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This demonstration is utter nonsense, because it's musing a critical step. By the automorphism of color permutations, all possible non-trivial reaponses are equ
by alisonkisk 5y ago
This demonstration is utter nonsense, because it's musing a critical step. By the automorphism of color permutations, all possible non-trivial reaponses are equivalent and prove nothing.
What's missing is that the prover has to send you something like a salted hash of every edge coloring (called a "cryptographic commitment") in advance, so that you can verify that the revealed coloring matches what was already committed.
But since the number of distinct single edge colorings is tiny (6), you could crack the hash easily, so you need a fancier commitment protocol than just a hash, perhaps some sort of mutually trusted hashing oracle that only lets you send one query per proof.
- kevinwang 5y agoWhy do you need something fancier? Couldn't you commit to n=1..6 by hashing a k*n where k is a random integer?