Teaser 2590: Moving downwards
From The Sunday Times, 13th May 2012 [link] [link]
A three-by-three array of the nine digits 1 – 9 is said to be “downward moving” if each digit is less than the digit to the east of it and to the south of it — see, for example, the keypad on a mobile phone.
Megan has produced a downward moving array that is different from that on a mobile phone. If she were to tell you the sum of the digits in its left-hand column and whether or not the central digit was even, then you should be able to work out Megan’s array.
What is Megan’s array?
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Jim Randell 10:09 am on 14 January 2025 Permalink |
This Python program uses the [[
SubstitutedExpression]] solver from the enigma.py library to generate all possible “downward moving” arrays. And then uses the [[filter_unique()]] function to find solutions that are unique by the required criteria.It runs in 74ms. (Internal runtime is 9ms).
from enigma import (SubstitutedExpression, irange, filter_unique, unpack, chunk, printf) # for a "downward moving" grid the rows and columns are in ascending order # consider the grid: # A B C # D E F # G H I exprs = [ "A < B < C", "D < E < F", "G < H < I", "A < D < G", "B < E < H", "C < F < I", ] # make a solver to find downward moving grids p = SubstitutedExpression( exprs, digits=irange(1, 9), answer="(A, B, C, D, E, F, G, H, I)", ) # look for a non-standard layout, unique by <f> non_standard = lambda x: x != (1, 2, 3, 4, 5, 6, 7, 8, 9) # f = (sum of digits in LH column, central digit parity) f = unpack(lambda A, B, C, D, E, F, G, H, I: (A + D + G, E % 2)) rs = filter_unique(p.answers(verbose=0), f, st=non_standard).unique # output solution(s) for r in rs: printf("{r}", r=list(chunk(r, 3)))Solution: Megan’s downward moving array looks like this:
Of the 42 possible downward moving arrays, this is the only one that is unique by LH column sum (11) and parity of the central digit (even).
(Of course moving from left-to-right and top-to-bottom the numbers are increasing, so maybe this should be considered an “upward moving” array).
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Ruud 8:27 am on 16 January 2025 Permalink |
import itertools import collections collect = collections.defaultdict(list) for a in itertools.permutations(range(1, 10)): if ( all(a[row * 3] < a[row * 3 + 1] < a[row * 3 + 2] for row in range(3)) and all(a[col] < a[col + 3] < a[col + 6] for col in range(3)) and a != tuple(range(1, 10)) ): collect[a[0] + a[3] + a[6], a[4] % 2].append(a) for ident, solutions in collect.items(): if len(solutions) == 1: for row in range(3): for col in range(3): print(solutions[0][row * 3 + col], end="") print()LikeLike
GeoffR 9:28 am on 18 January 2025 Permalink |
# A B C # D E F # G H I from itertools import permutations from collections import defaultdict MEG = defaultdict(list) digits = set(range(1, 10)) for p1 in permutations(digits, 6): A, B, C, D, E, F = p1 if A < B < C and D < E < F and \ A < D and B < E and C < F: q1 = digits.difference(p1) for p2 in permutations(q1): G, H, I = p2 if G < H < I: if A < D < G and B < E < H and C < F < I: MEG[(A + D + G, E % 2)].append([A, B, C, D, E, F, G, H, I]) # check E is even for Meghans array as 5th element(E) # ..of standard telephone array is odd i.e. 5 for k, v in MEG.items(): if len(v) == 1 and v[0][4] % 2 == 0: print("Meghans array: ") print(v[0][0], v[0][1], v[0][2]) print(v[0][3], v[0][4], v[0][5]) print(v[0][6], v[0][7], v[0][8]) ''' Meghans array: 1 2 5 3 4 6 7 8 9 '''LikeLike