All the single eights
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Given a non-empty rectangular array of integers from 0 to 9, output the amount of cells that are 8 and do not have a neighbour that is 8. Neighbouring is here understood in the Moore sense, that is, including diagonals. So each cell has 8 neighbours, except for cells at the edges of the array.
For example, given the input
8 4 5 6 5
9 3 8 4 8
0 8 6 1 5
6 7 9 8 2
8 8 7 4 2
the output should be 3. The three qualifying cells would be the following, marked with an asterisk (but only the amount of such entries should be output):
* 4 5 6 5
9 3 8 4 *
0 8 6 1 5
6 7 9 * 2
8 8 7 4 2
Additional rules
You can optionally take two numbers defining the size of the array as additional inputs.
Input can be taken by any reasonable means. The format is flexible as usual. For example, it can be a 2D character array, or a list of lists of numbers, or a flat list.
Programs or functions are allowed, in any programming language. Standard loopholes are forbidden.
Shortest code in bytes wins.
Test cases
Input:
8 4 5 6 5
9 3 8 4 8
0 8 6 1 5
6 7 9 8 2
8 8 7 4 2
Output:
3
Input
8 8
2 3
Output:
0
Input:
5 3 4
2 5 2
Output:
0
Input:
5 8 3 8
Output:
2
Input:
8
0
8
Output:
2.
Input:
4 2 8 5
2 6 1 8
8 5 5 8
Output:
1
Input:
4 5 4 3 8 1 8 2
8 2 7 7 8 3 9 3
9 8 7 8 5 4 2 8
4 5 0 2 1 8 6 9
1 5 4 3 4 5 6 1
Output
3.
Input:
8
Output:
1
Input:
8 5 8 1 6 8 7 7
9 9 2 8 2 7 8 3
2 8 4 9 7 3 2 7
9 2 9 7 1 9 5 6
6 9 8 7 3 1 5 2
1 9 9 7 1 8 8 2
3 5 6 8 1 4 7 5
Output:
4.
Input:
8 1 8
2 5 7
8 0 1
Output:
3.
Inputs in MATLAB format:
[8 4 5 6 5; 9 3 8 4 8; 0 8 6 1 5; 6 7 9 8 2; 8 8 7 4 2]
[8 8; 2 3]
[5 3 4; 2 5 2]
[5 8 3 8]
[8; 0; 8]
[4 2 8 5; 2 6 1 8; 8 5 5 8]
[4 5 4 3 8 1 8 2; 8 2 7 7 8 3 9 3; 9 8 7 8 5 4 2 8; 4 5 0 2 1 8 6 9; 1 5 4 3 4 5 6 1]
[8]
[8 5 8 1 6 8 7 7; 9 9 2 8 2 7 8 3; 2 8 4 9 7 3 2 7; 9 2 9 7 1 9 5 6; 6 9 8 7 3 1 5 2; 1 9 9 7 1 8 8 2; 3 5 6 8 1 4 7 5]
[8 1 8; 2 5 7; 8 0 1]
Inputs in Python format:
[[8, 4, 5, 6, 5], [9, 3, 8, 4, 8], [0, 8, 6, 1, 5], [6, 7, 9, 8, 2], [8, 8, 7, 4, 2]]
[[8, 8], [2, 3]]
[[5, 3, 4], [2, 5, 2]]
[[5, 8, 3, 8]]
[[8], [0], [8]]
[[4, 2, 8, 5], [2, 6, 1, 8], [8, 5, 5, 8]]
[[4, 5, 4, 3, 8, 1, 8, 2], [8, 2, 7, 7, 8, 3, 9, 3], [9, 8, 7, 8, 5, 4, 2, 8], [4, 5, 0, 2, 1, 8, 6, 9], [1, 5, 4, 3, 4, 5, 6, 1]]
[[8]]
[[8, 5, 8, 1, 6, 8, 7, 7], [9, 9, 2, 8, 2, 7, 8, 3], [2, 8, 4, 9, 7, 3, 2, 7], [9, 2, 9, 7, 1, 9, 5, 6], [6, 9, 8, 7, 3, 1, 5, 2], [1, 9, 9, 7, 1, 8, 8, 2], [3, 5, 6, 8, 1, 4, 7, 5]]
[[8, 1, 8], [2, 5, 7], [8, 0, 1]]
Outputs:
3, 0, 0, 2, 2, 1, 3, 1, 4, 3
code-golf array-manipulation integer
asked 2 days ago
Luis Mendo
73.6k885289
add a comment |
up vote
20
down vote
favorite
Given a non-empty rectangular array of integers from 0 to 9, output the amount of cells that are 8 and do not have a neighbour that is 8. Neighbouring is here understood in the Moore sense, that is, including diagonals. So each cell has 8 neighbours, except for cells at the edges of the array.
For example, given the input
8 4 5 6 5
9 3 8 4 8
0 8 6 1 5
6 7 9 8 2
8 8 7 4 2
the output should be 3. The three qualifying cells would be the following, marked with an asterisk (but only the amount of such entries should be output):
* 4 5 6 5
9 3 8 4 *
0 8 6 1 5
6 7 9 * 2
8 8 7 4 2
Additional rules
You can optionally take two numbers defining the size of the array as additional inputs.
Input can be taken by any reasonable means. The format is flexible as usual. For example, it can be a 2D character array, or a list of lists of numbers, or a flat list.
Programs or functions are allowed, in any programming language. Standard loopholes are forbidden.
Shortest code in bytes wins.
Test cases
Input:
8 4 5 6 5
9 3 8 4 8
0 8 6 1 5
6 7 9 8 2
8 8 7 4 2
Output:
3
Input
8 8
2 3
Output:
0
Input:
5 3 4
2 5 2
Output:
0
Input:
5 8 3 8
Output:
2
Input:
8
0
8
Output:
2.
Input:
4 2 8 5
2 6 1 8
8 5 5 8
Output:
1
Input:
4 5 4 3 8 1 8 2
8 2 7 7 8 3 9 3
9 8 7 8 5 4 2 8
4 5 0 2 1 8 6 9
1 5 4 3 4 5 6 1
Output
3.
Input:
8
Output:
1
Input:
8 5 8 1 6 8 7 7
9 9 2 8 2 7 8 3
2 8 4 9 7 3 2 7
9 2 9 7 1 9 5 6
6 9 8 7 3 1 5 2
1 9 9 7 1 8 8 2
3 5 6 8 1 4 7 5
Output:
4.
Input:
8 1 8
2 5 7
8 0 1
Output:
3.
Inputs in MATLAB format:
[8 4 5 6 5; 9 3 8 4 8; 0 8 6 1 5; 6 7 9 8 2; 8 8 7 4 2]
[8 8; 2 3]
[5 3 4; 2 5 2]
[5 8 3 8]
[8; 0; 8]
[4 2 8 5; 2 6 1 8; 8 5 5 8]
[4 5 4 3 8 1 8 2; 8 2 7 7 8 3 9 3; 9 8 7 8 5 4 2 8; 4 5 0 2 1 8 6 9; 1 5 4 3 4 5 6 1]
[8]
[8 5 8 1 6 8 7 7; 9 9 2 8 2 7 8 3; 2 8 4 9 7 3 2 7; 9 2 9 7 1 9 5 6; 6 9 8 7 3 1 5 2; 1 9 9 7 1 8 8 2; 3 5 6 8 1 4 7 5]
[8 1 8; 2 5 7; 8 0 1]
Inputs in Python format:
[[8, 4, 5, 6, 5], [9, 3, 8, 4, 8], [0, 8, 6, 1, 5], [6, 7, 9, 8, 2], [8, 8, 7, 4, 2]]
[[8, 8], [2, 3]]
[[5, 3, 4], [2, 5, 2]]
[[5, 8, 3, 8]]
[[8], [0], [8]]
[[4, 2, 8, 5], [2, 6, 1, 8], [8, 5, 5, 8]]
[[4, 5, 4, 3, 8, 1, 8, 2], [8, 2, 7, 7, 8, 3, 9, 3], [9, 8, 7, 8, 5, 4, 2, 8], [4, 5, 0, 2, 1, 8, 6, 9], [1, 5, 4, 3, 4, 5, 6, 1]]
[[8]]
[[8, 5, 8, 1, 6, 8, 7, 7], [9, 9, 2, 8, 2, 7, 8, 3], [2, 8, 4, 9, 7, 3, 2, 7], [9, 2, 9, 7, 1, 9, 5, 6], [6, 9, 8, 7, 3, 1, 5, 2], [1, 9, 9, 7, 1, 8, 8, 2], [3, 5, 6, 8, 1, 4, 7, 5]]
[[8, 1, 8], [2, 5, 7], [8, 0, 1]]
Outputs:
3, 0, 0, 2, 2, 1, 3, 1, 4, 3
code-golf array-manipulation integer
asked 2 days ago
Luis Mendo
73.6k885289
15
If you like it then you should have put a vote on it
– Luis Mendo
2 days ago
When I read "cells that equal 8", for a moment I thought you meant that a cell could be larger than a 1x1 chuck (NxN) of the grid. Should probably rephrase that to "cells that are 8" to clarify no math needed. =P
– Tezra
yesterday
@Tezra Edited. I find the new wording a little less natural, but I’m not a native speaker so I’ll trust your criterion
– Luis Mendo
yesterday
add a comment |
up vote
20
down vote
favorite
up vote
20
down vote
favorite
Given a non-empty rectangular array of integers from 0 to 9, output the amount of cells that are 8 and do not have a neighbour that is 8. Neighbouring is here understood in the Moore sense, that is, including diagonals. So each cell has 8 neighbours, except for cells at the edges of the array.
For example, given the input
8 4 5 6 5
9 3 8 4 8
0 8 6 1 5
6 7 9 8 2
8 8 7 4 2
the output should be 3. The three qualifying cells would be the following, marked with an asterisk (but only the amount of such entries should be output):
* 4 5 6 5
9 3 8 4 *
0 8 6 1 5
6 7 9 * 2
8 8 7 4 2
Additional rules
You can optionally take two numbers defining the size of the array as additional inputs.
Input can be taken by any reasonable means. The format is flexible as usual. For example, it can be a 2D character array, or a list of lists of numbers, or a flat list.
Programs or functions are allowed, in any programming language. Standard loopholes are forbidden.
Shortest code in bytes wins.
Test cases
Input:
8 4 5 6 5
9 3 8 4 8
0 8 6 1 5
6 7 9 8 2
8 8 7 4 2
Output:
3
Input
8 8
2 3
Output:
0
Input:
5 3 4
2 5 2
Output:
0
Input:
5 8 3 8
Output:
2
Input:
8
0
8
Output:
2.
Input:
4 2 8 5
2 6 1 8
8 5 5 8
Output:
1
Input:
4 5 4 3 8 1 8 2
8 2 7 7 8 3 9 3
9 8 7 8 5 4 2 8
4 5 0 2 1 8 6 9
1 5 4 3 4 5 6 1
Output
3.
Input:
8
Output:
1
Input:
8 5 8 1 6 8 7 7
9 9 2 8 2 7 8 3
2 8 4 9 7 3 2 7
9 2 9 7 1 9 5 6
6 9 8 7 3 1 5 2
1 9 9 7 1 8 8 2
3 5 6 8 1 4 7 5
Output:
4.
Input:
8 1 8
2 5 7
8 0 1
Output:
3.
Inputs in MATLAB format:
[8 4 5 6 5; 9 3 8 4 8; 0 8 6 1 5; 6 7 9 8 2; 8 8 7 4 2]
[8 8; 2 3]
[5 3 4; 2 5 2]
[5 8 3 8]
[8; 0; 8]
[4 2 8 5; 2 6 1 8; 8 5 5 8]
[4 5 4 3 8 1 8 2; 8 2 7 7 8 3 9 3; 9 8 7 8 5 4 2 8; 4 5 0 2 1 8 6 9; 1 5 4 3 4 5 6 1]
[8]
[8 5 8 1 6 8 7 7; 9 9 2 8 2 7 8 3; 2 8 4 9 7 3 2 7; 9 2 9 7 1 9 5 6; 6 9 8 7 3 1 5 2; 1 9 9 7 1 8 8 2; 3 5 6 8 1 4 7 5]
[8 1 8; 2 5 7; 8 0 1]
Inputs in Python format:
[[8, 4, 5, 6, 5], [9, 3, 8, 4, 8], [0, 8, 6, 1, 5], [6, 7, 9, 8, 2], [8, 8, 7, 4, 2]]
[[8, 8], [2, 3]]
[[5, 3, 4], [2, 5, 2]]
[[5, 8, 3, 8]]
[[8], [0], [8]]
[[4, 2, 8, 5], [2, 6, 1, 8], [8, 5, 5, 8]]
[[4, 5, 4, 3, 8, 1, 8, 2], [8, 2, 7, 7, 8, 3, 9, 3], [9, 8, 7, 8, 5, 4, 2, 8], [4, 5, 0, 2, 1, 8, 6, 9], [1, 5, 4, 3, 4, 5, 6, 1]]
[[8]]
[[8, 5, 8, 1, 6, 8, 7, 7], [9, 9, 2, 8, 2, 7, 8, 3], [2, 8, 4, 9, 7, 3, 2, 7], [9, 2, 9, 7, 1, 9, 5, 6], [6, 9, 8, 7, 3, 1, 5, 2], [1, 9, 9, 7, 1, 8, 8, 2], [3, 5, 6, 8, 1, 4, 7, 5]]
[[8, 1, 8], [2, 5, 7], [8, 0, 1]]
Outputs:
3, 0, 0, 2, 2, 1, 3, 1, 4, 3
code-golf array-manipulation integer
asked 2 days ago
Luis Mendo
73.6k885289
Given a non-empty rectangular array of integers from 0 to 9, output the amount of cells that are 8 and do not have a neighbour that is 8. Neighbouring is here understood in the Moore sense, that is, including diagonals. So each cell has 8 neighbours, except for cells at the edges of the array.
For example, given the input
8 4 5 6 5
9 3 8 4 8
0 8 6 1 5
6 7 9 8 2
8 8 7 4 2
the output should be 3. The three qualifying cells would be the following, marked with an asterisk (but only the amount of such entries should be output):
* 4 5 6 5
9 3 8 4 *
0 8 6 1 5
6 7 9 * 2
8 8 7 4 2
Additional rules
You can optionally take two numbers defining the size of the array as additional inputs.
Input can be taken by any reasonable means. The format is flexible as usual. For example, it can be a 2D character array, or a list of lists of numbers, or a flat list.
Programs or functions are allowed, in any programming language. Standard loopholes are forbidden.
Shortest code in bytes wins.
Test cases
Input:
8 4 5 6 5
9 3 8 4 8
0 8 6 1 5
6 7 9 8 2
8 8 7 4 2
Output:
3
Input
8 8
2 3
Output:
0
Input:
5 3 4
2 5 2
Output:
0
Input:
5 8 3 8
Output:
2
Input:
8
0
8
Output:
2.
Input:
4 2 8 5
2 6 1 8
8 5 5 8
Output:
1
Input:
4 5 4 3 8 1 8 2
8 2 7 7 8 3 9 3
9 8 7 8 5 4 2 8
4 5 0 2 1 8 6 9
1 5 4 3 4 5 6 1
Output
3.
