Hi there fellow coders,
I'm looking to find a way to fill a pre-built dynamic blank table with numbering (and colouring if possible) like so:
As you can see the numbering is ascending order diagonally. I know there's probably some way to calculate the number based on the tables td index but can't quite figure out how to do that for every column diagonally. Any help would be appreciated.
Update: Ok back from my Holidays. Thanks to all you clever people for your replies. As I'm sure you've all had to experience the pain in the neck clients can be, I've been told the spec has changed(again). This being the case I've had to put the grid/matrix into a database and output using a pivot table. Every square has to be customizable color-wise.
Nothing is going to waste though I have learnt quite a few nifty new javascript/jquery tricks from your responses I didn't know about before, so thanks, and I'll be sure to pay it forward :)
Here's what I came up with in the end.
Given you said "colouring if possible" I'll provide an example solution that doesn't do colours quite the way you want (it does it in a way that I found easier to code and more attractive to look at) but which does handle all the numbering correctly for varying table sizes.
The function below assumes the table already exists; in this demo I've included code that generates a table to whatever size you specify and then calls the function below to do the numbering and colours.
function numberDiagonally(tableId) {
var rows = document.getElementById(tableId).rows,
numRows = rows.length,
numCols = rows[0].cells.length,
sq = numRows + numCols - 1,
d, x, y,
i = 1,
dc,
c = -1,
colors = ["green","yellow","orange","red"];
diagonalLoop:
for (d = 0; d < sq; d++) {
dc = "diagonal" + d;
for (y = d, x = 0; y >= 0; y--, x++) {
if (x === numCols)
continue diagonalLoop;
if (y < numRows)
$(rows[y].cells[x]).html(i++).addClass(dc);
}
}
for (d = 0; d < sq; d++)
$(".diagonal" + d).css("background-color", colors[c=(c+1)%colors.length]);
}
Demo: http://jsfiddle.net/7NZt3/2
The general idea I came up with was to imagine a square twice as big as whichever of the x and y dimensions is bigger and then use a loop to create diagonals from the left edge of that bounding square going up and to the right - i.e., in the order you want the numbers. EDIT: Why twice as big as longer side? Because that's the first thing that came into my head when I started coding it and it worked (note that the variable i that holds the numbers that get displayed is not incremented for the imaginary cells). Now that I've had time to think, I realise that my sq variable can be set precisely to one less than the number of rows plus the columns - a number that ends up rather smaller for non-square tables. Code above and fiddle updated accordingly.
Note that the background colours could be set directly in the first loop, but instead I opted to assign classes and set the loops for each class later. Seemed like a good idea at the time because it meant individual diagonals could be easily selected in jQuery with a single class selector.
Explaining exactly how the rest works is left as an exercise for the reader...
UPDATE - this version does the colouring more like you asked for: http://jsfiddle.net/7NZt3/1/ (in my opinion not as pretty, but each to his own).
This fiddle populates an existing table with numbers and colors. It is not limited to being a 5x5 table. I didn't understand the logic of 15 being orange rather than yellow, so I simply grouped the diagonal cells into color regions.
// we're assuming the table exists
var $table = $('table'),
// cache the rows for quicker access
$rows = $table.find('tr'),
// determine number of rows
_rows = $rows.length,
// determine number of cells per row
_cols = $rows.first().children().length,
// determine total number of cells
max = _rows * _cols,
// current diagonal offset (for coloring)
d = 1,
// current row
r = 0,
// current cell
c = 0;
for (var i=1; i <= max; i++) {
// identify and fill the cell we're targeting
$rows.eq(r).children().eq(c)
.addClass('d' + d)
.text(i);
if (i < max/2) {
// in the first half we make a "line-break" by
// moving one row down and resetting to first cell
if (!r) {
r = c + 1;
c = 0;
d++;
continue;
}
} else {
// in the second half our "line-break" changes to
// moving to the last row and one cell to the right
if (c + 1 == _cols) {
c = 1 + r;
r = _rows -1;
d++;
continue;
}
}
r--;
c++;
}
Here's a jsFiddle that does what you asked for - http://jsfiddle.net/jaspermogg/MzNr8/8/
I took the liberty of making it a little bit user-customisable; it's interesting to see how long it takes the browser to render a 1000x1000 table using this method :-D
Assuming that each cell has an id of [column]x[row], here are teh codez for how to fill in the numbers of a square table of side length sidelength.
