Develop Reference

How to push both sum of even and odd result from for loop into an array?

Use for loop to iterate from 0 to 100 and print the sum of all evens and the sum of all odds. Print sum of evens and sum of odds as array
Output: [2550, 2500]
let sumOfEven = 0;
let EvenOddArr = [];
for (let i = 0; i <= 100; i += 2) {
sumOfEven += i;
}
console.log(sumOfEven);
let sumOfOdd = 0;
for (let i = 1; i <= 100; i += 2) {
sumOfOdd += i;
}
console.log(sumOfOdd);
console.log(EvenOddArr);

You could take the remainder of two as index for the array.
const evenOddArr = [0, 0];
for (let i = 0; i <= 100; i++) evenOddArr[i % 2] += i;
console.log(evenOddArr);

You're nearly there - all you need is a couple of pushes
let sumOfEven = 0;
let EvenOddArr = [];
for (let i = 0; i <= 100; i += 2) {
sumOfEven += i;
}
EvenOddArr.push(sumOfEven)
let sumOfOdd = 0;
for (let i = 1; i <= 100; i += 2) {
sumOfOdd += i;
}
EvenOddArr.push(sumOfOdd)
console.log(EvenOddArr);

console.log(Array(101).fill().reduce((a,_,i)=>(a[i%2]+=i,a),[0,0]))

Here is an alternative for when you have studied JS a bit more
let sumArr = Array.from({ length: 101 })
.reduce((acc,_,i) => (acc[i % 2] += i, acc), [0, 0]);
console.log(sumArr);

An easy-to-understand version:
let sumOfEven = 0;
let sumOfOdd = 0;
for (let i = 0; i <= 100; i++) {
if (i % 2 === 0) {
sumOfEven += i;
} else {
sumOfOdd += i;
}
}
let evenOddArr = [sumOfEven, sumOfOdd];
console.log(evenOddArr);

Comparing numerical values of an array against eachother javascript

let arrayOfNumbers = [1, 2, 3, 4, 5]
What would be the best way to compare the numbers against each other ?
For instance, comparing 1 to 2 then 2 to 3 then 3 to four, and so on ?
function t(a) {
let t = 0
for (let i = 0; i < a.length; i++) {
if (a[t] > a[t + 1]) {
console.log('down')
} else if (a[t] < a[t + 1]) {
console.log('up')
} else if (a[t] === a[t + 1]) {
console.log('no change')
}
t++
}
}

You could start from index one and check the previous value.
function t(a) {
for (let i = 1; i < a.length; i++) {
if (a[i - 1] > a[i]) console.log('down');
else if (a[i - 1] < a[i]) console.log('up');
else console.log('no change');
}
}
t([0, 1, 3, 2, 4, 4, 2]);

If you want to pick up trend then you can do a linear regression.
function linearRegression(y, x) {
var lr = {};
var n = y.length;
var sum_x = 0;
var sum_y = 0;
var sum_xy = 0;
var sum_xx = 0;
var sum_yy = 0;
for (var i = 0; i < y.length; i++) {
sum_x += x[i];
sum_y += y[i];
sum_xy += (x[i] * y[i]);
sum_xx += (x[i] * x[i]);
sum_yy += (y[i] * y[i]);
}
lr['slope'] = (n * sum_xy - sum_x * sum_y) / (n * sum_xx - sum_x * sum_x);
lr['intercept'] = (sum_y - lr.slope * sum_x) / n;
lr['r2'] = Math.pow((n * sum_xy - sum_x * sum_y) / Math.sqrt((n * sum_xx - sum_x * sum_x) * (n * sum_yy - sum_y * sum_y)), 2);
return lr;
}
function find_next(arr) {
var y = [];
for (var i = 0; i < arr.length; i++) {
y.push(i);
}
var reg = linearRegression(y, arr)
// y = a*x + b
// x = (y-b)/a
var next = (arr.length - reg.intercept) / reg.slope
return next
}
console.log(find_next([1, 2, 3, 4, 5]))
console.log(find_next([0, 1, 3, 2, 4, 4, 2]))

