[4,9],[4,10],[4,11],[4,12],[4,13],[4,14],[4,15],[4,16],[4,17],[4,18],[4,19],[4,20],[5,5],[5,6],[5,7],[5,8],[5,9],[5,10],[5,11],[5,12],[5,13],[5,14],[5,15],[5,16],[5,17],[5,18],[5,19],[5,20],[6,6],[6,7],[6,8],[6,9],[6,10],[6,11],[6,12],[6,13],[6,14],[6,15],[6,16],[6,17],[6,18],[6,19],[6,20],[7,7],[7,8],[7,9],[7,10],[7,11],[7,12],[7,13],[7,14],[7,15],[7,16],[7,17],[7,18],[7,19],[7,20],[8,8],[8,9],[8,10],[8,11],[8,12],[8,13],[8,14],[8,15],[8,16],[8,17],[8,18],[8,19],[8,20],[9,9],[9,10],[9,11],[9,12],[9,13],[9,14],[9,15],[9,16],[9,17],[9,18],[9,19],[9,20],[10,10],[10,11],[10,12],[10,13],[10,14],[10,15],[10,16],[10,17],[10,18],[10,19],[10,20],[11,11],[11,12],[11,13],[11,14],[11,15],[11,16],[11,17],[11,18],[11,19],[11,20],[12,12],[12,13],[12,14],[12,15],[12,16],[12,17],[12,18],[12,19],[12,20],[13,13],[13,14],[13,15],[13,16],[13,17],[13,18],[13,19],[13,20],[14,14],[14,15],[14,16],[14,17],[14,18],[14,19],[14,20],[15,15],[15,16],[15,17],[15,18],[15,19],[15,20],[16,16],[16,17],[16,18],[16,19],[16,20],[17,17],[17,18],[17,19],[17,20],[18,18],[18,19],[18,20],[19,19],[19,20],[20,20],[20,21],[21,21],[21,22],[22,22],[22,23],[23,23],[23,24],[24,24],[24,25],[25,25],[25,26],[26,26],[26,27],[27,27],[27,28],[28,28],[28,29],[29,29]]

Any other pair of integer would be an invalid badminton score.

Challenge rules:

I/O is flexible, so:
- You can take the input as a list of two numbers; two separated numbers through STDIN or function parameters; two strings; etc.
- Output will be three distinct and unique values of your own choice. Can be integers (i.e. [0,1,2], [1,2,3], [-1,0,1], etc.); can be Booleans (i.e. [true,false,undefined/null/empty]); can be characters/strings (i.e. ["valid & ended","valid","invalid"]); etc.
- Please specify the I/O you've used in your answer!

You are allowed to take the input-integers pre-ordered from lowest to highest or vice-versa.

The input integers can be negative, in which case they are of course invalid.

General rules:

This is code-golf, 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.

If possible, please add a link with a test for your code (i.e. TIO).

Also, adding an explanation for your answer is highly recommended.

Test cases:

These test cases are valid, and the set has ended:

These test cases are valid, but the set is still in play:

These test cases are invalid:

edited Mar 26 at 16:03

asked Mar 26 at 14:59

Kevin Cruijssen

42.3k570217

add a comment |

Introduction:

Challenge:

So these are all possible input-pairs (in either order) indicating it's a valid badminton score AND the set has ended:

[[0,21],[1,21],[2,21],[3,21],[4,21],[5,21],[6,21],[7,21],[8,21],[9,21],[10,21],[11,21],[12,21],[13,21],[14,21],[15,21],[16,21],[17,21],[18,21],[19,21],[20,22],[21,23],[22,24],[23,25],[24,26],[25,27],[26,28],[27,29],[28,30],[29,30]]

And these are all possible input-pairs (in either order) indicating it's a valid badminton score BUT the set is still in play:

[[0,0],[0,1],[0,2],[0,3],[0,4],[0,5],[0,6],[0,7],[0,8],[0,9],[0,10],[0,11],[0,12],[0,13],[0,14],[0,15],[0,16],[0,17],[0,18],[0,19],[0,20],[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[1,7],[1,8],[1,9],[1,10],[1,11],[1,12],[1,13],[1,14],[1,15],[1,16],[1,17],[1,18],[1,19],[1,20],[2,2],[2,3],[2,4],[2,5],[2,6],[2,7],[2,8],[2,9],[2,10],[2,11],[2,12],[2,13],[2,14],[2,15],[2,16],[2,17],[2,18],[2,19],[2,20],[3,3],[3,4],[3,5],[3,6],[3,7],[3,8],[3,9],[3,10],[3,11],[3,12],[3,13],[3,14],[3,15],[3,16],[3,17],[3,18],[3,19],[3,20],[4,4],[4,5],[4,6],[4,7],[4,8],[4,9],[4,10],[4,11],[4,12],[4,13],[4,14],[4,15],[4,16],[4,17],[4,18],[4,19],[4,20],[5,5],[5,6],[5,7],[5,8],[5,9],[5,10],[5,11],[5,12],[5,13],[5,14],[5,15],[5,16],[5,17],[5,18],[5,19],[5,20],[6,6],[6,7],[6,8],[6,9],[6,10],[6,11],[6,12],[6,13],[6,14],[6,15],[6,16],[6,17],[6,18],[6,19],[6,20],[7,7],[7,8],[7,9],[7,10],[7,11],[7,12],[7,13],[7,14],[7,15],[7,16],[7,17],[7,18],[7,19],[7,20],[8,8],[8,9],[8,10],[8,11],[8,12],[8,13],[8,14],[8,15],[8,16],[8,17],[8,18],[8,19],[8,20],[9,9],[9,10],[9,11],[9,12],[9,13],[9,14],[9,15],[9,16],[9,17],[9,18],[9,19],[9,20],[10,10],[10,11],[10,12],[10,13],[10,14],[10,15],[10,16],[10,17],[10,18],[10,19],[10,20],[11,11],[11,12],[11,13],[11,14],[11,15],[11,16],[11,17],[11,18],[11,19],[11,20],[12,12],[12,13],[12,14],[12,15],[12,16],[12,17],[12,18],[12,19],[12,20],[13,13],[13,14],[13,15],[13,16],[13,17],[13,18],[13,19],[13,20],[14,14],[14,15],[14,16],[14,17],[14,18],[14,19],[14,20],[15,15],[15,16],[15,17],[15,18],[15,19],[15,20],[16,16],[16,17],[16,18],[16,19],[16,20],[17,17],[17,18],[17,19],[17,20],[18,18],[18,19],[18,20],[19,19],[19,20],[20,20],[20,21],[21,21],[21,22],[22,22],[22,23],[23,23],[23,24],[24,24],[24,25],[25,25],[25,26],[26,26],[26,27],[27,27],[27,28],[28,28],[28,29],[29,29]]

Any other pair of integer would be an invalid badminton score.

Challenge rules:

I/O is flexible, so:
- You can take the input as a list of two numbers; two separated numbers through STDIN or function parameters; two strings; etc.
- Output will be three distinct and unique values of your own choice. Can be integers (i.e. [0,1,2], [1,2,3], [-1,0,1], etc.); can be Booleans (i.e. [true,false,undefined/null/empty]); can be characters/strings (i.e. ["valid & ended","valid","invalid"]); etc.
- Please specify the I/O you've used in your answer!

You are allowed to take the input-integers pre-ordered from lowest to highest or vice-versa.

The input integers can be negative, in which case they are of course invalid.

General rules:

This is code-golf, 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.

If possible, please add a link with a test for your code (i.e. TIO).

Also, adding an explanation for your answer is highly recommended.

Test cases:

These test cases are valid, and the set has ended:

These test cases are valid, but the set is still in play:

These test cases are invalid:

edited Mar 26 at 16:03

asked Mar 26 at 14:59

Kevin Cruijssen

42.3k570217

Introduction:

Challenge:

So these are all possible input-pairs (in either order) indicating it's a valid badminton score AND the set has ended:

[[0,21],[1,21],[2,21],[3,21],[4,21],[5,21],[6,21],[7,21],[8,21],[9,21],[10,21],[11,21],[12,21],[13,21],[14,21],[15,21],[16,21],[17,21],[18,21],[19,21],[20,22],[21,23],[22,24],[23,25],[24,26],[25,27],[26,28],[27,29],[28,30],[29,30]]

And these are all possible input-pairs (in either order) indicating it's a valid badminton score BUT the set is still in play:

[[0,0],[0,1],[0,2],[0,3],[0,4],[0,5],[0,6],[0,7],[0,8],[0,9],[0,10],[0,11],[0,12],[0,13],[0,14],[0,15],[0,16],[0,17],[0,18],[0,19],[0,20],[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[1,7],[1,8],[1,9],[1,10],[1,11],[1,12],[1,13],[1,14],[1,15],[1,16],[1,17],[1,18],[1,19],[1,20],[2,2],[2,3],[2,4],[2,5],[2,6],[2,7],[2,8],[2,9],[2,10],[2,11],[2,12],[2,13],[2,14],[2,15],[2,16],[2,17],[2,18],[2,19],[2,20],[3,3],[3,4],[3,5],[3,6],[3,7],[3,8],[3,9],[3,10],[3,11],[3,12],[3,13],[3,14],[3,15],[3,16],[3,17],[3,18],[3,19],[3,20],[4,4],[4,5],[4,6],[4,7],[4,8],[4,9],[4,10],[4,11],[4,12],[4,13],[4,14],[4,15],[4,16],[4,17],[4,18],[4,19],[4,20],[5,5],[5,6],[5,7],[5,8],[5,9],[5,10],[5,11],[5,12],[5,13],[5,14],[5,15],[5,16],[5,17],[5,18],[5,19],[5,20],[6,6],[6,7],[6,8],[6,9],[6,10],[6,11],[6,12],[6,13],[6,14],[6,15],[6,16],[6,17],[6,18],[6,19],[6,20],[7,7],[7,8],[7,9],[7,10],[7,11],[7,12],[7,13],[7,14],[7,15],[7,16],[7,17],[7,18],[7,19],[7,20],[8,8],[8,9],[8,10],[8,11],[8,12],[8,13],[8,14],[8,15],[8,16],[8,17],[8,18],[8,19],[8,20],[9,9],[9,10],[9,11],[9,12],[9,13],[9,14],[9,15],[9,16],[9,17],[9,18],[9,19],[9,20],[10,10],[10,11],[10,12],[10,13],[10,14],[10,15],[10,16],[10,17],[10,18],[10,19],[10,20],[11,11],[11,12],[11,13],[11,14],[11,15],[11,16],[11,17],[11,18],[11,19],[11,20],[12,12],[12,13],[12,14],[12,15],[12,16],[12,17],[12,18],[12,19],[12,20],[13,13],[13,14],[13,15],[13,16],[13,17],[13,18],[13,19],[13,20],[14,14],[14,15],[14,16],[14,17],[14,18],[14,19],[14,20],[15,15],[15,16],[15,17],[15,18],[15,19],[15,20],[16,16],[16,17],[16,18],[16,19],[16,20],[17,17],[17,18],[17,19],[17,20],[18,18],[18,19],[18,20],[19,19],[19,20],[20,20],[20,21],[21,21],[21,22],[22,22],[22,23],[23,23],[23,24],[24,24],[24,25],[25,25],[25,26],[26,26],[26,27],[27,27],[27,28],[28,28],[28,29],[29,29]]

Any other pair of integer would be an invalid badminton score.

