Whose neighbours are hostile?

10

Introduction

For the purposes of this challenge, we will define the neighbours of an element \$E\$ in a square matrix \$A\$ (such that \$E=A_{i,j}\$) as all the entries of \$A\$ that are immediately adjacent diagonally, horizontally or vertically to \$E\$ (i.e. they "surround" \$E\$, without wrapping around).

For pedants, a formal definition of the neighbours of \$A_{i,\:j}\$ for an \$n\times n\$ matix \$A\$ is (0-indexed): $$N_{i,\:j}=\{A_{a,\:b}\mid(a,b)\in E_{i,\:j}\:\cap\:([0,\:n)\:\cap\:\Bbb{Z})^2\}$$ where $$E_{i,\:j}=\{i-1,\:i,\:i+1\}\times \{j-1,\:j,\:j+1\} \text{ \\ } \{i,\:j\}$$

Let's say that the element at index \$i,\:j\$ lives in hostility if it is coprime to all its neighbours (that is, \$\gcd(A_{i,\:j},\:n)=1\:\forall\:n\in N_{i,\:j}\$). Sadly, this poor entry can't borrow even a cup of sugar from its rude nearby residents...

Task

Enough stories: Given a square matrix \$M\$ of positive integers, output one of the following:

  • A flat list of elements (deduplicated or not) indicating all entries that occupy some indices \$i,j\$ in \$M\$ such that the neighbours \$N_{i,\:j}\$ are hostile.
  • A boolean matrix with \$1\$s at positions where the neighbours are hostile and \$0\$ otherwise (you can choose any other consistent values in place of \$0\$ and \$1\$).
  • The list of pairs of indices \$i,\:j\$ that represent hostile neighbourhoods.

Reference Implementation in Physica – supports Python syntax as well for I/O. You can take input and provide output through any standard method and in any reasonable format, while taking note that these loopholes are forbidden by default. This is code-golf, so the shortest code in bytes (in every language) wins!

Moreover, you can take the matrix size as input too and additionally can take the matrix as a flat list since it will always be square.

Example

Consider the following matrix:

$$\left(\begin{matrix} 64 & 10 & 14 \\ 27 & 22 & 32 \\ 53 & 58 & 36 \\ \end{matrix}\right)$$

The corresponding neighbours of each element are:

i j – E  -> Neighbours                          | All coprime to E?
                                                |
0 0 – 64 -> {10; 27; 22}                        | False
0 1 – 10 -> {64; 14; 27; 22; 32}                | False
0 2 – 14 -> {10; 22; 32}                        | False
1 0 – 27 -> {64; 10; 22; 53; 58}                | True
1 1 – 22 -> {64; 10; 14; 27; 32; 53; 58; 36}    | False
1 2 – 32 -> {10; 14; 22; 58; 36}                | False
2 0 – 53 -> {27; 22; 58}                        | True
2 1 – 58 -> {27; 22; 32; 53; 36}                | False
2 2 – 36 -> {22; 32; 58}                        | False

And thus the output must be one of the following:

  • {27; 53}
  • {{0; 0; 0}; {1; 0; 0}; {1; 0; 0}}
  • {(1; 0); (2; 0)}

Test cases

Input –> Version 1 | Version 2 | Version 3

[[36, 94], [24, 69]] ->
    []
    [[0, 0], [0, 0]]
    []
[[38, 77, 11], [17, 51, 32], [66, 78, 19]] –>
    [38, 19]
    [[1, 0, 0], [0, 0, 0], [0, 0, 1]]
    [(0, 0), (2, 2)]
[[64, 10, 14], [27, 22, 32], [53, 58, 36]] ->
    [27, 53]
    [[0, 0, 0], [1, 0, 0], [1, 0, 0]]
    [(1, 0), (2, 0)]
[[9, 9, 9], [9, 3, 9], [9, 9, 9]] ->
    []
    [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
    []
[[1, 1, 1], [1, 1, 1], [1, 1, 1]] ->
    [1, 1, 1, 1, 1, 1, 1, 1, 1] or [1]
    [[1, 1, 1], [1, 1, 1], [1, 1, 1]]
    [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]
[[35, 85, 30, 71], [10, 54, 55, 73], [80, 78, 47, 2], [33, 68, 62, 29]] ->
    [71, 73, 47, 29]
    [[0, 0, 0, 1], [0, 0, 0, 1], [0, 0, 1, 0], [0, 0, 0, 1]]
    [(0, 3), (1, 3), (2, 2), (3, 3)]

Mr. Xcoder

Posted 2018-09-16T12:16:12.910

Reputation: 39 774

Borrowing stuff from hostile neighbors? For some reason, this reminds me of Jeff Minter's game Hover Bovver...

