How to select/group together concentric elements of a 2d matrix?
Following is the code for creating a 2d array.
function Create2DArray(rows) {
var arr = [];
for (var i=0;i<rows;i++) {
arr[i] = [];
}
return arr;
}
imagine an element at the centre of the matrix with address i,j.
How can I separate out the cells in the following manner?
such that only the elements highlighted are grouped together / separated out.
It would be great if a mathematical algorithm, rather than looping a lot. (as this will be more efficient)
NOTE
The above approach works only if the matrix is of odd size i.e. 3 x 3 , 5 x 5, 7 x 7 etc... So if there is a way to implement this for even numbered matrices greater than 3, please suggest that too (for matrices greater than or equal to 4 there is a subset of odd sized matrix which we can use of to implement the above algorithm & discard the remaining)..
Please use math/ indices to cut out concentric cells, rather than using traditional iterators, or anything similar, as this allows faster calculations (...as you can start from the central address (or index) and decrement/increment out) (if possible, if not possible you can suggest any approach)
the possible output can be either an array or an object of corresponding cell addresses
like spanning like this:
index/key 1-> level 1 concentric cells
index/key 2-> level 2 concentric cells
index/key 3-> level 3 concentric cells
...
or possible output can be just a way of traversing each successive levels rather than grouping out the concentric cells; Like if concentric cells in level one is traversed the say alert(level 1)... then after level two alert(level 2) and so on...
You could start from the center point (can be any point with row/column) and then you loop each row and column and do the checking based on the space from that center point.
function genMatrix(rows, cols) {
return Array.from(Array(rows), () => {
return Array(cols).fill(0)
})
}
function select(matrix, [cRow, cCol], space = 0) {
for (let i = 0; i < matrix.length; i++) {
for (let j = 0; j < matrix[i].length; j++) {
if (space === 0) {
// for center point
if (i == cRow && j == cCol) {
matrix[i][j] = 3
}
} else {
// for horizontal points
if (i == cRow - space || i == cRow + space) {
if (j <= cCol + space && j >= cCol - space) {
matrix[i][j] = 3
}
}
// for vertical points
if (j == cCol - space || j == cCol + space) {
if (i <= cRow + space && i >= cRow - space) {
matrix[i][j] = 3
}
}
}
}
}
return matrix
}
const center = [3, 3];
const result = [
select(genMatrix(7, 7), center),
select(genMatrix(7, 7), center, 1),
select(genMatrix(7, 7), center, 2),
select(genMatrix(7, 7), center, 3)
]
// for the demo
result.forEach(matrix => {
matrix.forEach(row => console.log(JSON.stringify(row)))
console.log('-'.repeat(20))
})
Related
given matrix / 2D array
1 2 3
4 5 6
7 8 9
we have to print
7 4 1 5 9 6 3 in javascript without using inbuilt functions
Here's a hint.
Assuming your matrix is presented as a 1-d array of integers:
m = [1,2,3,4,5,6,7,8,9];
which is logically assumed to be a "square" size (1x1, 2x2, 3x3, 4x4, etc...)
So the above array is logically assumed to be:
1 2 3
4 5 6
7 8 9
Let's use a helper function to help us obtain any that value at any given (x,y) position. With (0,0), being the top left value of the matrix
let getVal = (m, col, row) => {
let rowlen = Math.round(Math.sqrt(m.length));
let index = row * rowlen + col;
return m[index];
}
Hence, getVal(m, 0, 0) returns 1 for the top left corner and getVal(m,2,2) returns 9 for the value in the bottom right corner.
Now with that little helper function provided, do you think you can implement the three for loops for "going up from the bottom left", "diagonal from top left to bottom right", and "going up from the bottom right"?
var output= "";
var length = matrix.length;
// console.log(length)
for(var i = length - 1; i >= 0; i--){
output= output+ matrix[i][0] + " ";
}
// console.log(bag);
for(var j = 1; j < length; j++){
output= output+ matrix[j][j] + " ";
}
// console.log(bag);
for(var k = length - 2; k >= 0; k--){
output= output+ matrix[k][length - 1] + " ";
}
console.log(output);
}
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.
Here is an interactive page describing the problem and an academic paper going over the mathematics.
The problem can be roughly described as follows.
Given an arbitrary-length array of boolean values representing n adjacent urinals, with values of true indicating occupied and values of false indicating vacant, how would you construct an algorithm to populate this array, given any configuration, while:
Maximizing the 'privacy' of each occupant by keeping one as far as possible from other urinators on either side.
Maintaining this privacy for as long as possible by ensuring the configuration becomes saturated at the last possible time.
Faced with multiple suboptimal choices, prioritizing urinals without an adjacent urinal on either side over a merely unoccupied adjacent urinal.
I marked this javascript for simplicity, but any code or pseudo-code would be fine.
var urinals = Array
.apply(null, new Array(n))
.map(Boolean.prototype.valueOf,false);
edit - found a related problem here:
Optimal Seating Arrangement Algorithm
As close as I have to a solution:
var urinalFinder = function(urinals){
var gaps = new Array(), last = null;
for(var i = 0; i < urinals.length; i++){
last = gaps.length ? gaps[gaps.length - 1] : 0;
if(last < 0 && !urinals[i] || last > 0 && !!urinals[i] || last == 0)
gaps.push(0); // push if new sequence of vacant or occupied
// negatives are occupied count & positives vacant count
gaps[gaps.length - 1] += !!urinals[i] ? -1 : 1;
}
// find the first index of the largest gap
var maxGapSize = Math.max.apply(Math, gaps),
maxGapGapsIdx = gaps.indexOf(maxGapSize),
isFirst = maxGapGapsIdx === 0,
isLast = maxGapGapsIdx === gaps.length - 1,
maxGapIdx = 0;
if(maxGapSize < 1) return false; // no gaps available
var gapPoint = maxGapSize > 3
? Math.ceil(maxGapSize / 3) // per xkcd suggestion
: isFirst && maxGapSize === 2
? 1
: isLast && maxGapSize === 2 ? 2 : Math.ceil(maxGapSize / 2);
// find where our chosen gap begins in input array
for(var i = 0; i < maxGapGapsIdx; i++)
maxGapIdx += Math.abs(gaps[i]);
var result = maxGapIdx + gapPoint - 1; // arrays are zero-indexed
return result;
};
For example, applied to filling an array of 9 vacant spaces will fill them like this:
var foo = [0,0,0,0,0,0,0,0,0]; // nine values
for(var i = 0; i < foo.length; i++)
foo[urinalFinder(foo)] = i+1;
[4, 6, 1, 7, 2, 8, 3, 9, 5]
Does not always produce optimal results (sometimes a different placement could allow saturation a few moves later) and does not favor end urinals, but does a pretty good job fanning values around and keeping a minimum buffer for just about as long as possible.
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