Project Euler/28

Spiral Diagonals Sum Analysis

This page provides a mathematical analysis and algorithm to compute the sum of numbers on the diagonals of an $n \times n$ spiral grid, starting from 1 at the center and moving clockwise. The problem is to find the sum for a $1001 \times 1001$ spiral, given that the sum for a $5 \times 5$ spiral is 101.

Understanding the 5×5 Spiral

Consider the $5 \times 5$ spiral provided as an example:

21 22 23 24 25
20  7  8  9 10
19  6  1  2 11
18  5  4  3 12
17 16 15 14 13

The diagonals are:

Main diagonal (top-left to bottom-right): 21, 7, 1, 3, 13.
Anti-diagonal (top-right to bottom-left): 25, 9, 1, 5, 17.

The sum of these numbers is:

Main diagonal: $$ 21 + 7 + 1 + 3 + 13 = 45 $$
Anti-diagonal: $$ 25 + 9 + 1 + 5 + 17 = 57 $$

Total: $$ 45 + 57 = 102 $$ . Since the center (1) is counted twice, subtract it once: $$ 102 - 1 = 101 $$ , which matches the given sum.

Analyzing the Spiral Pattern

The spiral starts at 1 in the center and moves outward in layers, clockwise. For a $5 \times 5$ grid:

Layer 0: Center (1).
Layer 1: $3 \times 3$ grid (numbers 2 to 9).
Layer 2: $5 \times 5$ grid (numbers 10 to 25).

For an $n \times n$ grid where $$ n $$ is odd ( $$ n = 2k + 1 $$ ), the number of layers $$ k $$ is:

$$ n = 5 $$ , so $$ 5 = 2k + 1 $$ , $$ k = 2 $$ .
$$ n = 1001 $$ , so $$ 1001 = 2k + 1 $$ , $$ k = 500 $$ .

Each layer $$ m $$ forms a square of side length $$ 2m + 1 $$ . The numbers on the diagonals are the corners of these layers.

Number of Elements per Layer

Layer 0: 1 number.
Layer 1: $3 \times 3 = 9$ numbers, minus the center, so 8 numbers.
Layer 2: $5 \times 5 = 25$ numbers, minus the inner $3 \times 3$ (9 numbers), so $$ 25 - 9 = 16 $$ numbers.

For layer $$ m $$ , the side length is $$ 2m + 1 $$ , so the grid has $$ (2m + 1)^2 $$ numbers. The inner grid (layer $$ m-1 $$ ) has $$ (2m - 1)^2 $$ numbers. The numbers in layer $$ m $$ :

$$ (2m + 1)^2 - (2m - 1)^2 = (4m^2 + 4m + 1) - (4m^2 - 4m + 1) = 8m $$

Total numbers up to layer $$ m $$ :

$1 + \sum_{i=1}^m 8i = 1 + 8 \cdot \frac{m(m+1)}{2} = 1 + 4m(m+1)$

The last number in layer $$ m $$ (top-right corner) is $$ (2m + 1)^2 $$ .

Corner Numbers on Diagonals

For layer $$ m $$ , the corners are:

Top-right: $$ (2m + 1)^2 $$
Top-left: $$ (2m + 1)^2 - 2m $$
Bottom-left: $$ (2m + 1)^2 - 4m $$
Bottom-right: $$ (2m + 1)^2 - 6m $$

Verify for $5 \times 5$ ( $$ k = 2 $$ ):

Layer 1 ($ m = 1 $):
- Top-right: $(2 \times 1 + 1)^2 = 9$
- Top-left: $9 - 2 \times 1 = 7$
- Bottom-left: $9 - 4 \times 1 = 5$
- Bottom-right: $9 - 6 \times 1 = 3$
Layer 2 ($ m = 2 $):
- Top-right: $(2 \times 2 + 1)^2 = 25$
- Top-left: $25 - 2 \times 2 = 21$
- Bottom-left: $25 - 4 \times 2 = 17$
- Bottom-right: $25 - 6 \times 2 = 13$

These match the grid.

Sum of Diagonals for 1001×1001 Spiral

For $$ n = 1001 $$ , $$ k = 500 $$ . Sum the corners from layer 1 to 500, plus the center (1).

Sum of corners in layer $$ m $$ :

$$ 4(2m + 1)^2 - (2m + 4m + 6m) = 16m^2 + 4m + 4 $$

Total sum from layer 1 to 500:

$\sum_{m=1}^{500} (16m^2 + 4m + 4) = 16 \sum_{m=1}^{500} m^2 + 4 \sum_{m=1}^{500} m + 4 \sum_{m=1}^{500} 1$

Using:

$\sum_{m=1}^n m^2 = \frac{n(n+1)(2n+1)}{6}$
$\sum_{m=1}^n m = \frac{n(n+1)}{2}$
$\sum_{m=1}^n 1 = n$

For $$ n = 500 $$ :

$\sum_{m=1}^{500} m^2 = \frac{500 \cdot 501 \cdot 1001}{6} = 41791750$
$\sum_{m=1}^{500} m = \frac{500 \cdot 501}{2} = 125250$
$\sum_{m=1}^{500} 1 = 500$

Total:

$16 \cdot 41791750 + 4 \cdot 125250 + 4 \cdot 500 = 669171000$

Add the center: $669171000 + 1 = \mbox{the final answer}$ .

Algorithm to Compute the Sum

Below is a pseudocode algorithm to compute the sum efficiently:

function sumDiagonals(n):
    k = (n - 1) / 2
    sum = 1  # Center
    for m from 1 to k:
        sum += 16 * m * m + 4 * m + 4
    return sum

n = 1001
result = sumDiagonals(n)
print(result)

This algorithm avoids generating the entire spiral, focusing on the diagonal numbers.

Final Answer

The sum of the numbers on the diagonals of a $1001 \times 1001$ spiral is

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