Challenge rules:

I/O is flexible, so:
- You can take the input as a list of two numbers; two separated numbers through STDIN or function parameters; two strings; etc.
- Output will be three distinct and unique values of your own choice. Can be integers (i.e. [0,1,2], [1,2,3], [-1,0,1], etc.); can be Booleans (i.e. [true,false,undefined/null/empty]); can be characters/strings (i.e. ["valid & ended","valid","invalid"]); etc.
- Please specify the I/O you've used in your answer!

You are allowed to take the input-integers pre-ordered from lowest to highest or vice-versa.

The input integers can be negative, in which case they are of course invalid.

General rules:

This is code-golf, 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.

If possible, please add a link with a test for your code (i.e. TIO).

Also, adding an explanation for your answer is highly recommended.

Test cases:

These test cases are valid, and the set has ended:

These test cases are valid, but the set is still in play:

These test cases are invalid:

code-golf number integer

edited Mar 26 at 16:03

asked Mar 26 at 14:59

Kevin Cruijssen

42.3k570217

edited Mar 26 at 16:03

asked Mar 26 at 14:59

Kevin Cruijssen

42.3k570217

edited Mar 26 at 16:03

asked Mar 26 at 14:59

Kevin Cruijssen

42.3k570217

asked Mar 26 at 14:59

Kevin Cruijssen

42.3k570217

asked Mar 26 at 14:59

Kevin Cruijssen

42.3k570217

add a comment |

13 Answers
13

active

oldest

votes

Stax, 20 bytes

ÇåπßéD╩¬7▼ß▌ΣU¬í╡S┤╘

Run and debug it

It takes input in the same format as the examples. 0 means there's a valid winner. 1 means the game is in progress. -1 means invalid score.

In pseudo-code, with ordered inputs x and y, the algorithm is

sign(clamp(x + 2, 21, 30) - y) | (x < 0 || x >= 30 ? 0 : -1)

sign means numeric sign (-1, 0, or 1)

clamp forces its first argument into the specified half-open interval

answered 2 days ago

recursive

5,7191322

add a comment |

Python 2, 97 95 75 72 71 70 69 64 55 54 52 51 50 48 bytes

lambda a,b:(b-61<~a<a>b/22*b-3)*~(19<b-(b<30)>a)

Try it online!

Takes input as pre-ordered a,b.

Returns -2, -1, 0 for ended, in play, invalid.

_{-1 byte, thanks to Kevin Cruijssen}

Left part (b-61<~a<a>b/22*b-3) is a validity-check, and right part (19<b-(b<30)>a) is a check for game ended.

edited Mar 27 at 11:14

answered Mar 26 at 15:46

TFeld

16.4k21451

add a comment |

Python 2, 47 bytes

lambda a,b:[61>60-a>b<3+max(19,a)for b in-~b,b]

Try it online!

Outputs a list of two Booleans. Thanks to TFeld for writing a test suite in their answer that made it easy to check my solution.

ended: [False, True]

going: [True, True]

invalid: [False, False]

The key insight is that a valid score ends the game exactly if increasing the higher value b makes the score invalid. So, we just code up the validity condition, and check it for (a,b+1) in addition to (a,b) to see if the game has ended.

Validity is checked via three conditions that are chained together:

b<3+max(19,a): Checks that the higher score b isn't past winning, with either b<=21 or b<=a+2 (win by two)

60-a>b: Equivalent to a+b<=59, ensuring the score isn't above (29,30)

61>60-a: Equivalent to a>=0, ensures the lower score is non-negative

Python 2, 44 bytes

lambda a,b:[b-61<~a<a>b/22*b-3for b in-~b,b]

Try it online!

An improved validity check by TFeld saves 3 bytes. The main idea is to branch on "overtime" b>21 with b/22*b which effectively sets below-21 scores to zero, whereas I'd branched on a>19 with the longer max(19,a).

Python 2, 43 bytes

lambda a,b:a>>99|cmp(2+max(19,a)%30-a/29,b)

Try it online!

Outputs:

ended: 0

going: -1

invalid: 1

Assumes that the inputs are not below $-2^{99}$.

edited Mar 28 at 0:22

answered Mar 27 at 0:18

xnor

93.3k18190448

1

$begingroup$
Using my newest validity-check (b-61<~a<a>b/22*b-3), you can save 3 bytes.
$endgroup$
– TFeld
Mar 27 at 11:17

1

$begingroup$
+1 byte to make your second solution work for all inputs: lambda a,b:-(a<0)|cmp(2+max(19,a)%30-a/29,b)
$endgroup$
– TFeld
Mar 27 at 12:51

add a comment |

JavaScript (ES6), 55 53 48 bytes

^{Thanks to @KevinCruijssen for noticing that I was not fully assuming $ale b$ (saving 5 bytes)}

Takes input as (a)(b) with $ale b$. Returns $0$ (valid), $1$ (ended) or $2$ (invalid).

a=>b=>a<0|a>29|b>30|b>21&b-a>2?2:b>20&b-a>1|b>29

Try it online!

edited Mar 26 at 16:28

answered Mar 26 at 16:10

Arnauld

80.4k797333

add a comment |

C# (Visual C# Interactive Compiler), 53 52 bytes

a=>b=>b<0|a-b>2&a>21|b>29|a>30?3:a>20&a-b>1|a>29?1:2

Called as f(max)(min). Returns 3 for invalid, 1 for finished, 2 for ongoing.

Saved 1 byte thanks to Kevin Cruijjsen

Try it online!

edited Mar 26 at 18:20

answered Mar 26 at 17:32

Embodiment of Ignorance

2,778127

add a comment |

Jelly, 25 bytes

»19«28‘<‘×+2>ɗʋ⁹×,%Ƒ“œþ‘ɗ

Try it online!

Left argument: minimum. Right argument: maximum.

Invalid: 0. Ongoing: 1. Ended: 2.

Mathematically, this works as below (the left argument is $x$, the right is $y$):

$$[a]=cases{acolon1\lnot acolon0}\otimes(a,b)=(abmod30,bbmod31)\x,yinmathbb Z\X:=min(max(x+1,20),29)\p:=(x,y)\([X<y]+1)[X+2>y][p=otimes p]$$

Explanation:

»19«28‘<‘×+2>ɗʋ⁹×,%Ƒ“œþ‘ɗ Left argument: x, Right argument: y

»19«28‘                   X := Bound x + 1 in [20, 29]:

»19                         X := max(x, 19).

   «28                      X := min(X, 28).

      ‘                     X := X + 1.

       <‘×+2>ɗʋ⁹          X := If X + 2 <= y, then 0, else if X < y, then 2, else 1:

       <                    t := If X < y, then 1, else 0.

        ‘                   t := t + 1.

          +2>ɗ              u := Check if X + 2 > y:

          +2                  u := X + 2.

            >                 u := If u > y, then 1, else 0.

         ×                  X := t * u.

                 ,%Ƒ“œþ‘ɗ z := If x mod 30 = x and y mod 31 = y, then 1, else 0:

                 ,          z := (x, y).

                  % “œþ‘    m := z mod (30, 31) = (x mod 30, y mod 31).

                   Ƒ        z := If z = m, then 1, else 0.

                ×         X * z.

edited Mar 27 at 13:43

answered Mar 26 at 17:15

Erik the Outgolfer

32.9k429106

1

$begingroup$
@KevinCruijssen Added one.
$endgroup$
– Erik the Outgolfer
Mar 27 at 13:09

add a comment |

VDM-SL, 80 bytes

f(i,j)==if(j-i>2and j>21)or(i<0or i=30or j>30)then{}else{(j>20and j-i>1or j=30)}

This function takes the scores ordered in ascending order and returns the empty set if the score is invalid or the set containing whether the set is complete (so {true} if the set is complete and valid and {false} if the set is incomplete and valid)

A full program to run might look like this:

functions

f:int*int+>set of bool

f(i,j)==if(j-i>2and j>21)or(i<0or i=30or j>30)then{}else{(j>20and j-i>1or j=30)}

Explanation:

if(j-i>2 and j>21)             /*if scores are too far apart*/

or(i<0 or i=30 or j>30)        /*or scores not in a valid range*/

then {}                        /*return the empty set*/

else{                       }  /*else return the set containing...*/

     (j>20 and j-i>1 or j=30)  /*if the set is complete*/

answered Mar 26 at 17:22

Expired Data

52313

add a comment |

Java (JDK), 59 48 bytes

a->b->b<0|b>29|a>b+2&a>21|a>30?0:a<21|a<30&a<b+2

Try it online!

Returns an Object, which is the Integer 0 for invalid games and the Booleans true and false for valid ongoing games and for valid finished games respectively. Takes the score ordered (and curried), with the higher score first.

-2 bytes by inverting the end-of-match check.
-11 bytes by currying, using bitwise operators, and some return type autoboxing trickery - thanks to @KevinCruijssen

Ungolfed

a->                      // Curried: Target type IntFunction<IntFunction<Object>>

    b->                  // Target type IntFunction<Object>

                         // Invalid if:

            b<0          //    Any score is negative

          | b > 29       //    Both scores above 29

          |   a > b + 2  //    Lead too big

            & a > 21     //        and leader has at least 21 points

          | a > 30       //    Anyone has 31 points

        ? 0              // If invalid, return 0 (autoboxed to Integer)

                         // If valid, return whether the game is ongoing (autoboxed to Boolean)

                         // Ongoing if:

        :   a < 21       //    Nobody has 21 points

          |   a < 30     //    Leader has fewer than 30 points

            & a < b + 2  //        and lead is small

edited Mar 26 at 21:57

answered Mar 26 at 20:08

Sara J

555210

add a comment |

Retina 0.8.2, 92 bytes

d+

$*

^(1{0,19},1{21}|(1{20,28}),112|1{29},1{30})$|^(1*,1{0,20}|(1{0,28}),1?4)$|.+

$#1$#3

Try it online! Link includes test cases. Takes input in ascending order. Explanation: The first stage simply converts from decimal to unary so that the scores can be properly compared. The second stage contains six alternate patterns, grouped into three groups so that three distinct values can be output, which are 10 for win, 01 for ongoing and 00 for illegal. The patterns are:

Against 0-19, a score of 21 is a win

Against 20-28, a score of +2 is a win

Against 29, a score of 30 is a win

Against any (lower) score, a score of 0-20 is ongoing

Against a score of up to 28, a score of +1 is ongoing

Anything else (including negative scores) is illegal

answered Mar 26 at 20:09

Neil

82.6k745179

add a comment |

APL (Dyalog Unicode), 35 bytes^SBCS

Infix tacit function where ended is 2, ongoing is 1, invalid is 0, smaller and larger scores are left.

(,≡30 31|,)×(⊢<2+X)×1+⊢>X←29⌊20⌈1+⊣

Try it online!