Input:
8
Output:
1
Input:
8 5 8 1 6 8 7 7
9 9 2 8 2 7 8 3
2 8 4 9 7 3 2 7
9 2 9 7 1 9 5 6
6 9 8 7 3 1 5 2
1 9 9 7 1 8 8 2
3 5 6 8 1 4 7 5
Output:
4.
Input:
8 1 8
2 5 7
8 0 1
Output:
3.
Inputs in MATLAB format:
[8 4 5 6 5; 9 3 8 4 8; 0 8 6 1 5; 6 7 9 8 2; 8 8 7 4 2]
[8 8; 2 3]
[5 3 4; 2 5 2]
[5 8 3 8]
[8; 0; 8]
[4 2 8 5; 2 6 1 8; 8 5 5 8]
[4 5 4 3 8 1 8 2; 8 2 7 7 8 3 9 3; 9 8 7 8 5 4 2 8; 4 5 0 2 1 8 6 9; 1 5 4 3 4 5 6 1]
[8]
[8 5 8 1 6 8 7 7; 9 9 2 8 2 7 8 3; 2 8 4 9 7 3 2 7; 9 2 9 7 1 9 5 6; 6 9 8 7 3 1 5 2; 1 9 9 7 1 8 8 2; 3 5 6 8 1 4 7 5]
[8 1 8; 2 5 7; 8 0 1]
Inputs in Python format:
[[8, 4, 5, 6, 5], [9, 3, 8, 4, 8], [0, 8, 6, 1, 5], [6, 7, 9, 8, 2], [8, 8, 7, 4, 2]]
[[8, 8], [2, 3]]
[[5, 3, 4], [2, 5, 2]]
[[5, 8, 3, 8]]
[[8], [0], [8]]
[[4, 2, 8, 5], [2, 6, 1, 8], [8, 5, 5, 8]]
[[4, 5, 4, 3, 8, 1, 8, 2], [8, 2, 7, 7, 8, 3, 9, 3], [9, 8, 7, 8, 5, 4, 2, 8], [4, 5, 0, 2, 1, 8, 6, 9], [1, 5, 4, 3, 4, 5, 6, 1]]
[[8]]
[[8, 5, 8, 1, 6, 8, 7, 7], [9, 9, 2, 8, 2, 7, 8, 3], [2, 8, 4, 9, 7, 3, 2, 7], [9, 2, 9, 7, 1, 9, 5, 6], [6, 9, 8, 7, 3, 1, 5, 2], [1, 9, 9, 7, 1, 8, 8, 2], [3, 5, 6, 8, 1, 4, 7, 5]]
[[8, 1, 8], [2, 5, 7], [8, 0, 1]]
Outputs:
3, 0, 0, 2, 2, 1, 3, 1, 4, 3
code-golf array-manipulation integer
code-golf array-manipulation integer
asked 2 days ago
Luis Mendo
73.6k885289
asked 2 days ago
Luis Mendo
73.6k885289
edited yesterday
asked 2 days ago
Luis Mendo
73.6k885289
asked 2 days ago
Luis Mendo
73.6k885289
asked 2 days ago
Luis Mendo
73.6k885289
73.6k885289
15
If you like it then you should have put a vote on it
– Luis Mendo
2 days ago
When I read "cells that equal 8", for a moment I thought you meant that a cell could be larger than a 1x1 chuck (NxN) of the grid. Should probably rephrase that to "cells that are 8" to clarify no math needed. =P
– Tezra
yesterday
@Tezra Edited. I find the new wording a little less natural, but I’m not a native speaker so I’ll trust your criterion
– Luis Mendo
yesterday
add a comment |
15
If you like it then you should have put a vote on it
– Luis Mendo
2 days ago
When I read "cells that equal 8", for a moment I thought you meant that a cell could be larger than a 1x1 chuck (NxN) of the grid. Should probably rephrase that to "cells that are 8" to clarify no math needed. =P
– Tezra
yesterday
@Tezra Edited. I find the new wording a little less natural, but I’m not a native speaker so I’ll trust your criterion
– Luis Mendo
yesterday
15
15
If you like it then you should have put a vote on it
– Luis Mendo
2 days ago
If you like it then you should have put a vote on it
– Luis Mendo
2 days ago
When I read "cells that equal 8", for a moment I thought you meant that a cell could be larger than a 1x1 chuck (NxN) of the grid. Should probably rephrase that to "cells that are 8" to clarify no math needed. =P
– Tezra
yesterday
When I read "cells that equal 8", for a moment I thought you meant that a cell could be larger than a 1x1 chuck (NxN) of the grid. Should probably rephrase that to "cells that are 8" to clarify no math needed. =P
– Tezra
yesterday
@Tezra Edited. I find the new wording a little less natural, but I’m not a native speaker so I’ll trust your criterion
– Luis Mendo
yesterday
@Tezra Edited. I find the new wording a little less natural, but I’m not a native speaker so I’ll trust your criterion
– Luis Mendo
yesterday
add a comment |
15 Answers
15
active
oldest
votes
up vote
8
down vote
R, 117 63 59 bytes
function(m)sum(colSums(as.matrix(dist(which(m==8,T)))<2)<2)
Try it online!
dist computes distances (default is Euclidean) among rows of a matrix. which with second argument TRUE returns the coordinates where the predicate is true.
Coordinates are neighbours if the distance between them is not more than the square root of 2, but the inner <2 is good enough because the possible distance jumps from sqrt(2) ro 2.
answered yesterday
ngm
2,92923
it's a shame numerical imprecision doesn't allowcolSums()^2<=2to work.
– Giuseppe
yesterday
@Giuseppe of course there are only a few possible distances andsqrt(2)jumps to2(e.g.sort(c(dist(expand.grid(1:6,1:6))), decreasing = TRUE))) so we were being too clever there.
– ngm
23 hours ago
add a comment |
up vote
6
down vote
APL (Dyalog Classic), 29 28 25 bytes
≢∘⍸16=2⊥¨3,⌿3,/8=(⍉0,⌽)⍣4
Try it online!
answered 2 days ago
ngn
6,15312459
Note: 0 index origin is not even needed.
– Zacharý
yesterday
@Zacharý I always use it as a default, to avoid surprises.
– ngn
yesterday
Ah, so like others with1(except not explicitly set). That makes sense.
– Zacharý
yesterday
add a comment |
up vote
5
down vote
Jelly, 18 15 bytes
8=µ+Ż+ḊZµ⁺ỊṖḋµS
Try it online!
How it works
8=µ+Ż+ḊZµ⁺ỊṖḋµS Main link (monad). Input: digit matrix
8= 1) Convert elements by `x == 8`
µ 2) New chain:
+Ż+Ḋ x + [0,*x] + x[1:] (missing elements are considered 0)
Effectively, vertical convolution with [1,1,1]
Z Transpose
µ⁺ 3) Start new chain, apply 2) again
ỊṖ Convert elements by `|x| <= 1` and remove last row
ḋ Row-wise dot product with result of 1)
µS 4) Sum
Previous solution, 18 bytes
æc7B¤ZḊṖ
8=µÇÇỊḋµS
Try it online!
Wanted to share another approach, though this is 1 byte longer than Jonathan Allan's solution.
How it works
æc7B¤ZḊṖ Auxiliary link (monad). Input: integer matrix
æc7B¤ Convolution with [1,1,1] on each row
ZḊṖ Zip (transpose), remove first and last elements
8=µÇÇỊḋµS Main link (monad). Input: digit matrix
8= Convert 8 to 1, anything else to 0 (*A)
怀 Apply aux.link twice (effective convolution with [[1,1,1]]*3)
Ịḋ Convert to |x|<=1, then row-wise dot product with A
µS Sum the result
answered 2 days ago
Bubbler
4,859750
add a comment |
up vote
4
down vote
JavaScript (Node.js), 88 85 bytes
a=>g=(x,y=0,c=0)=>a.map(p=>p.map((q,j)=>c+=q-8?0:1/x?(x-j)**2+y*y<3:g(j,-y)<2)&--y)|c
Try it online!
Thank Arnauld for 2 bytes
answered 2 days ago
l4m2
4,2341432
add a comment |
up vote
4
down vote
J,43, 40 37 bytes
-3 bytes thanks to Bubbler
1#.1#.3 3(16=8#.@:=,);._3(0|:@,|.)^:4
Try it online!
Explanation:
The first part of the algorithm assures that we can apply a 3x3 sliding window to he input. This is achieved by prepending a row of zeroes and 90 degrees rotation, repeated 4 times.
1#.1#.3 3(16=8#.@:=,);._3(0|:@,|.)^:4
( )^:4 - repeat 4 times
0|:@,|. - reverse, prepend wit a row of 0 and transpose
;._3 - cut the input (already outlined with zeroes)
3 3 - into matrices with size 3x3
( ) - and for each matrix do
, - ravel (flatten)
8 = - check if each item equals 8
#.@: - and convert the list of 1s and 0s to a decimal
16= - is equal to 16?
1#. - add (the result has the shape of the input)
1#. - add again
answered 2 days ago
Galen Ivanov
5,82711032
1
37 bytes using@:and moving|.. Note that@in place of@:doesn't work.
– Bubbler
2 days ago
@Bubbler Thank you!
– Galen Ivanov
2 days ago
This is nice. Probably worth adding at least a high level explanation of how it works, if not a code breakdown. It took me 10m or so to figure it out. Also, it's interesting how much shorter the APL version (which uses the same approach) is. Looks like that's mostly the result of digraphs instead of single char symbols...
– Jonah
yesterday
@Jonah I'll add an explanation. For comparison with APL you can look at the revisions of ngn's solution, especially the 28 byte version
– Galen Ivanov
yesterday
1
@Jonah Explanation added
– Galen Ivanov
yesterday
add a comment |
up vote
3
down vote
Retina 0.8.2, 84 bytes
.+
_$&_
m`(?<!(?(1).)^(?<-1>.)*.?.?8.*¶(.)*.|8)8(?!8|.(.)*¶.*8.?.?(?<-2>.)*$(?(2).))
Try it online! Explanation:
.+
_$&_
Wrap each line in non-8 characters so that all 8s have at least one character on each side.
m`
This is the last stage, so counting matches is implied. The m modifier makes the ^ and $ characters match at the start or end of any line.
(?<!...|8)
Don't match a character directly after an 8, or...
(?(1).)^(?<-1>.)*.?.?8.*¶(.)*.
... a character below an 8; the (?(1).)^(?<-1>.)* matches the same column as the ¶(.)* on the next line, but the .?.? allows the 8 to be 1 left or right of the character after the . on the next line.
8
Match 8s.
(?!8|...)
Don't match an 8 immediately before an 8, or...
.(.)*¶.*8.?.?(?<-2>.)*$(?(2).)
... a character with an 8 in the line below; again, the (?<-2>.)*$(?(2).) matches the same column as the (.)*¶ on the previous line, but the .?.? allows the 8 to be 1 left or right of the 8 before the . on the previous line.
answered 2 days ago
Neil
77.7k744174
add a comment |
up vote
3
down vote
J, 42 bytes
[:+/@,8=]+[:+/(<:3 3#:4-.~i.9)|.!.0(_*8&=)
Try it online!
explanation
The high-level approach here is similar to the one used in the classic APL solution to the game of life: https://www.youtube.com/watch?v=a9xAKttWgP4.
In that solution, we shift our matrix in the 8 possible neighbor directions, creating 8 duplicates of the input, stack them up, and then add the "planes" together to get our neighbor counts.
Here, we use a "multiply by infinity" trick to adapt the solution for this problem.
[: +/@, 8 = ] + [: +/ (neighbor deltas) (|.!.0) _ * 8&= NB.
NB.
[: +/@, NB. the sum after flattening
8 = NB. a 0 1 matrix created by
NB. elmwise testing if 8
NB. equals the matrix
(the matrix to test for equality with 8 ) NB. defined by...
] + NB. the original input plus
[: +/ NB. the elmwise sum of 8
NB. matrices defined by
_ * NB. the elmwise product of
NB. infinity and
8&= NB. the matrix which is 1
NB. where the input is 8
NB. and 0 elsewhere, thus
NB. creating an infinity-0
NB. matrix
(|.!.0) NB. then 2d shifting that
NB. matrix in the 8 possible
NB. "neighbor" directions
(neighbor deltas) NB. defined by the "neighbor
NB. deltas" (see below)
NB. QED.
NB. ***********************
NB. The rest of the
NB. explanation merely
NB. breaks down the neighbor
NB. delta construction.
(neighbor deltas ) NB. the neighbor deltas are
NB. merely the cross product
NB. of _1 0 1 with itself,
NB. minus "0 0"
(<: 3 3 #: 4 -.~ i.9) NB. to create that...
<: NB. subtract one from
3 3 #: NB. the base 3 rep of
i.9 NB. the numbers 0 - 8
4 -.~ NB. minus the number 4
NB.
NB. All of which produces
NB. the eight "neighbor"
NB. deltas:
NB.
NB. _1 _1
NB. _1 0
NB. _1 1
NB. 0 _1
NB. 0 1
NB. 1 _1
NB. 1 0
NB. 1 1
answered 2 days ago
Jonah
1,841816
1
You have forgotten to remove a space between~and>
– Galen Ivanov
2 days ago
@GalenIvanov Fixed now. Thank you.
– Jonah
2 days ago
add a comment |
up vote
2
down vote
Jelly, 17 bytes
=8ŒṪµŒcZIỊȦƲƇẎ⁸ḟL
Try it online! Or see the test-suite.
How?
=8ŒṪµŒcZIỊȦƲƇẎ⁸ḟL - Link: list of lists of integers (digits)
=8 - equals 8?
ŒṪ - multidimensional truthy indices (pairs of row & column indices of 8s)
µ - start a new monadic chain
Œc - all pairs (of the index-pairs)
Ƈ - filter keep if: (keep those that represent adjacent positions)
Ʋ - last four links as a monad:
Z - transpose
I - incremental differences
Ị - insignificant? (abs(x) <= 1)
Ȧ - all?
Ẏ - tighten (to a list of all adjacent 8's index-pairs, at least once each)
⁸ - chain's left argument (all the index-pairs again)
ḟ - filter discard (remove those found to be adjacent to another)
L - length (of the remaining pairs of indices of single 8s)
answered 2 days ago
Jonathan Allan
49.9k534163
add a comment |
up vote
2
down vote
Python 2, 130 bytes
lambda a:sum(sum(u[~-c*(c>0):c+2].count(8)for u in a[~-r*(r>0):r+2])*8==8==a[r][c]for r in range(len(a))for c in range(len(a[0])))
Try it online!
answered 2 days ago
Chas Brown
4,6361519
Seems shorter if take length from args
– l4m2
2 days ago
add a comment |
up vote
2
down vote
Java 8, 181 bytes
(M,R,C)->{int z=0,c,f,t;for(;R-->0;)for(c=C;c-->0;z+=f==1?1:0)for(f=0,t=9;M[R][c]==8&t-->0;)try{f+=M[t<3?R-1:t>5?R+1:R][t%3<1?c-1:t%3>1?c+1:c]==8?1:0;}catch(Exception e){}return z;}
Takes the dimensions as additional parameters R (amount of rows) and C (amount of columns).
