This is an interview question:
What is the minimum representation in bits of two positions on an 8x8 chessboard?
I found the answer http://www.careercup.com/question?id=4981467352399872
But I am unable to understand what the author is trying to convey when she says:
You can represent 2^n values with n bits. However, you can represent 2^n + 2^(n-1) + 2^(n-2) + ... 1 = 2^(n+1) - 1 values with atmost n bits. So you can represent 2^11 - 1 = 2047 different values using just 10 bits.
I am not seeking an explanation of what the author is suggesting in his answer, but I am more interested in solving the problem itself. As far as I can think, since there are 64C2 = 2016 ways to represent two pieces on an 8x8 board, the minimum number of bits required should be 11. But someone suggested that one can use just 10 bits to represent the board. How?
The author is saying that you can represent the positions using 5, 6, 7, 8, 9 and 10 bit values.
In binary 2016 is 11111100000 (1024 + 512+ 256 + 128 + 64 + 32)
5 bits (00000 - 11111) represent 32 positions
6 bits (000000 - 111111) represent 64 positions
7 bits (0000000 - 1111111) represent 128 positions
8 bits (00000000 - 11111111) represent 256 positions
9 bits (000000000 - 111111111) represent 512 positions
10 bits (0000000000 - 1111111111) represent 1024 positions
A total of 2016 positions.
This could be implemented in languages with bit collections, e.g. C++ bitset, which has a size function to get the length.
Here's an example for a 2x2 board which will hopefully explain this better.
For a 2x2 board, there are 4C2 (6) positions
.x x. .. xx .x x.
.x x. xx .. x. .x
so you could use 3 bits 000, 001, 010, 011, 100, 101 and 110
But 6 is binary 110 (4+2) so you could use 1 bit (0-1) for 2 of the positions and 2 bits (00, 01, 10, 11) for the remaining 4. So the positions are:
0, 1, 00, 01, 10, 11.