BartMassey

# Once you know the Cartesian (x, y) coordinate of spiral position i in Part 1,
# you can compute Part 2 really easily using just a map from coordinates to int.
# map[(0,0)] = 1, and then you go around the spiral. At each new coordinate, you look
# at its nine neighbors with a pair of for loops. Most of them (including the new position)
# won't be in there.

# Here's a Python implementation to show what I mean.

# Essentially a Fibonacci-style dynamic
# programming calculation but with 2-3 neighbors.

imjasonh

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