#### Project Euler 28 Problem Statement

Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:

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

It can be verified that the sum of both diagonals is 101.

What is the sum of both diagonals in a 1001 by 1001 spiral formed in the same way?

#### Solution

#### Basic approach: Using a loop to count corners

The “corners”, which form the two principal diagonals, produce a simple series: (3, 5, 7, 9), (13, 17, 21, 25), (31, 37, 43, 49), …

This can be quantified as (* n^{2} – 3n + 3, n^{2} – 2n + 2, n^{2} – n + 1, n^{2}*) and summing them together yields

*; the sum of the corners for each odd-length square:*

**4n**^{2}– 6n + 63 + 5 + 7 + 9 =

**24**, 13 + 17 + 21 + 25 =

**76**, 31 + 37 + 43 + 49 =

**160**, …

sum, size = 1, 1001 for n in xrange(3, size+1, 2): sum += 4*n*n - 6*n + 6 print "Answer to PE28 = ",sum

#### Improved approach: Closed form summation

If we rewrite the `for`

loop as a summation we will have:

*2i+1* is every odd number, starting with 3, until we reach the size of the square. This will take (n-1)/2 iterations.

This simplifies further:

Finally, we can express this summation as a closed form equation by using the algebra of summation notation (Wikipedia or Project Euler Problem 6):

Or, factored if you prefer:

Let’s test this equation with the example in the problem statement, *n* = 5. Remember *n*≥3 and odd.

#### HackerRank version

The Hackerrank Project Euler 28 version of this problem runs 100,000 test cases and extends the size of the square from 1001 to any odd n, where 1 ≤ n ≤ 10^{18}. So any iterative approach will exceed their time limit. This solution works perfectly.

#### Python 2.7 Source

#### Last Word

Reference: The On-Line Encyclopedia of Integer Sequences (OEIS) A114254: Sum of all terms on the two principal diagonals of a 2n+1 X 2n+1 square spiral.