Implements Erik the Outgolfer's mathematical formulas combined into

$$X:=min(max(x+1,20),29)\ ([X< y]+1)[X+2>y][(x,y)=(xbmod30,ybmod31)]$$
rearranged (as if traditional mathematical notation had vectorisation and inline assignments) to

$$[(x,y)=(x,y)bmod(30,31)]×[y<2+X]×(1+[y< (X:=min(29,max(20,1+x)))])$$

and translated directly to APL (which is strictly right-associative, so we avoid some parentheses):

$$((x,y)≡30 31|x,y)×(y<2+X)×1+y>X←29⌊20⌈1 +x$$

This can be converted into a tacit function simply by substituting $⊣$ for $x$ and $⊢$ for $y$, symbolising the left and right arguments rather than the two variables:

$$((⊣,⊢)≡30 31|⊣,⊢)×(⊣<2+X)×1+⊢>X←29⌊20⌈1+⊣$$

Now $⊣⎕⊢$ is equivalent to $⎕$ for any infix function $⎕$, so we can simplify to

$$(,≡30 31|,)×(⊣<2+X)×1+⊢>X←29⌊20⌈1+⊣$$

which is our solution; (,≡30 31|,)×(⊢<2+X)×1+⊢>X←29⌊20⌈1+⊣:

⊣ the left argument; $x$
1+ one plus that; $1+x$
20⌈ maximum of 20 and that; $max(20,…)$
29⌊ minimum of 29 and that; $min(29,…)$
X← assign that to X; $X:=…$
⊢> is the right argument greater (0/1)?; $[y>…]$
1+ add one; $1+…$
(…)× multiply the following by that; $(…)×…$

2+X two plus X; $2+X$

⊢< is the right argument less than that (0/1); $[y<…]$
(…)× multiply the following by that; $(…)×…$

, concatenate the arguments; $(x,y)$

30 31| remainders when divided by these numbers; $…mod(30,31)$

,≡ are the concatenated arguments identical to that (0/1)?; $[(x,y)=…]$

edited Mar 28 at 11:01

answered Mar 28 at 10:52

Adám

28.9k276207

add a comment |

x86 Assembly, 42 Bytes

Takes input in ECX and EDX registers. Note that ECX must be greater than EDX.

Outputs into EAX, where 0 means the game's still on, 1 representing the game being over and -1 (aka FFFFFFFF) representing an invalid score.

31 C0 83 F9 1E 77 1F 83 FA 1D 77 1A 83 F9 15 7C 

18 83 F9 1E 74 12 89 CB 29 D3 83 FB 02 74 09 7C 

08 83 F9 15 74 02 48 C3 40 C3

Or, more readable in Intel Syntax:

check:

    XOR EAX, EAX

    CMP ECX, 30     ; check i_1 against 30

    JA .invalid     ; if >, invalid.

    CMP EDX, 29     ; check i_2 against 29

    JA .invalid     ; if >, invalid.

    CMP ECX, 21     ; check i_1 against 21

    JL .runi        ; if <, running.

    CMP ECX, 30     ; check i_1 against 30

    JE .over        ; if ==, over.

    MOV EBX, ECX

    SUB EBX, EDX    ; EBX = i_1 - i_2

    CMP EBX, 2      ; check EBX against 2

    JE .over        ; if ==, over.

    JL .runi        ; if <, running.

                    ; if >, keep executing!

    CMP ECX, 21     ; check i_1 against 21

    JE .over        ; if ==, over.

                    ; otherwise, it's invalid.

    ; fallthrough!

    .invalid:

        DEC EAX     ; EAX = -1

        RETN

    .over:

        INC EAX     ; EAX = 1

    ; fallthrough!

    .runi:

        RETN        ; EAX = 0 or 1

Fun fact: this function almost follows the C Calling Convention's rules on which registers to preserve, except I had to clobber EBX to save some bytes on stack usage.

Optional (not included in byte-count)

By adding the following 6 bytes directly before start of the code above, you can pass ECX and EDX unordered:

39 D1 7D 02 87 CA

Which is the following in readable Intel Syntax:

CMP ECX, EDX

JGE check

XCHG ECX, EDX

answered Mar 28 at 21:46

Fayti1703

513

add a comment |

APL (Dyalog Unicode), 33 32 bytes^SBCS

{h⍵+1 0}+h←(⊢≡31 30|⊢)×21 2∨.≥-

Try it online!

in: a pair in descending order

out: 2=ongoing, 1=ended, 0=invalid

tests stolen from Adám's answer

edited Apr 1 at 2:25

answered Apr 1 at 2:09

ngn

7,39612660

add a comment |

Bash 4+, 97 89 91 88 bytes

Assume that inputs are ascending. Used concepts from VDM-SL answer. Try it Online

z==0 - game in progress
z==1 - game completed
z==2 - invalid

-8 by bracket cleanup from (( & | )) conditions
+2 fixing a bug, thanks to Kevin Cruijssen
-3 logic improvements by Kevin Cruijssen

i=$1 j=$2 z=0

((j-i>2&j>21|i<0|i>29|j>30?z=2:0))

((z<1&(j>20&j-i>1|j>29)?z=1:0))

echo $z

edited Apr 3 at 8:46

answered Mar 28 at 13:40

roblogic

1515

1

$begingroup$
Your 89 bytes version seems to output 1 instead of 2 for 0 30. Your 97 bytes version was working correctly, so if you're unable to fix it, you can always rollback. Upvoted for that 97 version. :)
$endgroup$
– Kevin Cruijssen
Mar 28 at 13:59

1

$begingroup$
This should fix it, and golf a byte at the same time. :)
$endgroup$
– Kevin Cruijssen
Mar 28 at 14:05

1

$begingroup$
I fixed it, but yours was better! Hard to keep up :P
$endgroup$
– roblogic
Mar 28 at 14:14

$begingroup$
Bug at 29 30 :( it should be "completed"
$endgroup$
– roblogic
Apr 3 at 8:39

1

$begingroup$
Ah oops.. the i>29 should be j>29 in the second ternary to fix it.
$endgroup$
– Kevin Cruijssen
Apr 3 at 8:42

add a comment |

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13 Answers
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13 Answers
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votes

Stax, 20 bytes

ÇåπßéD╩¬7▼ß▌ΣU¬í╡S┤╘

Run and debug it

It takes input in the same format as the examples. 0 means there's a valid winner. 1 means the game is in progress. -1 means invalid score.

In pseudo-code, with ordered inputs x and y, the algorithm is

sign(clamp(x + 2, 21, 30) - y) | (x < 0 || x >= 30 ? 0 : -1)

sign means numeric sign (-1, 0, or 1)

clamp forces its first argument into the specified half-open interval

answered 2 days ago

recursive

5,7191322

add a comment |

Stax, 20 bytes

ÇåπßéD╩¬7▼ß▌ΣU¬í╡S┤╘

Run and debug it

It takes input in the same format as the examples. 0 means there's a valid winner. 1 means the game is in progress. -1 means invalid score.

In pseudo-code, with ordered inputs x and y, the algorithm is

sign(clamp(x + 2, 21, 30) - y) | (x < 0 || x >= 30 ? 0 : -1)

sign means numeric sign (-1, 0, or 1)

clamp forces its first argument into the specified half-open interval

answered 2 days ago

recursive

5,7191322

add a comment |

Stax, 20 bytes

ÇåπßéD╩¬7▼ß▌ΣU¬í╡S┤╘

Run and debug it

It takes input in the same format as the examples. 0 means there's a valid winner. 1 means the game is in progress. -1 means invalid score.

In pseudo-code, with ordered inputs x and y, the algorithm is

sign(clamp(x + 2, 21, 30) - y) | (x < 0 || x >= 30 ? 0 : -1)

sign means numeric sign (-1, 0, or 1)

clamp forces its first argument into the specified half-open interval

answered 2 days ago

recursive

5,7191322

Stax, 20 bytes

ÇåπßéD╩¬7▼ß▌ΣU¬í╡S┤╘

Run and debug it

It takes input in the same format as the examples. 0 means there's a valid winner. 1 means the game is in progress. -1 means invalid score.

In pseudo-code, with ordered inputs x and y, the algorithm is

sign(clamp(x + 2, 21, 30) - y) | (x < 0 || x >= 30 ? 0 : -1)

sign means numeric sign (-1, 0, or 1)

clamp forces its first argument into the specified half-open interval

answered 2 days ago

recursive

5,7191322

answered 2 days ago

recursive

5,7191322

answered 2 days ago

recursive

5,7191322

answered 2 days ago

recursive

5,7191322

add a comment |

Python 2, 97 95 75 72 71 70 69 64 55 54 52 51 50 48 bytes

lambda a,b:(b-61<~a<a>b/22*b-3)*~(19<b-(b<30)>a)

Try it online!

Takes input as pre-ordered a,b.

Returns -2, -1, 0 for ended, in play, invalid.

_{-1 byte, thanks to Kevin Cruijssen}

Left part (b-61<~a<a>b/22*b-3) is a validity-check, and right part (19<b-(b<30)>a) is a check for game ended.

edited Mar 27 at 11:14

answered Mar 26 at 15:46

TFeld

16.4k21451

add a comment |

Python 2, 97 95 75 72 71 70 69 64 55 54 52 51 50 48 bytes

lambda a,b:(b-61<~a<a>b/22*b-3)*~(19<b-(b<30)>a)

Try it online!

Takes input as pre-ordered a,b.

Returns -2, -1, 0 for ended, in play, invalid.

_{-1 byte, thanks to Kevin Cruijssen}

Left part (b-61<~a<a>b/22*b-3) is a validity-check, and right part (19<b-(b<30)>a) is a check for game ended.

edited Mar 27 at 11:14

answered Mar 26 at 15:46

TFeld

16.4k21451

add a comment |

Python 2, 97 95 75 72 71 70 69 64 55 54 52 51 50 48 bytes

lambda a,b:(b-61<~a<a>b/22*b-3)*~(19<b-(b<30)>a)

Try it online!

Takes input as pre-ordered a,b.

Returns -2, -1, 0 for ended, in play, invalid.

_{-1 byte, thanks to Kevin Cruijssen}

Left part (b-61<~a<a>b/22*b-3) is a validity-check, and right part (19<b-(b<30)>a) is a check for game ended.

edited Mar 27 at 11:14

answered Mar 26 at 15:46

TFeld

16.4k21451

Python 2, 97 95 75 72 71 70 69 64 55 54 52 51 50 48 bytes

lambda a,b:(b-61<~a<a>b/22*b-3)*~(19<b-(b<30)>a)

Try it online!

Takes input as pre-ordered a,b.

Returns -2, -1, 0 for ended, in play, invalid.

_{-1 byte, thanks to Kevin Cruijssen}

Left part (b-61<~a<a>b/22*b-3) is a validity-check, and right part (19<b-(b<30)>a) is a check for game ended.

edited Mar 27 at 11:14

answered Mar 26 at 15:46

TFeld

16.4k21451

edited Mar 27 at 11:14

answered Mar 26 at 15:46

TFeld

16.4k21451

answered Mar 26 at 15:46

TFeld

16.4k21451

answered Mar 26 at 15:46

TFeld

16.4k21451

add a comment |

Python 2, 47 bytes

lambda a,b:[61>60-a>b<3+max(19,a)for b in-~b,b]

Try it online!