– Arnauld – 2018-09-16T13:17:22.637

Can we take the matrix size as input? – Delfad0r – 2018-09-17T05:32:54.527

@Delfad0r I always forget to mention that. Yes, you may take the matrix size as input. – Mr. Xcoder – 2018-09-17T07:19:08.050

Answers

3

APL (Dyalog), 17 bytes

1=⊢∨(×/∘,↓)⌺3 3÷⊢

Try it online! (credits to ngn for translating the test cases to APL)

Brief explanation

(×/∘,↓)⌺3 3 gets the product of each element with its neighbours.

Then I divide by the argument ÷⊢, so that each entry in the matrix has been mapped to the product of its neighbors.

Finally I take the gcd of the argument with this matrix ⊢∨, and check for equality with 1, 1=

Note, as with ngn's answer, this fails for some inputs due to a bug in the interpreter.

H.PWiz

Posted 2018-09-16T12:16:12.910

Reputation: 10 962

2

JavaScript (ES6), 121 bytes

Returns a matrix of Boolean values, where false means hostile.

m=>m.map((r,y)=>r.map((v,x)=>[...'12221000'].some((k,j,a)=>(g=(a,b)=>b?g(b,a%b):a>1)(v,(m[y+~-k]||0)[x+~-a[j+2&7]]||1))))

Try it online!

How?

The method used to isolate the 8 neighbors of each cell is similar to the one I described here.

Commented

m =>                            // m[] = input matrix
  m.map((r, y) =>               // for each row r[] at position y in m[]:
    r.map((v, x) =>             //   for each value v at position x in r[]:
      [...'12221000']           //     we consider all 8 neighbors
      .some((k, j, a) =>        //     for each k at position j in this array a[]:
        ( g = (a, b) =>         //       g is a function which takes 2 integers a and b
            b ?                 //       and recursively determines whether they are
              g(b, a % b)       //       coprime to each other
            :                   //       (returns false if they are, true if they're not)
              a > 1             //
        )(                      //       initial call to g() with:
          v,                    //         the value of the current cell
          (m[y + ~-k] || 0)     //         and the value of the current neighbor
          [x + ~-a[j + 2 & 7]]  //
          || 1                  //         or 1 if this neighbor is undefined
  ))))                          //         (to make sure it's coprime with v)

Arnauld

Posted 2018-09-16T12:16:12.910

Reputation: 111 334

2

MATL, 22 bytes

tTT1&Ya3thYC5&Y)Zd1=A)

Input is a matrix. Output is all numbers with hostile neighbours.

Try it online! Or verify all test cases.

Explanation with worked example

Consider input [38, 77, 11; 17, 51, 32; 66, 78, 19] as an example. Stack contents are shown bottom to top.

t         % Implicit input. Duplicate
          % STACK: [38, 77, 11;
                    17, 51, 32;
                    66, 78, 19]
                   [38, 77, 11;
                    17, 51, 32;
                    66, 78, 19]
TT1&Ya    % Pad in the two dimensions with value 1 and width 1
          % STACK: [38, 77, 11;
                    17, 51, 32;
                    66, 78, 19]
                   [1,  1,  1,  1,  1;
                    1,  38, 77, 11, 1;
                    1,  17, 51, 32, 1;
                    1,  66, 78, 19, 1
                    1,  1,  1,  1,  1]
3thYC     % Convert each sliding 3×3 block into a column (in column-major order)
          % STACK: [38, 77, 11;
                    17, 51, 32;
                    66, 78, 19]
                   [ 1,  1,  1,  1, 38, 17,  1, 77, 51;
                     1,  1,  1, 38, 17, 66, 77, 51, 78;
                     1,  1,  1, 17, 66,  1, 51, 78,  1;
                     1, 38, 17,  1, 77, 51,  1, 11, 32;
                    38, 17, 66, 77, 51, 78, 11, 32, 19;
                    17, 66,  1, 51, 78,  1, 32, 19,  1;
                     1, 77, 51,  1, 11, 32,  1,  1,  1;
                    77, 51, 78, 11, 32, 19,  1,  1,  1;
                    51, 78,  1, 32, 19,  1,  1,  1,  1]
5&Y)      % Push 5th row (centers of the 3×3 blocks) and then the rest of the matrix
          % STACK: [38, 77, 11;
                    17, 51, 32;
                    66, 78, 19]
                   [38, 17, 66, 77, 51, 78, 11, 32, 19]
                   [ 1,  1,  1,  1, 38, 17,  1, 77, 51;
                     1,  1,  1, 38, 17, 66, 77, 51, 78;
                     1,  1,  1, 17, 66,  1, 51, 78,  1;
                     1, 38, 17,  1, 77, 51,  1, 11, 32;
                    17, 66,  1, 51, 78,  1, 32, 19,  1;
                     1, 77, 51,  1, 11, 32,  1,  1,  1;
                    77, 51, 78, 11, 32, 19,  1,  1,  1;
                    51, 78,  1, 32, 19,  1,  1,  1,  1]
Zd        % Greatest common divisor, element-wise with broadcast
          % STACK: [38, 77, 11;
                    17, 51, 32;
                    66, 78, 19]
                   [1,  1,  1,  1,  1,  1,  1,  1,  1;
                    1,  1,  1,  1, 17,  6, 11,  1,  1;
                    1,  1,  1,  1,  3,  1,  1,  2,  1;
                    1,  1,  1,  1,  1,  3,  1,  1,  1;
                    1,  1,  1,  1,  3,  1,  1,  1,  1;
                    1,  1,  3,  1,  1,  2,  1,  1,  1;
                    1, 17,  6, 11,  1,  1,  1,  1,  1;
                    1,  1,  1,  1,  1,  1,  1,  1,  1]
1=        % Compare with 1, element-wise. Gives true (1) or false (0)
          % STACK: [38, 77, 11;
                    17, 51, 32;
                    66, 78, 19]
                   [1, 1, 1, 1, 1, 1, 1, 1, 1;
                    1, 1, 1, 1, 0, 0, 0, 1, 1;
                    1, 1, 1, 1, 0, 1, 1, 0, 1;
                    1, 1, 1, 1, 1, 0, 1, 1, 1;
                    1, 1, 1, 1, 0, 1, 1, 1, 1;
                    1, 1, 0, 1, 1, 0, 1, 1, 1;
                    1, 0, 0, 0, 1, 1, 1, 1, 1;
                    1, 1, 1, 1, 1, 1, 1, 1, 1]
A         % All: true (1) for columns that do not contain 0
          % STACK: [38, 77, 11;
                    17, 51, 32;
                    66, 78, 19]
                   [1, 0, 0, 0, 0, 0, 0, 0, 1]
)         % Index (the matrix is read in column-major order). Implicit display
          % [38, 19]

