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