# # 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())