Find the Number With the Highest Sum of Neighbors

Question

The Challenge

Given a grid of numbers (10 <= N <= 99) Return number with the highest sum of the four numbers adjacent to it; that is the numbers above, below, right, and left of the number, but not itself.

1. The number itself does not count, only its four neighbors.
2. A number on the edge should be treated as though the missing number is a 0.
3. I will design the test to avoid ties.
4. Numbers will not repeat.
5. This is code-golf.

Example

Given

56 98 32 96
12 64 45 31
94 18 83 71

Return

18

A Real Test

Given

98 95 67 66 57 16 40 94 84 37
87 14 19 34 83 99 97 78 50 36
18 44 29 47 21 86 24 15 91 61
60 41 51 26 10 58 11 62 55 71
42 85 56 12 46 81 93 65 49 77
89 13 74 39 54 76 92 33 82 90
96 88 70 79 80 28 25 20 75 68
38 63 17 72 53 48 73 30 45 69
64 35 32 31 23 43 22 52 27 59

Return

13

Given

82 43 79 81 94 36 17 64 58
24 52 13 87 70 18 28 61 69
16 99 75 21 50 44 89 90 51
49 80 63 31 54 65 41 55 38
67 91 76 78 23 86 83 14 73
46 68 62 77 34 48 20 74 10
33 35 26 97 59 66 25 37 32
12 92 84 27 85 56 22 40 45
96 15 98 53 39 30 88 71 29
60 42 11 57 95 19 93 72 47

Return

15

Answer 1

MATL, 20 15 13 12 bytes

t1Y6Z+tuX>=)

Saved 5 bytes thanks to Emigna, 2 thanks to Giuseppe, and another thanks to Luis Mendo.
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Explanation

t1Y6Z+tuX>=)
t                  Duplicate the (implicit) input.
 1Y6               Push the array [0 1 0; 1 0 1; 0 1 0].
    Z+             Convolve with the input.
       uX>         Get the maximum unique element...
      t   =        ... and compare elementwise.
           )       Index into the input.

Answer 2

APL (Dyalog Unicode), 31 27 26 24 23 bytes^SBCS

-2 thanks to Cows quack. -1 thanks to ngn.

Anonymous tacit prefix function. Takes a matrix as argument. Assumes ⎕IO (Index Origin) to be 0, which is default on many systems.

{⊃⍒,(+/--/)⊢∘,⌺3 3⊢⍵}⊃,

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, ravel (flatten) the input

{…}⊃ pick an element from that according to the result of the following function:

 ⊢⍵ yield the argument (separates 3 3 from ⍵)

 …⌺3 3 apply the following function to each 3-by-3 neighbourhood:

  ⊢∘, ignore the edge-information in favour of the ravelled (flattened) neighbourhood

  (…) apply the following tacit function to those

   -/ the alternating sum (lit. right-associative minus reduction)

   +/- subtract that from the sum (this gives the sum of every other element)

 , ravel (flatten) that (the neighbourhood sums)

 ⍒ produce the indices which would sort that

 ⊃ pick the first (i.e. the index of the highest sum)

Answer 3

Jelly, 22 bytes

;€0ZṙØ+SṖ
,ZÇ€Z+$/ŒMœị

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Not having convolution built-ins like MATL and Dyalog Forgetting your language has convolution built-ins (thank you @dylnan) hurts, but we can do sort of okay without them, partially thanks to ŒM and œị. First, a helper function to compute neighbors in just one direction, which incidentally transposes the input:

;€0Z         Append a zero to each row, then transpose.
    ṙØ+S     Rotate by +1 and −1 (Ø+ = [1,-1]) and sum these.
        Ṗ    Pop the last row.

Visually, the computation is:

1 2 3   ;€0   1 2 3 0   Z   1 4 7   ṙØ+     2 5 8   0 0 0     S   2  5  8   Ṗ   2  5  8
4 5 6  ————→  4 5 6 0  ——→  2 5 8  ————→  [ 3 6 9 , 1 4 7 ]  ——→  4 10 16  ——→  4 10 16
7 8 9         7 8 9 0       3 6 9           0 0 0   2 5 8         2  5  8       2  5  8
                            0 0 0           1 4 7   3 6 9         4 10 16

Interpretation: cell (x, y) of this result is the sum of cell (y, x)'s horizontal neighbors. (For example, here we see that f(A)[2,3] = 16 = 7 + 9 = A[3,1] + A[3,3].)

Then, the main function:

,ZÇ€            Pair the input with its transpose and apply helper to both.
    Z+$/        Fold by Z+, i.e., turn [X,Y] into transpose(X)+Y.
                Now we've summed the horizontal and vertical neighbors for each cell.
        ŒM      Find the 2D index of the maximal value.
          œị    2D-index (into the original input).

