Inspired by Pi Han Goh

Logic Level 4

Cheryl Welcome, Albert and Bernard, to my birthday party, and I thank you for your gifts. To return the favor, as you entered my party, I privately made known to each of you a rational number of the form $n-\frac { 1 }{ { 2 }^{ k } } -\frac { 1 }{ { 2 }^{ k+r } }$ where n and k are positive integers and r is a non-negative integer; please consider it my gift to each of you. Your numbers are different from each other, and you have received no other information about these numbers or anyone’s knowledge about them beyond what I am now telling you. Let me ask, who of you has the larger number?

Albert : I don’t know.

Bernard: Neither do I.

Albert : Indeed, I still do not know.

Bernard : And still neither do I.

Cheryl : Well, it is no use to continue that way! I can tell you that no matter how long you continue that back-and-forth, you shall not come to know who has the larger number.

Albert : What interesting new information! But alas, I still do not know whose number is larger.

Bernard : And still also I do not know.

Albert : I continue not to know.

Bernard : I regret that I also do not know.

Cheryl : Let me say once again that no matter how long you continue truthfully to tell each other in succession that you do not yet know, you will not know who has the larger number.

Albert : Well, thank you very much for saving us from that tiresome trouble! But unfortunately, I still do not know who has the larger number.

Bernard : And also I remain in ignorance. However shall we come to know?

Cheryl : Well, in fact, no matter how long we three continue from now in the pattern we have followed so far—namely, the pattern in which you two state back-and-forth that still you do not yet know whose number is larger and then I tell you yet again that no further amount of that back-and-forth will enable you to know—then still after as much repetition of that pattern as we can stand, you will not know whose number is larger! Furthermore, I could make that same statement a second time, even after now that I have said it to you once, and it would still be true. And a third and fourth as well! Indeed, I could make that same pronouncement a hundred times altogether in succession (counting my first time as amongst the one hundred), and it would be true every time. And furthermore, even after my having said it altogether one hundred times in succession, you would still not know who has the larger number!

Albert: Such powerful new information! But I am very sorry to say that still I do not know whose number is larger.

Bernard : And also I do not know.

Albert : But wait! It suddenly comes upon me after Bernard’s last remark, that finally I know who has the larger number!

Bernard : Really? In that case, then I also know, and what is more, I know both of our numbers!

Albert : Well, now I also know them!

Now, who has more number of gifts?

Albert Bernard Both have equal number of gifts

1 solution

Kee Aun Ooi
Jun 25, 2015

Let us considered the another condition only have $n$ (different and positive integers), the first person getting know who have larger number will have smaller number.

But the condition replace with $-n$ (different and negative integers), the first person getting know who have larger number will have larger number.

Now, move to the reciprocal conditional $\frac {1}{n}$ ( $n$ is different and positive integers), the first person getting know who have larger number will have larger number.

Hence, in condition $\frac {1}{-n}$ , the first person getting know who have larger number will have smaller number.

The equation in birthday party only contain two type of conditions $n, \frac {1}{-n}$ , which is the first person getting know who have larger number will have smaller number.

$Albert$ getting know first, $Bernard$ will have larger number.

Inspired by Pi Han Goh

1 solution

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