326 lines
10 KiB
Markdown
326 lines
10 KiB
Markdown
---
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id: 594810f028c0303b75339ad7
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title: Zhang-Suen thinning algorithm
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challengeType: 1
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forumTopicId: 302347
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dashedName: zhang-suen-thinning-algorithm
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---
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# --description--
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This is an algorithm used to thin a black and white i.e. one bit per pixel images. For example, with an input image of:
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```js
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const testImage1 = [
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' ',
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'######### ######## ',
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'### #### #### #### ',
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'### ### ### ### ',
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'### #### ### ',
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'######### ### ',
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'### #### ### ### ',
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'### #### ### #### #### ### ',
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'### #### ### ######## ### ',
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' '
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];
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```
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It produces the thinned output:
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```js
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[ ' ',
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'######## ###### ',
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'# # ## ',
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'# # # ',
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'# # # ',
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'###### # # ',
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'# ## # ',
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'# # # ## ## # ',
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'# # #### ',
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' ' ];
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```
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## Algorithm
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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:
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$$\begin{array}{|c|c|c|}
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\\hline
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P9 & P2 & P3\\\\ \\hline
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P8 & \boldsymbol{P1} & P4\\\\ \\hline
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P7 & P6 & P5\\\\ \\hline
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\end{array}$$
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Obviously the boundary pixels of the image cannot have the full eight neighbours.
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- Define $A(P1)$ = the number of transitions from white to black, ($0 \to 1$) in the sequence P2, P3, P4, P5, P6, P7, P8, P9, P2. (Note the extra P2 at the end - it is circular).
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- Define $B(P1)$ = the number of black pixel neighbours of P1. ($= \\sum(P2 \ldots P9)$)
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**Step 1:**
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All pixels are tested and pixels satisfying all the following conditions (simultaneously) are just noted at this stage.
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1. The pixel is black and has eight neighbours
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2. $2 \le B(P1) \le 6$
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3. $A(P1) = 1$
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4. At least one of $P2$, $P4$ and $P6$ is white
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5. At least one of $P4$, $P6$ and $P8$ is white
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After iterating over the image and collecting all the pixels satisfying all step 1 conditions, all these condition satisfying pixels are set to white.
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**Step 2:**
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All pixels are again tested and pixels satisfying all the following conditions are just noted at this stage.
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1. The pixel is black and has eight neighbours
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2. $2 \le B(P1) \le 6$
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3. $A(P1) = 1$
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4. At least one of $P2$, $P4$ and $P8$ is white
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5. At least one of $P2$, $P6$ and $P8$ is white
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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.
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**Iteration:**
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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.
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# --instructions--
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Write a routine to perform Zhang-Suen thinning on the provided `image`, an array of strings, where each string represents single line of the image. In the string, `#` represents black pixel, and whitespace represents white pixel. Function should return thinned image, using the same representation.
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# --hints--
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`thinImage` should be a function.
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```js
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assert.equal(typeof thinImage, 'function');
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```
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`thinImage` should return an array.
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```js
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assert(Array.isArray(thinImage(_testImage1)));
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```
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`thinImage` should return an array of strings.
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```js
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assert.equal(typeof thinImage(_testImage1)[0], 'string');
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```
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`thinImage(testImage1)` should return a thinned image as in the example.
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```js
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assert.deepEqual(thinImage(_testImage1), expected1);
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```
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`thinImage(testImage2)` should return a thinned image.
