Project Euler Problem 28

Number spiral diagonals.

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

     43 44 45 46 47 48 49
     42 21 22 23 24 25 26
     41 20  7  8  9 10 27
     40 19  6  1  2 11 28
     39 18  5  4  3 12 29
     38 17 16 15 14 13 30
     37 36 35 34 33 32 31

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

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

Let f(n) is the function returning sum of diagonals for square with radius n (side of the square is 2n+1):

                7 8 9
                6   2
                5 4 3

Upper right corner is: (2n+1)², upper left is (2n+1)²-2n, bottom left is (2n+1)²-4n and the bottom right is (2n+1)²-6n So sum will be 4(2n+1)² - 12n.

So iterative f(n) = 4(2n+1)² - 12n + f(n-1).

And we are have generator for python and recursion for erlang:

Link to original description
Source code examples on Github

Python version

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#!/usr/bin/python

def t28(n):
    while n > 0:
        yield 4*(2*n+1)**2 - 12*n
        n -= 1
    yield 1

print "Answer: %s" % sum(t28(500))

Erlang version

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#!/usr/bin/env escript
%% -*- erlang -*-
%%! -smp enable -sname p28
% vim:syn=erlang

-mode(compile).

main(_) -> 
    io:format("Answer ~p ~n", [t28(500)]).

t28(0) -> 1;
t28(N) -> 4*(2*N+1)*(2*N+1) - 12*N + t28(N-1).

Also i found interesting explanation how to build non-iterative function f(n). Check it out here.