The cells are checked pretty similar as I did in my Fryer simulator answer.
Try it online.
Explanation:
(M,R,C)->{ // Method with integer-matrix as parameter & integer return
int z=0, // Result-counter, starting at 0
c,f,t; // Temp-integers, starting uninitialized
for(;R-->0;) // Loop over the rows:
for(c=C;c-->0 // Inner loop over the columns:
; // After every iteration:
z+=f==1? // If the flag-integer is exactly 1:
1 // Increase the result-counter by 1
:0) // Else: leave it the same
for(f=0, // Reset the flag to 0
t=9;M[R][c]==8& // If the current cell contains an 8:
t-->0;) // Inner loop `t` in the range (9, 0]:
try{f+= // Increase the flag by:
M[t<3? // If `t` is 0, 1, or 2:
R-1 // Look at the previous row
:t>5? // Else-if `t` is 6, 7, or 8:
R+1 // Look at the next row
: // Else (`t` is 3, 4, or 5):
R] // Look at the current row
[t%3<1? // If `t` is 0, 3, or 6:
c-1 // Look at the previous column
:t%3>1? // Else-if `t` is 2, 5, or 8:
c+1 // Look at the next column
: // Else (`t` is 1, 4, or 7):
c] // Look at the current column
==8? // If the digit in this cell is 8:
1 // Increase the flag-integer by 1
:0; // Else: leave it the same
}catch(Exception e){} // Catch and ignore ArrayIndexOutOfBoundsExceptions
// (try-catch saves bytes in comparison to if-checks)
return z;} // And finally return the counter
answered 2 days ago
Kevin Cruijssen
33.6k554179
add a comment |
up vote
2
down vote
Powershell, 121 bytes
param($a)(($b='='*4*($l=($a|% Le*)[0]))+($a|%{"!$_!"})+$b|sls "(?<=[^8]{3}.{$l}[^8])8(?=[^8].{$l}[^8]{3})" -a|% m*).Count
Less golfed test script:
$f = {
param($a)
$length=($a|% Length)[0]
$border='='*4*$length
$pattern="(?<=[^8]{3}.{$length}[^8])8(?=[^8].{$length}[^8]{3})"
$matches=$border+($a|%{"!$_!"})+$border |sls $pattern -a|% Matches
$matches.count
}
@(
,(3,"84565","93848","08615","67982","88742")
,(0,"88","23")
,(0,"534","252")
,(2,"5838")
,(2,"8","0","8")
,(1,"4285","2618","8558")
,(3,"45438182","82778393","98785428","45021869","15434561")
,(1,"8")
,(4,"85816877","99282783","28497327","92971956","69873152","19971882","35681475")
,(3,"818","257","801")
,(0,"")
) | % {
$expected,$a = $_
$result = &$f $a
"$($result-eq$expected): $result : $a"
}
Output:
True: 3 : 84565 93848 08615 67982 88742
True: 0 : 88 23
True: 0 : 534 252
True: 2 : 5838
True: 2 : 8 0 8
True: 1 : 4285 2618 8558
True: 3 : 45438182 82778393 98785428 45021869 15434561
True: 1 : 8
True: 4 : 85816877 99282783 28497327 92971956 69873152 19971882 35681475
True: 3 : 818 257 801
True: 0 :
Explanation:
First, the script calculates a length of the first string.
Second, it adds extra border to strings. Augmended reality string likes:
....=========!84565! !93848! !08615! !67982! !88742!===========....
represents the multiline string:
...=====
=======
!84565!
!93848!
!08615!
!67982!
!88742!
=======
========...
Note 1: the number of = is sufficient for a string of any length.
Note 2: a large number of = does not affect the search for eights.
Next, the regular expression (?<=[^8]{3}.{$l}[^8])8(?=[^8].{$l}[^8]{3}) looks for the digit 8 with the preceding non-eights (?<=[^8]{3}.{$l}[^8]) and the following non-eights (?=[^8].{$l}[^8]{3}):
.......
<<<....
<8>....
>>>....
.......
Finally, the number of matches is returned as a result.
answered yesterday
mazzy
1,715312
add a comment |
up vote
2
down vote
MATL, 21 17 10 bytes
8=t3Y6Z+>z
Try it online!
Thanks to Luis Mendo for help in chat, and for suggesting 2D convolution.
Explanation:
#implicit input, m
8= #equal to 8? matrix of 1 where m is 8, 0 otherwise
t #duplicate
3Y6 #push [1 1 1; 1 0 1; 1 1 1], "neighbor cells" for convolution
Z+ #2D convolution; each element is replaced by the number of neighbors that are 8
> #elementwise greater than -- matrix of 1s where an 8 is single, 0s otherwise
z #number of nonzero elements -- number of single eights
#implicit output
answered yesterday
Giuseppe
15.8k31051
You could save quite a few bytes using (2D-)convolution, if you are famiiar with the concept
– Luis Mendo
yesterday
1
@LuisMendo 2D convolution is one of those things where I don't understand 1D convolution either so there's no hope for me there...sounds like an opportunity to learn both!
– Giuseppe
yesterday
1
If you need help with that let me know in the chat room. Convolution is a very useful operation. If you want to learn convolution start with 1D. The generalization to 2D is immediate
– Luis Mendo
yesterday
add a comment |
up vote
2
down vote
Jelly, 12 bytes
œẹ8ạṀ¥þ`’Ạ€S
Try it online!
How it works
œẹ8ạṀ¥þ`’Ạ€S Main link. Argument: M (matrix)
œẹ8 Find all multidimensional indices of 8, yielding an array A of pairs.
þ` Table self; for all pairs [i, j] and [k, l] in A, call the link to the
left. Return the results as a matrix.
ạ Absolute difference; yield [|i - k|, |j - l|].
Ṁ Take the maximum.
’ Decrement all the maxmima, mapping 1 to 0.
Ạ€ All each; yield 1 for each row that contains no zeroes.
S Take the sum.
answered yesterday
Dennis♦
184k32293728
add a comment |
up vote
1
down vote
JavaScript (ES6), 106 bytes
a=>a.map((r,y)=>r.map((v,x)=>k+=v==8&[...'12221000'].every((d,i,v)=>(a[y+~-d]||0)[x+~-v[i+2&7]]^8)),k=0)|k
Try it online!
Bitwise approach, 110 bytes
a=>a.map(r=>r.map(v=>x=x*2|v==8,x=k=0)|x).map((n,y,b)=>a[0].map((_,x)=>(n^1<<x|b[y-1]|b[y+1])*2>>x&7||k++))&&k
Try it online!
answered 2 days ago
Arnauld
68.4k584289
Bitwise approach fail on[[7]]
– l4m2
2 days ago
@lm42 Oh, thanks. Now fixed.
– Arnauld
2 days ago
add a comment |
up vote
1
down vote
Clojure, 227 198 bytes
(fn[t w h](let[c #(for[y(range %3 %4)x(range % %2)][x y])e #(= 8(get-in t(reverse %)0))m(fn[[o p]](count(filter e(c(dec o)(+ o 2)(dec p)(+ p 2)))))](count(filter #(= 1(m %))(filter e(c 0 w 0 h))))))
Ouch. Definitely not the shortest here by any means. 54 bytes of parenthesis is killer. I'm still relatively happy with it though.
-29 bytes by creating a helper function that generates a range since I was doing that twice, changing the reduce to a (count (filter setup, and getting rid of the threading macro after golfing.
(defn count-single-eights [td-array width height]
; Define three helper functions. One generates a list of coords for a given range of dimensions, another checks if an eight is
; at the given coord, and the other counts how many neighbors around a coord are an eight
(letfn [(coords [x-min x-max y-min y-max]
(for [y (range y-min y-max)
x (range x-min x-max)]
[x y]))
(eight? [[x y]] (= 8 (get-in td-array [y x] 0)))
(n-eights-around [[cx cy]]
(count (filter eight?
(coords (dec cx) (+ cx 2), (dec cy) (+ cy 2)))))]
; Gen a list of each coord of the matrix
(->> (coords 0 width, 0 height)
; Remove any coords that don't contain an eight
(filter eight?)
; Then count how many "neighborhoods" only contain 1 eight
(filter #(= 1 (n-eights-around %)))
(count))))
(mapv #(count-single-eights % (count (% 0)) (count %))
test-cases)
=> [3 0 0 2 2 1 3 1 4 3]
Where test-cases is an array holding all the "Python test cases"
Try it online!
answered yesterday
Carcigenicate
2,24911224
add a comment |
15 Answers
15
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15 Answers
15
active
oldest
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up vote
8
down vote
R, 117 63 59 bytes
function(m)sum(colSums(as.matrix(dist(which(m==8,T)))<2)<2)
Try it online!
dist computes distances (default is Euclidean) among rows of a matrix. which with second argument TRUE returns the coordinates where the predicate is true.
Coordinates are neighbours if the distance between them is not more than the square root of 2, but the inner <2 is good enough because the possible distance jumps from sqrt(2) ro 2.
answered yesterday
ngm
2,92923
it's a shame numerical imprecision doesn't allowcolSums()^2<=2to work.
– Giuseppe
yesterday
@Giuseppe of course there are only a few possible distances andsqrt(2)jumps to2(e.g.sort(c(dist(expand.grid(1:6,1:6))), decreasing = TRUE))) so we were being too clever there.
– ngm
23 hours ago
add a comment |
up vote
8
down vote
R, 117 63 59 bytes
function(m)sum(colSums(as.matrix(dist(which(m==8,T)))<2)<2)
Try it online!
dist computes distances (default is Euclidean) among rows of a matrix. which with second argument TRUE returns the coordinates where the predicate is true.
Coordinates are neighbours if the distance between them is not more than the square root of 2, but the inner <2 is good enough because the possible distance jumps from sqrt(2) ro 2.
answered yesterday
ngm
2,92923
it's a shame numerical imprecision doesn't allowcolSums()^2<=2to work.
– Giuseppe
yesterday
@Giuseppe of course there are only a few possible distances andsqrt(2)jumps to2(e.g.sort(c(dist(expand.grid(1:6,1:6))), decreasing = TRUE))) so we were being too clever there.
– ngm
23 hours ago
add a comment |
up vote
8
down vote
up vote
8
down vote
R, 117 63 59 bytes
function(m)sum(colSums(as.matrix(dist(which(m==8,T)))<2)<2)
Try it online!
dist computes distances (default is Euclidean) among rows of a matrix. which with second argument TRUE returns the coordinates where the predicate is true.
Coordinates are neighbours if the distance between them is not more than the square root of 2, but the inner <2 is good enough because the possible distance jumps from sqrt(2) ro 2.
answered yesterday
ngm
2,92923
R, 117 63 59 bytes
function(m)sum(colSums(as.matrix(dist(which(m==8,T)))<2)<2)
Try it online!
dist computes distances (default is Euclidean) among rows of a matrix. which with second argument TRUE returns the coordinates where the predicate is true.
Coordinates are neighbours if the distance between them is not more than the square root of 2, but the inner <2 is good enough because the possible distance jumps from sqrt(2) ro 2.
answered yesterday
ngm
2,92923
edited 23 hours ago
answered yesterday
ngm
2,92923
answered yesterday
ngm
2,92923
answered yesterday
ngm
2,92923
2,92923
it's a shame numerical imprecision doesn't allowcolSums()^2<=2to work.
– Giuseppe
yesterday
@Giuseppe of course there are only a few possible distances andsqrt(2)jumps to2(e.g.sort(c(dist(expand.grid(1:6,1:6))), decreasing = TRUE))) so we were being too clever there.
– ngm
23 hours ago
add a comment |
it's a shame numerical imprecision doesn't allowcolSums()^2<=2to work.
– Giuseppe
yesterday
@Giuseppe of course there are only a few possible distances andsqrt(2)jumps to2(e.g.sort(c(dist(expand.grid(1:6,1:6))), decreasing = TRUE))) so we were being too clever there.
– ngm
23 hours ago
it's a shame numerical imprecision doesn't allow
colSums()^2<=2 to work.– Giuseppe
yesterday
it's a shame numerical imprecision doesn't allow
colSums()^2<=2 to work.– Giuseppe
yesterday
@Giuseppe of course there are only a few possible distances and
sqrt(2) jumps to 2 (e.g. sort(c(dist(expand.grid(1:6,1:6))), decreasing = TRUE))) so we were being too clever there.– ngm
23 hours ago
@Giuseppe of course there are only a few possible distances and
sqrt(2) jumps to 2 (e.g. sort(c(dist(expand.grid(1:6,1:6))), decreasing = TRUE))) so we were being too clever there.– ngm
23 hours ago
add a comment |
up vote
6
down vote
APL (Dyalog Classic), 29 28 25 bytes
≢∘⍸16=2⊥¨3,⌿3,/8=(⍉0,⌽)⍣4
Try it online!
answered 2 days ago
ngn
6,15312459
Note: 0 index origin is not even needed.
– Zacharý
yesterday
@Zacharý I always use it as a default, to avoid surprises.
– ngn
yesterday
Ah, so like others with1(except not explicitly set). That makes sense.
– Zacharý
yesterday
add a comment |
up vote
6
down vote
APL (Dyalog Classic), 29 28 25 bytes
≢∘⍸16=2⊥¨3,⌿3,/8=(⍉0,⌽)⍣4
Try it online!
answered 2 days ago
ngn
6,15312459
Note: 0 index origin is not even needed.
– Zacharý
yesterday
@Zacharý I always use it as a default, to avoid surprises.
– ngn
yesterday
Ah, so like others with1(except not explicitly set). That makes sense.
– Zacharý
yesterday
add a comment |
up vote
6
down vote
up vote
6
down vote
APL (Dyalog Classic), 29 28 25 bytes
≢∘⍸16=2⊥¨3,⌿3,/8=(⍉0,⌽)⍣4
Try it online!
answered 2 days ago
ngn
6,15312459
APL (Dyalog Classic), 29 28 25 bytes
≢∘⍸16=2⊥¨3,⌿3,/8=(⍉0,⌽)⍣4
Try it online!
answered 2 days ago
ngn
6,15312459
edited 2 days ago
answered 2 days ago
ngn
6,15312459
answered 2 days ago
ngn
6,15312459
answered 2 days ago
ngn
6,15312459
6,15312459
Note: 0 index origin is not even needed.
– Zacharý
yesterday
@Zacharý I always use it as a default, to avoid surprises.
– ngn
yesterday
Ah, so like others with1(except not explicitly set). That makes sense.
– Zacharý
yesterday
add a comment |
Note: 0 index origin is not even needed.
– Zacharý
yesterday
@Zacharý I always use it as a default, to avoid surprises.
– ngn
yesterday
Ah, so like others with1(except not explicitly set). That makes sense.
– Zacharý
yesterday
Note: 0 index origin is not even needed.
– Zacharý
yesterday
Note: 0 index origin is not even needed.
– Zacharý
yesterday
@Zacharý I always use it as a default, to avoid surprises.