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```js
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assert.deepEqual(thinImage(_testImage2), expected2);
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```
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# --seed--
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## --after-user-code--
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```js
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const _testImage1 = [
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' ',
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'######### ######## ',
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'### #### #### #### ',
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'### ### ### ### ',
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'### #### ### ',
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'######### ### ',
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'### #### ### ### ',
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'### #### ### #### #### ### ',
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'### #### ### ######## ### ',
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' '
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];
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const expected1 = [
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' ',
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'######## ###### ',
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'# # ## ',
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'# # # ',
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'# # # ',
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'###### # # ',
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'# ## # ',
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'# # # ## ## # ',
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'# # #### ',
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' '
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];
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const _testImage2 = [
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' ',
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' ################# ############# ',
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' ################## ################ ',
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' ################### ################## ',
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' ######## ####### ################### ',
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' ###### ####### ####### ###### ',
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' ###### ####### ####### ',
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' ################# ####### ',
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' ################ ####### ',
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' ################# ####### ',
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' ###### ####### ####### ',
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' ###### ####### ####### ',
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' ###### ####### ####### ###### ',
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' ######## ####### ################### ',
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' ######## ####### ###### ################## ###### ',
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' ######## ####### ###### ################ ###### ',
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' ######## ####### ###### ############# ###### ',
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' '];
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const expected2 = [
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' ',
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' ',
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' # ########## ####### ',
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' ## # #### # ',
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' # # ## ',
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' # # # ',
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' # # # ',
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' # # # ',
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' ############ # ',
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' # # # ',
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' # # # ',
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' # # # ',
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' # # # ',
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' # ## ',
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' # ############ ',
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' ### ### ',
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' ',
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' '
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];
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```
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## --seed-contents--
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```js
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function thinImage(image) {
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}
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const testImage1 = [
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' ',
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'######### ######## ',
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'### #### #### #### ',
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'### ### ### ### ',
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'### #### ### ',
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'######### ### ',
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'### #### ### ### ',
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'### #### ### #### #### ### ',
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'### #### ### ######## ### ',
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' '
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];
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```
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# --solutions--
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```js
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function Point(x, y) {
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this.x = x;
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this.y = y;
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}
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const ZhangSuen = (function () {
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function ZhangSuen() {
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}
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ZhangSuen.nbrs = [[0, -1], [1, -1], [1, 0], [1, 1], [0, 1], [-1, 1], [-1, 0], [-1, -1], [0, -1]];
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ZhangSuen.nbrGroups = [[[0, 2, 4], [2, 4, 6]], [[0, 2, 6], [0, 4, 6]]];
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ZhangSuen.toWhite = [];
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ZhangSuen.main = function (image) {
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ZhangSuen.grid = new Array(image);
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for (let r = 0; r < image.length; r++) {
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ZhangSuen.grid[r] = image[r].split('');
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}
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ZhangSuen.thinImage();
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return ZhangSuen.getResult();
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};
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ZhangSuen.thinImage = function () {
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let firstStep = false;
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let hasChanged;
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do {
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hasChanged = false;
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firstStep = !firstStep;
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for (let r = 1; r < ZhangSuen.grid.length - 1; r++) {
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for (let c = 1; c < ZhangSuen.grid[0].length - 1; c++) {
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if (ZhangSuen.grid[r][c] !== '#') {
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continue;
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}
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const nn = ZhangSuen.numNeighbors(r, c);
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if (nn < 2 || nn > 6) {
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continue;
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}
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if (ZhangSuen.numTransitions(r, c) !== 1) {
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continue;
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}
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if (!ZhangSuen.atLeastOneIsWhite(r, c, firstStep ? 0 : 1)) {
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continue;
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}
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ZhangSuen.toWhite.push(new Point(c, r));
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hasChanged = true;
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}
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}
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for (let i = 0; i < ZhangSuen.toWhite.length; i++) {
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const p = ZhangSuen.toWhite[i];
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ZhangSuen.grid[p.y][p.x] = ' ';
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}
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ZhangSuen.toWhite = [];
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} while ((firstStep || hasChanged));
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};
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ZhangSuen.numNeighbors = function (r, c) {
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let count = 0;
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for (let i = 0; i < ZhangSuen.nbrs.length - 1; i++) {
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if (ZhangSuen.grid[r + ZhangSuen.nbrs[i][1]][c + ZhangSuen.nbrs[i][0]] === '#') {
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count++;
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}
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}
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return count;
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};
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ZhangSuen.numTransitions = function (r, c) {
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let count = 0;
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for (let i = 0; i < ZhangSuen.nbrs.length - 1; i++) {
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if (ZhangSuen.grid[r + ZhangSuen.nbrs[i][1]][c + ZhangSuen.nbrs[i][0]] === ' ') {
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if (ZhangSuen.grid[r + ZhangSuen.nbrs[i + 1][1]][c + ZhangSuen.nbrs[i + 1][0]] === '#') {
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count++;
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}
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}
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}
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return count;
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};
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ZhangSuen.atLeastOneIsWhite = function (r, c, step) {
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let count = 0;
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const group = ZhangSuen.nbrGroups[step];
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for (let i = 0; i < 2; i++) {
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for (let j = 0; j < group[i].length; j++) {
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const nbr = ZhangSuen.nbrs[group[i][j]];
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if (ZhangSuen.grid[r + nbr[1]][c + nbr[0]] === ' ') {
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count++;
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break;
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}
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}
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}
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return count > 1;
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};
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ZhangSuen.getResult = function () {
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const result = [];
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for (let i = 0; i < ZhangSuen.grid.length; i++) {
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const row = ZhangSuen.grid[i].join('');
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result.push(row);
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}
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return result;
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};
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return ZhangSuen;
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}());
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function thinImage(image) {
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return ZhangSuen.main(image);
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}
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```
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