Sunday Times Teaser 3288 – Todd’s Password
by Henry C Warburton
Published Sunday September 28 2025 (link)
Todd privately tells each of four of his friends a letter of the password to his phone. All their letters have a different position in the password, but they do not know the position of their letter. He gives them all a list of 12 possible passwords, one of which is correct:
STEW, BOYS, PANS, STIG, NILE, LEER, STEM, WERE, GEMS, STAB, TEST, SPOT
He asks them all in turn if they know the password, and they all say no. He repeats this and gets the same response a number of times until the first friend to be asked says yes, and the rest no. Upon announcing that the first friend’s letter is in the second half of the alphabet, they all say yes.
What is Todd’s password?
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BRG permalink12345678910111213141516171819202122232425262728293031323334353637383940from collections import Counter# possible passwordspwds = "STEW BOYS PANS STIG NILE LEER STEM WERE GEMS STAB TEST SPOT".split()# progressively remove passwords that have letters in no other passwordsrnd, pl2p = 0, {}while True:rnd += 1# map letters to passwords which use theml2p = dict()for p in pwds:for ltr in p:l2p[ltr] = l2p.get(ltr, []) + [p]# if all passwords have been foundif not l2p or (l2p == pl2p):breakpl2p = l2p# collect passwords with letters that are not used elsewhere# and can hence be identified by one of these lettersmps = {ps[0] for l, ps in l2p.items() if len(ps) == 1}# the round on which the password is declared must have at least# one other possible password since the others don't declareif rnd > 4 and len(mps) > 1:mlps = [(l, ps[0]) for l, ps in l2p.items() if len(ps) == 1]# collect letters in these passwords that are used elsewhere ...c = Counter()for ltr, ps in mlps:c += Counter(set(ps).difference(ltr))# ... which must all be ambiguous since no-one else declaresif all(v > 1 for v in c.values()):if len(rs := [ps for ltr, ps in mlps if ltr > 'M']) == 1:print(f"*** the password is {rs[0]} ***")# remove passwords that can be identifiedprint(f"{rnd} removing {set(mps)}")for ps in mps:if ps in pwds:pwds.remove(ps)
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JohnZ permalink12345678910111213141516171819202122232425262728293031# set of possible passwordspw = {'STEW', 'BOYS', 'PANS', 'STIG', 'NILE', 'LEER', 'STEM', 'WERE', 'GEMS','STAB', 'TEST', 'SPOT'}# determine words to remove that have at least one letter that no other word# has because knowing such a letter gives immediate knowledge of the password.# loop removing words that have a unique letter(s) until all have been removedwhile pw:# make one string of all remaining passwordspwj = ''.join(pw)# find unique letters; place in alphabetical orderu = sorted(x for x in set(pwj) if pwj.count(x) == 1)# list of words to remove that contain a unique letter (or letters)r = [w for ltr in u for w in pw if ltr in w]pw -= (rs := set(r))# The password has a unique letter in the second half of the alphabet. There# must be a second word to remove at the same time with its unique letter in# the first half of the alphabet. The 2 words have three letters in common.if len(r) == 2 and u[1] >= "N" > u[0] and len(set(r[0] + r[1])) == 5:print("*** Password:", r[1], '***\n')print("remove", r[0: len(rs)], u)print("remaining",pw, '\n')