Struggling with while loop in this algorithm

I am writing a program to calculate euclidean distance and then display the lines based, with the below code:
function discreteFrechet(X, Y) {
var M = X.length;
var N = Y.length;
var S = [
[],
[]
];
var backpointers = [
[],
[]
];
var backpaths = [];
var idx;
var path = [];
var paths;
var back = [
[],
[]
];
for (i = 0; i < M; i++) {
for (j = 0; j < N; j++) {
S[i][j] = 0;
backpointers[i][j] = 0;
}
} /* populates S array */
/*sanity check*/
S[0, 0] = euclidian(X, Y, 0, 0);
opt1 = [-1, 0];
opt2 = [0, -1];
opt3 = [-1, -1];
backpaths.push(opt1);
backpaths.push(opt2);
backpaths.push(opt3);
/*backpaths populated*/
for (i = 0; i < M; i++) {
for (j = 0; j < N; j++) {
options = [];
if (i != 0 || j != 0) {
if (i > 0) {
options[0] = S[i - 1, j];
}
if (j > 0) {
options[1] = S[i - 1, j];
}
if (i > 0 && j > 0) {
options[2] = S[i - 1, j];
}
idx = Math.min(options);
backpointers[i][j] = idx;
S[i][j] = Math.max(options[idx], euclidian(X, Y, i, j));
}
}
}
console.log(S);
paths = [
[M - 1, N - 1]
];
path = [
[],
[]
];
path.push(paths);
//Create "path"
i = M - 1;
j = N - 1;
count = 0;
while ((path[path.length - 1][0] != 0) || (path[0][path[1].length - 1] != 0)) {
back[0][1] = backpaths[backpointers[i], [j]];
i += back[0];
j += back[1];
path.push([i, j]);
if (count > 1000) {
console.log("too many loops");
break;
}
count += 1;
}
path.push([0, 0]);
path.reverse();
//returns bottleneck and the path
}
As I am testing, I am running into a problem with an infinite while loop (hence the break statement) any help or suggestions would be greatly appreciated! The goal is to append indicies into the path element, such that I can then take those path indicies and the bottleneck and use them to plot with d3.

How many times needed to Sort a number in an Array Javascript

I want to count how many times needed for an array to be sorted
var array = [4,2,3,1]
var yourCounter = 0;
for (var i = 0; i < array.length; i++) {
for (var j = 1; j < array.length-j; j++)
if (array[j - 1] > array[j]) {
yourCounter++;
} }
it will return 4 , it should be 5
but if I input array [1,2,3] will correctly return 0 , and if I input array [3,2,1] it will correctly return 3

You could take the given code and swap the values while counting.
for (int i = 0; i < n; i++) {
for (int j = 0; j < n - 1; j++) {
// Swap adjacent elements if they are in decreasing order
if (a[j] > a[j + 1]) {
swap(a[j], a[j + 1]);
}
}
}
var array = [4, 2, 3, 1],
counter = 0,
i, j, n = array.length;
for (i = 0; i < n; i++) {
for (j = 0; j < n - 1; j++) {
if (array[j] > array[j + 1]) {
[array[j + 1], array[j]] = [array[j], array[j + 1]];
++counter;
}
}
}
console.log(counter);
console.log(array);

I found the solution
var a = [4,2,3,1]
function sortArray(a){
let swapCount = 0;
let swapOccurred = true;
let index = 0;
while (swapOccurred == true && index < a.length){
swapOccurred == false;
if (a[index] > a[index+1]){
let holder = a[index]
a[index] = a[index+1];
a[index+1] = holder;
swapOccurred == true;
swapCount ++;
index = -1;
}
index ++
}
function countSwaps(a) {
let swapCount = 0;
[a, swapCount] = sortArray(a)
console.log(swapCount)
}
return [a, swapCount]
}

my attempt at kernel-based image processing with p5.js

var imagething;
var kernel = [[1, 2, 1], [2, 4, 2], [1, 2, 1]];
var sum_of_elements = 0;
var mult_factor = 0.0625;
var bias = 0;
function preload() {
imagething = loadImage('data/image.png');
}
function setup() {
createCanvas(1000, 1000);
imagething.loadPixels();
image(imagething, 0, 0);
makeBlurred(imagething, 1);
image(imagething, 512, 0);
}
function imageIndex(img, x, y) {
return 4 * (x + y * img.width);
}
function makeBlurred() {
for(var i = 0; i < kernel.length; i++) {
for(var j = 0; j < kernel.length; j++) {
sum_of_elements += kernel[i][j];
}
}
for(var x = 0; x < imagething.width; x++) {
for(var y = 0; y < imagething.height; y++) {
var red = 0;
var green = 0;
var blue = 0;
for(var i = 0; i < kernel.length; i++) {
for(var j = 0; j < kernel.length; j++) {
var imageX = (x - kernel.length / 2 + i + imagething.width) % imagething.width;
var imageY = (y - kernel.length / 2 + j + imagething.height) % imagething.height;
var RGB = imageIndex(imagething, imageX, imageY);
var pix = imagething.pixels;
var R = pix[RGB];
var G = pix[RGB + 1];
var B = pix[RGB + 2];
console.log("Start: ",R,G,B);
red += (R * kernel[i][j]);
green += (G * kernel[i][j]);
blue += (B * kernel[i][j]);
}
}
var outR;
var outG;
var outB;
outR = Math.min(Math.max(Math.round(red * mult_factor), 0), 255);
outG = Math.min(Math.max(Math.round(green * mult_factor), 0), 255);
outB = Math.min(Math.max(Math.round(blue * mult_factor), 0), 255);
console.log("End: ", outR, outG, outB);
pix[RGB] = outR;
pix[RGB + 1] = outG;
pix[RGB + 2] = outB;
imagething.updatePixels();
}
}
}
So I am referencing this blog post on kernel-based image processing and I am trying to blur my image. I am using the blur matrix and I use the original image and it should turn out like this but it ends up like this so I must be doing something wrong. Help is much appreciated :)