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

P128: Tweaked bullet points in Haskell solution comments for consistency.

Author: nayuki

haskell/p128.hs

@@ -47,86 +47,86 @@ 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
  -         11
  -     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
  -         14
  -     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.
  -     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:
+ -   * 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
@@ -135,32 +135,32 @@ 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
  -     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
+ -   * {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.
  -     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:
+ -   * 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

python/p071.py

@@ -14,11 +14,11 @@
 # 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.
 # 

python/p145.py

@@ -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():