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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

Your task:

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
  • Add 2114+2147 to 1060
  • 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

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  • \$\begingroup\$ you are correct, thats a typo \$\endgroup\$ Commented Jun 23, 2016 at 22:33
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    \$\begingroup\$ 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 \$\endgroup\$ Commented Jun 23, 2016 at 23:09
  • \$\begingroup\$ @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. \$\endgroup\$ Commented Jun 24, 2016 at 7:53
  • \$\begingroup\$ actually thats correct. but then the test cases wont work properly \$\endgroup\$ Commented Jun 24, 2016 at 8:08
  • 2
    \$\begingroup\$ "I dont think the challenge should adjust for esolangs but maybe the other way around is more appropriate" that should be engraved in stone \$\endgroup\$ Commented Jun 24, 2016 at 8:43

6 Answers 6

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MATL, 36 34 bytes

tnq?`t&+stn:*sytn2/)+ 7M(6Lt3$)tnq

Input is a 2D array with ; as row separator

Try it online! Or verify all test cases.

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
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4
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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.

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  • 1
    \$\begingroup\$ You can remove a couple spaces between operators (...) + sum([i*j... -> ...)+sum([i*j...), but overall, great first post!!!! \$\endgroup\$ Commented Jun 24, 2016 at 19:04
  • \$\begingroup\$ oooh missed that. Thanks! \$\endgroup\$ Commented Jun 24, 2016 at 19:21
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    \$\begingroup\$ Also, ...]for ... works. You can remove at least 2 space like that. (end of list hits the for loop) \$\endgroup\$ Commented Jun 24, 2016 at 19:25
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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 }
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    \$\begingroup\$ Hey now, J isn't an esolang. \$\endgroup\$ Commented Jun 24, 2016 at 8:26
  • \$\begingroup\$ 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. \$\endgroup\$ Commented Jun 24, 2016 at 12:11
  • \$\begingroup\$ 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 \$\endgroup\$ Commented Jun 24, 2016 at 12:13
  • \$\begingroup\$ @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. \$\endgroup\$ Commented Jun 24, 2016 at 12:35
  • \$\begingroup\$ 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#. \$\endgroup\$ Commented Jun 24, 2016 at 15:16
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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
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2
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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.

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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
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