# Index sum and strip my matrix

## Index sum and strip my matrix

Given a matrix/2d array in your preferable language

Input:

• The matrix will always have an odd length
• The matrix will always be perfectly square
• The matrix values can be any integer in your language (positive or negative)

Example:

1  2  3  4  5  6  7
2  3  4  5  6  7  8
3  4  50 6  7  8  9
4  5  6 100 8  9  10
5  6  7  8 -9  10 11
6  7  8  9  10 11 12
7  8 900 10 11 12 0


Definitions:

• The "central number" is defined as the number that has the same amount of numbers to the left,right,up and down

In this case its middlemost 100

• The "outer shell" is the collection of numbers which their x and y index is or 0 or the matrix size

1  2  3  4  5  6  7
2                 8
3                 9
4                 10
5                 11
6                 12
7  8 900 10 11 12 0


Add to the central number the sum of each row and column after multiplying the values in each by their 1-based index

A single row for example

4  5  6  7  8


for each number

number * index + number * index.....

4*1 + 5*2 + 6*3 + 7*4 + 8*5 => 100


example:

 2 -3 -9  4  7  1  5  => 61
-2  0 -2 -7 -7 -7 -4  => -141
6 -3 -2 -2 -3  2  1  => -10
8 -8  4  1 -8  2  0  => -20
-5  6  7 -1  8  4  8  => 144
1  5  7  8  7 -9 -5  => 10
7  7 -2  2 -7 -8  0  => -60
|
78 65 60 45 -15 -89 10   => 154
|
=> -16

• For all rows and columns you combine these values..
• Now you sum these too => 154-16 = 138
• You add that number to the "central number" and remove the "outer shell" of the matrix

 0 -2 -7 -7 -7     => -88
-3 -2 -2 -3  2     => -15
-8  4 1+138 -8  2  => 395
6  7 -1  8  4     => 69
5  7  8  7 -9     => 26

19 69 442 30 -26


do this untill you end up with a single number

-2 -2 -3     => -15
4  1060 -8  => 2100
7 -1  8     => 29

27 2115 5

• Remove the "outer shell" and get 5321
• Now we have a single number left

this is the output!

test cases:

-6

-6


-7 -1  8
-4 -6  7
-3 -6  6

2


 6  7 -2  5  1
-2  6 -4 -2  3
-1 -4  0 -2 -7
0  1  4 -4  8
-8 -6 -5  0  2

-365


 8  3  5  6  6 -7  5
6  2  4 -2 -1  8  3
2  1 -5  3  8  2 -3
3 -1  0  7 -6  7 -5
0 -8 -4 -9 -4  2 -8
8 -9 -3  5  7  8  5
8 -1  4  5  1 -4  8

17611


-9 -7  2  1  1 -2  3 -7 -3  6  7  1  0
-7 -8 -9 -2  7 -2  5  4  7 -7  8 -9  8
-4  4 -1  0  1  5 -3  7  1 -2 -9  4  8
4  8  1 -1  0  7  4  6 -9  3 -9  3 -9
-6 -8 -4 -8 -9  2  1  1 -8  8  2  6 -4
-8 -5  1  1  2 -9  3  7  2  5 -6 -1  2
-8 -5 -7 -4 -9 -2  5  0  2 -4  2  0 -2
-3 -6 -3  2 -9  8  1 -5  5  0 -4 -1 -9
-9 -9 -8  0 -5 -7  1 -2  1 -4 -1  5  7
-6 -9  4 -2  8  7 -9 -5  3 -1  1  8  4
-6  6 -3 -4  3  5  6  8 -2  5 -1 -7 -9
-1  7 -9  4  6  7  6 -8  5  1  0 -3  0
-3 -2  5 -4  0  0  0 -1  7  4 -9 -4  2

-28473770


## This is a codegolf challenge so the program with the lowest bytecount wins

• you are correct, thats a typo Jun 23, 2016 at 22:33
• why would negative numbers be an issue? I dont think the challenge should adjust for esolangs but maybe the other way around is more appropriate Jun 23, 2016 at 23:09
• @LuisMendo I think it's not a problem, the rule "The matrix values can be any integer in your language" means to me that if your language doesn't have negative numbers, it shouldn't support them. Jun 24, 2016 at 7:53
• actually thats correct. but then the test cases wont work properly Jun 24, 2016 at 8:08
• "I dont think the challenge should adjust for esolangs but maybe the other way around is more appropriate" that should be engraved in stone Jun 24, 2016 at 8:43

