Day 25: CM, Page 12

Day 25: I've missed a couple of days. I'm now on pace for Nov 5, 2016.

The authors continue to examine the binary representation of the Josephus function. They observe that since the solution to the Josephus problem is a cyclic left bit shift, you might expect that m iterations of the function would return to the original number. However, this is not true because zeros are removed from the number during the shifting process. For example:

J(1101) = 1011
J(1011) = 0111 = 111
J(111) = 111

For a given number, you can find the convergent number by counting the number of 1's in the original number's binary representation. The point of convergence can be defined as 2^v(n) - 1 where v(n) is the numbers of 1's in n.

More on the Josephus problem tomorrow. The authors continue to generalize it even further.