up vote
5
down vote

favorite

(thanks to @JonathanFrech for the title)

Challenge

Write a program or function that, given a collection of positive integers $S$ and a positive integer $n$, finds as many non-overlapping subcollections $T_1, T_2, ..., T_k$ of $S$ as possible, such that the sum of every $T_m$ is equal to $n$.

Example

Given $S=lbrace1,1,2,2,3,3,3,5,6,8rbrace$ and $n=6$, you could take the following subcollections to get $k=3$:

$T_1=lbrace1,1,2,2rbrace$

$T_2=lbrace3,3rbrace$

$T_3=lbrace6rbrace$

And the remaining elements $lbrace3,5,8rbrace$ are not used. However, there is a way to get $k=4$ separate subcollections:

$T_1=lbrace1,2,3rbrace$

$T_2=lbrace1,5rbrace$

$T_3=lbrace3,3rbrace$

$T_4=lbrace6rbrace$

This leaves $lbrace2,8rbrace$ unused. There is no way to do this with 5 subcollections¹, so $k=4$ for this example.

Input

Input will be a collection of positive integers $S$, and a positive integer $n$. $S$ may be taken in any reasonable format: a comma- or newline-separated string, an actual array/list, etc. You may assume that $S$ is sorted (if your input format has a defined order).

Output

The output may be either:

the maximal number of subcollections $k$; or

any collection which has $k$ items (e.g., the list of subcollections $T$ itself), in any reasonable format.

Note that there is not guaranteed to be any subcollections with sum $n$ (i.e. some inputs will output 0, , etc.)

Test cases

n, S -> k
1, [1] -> 1
1, [2] -> 0
2, [1] -> 0
1, [1, 1] -> 2
2, [1, 1, 1] -> 1
3, [1, 1, 2, 2] -> 2
9, [1, 1, 3, 3, 5, 5, 7] -> 2
9, [1, 1, 3, 3, 5, 7, 7] -> 1
8, [1, 1, 2, 3, 4, 6, 6, 6] -> 2
6, [1, 1, 2, 2, 3, 3, 3, 5, 6, 8] -> 4
13, [2, 2, 2, 2, 3, 3, 4, 5, 5, 6, 8, 9, 9, 11, 12] -> 5
22, [1, 1, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 6, 6, 7, 7, 7, 8, 9, 9, 10, 13, 14, 14, 17, 18, 18, 21] -> 10

Scoring

This is code-golf, so the shortest program in bytes in each language wins. Good luck!

^{1. Mini-proof: the $8$ is obviously useless, but when you remove it, the total sum of $S$ becomes $26$; therefore, it can only contain up to $4$ non-overlapping subcollections whose sums are $6$.}

edited 4 hours ago

asked 5 hours ago

ETHproductions

43.9k572219

add a commentÂ |Â

up vote
5
down vote

favorite

(thanks to @JonathanFrech for the title)

Challenge

Example

Given $S=lbrace1,1,2,2,3,3,3,5,6,8rbrace$ and $n=6$, you could take the following subcollections to get $k=3$:

$T_1=lbrace1,1,2,2rbrace$

$T_2=lbrace3,3rbrace$

$T_3=lbrace6rbrace$

And the remaining elements $lbrace3,5,8rbrace$ are not used. However, there is a way to get $k=4$ separate subcollections:

$T_1=lbrace1,2,3rbrace$

$T_2=lbrace1,5rbrace$

$T_3=lbrace3,3rbrace$

$T_4=lbrace6rbrace$

This leaves $lbrace2,8rbrace$ unused. There is no way to do this with 5 subcollections¹, so $k=4$ for this example.

Input

Output

The output may be either:

the maximal number of subcollections $k$; or

any collection which has $k$ items (e.g., the list of subcollections $T$ itself), in any reasonable format.

Note that there is not guaranteed to be any subcollections with sum $n$ (i.e. some inputs will output 0, , etc.)

Test cases

n, S -> k
1, [1] -> 1
1, [2] -> 0
2, [1] -> 0
1, [1, 1] -> 2
2, [1, 1, 1] -> 1
3, [1, 1, 2, 2] -> 2
9, [1, 1, 3, 3, 5, 5, 7] -> 2
9, [1, 1, 3, 3, 5, 7, 7] -> 1
8, [1, 1, 2, 3, 4, 6, 6, 6] -> 2
6, [1, 1, 2, 2, 3, 3, 3, 5, 6, 8] -> 4
13, [2, 2, 2, 2, 3, 3, 4, 5, 5, 6, 8, 9, 9, 11, 12] -> 5
22, [1, 1, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 6, 6, 7, 7, 7, 8, 9, 9, 10, 13, 14, 14, 17, 18, 18, 21] -> 10

Scoring

This is code-golf, so the shortest program in bytes in each language wins. Good luck!

