Find the inside of a Loop

Question

Task

Given a ASCII diagram of a loop

e.g.

....................
......@@@@@.........
......@...@.........
.....@@...@@@@@.....
....@@........@.....
....@........@@.....
....@@@@@@@@@@......
....................
....................

And a location on the loop

e.g.

(7,1)

You must find the inside and outside of the loop

e.g.

00000000000000000000
00000011111000000000
00000011111000000000
00000111111111100000
00001111111111100000
00001111111111100000
00001111111111000000
00000000000000000000
00000000000000000000

Specifications

• You may take input for the diagram as a string separated by newlines or obvious equivalent

• You will receive a coordinate on the loop (0 or 1 indexed) as part of your input. You may place your origin at any place you wish. You may take this coordinate in (<row>, <column>), (<column>, <row>) or as the linear position on the string. You may receive these data in any reasonable method. All characters on the loop will be the same as the character at that index.

• Preferred output is a 2 dimensional array of truthy and falsy values, however strings of 1 and 0 separated by newlines or any obvious equivalent of the later two are accepted. The inside and outside must have different truth values but it does not matter which is which.

• A loop is defined as a group of characters such that they are all the same character (e.g. @) and so that every character in the loop has a path to the original character (The character at the coordinate of input) that only passes through that same character (Taxicab geometry No diagonals).

• The inside is all the loop itself and the places that cannot reach the edge of the diagram without crossing the loop.

• The outside is everywhere else

• This is code-golf

Test Cases

PasteBin

Answer 1

MATLAB, 163 159 146 78 bytes

function m=f(m,y,x);[~,i]=bwfill(m~=m(y,x),x,y,8);m=m*0;m(i)=1;m=bwfill(m,'h')

Thanks @rahnema1 for -66 bytes!!!

Now it does work on Try it online! BUT a few adjustments were needed, as MATLAB and Octave are not entirely compatible.

Explanation

First we make a binary image that just masks all the characters that are equal to the initial character. Then we determine the connected component the initial character is in.

% determine the connected component that is contains initial character

[~,i]=bwfill(m~=m(y,x),x,y,8);     % i contains the indices of the connected component
m=m*0;m(i)=1;                      % create an image of the connected component

After that we create an image of that connected component and apply fill all the "holes" in the image.

m=bwfill(m,'h')

Answer 2

MATLAB: 67 bytes

function A=f(A,r,c),A=bwlabel(A==A(r,c),4);A=imfill(A==A(r,c),'h');

A couple caveats:

• A is assumed to be a character array.
• Indices in MATLAB are 1-based, with rows indexed first. It is assumed these changes would be made to the function input (i.e. the question example would be called as output = f(A,2,8)).
• bwlabel and imfill are part of the Image Processing Toolbox.