Dominoes on a Board

Alright, time for a new puzzle. I have one I found recently I want to post, but some of the discussion first requires I post a puzzle I learned many years ago from my students in physics labs at UBC. It isn't a hard puzzle, which is why I never posted it, but the solution is very elegant, and it is one I like.

Consider an 8x8 chess board with two corner places deleted, in opposite corners. Attempt to tile this board with 2x1 dominoes. Prove or disprove whether such a tiling is possible.

Naturally such a tiling requires exactly (64-2)/2=31 dominoes to do this, as each domino will cover 2 squares and each square must be covered exactly once. Anyway, yeah, its not too tricky.

Josephus Solution

Time for the solution to the Josephus problem from last time. As is common for problems that are basically induction puzzles, its easiest to see what is going on by looking at a few cases.

First of all, you can just work through who wins in the cases N=1,2,3,4,5,6,7,8 and so on, you will see that the winners are 1,1,3,1,3,5,7,1. A bit of an interesting note is that all the winners are odd numbers, of course this note is instantly less interesting when you realize that all the even numbered individuals are killed on the first pass. Actually, if N is even, that brings up a good point, specifically that for N=2K you immediately reduce to the problem with K people, with #1 going first and a bit of a renumbering that person 2m-1 is now person m. But if this new K also happens to be even, then you can half it again. From here you can see that if the initial N is some power of 2, then person #1 will be the last one standing.

At this point, we can realize that whoevers turn it is when the total number of people is a power of two is the person who is going to win. Naturally, if the total is an even number, then that is person #1, but if N is equal to 2^k+M, for some M less than 2^k (or we would just use k+1 in the power instead), then once M people die, the next person on turn will live. Clearly since M is less than half of N, those first M people will die on the first pass around the table, so when M people have died it is the turn of person #2M+1.

This gives us the general solution for an arbitrary N. Let 2^k be the largest power of 2 less than or equal to N, and let M=N-2^k. The final person standing will be person #2M+1.

Josephus Problem

Ok, time for a new puzzle, I learned this one some time ago, it is a well known problem with a somewhat cool solution. This problem is well enough known as the "Josephus problem" that you can certainly just look it up along with its solution yourself if you hate my style of writing. The original problem involved N people who, having been captured by the enemy, are performing some sort of suicide pact, where they all kill eachother and there is one person standing at the end. There is a bunch of thematic bubble wrap to go with the puzzle, but I'm going to omit it just for the math. Anyway, here is the problem:

N individuals are in a circle, numbered 1 through N starting somewhere with 1 and increasing with each person to the left. Starting with person number 1, that person will kill the person to their left, and then the next person to the left will take a turn. This will continue until one person remains, which person is it?

Since that explanation is hard to do without an example, lets do one. Suppose N=5, then we start with 1 killing 2, 3 goes next, killing 4. 5 is next to take a turn, and 1 is to their left, so 1 dies and we only have 3 and 5 remaining. 3 goes next, killing 5 and the solution to the puzzle is 3.

So, in general, for an arbitrary N, who is the last person standing?