In JavaScript, I'm trying to find all matching coordinates of this 2D integer array:
inside this 2D integer array, with overlapping sub-arrays being counted:
Each image represents a 2D JavaScript integer array, and the black pixels correspond to 1, and the yellow pixels correspond to 0, but I depicted the arrays this way so that they would be easier to visualize.
So how can I find all matches of the array inside the array ?
Here's the function I'm trying to implement:
findAllMatchesOfOne2DArrayInsideAnother2DArray(containedArray, containingArray){
//find all matching coordinates of containedArray inside containingArray, and return a 2D array of coordinates
}
Here is a way to get all occurances of any given 2D array inside another 2D array. It is assumed that all subarrays have the same dimension (nothing like [[1,0,0],[1,0]]).
var x = [[0,1,0,0,0,0,0],
[1,1,1,0,0,1,0],
[0,1,0,0,1,1,1],
[0,0,0,0,0,1,0],
[0,0,1,0,0,0,0],
[0,1,1,1,0,0,0],
[0,0,1,0,0,0,0]];
var y = [[0,1,0],[1,1,1],[0,1,0]];
var res = [];
for (var i = 0; i < x.length - y.length + 1; i++) {
for (var k = 0; k < x[0].length - y[0].length + 1; k++) {
var count = 0;
for (var l = 0; l < y.length; l++) {
for (var m = 0; m < y[l].length; m++) {
count += Math.abs(y[l][m] - x[i + l][k + m]);
}
}
if (count === 0) {
res.push([i, k]);
}
}
}
The array res will contain the "coordinates" of the top-left corner of each match. I'm sure you'll find better performing algorithms but this one seems to work :)
Related
I have a string "ABCDEFGHIJKLMN" that I need to shuffle in a specific manner. To do that, I write the characters sequentially in columns bottom -> top and then left -> right (4 chars per column for example) until all characters are done. If the last column is not complete, then the empty spaces need to be on the bottom (this is very important). Like so:
D H L N
C G K M
B F J
A E I
The shuffle is accomplished by producing a new string reading the block of letters as we read text, in rows left -> right:
"DHLNCGKMBFJAEI"
The cases where the columns are not complete (word.size % column_height !=0) complicate things considerably.
I came up with a few solutions, but I'm not sure if there is a simpler (ie, shorter OR easier to read) and more elegant way of coding this problem. My solutions either have an ugly, separate block of code to handle the final incomplete column or seem way too complicated.
My question is, could it be done better?
If you don't want any spoilers and decide to try and figure it out for yourself, stop reading now. If you want to work from what I fiddled so far, then a working piece of code is
var result = "";
var str = "ABCDEFGHIJKLMN";
var nr_rows = 4;
var current_row = 4;
var columns = Math.floor(str.length / nr_rows);
var modulus_table = str.length % nr_rows;
var modulus_position = -1;
for (var i = 0; i < nr_rows; i++) {
for (var j = 0; j < columns; j++) {
result += str[current_row + j * nr_rows - 1];
}
if (modulus_table > 0) {
result += str[str.length + modulus_position];
modulus_table--;
modulus_position--;
}
current_row--;
}
console.log(result);
Moving on to arrays, the next example would loop through each character, placing it correctly in a matrix-like array, but it doesn't work. The array needs to be created another way. For another example of this issue, see How to create empty 2d array in javascript?. This would also need an ugly hack to fix the last characters on the last incomplete column aligning to the bottom instead of the top.
var result = [[],[]];
var str = "ABCDEFGHIJKLMN";
var nr_rows = 4;
var row = nr_rows - 1;
var column = 0;
for (var i = 0; i < str.length; i++) {
result[row][column] = str[i];
row--;
if (row < 0) {
row = nr_rows;
column++;
}
}
console.log(result);
This last method goes full matrix array, but it quickly becomes complicated, since it needs to loop through the array in 3 different directions. First, create a dummy array with the characters in the wrong place, but where the 'undefined' positions correspond to those that should be left empty. That is acomplished by populating the array 'rotated 90ยบ' from the reading orientation.
