11 KiB
11 KiB
| title | id | challengeType |
|---|---|---|
| Zhang-Suen thinning algorithm | 594810f028c0303b75339ad7 | 5 |
Description
################# #############
################## ################
################### ##################
######## ####### ###################
###### ####### ####### ######
###### ####### #######
################# #######
################ #######
################# #######
###### ####### #######
###### ####### #######
###### ####### ####### ######
######## ####### ###################
######## ####### ###### ################## ######
######## ####### ###### ################ ######
######## ####### ###### ############# ######
It produces the thinned output:
# ########## ####### ## # #### # # # ## # # # # # # # # # ############ # # # # # # # # # # # # # # ## # ############ ### ### </pre>Algorithm
Assume black pixels are one and white pixels zero, and that the input image is a rectangular N by M array of ones and zeroes. The algorithm operates on all black pixels P1 that can have eight neighbours. The neighbours are, in order, arranged as:
| P9 | P2 | P3 |
| P8 | P1 | P4 |
| P7 | P6 | P5 |
Define $A(P1)$ = the number of transitions from white to black, (0 -> 1) in the sequence P2,P3,P4,P5,P6,P7,P8,P9,P2. (Note the extra P2 at the end - it is circular).
Define $B(P1)$ = the number of black pixel neighbours of P1. ( = sum(P2 .. P9) )
Step 1:
All pixels are tested and pixels satisfying all the following conditions (simultaneously) are just noted at this stage. (0) The pixel is black and has eight neighbours (1) $2 <= B(P1) <= 6$ (2) $A(P1) = 1$ (3) At least one of P2 and P4 and P6 is white (4) At least one of P4 and P6 and P8 is white After iterating over the image and collecting all the pixels satisfying all step 1 conditions, all these condition satisfying pixels are set to white.Step 2:
All pixels are again tested and pixels satisfying all the following conditions are just noted at this stage. (0) The pixel is black and has eight neighbours (1) $2 <= B(P1) <= 6$ (2) $A(P1) = 1$ (3) At least one of P2 and P4 and '''P8''' is white (4) At least one of '''P2''' and P6 and P8 is white After iterating over the image and collecting all the pixels satisfying all step 2 conditions, all these condition satisfying pixels are again set to white. Iteration: If any pixels were set in this round of either step 1 or step 2 then all steps are repeated until no image pixels are so changed.Task: Write a routine to perform Zhang-Suen thinning on an image matrix of ones and zeroes.
Instructions
Tests
tests:
- text: <code>thinImage</code> must be a function
testString: assert.equal(typeof thinImage, 'function', '<code>thinImage</code> must be a function');
- text: <code>thinImage</code> must return an array
testString: assert(Array.isArray(result), '<code>thinImage</code> must return an array');
- text: <code>thinImage</code> must return an array of strings
testString: assert.equal(typeof result[0], 'string', '<code>thinImage</code> must return an array of strings');
- text: <code>thinImage</code> must return an array of strings
testString: assert.deepEqual(result, expected, '<code>thinImage</code> must return an array of strings');
Challenge Seed
const testImage = [
' ',
' ################# ############# ',
' ################## ################ ',
' ################### ################## ',
' ######## ####### ################### ',
' ###### ####### ####### ###### ',
' ###### ####### ####### ',
' ################# ####### ',
' ################ ####### ',
' ################# ####### ',
' ###### ####### ####### ',
' ###### ####### ####### ',
' ###### ####### ####### ###### ',
' ######## ####### ################### ',
' ######## ####### ###### ################## ###### ',
' ######## ####### ###### ################ ###### ',
' ######## ####### ###### ############# ###### ',
' '];
function thinImage(image) {
// Good luck!