– ngn
yesterday
@Zacharý I always use it as a default, to avoid surprises.
– ngn
yesterday
Ah, so like others with
1 (except not explicitly set). That makes sense.– Zacharý
yesterday
Ah, so like others with
1 (except not explicitly set). That makes sense.– Zacharý
yesterday
add a comment |
up vote
5
down vote
Jelly, 18 15 bytes
8=µ+Ż+ḊZµ⁺ỊṖḋµS
Try it online!
How it works
8=µ+Ż+ḊZµ⁺ỊṖḋµS Main link (monad). Input: digit matrix
8= 1) Convert elements by `x == 8`
µ 2) New chain:
+Ż+Ḋ x + [0,*x] + x[1:] (missing elements are considered 0)
Effectively, vertical convolution with [1,1,1]
Z Transpose
µ⁺ 3) Start new chain, apply 2) again
ỊṖ Convert elements by `|x| <= 1` and remove last row
ḋ Row-wise dot product with result of 1)
µS 4) Sum
Previous solution, 18 bytes
æc7B¤ZḊṖ
8=µÇÇỊḋµS
Try it online!
Wanted to share another approach, though this is 1 byte longer than Jonathan Allan's solution.
How it works
æc7B¤ZḊṖ Auxiliary link (monad). Input: integer matrix
æc7B¤ Convolution with [1,1,1] on each row
ZḊṖ Zip (transpose), remove first and last elements
8=µÇÇỊḋµS Main link (monad). Input: digit matrix
8= Convert 8 to 1, anything else to 0 (*A)
怀 Apply aux.link twice (effective convolution with [[1,1,1]]*3)
Ịḋ Convert to |x|<=1, then row-wise dot product with A
µS Sum the result
answered 2 days ago
Bubbler
4,859750
add a comment |
up vote
5
down vote
Jelly, 18 15 bytes
8=µ+Ż+ḊZµ⁺ỊṖḋµS
Try it online!
How it works
8=µ+Ż+ḊZµ⁺ỊṖḋµS Main link (monad). Input: digit matrix
8= 1) Convert elements by `x == 8`
µ 2) New chain:
+Ż+Ḋ x + [0,*x] + x[1:] (missing elements are considered 0)
Effectively, vertical convolution with [1,1,1]
Z Transpose
µ⁺ 3) Start new chain, apply 2) again
ỊṖ Convert elements by `|x| <= 1` and remove last row
ḋ Row-wise dot product with result of 1)
µS 4) Sum
Previous solution, 18 bytes
æc7B¤ZḊṖ
8=µÇÇỊḋµS
Try it online!
Wanted to share another approach, though this is 1 byte longer than Jonathan Allan's solution.
How it works
æc7B¤ZḊṖ Auxiliary link (monad). Input: integer matrix
æc7B¤ Convolution with [1,1,1] on each row
ZḊṖ Zip (transpose), remove first and last elements
8=µÇÇỊḋµS Main link (monad). Input: digit matrix
8= Convert 8 to 1, anything else to 0 (*A)
怀 Apply aux.link twice (effective convolution with [[1,1,1]]*3)
Ịḋ Convert to |x|<=1, then row-wise dot product with A
µS Sum the result
answered 2 days ago
Bubbler
4,859750
add a comment |
up vote
5
down vote
up vote
5
down vote
Jelly, 18 15 bytes
8=µ+Ż+ḊZµ⁺ỊṖḋµS
Try it online!
How it works
8=µ+Ż+ḊZµ⁺ỊṖḋµS Main link (monad). Input: digit matrix
8= 1) Convert elements by `x == 8`
µ 2) New chain:
+Ż+Ḋ x + [0,*x] + x[1:] (missing elements are considered 0)
Effectively, vertical convolution with [1,1,1]
Z Transpose
µ⁺ 3) Start new chain, apply 2) again
ỊṖ Convert elements by `|x| <= 1` and remove last row
ḋ Row-wise dot product with result of 1)
µS 4) Sum
Previous solution, 18 bytes
æc7B¤ZḊṖ
8=µÇÇỊḋµS
Try it online!
Wanted to share another approach, though this is 1 byte longer than Jonathan Allan's solution.
How it works
æc7B¤ZḊṖ Auxiliary link (monad). Input: integer matrix
æc7B¤ Convolution with [1,1,1] on each row
ZḊṖ Zip (transpose), remove first and last elements
8=µÇÇỊḋµS Main link (monad). Input: digit matrix
8= Convert 8 to 1, anything else to 0 (*A)
怀 Apply aux.link twice (effective convolution with [[1,1,1]]*3)
Ịḋ Convert to |x|<=1, then row-wise dot product with A
µS Sum the result
answered 2 days ago
Bubbler
4,859750
Jelly, 18 15 bytes
8=µ+Ż+ḊZµ⁺ỊṖḋµS
Try it online!
How it works
8=µ+Ż+ḊZµ⁺ỊṖḋµS Main link (monad). Input: digit matrix
8= 1) Convert elements by `x == 8`
µ 2) New chain:
+Ż+Ḋ x + [0,*x] + x[1:] (missing elements are considered 0)
Effectively, vertical convolution with [1,1,1]
Z Transpose
µ⁺ 3) Start new chain, apply 2) again
ỊṖ Convert elements by `|x| <= 1` and remove last row
ḋ Row-wise dot product with result of 1)
µS 4) Sum
Previous solution, 18 bytes
æc7B¤ZḊṖ
8=µÇÇỊḋµS
Try it online!
Wanted to share another approach, though this is 1 byte longer than Jonathan Allan's solution.
How it works
æc7B¤ZḊṖ Auxiliary link (monad). Input: integer matrix
æc7B¤ Convolution with [1,1,1] on each row
ZḊṖ Zip (transpose), remove first and last elements
8=µÇÇỊḋµS Main link (monad). Input: digit matrix
8= Convert 8 to 1, anything else to 0 (*A)
怀 Apply aux.link twice (effective convolution with [[1,1,1]]*3)
Ịḋ Convert to |x|<=1, then row-wise dot product with A
µS Sum the result
answered 2 days ago
Bubbler
4,859750
edited 2 days ago
answered 2 days ago
Bubbler
4,859750
answered 2 days ago
Bubbler
4,859750
answered 2 days ago
Bubbler
4,859750
4,859750
add a comment |
add a comment |
up vote
4
down vote
JavaScript (Node.js), 88 85 bytes
a=>g=(x,y=0,c=0)=>a.map(p=>p.map((q,j)=>c+=q-8?0:1/x?(x-j)**2+y*y<3:g(j,-y)<2)&--y)|c
Try it online!
Thank Arnauld for 2 bytes
answered 2 days ago
l4m2
4,2341432
add a comment |
up vote
4
down vote
JavaScript (Node.js), 88 85 bytes
a=>g=(x,y=0,c=0)=>a.map(p=>p.map((q,j)=>c+=q-8?0:1/x?(x-j)**2+y*y<3:g(j,-y)<2)&--y)|c
Try it online!
Thank Arnauld for 2 bytes
answered 2 days ago
l4m2
4,2341432
add a comment |
up vote
4
down vote
up vote
4
down vote
JavaScript (Node.js), 88 85 bytes
a=>g=(x,y=0,c=0)=>a.map(p=>p.map((q,j)=>c+=q-8?0:1/x?(x-j)**2+y*y<3:g(j,-y)<2)&--y)|c
Try it online!
Thank Arnauld for 2 bytes
answered 2 days ago
l4m2
4,2341432
JavaScript (Node.js), 88 85 bytes
a=>g=(x,y=0,c=0)=>a.map(p=>p.map((q,j)=>c+=q-8?0:1/x?(x-j)**2+y*y<3:g(j,-y)<2)&--y)|c
Try it online!
Thank Arnauld for 2 bytes
answered 2 days ago
l4m2
4,2341432
edited 2 days ago
answered 2 days ago
l4m2
4,2341432
answered 2 days ago
l4m2
4,2341432
answered 2 days ago
l4m2
4,2341432
4,2341432
add a comment |
add a comment |
up vote
4
down vote
J,43, 40 37 bytes
-3 bytes thanks to Bubbler
1#.1#.3 3(16=8#.@:=,);._3(0|:@,|.)^:4
Try it online!
Explanation:
The first part of the algorithm assures that we can apply a 3x3 sliding window to he input. This is achieved by prepending a row of zeroes and 90 degrees rotation, repeated 4 times.
1#.1#.3 3(16=8#.@:=,);._3(0|:@,|.)^:4
( )^:4 - repeat 4 times
0|:@,|. - reverse, prepend wit a row of 0 and transpose
;._3 - cut the input (already outlined with zeroes)
3 3 - into matrices with size 3x3
( ) - and for each matrix do
, - ravel (flatten)
8 = - check if each item equals 8
#.@: - and convert the list of 1s and 0s to a decimal
16= - is equal to 16?
1#. - add (the result has the shape of the input)
1#. - add again
answered 2 days ago
Galen Ivanov
5,82711032
1
37 bytes using@:and moving|.. Note that@in place of@:doesn't work.
– Bubbler
2 days ago
@Bubbler Thank you!
– Galen Ivanov
2 days ago
This is nice. Probably worth adding at least a high level explanation of how it works, if not a code breakdown. It took me 10m or so to figure it out. Also, it's interesting how much shorter the APL version (which uses the same approach) is. Looks like that's mostly the result of digraphs instead of single char symbols...
– Jonah
yesterday
@Jonah I'll add an explanation. For comparison with APL you can look at the revisions of ngn's solution, especially the 28 byte version
– Galen Ivanov
yesterday
1
@Jonah Explanation added
– Galen Ivanov
yesterday
add a comment |
up vote
4
down vote
J,43, 40 37 bytes
-3 bytes thanks to Bubbler
1#.1#.3 3(16=8#.@:=,);._3(0|:@,|.)^:4
Try it online!
Explanation:
The first part of the algorithm assures that we can apply a 3x3 sliding window to he input. This is achieved by prepending a row of zeroes and 90 degrees rotation, repeated 4 times.
1#.1#.3 3(16=8#.@:=,);._3(0|:@,|.)^:4
( )^:4 - repeat 4 times
0|:@,|. - reverse, prepend wit a row of 0 and transpose
;._3 - cut the input (already outlined with zeroes)
3 3 - into matrices with size 3x3
( ) - and for each matrix do
, - ravel (flatten)
8 = - check if each item equals 8
#.@: - and convert the list of 1s and 0s to a decimal
16= - is equal to 16?
1#. - add (the result has the shape of the input)
1#. - add again
answered 2 days ago
Galen Ivanov
5,82711032
1
37 bytes using@:and moving|.. Note that@in place of@:doesn't work.
– Bubbler
2 days ago
@Bubbler Thank you!
– Galen Ivanov
2 days ago
This is nice. Probably worth adding at least a high level explanation of how it works, if not a code breakdown. It took me 10m or so to figure it out. Also, it's interesting how much shorter the APL version (which uses the same approach) is. Looks like that's mostly the result of digraphs instead of single char symbols...
– Jonah
yesterday
@Jonah I'll add an explanation. For comparison with APL you can look at the revisions of ngn's solution, especially the 28 byte version
– Galen Ivanov
yesterday
1
@Jonah Explanation added
– Galen Ivanov
yesterday
add a comment |
up vote
4
down vote
up vote
4
down vote
J,43, 40 37 bytes
-3 bytes thanks to Bubbler
1#.1#.3 3(16=8#.@:=,);._3(0|:@,|.)^:4
Try it online!
Explanation:
The first part of the algorithm assures that we can apply a 3x3 sliding window to he input. This is achieved by prepending a row of zeroes and 90 degrees rotation, repeated 4 times.
1#.1#.3 3(16=8#.@:=,);._3(0|:@,|.)^:4
( )^:4 - repeat 4 times
0|:@,|. - reverse, prepend wit a row of 0 and transpose
;._3 - cut the input (already outlined with zeroes)
3 3 - into matrices with size 3x3
( ) - and for each matrix do
, - ravel (flatten)
8 = - check if each item equals 8
#.@: - and convert the list of 1s and 0s to a decimal
16= - is equal to 16?
1#. - add (the result has the shape of the input)
1#. - add again
answered 2 days ago
Galen Ivanov
5,82711032
J,43, 40 37 bytes
-3 bytes thanks to Bubbler
1#.1#.3 3(16=8#.@:=,);._3(0|:@,|.)^:4
Try it online!
Explanation:
The first part of the algorithm assures that we can apply a 3x3 sliding window to he input. This is achieved by prepending a row of zeroes and 90 degrees rotation, repeated 4 times.
1#.1#.3 3(16=8#.@:=,);._3(0|:@,|.)^:4
( )^:4 - repeat 4 times
0|:@,|. - reverse, prepend wit a row of 0 and transpose
;._3 - cut the input (already outlined with zeroes)
3 3 - into matrices with size 3x3
( ) - and for each matrix do
, - ravel (flatten)
8 = - check if each item equals 8
#.@: - and convert the list of 1s and 0s to a decimal
16= - is equal to 16?
1#. - add (the result has the shape of the input)
1#. - add again
answered 2 days ago
Galen Ivanov
5,82711032
edited yesterday
answered 2 days ago
Galen Ivanov
5,82711032
answered 2 days ago
Galen Ivanov
5,82711032
answered 2 days ago
Galen Ivanov
5,82711032
5,82711032
1
37 bytes using@:and moving|.. Note that@in place of@:doesn't work.
– Bubbler
2 days ago
@Bubbler Thank you!
– Galen Ivanov
2 days ago
This is nice. Probably worth adding at least a high level explanation of how it works, if not a code breakdown. It took me 10m or so to figure it out. Also, it's interesting how much shorter the APL version (which uses the same approach) is. Looks like that's mostly the result of digraphs instead of single char symbols...
– Jonah
yesterday
@Jonah I'll add an explanation. For comparison with APL you can look at the revisions of ngn's solution, especially the 28 byte version
– Galen Ivanov
yesterday
1
@Jonah Explanation added
– Galen Ivanov
yesterday
add a comment |
1
37 bytes using@:and moving|.. Note that@in place of@:doesn't work.
– Bubbler
2 days ago
@Bubbler Thank you!
– Galen Ivanov
2 days ago
This is nice. Probably worth adding at least a high level explanation of how it works, if not a code breakdown. It took me 10m or so to figure it out. Also, it's interesting how much shorter the APL version (which uses the same approach) is. Looks like that's mostly the result of digraphs instead of single char symbols...
– Jonah
yesterday
@Jonah I'll add an explanation. For comparison with APL you can look at the revisions of ngn's solution, especially the 28 byte version
– Galen Ivanov
yesterday
1
@Jonah Explanation added
– Galen Ivanov
yesterday
1
1
37 bytes using
@: and moving |.. Note that @ in place of @: doesn't work.– Bubbler
2 days ago
37 bytes using
@: and moving |.. Note that @ in place of @: doesn't work.– Bubbler
2 days ago
@Bubbler Thank you!
– Galen Ivanov
2 days ago
@Bubbler Thank you!
– Galen Ivanov
2 days ago
This is nice. Probably worth adding at least a high level explanation of how it works, if not a code breakdown. It took me 10m or so to figure it out. Also, it's interesting how much shorter the APL version (which uses the same approach) is. Looks like that's mostly the result of digraphs instead of single char symbols...
