nice

I meant that the people run the algorithm together. I understand if you want to call that cheating.

I think that adding this “generate a random permutation” part improves the more naive algorithm of “use the following given permutation…” (as obviously the adverdary can arrange the boxes in such a way as to fail their system).

I agree that letting all the men generate a random permutation together is similar to having a pre-shared secret.

And yes – It is a great riddle!

Well, if the algorithm says “generate a random permutation” in the beginning, everybody will draw independently (not the same h by my notation), and have %30 chance of drawing no big cycles. Therefore it would require for all P*h to have no big cycles which would happen with probability of (0.3)^100. And if it’s not independent (i.e. an Oracle draws them together the same random permutation) that’s information sharing.

You could deviate the question a bit by saying that the people are allowed to pass information of some size, let’s say X bits. Then the solution would be, generate 2^X random permutations (apriori, known to everybody including adversary). Then the first man draws a number between 0 and 2^X-1 and transmit that number (exactly X bits). Then everybody knows which permutation to use. I wonder what’s the minimal X so that the adversary cannot screw up permutation f*h for all possible h.

Great riddle overall!

As I explained in the solution above: “The men first decide upon a random one-to-one matching from their names to the numbers 1-100 (this selection being random, eliminates possible efforts from the part of the organizer of the boxes to fail the system). ”

I think this is exactly what Matan meant (and if I understood you correctly then its what you mean as well).

My view of this is a little different (instead of saying the men have a shared secret, I say the algorithm uses random – kind of the same). The men have no apriori secret (i.e. the adversary knows everything in advance) but the algorithm’s first step is “generate a random permutation”.

Kind of a philosophical issue…

I came up with something a little bit better, where the secret relies not on the whole algorithm – the men come up with a random permutation of their own, let that be h, and rather than following P, they follow P*h. The same calculation applies and they have 30% chance of saving, but this time they can tell the adversary the algorithm, but they need to keep their permutation secret (kind of like Kerkhoff’s cipher design rule).

I also like the “Warning! Very Hard!” logo, and I think I’ll keep it.

Anyways – if it is too easy for you, you can always try to solve it with 1000 men instead of 100!

guess I’ll check my constants and get back to you.

You raised an interesting question, and I still have to decide how I want solutions to be posted. I think what I will do is publish a “solution post”, which will be password protected with a known password (such as “solution”). That way everyone will be able to see the solutions but no one will see them accidentally.

Anyway, I am going to wait for a couple of weeks since the time I publish a riddle and the time I publish its solution (otherwise the temptation to read the solution is to big).

If the names are put in the boxes randomly, then I solved it!!!
What a cool riddle.

I want to comment about the solution, but is there any way to put text in spoiler tags? I don’t want to ruin the solution for people reading the comments…