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Transfinite epistemic logic puzzle challenge: Cheryl's birthday on steroids
- fsk 11y agotl;dr version He constructs a weird order on the rational numbers, and they play the game on that ordering. He also has steps for limit ordinals. For example: Cheryl: I have given each of you an ordinal. I didn't give both of you the same ordinal. Do you know if you have the smallest one? Bob: I can't prove I have the smallest. (I.e., I don't have 0.) Albert: With that information, I can't prove I have the smallest. (I.e., I don't have 1.) Bob: I don't have 2. Cheryl: You can go on forever without figuring out who has smaller. Bob: I don't have the smallest. (I.e., I don't have omega.) Albert: I don't have the smallest. (I.e., I don't have omega+1). Cheryl: You can do this 100 times and it wouldn't help. Bob: I don't have the smallest. (I.e., I don't have 100 * omega.) Albert: I know I have the smallest now! (I.e., I have 100 * omega+1) The weird ordering he places on the rational numbers is equivalent to playing this way with ordinals.
- AllTalk 11y agoI think it is just the usual order on the rational numbers, but since he uses only some of the rationals, it does amount to using ordinals. But, I think your solution is off by at least a factor of omega.
- fsk 11y agoI see, he has limit ordinals in his embedded order with the sequence 1 - (1/2)^n, with 1 being the equivalent of omega. You probably even could get an embedded omega squared with 1 - (1/2)^n - (1/3)^m with m>n So 1/2 = omega 3/4 = 2 * omega 7/8 = 3 * omega and 1 is omega * omega He might have omega squared in his version, but I didn't read it that carefully. Now play the version where the secret ordinal is an uncountable ordinal or a strongly inaccessible cardinal.