– Jonah
yesterday
This is nice. Probably worth adding at least a high level explanation of how it works, if not a code breakdown. It took me 10m or so to figure it out. Also, it's interesting how much shorter the APL version (which uses the same approach) is. Looks like that's mostly the result of digraphs instead of single char symbols...
– Jonah
yesterday
@Jonah I'll add an explanation. For comparison with APL you can look at the revisions of ngn's solution, especially the 28 byte version
– Galen Ivanov
yesterday
@Jonah I'll add an explanation. For comparison with APL you can look at the revisions of ngn's solution, especially the 28 byte version
– Galen Ivanov
yesterday
1
1
@Jonah Explanation added
– Galen Ivanov
yesterday
@Jonah Explanation added
– Galen Ivanov
yesterday
add a comment |
up vote
3
down vote
Retina 0.8.2, 84 bytes
.+
_$&_
m`(?<!(?(1).)^(?<-1>.)*.?.?8.*¶(.)*.|8)8(?!8|.(.)*¶.*8.?.?(?<-2>.)*$(?(2).))
Try it online! Explanation:
.+
_$&_
Wrap each line in non-8 characters so that all 8s have at least one character on each side.
m`
This is the last stage, so counting matches is implied. The m modifier makes the ^ and $ characters match at the start or end of any line.
(?<!...|8)
Don't match a character directly after an 8, or...
(?(1).)^(?<-1>.)*.?.?8.*¶(.)*.
... a character below an 8; the (?(1).)^(?<-1>.)* matches the same column as the ¶(.)* on the next line, but the .?.? allows the 8 to be 1 left or right of the character after the . on the next line.
8
Match 8s.
(?!8|...)
Don't match an 8 immediately before an 8, or...
.(.)*¶.*8.?.?(?<-2>.)*$(?(2).)
... a character with an 8 in the line below; again, the (?<-2>.)*$(?(2).) matches the same column as the (.)*¶ on the previous line, but the .?.? allows the 8 to be 1 left or right of the 8 before the . on the previous line.
answered 2 days ago
Neil
77.7k744174
add a comment |
up vote
3
down vote
Retina 0.8.2, 84 bytes
.+
_$&_
m`(?<!(?(1).)^(?<-1>.)*.?.?8.*¶(.)*.|8)8(?!8|.(.)*¶.*8.?.?(?<-2>.)*$(?(2).))
Try it online! Explanation:
.+
_$&_
Wrap each line in non-8 characters so that all 8s have at least one character on each side.
m`
This is the last stage, so counting matches is implied. The m modifier makes the ^ and $ characters match at the start or end of any line.
(?<!...|8)
Don't match a character directly after an 8, or...
(?(1).)^(?<-1>.)*.?.?8.*¶(.)*.
... a character below an 8; the (?(1).)^(?<-1>.)* matches the same column as the ¶(.)* on the next line, but the .?.? allows the 8 to be 1 left or right of the character after the . on the next line.
8
Match 8s.
(?!8|...)
Don't match an 8 immediately before an 8, or...
.(.)*¶.*8.?.?(?<-2>.)*$(?(2).)
... a character with an 8 in the line below; again, the (?<-2>.)*$(?(2).) matches the same column as the (.)*¶ on the previous line, but the .?.? allows the 8 to be 1 left or right of the 8 before the . on the previous line.
answered 2 days ago
Neil
77.7k744174
add a comment |
up vote
3
down vote
up vote
3
down vote
Retina 0.8.2, 84 bytes
.+
_$&_
m`(?<!(?(1).)^(?<-1>.)*.?.?8.*¶(.)*.|8)8(?!8|.(.)*¶.*8.?.?(?<-2>.)*$(?(2).))
Try it online! Explanation:
.+
_$&_
Wrap each line in non-8 characters so that all 8s have at least one character on each side.
m`
This is the last stage, so counting matches is implied. The m modifier makes the ^ and $ characters match at the start or end of any line.
(?<!...|8)
Don't match a character directly after an 8, or...
(?(1).)^(?<-1>.)*.?.?8.*¶(.)*.
... a character below an 8; the (?(1).)^(?<-1>.)* matches the same column as the ¶(.)* on the next line, but the .?.? allows the 8 to be 1 left or right of the character after the . on the next line.
8
Match 8s.
(?!8|...)
Don't match an 8 immediately before an 8, or...
.(.)*¶.*8.?.?(?<-2>.)*$(?(2).)
... a character with an 8 in the line below; again, the (?<-2>.)*$(?(2).) matches the same column as the (.)*¶ on the previous line, but the .?.? allows the 8 to be 1 left or right of the 8 before the . on the previous line.
answered 2 days ago
Neil
77.7k744174
Retina 0.8.2, 84 bytes
.+
_$&_
m`(?<!(?(1).)^(?<-1>.)*.?.?8.*¶(.)*.|8)8(?!8|.(.)*¶.*8.?.?(?<-2>.)*$(?(2).))
Try it online! Explanation:
.+
_$&_
Wrap each line in non-8 characters so that all 8s have at least one character on each side.
m`
This is the last stage, so counting matches is implied. The m modifier makes the ^ and $ characters match at the start or end of any line.
(?<!...|8)
Don't match a character directly after an 8, or...
(?(1).)^(?<-1>.)*.?.?8.*¶(.)*.
... a character below an 8; the (?(1).)^(?<-1>.)* matches the same column as the ¶(.)* on the next line, but the .?.? allows the 8 to be 1 left or right of the character after the . on the next line.
8
Match 8s.
(?!8|...)
Don't match an 8 immediately before an 8, or...
.(.)*¶.*8.?.?(?<-2>.)*$(?(2).)
... a character with an 8 in the line below; again, the (?<-2>.)*$(?(2).) matches the same column as the (.)*¶ on the previous line, but the .?.? allows the 8 to be 1 left or right of the 8 before the . on the previous line.
answered 2 days ago
Neil
77.7k744174
answered 2 days ago
Neil
77.7k744174
answered 2 days ago
Neil
77.7k744174
answered 2 days ago
Neil
77.7k744174
77.7k744174
add a comment |
add a comment |
up vote
3
down vote
J, 42 bytes
[:+/@,8=]+[:+/(<:3 3#:4-.~i.9)|.!.0(_*8&=)
Try it online!
explanation
The high-level approach here is similar to the one used in the classic APL solution to the game of life: https://www.youtube.com/watch?v=a9xAKttWgP4.
In that solution, we shift our matrix in the 8 possible neighbor directions, creating 8 duplicates of the input, stack them up, and then add the "planes" together to get our neighbor counts.
Here, we use a "multiply by infinity" trick to adapt the solution for this problem.
[: +/@, 8 = ] + [: +/ (neighbor deltas) (|.!.0) _ * 8&= NB.
NB.
[: +/@, NB. the sum after flattening
8 = NB. a 0 1 matrix created by
NB. elmwise testing if 8
NB. equals the matrix
(the matrix to test for equality with 8 ) NB. defined by...
] + NB. the original input plus
[: +/ NB. the elmwise sum of 8
NB. matrices defined by
_ * NB. the elmwise product of
NB. infinity and
8&= NB. the matrix which is 1
NB. where the input is 8
NB. and 0 elsewhere, thus
NB. creating an infinity-0
NB. matrix
(|.!.0) NB. then 2d shifting that
NB. matrix in the 8 possible
NB. "neighbor" directions
(neighbor deltas) NB. defined by the "neighbor
NB. deltas" (see below)
NB. QED.
NB. ***********************
NB. The rest of the
NB. explanation merely
NB. breaks down the neighbor
NB. delta construction.
(neighbor deltas ) NB. the neighbor deltas are
NB. merely the cross product
NB. of _1 0 1 with itself,
NB. minus "0 0"
(<: 3 3 #: 4 -.~ i.9) NB. to create that...
<: NB. subtract one from
3 3 #: NB. the base 3 rep of
i.9 NB. the numbers 0 - 8
4 -.~ NB. minus the number 4
NB.
NB. All of which produces
NB. the eight "neighbor"
NB. deltas:
NB.
NB. _1 _1
NB. _1 0
NB. _1 1
NB. 0 _1
NB. 0 1
NB. 1 _1
NB. 1 0
NB. 1 1
answered 2 days ago
Jonah
1,841816
1
You have forgotten to remove a space between~and>
– Galen Ivanov
2 days ago
@GalenIvanov Fixed now. Thank you.
– Jonah
2 days ago
add a comment |
up vote
3
down vote
J, 42 bytes
[:+/@,8=]+[:+/(<:3 3#:4-.~i.9)|.!.0(_*8&=)
Try it online!
explanation
The high-level approach here is similar to the one used in the classic APL solution to the game of life: https://www.youtube.com/watch?v=a9xAKttWgP4.
In that solution, we shift our matrix in the 8 possible neighbor directions, creating 8 duplicates of the input, stack them up, and then add the "planes" together to get our neighbor counts.
Here, we use a "multiply by infinity" trick to adapt the solution for this problem.
[: +/@, 8 = ] + [: +/ (neighbor deltas) (|.!.0) _ * 8&= NB.
NB.
[: +/@, NB. the sum after flattening
8 = NB. a 0 1 matrix created by
NB. elmwise testing if 8
NB. equals the matrix
(the matrix to test for equality with 8 ) NB. defined by...
] + NB. the original input plus
[: +/ NB. the elmwise sum of 8
NB. matrices defined by
_ * NB. the elmwise product of
NB. infinity and
8&= NB. the matrix which is 1
NB. where the input is 8
NB. and 0 elsewhere, thus
NB. creating an infinity-0
NB. matrix
(|.!.0) NB. then 2d shifting that
NB. matrix in the 8 possible
NB. "neighbor" directions
(neighbor deltas) NB. defined by the "neighbor
NB. deltas" (see below)
NB. QED.
NB. ***********************
NB. The rest of the
NB. explanation merely
NB. breaks down the neighbor
NB. delta construction.
(neighbor deltas ) NB. the neighbor deltas are
NB. merely the cross product
NB. of _1 0 1 with itself,
NB. minus "0 0"
(<: 3 3 #: 4 -.~ i.9) NB. to create that...
<: NB. subtract one from
3 3 #: NB. the base 3 rep of
i.9 NB. the numbers 0 - 8
4 -.~ NB. minus the number 4
NB.
NB. All of which produces
NB. the eight "neighbor"
NB. deltas:
NB.
NB. _1 _1
NB. _1 0
NB. _1 1
NB. 0 _1
NB. 0 1
NB. 1 _1
NB. 1 0
NB. 1 1
answered 2 days ago
Jonah
1,841816
1
You have forgotten to remove a space between~and>
– Galen Ivanov
2 days ago
@GalenIvanov Fixed now. Thank you.
– Jonah
2 days ago
add a comment |
up vote
3
down vote
up vote
3
down vote
J, 42 bytes
[:+/@,8=]+[:+/(<:3 3#:4-.~i.9)|.!.0(_*8&=)
Try it online!
explanation
The high-level approach here is similar to the one used in the classic APL solution to the game of life: https://www.youtube.com/watch?v=a9xAKttWgP4.
In that solution, we shift our matrix in the 8 possible neighbor directions, creating 8 duplicates of the input, stack them up, and then add the "planes" together to get our neighbor counts.
Here, we use a "multiply by infinity" trick to adapt the solution for this problem.
[: +/@, 8 = ] + [: +/ (neighbor deltas) (|.!.0) _ * 8&= NB.
NB.
[: +/@, NB. the sum after flattening
8 = NB. a 0 1 matrix created by
NB. elmwise testing if 8
NB. equals the matrix
(the matrix to test for equality with 8 ) NB. defined by...
] + NB. the original input plus
[: +/ NB. the elmwise sum of 8
NB. matrices defined by
_ * NB. the elmwise product of
NB. infinity and
8&= NB. the matrix which is 1
NB. where the input is 8
NB. and 0 elsewhere, thus
NB. creating an infinity-0
NB. matrix
(|.!.0) NB. then 2d shifting that
NB. matrix in the 8 possible
NB. "neighbor" directions
(neighbor deltas) NB. defined by the "neighbor
NB. deltas" (see below)
NB. QED.
NB. ***********************
NB. The rest of the
NB. explanation merely
NB. breaks down the neighbor
NB. delta construction.
(neighbor deltas ) NB. the neighbor deltas are
NB. merely the cross product
NB. of _1 0 1 with itself,
NB. minus "0 0"
(<: 3 3 #: 4 -.~ i.9) NB. to create that...
<: NB. subtract one from
3 3 #: NB. the base 3 rep of
i.9 NB. the numbers 0 - 8
4 -.~ NB. minus the number 4
NB.
NB. All of which produces
NB. the eight "neighbor"
NB. deltas:
NB.
NB. _1 _1
NB. _1 0
NB. _1 1
NB. 0 _1
NB. 0 1
NB. 1 _1
NB. 1 0
NB. 1 1
answered 2 days ago
Jonah
1,841816
J, 42 bytes
[:+/@,8=]+[:+/(<:3 3#:4-.~i.9)|.!.0(_*8&=)
Try it online!
explanation
The high-level approach here is similar to the one used in the classic APL solution to the game of life: https://www.youtube.com/watch?v=a9xAKttWgP4.
In that solution, we shift our matrix in the 8 possible neighbor directions, creating 8 duplicates of the input, stack them up, and then add the "planes" together to get our neighbor counts.
Here, we use a "multiply by infinity" trick to adapt the solution for this problem.
[: +/@, 8 = ] + [: +/ (neighbor deltas) (|.!.0) _ * 8&= NB.
NB.
[: +/@, NB. the sum after flattening
8 = NB. a 0 1 matrix created by
NB. elmwise testing if 8
NB. equals the matrix
(the matrix to test for equality with 8 ) NB. defined by...
] + NB. the original input plus
[: +/ NB. the elmwise sum of 8
NB. matrices defined by
_ * NB. the elmwise product of
NB. infinity and
8&= NB. the matrix which is 1
NB. where the input is 8
NB. and 0 elsewhere, thus
NB. creating an infinity-0
NB. matrix
(|.!.0) NB. then 2d shifting that
NB. matrix in the 8 possible
NB. "neighbor" directions
(neighbor deltas) NB. defined by the "neighbor
NB. deltas" (see below)
NB. QED.
NB. ***********************
NB. The rest of the
NB. explanation merely
NB. breaks down the neighbor
NB. delta construction.
(neighbor deltas ) NB. the neighbor deltas are
NB. merely the cross product
NB. of _1 0 1 with itself,
NB. minus "0 0"
(<: 3 3 #: 4 -.~ i.9) NB. to create that...
<: NB. subtract one from
3 3 #: NB. the base 3 rep of
i.9 NB. the numbers 0 - 8
4 -.~ NB. minus the number 4
NB.
NB. All of which produces
NB. the eight "neighbor"
NB. deltas:
NB.
