Get diagonal values of spiral matrix - php

Question

I have a n*n spiral matrix.
if N = 4
then matrix :
7 8 9 10
6 1 2 11
5 4 3 12
16 15 14 13
if N = 3
7 8 9
6 1 2
5 4 3
I want to get the diagonal values of this spiral matrix.
In the n=4 case diagonal values would be 7,1,3,13,10,2,4,16
I can do this by storing this matrix in array and traversing for each diagonal value.
Is there any better way to get these values.

Answer 1

To get the numbers on the main diagonal, we can notice that the values are
1 = 1
1 + 2 = 3
1 + 2 + 4 = 7
1 + 2 + 4 + 6 = 13
So the general formula is 1 + (sum i = 0 to k of 2*i) for k = 0, 1, 2, ...
Simplifying this, we get k^2 + k + 1 for k = 0, 1, 2, ...
In PHP, we can generate these via something like this:
function mainDiagonal($n) {
$values = array();
for ($k = 0; $k < $n; $k++) {
$values[] = $k*$k + $k + 1;
}
return $values;
}
To get the numbers on the antidiagonal for even N we see:
2 = 2
2 + 2 = 4
2 + 2 + 6 = 10
2 + 2 + 6 + 6 = 16
If we continue this pattern for larger matrices we see the general formula is
sum i = 0 to k of floor(i/2)*4 + 2 for k = 0, 1, 2, ...
Similarly for odd N we find the formula is
1 + (sum i = 0 to k of ceil(i/2)*4) for k = 0, 1, 2, ...
In PHP, we can generate these via something like this:
function antiDiagonal($n) {
$values = array();
if ($n % 2 == 0) {
for ($k = 0; $k < $n; $k++) {
$accum = 0;
for ($j = 0; $j <= $k; $j++) {
$accum += floor($j/2)*4 + 2;
}
$values[] = $accum;
}
} else {
for ($k = 0; $k < $n; $k++) {
$accum = 1;
for ($j = 0; $j <= $k; $j++) {
$accum += ceil($j/2)*4;
}
$values[] = $accum;
}
}
return $values;
}
Notice that the maximum value of k is one less than the dimension of the matrix.
Combining these functions, we obtain:
array_unique(array_merge(mainDiagonal($n), antiDiagonal($n)))

Answer 2

The problem can be divided into 4 parts: Find the numbers along the diagonal spoke in each quadrant. There are four quadrants, so we have four spokes:
Northwest (NW) spoke
Northeast (NE) spoke
Southwest (SW) spoke
Southeast (SE) spoke
For example, in your illustration of Ulam spiral, when N is even.
NW spoke have 1, 7, ...
NE spoke have 2, 10, ...
SW spoke have 4, 16, ...
SE spoke have 3, 13, ...
The problem is further subdivided into two cases:
N is even.
N is odd.
Case 1: N is even
Here are the formulas for each spoke:
NW spoke: f(n) = 4*n*n + 2*n + 1
NE spoke: g(n) = 4*n*n + 4n + 2
SW spoke: h(n) = 4*n*n + 8*n + 4
SE spoke: i(n) = 4*n*n + 6*n + 3
where n = 0, 1, 2, ...
For 4x4 matrix, compute the following set:
{f(0), f(1), g(0), g(1), h(0), h(1), i(0), i(1)}
It yields the diagonal values:
{1, 7, 2, 10, 4, 16, 3, 13}
In general, for an NxN matrix, when N is even, compute the following set to get the diagonal values:
{ f(0), ..., f(N/2 - 1),
g(0), ..., g(N/2 - 1),
h(0), ..., h(N/2 - 1),
i(0), ..., i(N/2 - 1) }
Case 2: N is odd
In your illustration of Ulam spiral, when N is odd, the formulas for each spoke are:
NW spoke: f(n) = 4*n*n + 2*n + 1
NE spoke: g(n) = 4*n*n + 4*n + 1
SW spoke: h(n) = 4*n*n + 1
SE spoke: i(n) = 4*n*n - 2*n + 1
where n = 0, 1, 2, ...
Note that f(0) = g(0) = h(0) = i(0) = 1.
For 3x3, compute the following set:
{f(0), f(1), g(1), h(1), i(1)}
It yields the following diagonal values:
{1, 7, 9, 5, 3}.
In general, for an NxN matrix, when N is odd, compute the following set to get the diagonal values:
{ f(0), ..., f((N - 1)/2,
g(0), ..., g((N - 1)/2),
h(0), ..., h((N - 1)/2),
i(0), ..., i((N - 1)/2) }
PHP Code
Finally, here is a PHP program that demonstrates what I have discussed above.
<?php
function ulam_diag($N)
{
$result = array();
if ($N % 2 == 0) {
for ($n = 0; $n < $N / 2; $n++) {
$result[] = 4*$n*$n + 2*$n + 1;
$result[] = 4*$n*$n + 4*$n + 2;
$result[] = 4*$n*$n + 8*$n + 4;
$result[] = 4*$n*$n + 6*$n + 3;
}
} else {
$result[] = 1;
for ($n = 1; $n <= ($N - 1) / 2; $n++) {
$result[] = 4*$n*$n + 2*$n + 1;
$result[] = 4*$n*$n + 4*$n + 1;
$result[] = 4*$n*$n + 1;
$result[] = 4*$n*$n - 2*$n + 1;
}
}
sort($result);
return $result;
}
print_r(ulam_diag(4));
print_r(ulam_diag(3));
?>
Output:
Array
(
[0] => 1
[1] => 2
[2] => 3
[3] => 4
[4] => 7
[5] => 10
[6] => 13
[7] => 16
)
Array
(
[0] => 1
[1] => 3
[2] => 5
[3] => 7
[4] => 9
)
Here is the code one Ideone: http://ideone.com/F9jaC0
In case you are wondering how I arrived at the formulas, there are well established results for the four spokes of Ulam spiral. Here are the references:
https://oeis.org/A054569 (NW spoke in your illustration)
https://oeis.org/A016754 (NE spoke in your illustration)
https://oeis.org/A053755 (SW spoke in your illustration)
https://oeis.org/A054554 (SE spoke in your illustration)
The Ulam spirals in your illustrations are oriented differently from the popular representation of Ulam spirals, so I took these well known results and adjusted the offset of n for each formula, so that it works with your Ulam spiral. These adjustments are left as exercises to the reader. ;-)

Answer 3

Well for each row in a matrix we have two diagonal values. To obtain these two values i used two position(x1,y1) and (x2,y2) for main and anti diagonals.
Well I wrote this code:
<?php
function getSpiralDiagonal($spiralArr,$N){
$diagonalValueCount = $N*2;
$xIndexMainDiagonal = 0;
$yIndexMainDiagonal = 0;
$xIndexAntiDiagonal = 0;
$yIndexAntiDiagonal = $N-1;
while($diagonalValueCount > 0){
//checking for same position
if($yIndexMainDiagonal == $yIndexAntiDiagonal){
echo $spiralArr[$xIndexMainDiagonal][$yIndexMainDiagonal].'<br>';
}else{
echo $spiralArr[$xIndexMainDiagonal][$yIndexMainDiagonal].'<br>';
echo $spiralArr[$xIndexAntiDiagonal][$yIndexAntiDiagonal].'<br>';
}
$xIndexMainDiagonal++;
$yIndexMainDiagonal++;
$xIndexAntiDiagonal++;
$yIndexAntiDiagonal--;
$diagonalValueCount -= 2;
}
}
$spiralArr = array(array('7','8','9'),array('6','1','2'),array('5','4','3'));
getSpiralDiagonal($spiralArr,3);
?>