Luis Mendo

Posted 2018-09-16T12:16:12.910

Reputation: 87 464

Will this work if the matrix is bigger than 3x3? – Robert Fraser – 2018-09-17T04:42:37.353

@RobertFraser Yes, the procedure does not depend on matrix size. See last test case for example – Luis Mendo – 2018-09-17T09:23:24.797

1

APL (Dyalog Classic), 23 22 bytes

-1 byte thanks to @H.PWiz

{∧/1=1↓∨∘⊃⍨1⌈4⌽,⍵}⌺3 3

Try it online!

doesn't support matrices smaller than 3x3 due to a bug in the interpreter

ngn

Posted 2018-09-16T12:16:12.910

Reputation: 11 449

@H.PWiz that's very smart, do you wanna post it as your own? – ngn – 2018-09-16T18:03:28.893

Sure, you can also use (⊃∨⊢) -> ∨∘⊂⍨ I think – H.PWiz – 2018-09-16T18:04:20.793

1

Jelly, 24 bytes

Hmm, seems long.

ỊẠ€T
ŒJ_€`Ç€ḟ"J$ịFg"FÇịF

A monadic Link accepting a list of lists of positive integers which returns a list of each of the values which are in hostile neighbourhoods (version 1 with no de-duplication).

Try it online! Or see a test-suite.

How?

ỊẠ€T - Link 1: indices of items which only contain "insignificant" values: list of lists
Ị    - insignificant (vectorises) -- 1 if (-1<=value<=1) else 0 
  €  - for €ach:
 Ạ   -   all?
   T - truthy indices

ŒJ_€`Ç€ḟ"J$ịFg"FÇịF - Main Link: list of lists of positive integers, M
ŒJ                  - multi-dimensional indices
    `               - use as right argument as well as left...
   €                -   for €ach:
  _                 -     subtract (vectorises)
      €             - for €ach:
     Ç              -   call last Link (1) as a monad
          $         - last two links as a monad:
         J          -   range of length -> [1,2,3,...,n(elements)]
        "           -   zip with:
       ḟ            -     filter discard (remove the index of the item itself)
            F       - flatten M
           ị        - index into (vectorises) -- getting a list of lists of neighbours
               F    - flatten M
              "     - zip with:
             g      -   greatest common divisor
                Ç   - call last Link (1) as a monad
                  F - flatten M
                 ị  - index into

Jonathan Allan

Posted 2018-09-16T12:16:12.910

Reputation: 67 804

1

Python 2, 182 177 166 bytes

lambda a:[[all(gcd(t,a[i+v][j+h])<2for h in[-1,0,1]for v in[-1,0,1]if(h|v)*(i+v>-1<j+h<len(a)>i+v))for j,t in E(s)]for i,s in E(a)]
from fractions import*
E=enumerate

Try it online!

Outputs a list of lists with True/False entries.

Chas Brown

Posted 2018-09-16T12:16:12.910

Reputation: 8 959

1

Haskell, 95 bytes

m?n|l<-[0..n-1]=[a|i<-l,j<-l,a<-[m!!i!!j],2>sum[1|u<-l,v<-l,(i-u)^2+(j-v)^2<4,gcd(m!!u!!v)a>1]]

Try it online!

The function ? takes the matrix m as a list of lists and the matrix size n; it returns the list of entries in hostility.

Delfad0r

Posted 2018-09-16T12:16:12.910

Reputation: 1 688