is one of the inventors of assymetric encryption – probably the biggest paradigm shift in the history of cryptography. I just stumbled upon a quote by him which goes well with my recent post about research and willpower
“The way to get to the top of the heap in terms of developing original research is to be a fool, because only fools keep trying. You have idea number 1, you get excited, and it flops. Then you have idea number 2, you get excited, and it flops. Then you have idea number 99, you get excited, and it flops. Only a fool would be excited by the 100th idea, but it might take 100 ideas before one really pays off. Unless you’re foolish enough to be continually excited, you won’t have the motivation, you won’t have the energy to carry it through. God rewards fools.”
I wish I was more of a fool…
I just heard a nice riddle and wanted to share.
Prisoner A and prisoner B are given an integer between 1 and 100. Each prisoner only knows his own number, but also that the two numbers are consecutive. For example, if prisoner A got the number 60, he knows that prisoner B got either 59 or 61. At the end of every hour each prisoner can choose to guess the number of the other. If either prisoner guesses correctly, they both go free. However, if either prisoner guesses wrong, they both get executed (even if at the same time the other guessed correctly). Both prisoners can choose not to guess for as many hours as they like.
The two prisoners are in different cells and cannot communicate in any way. Also, they did not have time to coordinate a strategy in advance. Each can only assume that the other prisoner is intelligent. Can they always guess correctly? After how many hours?
I think it would be nice not to post solutions in the comments. Although you can just write the number of hours you got.
I just recalled a nice mathematical riddle. I can’t remember where I originally read it, but it was likely in one of the blogs that are in the blogroll to the right.
A game takes place where person A and person B are on the same team while person Z is their adversary. There are 100 boxes and 100 notes containing the numbers 1 to 100 (that is, each number is on exactly one note). The game goes as follows:
- First, only Z and A are in the room. Z places one note in each box.
- A sees the actions of Z and may afterwards pick two boxes and switch the notes in them (with each other). He may only perform one such switch.
- Z sees the actions of A and then chooses a number N between 1 and 100.
- A leaves the room while B enters it (they cannot exchange information during this process). Z tells B the number N.
- Finally, B needs to find the box that contains the note with the number N. For this purpose, B may open up to 50 boxes.
If A chooses the 50 boxes at random, his probability of success is obviously 0.5. Do A and B have a strategy for increasing their chances? Or does Z have a strategy for which 0.5 is the best possible?
Feel free to write the answer as a comment. However, I think that it would be nice not to provide a full explanation here (that is, only to write whether it’s 0.5 or what better probability you can get).
(Later edit: I seem to be getting senile! I just noticed that I wrote the same riddle in a post two years ago…)