//populate the cells with numbers according to the spec
function nums(){
var xpos = 0
var ypos = 0
var cellval = 1
for(i=0;i<2*sidelength;i++){
if(i >= sidelength){
ypos = sidelength - 1
xpos = 1 + i - sidelength
$('td#' + xpos + 'x' + ypos).text(cellval)
cellval = cellval + 1
while(xpos + 1 < sidelength){
ypos = ypos-1
xpos = xpos+1
$('td#' + xpos + 'x' + ypos).text(cellval)
cellval = cellval + 1
}
} else {
ypos = i
xpos = 0
$('td#' + xpos + 'x' + ypos).text(cellval)
cellval = cellval + 1
while(!(ypos-1 < 0)){
ypos = ypos-1
xpos = xpos+1
$('td#' + xpos + 'x' + ypos).text(cellval)
cellval = cellval + 1
}
}
}
}
And here they are for how to colour the bugger.
// color the cells according to the spec
function cols(){
if(+$('td#0x0').text() === 99){
return false
} else {
$('td').each(function(index, element){
if(+$(this).text() > 22)
{
$(this).attr("bgcolor", "red")
}
if(+$(this).text() <= 22)
{
$(this).attr("bgcolor", "orange")
}
if(+$(this).text() <= 14)
{
$(this).attr("bgcolor", "yellow")
}
if(+$(this).text() <= 6)
{
$(this).attr("bgcolor", "green")
}
})
}
}
Enjoy, eh? :-)
Related
The question is as follows:
The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
P A H N
A P L S I I G
Y I R
And then read line by line: "PAHNAPLSIIGYIR"
Write the code that will take a string and make this conversion given a number of rows:
string convert(string s, int numRows);
Example 1:
Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"
Example 2:
Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:
P I N
A L S I G
Y A H R
P I
I have written the following code, but I am stuck in terms of how to flag the row as one time to be downward moving, where I increment the start row, but when it's zigzagging back to the top, it should be decremented. I am unable to figure out the logic to make this work without affecting the downward movement. Any help would be appreciated.
const convert = (s, numRows) => {
let startRow = 0
let endRow = numRows - 1
let startColumn = 0
let endColumn = Math.floor((s.length / 2) - 1)
s = s.split('')
let results = []
// to setup the columns
for (let i = 0; i < numRows; i++) {
results.push([])
}
while (startRow <= endRow && startColumn <= endColumn && s.length) {
for (let i = startRow; i <= endRow; i++) {
results[i][startColumn] = s.shift()
}
for (let i = endRow - 1; i >= startRow; i--) {
results[i][startColumn + 1] = s.shift()
startColumn++
}
//this line seems to be the issue
startRow++
}
return results
}
console.log(convert('PAYPALISHIRING', 4))
I rewrote your while loop as follows where I simply walk a "zigzag" pattern! Hopefully, it is simple enough to understand.
let c=0, row=0,col=0, down=0;
while(c<s.length) {
results[row][col]=s[c];
if(down==0) { // moving down
row++;
if(row==numRows) {
down = 1;
col++;
row-=2;
}
} else { // moving up
row--;
col++;
if(row==0) {
down=0;
}
}
c++;
}
Ps. Above code does not handle numRows < 3 so you have to manage them before this loop.
My precalculus is a little rusty, but the logic behind this problem seems like a sine wave. I made a math error somewhere in creating the sin equation that prevents this from working (r never equals c with the current paramaters), but hopefully this will help if this is the direction you choose to go in.