Outputs a list of two Booleans. Thanks to TFeld for writing a test suite in their answer that made it easy to check my solution.

ended: [False, True]

going: [True, True]

invalid: [False, False]

Validity is checked via three conditions that are chained together:

b<3+max(19,a): Checks that the higher score b isn't past winning, with either b<=21 or b<=a+2 (win by two)

60-a>b: Equivalent to a+b<=59, ensuring the score isn't above (29,30)

61>60-a: Equivalent to a>=0, ensures the lower score is non-negative

Python 2, 44 bytes

lambda a,b:[b-61<~a<a>b/22*b-3for b in-~b,b]

Try it online!

Python 2, 43 bytes

lambda a,b:a>>99|cmp(2+max(19,a)%30-a/29,b)

Try it online!

Outputs:

ended: 0

going: -1

invalid: 1

Assumes that the inputs are not below $-2^{99}$.

edited Mar 28 at 0:22

answered Mar 27 at 0:18

xnor

93.3k18190448

1

$begingroup$
Using my newest validity-check (b-61<~a<a>b/22*b-3), you can save 3 bytes.
$endgroup$
– TFeld
Mar 27 at 11:17

1

$begingroup$
+1 byte to make your second solution work for all inputs: lambda a,b:-(a<0)|cmp(2+max(19,a)%30-a/29,b)
$endgroup$
– TFeld
Mar 27 at 12:51

add a comment |

Python 2, 47 bytes

lambda a,b:[61>60-a>b<3+max(19,a)for b in-~b,b]

Try it online!

Outputs a list of two Booleans. Thanks to TFeld for writing a test suite in their answer that made it easy to check my solution.

ended: [False, True]

going: [True, True]

invalid: [False, False]

Validity is checked via three conditions that are chained together:

b<3+max(19,a): Checks that the higher score b isn't past winning, with either b<=21 or b<=a+2 (win by two)

60-a>b: Equivalent to a+b<=59, ensuring the score isn't above (29,30)

61>60-a: Equivalent to a>=0, ensures the lower score is non-negative

Python 2, 44 bytes

lambda a,b:[b-61<~a<a>b/22*b-3for b in-~b,b]

Try it online!

Python 2, 43 bytes

lambda a,b:a>>99|cmp(2+max(19,a)%30-a/29,b)

Try it online!

Outputs:

ended: 0

going: -1

invalid: 1

Assumes that the inputs are not below $-2^{99}$.

edited Mar 28 at 0:22

answered Mar 27 at 0:18

xnor

93.3k18190448

1

$begingroup$
Using my newest validity-check (b-61<~a<a>b/22*b-3), you can save 3 bytes.
$endgroup$
– TFeld
Mar 27 at 11:17

1

$begingroup$
+1 byte to make your second solution work for all inputs: lambda a,b:-(a<0)|cmp(2+max(19,a)%30-a/29,b)
$endgroup$
– TFeld
Mar 27 at 12:51

add a comment |

Python 2, 47 bytes

lambda a,b:[61>60-a>b<3+max(19,a)for b in-~b,b]

Try it online!

Outputs a list of two Booleans. Thanks to TFeld for writing a test suite in their answer that made it easy to check my solution.

ended: [False, True]

going: [True, True]

invalid: [False, False]

Validity is checked via three conditions that are chained together:

b<3+max(19,a): Checks that the higher score b isn't past winning, with either b<=21 or b<=a+2 (win by two)

60-a>b: Equivalent to a+b<=59, ensuring the score isn't above (29,30)

61>60-a: Equivalent to a>=0, ensures the lower score is non-negative

Python 2, 44 bytes

lambda a,b:[b-61<~a<a>b/22*b-3for b in-~b,b]

Try it online!

Python 2, 43 bytes

lambda a,b:a>>99|cmp(2+max(19,a)%30-a/29,b)

Try it online!

Outputs:

ended: 0

going: -1

invalid: 1

Assumes that the inputs are not below $-2^{99}$.

edited Mar 28 at 0:22

answered Mar 27 at 0:18

xnor

93.3k18190448

Python 2, 47 bytes

lambda a,b:[61>60-a>b<3+max(19,a)for b in-~b,b]

Try it online!

Outputs a list of two Booleans. Thanks to TFeld for writing a test suite in their answer that made it easy to check my solution.

ended: [False, True]

going: [True, True]

invalid: [False, False]

Validity is checked via three conditions that are chained together:

b<3+max(19,a): Checks that the higher score b isn't past winning, with either b<=21 or b<=a+2 (win by two)

60-a>b: Equivalent to a+b<=59, ensuring the score isn't above (29,30)

61>60-a: Equivalent to a>=0, ensures the lower score is non-negative

Python 2, 44 bytes

lambda a,b:[b-61<~a<a>b/22*b-3for b in-~b,b]

Try it online!

Python 2, 43 bytes

lambda a,b:a>>99|cmp(2+max(19,a)%30-a/29,b)

Try it online!

Outputs:

ended: 0

going: -1

invalid: 1

Assumes that the inputs are not below $-2^{99}$.

edited Mar 28 at 0:22

answered Mar 27 at 0:18

xnor

93.3k18190448

edited Mar 28 at 0:22

answered Mar 27 at 0:18

xnor

93.3k18190448

answered Mar 27 at 0:18

xnor

93.3k18190448

answered Mar 27 at 0:18

xnor

93.3k18190448

1

$begingroup$
Using my newest validity-check (b-61<~a<a>b/22*b-3), you can save 3 bytes.
$endgroup$
– TFeld
Mar 27 at 11:17

1

$begingroup$
+1 byte to make your second solution work for all inputs: lambda a,b:-(a<0)|cmp(2+max(19,a)%30-a/29,b)
$endgroup$
– TFeld
Mar 27 at 12:51

add a comment |

1

$begingroup$
Using my newest validity-check (b-61<~a<a>b/22*b-3), you can save 3 bytes.
$endgroup$
– TFeld
Mar 27 at 11:17

1

$begingroup$
+1 byte to make your second solution work for all inputs: lambda a,b:-(a<0)|cmp(2+max(19,a)%30-a/29,b)
$endgroup$
– TFeld
Mar 27 at 12:51

Using my newest validity-check (b-61<~a<a>b/22*b-3), you can save 3 bytes.

– TFeld
Mar 27 at 11:17

+1 byte to make your second solution work for all inputs: lambda a,b:-(a<0)|cmp(2+max(19,a)%30-a/29,b)

– TFeld
Mar 27 at 12:51

add a comment |

JavaScript (ES6), 55 53 48 bytes

^{Thanks to @KevinCruijssen for noticing that I was not fully assuming $ale b$ (saving 5 bytes)}

Takes input as (a)(b) with $ale b$. Returns $0$ (valid), $1$ (ended) or $2$ (invalid).

a=>b=>a<0|a>29|b>30|b>21&b-a>2?2:b>20&b-a>1|b>29

Try it online!

edited Mar 26 at 16:28

answered Mar 26 at 16:10

Arnauld

80.4k797333

add a comment |

JavaScript (ES6), 55 53 48 bytes

^{Thanks to @KevinCruijssen for noticing that I was not fully assuming $ale b$ (saving 5 bytes)}

Takes input as (a)(b) with $ale b$. Returns $0$ (valid), $1$ (ended) or $2$ (invalid).

a=>b=>a<0|a>29|b>30|b>21&b-a>2?2:b>20&b-a>1|b>29

Try it online!

edited Mar 26 at 16:28

answered Mar 26 at 16:10

Arnauld

80.4k797333

add a comment |

JavaScript (ES6), 55 53 48 bytes

^{Thanks to @KevinCruijssen for noticing that I was not fully assuming $ale b$ (saving 5 bytes)}

Takes input as (a)(b) with $ale b$. Returns $0$ (valid), $1$ (ended) or $2$ (invalid).

a=>b=>a<0|a>29|b>30|b>21&b-a>2?2:b>20&b-a>1|b>29

Try it online!

edited Mar 26 at 16:28

answered Mar 26 at 16:10

Arnauld

80.4k797333

JavaScript (ES6), 55 53 48 bytes

^{Thanks to @KevinCruijssen for noticing that I was not fully assuming $ale b$ (saving 5 bytes)}

Takes input as (a)(b) with $ale b$. Returns $0$ (valid), $1$ (ended) or $2$ (invalid).

a=>b=>a<0|a>29|b>30|b>21&b-a>2?2:b>20&b-a>1|b>29

Try it online!

edited Mar 26 at 16:28

answered Mar 26 at 16:10

Arnauld

80.4k797333

edited Mar 26 at 16:28

answered Mar 26 at 16:10

Arnauld

80.4k797333

answered Mar 26 at 16:10

Arnauld

80.4k797333

answered Mar 26 at 16:10

Arnauld

80.4k797333

add a comment |

C# (Visual C# Interactive Compiler), 53 52 bytes

a=>b=>b<0|a-b>2&a>21|b>29|a>30?3:a>20&a-b>1|a>29?1:2

Called as f(max)(min). Returns 3 for invalid, 1 for finished, 2 for ongoing.

Saved 1 byte thanks to Kevin Cruijjsen

Try it online!

edited Mar 26 at 18:20

answered Mar 26 at 17:32

Embodiment of Ignorance

2,778127

add a comment |

C# (Visual C# Interactive Compiler), 53 52 bytes

a=>b=>b<0|a-b>2&a>21|b>29|a>30?3:a>20&a-b>1|a>29?1:2

Called as f(max)(min). Returns 3 for invalid, 1 for finished, 2 for ongoing.

Saved 1 byte thanks to Kevin Cruijjsen

Try it online!

edited Mar 26 at 18:20

answered Mar 26 at 17:32

Embodiment of Ignorance

2,778127

add a comment |

C# (Visual C# Interactive Compiler), 53 52 bytes

a=>b=>b<0|a-b>2&a>21|b>29|a>30?3:a>20&a-b>1|a>29?1:2

Called as f(max)(min). Returns 3 for invalid, 1 for finished, 2 for ongoing.

Saved 1 byte thanks to Kevin Cruijjsen

Try it online!

edited Mar 26 at 18:20

answered Mar 26 at 17:32

Embodiment of Ignorance

2,778127

C# (Visual C# Interactive Compiler), 53 52 bytes

a=>b=>b<0|a-b>2&a>21|b>29|a>30?3:a>20&a-b>1|a>29?1:2

Called as f(max)(min). Returns 3 for invalid, 1 for finished, 2 for ongoing.

Saved 1 byte thanks to Kevin Cruijjsen

Try it online!

edited Mar 26 at 18:20

answered Mar 26 at 17:32

Embodiment of Ignorance

2,778127

edited Mar 26 at 18:20

answered Mar 26 at 17:32

Embodiment of Ignorance

2,778127

answered Mar 26 at 17:32

Embodiment of Ignorance

2,778127

answered Mar 26 at 17:32

Embodiment of Ignorance

2,778127

add a comment |

Jelly, 25 bytes

»19«28‘<‘×+2>ɗʋ⁹×,%Ƒ“œþ‘ɗ

Try it online!