Without this first step, the empty positions would be stacked at the bottom instead of the top.
A second pass is required to re-write the caracters in the correct places, skipping any holes in the matrix using the 'undefined' value. This check is made for every position and there is no separate block of code to handle an incomplete last line.
A third pass then reads every character in order to form the final shuffled string. All this seems way too complicated and confusing.
// matrix populated top->bottom and left->right
// with the characters in the wrong place
// but the undefined postions in the correct place of the empty positions
var matrix = [];
var str = "ABCDEFGHIJKLMN";
var rows = 4;
var columns = Math.ceil(str.length / rows);
var k = 0;
for (var i = 0; i < rows; i++) {
matrix[i] = [];
for (var j = columns - 1; j >= 0; j--) {
matrix[i][j] = str[k];
k++;
}
}
// populate the matrix with the chars in the correct place and the 'undefined' positions left empty
var k = 0;
for (var i = 0; i < rows; i++) {
for (var j = 0; j < columns; j++) {
if (matrix[i][j] != undefined) {
matrix[i][j] = str[k];
k++;
}
}
}
// read matrix in correct direction and send to string, skipping empty positions
var result = "";
for (var j = columns - 1; j >= 0; j--) {
for (var i = 0; i < rows; i++) {
if (matrix[i][j] != undefined) {
result += matrix[i][j];
}
}
}
console.log(result);
What if you just split/reverse the array into column groups, and convert to rows?
const result = str.match(/.{1,4}/g) // split string into groups of 4
.map(i => i.split('').reverse()) // reverse each group (bottom to top, and solves the last col issue)
.reduce((res, col) => { // reduce the groups into rows
col.forEach((c, i) => res[i] += c) // concat each string char to the right row
return res
}, ['','','','']) // initialise empty strings per row
.join('') // join the rows up
Fiddle here
If you wish to return a string, I don't see why any intermediate result should use an array when it doesn't have to. The following could use one less array, but it's convenient to use split and an array to control the while loop rather than mutate the string.
The idea is to fill the strings from the bottom up until the column is full, then keep adding from the bottom of each column until it runs out of characters to assign. The row to start filling from is based on how many characters are left and how many rows there are.
Rather than building strings, it could build arrays but then generating a string requires multiple joins.
It can also produce results where there are insufficient slots for all the characters, so a result using 9 characters from 10 or more using a 3x3 "matrix" (see last example).
function verticalShuffle(s, rows, cols) {
var result = [''];
s = s.split('');
while (s.length && result[0].length < cols) {
for (var i = (rows < s.length? rows : s.length) -1 ; i>=0; i--) {
if (!result[i]) result[i] = '';
result[i] += s.splice(0,1)[0] || '';
}
}
return result.join('');
}
var s = 'ABCDEFGHIJKLMN';
console.log(verticalShuffle(s, 4, 4)); // DHLNCGKMBFJAEI
console.log(verticalShuffle(s, 6, 3)); // FLNEKMDJCIBHAG
// Only use 9 characters
console.log(verticalShuffle(s, 3, 3)); // CFIBEHADG
This uses plain ed3 functionality that will run in any browser. I don't see the point of restricting it to ECMAScript 2015 or later hosts.
If interpret Question correctly, you can use for loop, String.prototype.slice() to to populate arrays with characters of string. Use Array.prototype.pop() within recursive function to get last element of array until each array .length is 0.
