23
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Challenge:

Given a square input matrix A, pad the matrix with one row and one column on all four sides.

  • The value of each element in the top and bottom row should be the sum of the elements in each corresponding column.
  • The value of each element in the left and right column should be the sum of the elements in each corresponding row.
  • The value of the elements in the top left, and bottom right corner should be the sum of the elements on the diagonal
  • The value of the elements in the top right, and bottom left corner should be the sum of the elements in the anti-diagonal.

Example:

A = 
1   5   3
3   2   4
2   5   5

Output:
 8    6   12   12    7
 9    1    5    3    9
 9    3    2    4    9
12    2    5    5   12
 7    6   12   12    8

Explanation:

The top left and bottom right elements are the sum of the diagonal 1+2+5=8. The top right and bottom left elements are the sum of the anti-diagonal 2+2+3=7.

The top and bottom row (except the corners) are the sum of each of the columns in A: 1+3+2=6, 5+2+5=12 and 3+4+5=12. Similarly, the left and right column (except the corners) are the sum of each of the rows of A: 1+5+3=9, 3+2+4=9 and 2+5+5=12.

Input:

  • A non-empty square matrix, with non-negative integers.
  • Optional format

Output:

  • The matrix padded as explained above
  • Optional format, but it must be the same as the input format

Test cases:

Use the submissions in this challenge if you want to convert the input format to a more suitable one (for instance [[1, 5],[0, 2]]).

0
----------------
0 0 0
0 0 0
0 0 0

1 5
0 2
----------------
3 1 7 5
6 1 5 6
2 0 2 2
5 1 7 3

17   24    1    8   15
23    5    7   14   16
 4    6   13   20   22
10   12   19   21    3
11   18   25    2    9 
----------------
65   65   65   65   65   65   65
65   17   24    1    8   15   65
65   23    5    7   14   16   65
65    4    6   13   20   22   65
65   10   12   19   21    3   65
65   11   18   25    2    9   65
65   65   65   65   65   65   65

15    1    2   12
 4   10    9    7
 8    6    5   11
 3   13   14    0
----------------
30   30   30   30   30   30
30   15    1    2   12   30
30    4   10    9    7   30
30    8    6    5   11   30
30    3   13   14    0   30
30   30   30   30   30   30

This is , so the shortest solution in each language wins. Explanations are highly encouraged.

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  • 2
    \$\begingroup\$ Is that to check magic squares? \$\endgroup\$ – mdahmoune Jul 4 '17 at 12:19
  • \$\begingroup\$ Just checking is quite a bit easier, but it's indeed easy to see if a square is magic this way, yes :-) \$\endgroup\$ – Stewie Griffin Jul 4 '17 at 12:33

12 Answers 12

5
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Octave, 64 bytes

Thanks to Tom Carpenter for both saving 4 bytes, and correcting an error I had in the original code!

@(a)[b=(t=@trace)(a),c=sum(a),d=t(flip(a));z=sum(a,2),a,z;d,c,b]

Try it online!

Explanation:

@(a)                 % Anonymous function that takes the matrix 'a' as input
 [ ... ]             % Concatenate everything inside to a single matrix
  b=(t=@trace)(a),   % Clever trick by Tom Carpenter. Save a function handle 
                     % for 't=trace', and call it with 'a' as input
                     % Save the result in the variable 'b'
  c=sum(a)           % Sum of all columns of 'a'
  d=t(flip(a));      % Save the trace of 'a' flipped as a variable 'd', while 
                     % concatenating [b,c,d] horizontally at the same time, creating the 
                     % first row of the output
  z=sum(a,2)         % The sum of each row of the input, and store it in a variable 'z'
  ,a,z;              % Concatenate it with 'a' and 'z' again, to create the middle part of the output
 d,c,b]              % Add [d,c,b] to complete the bottom row

Note, I wrote this long after I posted the challenge.

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4
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Jelly, 27 bytes

,UŒDḢ$S$€,Ṛ$j€SW€jSẋ¥€2$j"$

Try it online!