^{1. Mini-proof: the $8$ is obviously useless, but when you remove it, the total sum of $S$ becomes $26$; therefore, it can only contain up to $4$ non-overlapping subcollections whose sums are $6$.}

edited 4 hours ago

asked 5 hours ago

ETHproductions

43.9k572219

add a commentÂ |Â

up vote
5
down vote

favorite

(thanks to @JonathanFrech for the title)

Challenge

Example

Given $S=lbrace1,1,2,2,3,3,3,5,6,8rbrace$ and $n=6$, you could take the following subcollections to get $k=3$:

$T_1=lbrace1,1,2,2rbrace$

$T_2=lbrace3,3rbrace$

$T_3=lbrace6rbrace$

And the remaining elements $lbrace3,5,8rbrace$ are not used. However, there is a way to get $k=4$ separate subcollections:

$T_1=lbrace1,2,3rbrace$

$T_2=lbrace1,5rbrace$

$T_3=lbrace3,3rbrace$

$T_4=lbrace6rbrace$

This leaves $lbrace2,8rbrace$ unused. There is no way to do this with 5 subcollections¹, so $k=4$ for this example.

Input

Output

The output may be either:

the maximal number of subcollections $k$; or

any collection which has $k$ items (e.g., the list of subcollections $T$ itself), in any reasonable format.

Note that there is not guaranteed to be any subcollections with sum $n$ (i.e. some inputs will output 0, , etc.)

Test cases

n, S -> k
1, [1] -> 1
1, [2] -> 0
2, [1] -> 0
1, [1, 1] -> 2
2, [1, 1, 1] -> 1
3, [1, 1, 2, 2] -> 2
9, [1, 1, 3, 3, 5, 5, 7] -> 2
9, [1, 1, 3, 3, 5, 7, 7] -> 1
8, [1, 1, 2, 3, 4, 6, 6, 6] -> 2
6, [1, 1, 2, 2, 3, 3, 3, 5, 6, 8] -> 4
13, [2, 2, 2, 2, 3, 3, 4, 5, 5, 6, 8, 9, 9, 11, 12] -> 5
22, [1, 1, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 6, 6, 7, 7, 7, 8, 9, 9, 10, 13, 14, 14, 17, 18, 18, 21] -> 10

Scoring

This is code-golf, so the shortest program in bytes in each language wins. Good luck!

^{1. Mini-proof: the $8$ is obviously useless, but when you remove it, the total sum of $S$ becomes $26$; therefore, it can only contain up to $4$ non-overlapping subcollections whose sums are $6$.}

edited 4 hours ago

asked 5 hours ago

ETHproductions

43.9k572219

(thanks to @JonathanFrech for the title)

Challenge

Example

Given $S=lbrace1,1,2,2,3,3,3,5,6,8rbrace$ and $n=6$, you could take the following subcollections to get $k=3$:

$T_1=lbrace1,1,2,2rbrace$

$T_2=lbrace3,3rbrace$

$T_3=lbrace6rbrace$

And the remaining elements $lbrace3,5,8rbrace$ are not used. However, there is a way to get $k=4$ separate subcollections:

$T_1=lbrace1,2,3rbrace$

$T_2=lbrace1,5rbrace$

$T_3=lbrace3,3rbrace$

$T_4=lbrace6rbrace$

This leaves $lbrace2,8rbrace$ unused. There is no way to do this with 5 subcollections¹, so $k=4$ for this example.

Input

Output

The output may be either:

the maximal number of subcollections $k$; or

any collection which has $k$ items (e.g., the list of subcollections $T$ itself), in any reasonable format.

Note that there is not guaranteed to be any subcollections with sum $n$ (i.e. some inputs will output 0, , etc.)

Test cases

n, S -> k
1, [1] -> 1
1, [2] -> 0
2, [1] -> 0
1, [1, 1] -> 2
2, [1, 1, 1] -> 1
3, [1, 1, 2, 2] -> 2
9, [1, 1, 3, 3, 5, 5, 7] -> 2
9, [1, 1, 3, 3, 5, 7, 7] -> 1
8, [1, 1, 2, 3, 4, 6, 6, 6] -> 2
6, [1, 1, 2, 2, 3, 3, 3, 5, 6, 8] -> 4
13, [2, 2, 2, 2, 3, 3, 4, 5, 5, 6, 8, 9, 9, 11, 12] -> 5
22, [1, 1, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 6, 6, 7, 7, 7, 8, 9, 9, 10, 13, 14, 14, 17, 18, 18, 21] -> 10

Scoring

This is code-golf, so the shortest program in bytes in each language wins. Good luck!