/*If x-axis is position in string, and y-axis is row number...
n=number of rows
Equation for a sin curve: y = A sin(B(x + C)) + D
D=vertical shift (y value of mid point)
D=median of 1 and n
n: median:
1 1
2 1.5
3 2
4 2.5
5 3
6 3.5
7 4
median=(n+1)/2
D=(n+1)/2
A=amplitude (from the mid-point, how high does the curve go)
median + amplitude = number of rows
amplitude = number of rows - median
A=n-D
C=phase shift
Phase shift for a sin curve starting at its lowest point: 3π/2
(so at time 1, row number is 1, and curve goes up from there)
C=3π/2
Period is 2π/B
n p
3 4
4 6
5 8
6 10
period=2(n-1)
2(n-1)=2π/B
B(2(n-1)=2π
B=2π/2(n-1)
B=π/(n-1)
Variables:
s = string
n = number of rows
c = current row number being evaluated
p = position in string
r = row number
*/
var output='';
function convert(s,n) {
D=(n+1)/2
A=n-D
C=(3*Math.PI)/2
B=Math.PI/(n-1)
for (c=1;c<=n;c++) { //loop from 1st row to number of rows
for (p=1;p<=s.length;p++) { //loop from 1st to last character in string
r=A*Math.sin(B*(p+C))+D //calculate the row this character belongs in
if (r==c) { output+= s.charAt(r) } //if the character belongs in this row, add it to the output variable. (minus one because character number 1 is at position 0)
}}
//do something with output here
}
var image = new SimpleImage("lena.png");
var col = [];
var uniqcol = [];
for (var px of image.values()){
col.push([px.getRed,px.getGreen,px.getBlue]);
if(uniqcol.includes([px.getRed +- 1, px.getGreen +- 1, px.getBlue +- 1]) ){
print('not unique');
}else{
uniqcol.push([px.getRed,px.getGreen,px.getBlue]);
}
}
I would like to count the number of unique pixels within an image. A unique pixel being one which RGB values are not within 1 to anothers pixels. I have the above code but it does not work. I think the issue that I am having is with checking that the RGB values are either +1 or -1 from the selected pixel px value. If a unique pixel is found, id like to add to the the uniqcol array. Is there any other way to count the unique pixels, or a way to check that the RGB values are within 1 from the selected px value?
Thanks.
This tests each component to see if it's within 1 by subtracting the two, taking the absolute value, and checking if it's less than 2.
This is probably super inefficient. For each pixel you're iterating a potentially massive array until you get a match, or worst case, you don't find a match.
var image = new SimpleImage("lena.png");
var col = [];
var uniqcol = [];
for (var px of image.values()){
var found = uniqcol.find(function (el) {
return
Math.abs(el[0] - px.getRed) < 2 &&
Math.abs(el[1] - px.getGreen) < 2 &&
Math.abs(el[2] - px.getBlue) < 2;
});
if (!found) {
uniqcol.push([px.getRed,px.getGreen,px.getBlue]);
} else {
print('not unique');
}
}
Here's another approach that uses memoization. It should be a lot faster at the expense of storing a separate lookup structure.
Edit - I deleted this approach because it can fail. It's probably possible to do but quite tricky.
You need to check for all the different pixel values, putting +- will not match a range of values. .includes() looks for exact matches.
for (var px of image.values()) {
col.push([px.getRed,px.getGreen,px.getBlue]);
var found = false;
for (dRed of [-1, 0, +1]) {
for (dGreen of [-1, 0, +1]) {
for (dBlue of [-1, 0, +1]) {
if (uniqcol.includes([px.getRed + dRed, px.getGreen + dGreen, px.getBlue + dBlue]) {
found = true;
print("not unique");
break;
}
}
if (found) {
break;
}
if (found) {
break;
}
}
if (!found) {
uniqcol.push([px.getRed,px.getGreen,px.getBlue]);
}
}
This is probably not a very efficient way to do it, since it will search the entire image 9 times for each pixel. It would probably be better to loop through all the pixels, testing if all the colors are within a range of the current pixel:
if (px.getRed >= curPixel.getRed - 1 && px.getRed <= curPixel.getRed + 1 &&
px.getGreen >= curPixel.getGreen - 1 && px.getGreen <= curPixel.getGreen + 1 &&
px.getBlue >= curPixel.getBlue - 1 && px.getBlue <= curPixel.getBlue + 1)
A really efficient algorithm would involve sorting all the pixels (nested arrays of red, blue, and green values would be a good structure), then searching this. But that's more a topic for CodeReview.stackexchange.com.
