# Bye (for now?)

SPMars
Posers and Puzzles 20 Sep '06 16:13
1. 20 Sep '06 16:13
Got too much work on at the moment, so am having a break from RHP for at least the next few months.

Since this is the only forum I have really posted in I thought I'd say bye and leave you with a couple of (hard) puzzles!!

1) A knight starts at the corner of a chessboard, and at each time t = 1, 2, 3, 4, ... it moves once, with all possible moves equally likely. Let T be the first time the knight returns to its starting corner. What is the expected value of T?

2) A monkey is typing on a typewriter, with each letter A, B, C, ... , Z equally likely. How long before you expect to see the sequence ABRACADABRA?

Enjoy!
2. 20 Sep '06 18:16
An ugly incomplete solution for question 1

I can solve it with a system of 64 equations in 64 variables, in some mathematical software. Basically we need to write a number in each square of the chessboard, such that at a1 we have 0, and at each other square the number is 1 more than the average of the numbers which are at a knight's-move distance. Then the solution is the number at b3 (or c2) plus 1. (I think)
3. 20 Sep '06 18:18
I managed to solve problem 2 in the binary case - the monkey types 0 or 1, and we wait for a specific string of 0's and 1's. But I didn't solve yet the original problem. Maybe you should try the binary case first.
4. PBE6
Bananarama
20 Sep '06 19:32
Originally posted by SPMars
Got too much work on at the moment, so am having a break from RHP for at least the next few months.

Since this is the only forum I have really posted in I thought I'd say bye and leave you with a couple of (hard) puzzles!!

1) A knight starts at the corner of a chessboard, and at each time t = 1, 2, 3, 4, ... it moves once, with all possible moves equall ...[text shortened]... ... , Z equally likely. How long before you expect to see the sequence ABRACADABRA?

Enjoy!
Good luck SPMars! I always enjoyed your posts and analyses. Take care! We'll work on these in the meantime.
5. 22 Sep '06 03:06
Originally posted by David113
I managed to solve problem 2 in the binary case - the monkey types 0 or 1, and we wait for a specific string of 0's and 1's. But I didn't solve yet the original problem. Maybe you should try the binary case first.
it's just 26^11 is it not?
6. 22 Sep '06 10:311 edit
No. The waiting time depends on which string of 0's and 1's you are waiting for.

Example:

If you wait for 01 or 10 then the expected waiting time is 4; but if you wait for 00 or 11 then the expected waiting time is 6.
7. AThousandYoung
iPlay el Degüello!
23 Sep '06 02:38
Originally posted by SPMars
Got too much work on at the moment, so am having a break from RHP for at least the next few months.

Since this is the only forum I have really posted in I thought I'd say bye and leave you with a couple of (hard) puzzles!!

1) A knight starts at the corner of a chessboard, and at each time t = 1, 2, 3, 4, ... it moves once, with all possible moves equall ...[text shortened]... ... , Z equally likely. How long before you expect to see the sequence ABRACADABRA?

Enjoy!
2) 26^11 letters is how many I'd expect before that sequence occurred.
8. AThousandYoung
iPlay el Degüello!
23 Sep '06 02:39
Originally posted by David113
No. The waiting time depends on which string of 0's and 1's you are waiting for.

Example:

If you wait for 01 or 10 then the expected waiting time is 4; but if you wait for 00 or 11 then the expected waiting time is 6.
What? Why?
9. 23 Sep '06 18:41
See proof here:

http://www.qbyte.org/puzzles/p082s.html