^{1. Mini-proof: the $8$ is obviously useless, but when you remove it, the total sum of $S$ becomes $26$; therefore, it can only contain up to $4$ non-overlapping subcollections whose sums are $6$.}

code-golf number combinatorics

edited 4 hours ago

asked 5 hours ago

ETHproductions

43.9k572219

edited 4 hours ago

asked 5 hours ago

ETHproductions

43.9k572219

edited 4 hours ago

asked 5 hours ago

ETHproductions

43.9k572219

asked 5 hours ago

ETHproductions

43.9k572219

asked 5 hours ago

ETHproductions

43.9k572219

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
1
down vote

Jelly, 17 bytes

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã¢Â‚Â¬Ã¡ÂºÂŽÃ…Â’PÃ¢Â‚Â¬Ã¡ÂºÂŽÃ‚Â§;EÃ‰Â—Ã†Â‡Ã¡ÂºÂˆÃ¡Â¹Â€

Try it online!

Outputs $k$ if there is a $T$, $0$ otherwise. Very slow.

edited 4 hours ago

answered 4 hours ago

Erik the Outgolfer

29.9k42899

add a commentÂ |Â

up vote
1
down vote

Jelly, Â 17Â 11 bytes

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã‚Â§=Ã‰Â—Ã¢Â‚Â¬Ã‚Â§FÃ¡Â¹Â€

A dyadic link accepting a list and an integer which yields the maximal count, $k$.

Try it online! Or see the test-suite -- it's too slow for the larger test cases :(

How?

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã‚Â§=Ã‰Â—Ã¢Â‚Â¬Ã‚Â§FÃ¡Â¹Â€ - Link: list L, integer I
Ã…Â’! - all permutations
 Ã¢Â‚Â¬ - for Ã¢Â‚Â¬ach:
 Ã‰Â— - last three links as a dyad - i.e. f(permutation, I)
 Ã…Â’Ã¡Â¹Â– - partitions (all ways to slice up the permutation)
 Ã‚Â§ - sum each (vectorises at depth 1, so this gets the sums of the parts of
 - the partitions)
 = - equals? (==I) -> 1 if so 0 otherwise 
 Ã‚Â§ - sum each (vectorises at depth 1, so this gets the counts of parts equal to I
 - for each partition of each permutation) 
 F - flatten
 Ã¡Â¹Â€ - maximum

edited 40 mins ago

answered 1 hour ago

Jonathan Allan

49.5k534163

add a commentÂ |Â

up vote
0
down vote

JavaScript (Node.js), 115 bytes

k=>f=([i,...s])=>i?Math.max(...[i,...s].map((_,j)=>(e[j]=~~e[j]+i,_=f(s),e[j]-=i,_))):e.filter(_=>_==k).length;e=

Try it online!

edited 3 hours ago

answered 3 hours ago

l4m2

4,0381431

add a commentÂ |Â

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Name

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
1
down vote

Jelly, 17 bytes

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã¢Â‚Â¬Ã¡ÂºÂŽÃ…Â’PÃ¢Â‚Â¬Ã¡ÂºÂŽÃ‚Â§;EÃ‰Â—Ã†Â‡Ã¡ÂºÂˆÃ¡Â¹Â€

Try it online!

Outputs $k$ if there is a $T$, $0$ otherwise. Very slow.

edited 4 hours ago

answered 4 hours ago

Erik the Outgolfer

29.9k42899

add a commentÂ |Â

up vote
1
down vote

Jelly, 17 bytes

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã¢Â‚Â¬Ã¡ÂºÂŽÃ…Â’PÃ¢Â‚Â¬Ã¡ÂºÂŽÃ‚Â§;EÃ‰Â—Ã†Â‡Ã¡ÂºÂˆÃ¡Â¹Â€

Try it online!

Outputs $k$ if there is a $T$, $0$ otherwise. Very slow.

edited 4 hours ago

answered 4 hours ago

Erik the Outgolfer

29.9k42899

add a commentÂ |Â

up vote
1
down vote

Jelly, 17 bytes

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã¢Â‚Â¬Ã¡ÂºÂŽÃ…Â’PÃ¢Â‚Â¬Ã¡ÂºÂŽÃ‚Â§;EÃ‰Â—Ã†Â‡Ã¡ÂºÂˆÃ¡Â¹Â€

Try it online!

