10 KiB
10 KiB
| id | title | challengeType | forumTopicId | dashedName |
|---|---|---|---|---|
| 594810f028c0303b75339ad7 | Zhang-Suen thinning algorithm | 1 | 302347 | 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:
const testImage1 = [
' ',
'######### ######## ',
'### #### #### #### ',
'### ### ### ### ',
'### #### ### ',
'######### ### ',
'### #### ### ### ',
'### #### ### #### #### ### ',
'### #### ### ######## ### ',
' '
];
It produces the thinned output:
[ ' ',
'######## ###### ',
'# # ## ',
'# # # ',
'# # # ',
'###### # # ',
'# ## # ',
'# # # ## ## # ',
'# # #### ',
' ' ];
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);
}
```