#
# Solution to Project Euler problem 128
# Copyright (c) Project Nayuki. All rights reserved.
#
# https://www.nayuki.io/page/project-euler-solutions
# https://github.com/nayuki/Project-Euler-solutions
#
import eulerlib, itertools
# Let's do mathematical analysis to drastically reduce the amount of
# logic we need to implement and calculation the computer needs to do.
# We begin with a couple of definitions.
#
# Ring number: Each cell belongs in a hexagonal ring,
# numbered starting from 0 at the center like this:
# 3
# 3 3
# 3 2 3
# 3 2 2 3
# 2 1 2
# 3 1 1 3
# 2 0 2
# 3 1 1 3
# 2 1 2
# 3 2 2 3
# 3 2 3
# 3 3
# 3
#
# Corner/edge cell: Within a ring, each cell is
# either a corner cell or an edge cell, as shown:
# C
# E E
# E C E
# C E E C
# C C C
# E C C E
# E C E
# E C C E
# C C C
# C E E C
# E C E
# E E
# C
#
# Basic observations:
# - Except for the degenerate ring 0, each ring k has 6k cells.
# The kth ring has exactly 6 corner cells and 6(k - 1) edge cells.
# - In the code we will skip the PD (prime difference) calculation for
# rings 0 and 1 because the existence of ring 0 breaks many patterns.
# - Doing the PD calculation for rings 0 and 1 by hand (n = 1 to 7
# inclusive), we find that PD(n) = 3 for and only for n = 1, 2.
#
# Now let's analyze the characteristics of all cells in rings 2 or above.
# It's hard to justify these assertions rigorously, but they are true from
# looking at the spiral diagram.
#
# - Corner cells along the upward vertical direction and the edge cells
# immediately to the right of this vertical column are the most interesting,
# so we will save these cases for last.
#
# - Claim: Except for cells immediately right of the upward corner column,
# no edge cell satisfies PD(n) = 3. Proof: Take an arbitrary edge cell n
# not immediately to the right of the upward corner column...
# - The two neighbors in the same ring have a difference of 1 compared to n,
# which is not a prime number.
# - The two neighbors in the previous (inward) ring are consecutive numbers,
# so exactly one of them has an even absolute difference with n. Because
# n is in ring 2 or above, the difference with any neighboring number in the
# previous ring is at least 6. Thus an even number greater than 2 is not prime.
# - Similarly, the two neighbors in the next (outward) ring are consecutive numbers.
# One of them has an even difference with n, and this number is also at least 6,
# so one neighbor is definitely not prime.
# - Therefore with at least 4 neighbors that do not have a prime difference, PD(n) <= 2.
# Example of an edge cell n = 11 in ring 2, which is straight left of the origin:
# 10
# 24 03
# 11
# 25 04
# 12
#
# - Claim: No corner cell in the other 5 directions satisfies PD(n) = 3.
# Proof: Take an arbitrary corner cell n in the non-upward direction...
# - Two of its neighbors (in the same ring) have a difference of 1,
# which is not prime.
# - One neighbor is in the previous ring (inward) while three neighbors
# are in the next ring (outward).
# - Let the inner ring neighbor be k and the outer ring's middle neighbor
# be m. The three outer ring neighbors are {m - 1, m, m + 1}.
# - Then n - k + 6 = m - n. Also, {m - 1, m + 1} have the same parity,
# and {k, m} have the same other parity.
# - Either both {|k - n|, |m - n|} are even or both {|m - 1 - n|, |m + 1 - n|} are even.
# In any case, all these differences are at least 6, so the even numbers are not prime.
# - Therefore with at least 4 neighbors that do not have a prime difference, PD(n) <= 2.
# Example of a corner cell n = 14 in ring 2, which is straight below the origin:
# 05
# 13 15
# 14
# 28 30
# 29
#
# - Now let's consider an arbitrary upward corner cell n in ring k, with k >= 2.
# We shall give variables to all its neighbors like this:
# d
# e f
# n
# b c
# a
# - a is in the previous ring, {b, c} are in the same ring as n,
# and {d, e, f} are in the next ring.
# - Equations derived from the structure of the hexagonal spiral:
# n = 3k(k - 1) + 2.
# a = n - 6(k - 1).
# b = n + 1.
# c = n + 6k - 1 = d - 1.
# d = n + 6k.
# e = n + 6k + 1 = d + 1.
# f = n + 6k + 6(k + 1) - 1 = n + 12k + 5.
# - Hence we get these absolute differences with n:
# |a - n| = 6(k - 1). (Not prime because it's a multiple of 6)
# |b - n| = 1. (Not prime)
# |c - n| = 6k - 1. (Possibly prime)
# |d - n| = 6k. (Not prime because it's a multiple of 6)
# |e - n| = 6k + 1. (Possibly prime)
# |f - n| = 12k + 5. (Possibly prime)
# - Therefore for each k >= 2, we need to count how many numbers
# in the set {6k - 1, 6k + 1, 12k + 5} are prime.
# Example of a corner cell n = 8 in ring 2, which is straight above the origin:
# 20
# 21 37
# 08
# 09 19
# 02
#
# - Finally let's consider an arbitrary edge cell immediately to the right of the
# upward vertical column. Suppose the cell's value is n and it is in ring k,
# with k >= 2. Give variables to all its neighbors like this:
# f
# c e
# n
# a d
# b
# - {a, b} are in the previous ring, {c, d} are in the current ring, and {e, f} are in
# the next ring. The ascending ordering of all these numbers is (a, b, c, d, n, e, f).
# - Equations derived from the structure of the hexagonal spiral:
# n = 3k(k + 1) + 1.
# a = n - 6k - 6(k - 1) + 1 = n - 12k + 7.
# b = n - 6k.
# c = n - 6k + 1.
# d = n - 1.
# e = n + 6(k + 1) - 1 = n + 6k + 5.
# f = n + 6(k + 1).
# - Hence we get these absolute differences with n:
# |a - n| = 12k - 7. (Possibly prime)
# |b - n| = 6k. (Not prime because it's a multiple of 6)
# |c - n| = 6k - 1. (Possibly prime)
# |d - n| = 1. (Not prime)
# |e - n| = 6k + 5. (Possibly prime)
# |f - n| = 6(k + 1). (Not prime because it's a multiple of 6)
# - Therefore for each k >= 2, we need to count how many numbers
# in the set {6k - 1, 6k + 5, 12k - 7} are prime.
# Example of an edge cell n = 19 in ring 2:
# 37
# 08 36
# 19
# 02 18
# 07
def compute():
TARGET = 2000 # Must be at least 3
count = 2 # Because n = 1 and 2 satisfy PD(n) = 3
for ring in itertools.count(2):
if all(map(eulerlib.is_prime, (ring * 6 - 1, ring * 6 + 1, ring * 12 + 5))):
count += 1
if count == TARGET:
return str(ring * (ring - 1) * 3 + 2)
if all(map(eulerlib.is_prime, (ring * 6 - 1, ring * 6 + 5, ring * 12 - 7))):
count += 1
if count == TARGET:
return str(ring * (ring + 1) * 3 + 1)
if __name__ == "__main__":
print(compute())