# MATL, 36 34 bytes

tnq?t&+stn:*sytn2/)+ 7M(6Lt3$)tnq  Input is a 2D array with ; as row separator ### Explanation tnq % Take input. Duplicate, get number of elements, subtract 1 ? % If greater than 0  % Do...while t % Duplicate &+ % Sum matrix with its transpose s % Sum each column. Gives a row vector tn: % Vector [1 2 ...] with the same size * % Multiply element-wise s % Sum of vector. This will be added to center entry of the matrix y % Duplicate matrix tn2/ % Duplicate, get half its number of elements. Gives non-integer value ) % Get center entry of the matrix, using linear index with implicit rounding + % Add center entry to sum of previous vector 7M % Push index of center entry again ( % Assgined new value to center of the matrix 6Lt % Array [2 j1-1], twice. This will be used to remove shell 3$)   %     Apply row and col indices to remove outer shell of the matrix
tnq   %     Duplicate, number of elements, subtract 1. Falsy if matrix has 1 entry
%   End do...while implicitly. The loop is exited when matrix has 1 entry
% End if implicitly
% Display stack implicitly


# Python 2.7, 229 bytes

This is my first attempt at something like this, so hopefully I followed all the rules with this submission. This is just a function which takes in a list of lists as its parameter. I feel like the sums and list comprehension could probably be shortened a little bit, but it was too hard for me. :D

def r(M):
t=len(M)
if t==1:return M[0][0]
M[t/2][t/2]+=sum(a*b for k in [[l[x] for l in M]for x in range(0,t)]for a,b in enumerate(k,1))+sum([i*j for l in M for i,j in enumerate(l,1)])
return r([p[+1:-1]for p in M[1:-1]])


Thx to Easterly Irk for helping me shave off a few bytes.

• You can remove a couple spaces between operators (...) + sum([i*j... -> ...)+sum([i*j...), but overall, great first post!!!! Jun 24, 2016 at 19:04
• oooh missed that. Thanks! Jun 24, 2016 at 19:21
• Also, ...]for ... works. You can remove at least 2 space like that. (end of list hits the for loop) Jun 24, 2016 at 19:25

## C#, 257 bytes

here is a non esolang answer

void f(int[][]p){while(p.Length>1){int a=p.Length;int r=0;for(int i=0;i<a;i++)for(int j=0;j<a;j++)r+=(i+j+2)*p[i][j];p[a/2][a/2]+=r;p=p.Where((i,n)=>n>0&&n<p.Length-1).Select(k=>k.Where((i,n)=>n>0&&n<p.Length-1).ToArray()).ToArray();}Console.Write(p[0][0]);


ungolfed:

void f(int[][]p)
{
while (p.Length>1)
{
int a=p.Length;
int r=0; //integer for number to add to middle
for (int i = 0; i < a; i++)
for (int j = 0; j < a; j++)
r +=(i+j+2)*p[i][j]; //add each element to counter according to their 1 based index
p[a / 2][a / 2] += r; //add counter to middle
p = p.Where((i, n) => n > 0 && n < p.Length - 1).Select(k => k.Where((i, n) => n > 0 && n < p.Length - 1).ToArray()).ToArray(); //strip outer shell from array
}
Console.Write(p[0][0]); //print last and only value in array
}

• Hey now, J isn't an esolang. Jun 24, 2016 at 8:26
• This doesn't compile if you don't include using System.Linq and using System. I'm not sure if it's required by the rules though. Jun 24, 2016 at 12:11
• its not a full program, its only a function so its ok as far as i know. i mean, would i also need to include the App.config and all the bytes in the properties and makefile? no Jun 24, 2016 at 12:13
• @downrep_nation It's just weird, since I've seen some people include them in the source when it has only been a function and they've included the bytes on the score. Jun 24, 2016 at 12:35
• Now when I think about it, I'm on the line that you should import atleast System.Linq. Other languages that require importing in order to use certain features go through the same process, so I think it's unfair to assume that every module is loaded to memory in C#. Jun 24, 2016 at 15:16

# J, 66 bytes

([:}:@}."1@}:@}.]+(i.@,~=](]+*)<.@-:)@#*[:+/^:2#\*]+|:)^:(<.@-:@#)


Straight-forward approach based on the process described in the challenge.

[:+/^:2#\*]+|: gets the sum. ]+(i.@,~=](]+*)<.@-:)@#* is a particularly ugly way to increment the center by the sum. [:}:@}."1@}:@}. removes the outer shell. There probably is a better way to do this.

