Sunday Times Teaser 2439
Published 21st June 2009
Six cards are numbered 1 to 6. One card is placed on the forehead of each of three expert logicians. Each can see the other numbers, but not his own. I asked each in turn if his number was higher, lower or the same as the average of all three. They replied as follows:
Alf: I don’t know.
Bert: I don’t know.
Charlie: I don’t know.
If I now told you the product of the three numbers, you could work out which number each holds.
What numbers did Alf, Bert and Charlie hold (in that order)?
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BRG permalink12345678910111213141516171819202122232425262728293031323334353637383940from itertools import permutationsfrom collections import defaultdict# three way comparecmp = lambda x, y: (y < x) - (x < y)# return triples from <tps> for which <a> cannot decide whether# his number is less, equal or greater than the triple's averagedef filter(tps, a, b, c):results = set()# consider each triple from the first person's point of viewfor fp in tps:# inspect those triples for which the first person# sees the second person values <b> and <c>ts = (sp for sp in tps if sp[b] == fp[b] and sp[c] == fp[c])# the answer will be 'don't know' if the results are not# all less, all equal or all more than the triple's averageif len(set(cmp(3 * t[a], sum(t)) for t in ts)) > 1:results.add(fp)return results# the set of all triples to be connsideredtriples = set(permutations(range(1, 7), 3))# filter the triples seen by Alf, Bert and Charlie in turnfor i in range(3):triples = filter(triples, i, (i + 1) % 3, (i + 2) % 3)# collect triples with the same productp2t = defaultdict(set)for a, b, c in triples:p2t[a * b * c].add((a, b, c))# look for triples with a unique productfor p, t in p2t.items():if len(t) == 1:Alf, Bert, Charlie = t.pop()print(f"{Alf = } {Bert = } {Charlie = }")