Left argument: minimum. Right argument: maximum.

Invalid: 0. Ongoing: 1. Ended: 2.

Mathematically, this works as below (the left argument is $x$, the right is $y$):

$$[a]=cases{acolon1\lnot acolon0}\otimes(a,b)=(abmod30,bbmod31)\x,yinmathbb Z\X:=min(max(x+1,20),29)\p:=(x,y)\([X<y]+1)[X+2>y][p=otimes p]$$

Explanation:

»19«28‘<‘×+2>ɗʋ⁹×,%Ƒ“œþ‘ɗ Left argument: x, Right argument: y

»19«28‘                   X := Bound x + 1 in [20, 29]:

»19                         X := max(x, 19).

   «28                      X := min(X, 28).

      ‘                     X := X + 1.

       <‘×+2>ɗʋ⁹          X := If X + 2 <= y, then 0, else if X < y, then 2, else 1:

       <                    t := If X < y, then 1, else 0.

        ‘                   t := t + 1.

          +2>ɗ              u := Check if X + 2 > y:

          +2                  u := X + 2.

            >                 u := If u > y, then 1, else 0.

         ×                  X := t * u.

                 ,%Ƒ“œþ‘ɗ z := If x mod 30 = x and y mod 31 = y, then 1, else 0:

                 ,          z := (x, y).

                  % “œþ‘    m := z mod (30, 31) = (x mod 30, y mod 31).

                   Ƒ        z := If z = m, then 1, else 0.

                ×         X * z.

edited Mar 27 at 13:43

answered Mar 26 at 17:15

Erik the Outgolfer

32.9k429106

1

$begingroup$
@KevinCruijssen Added one.
$endgroup$
– Erik the Outgolfer
Mar 27 at 13:09

add a comment |

Jelly, 25 bytes

»19«28‘<‘×+2>ɗʋ⁹×,%Ƒ“œþ‘ɗ

Try it online!

Left argument: minimum. Right argument: maximum.

Invalid: 0. Ongoing: 1. Ended: 2.

Mathematically, this works as below (the left argument is $x$, the right is $y$):

$$[a]=cases{acolon1\lnot acolon0}\otimes(a,b)=(abmod30,bbmod31)\x,yinmathbb Z\X:=min(max(x+1,20),29)\p:=(x,y)\([X<y]+1)[X+2>y][p=otimes p]$$

Explanation:

»19«28‘<‘×+2>ɗʋ⁹×,%Ƒ“œþ‘ɗ Left argument: x, Right argument: y

»19«28‘                   X := Bound x + 1 in [20, 29]:

»19                         X := max(x, 19).

   «28                      X := min(X, 28).

      ‘                     X := X + 1.

       <‘×+2>ɗʋ⁹          X := If X + 2 <= y, then 0, else if X < y, then 2, else 1:

       <                    t := If X < y, then 1, else 0.

        ‘                   t := t + 1.

          +2>ɗ              u := Check if X + 2 > y:

          +2                  u := X + 2.

            >                 u := If u > y, then 1, else 0.

         ×                  X := t * u.

                 ,%Ƒ“œþ‘ɗ z := If x mod 30 = x and y mod 31 = y, then 1, else 0:

                 ,          z := (x, y).

                  % “œþ‘    m := z mod (30, 31) = (x mod 30, y mod 31).

                   Ƒ        z := If z = m, then 1, else 0.

                ×         X * z.

edited Mar 27 at 13:43

answered Mar 26 at 17:15

Erik the Outgolfer

32.9k429106

1

$begingroup$
@KevinCruijssen Added one.
$endgroup$
– Erik the Outgolfer
Mar 27 at 13:09

add a comment |

Jelly, 25 bytes

»19«28‘<‘×+2>ɗʋ⁹×,%Ƒ“œþ‘ɗ

Try it online!

Left argument: minimum. Right argument: maximum.

Invalid: 0. Ongoing: 1. Ended: 2.

Mathematically, this works as below (the left argument is $x$, the right is $y$):

$$[a]=cases{acolon1\lnot acolon0}\otimes(a,b)=(abmod30,bbmod31)\x,yinmathbb Z\X:=min(max(x+1,20),29)\p:=(x,y)\([X<y]+1)[X+2>y][p=otimes p]$$

Explanation:

»19«28‘<‘×+2>ɗʋ⁹×,%Ƒ“œþ‘ɗ Left argument: x, Right argument: y

»19«28‘                   X := Bound x + 1 in [20, 29]:

»19                         X := max(x, 19).

   «28                      X := min(X, 28).

      ‘                     X := X + 1.

       <‘×+2>ɗʋ⁹          X := If X + 2 <= y, then 0, else if X < y, then 2, else 1:

       <                    t := If X < y, then 1, else 0.

        ‘                   t := t + 1.

          +2>ɗ              u := Check if X + 2 > y:

          +2                  u := X + 2.

            >                 u := If u > y, then 1, else 0.

         ×                  X := t * u.

                 ,%Ƒ“œþ‘ɗ z := If x mod 30 = x and y mod 31 = y, then 1, else 0:

                 ,          z := (x, y).

                  % “œþ‘    m := z mod (30, 31) = (x mod 30, y mod 31).

                   Ƒ        z := If z = m, then 1, else 0.

                ×         X * z.

edited Mar 27 at 13:43

answered Mar 26 at 17:15

Erik the Outgolfer

32.9k429106

Jelly, 25 bytes

»19«28‘<‘×+2>ɗʋ⁹×,%Ƒ“œþ‘ɗ

Try it online!

Left argument: minimum. Right argument: maximum.

Invalid: 0. Ongoing: 1. Ended: 2.

Mathematically, this works as below (the left argument is $x$, the right is $y$):

$$[a]=cases{acolon1\lnot acolon0}\otimes(a,b)=(abmod30,bbmod31)\x,yinmathbb Z\X:=min(max(x+1,20),29)\p:=(x,y)\([X<y]+1)[X+2>y][p=otimes p]$$

Explanation:

»19«28‘<‘×+2>ɗʋ⁹×,%Ƒ“œþ‘ɗ Left argument: x, Right argument: y

»19«28‘                   X := Bound x + 1 in [20, 29]:

»19                         X := max(x, 19).

   «28                      X := min(X, 28).

      ‘                     X := X + 1.

       <‘×+2>ɗʋ⁹          X := If X + 2 <= y, then 0, else if X < y, then 2, else 1:

       <                    t := If X < y, then 1, else 0.

        ‘                   t := t + 1.

          +2>ɗ              u := Check if X + 2 > y:

          +2                  u := X + 2.

            >                 u := If u > y, then 1, else 0.

         ×                  X := t * u.

                 ,%Ƒ“œþ‘ɗ z := If x mod 30 = x and y mod 31 = y, then 1, else 0:

                 ,          z := (x, y).

                  % “œþ‘    m := z mod (30, 31) = (x mod 30, y mod 31).

                   Ƒ        z := If z = m, then 1, else 0.

                ×         X * z.

edited Mar 27 at 13:43

answered Mar 26 at 17:15

Erik the Outgolfer

32.9k429106

edited Mar 27 at 13:43

answered Mar 26 at 17:15

Erik the Outgolfer

32.9k429106

answered Mar 26 at 17:15

Erik the Outgolfer

32.9k429106

answered Mar 26 at 17:15

Erik the Outgolfer

32.9k429106

1

$begingroup$
@KevinCruijssen Added one.
$endgroup$
– Erik the Outgolfer
Mar 27 at 13:09

add a comment |

1

$begingroup$
@KevinCruijssen Added one.
$endgroup$
– Erik the Outgolfer
Mar 27 at 13:09

@KevinCruijssen Added one.

– Erik the Outgolfer
Mar 27 at 13:09

add a comment |

VDM-SL, 80 bytes

f(i,j)==if(j-i>2and j>21)or(i<0or i=30or j>30)then{}else{(j>20and j-i>1or j=30)}

A full program to run might look like this:

functions

f:int*int+>set of bool

f(i,j)==if(j-i>2and j>21)or(i<0or i=30or j>30)then{}else{(j>20and j-i>1or j=30)}

Explanation:

if(j-i>2 and j>21)             /*if scores are too far apart*/

or(i<0 or i=30 or j>30)        /*or scores not in a valid range*/

then {}                        /*return the empty set*/

else{                       }  /*else return the set containing...*/

     (j>20 and j-i>1 or j=30)  /*if the set is complete*/

answered Mar 26 at 17:22

Expired Data

52313

add a comment |

VDM-SL, 80 bytes

f(i,j)==if(j-i>2and j>21)or(i<0or i=30or j>30)then{}else{(j>20and j-i>1or j=30)}

A full program to run might look like this:

functions

f:int*int+>set of bool

f(i,j)==if(j-i>2and j>21)or(i<0or i=30or j>30)then{}else{(j>20and j-i>1or j=30)}

Explanation:

if(j-i>2 and j>21)             /*if scores are too far apart*/

or(i<0 or i=30 or j>30)        /*or scores not in a valid range*/

then {}                        /*return the empty set*/

else{                       }  /*else return the set containing...*/

     (j>20 and j-i>1 or j=30)  /*if the set is complete*/

answered Mar 26 at 17:22

Expired Data

52313

add a comment |

VDM-SL, 80 bytes

f(i,j)==if(j-i>2and j>21)or(i<0or i=30or j>30)then{}else{(j>20and j-i>1or j=30)}

A full program to run might look like this:

functions

f:int*int+>set of bool

f(i,j)==if(j-i>2and j>21)or(i<0or i=30or j>30)then{}else{(j>20and j-i>1or j=30)}

Explanation:

if(j-i>2 and j>21)             /*if scores are too far apart*/

or(i<0 or i=30 or j>30)        /*or scores not in a valid range*/

then {}                        /*return the empty set*/

else{                       }  /*else return the set containing...*/

     (j>20 and j-i>1 or j=30)  /*if the set is complete*/

answered Mar 26 at 17:22

Expired Data

52313

VDM-SL, 80 bytes

f(i,j)==if(j-i>2and j>21)or(i<0or i=30or j>30)then{}else{(j>20and j-i>1or j=30)}

A full program to run might look like this:

functions

f:int*int+>set of bool

f(i,j)==if(j-i>2and j>21)or(i<0or i=30or j>30)then{}else{(j>20and j-i>1or j=30)}

Explanation:

if(j-i>2 and j>21)             /*if scores are too far apart*/

or(i<0 or i=30 or j>30)        /*or scores not in a valid range*/

then {}                        /*return the empty set*/

else{                       }  /*else return the set containing...*/

     (j>20 and j-i>1 or j=30)  /*if the set is complete*/

answered Mar 26 at 17:22

Expired Data

52313

answered Mar 26 at 17:22

Expired Data

52313

answered Mar 26 at 17:22

Expired Data

52313

answered Mar 26 at 17:22

Expired Data

52313

add a comment |

Java (JDK), 59 48 bytes

a->b->b<0|b>29|a>b+2&a>21|a>30?0:a<21|a<30&a<b+2

Try it online!