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  • 2
    \$\begingroup\$ I want to know how this works, please. \$\endgroup\$ – Comrade SparklePony Jul 4 '17 at 9:51
  • \$\begingroup\$ @ComradeSparklePony I don't know if I have time to write an explanation right now, sorry. \$\endgroup\$ – Erik the Outgolfer Jul 4 '17 at 9:53
  • \$\begingroup\$ It's fine, don't worry. \$\endgroup\$ – Comrade SparklePony Jul 4 '17 at 9:55
4
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MATL, 27 26 bytes

,!tXswyv]GXds5L(PGPXds5L(P

Try it online! Or verify all test cases.

Explanation

,        % Do the following twice
  !      %   Tranpose. Takes input implititly in the first iteration
  t      %   Duplicate
  Xs     %   Row vector with the sum of each column
  wy     %   Push a copy to the bottom of the stack
  v      %   Concatenate stack vertically. This attaches the sum of
         %   each row (first iteration) and column (second), leaving 
         %   the matrix with the correct orientation (transposed twice)
]        % End
G        % Push input again
Xds      % Column vector with the diagonal of the matrix. Sum of vector
5L(      % Write that into first and last entries of the result matrix
         % matrix; that is, its upper-left and lower-right corners
P        % Flip result matrix vertically
GP       % Push input matrix vertically flipped
Xds      % Diagonal, sum. Since the input has been vertically flipped,
         % this gives the sum of the anti-diagonal of the input.
5L(      % Write that into the upper-left and lower-right corners of
         % the verticallly flipped version of the result matrix
P        % Flip vertically again, to restore initial orientation
         % Implicitly display
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  • \$\begingroup\$ Of course MATL is designed to work with matrices unlike Jelly. >_> \$\endgroup\$ – Erik the Outgolfer Jul 4 '17 at 11:33
  • \$\begingroup\$ @EriktheOutgolfer But your answer has more euros! \$\endgroup\$ – Luis Mendo Jul 4 '17 at 11:38
  • 3
    \$\begingroup\$ Yeah it has euros dollars and yens...unfortunately that's not the winning criterion here. D: \$\endgroup\$ – Erik the Outgolfer Jul 4 '17 at 11:43
3
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APL (Dyalog), 37 bytes

(d,+⌿,d∘⌽)⍪(+/,⊢,+/)⍪d∘⌽,+⌿,d←+/1 1∘⍉

Try it online!

1 1∘⍉ diagonal (lit. collapse both axes into one)

d← store that function as d and apply it to the argument

+⌿ prepend the column sums

d∘⌽, prepend d applied to the reversed argument

()⍪ stack the following on top:

+/,⊢,+/ row sums, the unmodified argument, row sums

()⍪ stack the following on top:

d,+⌿,d∘⌽ applied to the argument, column sums, d applied to the reversed argument

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3
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Jelly, 26 bytes

ŒDµḊṖѵ€1¦ŒḌU
S;;S
Ç€Zµ⁺ÑÑ

Try it online!

Looks surprisingly different from Erik's solution.

I finally managed to understand how ¦ works (by debugging through Jelly's code, lol). Too bad it requires a to work with Ç in my case.

Explanation

The code uses three links. The first helper link pads a vector with its sum on both ends, the second helper link fixes two corners of the matrix and the main link calls these appropriately.

Ç€Zµ⁺ÑÑ    Main link. Argument: M (matrix)
Ç            Call the first helper link (pad row with sums)...
 €           ...on each row of the matrix.
  Z          Transpose, so that the second invocation uses the columns.
   µ         Begin a new monadic chain.
    ⁺        Repeat the previous chain (everything up to here).
     ÑÑ      Call the second helper link twice on the whole matrix.

S;;S    First helper link. Argument: v (1-dimensional list)
S         Sum the argument list.
 ;        Append the argument list to the sum.
  ;       Append...
   S      ...the sum of the argument list.