NB. _1 _1
NB. _1 0
NB. _1 1
NB. 0 _1
NB. 0 1
NB. 1 _1
NB. 1 0
NB. 1 1
answered 2 days ago
Jonah
1,841816
edited yesterday
answered 2 days ago
Jonah
1,841816
answered 2 days ago
Jonah
1,841816
answered 2 days ago
Jonah
1,841816
1,841816
1
You have forgotten to remove a space between~and>
– Galen Ivanov
2 days ago
@GalenIvanov Fixed now. Thank you.
– Jonah
2 days ago
add a comment |
1
You have forgotten to remove a space between~and>
– Galen Ivanov
2 days ago
@GalenIvanov Fixed now. Thank you.
– Jonah
2 days ago
1
1
You have forgotten to remove a space between
~ and >– Galen Ivanov
2 days ago
You have forgotten to remove a space between
~ and >– Galen Ivanov
2 days ago
@GalenIvanov Fixed now. Thank you.
– Jonah
2 days ago
@GalenIvanov Fixed now. Thank you.
– Jonah
2 days ago
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up vote
2
down vote
Jelly, 17 bytes
=8ŒṪµŒcZIỊȦƲƇẎ⁸ḟL
Try it online! Or see the test-suite.
How?
=8ŒṪµŒcZIỊȦƲƇẎ⁸ḟL - Link: list of lists of integers (digits)
=8 - equals 8?
ŒṪ - multidimensional truthy indices (pairs of row & column indices of 8s)
µ - start a new monadic chain
Œc - all pairs (of the index-pairs)
Ƈ - filter keep if: (keep those that represent adjacent positions)
Ʋ - last four links as a monad:
Z - transpose
I - incremental differences
Ị - insignificant? (abs(x) <= 1)
Ȧ - all?
Ẏ - tighten (to a list of all adjacent 8's index-pairs, at least once each)
⁸ - chain's left argument (all the index-pairs again)
ḟ - filter discard (remove those found to be adjacent to another)
L - length (of the remaining pairs of indices of single 8s)
answered 2 days ago
Jonathan Allan
49.9k534163
add a comment |
up vote
2
down vote
Jelly, 17 bytes
=8ŒṪµŒcZIỊȦƲƇẎ⁸ḟL
Try it online! Or see the test-suite.
How?
=8ŒṪµŒcZIỊȦƲƇẎ⁸ḟL - Link: list of lists of integers (digits)
=8 - equals 8?
ŒṪ - multidimensional truthy indices (pairs of row & column indices of 8s)
µ - start a new monadic chain
Œc - all pairs (of the index-pairs)
Ƈ - filter keep if: (keep those that represent adjacent positions)
Ʋ - last four links as a monad:
Z - transpose
I - incremental differences
Ị - insignificant? (abs(x) <= 1)
Ȧ - all?
Ẏ - tighten (to a list of all adjacent 8's index-pairs, at least once each)
⁸ - chain's left argument (all the index-pairs again)
ḟ - filter discard (remove those found to be adjacent to another)
L - length (of the remaining pairs of indices of single 8s)
answered 2 days ago
Jonathan Allan
49.9k534163
add a comment |
up vote
2
down vote
up vote
2
down vote
Jelly, 17 bytes
=8ŒṪµŒcZIỊȦƲƇẎ⁸ḟL
Try it online! Or see the test-suite.
How?
=8ŒṪµŒcZIỊȦƲƇẎ⁸ḟL - Link: list of lists of integers (digits)
=8 - equals 8?
ŒṪ - multidimensional truthy indices (pairs of row & column indices of 8s)
µ - start a new monadic chain
Œc - all pairs (of the index-pairs)
Ƈ - filter keep if: (keep those that represent adjacent positions)
Ʋ - last four links as a monad:
Z - transpose
I - incremental differences
Ị - insignificant? (abs(x) <= 1)
Ȧ - all?
Ẏ - tighten (to a list of all adjacent 8's index-pairs, at least once each)
⁸ - chain's left argument (all the index-pairs again)
ḟ - filter discard (remove those found to be adjacent to another)
L - length (of the remaining pairs of indices of single 8s)
answered 2 days ago
Jonathan Allan
49.9k534163
Jelly, 17 bytes
=8ŒṪµŒcZIỊȦƲƇẎ⁸ḟL
Try it online! Or see the test-suite.
How?
=8ŒṪµŒcZIỊȦƲƇẎ⁸ḟL - Link: list of lists of integers (digits)
=8 - equals 8?
ŒṪ - multidimensional truthy indices (pairs of row & column indices of 8s)
µ - start a new monadic chain
Œc - all pairs (of the index-pairs)
Ƈ - filter keep if: (keep those that represent adjacent positions)
Ʋ - last four links as a monad:
Z - transpose
I - incremental differences
Ị - insignificant? (abs(x) <= 1)
Ȧ - all?
Ẏ - tighten (to a list of all adjacent 8's index-pairs, at least once each)
⁸ - chain's left argument (all the index-pairs again)
ḟ - filter discard (remove those found to be adjacent to another)
L - length (of the remaining pairs of indices of single 8s)
answered 2 days ago
Jonathan Allan
49.9k534163
edited 2 days ago
answered 2 days ago
Jonathan Allan
49.9k534163
answered 2 days ago
Jonathan Allan
49.9k534163
answered 2 days ago
Jonathan Allan
49.9k534163
49.9k534163
add a comment |
add a comment |
up vote
2
down vote
Python 2, 130 bytes
lambda a:sum(sum(u[~-c*(c>0):c+2].count(8)for u in a[~-r*(r>0):r+2])*8==8==a[r][c]for r in range(len(a))for c in range(len(a[0])))
Try it online!
answered 2 days ago
Chas Brown
4,6361519
Seems shorter if take length from args
– l4m2
2 days ago
add a comment |
up vote
2
down vote
Python 2, 130 bytes
lambda a:sum(sum(u[~-c*(c>0):c+2].count(8)for u in a[~-r*(r>0):r+2])*8==8==a[r][c]for r in range(len(a))for c in range(len(a[0])))
Try it online!
answered 2 days ago
Chas Brown
4,6361519
Seems shorter if take length from args
– l4m2
2 days ago
add a comment |
up vote
2
down vote
up vote
2
down vote
Python 2, 130 bytes
lambda a:sum(sum(u[~-c*(c>0):c+2].count(8)for u in a[~-r*(r>0):r+2])*8==8==a[r][c]for r in range(len(a))for c in range(len(a[0])))
Try it online!
answered 2 days ago
Chas Brown
4,6361519
Python 2, 130 bytes
lambda a:sum(sum(u[~-c*(c>0):c+2].count(8)for u in a[~-r*(r>0):r+2])*8==8==a[r][c]for r in range(len(a))for c in range(len(a[0])))
Try it online!
answered 2 days ago
Chas Brown
4,6361519
edited 2 days ago
answered 2 days ago
Chas Brown
4,6361519
answered 2 days ago
Chas Brown
4,6361519
answered 2 days ago
Chas Brown
4,6361519
4,6361519
Seems shorter if take length from args
– l4m2
2 days ago
add a comment |
Seems shorter if take length from args
– l4m2
2 days ago
Seems shorter if take length from args
– l4m2
2 days ago
Seems shorter if take length from args
– l4m2
2 days ago
add a comment |
up vote
2
down vote
Java 8, 181 bytes
(M,R,C)->{int z=0,c,f,t;for(;R-->0;)for(c=C;c-->0;z+=f==1?1:0)for(f=0,t=9;M[R][c]==8&t-->0;)try{f+=M[t<3?R-1:t>5?R+1:R][t%3<1?c-1:t%3>1?c+1:c]==8?1:0;}catch(Exception e){}return z;}
Takes the dimensions as additional parameters R (amount of rows) and C (amount of columns).
The cells are checked pretty similar as I did in my Fryer simulator answer.
Try it online.
Explanation:
(M,R,C)->{ // Method with integer-matrix as parameter & integer return
int z=0, // Result-counter, starting at 0
c,f,t; // Temp-integers, starting uninitialized
for(;R-->0;) // Loop over the rows:
for(c=C;c-->0 // Inner loop over the columns:
; // After every iteration:
z+=f==1? // If the flag-integer is exactly 1:
1 // Increase the result-counter by 1
:0) // Else: leave it the same
for(f=0, // Reset the flag to 0
t=9;M[R][c]==8& // If the current cell contains an 8:
t-->0;) // Inner loop `t` in the range (9, 0]:
try{f+= // Increase the flag by:
M[t<3? // If `t` is 0, 1, or 2:
R-1 // Look at the previous row
:t>5? // Else-if `t` is 6, 7, or 8:
R+1 // Look at the next row
: // Else (`t` is 3, 4, or 5):
R] // Look at the current row
[t%3<1? // If `t` is 0, 3, or 6:
c-1 // Look at the previous column
:t%3>1? // Else-if `t` is 2, 5, or 8:
c+1 // Look at the next column
: // Else (`t` is 1, 4, or 7):
c] // Look at the current column
==8? // If the digit in this cell is 8:
1 // Increase the flag-integer by 1
:0; // Else: leave it the same
}catch(Exception e){} // Catch and ignore ArrayIndexOutOfBoundsExceptions
// (try-catch saves bytes in comparison to if-checks)
return z;} // And finally return the counter
answered 2 days ago
Kevin Cruijssen
33.6k554179
add a comment |
up vote
2
down vote
Java 8, 181 bytes
(M,R,C)->{int z=0,c,f,t;for(;R-->0;)for(c=C;c-->0;z+=f==1?1:0)for(f=0,t=9;M[R][c]==8&t-->0;)try{f+=M[t<3?R-1:t>5?R+1:R][t%3<1?c-1:t%3>1?c+1:c]==8?1:0;}catch(Exception e){}return z;}
Takes the dimensions as additional parameters R (amount of rows) and C (amount of columns).
The cells are checked pretty similar as I did in my Fryer simulator answer.
Try it online.
Explanation:
(M,R,C)->{ // Method with integer-matrix as parameter & integer return
int z=0, // Result-counter, starting at 0
c,f,t; // Temp-integers, starting uninitialized
for(;R-->0;) // Loop over the rows:
for(c=C;c-->0 // Inner loop over the columns:
; // After every iteration:
z+=f==1? // If the flag-integer is exactly 1:
1 // Increase the result-counter by 1
:0) // Else: leave it the same
for(f=0, // Reset the flag to 0
t=9;M[R][c]==8& // If the current cell contains an 8:
t-->0;) // Inner loop `t` in the range (9, 0]:
try{f+= // Increase the flag by:
M[t<3? // If `t` is 0, 1, or 2:
R-1 // Look at the previous row
:t>5? // Else-if `t` is 6, 7, or 8:
R+1 // Look at the next row
: // Else (`t` is 3, 4, or 5):
R] // Look at the current row
[t%3<1? // If `t` is 0, 3, or 6:
c-1 // Look at the previous column
:t%3>1? // Else-if `t` is 2, 5, or 8:
c+1 // Look at the next column
: // Else (`t` is 1, 4, or 7):
c] // Look at the current column
==8? // If the digit in this cell is 8:
1 // Increase the flag-integer by 1
:0; // Else: leave it the same
}catch(Exception e){} // Catch and ignore ArrayIndexOutOfBoundsExceptions
// (try-catch saves bytes in comparison to if-checks)
return z;} // And finally return the counter
answered 2 days ago
Kevin Cruijssen
33.6k554179
add a comment |
up vote
2
down vote
up vote
2
down vote
Java 8, 181 bytes
(M,R,C)->{int z=0,c,f,t;for(;R-->0;)for(c=C;c-->0;z+=f==1?1:0)for(f=0,t=9;M[R][c]==8&t-->0;)try{f+=M[t<3?R-1:t>5?R+1:R][t%3<1?c-1:t%3>1?c+1:c]==8?1:0;}catch(Exception e){}return z;}
Takes the dimensions as additional parameters R (amount of rows) and C (amount of columns).
The cells are checked pretty similar as I did in my Fryer simulator answer.
Try it online.
Explanation:
(M,R,C)->{ // Method with integer-matrix as parameter & integer return
int z=0, // Result-counter, starting at 0
c,f,t; // Temp-integers, starting uninitialized
for(;R-->0;) // Loop over the rows:
for(c=C;c-->0 // Inner loop over the columns:
; // After every iteration:
z+=f==1? // If the flag-integer is exactly 1:
1 // Increase the result-counter by 1
:0) // Else: leave it the same
for(f=0, // Reset the flag to 0
t=9;M[R][c]==8& // If the current cell contains an 8:
t-->0;) // Inner loop `t` in the range (9, 0]:
try{f+= // Increase the flag by:
M[t<3? // If `t` is 0, 1, or 2:
R-1 // Look at the previous row
:t>5? // Else-if `t` is 6, 7, or 8:
R+1 // Look at the next row
: // Else (`t` is 3, 4, or 5):
R] // Look at the current row
[t%3<1? // If `t` is 0, 3, or 6:
c-1 // Look at the previous column
:t%3>1? // Else-if `t` is 2, 5, or 8:
c+1 // Look at the next column
: // Else (`t` is 1, 4, or 7):
c] // Look at the current column
==8? // If the digit in this cell is 8:
1 // Increase the flag-integer by 1
:0; // Else: leave it the same
}catch(Exception e){} // Catch and ignore ArrayIndexOutOfBoundsExceptions
// (try-catch saves bytes in comparison to if-checks)
return z;} // And finally return the counter
answered 2 days ago
Kevin Cruijssen
33.6k554179
Java 8, 181 bytes
(M,R,C)->{int z=0,c,f,t;for(;R-->0;)for(c=C;c-->0;z+=f==1?1:0)for(f=0,t=9;M[R][c]==8&t-->0;)try{f+=M[t<3?R-1:t>5?R+1:R][t%3<1?c-1:t%3>1?c+1:c]==8?1:0;}catch(Exception e){}return z;}
Takes the dimensions as additional parameters R (amount of rows) and C (amount of columns).
The cells are checked pretty similar as I did in my Fryer simulator answer.
Try it online.