-2 bytes by inverting the end-of-match check.
-11 bytes by currying, using bitwise operators, and some return type autoboxing trickery - thanks to @KevinCruijssen

Ungolfed

a->                      // Curried: Target type IntFunction<IntFunction<Object>>

    b->                  // Target type IntFunction<Object>

                         // Invalid if:

            b<0          //    Any score is negative

          | b > 29       //    Both scores above 29

          |   a > b + 2  //    Lead too big

            & a > 21     //        and leader has at least 21 points

          | a > 30       //    Anyone has 31 points

        ? 0              // If invalid, return 0 (autoboxed to Integer)

                         // If valid, return whether the game is ongoing (autoboxed to Boolean)

                         // Ongoing if:

        :   a < 21       //    Nobody has 21 points

          |   a < 30     //    Leader has fewer than 30 points

            & a < b + 2  //        and lead is small

edited Mar 26 at 21:57

answered Mar 26 at 20:08

Sara J

555210

add a comment |

Java (JDK), 59 48 bytes

a->b->b<0|b>29|a>b+2&a>21|a>30?0:a<21|a<30&a<b+2

Try it online!

-2 bytes by inverting the end-of-match check.
-11 bytes by currying, using bitwise operators, and some return type autoboxing trickery - thanks to @KevinCruijssen

Ungolfed

a->                      // Curried: Target type IntFunction<IntFunction<Object>>

    b->                  // Target type IntFunction<Object>

                         // Invalid if:

            b<0          //    Any score is negative

          | b > 29       //    Both scores above 29

          |   a > b + 2  //    Lead too big

            & a > 21     //        and leader has at least 21 points

          | a > 30       //    Anyone has 31 points

        ? 0              // If invalid, return 0 (autoboxed to Integer)

                         // If valid, return whether the game is ongoing (autoboxed to Boolean)

                         // Ongoing if:

        :   a < 21       //    Nobody has 21 points

          |   a < 30     //    Leader has fewer than 30 points

            & a < b + 2  //        and lead is small

edited Mar 26 at 21:57

answered Mar 26 at 20:08

Sara J

555210

add a comment |

Java (JDK), 59 48 bytes

a->b->b<0|b>29|a>b+2&a>21|a>30?0:a<21|a<30&a<b+2

Try it online!

-2 bytes by inverting the end-of-match check.
-11 bytes by currying, using bitwise operators, and some return type autoboxing trickery - thanks to @KevinCruijssen

Ungolfed

a->                      // Curried: Target type IntFunction<IntFunction<Object>>

    b->                  // Target type IntFunction<Object>

                         // Invalid if:

            b<0          //    Any score is negative

          | b > 29       //    Both scores above 29

          |   a > b + 2  //    Lead too big

            & a > 21     //        and leader has at least 21 points

          | a > 30       //    Anyone has 31 points

        ? 0              // If invalid, return 0 (autoboxed to Integer)

                         // If valid, return whether the game is ongoing (autoboxed to Boolean)

                         // Ongoing if:

        :   a < 21       //    Nobody has 21 points

          |   a < 30     //    Leader has fewer than 30 points

            & a < b + 2  //        and lead is small

edited Mar 26 at 21:57

answered Mar 26 at 20:08

Sara J

555210

Java (JDK), 59 48 bytes

a->b->b<0|b>29|a>b+2&a>21|a>30?0:a<21|a<30&a<b+2

Try it online!

-2 bytes by inverting the end-of-match check.
-11 bytes by currying, using bitwise operators, and some return type autoboxing trickery - thanks to @KevinCruijssen

Ungolfed

a->                      // Curried: Target type IntFunction<IntFunction<Object>>

    b->                  // Target type IntFunction<Object>

                         // Invalid if:

            b<0          //    Any score is negative

          | b > 29       //    Both scores above 29

          |   a > b + 2  //    Lead too big

            & a > 21     //        and leader has at least 21 points

          | a > 30       //    Anyone has 31 points

        ? 0              // If invalid, return 0 (autoboxed to Integer)

                         // If valid, return whether the game is ongoing (autoboxed to Boolean)

                         // Ongoing if:

        :   a < 21       //    Nobody has 21 points

          |   a < 30     //    Leader has fewer than 30 points

            & a < b + 2  //        and lead is small

edited Mar 26 at 21:57

answered Mar 26 at 20:08

Sara J

555210

edited Mar 26 at 21:57

answered Mar 26 at 20:08

Sara J

555210

answered Mar 26 at 20:08

Sara J

555210

answered Mar 26 at 20:08

Sara J

555210

add a comment |

Retina 0.8.2, 92 bytes

d+

$*

^(1{0,19},1{21}|(1{20,28}),112|1{29},1{30})$|^(1*,1{0,20}|(1{0,28}),1?4)$|.+

$#1$#3

Against 0-19, a score of 21 is a win

Against 20-28, a score of +2 is a win

Against 29, a score of 30 is a win

Against any (lower) score, a score of 0-20 is ongoing

Against a score of up to 28, a score of +1 is ongoing

Anything else (including negative scores) is illegal

answered Mar 26 at 20:09

Neil

82.6k745179

add a comment |

Retina 0.8.2, 92 bytes

d+

$*

^(1{0,19},1{21}|(1{20,28}),112|1{29},1{30})$|^(1*,1{0,20}|(1{0,28}),1?4)$|.+

$#1$#3

Against 0-19, a score of 21 is a win

Against 20-28, a score of +2 is a win

Against 29, a score of 30 is a win

Against any (lower) score, a score of 0-20 is ongoing

Against a score of up to 28, a score of +1 is ongoing

Anything else (including negative scores) is illegal

answered Mar 26 at 20:09

Neil

82.6k745179

add a comment |

Retina 0.8.2, 92 bytes

d+

$*

^(1{0,19},1{21}|(1{20,28}),112|1{29},1{30})$|^(1*,1{0,20}|(1{0,28}),1?4)$|.+

$#1$#3

Against 0-19, a score of 21 is a win

Against 20-28, a score of +2 is a win

Against 29, a score of 30 is a win

Against any (lower) score, a score of 0-20 is ongoing

Against a score of up to 28, a score of +1 is ongoing

Anything else (including negative scores) is illegal

answered Mar 26 at 20:09

Neil

82.6k745179

Retina 0.8.2, 92 bytes

d+

$*

^(1{0,19},1{21}|(1{20,28}),112|1{29},1{30})$|^(1*,1{0,20}|(1{0,28}),1?4)$|.+

$#1$#3

Against 0-19, a score of 21 is a win

Against 20-28, a score of +2 is a win

Against 29, a score of 30 is a win

Against any (lower) score, a score of 0-20 is ongoing

Against a score of up to 28, a score of +1 is ongoing

Anything else (including negative scores) is illegal

answered Mar 26 at 20:09

Neil

82.6k745179

answered Mar 26 at 20:09

Neil

82.6k745179

answered Mar 26 at 20:09

Neil

82.6k745179

answered Mar 26 at 20:09

Neil

82.6k745179

add a comment |

APL (Dyalog Unicode), 35 bytes^SBCS

Infix tacit function where ended is 2, ongoing is 1, invalid is 0, smaller and larger scores are left.

(,≡30 31|,)×(⊢<2+X)×1+⊢>X←29⌊20⌈1+⊣

Try it online!

Implements Erik the Outgolfer's mathematical formulas combined into

$$X:=min(max(x+1,20),29)\ ([X< y]+1)[X+2>y][(x,y)=(xbmod30,ybmod31)]$$
rearranged (as if traditional mathematical notation had vectorisation and inline assignments) to

$$[(x,y)=(x,y)bmod(30,31)]×[y<2+X]×(1+[y< (X:=min(29,max(20,1+x)))])$$

and translated directly to APL (which is strictly right-associative, so we avoid some parentheses):

$$((x,y)≡30 31|x,y)×(y<2+X)×1+y>X←29⌊20⌈1 +x$$

This can be converted into a tacit function simply by substituting $⊣$ for $x$ and $⊢$ for $y$, symbolising the left and right arguments rather than the two variables:

$$((⊣,⊢)≡30 31|⊣,⊢)×(⊣<2+X)×1+⊢>X←29⌊20⌈1+⊣$$

Now $⊣⎕⊢$ is equivalent to $⎕$ for any infix function $⎕$, so we can simplify to

$$(,≡30 31|,)×(⊣<2+X)×1+⊢>X←29⌊20⌈1+⊣$$

which is our solution; (,≡30 31|,)×(⊢<2+X)×1+⊢>X←29⌊20⌈1+⊣:

edited Mar 28 at 11:01

answered Mar 28 at 10:52

Adám

28.9k276207

add a comment |

APL (Dyalog Unicode), 35 bytes^SBCS

Infix tacit function where ended is 2, ongoing is 1, invalid is 0, smaller and larger scores are left.

(,≡30 31|,)×(⊢<2+X)×1+⊢>X←29⌊20⌈1+⊣

Try it online!

Implements Erik the Outgolfer's mathematical formulas combined into

$$X:=min(max(x+1,20),29)\ ([X< y]+1)[X+2>y][(x,y)=(xbmod30,ybmod31)]$$
rearranged (as if traditional mathematical notation had vectorisation and inline assignments) to

$$[(x,y)=(x,y)bmod(30,31)]×[y<2+X]×(1+[y< (X:=min(29,max(20,1+x)))])$$

and translated directly to APL (which is strictly right-associative, so we avoid some parentheses):

$$((x,y)≡30 31|x,y)×(y<2+X)×1+y>X←29⌊20⌈1 +x$$

This can be converted into a tacit function simply by substituting $⊣$ for $x$ and $⊢$ for $y$, symbolising the left and right arguments rather than the two variables:

$$((⊣,⊢)≡30 31|⊣,⊢)×(⊣<2+X)×1+⊢>X←29⌊20⌈1+⊣$$

Now $⊣⎕⊢$ is equivalent to $⎕$ for any infix function $⎕$, so we can simplify to

$$(,≡30 31|,)×(⊣<2+X)×1+⊢>X←29⌊20⌈1+⊣$$

which is our solution; (,≡30 31|,)×(⊢<2+X)×1+⊢>X←29⌊20⌈1+⊣:

edited Mar 28 at 11:01

answered Mar 28 at 10:52

Adám

28.9k276207

add a comment |

APL (Dyalog Unicode), 35 bytes^SBCS

Infix tacit function where ended is 2, ongoing is 1, invalid is 0, smaller and larger scores are left.

(,≡30 31|,)×(⊢<2+X)×1+⊢>X←29⌊20⌈1+⊣

Try it online!