I have a 2D array, something like the following:
[1.11, 23]
[2.22, 52]
[3.33, 61]
...
Where the array is ordered by the first value in each row.
I am trying to find a value within the array that is close to the search value - within a certain sensitivity. The way this is set up, and the value of the sensitivity, ensure only one possible match within the array.
The search value is the current x-pos of the mouse. The search is called on mousemove, and so is being called often.
Originally I had the following (using a start-to-end for loop):
for(var i = 0; i < arr.length; i++){
if(Math.abs(arr[i][0] - x) <= sensitivity){
hit = true;
break;
}
}
And it works like a charm. So far, I've only been using small data sets, so there is no apparent lag using this method. But, eventually I will be using much larger data sets, and so want to switch this to a Binary Search:
var a = 0;
var b = arr.length - 1;
var c = 0;
while(a < b){
c = Math.floor((a + b) / 2);
if(Math.abs(arr[c][0] - x) <= sensitivity){
hit = true;
break;
}else if(arr[c][0] < x){
a = c;
}else{
b = c;
}
}
This works well, for all of 2 seconds, and then it hangs to the point where I need to restart my browser. I've used binary searches plenty in the past, and cannot for the life of me figure out why this one isn't working properly.
EDIT 1
var sensitivity = (width / arr.length) / 2.001
The points in the array are equidistant, and so this sensitivity ensures that there is no ambiguous 1/2-way point in between two arr values. You are either in one or the other.
Values are created dynamically at page load, but look exactly like what I've mentioned above. The x-values have more significant figures, and the y values are all over the place, but there is no significant difference between the small sample I provided and the generated one.
EDIT 2
Printed a list that was dynamically created:
[111.19999999999999, 358.8733333333333]
[131.4181818181818, 408.01333333333326]
[151.63636363636363, 249.25333333333327]
[171.85454545454544, 261.01333333333326]
[192.07272727272726, 298.39333333333326]
[212.29090909090908, 254.2933333333333]
[232.5090909090909, 308.47333333333324]
[252.72727272727272, 331.1533333333333]
[272.94545454545454, 386.1733333333333]
[293.16363636363633, 384.9133333333333]
[313.3818181818182, 224.05333333333328]
[333.6, 284.53333333333325]
[353.81818181818187, 278.2333333333333]
[374.0363636363637, 391.63333333333327]
[394.25454545454556, 322.33333333333326]
[414.4727272727274, 300.9133333333333]
[434.69090909090926, 452.95333333333326]
[454.9090909090911, 327.7933333333333]
[475.12727272727295, 394.9933333333332]
[495.3454545454548, 451.27333333333326]
[515.5636363636366, 350.89333333333326]
[535.7818181818185, 308.47333333333324]
[556.0000000000003, 395.83333333333326]
[576.2181818181822, 341.23333333333323]
[596.436363636364, 371.47333333333324]
[616.6545454545459, 436.9933333333333]
[636.8727272727277, 280.7533333333333]
[657.0909090909096, 395.4133333333333]
[677.3090909090914, 433.21333333333325]
[697.5272727272733, 355.09333333333325]
[717.7454545454551, 333.2533333333333]
[737.963636363637, 255.55333333333328]
[758.1818181818188, 204.7333333333333]
[778.4000000000007, 199.69333333333327]
[798.6181818181825, 202.63333333333327]
[818.8363636363644, 253.87333333333328]
[839.0545454545462, 410.5333333333333]
[859.272727272728, 345.85333333333324]
[879.4909090909099, 305.11333333333323]
[899.7090909090917, 337.8733333333333]
[919.9272727272736, 351.3133333333333]
[940.1454545454554, 324.01333333333326]
[960.3636363636373, 331.57333333333327]
[980.5818181818191, 447.4933333333333]
[1000.800000000001, 432.3733333333333]
As you can see, it is ordered by the first value in each row, ascending.
SOLUTION
Changing the condition to
while(a < b)
and
var b = positions.length;
and
else if(arr[c][0] < x){
a = c + 1;
}
did the trick.