Explanation:
(M,R,C)->{ // Method with integer-matrix as parameter & integer return
int z=0, // Result-counter, starting at 0
c,f,t; // Temp-integers, starting uninitialized
for(;R-->0;) // Loop over the rows:
for(c=C;c-->0 // Inner loop over the columns:
; // After every iteration:
z+=f==1? // If the flag-integer is exactly 1:
1 // Increase the result-counter by 1
:0) // Else: leave it the same
for(f=0, // Reset the flag to 0
t=9;M[R][c]==8& // If the current cell contains an 8:
t-->0;) // Inner loop `t` in the range (9, 0]:
try{f+= // Increase the flag by:
M[t<3? // If `t` is 0, 1, or 2:
R-1 // Look at the previous row
:t>5? // Else-if `t` is 6, 7, or 8:
R+1 // Look at the next row
: // Else (`t` is 3, 4, or 5):
R] // Look at the current row
[t%3<1? // If `t` is 0, 3, or 6:
c-1 // Look at the previous column
:t%3>1? // Else-if `t` is 2, 5, or 8:
c+1 // Look at the next column
: // Else (`t` is 1, 4, or 7):
c] // Look at the current column
==8? // If the digit in this cell is 8:
1 // Increase the flag-integer by 1
:0; // Else: leave it the same
}catch(Exception e){} // Catch and ignore ArrayIndexOutOfBoundsExceptions
// (try-catch saves bytes in comparison to if-checks)
return z;} // And finally return the counter
answered 2 days ago
Kevin Cruijssen
33.6k554179
edited 2 days ago
answered 2 days ago
Kevin Cruijssen
33.6k554179
answered 2 days ago
Kevin Cruijssen
33.6k554179
answered 2 days ago
Kevin Cruijssen
33.6k554179
33.6k554179
add a comment |
add a comment |
up vote
2
down vote
Powershell, 121 bytes
param($a)(($b='='*4*($l=($a|% Le*)[0]))+($a|%{"!$_!"})+$b|sls "(?<=[^8]{3}.{$l}[^8])8(?=[^8].{$l}[^8]{3})" -a|% m*).Count
Less golfed test script:
$f = {
param($a)
$length=($a|% Length)[0]
$border='='*4*$length
$pattern="(?<=[^8]{3}.{$length}[^8])8(?=[^8].{$length}[^8]{3})"
$matches=$border+($a|%{"!$_!"})+$border |sls $pattern -a|% Matches
$matches.count
}
@(
,(3,"84565","93848","08615","67982","88742")
,(0,"88","23")
,(0,"534","252")
,(2,"5838")
,(2,"8","0","8")
,(1,"4285","2618","8558")
,(3,"45438182","82778393","98785428","45021869","15434561")
,(1,"8")
,(4,"85816877","99282783","28497327","92971956","69873152","19971882","35681475")
,(3,"818","257","801")
,(0,"")
) | % {
$expected,$a = $_
$result = &$f $a
"$($result-eq$expected): $result : $a"
}
Output:
True: 3 : 84565 93848 08615 67982 88742
True: 0 : 88 23
True: 0 : 534 252
True: 2 : 5838
True: 2 : 8 0 8
True: 1 : 4285 2618 8558
True: 3 : 45438182 82778393 98785428 45021869 15434561
True: 1 : 8
True: 4 : 85816877 99282783 28497327 92971956 69873152 19971882 35681475
True: 3 : 818 257 801
True: 0 :
Explanation:
First, the script calculates a length of the first string.
Second, it adds extra border to strings. Augmended reality string likes:
....=========!84565! !93848! !08615! !67982! !88742!===========....
represents the multiline string:
...=====
=======
!84565!
!93848!
!08615!
!67982!
!88742!
=======
========...
Note 1: the number of = is sufficient for a string of any length.
Note 2: a large number of = does not affect the search for eights.
Next, the regular expression (?<=[^8]{3}.{$l}[^8])8(?=[^8].{$l}[^8]{3}) looks for the digit 8 with the preceding non-eights (?<=[^8]{3}.{$l}[^8]) and the following non-eights (?=[^8].{$l}[^8]{3}):
.......
<<<....
<8>....
>>>....
.......
Finally, the number of matches is returned as a result.
answered yesterday
mazzy
1,715312
add a comment |
up vote
2
down vote
Powershell, 121 bytes
param($a)(($b='='*4*($l=($a|% Le*)[0]))+($a|%{"!$_!"})+$b|sls "(?<=[^8]{3}.{$l}[^8])8(?=[^8].{$l}[^8]{3})" -a|% m*).Count
Less golfed test script:
$f = {
param($a)
$length=($a|% Length)[0]
$border='='*4*$length
$pattern="(?<=[^8]{3}.{$length}[^8])8(?=[^8].{$length}[^8]{3})"
$matches=$border+($a|%{"!$_!"})+$border |sls $pattern -a|% Matches
$matches.count
}
@(
,(3,"84565","93848","08615","67982","88742")
,(0,"88","23")
,(0,"534","252")
,(2,"5838")
,(2,"8","0","8")
,(1,"4285","2618","8558")
,(3,"45438182","82778393","98785428","45021869","15434561")
,(1,"8")
,(4,"85816877","99282783","28497327","92971956","69873152","19971882","35681475")
,(3,"818","257","801")
,(0,"")
) | % {
$expected,$a = $_
$result = &$f $a
"$($result-eq$expected): $result : $a"
}
Output:
True: 3 : 84565 93848 08615 67982 88742
True: 0 : 88 23
True: 0 : 534 252
True: 2 : 5838
True: 2 : 8 0 8
True: 1 : 4285 2618 8558
True: 3 : 45438182 82778393 98785428 45021869 15434561
True: 1 : 8
True: 4 : 85816877 99282783 28497327 92971956 69873152 19971882 35681475
True: 3 : 818 257 801
True: 0 :
Explanation:
First, the script calculates a length of the first string.
Second, it adds extra border to strings. Augmended reality string likes:
....=========!84565! !93848! !08615! !67982! !88742!===========....
represents the multiline string:
...=====
=======
!84565!
!93848!
!08615!
!67982!
!88742!
=======
========...
Note 1: the number of = is sufficient for a string of any length.
Note 2: a large number of = does not affect the search for eights.
Next, the regular expression (?<=[^8]{3}.{$l}[^8])8(?=[^8].{$l}[^8]{3}) looks for the digit 8 with the preceding non-eights (?<=[^8]{3}.{$l}[^8]) and the following non-eights (?=[^8].{$l}[^8]{3}):
.......
<<<....
<8>....
>>>....
.......
Finally, the number of matches is returned as a result.
answered yesterday
mazzy
1,715312
add a comment |
up vote
2
down vote
up vote
2
down vote
Powershell, 121 bytes
param($a)(($b='='*4*($l=($a|% Le*)[0]))+($a|%{"!$_!"})+$b|sls "(?<=[^8]{3}.{$l}[^8])8(?=[^8].{$l}[^8]{3})" -a|% m*).Count
Less golfed test script:
$f = {
param($a)
$length=($a|% Length)[0]
$border='='*4*$length
$pattern="(?<=[^8]{3}.{$length}[^8])8(?=[^8].{$length}[^8]{3})"
$matches=$border+($a|%{"!$_!"})+$border |sls $pattern -a|% Matches
$matches.count
}
@(
,(3,"84565","93848","08615","67982","88742")
,(0,"88","23")
,(0,"534","252")
,(2,"5838")
,(2,"8","0","8")
,(1,"4285","2618","8558")
,(3,"45438182","82778393","98785428","45021869","15434561")
,(1,"8")
,(4,"85816877","99282783","28497327","92971956","69873152","19971882","35681475")
,(3,"818","257","801")
,(0,"")
) | % {
$expected,$a = $_
$result = &$f $a
"$($result-eq$expected): $result : $a"
}
Output:
True: 3 : 84565 93848 08615 67982 88742
True: 0 : 88 23
True: 0 : 534 252
True: 2 : 5838
True: 2 : 8 0 8
True: 1 : 4285 2618 8558
True: 3 : 45438182 82778393 98785428 45021869 15434561
True: 1 : 8
True: 4 : 85816877 99282783 28497327 92971956 69873152 19971882 35681475
True: 3 : 818 257 801
True: 0 :
Explanation:
First, the script calculates a length of the first string.
Second, it adds extra border to strings. Augmended reality string likes:
....=========!84565! !93848! !08615! !67982! !88742!===========....
represents the multiline string:
...=====
=======
!84565!
!93848!
!08615!
!67982!
!88742!
=======
========...
Note 1: the number of = is sufficient for a string of any length.
Note 2: a large number of = does not affect the search for eights.
Next, the regular expression (?<=[^8]{3}.{$l}[^8])8(?=[^8].{$l}[^8]{3}) looks for the digit 8 with the preceding non-eights (?<=[^8]{3}.{$l}[^8]) and the following non-eights (?=[^8].{$l}[^8]{3}):
.......
<<<....
<8>....
>>>....
.......
Finally, the number of matches is returned as a result.
answered yesterday
mazzy
1,715312
Powershell, 121 bytes
param($a)(($b='='*4*($l=($a|% Le*)[0]))+($a|%{"!$_!"})+$b|sls "(?<=[^8]{3}.{$l}[^8])8(?=[^8].{$l}[^8]{3})" -a|% m*).Count
Less golfed test script:
$f = {
param($a)
$length=($a|% Length)[0]
$border='='*4*$length
$pattern="(?<=[^8]{3}.{$length}[^8])8(?=[^8].{$length}[^8]{3})"
$matches=$border+($a|%{"!$_!"})+$border |sls $pattern -a|% Matches
$matches.count
}
@(
,(3,"84565","93848","08615","67982","88742")
,(0,"88","23")
,(0,"534","252")
,(2,"5838")
,(2,"8","0","8")
,(1,"4285","2618","8558")
,(3,"45438182","82778393","98785428","45021869","15434561")
,(1,"8")
,(4,"85816877","99282783","28497327","92971956","69873152","19971882","35681475")
,(3,"818","257","801")
,(0,"")
) | % {
$expected,$a = $_
$result = &$f $a
"$($result-eq$expected): $result : $a"
}
Output:
True: 3 : 84565 93848 08615 67982 88742
True: 0 : 88 23
True: 0 : 534 252
True: 2 : 5838
True: 2 : 8 0 8
True: 1 : 4285 2618 8558
True: 3 : 45438182 82778393 98785428 45021869 15434561
True: 1 : 8
True: 4 : 85816877 99282783 28497327 92971956 69873152 19971882 35681475
True: 3 : 818 257 801
True: 0 :
Explanation:
First, the script calculates a length of the first string.
Second, it adds extra border to strings. Augmended reality string likes:
....=========!84565! !93848! !08615! !67982! !88742!===========....
represents the multiline string:
...=====
=======
!84565!
!93848!
!08615!
!67982!
!88742!
=======
========...
Note 1: the number of = is sufficient for a string of any length.
Note 2: a large number of = does not affect the search for eights.
Next, the regular expression (?<=[^8]{3}.{$l}[^8])8(?=[^8].{$l}[^8]{3}) looks for the digit 8 with the preceding non-eights (?<=[^8]{3}.{$l}[^8]) and the following non-eights (?=[^8].{$l}[^8]{3}):
.......
<<<....
<8>....
>>>....
.......
Finally, the number of matches is returned as a result.
answered yesterday
mazzy
1,715312
answered yesterday
mazzy
1,715312
answered yesterday
mazzy
1,715312
answered yesterday
mazzy
1,715312
1,715312
add a comment |
add a comment |
up vote
2
down vote
MATL, 21 17 10 bytes
8=t3Y6Z+>z
Try it online!
Thanks to Luis Mendo for help in chat, and for suggesting 2D convolution.
Explanation:
#implicit input, m
8= #equal to 8? matrix of 1 where m is 8, 0 otherwise
t #duplicate
3Y6 #push [1 1 1; 1 0 1; 1 1 1], "neighbor cells" for convolution
Z+ #2D convolution; each element is replaced by the number of neighbors that are 8
> #elementwise greater than -- matrix of 1s where an 8 is single, 0s otherwise
z #number of nonzero elements -- number of single eights
#implicit output
answered yesterday
Giuseppe
15.8k31051
You could save quite a few bytes using (2D-)convolution, if you are famiiar with the concept
– Luis Mendo
yesterday
1
@LuisMendo 2D convolution is one of those things where I don't understand 1D convolution either so there's no hope for me there...sounds like an opportunity to learn both!
– Giuseppe
yesterday
1
If you need help with that let me know in the chat room. Convolution is a very useful operation. If you want to learn convolution start with 1D. The generalization to 2D is immediate
– Luis Mendo
yesterday
add a comment |
up vote
2
down vote
MATL, 21 17 10 bytes
8=t3Y6Z+>z
Try it online!
Thanks to Luis Mendo for help in chat, and for suggesting 2D convolution.
Explanation:
#implicit input, m
8= #equal to 8? matrix of 1 where m is 8, 0 otherwise
t #duplicate
3Y6 #push [1 1 1; 1 0 1; 1 1 1], "neighbor cells" for convolution
Z+ #2D convolution; each element is replaced by the number of neighbors that are 8
> #elementwise greater than -- matrix of 1s where an 8 is single, 0s otherwise
z #number of nonzero elements -- number of single eights
#implicit output
answered yesterday
Giuseppe
15.8k31051
You could save quite a few bytes using (2D-)convolution, if you are famiiar with the concept
– Luis Mendo
yesterday
1
@LuisMendo 2D convolution is one of those things where I don't understand 1D convolution either so there's no hope for me there...sounds like an opportunity to learn both!
– Giuseppe
yesterday
1
If you need help with that let me know in the chat room. Convolution is a very useful operation. If you want to learn convolution start with 1D. The generalization to 2D is immediate
– Luis Mendo
yesterday
add a comment |
up vote
2
down vote
up vote
2
down vote
MATL, 21 17 10 bytes
8=t3Y6Z+>z
Try it online!
Thanks to Luis Mendo for help in chat, and for suggesting 2D convolution.
Explanation:
#implicit input, m
8= #equal to 8? matrix of 1 where m is 8, 0 otherwise
t #duplicate
3Y6 #push [1 1 1; 1 0 1; 1 1 1], "neighbor cells" for convolution
Z+ #2D convolution; each element is replaced by the number of neighbors that are 8
> #elementwise greater than -- matrix of 1s where an 8 is single, 0s otherwise
z #number of nonzero elements -- number of single eights
#implicit output
answered yesterday
Giuseppe
15.8k31051
MATL, 21 17 10 bytes
8=t3Y6Z+>z
Try it online!
Thanks to Luis Mendo for help in chat, and for suggesting 2D convolution.
Explanation:
#implicit input, m
8= #equal to 8? matrix of 1 where m is 8, 0 otherwise
t #duplicate
3Y6 #push [1 1 1; 1 0 1; 1 1 1], "neighbor cells" for convolution
Z+ #2D convolution; each element is replaced by the number of neighbors that are 8
> #elementwise greater than -- matrix of 1s where an 8 is single, 0s otherwise
z #number of nonzero elements -- number of single eights
#implicit output
answered yesterday
Giuseppe
15.8k31051
edited yesterday
answered yesterday
Giuseppe
15.8k31051
answered yesterday
Giuseppe
15.8k31051
answered yesterday
Giuseppe
15.8k31051
15.8k31051
You could save quite a few bytes using (2D-)convolution, if you are famiiar with the concept
– Luis Mendo
yesterday
1
@LuisMendo 2D convolution is one of those things where I don't understand 1D convolution either so there's no hope for me there...sounds like an opportunity to learn both!
– Giuseppe
yesterday
1
If you need help with that let me know in the chat room. Convolution is a very useful operation. If you want to learn convolution start with 1D. The generalization to 2D is immediate
– Luis Mendo
yesterday
add a comment |
You could save quite a few bytes using (2D-)convolution, if you are famiiar with the concept
– Luis Mendo
yesterday
1
@LuisMendo 2D convolution is one of those things where I don't understand 1D convolution either so there's no hope for me there...sounds like an opportunity to learn both!
