Project Euler Problem 28 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 of each sub-square (shown above in red) form the two principal diagonals and produce a simple series: (3, 5, 7, 9), (13, 17, 21, 25), (31, 37, 43, 49), …
This can be quantified as the quad (n2 – 3n + 3, n2 – 2n + 2, n2 – n + 1, n2) and summing them together yields 4n2 – 6n + 6; the sum of the corners for each odd-size square:
sum, size = 1, 1001
for n in xrange(3, size+1, 2):
sum+= 4*n*n - 6*n + 6
print "Answer to PE28 =", sum
This simple, O(n), iterative approach will not solve the larger sizes specified in the HackerRank version of this problem; a better, O(1), solution must be found.
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
HackerRank requires us to run 10,000 test cases with an odd N, 1 ≤ N < 1018. Don’t forget to mod the result by 1000000007.
Python Source Code
N = int(input("Enter an odd value for N? "))
s = (N-1) // 2
print ("Sum of both diagonals is", (16*s**3 + 30*s**2 + 26*s + 3) // 3)
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.