To create array
[
["D","H","L","N"],
["C","G","K","M"],
["B","F","J"],
["A","E","I"]
]
from vertically inverted string you can use for loop, String.prototype.slice() to set array of arrays containing elements having .length 4, or 3 once .length of parent array is 2, having been set with two arrays containing four elements
var str = "ABCDEFGHIJKLMN";
function fnVerticalInvert(str, arr, res) {
if (!arr && !res) {
arr = []; res = "";
}
if (str) {
for (var i = 0; i < str.length; i += 4) {
arr.push([].slice.call(str.slice(i, i + 4)));
}
}
for (var i = 0; i < arr.length; i++) {
if (arr[i].length) {
res += arr[i].pop()
}
}
if (arr.some(function(curr) {return curr.length}))
return fnVerticalInvert(null, arr, res);
for (var i = 0, l = 4, j = 0, n = l - 1, k; i < l; i++, j += l) {
if (i === l / 2) k = j;
arr[i] = [].slice.call(res.slice(!k ? j : k, !k ? j + l : k + n));
if (k) k += n;
}
return {str: res, arr:arr};
};
var res = fnVerticalInvert(str);
console.log(res);
For example:
From
[x0,x1,x2,x3] - x0 compares itself to x1,x2 and x3.
x1 compares itself to x0, x2 and x3. And so on...
To
[x0,x1,x2,x3] - x0 compares itself to x1,x2 and x3.
x1 compares itself to x2 and x3 only.
x2 only compares itself to x3.
x3 does not need to do any comparison at all.
Essentially I'm looking to traverse an array one way only, with every element behind the current element ignored.
for (var i = 0; i < boids.length; i++) {
//boids is an array containing elements "boid"
var d = distSquared(this.position.x, this.position.y, boids[i].position.x, boids[i].position.y);
//get distance between current boid and all other boids in array.
//Looking to change this to get distance between current boid and all other boids infront of this element in the array.
if (boids[i] == this) { //if the boid being compared is to its own self, skip it.
continue;
}
}
How would I go about implementing such a structure?
What you are looking for is something like this:
var a = [1,2,3,4];
for(var i=0; i<a.length; i++){
for(var j=i+1; j<a.length; j++){
console.log(i,j)
}
}
You can use Array.prototype.filter() to exclude an element or elements from an array
var arr = [1,2,3,4,5];
for (var i = 0; i < arr.length; i++) {
var curr = arr[i];
var not = arr.filter(function(_, index) {
return index != i
});
console.log(`curr:${curr}`);
not.forEach(function(el) {
console.log(`${curr} is not ${el}`)
})
}
You might do like this. Shaping up how to compare is up to you.
var arr = [1,2,3,4,5],
result = arr.map((e,i,a) => a.slice(i+1).map(n => e-n));
console.log(result);
I used Array.prototype.indexOf() to solved my issue.
var j = boids.indexOf(this);
//get index of current element i array.
if (j < boids.length) {
// if the index is lower than the overall length of the array (e.g. check traverse the array to elements infront of the current element, ignoring those behind)
//Perform calculations here
j++; //increment j
}
I'm writing a simple snakes game using JavaScript and HTML5 canvas.
I have a Multidimensional array that hold snake block like this:
snake=[[1,1],[1,2]];
and set it on arrayMap using (snake.indexOf([i],[j])!=-1) then draw arrayMap on canvas.
for (var i = 0; i < blocksHeightCount; i++) {
for (var j = 0; j < blocksWidthCount; j++) {
if ((snake.indexOf(i,j)!=-1)||
(walls.indexOf(i,j)!=-1)||
(foods.indexOf(i,j)!=-1)) {
arrayMap[i][j]=1;
} else {
arrayMap[i][j]=0;
}
}
}
for (var i = 0; i < blocksHeightCount; i++) {
for (var j = 0; j < blocksWidthCount; j++) {
Block = arrayMap[i][j];
if (Block!=0){
ctx.fillStyle = (Block != 9) ? colors[Block]
: "#bdc3c7";
ctx.fillRect(j * cubeWidth, i * cubeHeight
, cubeWidth-.4,cubeHeight-.4);
}
}
}
the problem is indexOf isn't working when I set array on it!