– Giuseppe
yesterday
1
If you need help with that let me know in the chat room. Convolution is a very useful operation. If you want to learn convolution start with 1D. The generalization to 2D is immediate
– Luis Mendo
yesterday
You could save quite a few bytes using (2D-)convolution, if you are famiiar with the concept
– Luis Mendo
yesterday
You could save quite a few bytes using (2D-)convolution, if you are famiiar with the concept
– Luis Mendo
yesterday
1
1
@LuisMendo 2D convolution is one of those things where I don't understand 1D convolution either so there's no hope for me there...sounds like an opportunity to learn both!
– Giuseppe
yesterday
@LuisMendo 2D convolution is one of those things where I don't understand 1D convolution either so there's no hope for me there...sounds like an opportunity to learn both!
– Giuseppe
yesterday
1
1
If you need help with that let me know in the chat room. Convolution is a very useful operation. If you want to learn convolution start with 1D. The generalization to 2D is immediate
– Luis Mendo
yesterday
If you need help with that let me know in the chat room. Convolution is a very useful operation. If you want to learn convolution start with 1D. The generalization to 2D is immediate
– Luis Mendo
yesterday
add a comment |
up vote
2
down vote
Jelly, 12 bytes
œẹ8ạṀ¥þ`’Ạ€S
Try it online!
How it works
œẹ8ạṀ¥þ`’Ạ€S Main link. Argument: M (matrix)
œẹ8 Find all multidimensional indices of 8, yielding an array A of pairs.
þ` Table self; for all pairs [i, j] and [k, l] in A, call the link to the
left. Return the results as a matrix.
ạ Absolute difference; yield [|i - k|, |j - l|].
Ṁ Take the maximum.
’ Decrement all the maxmima, mapping 1 to 0.
Ạ€ All each; yield 1 for each row that contains no zeroes.
S Take the sum.
answered yesterday
Dennis♦
184k32293728
add a comment |
up vote
2
down vote
Jelly, 12 bytes
œẹ8ạṀ¥þ`’Ạ€S
Try it online!
How it works
œẹ8ạṀ¥þ`’Ạ€S Main link. Argument: M (matrix)
œẹ8 Find all multidimensional indices of 8, yielding an array A of pairs.
þ` Table self; for all pairs [i, j] and [k, l] in A, call the link to the
left. Return the results as a matrix.
ạ Absolute difference; yield [|i - k|, |j - l|].
Ṁ Take the maximum.
’ Decrement all the maxmima, mapping 1 to 0.
Ạ€ All each; yield 1 for each row that contains no zeroes.
S Take the sum.
answered yesterday
Dennis♦
184k32293728
add a comment |
up vote
2
down vote
up vote
2
down vote
Jelly, 12 bytes
œẹ8ạṀ¥þ`’Ạ€S
Try it online!
How it works
œẹ8ạṀ¥þ`’Ạ€S Main link. Argument: M (matrix)
œẹ8 Find all multidimensional indices of 8, yielding an array A of pairs.
þ` Table self; for all pairs [i, j] and [k, l] in A, call the link to the
left. Return the results as a matrix.
ạ Absolute difference; yield [|i - k|, |j - l|].
Ṁ Take the maximum.
’ Decrement all the maxmima, mapping 1 to 0.
Ạ€ All each; yield 1 for each row that contains no zeroes.
S Take the sum.
answered yesterday
Dennis♦
184k32293728
Jelly, 12 bytes
œẹ8ạṀ¥þ`’Ạ€S
Try it online!
How it works
œẹ8ạṀ¥þ`’Ạ€S Main link. Argument: M (matrix)
œẹ8 Find all multidimensional indices of 8, yielding an array A of pairs.
þ` Table self; for all pairs [i, j] and [k, l] in A, call the link to the
left. Return the results as a matrix.
ạ Absolute difference; yield [|i - k|, |j - l|].
Ṁ Take the maximum.
’ Decrement all the maxmima, mapping 1 to 0.
Ạ€ All each; yield 1 for each row that contains no zeroes.
S Take the sum.
answered yesterday
Dennis♦
184k32293728
edited yesterday
answered yesterday
Dennis♦
184k32293728
answered yesterday
Dennis♦
184k32293728
answered yesterday
Dennis♦
184k32293728
184k32293728
add a comment |
add a comment |
up vote
1
down vote
JavaScript (ES6), 106 bytes
a=>a.map((r,y)=>r.map((v,x)=>k+=v==8&[...'12221000'].every((d,i,v)=>(a[y+~-d]||0)[x+~-v[i+2&7]]^8)),k=0)|k
Try it online!
Bitwise approach, 110 bytes
a=>a.map(r=>r.map(v=>x=x*2|v==8,x=k=0)|x).map((n,y,b)=>a[0].map((_,x)=>(n^1<<x|b[y-1]|b[y+1])*2>>x&7||k++))&&k
Try it online!
answered 2 days ago
Arnauld
68.4k584289
Bitwise approach fail on[[7]]
– l4m2
2 days ago
@lm42 Oh, thanks. Now fixed.
– Arnauld
2 days ago
add a comment |
up vote
1
down vote
JavaScript (ES6), 106 bytes
a=>a.map((r,y)=>r.map((v,x)=>k+=v==8&[...'12221000'].every((d,i,v)=>(a[y+~-d]||0)[x+~-v[i+2&7]]^8)),k=0)|k
Try it online!
Bitwise approach, 110 bytes
a=>a.map(r=>r.map(v=>x=x*2|v==8,x=k=0)|x).map((n,y,b)=>a[0].map((_,x)=>(n^1<<x|b[y-1]|b[y+1])*2>>x&7||k++))&&k
Try it online!
answered 2 days ago
Arnauld
68.4k584289
Bitwise approach fail on[[7]]
– l4m2
2 days ago
@lm42 Oh, thanks. Now fixed.
– Arnauld
2 days ago
add a comment |
up vote
1
down vote
up vote
1
down vote
JavaScript (ES6), 106 bytes
a=>a.map((r,y)=>r.map((v,x)=>k+=v==8&[...'12221000'].every((d,i,v)=>(a[y+~-d]||0)[x+~-v[i+2&7]]^8)),k=0)|k
Try it online!
Bitwise approach, 110 bytes
a=>a.map(r=>r.map(v=>x=x*2|v==8,x=k=0)|x).map((n,y,b)=>a[0].map((_,x)=>(n^1<<x|b[y-1]|b[y+1])*2>>x&7||k++))&&k
Try it online!
answered 2 days ago
Arnauld
68.4k584289
JavaScript (ES6), 106 bytes
a=>a.map((r,y)=>r.map((v,x)=>k+=v==8&[...'12221000'].every((d,i,v)=>(a[y+~-d]||0)[x+~-v[i+2&7]]^8)),k=0)|k
Try it online!
Bitwise approach, 110 bytes
a=>a.map(r=>r.map(v=>x=x*2|v==8,x=k=0)|x).map((n,y,b)=>a[0].map((_,x)=>(n^1<<x|b[y-1]|b[y+1])*2>>x&7||k++))&&k
Try it online!
answered 2 days ago
Arnauld
68.4k584289
edited 2 days ago
answered 2 days ago
Arnauld
68.4k584289
answered 2 days ago
Arnauld
68.4k584289
answered 2 days ago
Arnauld
68.4k584289
68.4k584289
Bitwise approach fail on[[7]]
– l4m2
2 days ago
@lm42 Oh, thanks. Now fixed.
– Arnauld
2 days ago
add a comment |
Bitwise approach fail on[[7]]
– l4m2
2 days ago
@lm42 Oh, thanks. Now fixed.
– Arnauld
2 days ago
Bitwise approach fail on
[[7]]– l4m2
2 days ago
Bitwise approach fail on
[[7]]– l4m2
2 days ago
@lm42 Oh, thanks. Now fixed.
– Arnauld
2 days ago
@lm42 Oh, thanks. Now fixed.
– Arnauld
2 days ago
add a comment |
up vote
1
down vote
Clojure, 227 198 bytes
(fn[t w h](let[c #(for[y(range %3 %4)x(range % %2)][x y])e #(= 8(get-in t(reverse %)0))m(fn[[o p]](count(filter e(c(dec o)(+ o 2)(dec p)(+ p 2)))))](count(filter #(= 1(m %))(filter e(c 0 w 0 h))))))
Ouch. Definitely not the shortest here by any means. 54 bytes of parenthesis is killer. I'm still relatively happy with it though.
-29 bytes by creating a helper function that generates a range since I was doing that twice, changing the reduce to a (count (filter setup, and getting rid of the threading macro after golfing.
(defn count-single-eights [td-array width height]
; Define three helper functions. One generates a list of coords for a given range of dimensions, another checks if an eight is
; at the given coord, and the other counts how many neighbors around a coord are an eight
(letfn [(coords [x-min x-max y-min y-max]
(for [y (range y-min y-max)
x (range x-min x-max)]
[x y]))
(eight? [[x y]] (= 8 (get-in td-array [y x] 0)))
(n-eights-around [[cx cy]]
(count (filter eight?
(coords (dec cx) (+ cx 2), (dec cy) (+ cy 2)))))]
; Gen a list of each coord of the matrix
(->> (coords 0 width, 0 height)
; Remove any coords that don't contain an eight
(filter eight?)
; Then count how many "neighborhoods" only contain 1 eight
(filter #(= 1 (n-eights-around %)))
(count))))
(mapv #(count-single-eights % (count (% 0)) (count %))
test-cases)
=> [3 0 0 2 2 1 3 1 4 3]
Where test-cases is an array holding all the "Python test cases"
Try it online!
answered yesterday
Carcigenicate
2,24911224
add a comment |
up vote
1
down vote
Clojure, 227 198 bytes
(fn[t w h](let[c #(for[y(range %3 %4)x(range % %2)][x y])e #(= 8(get-in t(reverse %)0))m(fn[[o p]](count(filter e(c(dec o)(+ o 2)(dec p)(+ p 2)))))](count(filter #(= 1(m %))(filter e(c 0 w 0 h))))))
Ouch. Definitely not the shortest here by any means. 54 bytes of parenthesis is killer. I'm still relatively happy with it though.
-29 bytes by creating a helper function that generates a range since I was doing that twice, changing the reduce to a (count (filter setup, and getting rid of the threading macro after golfing.
(defn count-single-eights [td-array width height]
; Define three helper functions. One generates a list of coords for a given range of dimensions, another checks if an eight is
; at the given coord, and the other counts how many neighbors around a coord are an eight
(letfn [(coords [x-min x-max y-min y-max]
(for [y (range y-min y-max)
x (range x-min x-max)]
[x y]))
(eight? [[x y]] (= 8 (get-in td-array [y x] 0)))
(n-eights-around [[cx cy]]
(count (filter eight?
(coords (dec cx) (+ cx 2), (dec cy) (+ cy 2)))))]
; Gen a list of each coord of the matrix
(->> (coords 0 width, 0 height)
; Remove any coords that don't contain an eight
(filter eight?)
; Then count how many "neighborhoods" only contain 1 eight
(filter #(= 1 (n-eights-around %)))
(count))))
(mapv #(count-single-eights % (count (% 0)) (count %))
test-cases)
=> [3 0 0 2 2 1 3 1 4 3]
Where test-cases is an array holding all the "Python test cases"
Try it online!
answered yesterday
Carcigenicate
2,24911224
add a comment |
up vote
1
down vote
up vote
1
down vote
Clojure, 227 198 bytes
(fn[t w h](let[c #(for[y(range %3 %4)x(range % %2)][x y])e #(= 8(get-in t(reverse %)0))m(fn[[o p]](count(filter e(c(dec o)(+ o 2)(dec p)(+ p 2)))))](count(filter #(= 1(m %))(filter e(c 0 w 0 h))))))
Ouch. Definitely not the shortest here by any means. 54 bytes of parenthesis is killer. I'm still relatively happy with it though.
-29 bytes by creating a helper function that generates a range since I was doing that twice, changing the reduce to a (count (filter setup, and getting rid of the threading macro after golfing.
(defn count-single-eights [td-array width height]
; Define three helper functions. One generates a list of coords for a given range of dimensions, another checks if an eight is
; at the given coord, and the other counts how many neighbors around a coord are an eight
(letfn [(coords [x-min x-max y-min y-max]
(for [y (range y-min y-max)
x (range x-min x-max)]
[x y]))
(eight? [[x y]] (= 8 (get-in td-array [y x] 0)))
(n-eights-around [[cx cy]]
(count (filter eight?
(coords (dec cx) (+ cx 2), (dec cy) (+ cy 2)))))]
; Gen a list of each coord of the matrix
(->> (coords 0 width, 0 height)
; Remove any coords that don't contain an eight
(filter eight?)
; Then count how many "neighborhoods" only contain 1 eight
(filter #(= 1 (n-eights-around %)))
(count))))
(mapv #(count-single-eights % (count (% 0)) (count %))
test-cases)
=> [3 0 0 2 2 1 3 1 4 3]
Where test-cases is an array holding all the "Python test cases"
Try it online!
answered yesterday
Carcigenicate
2,24911224
Clojure, 227 198 bytes
(fn[t w h](let[c #(for[y(range %3 %4)x(range % %2)][x y])e #(= 8(get-in t(reverse %)0))m(fn[[o p]](count(filter e(c(dec o)(+ o 2)(dec p)(+ p 2)))))](count(filter #(= 1(m %))(filter e(c 0 w 0 h))))))
Ouch. Definitely not the shortest here by any means. 54 bytes of parenthesis is killer. I'm still relatively happy with it though.
-29 bytes by creating a helper function that generates a range since I was doing that twice, changing the reduce to a (count (filter setup, and getting rid of the threading macro after golfing.
(defn count-single-eights [td-array width height]
; Define three helper functions. One generates a list of coords for a given range of dimensions, another checks if an eight is
; at the given coord, and the other counts how many neighbors around a coord are an eight
(letfn [(coords [x-min x-max y-min y-max]
(for [y (range y-min y-max)
x (range x-min x-max)]
[x y]))
(eight? [[x y]] (= 8 (get-in td-array [y x] 0)))
(n-eights-around [[cx cy]]
(count (filter eight?
(coords (dec cx) (+ cx 2), (dec cy) (+ cy 2)))))]
; Gen a list of each coord of the matrix
(->> (coords 0 width, 0 height)
; Remove any coords that don't contain an eight
(filter eight?)
; Then count how many "neighborhoods" only contain 1 eight
(filter #(= 1 (n-eights-around %)))
(count))))
(mapv #(count-single-eights % (count (% 0)) (count %))
test-cases)
=> [3 0 0 2 2 1 3 1 4 3]
Where test-cases is an array holding all the "Python test cases"
Try it online!
answered yesterday
Carcigenicate
2,24911224
edited yesterday
answered yesterday
Carcigenicate
2,24911224
answered yesterday
Carcigenicate
2,24911224
answered yesterday
Carcigenicate
2,24911224
2,24911224
add a comment |
add a comment |
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15
If you like it then you should have put a vote on it
– Luis Mendo
2 days ago
When I read "cells that equal 8", for a moment I thought you meant that a cell could be larger than a 1x1 chuck (NxN) of the grid. Should probably rephrase that to "cells that are 8" to clarify no math needed. =P
– Tezra
yesterday
@Tezra Edited. I find the new wording a little less natural, but I’m not a native speaker so I’ll trust your criterion
– Luis Mendo
yesterday