# All the single eights

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


## Test cases

1. 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

2. Input

8 8
2 3


Output: 0

3. Input:

5 3 4
2 5 2


Output: 0

4. Input:

5 8 3 8


Output: 2

5. Input:

8
0
8


Output: 2.

6. Input:

4 2 8 5
2 6 1 8
8 5 5 8


Output: 1

7. 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.

8. Input:

8


Output: 1

9. 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.

10. 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 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]]
[, , ]
[[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, 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

• If you like it then you should have put a vote on it – Luis Mendo Nov 11 '18 at 19:52
• 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 Nov 12 '18 at 17:46
• @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 Nov 12 '18 at 18:57

# MATL, 2117 10 bytes

8=t3Y6Z+>z


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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

• You could save quite a few bytes using (2D-)convolution, if you are famiiar with the concept – Luis Mendo Nov 12 '18 at 17:31
• @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 Nov 12 '18 at 17:33
• 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 Nov 12 '18 at 17:39

# R, 117 63 59 bytes

function(m)sum(colSums(as.matrix(dist(which(m==8,T)))<2)<2)


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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.

• it's a shame numerical imprecision doesn't allow colSums()^2<=2 to work. – Giuseppe Nov 12 '18 at 22:28
• @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 Nov 13 '18 at 15:33

# APL (Dyalog Classic), 2928 25 bytes

≢∘⍸16=2⊥¨3,⌿3,/8=(⍉0,⌽)⍣4


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• Note: 0 index origin is not even needed. – Zacharý Nov 12 '18 at 16:36
• @Zacharý I always use it as a default, to avoid surprises. – ngn Nov 12 '18 at 20:03
• Ah, so like others with 1 (except not explicitly set). That makes sense. – Zacharý Nov 12 '18 at 20:05
• Surprised this doesn't use Stencil. Is there something that makes stencil inconvenient here? – lirtosiast Nov 20 '18 at 7:49
• @lirtosiast it's just longer with it :) – ngn Nov 20 '18 at 9:39

# Jelly, 18 15 bytes

8=µ+Ż+ḊZµ⁺ỊṖḋµS


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### 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


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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=           Convert 8 to 1, anything else to 0 (*A)
µÇÇ        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


# 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

# J,43, 40 37 bytes

-3 bytes thanks to Bubbler

1#.1#.3 3(16=8#.@:=,);._3(0|:@,|.)^:4


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## 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)

• 37 bytes using @: and moving |.. Note that @ in place of @: doesn't work. – Bubbler Nov 12 '18 at 10:26
• @Bubbler Thank you! – Galen Ivanov Nov 12 '18 at 11:07
• 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 Nov 12 '18 at 15:00
• @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 Nov 12 '18 at 15:16
• @Jonah Explanation added – Galen Ivanov Nov 12 '18 at 18:26

# 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.

# 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)
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)


# J, 42 bytes

[:+/@,8=]+[:+/(<:3 3#:4-.~i.9)|.!.0(_*8&=)


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# 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

• You have forgotten to remove a space between ~ and > – Galen Ivanov Nov 12 '18 at 11:12
• @GalenIvanov Fixed now. Thank you. – Jonah Nov 12 '18 at 13:58

# Java 8, 181157 156 bytes

(M,R,C)->{int z=0,c,f,t;for(;R-->0;)for(c=C;c-->0;z+=f>1?0:f)for(f=0,t=9;M[R][c]==8&t-->0;)try{f+=M[R+t/3-1][c+t%3-1]==8?1:0;}catch(Exception e){}return z;}


-24 bytes thanks to @OlivierGrégoire.

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 larger than 1:
0             //       Leave the result-counter the same by adding 0
:              //      Else:
f)            //       Add the flag-integer (either 0 or 1)
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[R+t/3-1]     //       If t is 0, 1, or 2: Look at the previous row
//       Else-if t is 6, 7, or 8: Look at the next row
//       Else (t is 3, 4, or 5): Look at the current row
[c+t%3-1]     //       If t is 0, 3, or 6: Look at the previous column
//       Else-if t is 2, 5, or 8: Look at the next column
//       Else (t is 1, 4, or 7): Look at the current column
==8?          //       And 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


# 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)))


Try it online!

• Seems shorter if take length from args – l4m2 Nov 12 '18 at 2:54

# Powershell, 121 bytes

param($a)(($b='='*4*($l=($a|% Le*)))+($a|%{"!$_!"})+$b|sls "(?<=[^8]{3}.{$l}[^8])8(?=[^8].{$l}[^8]{3})" -a|% m*).Count  Less golfed test script: $f = {

param($a)$length=($a|% Length)$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.

# 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.


# 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.map((_,x)=>(n^1<<x|b[y-1]|b[y+1])*2>>x&7||k++))&&k


Try it online!

• Bitwise approach fail on [] – l4m2 Nov 12 '18 at 2:52
• @lm42 Oh, thanks. Now fixed. – Arnauld Nov 12 '18 at 9:29

# 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!