Sunday Times Teaser 3341 – Sum of Squares
by Colin Vout
Sunday October 04 2026 (link)
A seamstress is making a patchwork blanket from knitted squares of two colours, red and orange, forming a rectangular pattern. She has already sewn eight rows of seven squares each. Row A has its red squares in columns 1, 2, 4, 6 & 7; B has 2, 3, 6 & 7; C has 3, 4, 5 & 7; D has 1, 2, 3 & 5; E has 1, 2, 3, 5 & 6; F has 1, 3, 4 & 7; G has 2, 4, 6 & 7; H has 4, 5 & 6. All other squares are orange.
She wants to assemble the rows (without reversing any left to right) into a rectangle, but choosing the order so that each column has exactly one run of exactly three consecutive reds, with all other reds having no red neighbours in the column, and with more reds in the top half than in the bottom half.
What should be the order of rows from top to bottom?
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BRG permalink123456789101112131415161718192021222324252627282930313233343536373839404142from itertools import permutations, groupbyfrom collections import defaultdict# dictionary of rows giving the one based# index of each red in the rowreds = {'A': (1, 2, 4, 6, 7),'B': (2, 3, 6, 7),'C': (3, 4, 5, 7),'D': (1, 2, 3, 5),'E': (1, 2, 3, 5, 6),'F': (1, 3, 4, 7),'G': (2, 4, 6, 7),'H': (4, 5, 6)}# convert to rows with 0 for orange and 1 for redirows = defaultdict(list)for n, rws in reds.items():irows[n] = [1 if c in rws else 0 for c in range(1, 8)]# check a columndef check(vs):d = defaultdict(list)# collect sequences of consecutive equal valuesfor k, v in groupby(vs):d[k].append(len([*v]))# red run lengths can only be 1 or 3if not set(d[1]).difference((1, 3)):# do we have three reds in a rowreturn 3 in d[1]# consider all permutations of the rowsfor rows in permutations(irows.items()):nms, rws = list(zip(*rows))# there are more reds in the top half# than in the bottom halfif (sum(sum(r) for r in rws[:4]) >sum(sum(r) for r in rws[4:])):# check the columnsif all(check(c) for c in zip(*rws)):print(''.join(nms))
