Sunday Puzzle thread

[Harry]

A place to discuss/post solutions to the Sunday puzzle. A bit late I know, but it will remind me to get ahead of the game next year. Harry

[Robb27]

@Harry

Given how you describe the puzzle, I am giving two answers:
1) which is in the spirit intended:
A and B cross — takes 2 min with running total 2
A returns — 1 min, running total 3
C and D cross — 8 min, running total 11
B returns — 2 min, running total 13
A and B cross — 2 min, running total 15

2) As it WASN’t stated that they had to cross at night (!) then they just wait for the morning. They cross in the light so the torch isn’t needed and they can go in any paired order they like! From above, C+D = 8 + A+B =2 gives a crossing of 10 mins with no torch in daylight.

I do like the second solution. Speaks to a lateral way of thinking that might appeal to our sponsors at GCHQ. Mind you I think they would also like the first answer!!Harry

[ByteInBits]

Bridge of sighs? (Sunday 7th December by Harry)

Yes it is possible for all to cross in 15 minutes, here is how:

Alf, Bill, Carl and Dick are needing to cross.
Alf takes charge and switchs the light on.
He then tells Bill to cross with him.
Alf then returns alone, gives torch to Carl,
he tells Carl and Dick to cross over together and asks
them to give the torch to Bill and send Bill back to him
Alf and Bill then cross over together again.
Now everyone is across and, as if on cue, the torch flickers and goes out!

Or if you prefer:
Label the people A, B, C and D.

1. A and B cross together : Torch time = 2 minutes
2. A returns with the torch : Torch time = 1 minute
3. C and D cross together : Torch time = 8 minutes
4. B returns with the torch : Torch time = 2 minutes
5. A and B cross together again : Torch time = 2 minutes

Total time: 2 + 1 + 8 + 2 + 2 = 15 minutes.

[solveded Monday evening 8/12/25]

[Crackerjack_404]

Fun thing about this week’s bridge and torch puzzle, it’s easy to see that you can in fact get the people across in 15 mins, but we can also model it as a graph where the possible state (which person is on which side) becomes a node, and valid movement as an edge between the nodes, with the time it takes to cross as the weights. Then, the whole thing becomes a shortest-path problem which can be solved using something like Dijkstra’s or A*.

The good thing about generalising it is that we can now have any number of people with any crossing times, which is pretty cool!

[Crackerjack_404]

14/12 puzzle:

Assuming they have plenty of keys and padlocks, Boris puts the ring in a box and locks it. He then puts that box inside another locked box, and repeats until the parcel is heavy enough to be an absolute pain to steal or open quickly, but still small enough for the postal service of Kleptopia to deliver. The ring arrives safely by default. If an attacker is sufficiently confused, exhausted, or emotionally compromised, hopefully the system is pretty secure.

Or… if Boris prefers mathematics to cardboard, he could also use public-key cryptography to avoid ever sending a secret key at all.

(Bit of an etymology aside- “Kleptopia” combines Greek ‘kleptein’ (to steal) with -topia, meaning land/place)

Looking forward to the Christmas Miscellany!

[AndGiggles]

Let’s hope Boris picks the right ring to initiate their relationship: it is really integral! Personally, because I am very poor, I would send a ring of polynomials with integer coefficients (for the purpose of this joke all rings are commutative).

That is indeed the key to this particular conundrum! Harry

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