Your binary search seems to be a bit off: try this.
var arr = [[1,0],[3,0],[5,0]];
var lo = 0;
var hi = arr.length;
var x = 5;
var sensitivity = 0.1;
while (lo < hi) {
var c = Math.floor((lo + hi) / 2);
if (Math.abs(arr[c][0] - x) <= sensitivity) {
hit = true;
console.log("FOUND " + c);
break;
} else if (x > arr[c][0]) {
lo = c + 1;
} else {
hi = c;
}
}
This is meant as a general reference to anyone implementing binary search.
Let:
lo be the smallest index that may possibly contain your value,
hi be one more than the largest index that may contain your value
If these conventions are followed, then binary search is simply:
while (lo < hi) {
var mid = (lo + hi) / 2;
if (query == ary[mid]) {
// do stuff
else if (query < ary[mid]) {
// query is smaller than mid
// so query can be anywhere between lo and (mid - 1)
// the upper bound should be adjusted
hi = mid;
else {
// query can be anywhere between (mid + 1) and hi.
// adjust the lower bound
lo = mid + 1;
}
I don't know your exact situation, but here's a way the code could crash:
1) Start with an array with two X values. This array will have a length of 2, so a = 0, b = 1, c = 0.
2) a < b, so the while loop executes.
3) c = floor((a + b) / 2) = floor(0.5) = 0.
4) Assume the mouse is not within sensitivity of the first X value, so the first if branch does not hit.
5) Assume our X values are to the right of our mouse, so the second if branch enters. This sets a = c, or 0, which it already is.
6) Thus, we get an endless loop.
I've been trying to implement a recursive backtracking maze generation algorithm in javascript. These were done after reading a great series of posts on the topic here
While the recursive version of the algorithm was a no brainer, the iterative equivalent has got me stumped.
I thought I understood the concept, but my implementation clearly produces incorrect results. I've been trying to pin down a bug that might be causing it, but I am beginning to believe that my problems are being caused by a failure in logic, but of course I am not seeing where.
My understanding of the iterative algorithm is as follows:
A stack is created holding representations of cell states.
Each representation holds the coordinates of that particular cell, and a list of directions to access adjacent cells.
While the stack isn't empty iterate through the directions on the top of the stack, testing adjacent cells.
If a valid cell is found place it at the top of the stack and continue with that cell.
Here is my recursive implementation ( note: keydown to step forward ): http://jsbin.com/urilan/14
And here is my iterative implementation ( once again, keydown to step forward ): http://jsbin.com/eyosij/2
Thanks for the help.
edit: I apologize if my question wasn't clear. I will try to further explain my problem.
When running the iterative solution various unexpected behaviors occur. First and foremost, the algorithm doesn't exhaust all available options before backtracking. Rather, it appears to be selecting cells at a random when there is one valid cell left. Overall however, the movement doesn't appear to be random.
var dirs = [ 'N', 'W', 'E', 'S' ];
var XD = { 'N': 0, 'S':0, 'E':1, 'W':-1 };
var YD = { 'N':-1, 'S':1, 'E':0, 'W': 0 };
function genMaze(){
var dirtemp = dirs.slice().slice(); //copies 'dirs' so its not overwritten or altered
var path = []; // stores path traveled.
var stack = [[0,0, shuffle(dirtemp)]]; //Stack of instances. Each subarray in 'stacks' represents a cell
//and its current state. That is, its coordinates, and which adjacent cells have been
//checked. Each time it checks an adjacent cell a direction value is popped from
//from the list
while ( stack.length > 0 ) {
var current = stack[stack.length-1]; // With each iteration focus is to be placed on the newest cell.
var x = current[0], y = current[1], d = current[2];
var sLen = stack.length; // For testing whether there is a newer cell in the stack than the current.
path.push([x,y]); // Store current coordinates in the path
while ( d.length > 0 ) {
if( stack.length != sLen ){ break;}// If there is a newer cell in stack, break and then continue with that cell
else {
var cd = d.pop();
var nx = x + XD[ cd ];
var ny = y + YD[ cd ];
if ( nx >= 0 && ny >= 0 && nx < w && ny < h && !cells[nx][ny] ){
dtemp = dirs.slice().slice();
cells[nx][ny] = 1;
stack.push( [ nx, ny, shuffle(dtemp) ] ); //add new cell to the stack with new list of directions.
