Question

Well I want to generate pseudo-random bits for a mathematical problem. The Bit size is user fed. But unlike normal problems there’s a small twist here. The mathematical problem also gives valid results if the bits begin with 0’s and literally match the size input by user.
Example: For User Input 4, all the following are valid outputs:

-> 0000
-> 0001
-> 1000
-> 0100

etc.
I am using this function to generate the numbers now:

int randomChoice = choice.Next(1 << (input - 1), (1 << input) - 1);
bitSize = randomChoice;

Is there any other way of random bit generation that may begin with 0’s as long as it follows the literal bit length in C# without going into complex array manipulation?

Answer 1

The following line:

int randomChoice = choice.Next(1 << (input - 1), (1 << input) - 1)

for input = 4 is equivalent to:

int randomChoice = choice.Next(8, 15);

Which will per MSDN documentation pick one of the following values: 8, 9, 10, 11, 12, 13, 14 – randomly. Notice that the value 15 will not be picked as the documentation states clearly that the lower bound (8) is )inclusive_ while the upper bound (15) is exclusive.

Those values in binary are:

8 : 1000
9 : 1001
10: 1010
11: 1011
12: 1100
13: 1101
14: 1110

I believe what @Jon Skeet was saying is that you are generating random numbers that always have the MSB (left-most bit) set. And are thus missing out on all values that fit in 4 bits that would start with 0 (have the MSB unset).

I would also point out that you should probably include value 15 here, since that also fits in the 4 bits. So, with the two modifications, the line of code should read:

int randomChoice = choice.Next(0, (1 << input));

That would generate all values from 0 to 15, inclusive:

0 : 0000
1 : 0001
2 : 0010
3 : 0011
    ...
13: 1101
14: 1110
15: 1111

Is that what you’re looking for?