Problem 61 deals with figurate numbers. Here is the problem statement.
Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:
Triangle P3,n=n(n+1)/2 1, 3, 6, 10, 15, … Square P4,n=n2 1, 4, 9, 16, 25, … Pentagonal P5,n=n(3n−1)/2 1, 5, 12, 22, 35, … Hexagonal P6,n=n(2n−1) 1, 6, 15, 28, 45, … Heptagonal P7,n=n(5n−3)/2 1, 7, 18, 34, 55, … Octagonal P8,n=n(3n−2) 1, 8, 21, 40, 65, … The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three interesting properties.
- The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first).
- Each polygonal type: triangle (P3,127=8128), square (P4,91=8281), and pentagonal (P5,44=2882), is represented by a different number in the set.
- This is the only set of 4-digit numbers with this property.
Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set.
The statement at the end — “only ordered set“ — threw me off a bit, because I took it to mean that the six numbers would be triangular, square, pentagonal and so on in that order. There is no such set, and so I had to go back to the drawing board again.
It was then that point 2 in the problem statement dawned on me. 8128, 2882, and 8281 are cyclical, but 2882 is pentagonal. This makes the problem more interesting. The set of six numbers could be any permutation of the six types of figurates we are dealing with. Thankfully, since we know that there exists only one such set of six numbers, we can stop executing when we reach that set.
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from itertools import permutations def is_cyclical(i, j): return str(i)[2:] == str(j)[:2] def append_to_list(l, e): return_list = [] for i in l: for j in e: if is_cyclical(i[-1], j[0]): k = i[:] k.append(j[0]) return_list.append(k) return return_list def sum_of_cyclical_figurate(): perms = list(permutations([3, 4, 5, 6, 7, 8])) for p in perms: cy1 = append_to_list(figurates[p[0]], figurates[p[1]]) cy2 = append_to_list(cy1, figurates[p[2]]) cy3 = append_to_list(cy2, figurates[p[3]]) cy4 = append_to_list(cy3, figurates[p[4]]) cy5 = append_to_list(cy4, figurates[p[5]]) for x in cy5: if is_cyclical(x[-1], x[0]): return sum(x) figurates = [[]] * 9 figurates[3] = [[n] for n in [i * (i + 1) // 2 for i in range(3, 200)] if 999 < n < 10000] figurates[4] = [[n] for n in [i ** 2 for i in range(3, 100)] if 999 < n < 10000] figurates[5] = [[n] for n in [i * (3 * i - 1) // 2 for i in range(5, 100)] if 999 < n < 10000] figurates[6] = [[n] for n in [i * (2 * i - 1) for i in range(6, 100)] if 999 < n < 10000] figurates[7] = [[n] for n in [i * (5 * i - 3) // 2 for i in range(7, 100)] if 999 < n < 10000] figurates[8] = [[n] for n in [i * (3 * i - 2) for i in range(8, 100)] if 999 < n < 10000] print(sum_of_cyclical_figurate()) |