Answer 4

Jelly, 18 bytes

5BæcµḊṖ)
ZÇZ+ÇŒMœị

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The helper function finds the neighbors of each element in each row. The main function does this to the rows and the columns then finds the element that has the maximum neighborhood sum.

5BæcµḊṖ)
5B           bin(5): 1,0,1
  æc         Convolve with [[1,2,9],[other rows]] (for example): [[1,2,10,2,9],...]
    µ        New chain.
       )     Apply this to each element:
     Ḋ       Remove the first element.
      Ṗ      Remove the last element.
             Gives [[2,10,2],...]

ZÇZ+ÇŒMœị   
Z            Zip the matrix
 Ç           Apply helper function
  Z          Zip again. Yields the sum of the vertical neighbors of each element.
   +         Add:
    Ç        The sum of each element's horizontal neighbors.
     ŒM      Find the multidimensional index of the maximal element.
       œị    Index back into the original matrix.

Answer 5

Wolfram Language (Mathematica), 58 bytes

Im@Last[Union@@ListConvolve[{a={0,1,0},{1,I,1},a},#,2,0]]&

Convolve the matrix with \$ \left( \begin{smallmatrix} 0 & 1 & 0 \\ 1 & i & 1 \\ 0 & 1 & 0 \end{smallmatrix} \right) \$, take the element with the largest real part, and take its imaginary part.

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Answer 6

Python 2, 127 bytes

def f(a):n=len(a[0]);b=sum(a,[])+[0]*n;print b[max(range(len(b)-n),key=lambda i:b[i-1]*(i%n>0)+b[i+1]*(i%n<n-1)+b[i-n]+b[i+n])]

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Answer 7

Stencil, 1 + 10 = 11 bytes (non-competing)

Command-line option: 1 compute 1 generation

y⊃⍨⊃⍒,
+/N

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y from the flattened original input
⊃⍨ pick
⊃ the first
⍒ in descending order
, of the flattened

+/ sums of
N the von neumanN neighbourhoods without selves

Answer 8

JavaScript (ES6), 94 bytes

a=>a.map(p=m=(r,y)=>p=r.map((v,x)=>(s=~r[x-1]+~p[x]+~(a[y+1]||0)[x]+~r[x+1])>m?v:(m=s,o=v)))|o

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How?

Instead of looking for the maximum of the sum of the 4 neighbors, we look for the minimum m of the sum s of their one-complements. This allows us to process undefined values just like zeros, because:

~undefined === -1
~0 === -1

The inner map() is written in such a way that it does not alter the content of the row r. Therefore, we can save its result in p in order to test top neighbors in the next iteration.

We use:

• ~r[x-1] for the left cell
• ~r[x+1] for the right cell
• ~p[x] for the top cell
• ~(a[y+1]||0)[x] for the bottom cell

Answer 9

K (ngn/k), 43 40 bytes

{(,/x)@*>,/(+g@+x)+(g:{(+':x)-|-':|x})x}

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Answer 10

Javascript ES6, 170 bytes

c=g=>{s=0;n=0;f=m=>m||[];for(i=0;i<g.length;i++)for(j=0;j<g[i].length;j++){s=~~f(g[i-1])[j]+~~f(g[i+1])[j]+~~f(g[i])[j-1]+~~f(g[i])[j+1];b=s>n?g[i][j]+!(n=s):b;}return b}

Answer 11

Java 8, 187 bytes

m->{int r=0,l=m.length,i=l,L,j,S=0,s;for(;i-->0;)for(L=j=m[i].length;j-->0;)if((s=(i<1?0:m[i-1][j])+(i<l-1?m[i+1][j]:0)+(j<1?0:m[i][j-1])+(j<L-1?m[i][j+1]:0))>S){S=s;r=m[i][j];}return r;}

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Explanation:

m->{                           // Method with integer-matrix parameter and integer return
  int r=0,                     //  Result-integer, starting at 0
      l=m.length,              //  Amount of rows
      i=l,                     //  Rows index integer
      L,                       //  Amount of columns
      j,                       //  Column index integer
      S=0,                     //  Largest sum of four cells
      s;                       //  Current sum of four cells
  for(;i-->0;)                 //  Loop over the rows
    for(L=j=m[i].length;j-->0;)//   Inner loop over the columns
      if((s=                   //    Set the current sum to: the sum of:
           (i<1?0:m[i-1][j])   //     Value of the cell to the left, or 0 if out of bounds
          +(i<l-1?m[i+1][j]:0) //     Value of the cell to the right, or 0 if out of bounds
          +(j<1?0:m[i][j-1])   //     Value of the cell down, or 0 if out of bounds
          +(j<L-1?m[i][j+1]:0))//     Value of the cell up, or 0 if out of bounds
         >S){                  //    If this current sum is larger than the largest sum:
        S=s;                   //     Replace the largest sum with this current sum
        r=m[i][j];}            //     And set the result to the current cell
  return r;}                   //  Return the result