Outputs $k$ if there is a $T$, $0$ otherwise. Very slow.

edited 4 hours ago

answered 4 hours ago

Erik the Outgolfer

29.9k42899

Jelly, 17 bytes

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã¢Â‚Â¬Ã¡ÂºÂŽÃ…Â’PÃ¢Â‚Â¬Ã¡ÂºÂŽÃ‚Â§;EÃ‰Â—Ã†Â‡Ã¡ÂºÂˆÃ¡Â¹Â€

Try it online!

Outputs $k$ if there is a $T$, $0$ otherwise. Very slow.

edited 4 hours ago

answered 4 hours ago

Erik the Outgolfer

29.9k42899

edited 4 hours ago

answered 4 hours ago

Erik the Outgolfer

29.9k42899

answered 4 hours ago

Erik the Outgolfer

29.9k42899

answered 4 hours ago

Erik the Outgolfer

29.9k42899

add a commentÂ |Â

up vote
1
down vote

Jelly, Â 17Â 11 bytes

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã‚Â§=Ã‰Â—Ã¢Â‚Â¬Ã‚Â§FÃ¡Â¹Â€

A dyadic link accepting a list and an integer which yields the maximal count, $k$.

Try it online! Or see the test-suite -- it's too slow for the larger test cases :(

How?

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã‚Â§=Ã‰Â—Ã¢Â‚Â¬Ã‚Â§FÃ¡Â¹Â€ - Link: list L, integer I
Ã…Â’! - all permutations
 Ã¢Â‚Â¬ - for Ã¢Â‚Â¬ach:
 Ã‰Â— - last three links as a dyad - i.e. f(permutation, I)
 Ã…Â’Ã¡Â¹Â– - partitions (all ways to slice up the permutation)
 Ã‚Â§ - sum each (vectorises at depth 1, so this gets the sums of the parts of
 - the partitions)
 = - equals? (==I) -> 1 if so 0 otherwise 
 Ã‚Â§ - sum each (vectorises at depth 1, so this gets the counts of parts equal to I
 - for each partition of each permutation) 
 F - flatten
 Ã¡Â¹Â€ - maximum

edited 40 mins ago

answered 1 hour ago

Jonathan Allan

49.5k534163

add a commentÂ |Â

up vote
1
down vote

Jelly, Â 17Â 11 bytes

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã‚Â§=Ã‰Â—Ã¢Â‚Â¬Ã‚Â§FÃ¡Â¹Â€

A dyadic link accepting a list and an integer which yields the maximal count, $k$.

Try it online! Or see the test-suite -- it's too slow for the larger test cases :(

How?

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã‚Â§=Ã‰Â—Ã¢Â‚Â¬Ã‚Â§FÃ¡Â¹Â€ - Link: list L, integer I
Ã…Â’! - all permutations
 Ã¢Â‚Â¬ - for Ã¢Â‚Â¬ach:
 Ã‰Â— - last three links as a dyad - i.e. f(permutation, I)
 Ã…Â’Ã¡Â¹Â– - partitions (all ways to slice up the permutation)
 Ã‚Â§ - sum each (vectorises at depth 1, so this gets the sums of the parts of
 - the partitions)
 = - equals? (==I) -> 1 if so 0 otherwise 
 Ã‚Â§ - sum each (vectorises at depth 1, so this gets the counts of parts equal to I
 - for each partition of each permutation) 
 F - flatten
 Ã¡Â¹Â€ - maximum

edited 40 mins ago

answered 1 hour ago

Jonathan Allan

49.5k534163

add a commentÂ |Â

up vote
1
down vote

Jelly, Â 17Â 11 bytes

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã‚Â§=Ã‰Â—Ã¢Â‚Â¬Ã‚Â§FÃ¡Â¹Â€

A dyadic link accepting a list and an integer which yields the maximal count, $k$.

Try it online! Or see the test-suite -- it's too slow for the larger test cases :(

How?

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã‚Â§=Ã‰Â—Ã¢Â‚Â¬Ã‚Â§FÃ¡Â¹Â€ - Link: list L, integer I
Ã…Â’! - all permutations
 Ã¢Â‚Â¬ - for Ã¢Â‚Â¬ach:
 Ã‰Â— - last three links as a dyad - i.e. f(permutation, I)
 Ã…Â’Ã¡Â¹Â– - partitions (all ways to slice up the permutation)
 Ã‚Â§ - sum each (vectorises at depth 1, so this gets the sums of the parts of
 - the partitions)
 = - equals? (==I) -> 1 if so 0 otherwise 
 Ã‚Â§ - sum each (vectorises at depth 1, so this gets the counts of parts equal to I
 - for each partition of each permutation) 
 F - flatten
 Ã¡Â¹Â€ - maximum

edited 40 mins ago

answered 1 hour ago

Jonathan Allan

49.5k534163

Jelly, Â 17Â 11 bytes

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã‚Â§=Ã‰Â—Ã¢Â‚Â¬Ã‚Â§FÃ¡Â¹Â€

A dyadic link accepting a list and an integer which yields the maximal count, $k$.