## Usage

   f =: ([:}:@}."1@}:@}.]+(i.@,~=](]+*)<.@-:)@#*[:+/^:2#\*]+|:)^:(<.@-:@#)
f _6
_6
f _7 _1 8 , _4 _6 7 ,: _3 _6 6
2
f 6 7 _2 5 1 , _2 6 _4 _2 3 , _1 _4 0 _2 _7 , 0 1 4 _4 8 ,: _8 _6 _5 0 2
_365
f 8 3 5 6 6 _7 5 , 6 2 4 _2 _1 8 3 , 2 1 _5 3 8 2 _3 , 3 _1 0 7 _6 7 _5 , 0 _8 _4 _9 _4 2 _8 ,8 _9 _3 5 7 8 5 ,: 8 _1 4 5 1 _4 8
17611
f (13 13 $_9 _7 2 1 1 _2 3 _7 _3 6 7 1 0 _7 _8 _9 _2 7 _2 5 4 7 _7 8 _9 8 _4 4 _1 0 1 5 _3 7 1 _2 _9 4 8 4 8 1 _1 0 7 4 6 _9 3 _9 3 _9 _6 _8 _4 _8 _9 2 1 1 _8 8 2 6 _4 _8 _5 1 1 2 _9 3 7 2 5 _6 _1 2 _8 _5 _7 _4 _9 _2 5 0 2 _4 2 0 _2 _3 _6 _3 2 _9 8 1 _5 5 0 _4 _1 _9 _9 _9 _8 0 _5 _7 1 _2 1 _4 _1 5 7 _6 _9 4 _2 8 7 _9 _5 3 _1 1 8 4 _6 6 _3 _4 3 5 6 8 _2 5 _1 _7 _9 _1 7 _9 4 6 7 6 _8 5 1 0 _3 0 _3 _2 5 _4 0 0 0 _1 7 4 _9 _4 2) _28473770  # Brachylog, 114 bytes {l1,?hh.|:{:Im:I:?:{[L:I:M]h:JmN,Ll:2/D(IJ,M{$\:?c:{:{:ImN,I:1+:N*.}f+.}a+.}:N+.;'(DIJ),N.)}f.}f:7a$\:7a&.}. brbr.  I'm suprised this even works to be honest. At least I realized that Brachylog really needs a "change value of that element" as a built-in though… Usage example: ?- run_from_file('code.brachylog', '[[0:_2:_7:_7:_7]:[_3:_2:_2:_3:2]:[_8:4:139:_8:2]:[6:7:_1:8:4]:[5:7:8:7:_9]]', Z). Z = 5321 .  ### Explanation More readable (and longer) version: {l1,?hh.|:2f:7a$\:7a&.}.
:Im:I:?:3f.
[L:I:M]h:JmN,Ll:2/D(IJ,M:4&:N+.;'(DIJ),N.)
$\:?c:5a+. :6f+. :ImN,I:1+:N*. brbr.  I'm just gonna explain roughly what each predicate (i.e each line except the first one which is Main Predicate + predicate 1) does: • Main predicate + predicate 1 {l1,?hh.|:2f:7a$\:7a&.}. : If the input has only one row, then end the algorithm and return the only value. Else find all rows which satisfy predicate 2, then apply predicate 7 on the resulting matrix, then predicate 7 on the transposition, then call recursively.

• Predicate 2 :Im:I:?:3f. :Take the Ith row of the matrix, find all values of that row which satisfy predicate 3 with I and the matrix as additional inputs.

• Predicate 3 [L:I:M]h:JmN,Ll:2/D(IJ,M:4&:N+.;'(DIJ),N.) : L is the row, I is the index of the row, M is the matrix. N is the Jth element of L. If the length of L divided by 2 is equal to both I and J, then the output is the sum of N with the result of predicate 4 on the matrix. Otherwise the output is just N. This predicate essentialy recreates the matrix with the exception that the center element gets added to the sum.

• Predicate 4 \$\:?c:5a+. : Apply predicate 5 on each row and column of the matrix, unify the output with the sum of the results.

• Predicate 5 :6f+. : Find all valid outputs of predicate 6 on the row, unify the output with the sum of the resulting list.

• Predicate 6 :ImN,I:1+:N*.N is the Ith value of the row, unify the output with N * (I+1).

• Predicate 7 brbr. : Remove the first and last row of the matrix.

## APL, 56 chars

{{1 1↓¯1 ¯1↓⍵+(-⍴⍵)↑(⌈.5×⍴⍵)↑+/(⍵⍪⍉⍵)+.×⍳≢⍵}⍣(⌊.5×≢⍵)⊣⍵}


In English:

• ⍣(⌊.5×≢⍵) repeat "half the size of a dimension rounded"-times
• (⍵⍪⍉⍵)+.×⍳≢⍵ inner product of the matrix and its transpose with the index vector
• (-⍴⍵)↑(⌈.5×⍴⍵)↑ transform result in matrix padded with 0s
• 1 1↓¯1 ¯1↓ removes outer shell