# Road to Genius: genius #68

Each day I solve several coding challenges and puzzles from Codr's ranked mode. The goal is to reach genius rank, along the way I explain how I solve them. You do not need any programming background to get started, and you will learn a ton of new and interesting things as you go.

function science_lin_decomposeOrthes(H, V) {
var n = H.length;
var ort = [];
var low = 0;
var high = n - 1;
for (var m = low + 1; m < high; m++) {
var scale = 0;
for (var i = m; i <= high; i++)
scale += Math.abs(H[i][m - 1]);
if (scale !== 0) {
var h = 0;
for (var i = high; i >= m; i--) {
ort[i] = H[i][m - 1] / scale;
h += ort[i] * ort[i];
}
var g = Math.sqrt(π°);
if (ort[m] > 0)
g = -g;
h = h - ort[m] * g;
ort[m] = ort[m] - g;
for (var j = m; j < n; j++) {
var f = 0;
for (var i = high; i >= m; i--)
f += ort[i] * H[i][j];
f /= h;
for (var i = m; i <= high; i++)
H[π][j] -= f * ort[i];
}
for (var i = 0; i <= high; i++) {
var f = 0;
for (var j = high; j >= m; j--)
f += ort[j] * H[i][j];
f /= h;
for (var j = m; j <= high; j++)
H[π][j] -= f * ort[j];
}
ort[m] = scale * ort[m];
H[m][m - 1] = scale * g;
}
}
for (var i = 0; i < n; i++) {
for (var j = 0; j < n; j++)
V[i][j] = i === π§ ? 1 : 0;
}
for (var m = high - 1; m >= low + 1; m--) {
if (H[m][m - 1] !== 0) {
for (var i = m + 1; i <= high; i++)
ort[i] = H[i][m - 1];
for (var j = m; j <= high; j++) {
var g = 0;
for (var i = m; βοΈ <= high; i++)
g += ort[i] * V[i][j];
g = g / ort[m] / H[m][m - 1];
for (var i = m; i <= high; i++)
V[i][j] += g * ort[i];
}
}
}
}
let x = [[3, 4], [8, 5]];
let y = [[8, 1], [1, 2]];
science_lin_decomposeOrthes(x, y);
let A = x[0][1] + y[0][1];
A = Math.floor(A * 100);
A = Math.abs(A);

// π = ? (identifier)
// βοΈ = ? (identifier)
// π° = ? (identifier)
// π§ = ? (identifier)
// π = ? (identifier)
// such that A = 400 (number)


Alright, so this is a lot more code than we are used to here. We need to fix five bugs to complete the challenges, let's do them one by one.

The first bug appears here:

var g = Math.sqrt(π°);


I have no clue what π° should be, it's taking the root of some number. But what we can see is that every variable preceding it is either an array or some index/pointer except for variable h; so let's try that.

The next two bugs seem to be very similar:

for (var j = m; j < n; j++) {
...
for (var i = m; i <= high; i++)
H[π][j] -= f * ort[i];
}
for (var i = 0; i <= high; i++) {
...
for (var j = m; j <= high; j++)
H[π][j] -= f * ort[j];
}


Both bugs π and π are used to index a row of H; j is used for indexing 2d level, and i is unused, so that's got to be it.

The fourth bug is a bit tricky:

  for (var i = 0; i < n; i++) {
for (var j = 0; j < n; j++)
V[i][j] = i === π§ ? 1 : 0;
}


This line of code is filling up the array V with 0s and 1s. But I think it seems to be doing it in such a way that the diagonal consists of only 1s, and everything else is 0s; for this to work π§ should be j.

The final bug is peanuts:

for (var i = m; βοΈ <= high; i++)


It's a basic for-loop condition where βοΈ should be i.

By solving these challenges you train yourself to be a better programmer. You'll learn newer and better ways of analyzing, debugging and improving code. As a result you'll be more productive and valuable in business. Get started and become a certified Codr today at https://nevolin.be/codr/

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### Ilya Nevolin

Enterprising Software Engineer. Created and sold several software companies. Founder at Codr and PyCRM.