// from here the code should break from the loop and start again with this latest addition being considered.
}
}
}
if (current[2].length === 0){stack.pop(); } //if all available directions have been tested, remove from stack
}
return path;
}
I hope that helps clear up the question for you. If it is still missing any substance please let me know.
Thanks again.
I'm not very good in javascript, but I try to implement your recursive code to iterative. You need to store For index on stack also. So code look like:
function genMaze(cx,cy) {
var dirtemp = dirs; //copies 'dirs' so its not overwritten
var path = []; // stores path traveled.
var stack = [[cx, cy, shuffle(dirtemp), 0]]; // we also need to store `for` indexer
while (stack.length > 0) {
var current = stack[stack.length - 1]; // With each iteration focus is to be placed on the newest cell.
var x = current[0], y = current[1], d = current[2], i = current[3];
if (i > d.length) {
stack.pop();
continue;
}
stack[stack.length - 1][3] = i + 1; // for next iteration
path.push([x, y]); // Store current coordinates in the path
cells[x][y] = 1;
var cd = d[i];
var nx = x + XD[cd];
var ny = y + YD[cd];
if (nx >= 0 && ny >= 0 && nx < w && ny < h && !cells[nx][ny]) {
dtemp = dirs;
stack.push([nx, ny, shuffle(dtemp), 0]);
}
}
return path;
}
Does this little code could also help ?
/**
Examples
var sum = tco(function(x, y) {
return y > 0 ? sum(x + 1, y - 1) :
y < 0 ? sum(x - 1, y + 1) :
x
})
sum(20, 100000) // => 100020
**/
function tco(f) {
var value, active = false, accumulated = []
return function accumulator() {
accumulated.push(arguments)
if (!active) {
active = true
while (accumulated.length) value = f.apply(this, accumulated.shift())
active = false
return value
}
}
}
Credits, explanations ans more infos are on github https://gist.github.com/1697037
Is has the benefit to not modifying your code, so it could be applied in other situations too. Hope that helps :)
I am working on looping through a few grid items and was hoping someone could help me find a way to figure out how to get to the items that I need and possibly I will better understand how to access the items in the grid. So here is the grid.
[0] [1] [2] [3] [4]
[5] [6] [7] [8] [9]
[10] [11] [12] [13] [14]
[15] [16] [17] [18] [19]
[20] [21] [22] [23] [24]
This is basically a 5x5 grid, however it could be any size but I just used this for the example. There are two ways I would like to loop through this. The first one being in this order:
0,1,2,3,4,9,14,19,24,23,22,21,20,15,10,5,6,7,8,13,18,17,16,11,12
Basically all that is doing is going around the outside starting from the top left. The next way I would want to loop through that is through the same exact values except in reverse order (basically inside out instead of outside in) and actually thinking about this now I could just loop through the first method backwards. If anyone can help me with this it would be great. I really want to learn more on how to loop through items in crazy arrangements like this.
This function
function n(i, w, h)
{
if (i < w)
return i;
if (i < w + h-1)
return w-1 + (i-w+1)*w;
if (i < w + h-1 + w-1)
return w-1 + (h-1)*w - (i - (w + h-1 - 1));
if (i < w + h-1 + w-1 + h-2)
return (h - (i - (w + w-1 + h-1 - 2)))*w;
var r = n(i - (w-1)*2 - (h-1)*2, w-2, h-2);
var x = r % (w-2);
var y = Math.floor(r / (w-2));
return (y+1)*w + (x+1);
}
accepts as input
i: Index of the item you're looking for
w: Width of the grid
h: Height of the grid
and returns the corresponding element of the grid assuming that clock-wise spiral traversal.
The implementation simply checks if we're on the top side (i<w), on the downward right side (i<w+h-1) and so on and for these cases it computes the cell element explicitly.
If we complete one single trip around the spiral then it calls recursively itself to find the element in the inner (w-2)*(h-2) grid and then extracts and adjusts the two coordinates considering the original grid size.