ŒDµḊṖѵ€1¦ŒḌU    Second helper link. Argument: M (matrix)
ŒD                 Get the diagonals of the matrix, starting with the main diagonal.
  µ                Begin a new monadic chain.
      µ€           Perform the following actions on each diagonal...
        1¦         ...and keep the result for the first item (main diagonal):
   Ḋ                 Remove the first item (incorrect top corner).
    Ṗ                Remove the last item (incorrect bottom corner).
     Ñ               Call the first helper link on the diagonal to pad it with its sum.
          ŒḌ       Convert the diagonals back to the matrix.
            U      Reverse each row, so that consecutive calls fix the other corners.
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3
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Python 3, 155 bytes

This is the suggestion of @LeakyNun, which saves 54 bytes. I then golfed it myself a bit.

def f(m):l=len(m);r=range(l);s=sum;b=[s(m[i][i]for i in r)];c=[s(m[i][l+~i]for i in r)];d=[*map(s,zip(*m))];return[b+d+c,*[[s(a),*a,s(a)]for a in m],c+d+b]

Try it online!

Initial solution - Python 3, 216 bytes

def f(m):l=len(m);r,s=range(l),sum;a,b,c,d=s(m[i][i]for i in r),s(m[i][l-i-1]for i in r),[s(m[i][j]for j in r)for i in r],[s(m[i][j]for i in r)for j in r];print([[a]+d+[b]]+[[c[i]]+m[i]+[c[i]]for i in r]+[[b]+d+[a]])

Try it online!

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  • \$\begingroup\$ 162 bytes \$\endgroup\$ – Leaky Nun Jul 4 '17 at 12:46
  • \$\begingroup\$ @LeakyNun Thanks. Was just updating with ~190 bytes, this is much shorter :P \$\endgroup\$ – Mr. Xcoder Jul 4 '17 at 12:52
2
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Python 2, 268 250 184 174 bytes

10 thanks to Stewie Griffin

from numpy import *
a,c,v,s=sum,trace,vstack,matrix(input())
l,r,d,e=a(s,0),a(s,1),c(s),c(fliplr(s))
print hstack((v(([[d]],r,[[e]])),v((l,s,l)),v(([[e]],r,[[d]])))).tolist()

Try it online!

Some explanations The input is uploaded as a matrix. First the code computes the sum of each column and each row using numpy.sum. Then it computes the sum of the diagonal by numpy.trace. After this it obtains the other diagonal by making a left-right flip on the matrix. Finally, it uses numpy.vstack and numpy.hstack to glue the pieces together.

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  • \$\begingroup\$ @StewieGriffin Ok, I just updated the code :) \$\endgroup\$ – mdahmoune Jul 4 '17 at 12:02
  • 1
    \$\begingroup\$ I believe this works for 174 tio.run/… \$\endgroup\$ – Stewie Griffin Jul 4 '17 at 12:19
2
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R, 129 bytes

pryr::f(t(matrix(c(d<-sum(diag(m)),c<-colSums(m),a<-sum(diag(m[(n<-nrow(m)):1,])),t(matrix(c(r<-rowSums(m),m,r),n)),a,c,d),n+2)))

An anonymous function that takes a square matrix as input. I'll post an explanation if there's interest.

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2
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PHP, 211 bytes

<?foreach($_GET as$l=>$r){$y=0;foreach($r as$k=>$c){$y+=$c;$x[$k]+=$c;$l-$k?:$d+=$c;($z=count($_GET))-1-$k-$l?:$h+=$c;}$o[]=[-1=>$y]+$r+[$z=>$y];}$o[]=[-1=>$h]+$x+[$z=>$d];print_r([-1=>[-1=>$d]+$x+[$z=>$h]]+$o);

Try it online!

Expanded

foreach($_GET as$l=>$r){
  $y=0; # sum for a row
  foreach($r as$k=>$c){
    $y+=$c; # add to sum for a row
    $x[$k]+=$c; # add to sum for a column and store in array
    $l-$k?:$d+=$c; # make the diagonal sum left to right
    ($z=count($_GET))-1-$k-$l?:$h+=$c; # make the diagonal sum right to left
  }
  $o[]=[-1=>$y]+$r+[$z=>$y]; # add to result array the actual row with sum of both sides
}
$o[]=[-1=>$h]+$x+[$z=>$d]; # add to result array the last array
print_r([-1=>[-1=>$d]+$x+[$z=>$h]]+$o); #output after adding the first array to the result array
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2
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Python 3, 125 bytes

from numpy import*
f=lambda m,t=trace,s=sum:c_[r_[t(m),s(m,1),t(m[::-1])],c_[s(m,0),m.T,s(m,0)].T,r_[t(m[::-1]),s(m,1),t(m)]]

Try it online!