Try it online! Or see the test-suite -- it's too slow for the larger test cases :(

How?

Ã…Â’!Ã…Â’Ã¡Â¹Â–Ã‚Â§=Ã‰Â—Ã¢Â‚Â¬Ã‚Â§FÃ¡Â¹Â€ - Link: list L, integer I
Ã…Â’! - all permutations
 Ã¢Â‚Â¬ - for Ã¢Â‚Â¬ach:
 Ã‰Â— - last three links as a dyad - i.e. f(permutation, I)
 Ã…Â’Ã¡Â¹Â– - partitions (all ways to slice up the permutation)
 Ã‚Â§ - sum each (vectorises at depth 1, so this gets the sums of the parts of
 - the partitions)
 = - equals? (==I) -> 1 if so 0 otherwise 
 Ã‚Â§ - sum each (vectorises at depth 1, so this gets the counts of parts equal to I
 - for each partition of each permutation) 
 F - flatten
 Ã¡Â¹Â€ - maximum

edited 40 mins ago

answered 1 hour ago

Jonathan Allan

49.5k534163

edited 40 mins ago

answered 1 hour ago

Jonathan Allan

49.5k534163

answered 1 hour ago

Jonathan Allan

49.5k534163

answered 1 hour ago

Jonathan Allan

49.5k534163

add a commentÂ |Â

up vote
0
down vote

JavaScript (Node.js), 115 bytes

k=>f=([i,...s])=>i?Math.max(...[i,...s].map((_,j)=>(e[j]=~~e[j]+i,_=f(s),e[j]-=i,_))):e.filter(_=>_==k).length;e=

Try it online!

edited 3 hours ago

answered 3 hours ago

l4m2

4,0381431

add a commentÂ |Â

up vote
0
down vote

JavaScript (Node.js), 115 bytes

k=>f=([i,...s])=>i?Math.max(...[i,...s].map((_,j)=>(e[j]=~~e[j]+i,_=f(s),e[j]-=i,_))):e.filter(_=>_==k).length;e=

Try it online!

edited 3 hours ago

answered 3 hours ago

l4m2

4,0381431

add a commentÂ |Â

up vote
0
down vote

JavaScript (Node.js), 115 bytes

k=>f=([i,...s])=>i?Math.max(...[i,...s].map((_,j)=>(e[j]=~~e[j]+i,_=f(s),e[j]-=i,_))):e.filter(_=>_==k).length;e=

Try it online!

edited 3 hours ago

answered 3 hours ago

l4m2

4,0381431

JavaScript (Node.js), 115 bytes

k=>f=([i,...s])=>i?Math.max(...[i,...s].map((_,j)=>(e[j]=~~e[j]+i,_=f(s),e[j]-=i,_))):e.filter(_=>_==k).length;e=

Try it online!

edited 3 hours ago

answered 3 hours ago

l4m2

4,0381431

edited 3 hours ago

answered 3 hours ago

l4m2

4,0381431

answered 3 hours ago

l4m2

4,0381431

answered 3 hours ago

l4m2

4,0381431

add a commentÂ |Â

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Seeking Substantial Subcollections

Challenge

Example

Input

Output

Test cases

Scoring

Challenge

Example

Input

Output

Test cases

Scoring

Challenge

Example

Input

Output

Test cases

Scoring

Challenge

Example

Input

Output

Test cases

Scoring

3 Answers 3

Jelly, 17 bytes

Jelly, Â 17Â 11 bytes

How?

JavaScript (Node.js), 115 bytes

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3 Answers 3

Jelly, 17 bytes

Jelly, 17 bytes

Jelly, 17 bytes

Jelly, 17 bytes

Jelly, Â 17Â 11 bytes

How?

Jelly, Â 17Â 11 bytes

How?

Jelly, Â 17Â 11 bytes

How?

Jelly, Â 17Â 11 bytes

How?

JavaScript (Node.js), 115 bytes

JavaScript (Node.js), 115 bytes

JavaScript (Node.js), 115 bytes

JavaScript (Node.js), 115 bytes

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