Implements Erik the Outgolfer's mathematical formulas combined into

$$X:=min(max(x+1,20),29)\ ([X< y]+1)[X+2>y][(x,y)=(xbmod30,ybmod31)]$$
rearranged (as if traditional mathematical notation had vectorisation and inline assignments) to

$$[(x,y)=(x,y)bmod(30,31)]×[y<2+X]×(1+[y< (X:=min(29,max(20,1+x)))])$$

and translated directly to APL (which is strictly right-associative, so we avoid some parentheses):

$$((x,y)≡30 31|x,y)×(y<2+X)×1+y>X←29⌊20⌈1 +x$$

This can be converted into a tacit function simply by substituting $⊣$ for $x$ and $⊢$ for $y$, symbolising the left and right arguments rather than the two variables:

$$((⊣,⊢)≡30 31|⊣,⊢)×(⊣<2+X)×1+⊢>X←29⌊20⌈1+⊣$$

Now $⊣⎕⊢$ is equivalent to $⎕$ for any infix function $⎕$, so we can simplify to

$$(,≡30 31|,)×(⊣<2+X)×1+⊢>X←29⌊20⌈1+⊣$$

which is our solution; (,≡30 31|,)×(⊢<2+X)×1+⊢>X←29⌊20⌈1+⊣:

edited Mar 28 at 11:01

answered Mar 28 at 10:52

Adám

28.9k276207

APL (Dyalog Unicode), 35 bytes^SBCS

Infix tacit function where ended is 2, ongoing is 1, invalid is 0, smaller and larger scores are left.

(,≡30 31|,)×(⊢<2+X)×1+⊢>X←29⌊20⌈1+⊣

Try it online!

Implements Erik the Outgolfer's mathematical formulas combined into

$$X:=min(max(x+1,20),29)\ ([X< y]+1)[X+2>y][(x,y)=(xbmod30,ybmod31)]$$
rearranged (as if traditional mathematical notation had vectorisation and inline assignments) to

$$[(x,y)=(x,y)bmod(30,31)]×[y<2+X]×(1+[y< (X:=min(29,max(20,1+x)))])$$

and translated directly to APL (which is strictly right-associative, so we avoid some parentheses):

$$((x,y)≡30 31|x,y)×(y<2+X)×1+y>X←29⌊20⌈1 +x$$

This can be converted into a tacit function simply by substituting $⊣$ for $x$ and $⊢$ for $y$, symbolising the left and right arguments rather than the two variables:

$$((⊣,⊢)≡30 31|⊣,⊢)×(⊣<2+X)×1+⊢>X←29⌊20⌈1+⊣$$

Now $⊣⎕⊢$ is equivalent to $⎕$ for any infix function $⎕$, so we can simplify to

$$(,≡30 31|,)×(⊣<2+X)×1+⊢>X←29⌊20⌈1+⊣$$

which is our solution; (,≡30 31|,)×(⊢<2+X)×1+⊢>X←29⌊20⌈1+⊣:

edited Mar 28 at 11:01

answered Mar 28 at 10:52

Adám

28.9k276207

edited Mar 28 at 11:01

answered Mar 28 at 10:52

Adám

28.9k276207

answered Mar 28 at 10:52

Adám

28.9k276207

answered Mar 28 at 10:52

Adám

28.9k276207

add a comment |

x86 Assembly, 42 Bytes

31 C0 83 F9 1E 77 1F 83 FA 1D 77 1A 83 F9 15 7C 

18 83 F9 1E 74 12 89 CB 29 D3 83 FB 02 74 09 7C 

08 83 F9 15 74 02 48 C3 40 C3

Or, more readable in Intel Syntax:

check:

    XOR EAX, EAX

    CMP ECX, 30     ; check i_1 against 30

    JA .invalid     ; if >, invalid.

    CMP EDX, 29     ; check i_2 against 29

    JA .invalid     ; if >, invalid.

    CMP ECX, 21     ; check i_1 against 21

    JL .runi        ; if <, running.

    CMP ECX, 30     ; check i_1 against 30

    JE .over        ; if ==, over.

    MOV EBX, ECX

    SUB EBX, EDX    ; EBX = i_1 - i_2

    CMP EBX, 2      ; check EBX against 2

    JE .over        ; if ==, over.

    JL .runi        ; if <, running.

                    ; if >, keep executing!

    CMP ECX, 21     ; check i_1 against 21

    JE .over        ; if ==, over.

                    ; otherwise, it's invalid.

    ; fallthrough!

    .invalid:

        DEC EAX     ; EAX = -1

        RETN

    .over:

        INC EAX     ; EAX = 1

    ; fallthrough!

    .runi:

        RETN        ; EAX = 0 or 1

Fun fact: this function almost follows the C Calling Convention's rules on which registers to preserve, except I had to clobber EBX to save some bytes on stack usage.

Optional (not included in byte-count)

By adding the following 6 bytes directly before start of the code above, you can pass ECX and EDX unordered:

39 D1 7D 02 87 CA

Which is the following in readable Intel Syntax:

CMP ECX, EDX

JGE check

XCHG ECX, EDX

answered Mar 28 at 21:46

Fayti1703

513

add a comment |

x86 Assembly, 42 Bytes

31 C0 83 F9 1E 77 1F 83 FA 1D 77 1A 83 F9 15 7C 

18 83 F9 1E 74 12 89 CB 29 D3 83 FB 02 74 09 7C 

08 83 F9 15 74 02 48 C3 40 C3

Or, more readable in Intel Syntax:

check:

    XOR EAX, EAX

    CMP ECX, 30     ; check i_1 against 30

    JA .invalid     ; if >, invalid.

    CMP EDX, 29     ; check i_2 against 29

    JA .invalid     ; if >, invalid.

    CMP ECX, 21     ; check i_1 against 21

    JL .runi        ; if <, running.

    CMP ECX, 30     ; check i_1 against 30

    JE .over        ; if ==, over.

    MOV EBX, ECX

    SUB EBX, EDX    ; EBX = i_1 - i_2

    CMP EBX, 2      ; check EBX against 2

    JE .over        ; if ==, over.

    JL .runi        ; if <, running.

                    ; if >, keep executing!

    CMP ECX, 21     ; check i_1 against 21

    JE .over        ; if ==, over.

                    ; otherwise, it's invalid.

    ; fallthrough!

    .invalid:

        DEC EAX     ; EAX = -1

        RETN

    .over:

        INC EAX     ; EAX = 1

    ; fallthrough!

    .runi:

        RETN        ; EAX = 0 or 1

Fun fact: this function almost follows the C Calling Convention's rules on which registers to preserve, except I had to clobber EBX to save some bytes on stack usage.

Optional (not included in byte-count)

By adding the following 6 bytes directly before start of the code above, you can pass ECX and EDX unordered:

39 D1 7D 02 87 CA

Which is the following in readable Intel Syntax:

CMP ECX, EDX

JGE check

XCHG ECX, EDX

answered Mar 28 at 21:46

Fayti1703

513

add a comment |

x86 Assembly, 42 Bytes

31 C0 83 F9 1E 77 1F 83 FA 1D 77 1A 83 F9 15 7C 

18 83 F9 1E 74 12 89 CB 29 D3 83 FB 02 74 09 7C 

08 83 F9 15 74 02 48 C3 40 C3

Or, more readable in Intel Syntax:

check:

    XOR EAX, EAX

    CMP ECX, 30     ; check i_1 against 30

    JA .invalid     ; if >, invalid.

    CMP EDX, 29     ; check i_2 against 29

    JA .invalid     ; if >, invalid.

    CMP ECX, 21     ; check i_1 against 21

    JL .runi        ; if <, running.

    CMP ECX, 30     ; check i_1 against 30

    JE .over        ; if ==, over.

    MOV EBX, ECX

    SUB EBX, EDX    ; EBX = i_1 - i_2

    CMP EBX, 2      ; check EBX against 2

    JE .over        ; if ==, over.

    JL .runi        ; if <, running.

                    ; if >, keep executing!

    CMP ECX, 21     ; check i_1 against 21

    JE .over        ; if ==, over.

                    ; otherwise, it's invalid.

    ; fallthrough!

    .invalid:

        DEC EAX     ; EAX = -1

        RETN

    .over:

        INC EAX     ; EAX = 1

    ; fallthrough!

    .runi:

        RETN        ; EAX = 0 or 1

Fun fact: this function almost follows the C Calling Convention's rules on which registers to preserve, except I had to clobber EBX to save some bytes on stack usage.

Optional (not included in byte-count)

By adding the following 6 bytes directly before start of the code above, you can pass ECX and EDX unordered:

39 D1 7D 02 87 CA

Which is the following in readable Intel Syntax:

CMP ECX, EDX

JGE check

XCHG ECX, EDX

answered Mar 28 at 21:46

Fayti1703

513

x86 Assembly, 42 Bytes

31 C0 83 F9 1E 77 1F 83 FA 1D 77 1A 83 F9 15 7C 

18 83 F9 1E 74 12 89 CB 29 D3 83 FB 02 74 09 7C 

08 83 F9 15 74 02 48 C3 40 C3

Or, more readable in Intel Syntax:

check:

    XOR EAX, EAX

    CMP ECX, 30     ; check i_1 against 30

    JA .invalid     ; if >, invalid.

    CMP EDX, 29     ; check i_2 against 29

    JA .invalid     ; if >, invalid.

    CMP ECX, 21     ; check i_1 against 21

    JL .runi        ; if <, running.

    CMP ECX, 30     ; check i_1 against 30

    JE .over        ; if ==, over.

    MOV EBX, ECX

    SUB EBX, EDX    ; EBX = i_1 - i_2

    CMP EBX, 2      ; check EBX against 2

    JE .over        ; if ==, over.

    JL .runi        ; if <, running.

                    ; if >, keep executing!

    CMP ECX, 21     ; check i_1 against 21

    JE .over        ; if ==, over.

                    ; otherwise, it's invalid.

    ; fallthrough!

    .invalid:

        DEC EAX     ; EAX = -1

        RETN

    .over:

        INC EAX     ; EAX = 1

    ; fallthrough!

    .runi:

        RETN        ; EAX = 0 or 1

Fun fact: this function almost follows the C Calling Convention's rules on which registers to preserve, except I had to clobber EBX to save some bytes on stack usage.

