Re turing and cryptanalzing bigrams in key covers used in layter naval enigma

From web:-

http://stackoverflow.com/questions/352670/weighted-random-selection-with-and-without-replacement

One of the fastest ways to make many with replacement samples from an unchanging list is the alias method. The core intuition is that we can create a set of equal-sized bins for the weighted list that can be indexed very efficiently through bit operations, to avoid a binary search. It will turn out that, done correctly, we will need to only store two items from the original list per bin, and thus can represent the split with a single percentage.

Let’s us take the example of five equally weighted choices, (a:1, b:1, c:1, d:1, e:1)

To create the alias lookup:

Normalize the weights such that they sum to 1.0. (a:0.2 b:0.2 c:0.2 d:0.2 e:0.2) This is the probability of choosing each weight.

Find the smallest power of 2 greater than or equal to the number of variables, and create this number of partitions, |p|. Each partition represents a probability mass of 1/|p|. In this case, we create 8 partitions, each able to contain 0.125.

Take the variable with the least remaining weight, and place as much of it’s mass as possible in an empty partition. In this example, we see that a fills the first partition. (p1{a|null,1.0},p2,p3,p4,p5,p6,p7,p8) with (a:0.075, b:0.2 c:0.2 d:0.2 e:0.2)

If the partition is not filled, take the variable with the most weight, and fill the partition with that variable.

Repeat steps 3 and 4, until none of the weight from the original partition need be assigned to the list.

For example, if we run another iteration of 3 and 4, we see

(p1{a|null,1.0},p2{a|b,0.6},p3,p4,p5,p6,p7,p8) with (a:0, b:0.15 c:0.2 d:0.2 e:0.2) left to be assigned

At runtime:

Get a U(0,1) random number, say binary 0.001100000

bitshift it lg2(p), finding the index partition. Thus, we shift it by 3, yielding 001.1, or position 1, and thus partition 2.

If the partition is split, use the decimal portion of the shifted random number to decide the split. In this case, the value is 0.5, and 0.5 < 0.6, so return a.

Here is some code and another explanation, but unfortunately it doesn't use the bitshifting technique, nor have I actually verified it.