It works fine when I set indexOf("i,j") but i need it to be array.
please help, thx
First solution : using Array.map
Each element of your arrays snake, walls and foods is an array with 2 elements. So to check if an [x,y] exists in one of the arrays you need a simple way to
compare two arrays [x1, y1] and [x2, y2]. Comparing arrays directly using the operator == will compare their references and not values (Thanks #Elena for the remarque). A way to compare values
would be to affect a hash to each array and compare hashes. By hash I mean a number which is unique for each array of type [x, y]. That could be x * blocksWidthCount + y
in your case and the code will be :
function getHash(x, y){
return x * blocksWidthCount + y;
}
var blockHashes = snake.concat(walls).concat(foods).map(function(cell) {
return getHash(cell[0], cell[1]);
}); // hashes of all blocks in one array
for (var i = 0; i < blocksHeightCount; i++) {
for (var j = 0; j < blocksWidthCount; j++) {
if (blockHashes.indexOf(getHash(i, j)) != -1) {
arrayMap[i][j]=1;
} else {
arrayMap[i][j]=0;
}
}
}
Second Solution Changing the way we see things
Instead of looping over all cells and verifying if every single cell is a block which gives a complexity of O(N * M) (N number of cells and M number of blocks).
We can do better simply by supposing that there is no block and then loop over blocks and mark them as blocks which is in O(N + M) !
function markBlock(cell){
arrayMap[cell[0]][cell[1]] = 1;
}
for (var i = 0; i < blocksHeightCount; i++)
for (var j = 0; j < blocksWidthCount; j++)
arrayMap[i][j] = 0;
snake.forEach(markBlock);
walls.forEach(markBlock);
foods.forEach(markBlock);
My goal is to make a randomly generated 2D Array in Javascript, that has an X amount of the same one character value while the rest of the values are equal to another character.
In this example, there are 10 rows and 10 columns for the 2D Array. 20 out of the possible 100 values of the Array should be equal to 'Y' (for yes) and the 80 others should be 'N' (for no). I want the 'Y's to be randomly placed all over the Array, and I absolute need exactly 20 of them to be 'Y's and the rest 'N's.
I had a less efficient way before, and I thought to try this approach, where after I define the Array, I make the first X amount of values a 'Y' and then the rest all 'N's. Then I shuffle the array, (using the shuffle from the underscore library) so that the 'Y's are all spread out randomly everywhere.
Is this an efficient way of getting what I need done? Are there any better solutions? I tried making a JSFiddle with my example, but the site appears to be down at the moment.
(I was unable to test my code yet to see if the shuffle worked correctly on my 2D array)
var rows = 10;
var cols = 10;
var elements = 20;
//Define Empty Array
var test = new Array(rows);
for (var k = 0; k < rows; k++)
{
test[k] = Array(cols);
}
var i = 1;
for (var x = 0; x < rows; x++)
{
for (var y = 0; y < cols; y++)
{
if (i <= elements)
{
test[x][y] = "Y";
}
else
{
test[x][y] = "N";
}
}
}
//Shuffle all those values so they're no longer in order
var shuffledTest = _.shuffle(test);
//Print in rows
for (var x = 0; x < rows; x++)
{
console.log(shuffledTest[x]);
}
A very simple solution is to first create an array, fill it with a number of "N"s, insert the "Y"s at random indexes, and then finally splitting it into the 2-dimensional array that you want:
var tmpArr = [], // Temporary 1-dimensional array to hold all values
arr = [], // The final 2-dimensional array
rows = 10,
cols = 10,
elements = 20; // Number of "Y"s
// 1. Fill temporary array with "N"s
for (var i = 0; i < rows * cols - elements; i += 1) {
tmpArr.push("N");
}
// 2. Insert "Y"s at random indexes in the temporary array
for (var i = 0; i < elements; i += 1) {
var index = Math.round(Math.random() * (tmpArr.length + 1));
tmpArr.splice(index, 0, "Y");
}
// 3. Split temporary array into 10 seperate arrays
// and insert them into the final array
for (var i = 0; i < rows; i += 1) {
var row = tmpArr.slice(i * cols, (i + 1) * cols);
arr.push(row);
}
JSBin to illustrate: http://jsbin.com/luyacora/1/edit
You can try this solution, it uses underscores range to create a pair of arrays to use as iterators, though their values don't matter.
