--- id: 594810f028c0303b75339ad7 title: Zhang-Suen thinning algorithm challengeType: 5 forumTopicId: 302347 dashedName: zhang-suen-thinning-algorithm --- # --description-- 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: ```js const testImage1 = [ ' ', '######### ######## ', '### #### #### #### ', '### ### ### ### ', '### #### ### ', '######### ### ', '### #### ### ### ', '### #### ### #### #### ### ', '### #### ### ######## ### ', ' ' ]; ``` It produces the thinned output: ```js [ ' ', '######## ###### ', '# # ## ', '# # # ', '# # # ', '###### # # ', '# ## # ', '# # # ## ## # ', '# # #### ', ' ' ]; ``` ## 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: $$\begin{array}{|c|c|c|} \\hline P9 & P2 & P3\\\\ \\hline P8 & \boldsymbol{P1} & P4\\\\ \\hline P7 & P6 & P5\\\\ \\hline \end{array}$$ Obviously the boundary pixels of the image cannot have the full eight neighbours. - 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). - Define $B(P1)$ = the number of black pixel neighbours of P1. ($= \\sum(P2 \ldots P9)$) **Step 1:** All pixels are tested and pixels satisfying all the following conditions (simultaneously) are just noted at this stage. 1. The pixel is black and has eight neighbours 2. $2 \le B(P1) \le 6$ 3. $A(P1) = 1$ 4. At least one of $P2$, $P4$ and $P6$ is white 5. At least one of $P4$, $P6$ 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. 1. The pixel is black and has eight neighbours 2. $2 \le B(P1) \le 6$ 3. $A(P1) = 1$ 4. At least one of $P2$, $P4$ and $P8$ is white 5. At least one of $P2$, $P6$ 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. # --instructions-- 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. # --hints-- `thinImage` should be a function. ```js assert.equal(typeof thinImage, 'function'); ``` `thinImage` should return an array. ```js assert(Array.isArray(thinImage(_testImage1))); ``` `thinImage` should return an array of strings. ```js assert.equal(typeof thinImage(_testImage1)[0], 'string'); ``` `thinImage(testImage1)` should return a thinned image as in the example. ```js assert.deepEqual(thinImage(_testImage1), expected1); ``` `thinImage(testImage2)` should return a thinned image. ```js assert.deepEqual(thinImage(_testImage2), expected2); ``` # --seed-- ## --after-user-code-- ```js const _testImage1 = [ ' ', '######### ######## ', '### #### #### #### ', '### ### ### ### ', '### #### ### ', '######### ### ', '### #### ### ### ', '### #### ### #### #### ### ', '### #### ### ######## ### ', ' ' ]; const expected1 = [ ' ', '######## ###### ', '# # ## ', '# # # ', '# # # ', '###### # # ', '# ## # ', '# # # ## ## # ', '# # #### ', ' ' ]; const _testImage2 = [ ' ', ' ################# ############# ', ' ################## ################ ', ' ################### ################## ', ' ######## ####### ################### ', ' ###### ####### ####### ###### ', ' ###### ####### ####### ', ' ################# ####### ', ' ################ ####### ', ' ################# ####### ', ' ###### ####### ####### ', ' ###### ####### ####### ', ' ###### ####### ####### ###### ', ' ######## ####### ################### ', ' ######## ####### ###### ################## ###### ', ' ######## ####### ###### ################ ###### ', ' ######## ####### ###### ############# ###### ', ' ']; const expected2 = [ ' ', ' ', ' # ########## ####### ', ' ## # #### # ', ' # # ## ', ' # # # ', ' # # # ', ' # # # ', ' ############ # ', ' # # # ', ' # # # ', ' # # # ', ' # # # ', ' # ## ', ' # ############ ', ' ### ### ', ' ', ' ' ]; ``` ## --seed-contents-- ```js function thinImage(image) { } const testImage1 = [ ' ', '######### ######## ', '### #### #### #### ', '### ### ### ### ', '### #### ### ', '######### ### ', '### #### ### ### ', '### #### ### #### #### ### ', '### #### ### ######## ### ', ' ' ]; ``` # --solutions-- ```js 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); } ```