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{-
- 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 qualified EulerLib
{-
- 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
-}
target = 2000 -- Must be at least 3
main = putStrLn (show ans)
ans = find 2 (target - 2) -- Already found 2 because n = 1 and 2 satisfy PD(n) = 3
find :: Integer -> Integer -> Integer
find ring remain = let
a = all EulerLib.isPrime [ring * 6 - 1, ring * 6 + 1, ring * 12 + 5]
b = all EulerLib.isPrime [ring * 6 - 1, ring * 6 + 5, ring * 12 - 7]
remain' = remain - (EulerLib.boolToInt a)
remain'' = remain' - (EulerLib.boolToInt b)
in if remain' == 0
then (ring * (ring - 1) * 3 + 2)
else if remain'' == 0
then (ring * (ring + 1) * 3 + 1)
else find (ring + 1) remain''