I'd asked whether my solution involving cosets of error correcting codes, had any use. Here is a use, sort of. Knuth in some exercise asked "if in a table of random numbers, there is a misprint -- does that matter, i.e. can it stop the numbers from being random?" He then presented a "proof" that the answer is "yes." Well, the Mensan problem is another such, rather more direct, "proof." This also yields a way, I suppose, to do "steganography." (And the bitstring we begin with can be either random or not, steganography still works.) If you wanted to make a harder version of the Mensan problem, you could make one that depends on hairier error correcting codes (and knowledge of their properties) than the Hamming codes. Two example good versions: A. suppose the Warden had 23 coins (12+11, 2-row array), and prisoner #1 had to flip at most 3 coins to deliver an 11-bit message to prisoner #2. B. suppose the Warden had 15 coins (7 and 8 in a 2-row array), and prisoner #1 had to flip at most 2 coins to deliver a 7-bit message to prisoner #2. For an even harder version of the problem, which also seems perhaps even more relevant to steganography... change the rules as follows: ENHANCED_MENSAN_PROBLEM(N,J,K,L): 1. prisoners #1 & #2 agree on a protocol. 2. prisoner #2 leaves. 3. Warden chooses N-bit string. 4. prisoner #1 flips up to J bits. 5. prisoner #1 leaves. 6. Warden flips up to K bits in attempt to ruin it. 7. prisoner #2 re-enters, observes pattern, and deduces message from prisoner #1, which could have been up to L possible messages. Anyone see how to solve that one (& good parameter sets N,J,K,L to suggest)? This is sort of an adversarial steganography model.