Slightly ungolfed:

import numpy as np

def f_expanded(m):
    return np.c_[np.r_[np.trace(m), np.sum(m, 1), np.trace(m[::-1])],
                 np.c_[np.sum(m, 0), m.T, np.sum(m, 0)].T,
                 np.r_[np.trace(m[::-1]), np.sum(m, 1), np.trace(m)]]

This takes input formatted as a numpy array, then uses the np.c_ and np.r_ indexing tools to build a new array in one go. np.trace and np.sumare used to calculate the sums along the diagonals and everywhere else, respectively. T is used to take the transpose before and after concatenating the sums because it's shorter than making all the arrays 2-dimensional and using np.r_. m[::-1] saves bytes when compared to rot90(m) or fliplr(m) for finding the trace for the second diagonal.

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  • \$\begingroup\$ Nice answer! Welcome to the site :) \$\endgroup\$ – DJMcMayhem Jul 6 '17 at 23:08
1
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JavaScript (ES6), 170 bytes

(a,m=g=>a.map((_,i)=>g(i)),s=x=>eval(x.join`+`))=>[[d=s(m(i=>a[i][i])),...c=m(i=>s(m(j=>a[j][i]))),g=s(m(i=>a[i][a.length-i-1]))],...a.map(b=>[r=s(b),...b,r]),[g,...c,d]]

Input and output is a 2D array of numbers.

Explained

(a,                             // input matrix: a
    m=g=>a.map((_,i)=>g(i)),    // helper func m: map by index
    s=x=>eval(x.join`+`)        // helper func s: array sum
) =>
[
    [
        d = s(m(i=>a[i][i])),           // diagonal sum: d
        ...c=m(i=>s(m(j=>a[j][i]))),    // column sums: c
        g = s(m(i=>a[i][a.length-i-1])) // antidiagonal sum: g
    ],
    ...a.map(b=>[r = s(b), ...b, r]),   // all rows with row sums on each end
    [g, ...c, d]                        // same as top row, with corners flipped
]

Test Snippet

Input/output has been formatted with newlines and tabs.

f=
(a,m=g=>a.map((_,i)=>g(i)),s=x=>eval(x.join`+`))=>[[d=s(m(i=>a[i][i])),...c=m(i=>s(m(j=>a[j][i]))),g=s(m(i=>a[i][a.length-i-1]))],...a.map(b=>[r=s(b),...b,r]),[g,...c,d]]

let tests=[[[0]],[[1,5],[0,2]],[[17,24,1,8,15],[23,5,7,14,16],[4,6,13,20,22],[10,12,19,21,3],[11,18,25,2,9]],[[15,1,2,12],[4,10,9,7],[8,6,5,11],[3,13,14,0]]];
<select id=S oninput="I.value=S.selectedIndex?tests[S.value-1].map(s=>s.join`\t`).join`\n`:''"><option>Tests<option>1<option>2<option>3<option>4</select> <button onclick="O.innerHTML=I.value.trim()?f(I.value.split`\n`.map(s=>s.trim().split(/\s+/g))).map(s=>s.join`\t`).join`\n`:''">Run</button><br><textarea rows=6 cols=50 id=I></textarea><pre id=O>

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0
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LOGO, 198 bytes

to g :v[:h reduce "+ :v]
op(se :h :v :h)
end
to f :s[:a reduce "+ map[item # ?]:s][:b reduce "+ map[item # reverse ?]:s][:c apply "map se "sum :s]
op `[[,:a ,@:c ,:b],@[map "g :s][,:b ,@:c ,:a]]
end

The function f takes in a matrix as a 2D list, then output the resulting matrix. g is helper function.

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