I have the code like below to return the number that makes the sum of each line the same.
let getnums = () => {
var numsarray = [];
var arrays = [1, 2, 3, 4, 5, 6, 7, 8, 9];
var len = arrays.length;
while (len > 0) {
var rans = Math.floor(Math.random() * len);
numsarray.push(arrays[rans]);
arrays.splice(rans, 1);
len = len - 1;
}
return numsarray;
};
let done = false
function test(num){
while (!done) {
let output = getnums();
if (
output[0] + output[1] + output[2] == 15 &&
output[1] + output[4] + output[7] == 15 &&
output[2] + output[5] + output[8] == 15 &&
output[0] + output[3] + output[6] == 15 &&
output[2] + output[4] + output[6] == 15 &&
output[3] + output[4] + output[5] == 15 &&
output[6] + output[7] + output[8] == 15 &&
output[0] + output[4] + output[8] == 15
) {
done = true
console.log(output)
}
}
}
test()The code works fine, but I am looking for ways to shorten the if statement.
My current code only will work result 3*3 because of the specific if statement , I am looking for a better and that will work for all different sizes:
E.g: 4*4, a function produces:
if(
output[0] + output[1] + output[2] +output[3]
output[4] + output[5] + output[6] +output[7]
....
)
I have already had idea or using loop to make randomnumber, but don't know how to make if statement.
Is it possible I could use loop or other methods to create the if statement?
Any suggestions or ideas?
Thank you!
Since you have perfect square matrices, you can use the below logic:
So, we add every element of the matrix twice and add diagonal elements one more time. Needlessly to say that the value at the center of the matrix will be added 4 times. We do get a total sum.
We need to compare this sum with the new desired sum derivation in the following way:
desired_sum * (R + C + 2)
where R is no. of rows, C is no. of columns and 2 is for the 2 diagonals. Since R = C for your case, it will be
desired_sum * (N * 2 + 2)
where N is the no. of rows or columns of the matrix.
Snippet:
var mat = [
[1, 1, 1, 1],
[1, 1, 1, 1],
[1, 1, 1, 1],
[1, 1, 1, 1]
];
function test(mat, N, desired_sum){
let sum = 0;
let d1_col = 0,d2_col = N - 1;
for(let i = 0; i < N; ++i, d1_col++, d2_col--){
for(let j = 0; j < N; ++j){
sum += 2 * mat[i][j];
if(d1_col == j) sum += mat[i][j];
if(d2_col == j) sum += mat[i][j];
}
}
return sum == desired_sum * (N * 2 + 2);
}
console.log(test(mat, mat.length, 4));For the 3x3 problem, you could take the following array of indices and check the sum.
For 4x4 arrays, take another array of indices.
const
getSum = indices => indices.reduce((sum, index) => sum + output[index], 0),
indices3 = [[0, 1, 2], [1, 4, 7], [2, 5, 8], [0, 3, 6], [2, 4, 6], [3, 4, 5], [6, 7, 8], [0, 4, 8]];
if (indices3.every(indices => getSum(indices) === 15)) {
// ...
}