This is much faster for big grids than just iterating and emulating the spiral walk, for example computing n(123121, 1234, 3012) is immediate while the complete grid has 3712808 elements and a walk of 123121 steps would require quite a long time.
You just need a way to represent the traversal pattern.
Given an NxM matrix (e.g. 5x5), the pattern is
GENERAL 5x5
------- -------
N right 5
M-1 down 4
N-1 left 4
M-2 up 3
N-2 right 3
M-3 down 2
N-3 left 2
M-4 up 1
N-4 right 1
This says move 5 to the right, 4 down, 4 left, 3 up, 3 right, 2 down, 2 left, 1 up, 1 right. The step size shifts after each two iterations.
So, you can track the current "step-size" and the current direction while decrementing N, M until you reach the end.
IMPORTANT: make sure to write down the pattern for a non-square matrix to see if the same pattern applies.
Here's the "walking" method. Less efficient, but it works.
var arr = new Array();
for(var n=0; n<25; n++) arr.push(n);
var coords = new Array();
var x = 0;
var y = 0;
for(var i=0; i<arr.length; i++) {
if( x > 4 ) {
x = 0;
y++;
}
coords[i] = {'x': x, 'y': y};
x++;
}
// okay, coords contain the coordinates of each item in arr
// need to move along the perimeter until a collision, then turn.
// start at 0,0 and move east.
var dir = 0; // 0=east, 1=south, 2=west, 3=north.
var curPos = {'x': 0, 'y': 0};
var resultList = new Array();
for(var x=0; x<arr.length; x++) {
// record the current position in results
var resultIndex = indexOfCoords(curPos, coords);
if(resultIndex > -1) {
resultList[x] = arr[resultIndex];
}
else {
resultList[x] = null;
}
// move the cursor to a valid position
var tempCurPos = movePos(curPos, dir);
var outOfBounds = isOutOfBounds(tempCurPos, coords);
var itemAtTempPos = arr[indexOfCoords(tempCurPos, coords)];
var posInList = resultList.indexOf( itemAtTempPos );
if(outOfBounds || posInList > -1) {
dir++;
if(dir > 3) dir=0;
curPos = movePos(curPos, dir);
}
else {
curPos = tempCurPos;
}
}
/* int indexOfCoords
*
* Searches coordList for a match to myCoords. If none is found, returns -1;
*/
function indexOfCoords(myCoords, coordsList) {
for(var i=0; i<coordsList.length; i++) {
if(myCoords.x == coordsList[i].x && myCoords.y == coordsList[i].y) return i;
}
return -1;
}
/* obj movePos
*
* Alters currentPosition by incrementing it 1 in the direction provided.
* Valid directions are 0=east, 1=south, 2=west, 3=north
* Returns the resulting coords as an object with x, y.
*/
function movePos(currentPosition, direction) {
var newPosition = {'x':currentPosition.x, 'y':currentPosition.y};
if(direction == 0) {
newPosition.x++;
}
else if(direction == 1) {
newPosition.y++;
}
else if(direction == 2) {
newPosition.x--;
}
else if(direction == 3) {
newPosition.y--;
}
return newPosition;
}
/* bool isOutOfBounds
*
* Compares the x and y coords of a given position to the min/max coords in coordList.
* Returns true if the provided position is outside the boundaries of coordsList.
*
* NOTE: This is just one, lazy way of doing this. There are others.
*/
function isOutOfBounds(position, coordsList) {
// get min/max
var minx=0, miny=0, maxx=0, maxy=0;
for(var i=0; i<coordsList.length; i++) {
if(coordsList[i].x > maxx) maxx = coordsList[i].x;
else if(coordsList[i].x < minx) minx = coordsList[i].x;
if(coordsList[i].y > maxy) maxy = coordsList[i].y;
else if(coordsList[i].y < miny) miny = coordsList[i].y;
}
if(position.x < minx || position.x > maxx || position.y < miny || position.y > maxy) return true;
else return false;
}
This will move through the grid in a direction until it hits the bounds or an item already in the results array, then turn clockwise. It's pretty rudimentary, but I think it would do the job. You could reverse it pretty simply.
Here's a working example: http://www.imakewidgets.com/test/walkmatrix.html