Optional (not included in byte-count)

By adding the following 6 bytes directly before start of the code above, you can pass ECX and EDX unordered:

39 D1 7D 02 87 CA

Which is the following in readable Intel Syntax:

CMP ECX, EDX

JGE check

XCHG ECX, EDX

answered Mar 28 at 21:46

Fayti1703

513

answered Mar 28 at 21:46

Fayti1703

513

answered Mar 28 at 21:46

Fayti1703

513

answered Mar 28 at 21:46

Fayti1703

513

add a comment |

APL (Dyalog Unicode), 33 32 bytes^SBCS

{h⍵+1 0}+h←(⊢≡31 30|⊢)×21 2∨.≥-

Try it online!

in: a pair in descending order

out: 2=ongoing, 1=ended, 0=invalid

tests stolen from Adám's answer

edited Apr 1 at 2:25

answered Apr 1 at 2:09

ngn

7,39612660

add a comment |

APL (Dyalog Unicode), 33 32 bytes^SBCS

{h⍵+1 0}+h←(⊢≡31 30|⊢)×21 2∨.≥-

Try it online!

in: a pair in descending order

out: 2=ongoing, 1=ended, 0=invalid

tests stolen from Adám's answer

edited Apr 1 at 2:25

answered Apr 1 at 2:09

ngn

7,39612660

add a comment |

APL (Dyalog Unicode), 33 32 bytes^SBCS

{h⍵+1 0}+h←(⊢≡31 30|⊢)×21 2∨.≥-

Try it online!

in: a pair in descending order

out: 2=ongoing, 1=ended, 0=invalid

tests stolen from Adám's answer

edited Apr 1 at 2:25

answered Apr 1 at 2:09

ngn

7,39612660

APL (Dyalog Unicode), 33 32 bytes^SBCS

{h⍵+1 0}+h←(⊢≡31 30|⊢)×21 2∨.≥-

Try it online!

in: a pair in descending order

out: 2=ongoing, 1=ended, 0=invalid

tests stolen from Adám's answer

edited Apr 1 at 2:25

answered Apr 1 at 2:09

ngn

7,39612660

edited Apr 1 at 2:25

answered Apr 1 at 2:09

ngn

7,39612660

answered Apr 1 at 2:09

ngn

7,39612660

answered Apr 1 at 2:09

ngn

7,39612660

add a comment |

Bash 4+, 97 89 91 88 bytes

Assume that inputs are ascending. Used concepts from VDM-SL answer. Try it Online

z==0 - game in progress
z==1 - game completed
z==2 - invalid

-8 by bracket cleanup from (( & | )) conditions
+2 fixing a bug, thanks to Kevin Cruijssen
-3 logic improvements by Kevin Cruijssen

i=$1 j=$2 z=0

((j-i>2&j>21|i<0|i>29|j>30?z=2:0))

((z<1&(j>20&j-i>1|j>29)?z=1:0))

echo $z

edited Apr 3 at 8:46

answered Mar 28 at 13:40

roblogic

1515

1

$begingroup$
Your 89 bytes version seems to output 1 instead of 2 for 0 30. Your 97 bytes version was working correctly, so if you're unable to fix it, you can always rollback. Upvoted for that 97 version. :)
$endgroup$
– Kevin Cruijssen
Mar 28 at 13:59

1

$begingroup$
This should fix it, and golf a byte at the same time. :)
$endgroup$
– Kevin Cruijssen
Mar 28 at 14:05

1

$begingroup$
I fixed it, but yours was better! Hard to keep up :P
$endgroup$
– roblogic
Mar 28 at 14:14

$begingroup$
Bug at 29 30 :( it should be "completed"
$endgroup$
– roblogic
Apr 3 at 8:39

1

$begingroup$
Ah oops.. the i>29 should be j>29 in the second ternary to fix it.
$endgroup$
– Kevin Cruijssen
Apr 3 at 8:42

add a comment |

Bash 4+, 97 89 91 88 bytes

Assume that inputs are ascending. Used concepts from VDM-SL answer. Try it Online

z==0 - game in progress
z==1 - game completed
z==2 - invalid

-8 by bracket cleanup from (( & | )) conditions
+2 fixing a bug, thanks to Kevin Cruijssen
-3 logic improvements by Kevin Cruijssen

i=$1 j=$2 z=0

((j-i>2&j>21|i<0|i>29|j>30?z=2:0))

((z<1&(j>20&j-i>1|j>29)?z=1:0))

echo $z

edited Apr 3 at 8:46

answered Mar 28 at 13:40

roblogic

1515

1

$begingroup$
Your 89 bytes version seems to output 1 instead of 2 for 0 30. Your 97 bytes version was working correctly, so if you're unable to fix it, you can always rollback. Upvoted for that 97 version. :)
$endgroup$
– Kevin Cruijssen
Mar 28 at 13:59

1

$begingroup$
This should fix it, and golf a byte at the same time. :)
$endgroup$
– Kevin Cruijssen
Mar 28 at 14:05

1

$begingroup$
I fixed it, but yours was better! Hard to keep up :P
$endgroup$
– roblogic
Mar 28 at 14:14

$begingroup$
Bug at 29 30 :( it should be "completed"
$endgroup$
– roblogic
Apr 3 at 8:39

1

$begingroup$
Ah oops.. the i>29 should be j>29 in the second ternary to fix it.
$endgroup$
– Kevin Cruijssen
Apr 3 at 8:42

add a comment |

Bash 4+, 97 89 91 88 bytes

Assume that inputs are ascending. Used concepts from VDM-SL answer. Try it Online

z==0 - game in progress
z==1 - game completed
z==2 - invalid

-8 by bracket cleanup from (( & | )) conditions
+2 fixing a bug, thanks to Kevin Cruijssen
-3 logic improvements by Kevin Cruijssen

i=$1 j=$2 z=0

((j-i>2&j>21|i<0|i>29|j>30?z=2:0))

((z<1&(j>20&j-i>1|j>29)?z=1:0))

echo $z

edited Apr 3 at 8:46

answered Mar 28 at 13:40

roblogic

1515

Bash 4+, 97 89 91 88 bytes

Assume that inputs are ascending. Used concepts from VDM-SL answer. Try it Online

z==0 - game in progress
z==1 - game completed
z==2 - invalid

-8 by bracket cleanup from (( & | )) conditions
+2 fixing a bug, thanks to Kevin Cruijssen
-3 logic improvements by Kevin Cruijssen

i=$1 j=$2 z=0

((j-i>2&j>21|i<0|i>29|j>30?z=2:0))

((z<1&(j>20&j-i>1|j>29)?z=1:0))

echo $z

edited Apr 3 at 8:46

answered Mar 28 at 13:40

roblogic

1515

edited Apr 3 at 8:46

answered Mar 28 at 13:40

roblogic

1515

answered Mar 28 at 13:40

roblogic

1515

answered Mar 28 at 13:40

roblogic

1515

1

$begingroup$
Your 89 bytes version seems to output 1 instead of 2 for 0 30. Your 97 bytes version was working correctly, so if you're unable to fix it, you can always rollback. Upvoted for that 97 version. :)
$endgroup$
– Kevin Cruijssen
Mar 28 at 13:59

1

$begingroup$
This should fix it, and golf a byte at the same time. :)
$endgroup$
– Kevin Cruijssen
Mar 28 at 14:05

1

$begingroup$
I fixed it, but yours was better! Hard to keep up :P
$endgroup$
– roblogic
Mar 28 at 14:14

$begingroup$
Bug at 29 30 :( it should be "completed"
$endgroup$
– roblogic
Apr 3 at 8:39

1

$begingroup$
Ah oops.. the i>29 should be j>29 in the second ternary to fix it.
$endgroup$
– Kevin Cruijssen
Apr 3 at 8:42

add a comment |

1

$begingroup$
Your 89 bytes version seems to output 1 instead of 2 for 0 30. Your 97 bytes version was working correctly, so if you're unable to fix it, you can always rollback. Upvoted for that 97 version. :)
$endgroup$
– Kevin Cruijssen
Mar 28 at 13:59

1

$begingroup$
This should fix it, and golf a byte at the same time. :)
$endgroup$
– Kevin Cruijssen
Mar 28 at 14:05

1

$begingroup$
I fixed it, but yours was better! Hard to keep up :P
$endgroup$
– roblogic
Mar 28 at 14:14

$begingroup$
Bug at 29 30 :( it should be "completed"
$endgroup$
– roblogic
Apr 3 at 8:39

1

$begingroup$
Ah oops.. the i>29 should be j>29 in the second ternary to fix it.
$endgroup$
– Kevin Cruijssen
Apr 3 at 8:42

Your 89 bytes version seems to output 1 instead of 2 for 0 30. Your 97 bytes version was working correctly, so if you're unable to fix it, you can always rollback. Upvoted for that 97 version. :)

– Kevin Cruijssen
Mar 28 at 13:59

This should fix it, and golf a byte at the same time. :)

– Kevin Cruijssen
Mar 28 at 14:05

I fixed it, but yours was better! Hard to keep up :P

– roblogic
Mar 28 at 14:14

Bug at 29 30 :( it should be "completed"

– roblogic
Apr 3 at 8:39

Ah oops.. the i>29 should be j>29 in the second ternary to fix it.

– Kevin Cruijssen
Apr 3 at 8:42

add a comment |

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

draft saved

draft discarded

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

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Valid Badminton Score?

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13 Answers 13

Stax, 20 bytes

Python 2, 97 95 75 72 71 70 69 64 55 54 52 51 50 48 bytes

Python 2, 47 bytes

Python 2, 44 bytes

Python 2, 43 bytes

JavaScript (ES6), 55 53 48 bytes

C# (Visual C# Interactive Compiler), 53 52 bytes

Jelly, 25 bytes

VDM-SL, 80 bytes

Explanation:

Java (JDK), 59 48 bytes

Ungolfed

Retina 0.8.2, 92 bytes

APL (Dyalog Unicode), 35 bytesSBCS

x86 Assembly, 42 Bytes

Optional (not included in byte-count)

APL (Dyalog Unicode), 33 32 bytesSBCS

Bash 4+, 97 89 91 88 bytes

Your Answer

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13 Answers 13

13 Answers 13

Stax, 20 bytes

Stax, 20 bytes

Stax, 20 bytes

Stax, 20 bytes

Python 2, 97 95 75 72 71 70 69 64 55 54 52 51 50 48 bytes

Python 2, 97 95 75 72 71 70 69 64 55 54 52 51 50 48 bytes

Python 2, 97 95 75 72 71 70 69 64 55 54 52 51 50 48 bytes

Python 2, 97 95 75 72 71 70 69 64 55 54 52 51 50 48 bytes

Python 2, 47 bytes

Python 2, 44 bytes

Python 2, 43 bytes

Python 2, 47 bytes

Python 2, 44 bytes

Python 2, 43 bytes

Python 2, 47 bytes

Python 2, 44 bytes

Python 2, 43 bytes

Python 2, 47 bytes

Python 2, 44 bytes

Python 2, 43 bytes

JavaScript (ES6), 55 53 48 bytes

JavaScript (ES6), 55 53 48 bytes

JavaScript (ES6), 55 53 48 bytes

JavaScript (ES6), 55 53 48 bytes

C# (Visual C# Interactive Compiler), 53 52 bytes

C# (Visual C# Interactive Compiler), 53 52 bytes

C# (Visual C# Interactive Compiler), 53 52 bytes

C# (Visual C# Interactive Compiler), 53 52 bytes

Jelly, 25 bytes

Jelly, 25 bytes

Jelly, 25 bytes

Jelly, 25 bytes

VDM-SL, 80 bytes

Explanation:

13 Answers
13

APL (Dyalog Unicode), 35 bytes^SBCS

APL (Dyalog Unicode), 33 32 bytes^SBCS

13 Answers
13

13 Answers
13

APL (Dyalog Unicode), 35 bytes^SBCS

APL (Dyalog Unicode), 35 bytes^SBCS

APL (Dyalog Unicode), 35 bytes^SBCS

APL (Dyalog Unicode), 35 bytes^SBCS

APL (Dyalog Unicode), 33 32 bytes^SBCS

APL (Dyalog Unicode), 33 32 bytes^SBCS

APL (Dyalog Unicode), 33 32 bytes^SBCS

APL (Dyalog Unicode), 33 32 bytes^SBCS