Transfinite epistemic logic puzzle challenge: Cheryl's birthday on steroids (jdh.hamkins.org) 2 points by AllTalk 11y ago ↗ HN
[–] fsk 11y ago ↗ tl;dr versionHe 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 ago ↗ I 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 ago ↗ I 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 with1 - (1/2)^n - (1/3)^m with m>nSo 1/2 = omega3/4 = 2 * omega7/8 = 3 * omegaand 1 is omega * omegaHe 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.
[–] AllTalk 11y ago ↗ I 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 ago ↗ I 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 with1 - (1/2)^n - (1/3)^m with m>nSo 1/2 = omega3/4 = 2 * omega7/8 = 3 * omegaand 1 is omega * omegaHe 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.
[–] fsk 11y ago ↗ I 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 with1 - (1/2)^n - (1/3)^m with m>nSo 1/2 = omega3/4 = 2 * omega7/8 = 3 * omegaand 1 is omega * omegaHe 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.
3 comments
[ 3.7 ms ] story [ 22.5 ms ] threadHe 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.
But, I think your solution is off by at least a factor 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.