### P71, P145: Tweaked bullet points in Python solution comments for consistency..

`P128: Tweaked bullet points in Haskell solution comments for consistency.`
parent 3635e913
 ... ... @@ -47,34 +47,34 @@ import qualified EulerLib - C - - Basic observations: - - Except for the degenerate ring 0, each ring k has 6k cells. - * 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 - * 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 - * 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 - * 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, - * 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, - * 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, - * 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. - * 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. - * 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 ... ... @@ -82,19 +82,19 @@ import qualified EulerLib - 25 04 - 12 - - - Claim: No corner cell in the other 5 directions satisfies PD(n) = 3. - * 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, - * 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 - * 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 - * 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, - * 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. - * 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. - * 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 ... ... @@ -102,16 +102,16 @@ import qualified EulerLib - 28 30 - 29 - - - Now let's consider an arbitrary upward corner cell n in ring k, with k >= 2. - * 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, - * 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: - * Equations derived from the structure of the hexagonal spiral: - n = 3k(k - 1) + 2. - a = n - 6(k - 1). - b = n + 1. ... ... @@ -119,14 +119,14 @@ import qualified EulerLib - 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: - * 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 - * 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 ... ... @@ -135,7 +135,7 @@ import qualified EulerLib - 09 19 - 02 - - - Finally let's consider an arbitrary edge cell immediately to the right of the - * 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 ... ... @@ -143,9 +143,9 @@ import qualified EulerLib - n - a d - b - - {a, b} are in the previous ring, {c, d} are in the current ring, and {e, f} are in - * {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: - * 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. ... ... @@ -153,14 +153,14 @@ import qualified EulerLib - 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: - * 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 - * 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 ... ...
 ... ... @@ -14,11 +14,11 @@ if sys.version_info.major == 2: # We consider each (integer) denominator d from 1 to 1000000 by brute force. # For a given d, what is the largest integer n such that n/d < 3/7? # # * If d is a multiple of 7, then the integer n' = (d / 7) * 3 satisfies n'/d = 3/7. # - If d is a multiple of 7, then the integer n' = (d / 7) * 3 satisfies n'/d = 3/7. # Hence we choose n = n' - 1 = (d / 7) * 3 - 1, so that n/d < 3/7. # Since (d / 7) * 3 is already an integer, it is equal to floor(d * 3 / 7), # which will unifie with the next case. Thus n = floor(d * 3 / 7) - 1. # * Otherwise d is not a multiple of 7, so choosing n = floor(d * 3 / 7) # - Otherwise d is not a multiple of 7, so choosing n = floor(d * 3 / 7) # will automatically satisfy n/d < 3/7, and be the largest possible n # due to the definition of the floor function. # ... ...
 ... ... @@ -15,10 +15,10 @@ # reversed number does not have leading zeros. # # Consider different cases for the number of digits n (from 1 to 9, but the arguments apply generally): # * n = 1: # - n = 1: # Clearly there are no solutions because the last digit is always even. # # * n = 0 mod 2: # - n = 0 mod 2: # We begin by proving that when a number is "reversible", the process of adding # the number to the reverse of itself will involve no carries in the arithmetic. # Normally a rigorous proof would require the use of mathematical induction, ... ... @@ -70,7 +70,7 @@ # (9,0). # Therefore by combinatorics, there are 20 * 30^(n/2 - 1) reversible n-digit numbers when n is even. # # * n = 1 mod 2: # - n = 1 mod 2: # Let's illustrate what happens with a 7-digit number abcdefg: # 0101010 # abcdefg ... ... @@ -93,7 +93,7 @@ # This is why we get the alternating pattern of carries in the adding process. # # The rest of the work is to enumerate the possibilities for each type of digit(s) in the number: # * Pairs of digits which take no carry and must generate a carry (20 choices): # - Pairs of digits which take no carry and must generate a carry (20 choices): # (9,8), (9,6), (9,4), (9,2), # (8,9), (8,7), (8,5), (8,3), # (7,8), (7,6), (7,4), ... ... @@ -103,7 +103,7 @@ # (3,8), # (2,9). # Note that the first and last digits fall into this category, and there are no 0s at all. # * Non-middle pairs of digits which take a carry and generate no carry (25 choices): # - Non-middle pairs of digits which take a carry and generate no carry (25 choices): # (0,0), (0,2), (0,4), (0,6), (0,8), # (1,1), (1,3), (1,5), (1,7), # (2,0), (2,2), (2,4), (2,6), ... ... @@ -113,7 +113,7 @@ # (6,0), (6,2), # (7,1), # (8,0). # * Middle single digit, which takes a carry and generates no carry (5 choices): 0, 1, 2, 3, 4. # - Middle single digit, which takes a carry and generates no carry (5 choices): 0, 1, 2, 3, 4. # All in all, there are 5 * 20^((n + 1)/4) * 25^((n - 3)/4) = 100 * 500^((n - 3)/4) # reversible n-digit numbers when n = 3 mod 4. def compute(): ... ...
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