Play around with the randomizer function to get an even distribution of 'y's
JSBIN: http://jsbin.com/yaletape/1/
var rows = _.range(0, 10, 0);
var columns = _.range(0, 10, 0);
function randomizer(mult){
return Math.floor((Math.random()*mult)+1);
}
var y_count = 0;
var matrix = _.map(rows, function(){
return _.map(columns, function(v, i){
var value;
var y_allowed = randomizer(3);
var current_y_count = 0;
if(y_count < 20 && current_y_count < y_allowed){
var rand = randomizer(5);
if(rand > 4){
value = 'y';
current_y_count++;
y_count++;
}
}
if(!value){
value = 'n';
}
return value;
});
});
//The above could be simplified to
var matrix = _.range(0,10,0).map(function(){
return _.range(0,10,0).map(function(){
//put the logic code from above here
});
});
Maybe shuflle a 2D array is not the best way. As #Zeb mentioned, here is some code that fill random positions with the 'Y' value. After that, the other positions are filled with 'N'.
http://plnkr.co/edit/avyKfgsgOSdAkRa1WOsk
var arr = [];
var cols = 10;
var rows = 10;
var positions = rows*cols; // 100
var YQty = 10; // only 10 'Y' are needed
// 'Y' values.
for(i = 0; i < YQty; i++)
{
do
{
x = parseInt(Math.random() * cols);
y = parseInt(Math.random() * rows);
filled = false;
if (typeof(arr[x]) == "undefined")
{
arr[x] = [];
}
if (typeof(arr[x][y]) == "undefined")
{
arr[x][y] = 'Y';
filled = true;
}
}
while (!filled);
}
// 'N' values.
for (x = 0; x < cols; x++)
{
if (typeof(arr[x]) == "undefined")
{
arr[x] = [];
}
for (y = 0; y < rows; y++)
{
if (arr[x][y] != 'Y')
{
arr[x][y] = 'N';
}
}
}
Shuffling the multidimensional array is not the best approach. Seeing as any sort is worse than linear time complexity. The easiest solution would be to create your multidimensional array and then set each index value to the char you want the 'rest' of the values to be. Then for 1 -> the number of other char value choose a random index and set that to the char.
Note: If the randomly picked spot has already been changed you need to choose a new one to make sure you have the right amount at the end.
I have an array something like this
a=[
[1,6,6,8,0],
[7,3,2,6],
[7,3,2]
]
Here, I need to find the maximum length of the inside array. And I need to replace other with '0'.
For example the first array has 5 elements, which is the maximum length, so I need to put '0' to the 2nd and 3rd array till reaches the maximum lengh (i.e 5). What should I do here?
var max = Math.max.apply(null, a.map(function(i) {
return i.length;
}));
var new_arr = a.map(function(i) {
var pad = max - i.length;
while (pad--) {
i.push(0);
}
return i;
});
http://jsfiddle.net/Mgfnf/1/
As you said, your task is split into two:
Find the maximum array length
Pad the arrays to meet that length
Both #1 and #2 can be achieved through a simple loop through the outer array. In the first, we keep a variable (let's call it maxLength) which will hold, yes, our max length.
var maxLength = 0;
//walk through the array
for (var i = 0; i < a.length; i += 1) {
//choose the larger
maxLength = Math.max(maxLength, a[i].length);
}
Now that we have the size we wish to expand to, we go over the outer loop, and on each sub-array, we push 0s until the lengths match:
//walk through the array
for (var j = 0; j < a.length; j += 1) {
//the length will increase by 1 on each push
while (a[j].length < maxLength) {
a[j].push(0);
}
}
And that's it.
try this
var i = 0,
max = 0;
while(i<a.length){ // find max
max = a[i].length > max ? a[i].length : max;
i++;
}
i = 0;
while(i<a.length){
for(var j=a[i].length;j<max;j++){
a[i][j] = 0;
}
i++;
}