Compute the Optimal Square Matrix

Question

The optimal matrix (for the rather narrow scope of this challenge) is obtained by "zipping" the elements from the corresponding rows and columns of a square matrix and getting the maximum of each pair.

For instance, given the following matrix:

4 5 6
1 7 2
7 3 0

You can combine it with its transpose to get: [[[4,5,6],[4,1,7]],[[1,7,2],[5,7,3]],[[7,3,0],[6,2,0]]]. If you zip each pair of lists, you obtain the following: [[(4,4),(5,1),(6,7)],[(1,5),(7,7),(2,3)],[(7,6),(3,2),(0,0)]]. The last step is to get the maximum of each pair to get the optimal matrix:

4 5 7
5 7 3
7 3 0

Your task is to output the optimal matrix of a square matrix given as input. The matrix will only contain integers. I/O can be done in any reasonable format. The shortest code in bytes (either in UTF-8 or in the language's custom encoding) wins!

Tests

[[172,29],[29,0]] -> [[172,29],[29,0]]
[[4,5,6],[1,7,2],[7,3,0]] -> [[4,5,7],[5,7,3],[7,3,0]]
[[1,2,3],[1,2,3],[1,2,3]] -> [[1,2,3],[2,2,3],[3,3,3]]
[[4,5,-6],[0,8,-12],[-2,2,4]] -> [[4,5,-2],[5,8,2],[-2,2,4]]

Answer 1

Jelly, 2 bytes

»Z

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How it works

»Z  Main link. Argument: M (integer matrix)

 Z  Zip the rows of M, transposing rows and columns.
»   Take the maxima of all corresponding integers.

Answer 2

Husk, 5 4 bytes

Whoop, never got to use ‡ before (or †):

S‡▲T

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Explanation

S  T -- apply the function to itself and itself transposed
 ‡▲  -- bi-vectorized maximum

Answer 3

Haskell, 40 bytes

z(z max)<*>foldr(z(:))e
e=[]:e
z=zipWith

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I would ungolf this as:

import Data.List
f m = zipWith (zipWith max) m (transpose m)

...which is so much more elegant.

Answer 4

Octave, 13 bytes

@(A)max(A,A')

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Answer 5

MATL, 6 bytes

t!2$X>

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Explanation:

t        % Duplicate the input.
!        % Transpose the duplicate.
2$X>     % Elementwise maximum of the two matrices.

Answer 6

APL (Dyalog Unicode), 3 bytes

Anonymous tacit prefix function.

⊢⌈⍉

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⊢ argument

⌈ ceiling'd with

⍉ transposed argument

Answer 7

JavaScript (ES6), 48 bytes

m=>m.map((r,y)=>r.map((v,x)=>v>(k=m[x][y])?v:k))

Test cases

let f =

m=>m.map((r,y)=>r.map((v,x)=>v>(k=m[x][y])?v:k))

console.log(JSON.stringify(f([[172,29],[29,0]]            ))) // -> [[172,29],[29,0]]
console.log(JSON.stringify(f([[4,5,6],[1,7,2],[7,3,0]]    ))) // -> [[4,5,7],[5,7,3],[7,3,0]]
console.log(JSON.stringify(f([[1,2,3],[1,2,3],[1,2,3]]    ))) // -> [[1,2,3],[2,2,3],[3,3,3]]
console.log(JSON.stringify(f([[4,5,-6],[0,8,-12],[-2,2,4]]))) // -> [[4,5,-2],[5,8,2],[-2,2,4]]

Answer 8

Japt, 12 10 8 bytes

Look, Ma, no transposing or zipping!

£XËwUgEY

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Answer 9

J, 4 bytes

Tacit prefix function.

>.|:

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>. ceiling [of the argument] with

|: the transposed argument

Answer 10

R, 23 bytes

function(A)pmax(A,t(A))

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This is equivalent to most other answers. However, R has two distinct max functions for the two common scenarios:

max and min return the maximum or minimum of all the values present in their arguments, as integer if all are logical or integer, as double if all are numeric, and character otherwise.

pmax and pmin take one or more vectors (or matrices) as arguments and return a single vector giving the ‘parallel’ maxima (or minima) of the vectors. The first element of the result is the maximum (minimum) of the first elements of all the arguments, the second element of the result is the maximum (minimum) of the second elements of all the arguments and so on. Shorter inputs (of non-zero length) are recycled if necessary.

Answer 11

CJam, 8 bytes

{_z..e>}

Anonymous block (function) that takes the input from the stack and replaces it by the output.

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Explanation

{      }    e# Define block
 _          e# Duplicate
  z         e# Zip
   .        e# Apply next operator to the two arrays, item by item
            e# (that is, to rows of the two matrices)
    .       e# Apply next operator to the two arrays, item by item
            e# (that is, to numbers of the two rows)
     e>     e# Maximum of two numbers

Answer 12

C (gcc), 79 77 bytes

• Saved two bytes thanks to Steadybox; only taking in one matrix dimension parameter as all matrices in this challenge are square.

j,i;f(A,n)int*A;{for(j=0;j<n*n;j++)printf("%d,",A[A[j]>A[i=j/n+j%n*n]?j:i]);}

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Takes a flat integer array A and the matrix dimension n (as the matrix has to be square) as input. Outputs a flat integer array string representation to stdout.

Answer 13

Clean, 58 bytes

import StdEnv,StdLib
@l=zipWith(zipWith max)(transpose l)l

I don't think this needs an explanation.

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Answer 14

Julia 0.6, 13 bytes

max. applies the function max elementwise to it's arugments.

a->max.(a,a')

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Answer 15

Python 2, 45 bytes

lambda k:[map(max,*c)for c in zip(k,zip(*k))]

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Thanks to totallyhuman for a few bytes saved.

Answer 16

05AB1E, 7 bytes

ø‚øεøεà

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Explanation

ø         # transpose input matrix
 ‚        # pair with original matrix
  ø       # zip together
   ε      # apply on each sublist ([[row],[transposed row]])
    ø     # zip
     ε    # apply on each sublist (pair of elements)
      à   # extract greatest element

Answer 17

Jelly, 7 bytes

żZZṀ€$€

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Answer 18

Mathematica, 30 bytes

-8 bytes thanks to Jonathan Frech.

Map@Max/@t/@t@{#,t@#}&
t=#&

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Answer 19

Pari/GP, 21 bytes

m->(m+m~+abs(m-m~))/2

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Answer 20

Wolfram Language (Mathematica), 23 bytes

A port of my Pari/GP answer.

(#+#+Abs[#-#])/2&

 is \[Transpose].

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