}
After Test
const imageForTests = [
' ',
' ################# ############# ',
' ################## ################ ',
' ################### ################## ',
' ######## ####### ################### ',
' ###### ####### ####### ###### ',
' ###### ####### ####### ',
' ################# ####### ',
' ################ ####### ',
' ################# ####### ',
' ###### ####### ####### ',
' ###### ####### ####### ',
' ###### ####### ####### ###### ',
' ######## ####### ################### ',
' ######## ####### ###### ################## ###### ',
' ######## ####### ###### ################ ###### ',
' ######## ####### ###### ############# ###### ',
' '];
const expected = [
' ',
' ',
' # ########## ####### ',
' ## # #### # ',
' # # ## ',
' # # # ',
' # # # ',
' # # # ',
' ############ # ',
' # # # ',
' # # # ',
' # # # ',
' # # # ',
' # ## ',
' # ############ ',
' ### ### ',
' ',
' '
];
const result = thinImage(imageForTests);
Solution
function Point(x, y) {
this.x = x;
this.y = y;
}
const ZhangSuen = (function () {
function ZhangSuen() {
}
ZhangSuen.nbrs = [[0, -1], [1, -1], [1, 0], [1, 1], [0, 1], [-1, 1], [-1, 0], [-1, -1], [0, -1]];
ZhangSuen.nbrGroups = [[[0, 2, 4], [2, 4, 6]], [[0, 2, 6], [0, 4, 6]]];
ZhangSuen.toWhite = [];
ZhangSuen.main = function (image) {
ZhangSuen.grid = new Array(image);
for (let r = 0; r < image.length; r++) {
ZhangSuen.grid[r] = image[r].split('');
}
ZhangSuen.thinImage();
return ZhangSuen.getResult();
};
ZhangSuen.thinImage = function () {
let firstStep = false;
let hasChanged;
do {
hasChanged = false;
firstStep = !firstStep;
for (let r = 1; r < ZhangSuen.grid.length - 1; r++) {
for (let c = 1; c < ZhangSuen.grid[0].length - 1; c++) {
if (ZhangSuen.grid[r][c] !== '#') {
continue;
}
const nn = ZhangSuen.numNeighbors(r, c);
if (nn < 2 || nn > 6) {
continue;
}
if (ZhangSuen.numTransitions(r, c) !== 1) {
continue;
}
if (!ZhangSuen.atLeastOneIsWhite(r, c, firstStep ? 0 : 1)) {
continue;
}
ZhangSuen.toWhite.push(new Point(c, r));
hasChanged = true;
}
}
for (let i = 0; i < ZhangSuen.toWhite.length; i++) {
const p = ZhangSuen.toWhite[i];
ZhangSuen.grid[p.y][p.x] = ' ';
}
ZhangSuen.toWhite = [];
} while ((firstStep || hasChanged));
};
ZhangSuen.numNeighbors = function (r, c) {
let count = 0;
for (let i = 0; i < ZhangSuen.nbrs.length - 1; i++) {
if (ZhangSuen.grid[r + ZhangSuen.nbrs[i][1]][c + ZhangSuen.nbrs[i][0]] === '#') {
count++;
}
}
return count;
};
ZhangSuen.numTransitions = function (r, c) {
let count = 0;
for (let i = 0; i < ZhangSuen.nbrs.length - 1; i++) {
if (ZhangSuen.grid[r + ZhangSuen.nbrs[i][1]][c + ZhangSuen.nbrs[i][0]] === ' ') {
if (ZhangSuen.grid[r + ZhangSuen.nbrs[i + 1][1]][c + ZhangSuen.nbrs[i + 1][0]] === '#') {
count++;
}
}
}
return count;
};
ZhangSuen.atLeastOneIsWhite = function (r, c, step) {
let count = 0;
const group = ZhangSuen.nbrGroups[step];
for (let i = 0; i < 2; i++) {
for (let j = 0; j < group[i].length; j++) {
const nbr = ZhangSuen.nbrs[group[i][j]];
if (ZhangSuen.grid[r + nbr[1]][c + nbr[0]] === ' ') {
count++;
break;
}
}
}
return count > 1;
};
ZhangSuen.getResult = function () {
const result = [];
for (let i = 0; i < ZhangSuen.grid.length; i++) {
const row = ZhangSuen.grid[i].join('');
result.push(row);
}
return result;
};
return ZhangSuen;
}());
function thinImage(image) {
return ZhangSuen.main(image);
}