Hashing binary numbers with don't care bits - data-structures

Question

How can I find out if a binary number is contained in a set, where it is possible that an element of the set has don’t care bits?
I thought about using hash table, but there is a need to duplicate the numbers with don’t care bits in the hash table in order to cover all the possibilities.
For example:
The set of numbers is:
0 00x1
1 10xx
2 110x
3 1010
4 11x1
5 0010
and the number is 0011, the result should be 0.

Answer 1

If number of digits of binary number are limited then you can duplicate those don't care bits and convert the binary numbers to integers then use these integers as keys for map and other as values.
Example
0 00x1
1 10xx
can be converted to
0001 0
0011 0
1000 1
1001 1
1010 1
1011 1
and saved as
i j
1 0
3 0
8 1
9 1
10 1
11 1
where i is the key and j is the value

Answer 2

Let's say you have the binary number 1xxx, that would match 8 numbers. So, do not go with duplicating for each option.
You have to keep the "do not care" bits somewhere. Use another number for this, set the "do not care" bits to 1. If we go over your example:
i x y
0 00x1 0010
1 10xx 0011
2 110x 0001
3 1010 0000
4 11x1 0010
5 0010 0000
And you need to decide what to use for x, 0 or 1. You can use any of them, once you keep the information in the second number it does not matter.
Now use bitwise operations:
if ((n ^ x[i]) | y[i]) == y[i] then match
This solution is based on checking the existence of any non-matching bits except do-not-care bits. (n xor x[i]) gives the non-matching bits, then or'ing it with y[i] should not be different than y[i].
If we go over your example, and assuming you choose 0 for x, the check becomes
i:0 -->> ((0011 ^ 0001) | 0010) == 0010 -->> match!
i:1 -->> ((0011 ^ 1000) | 0011) != 0011 -->> no match!
i:2 -->> ((0011 ^ 1100) | 0001) != 0001 -->> no match!
i:3 -->> ((0011 ^ 1010) | 0000) != 0001 -->> no match!
i:4 -->> ((0011 ^ 1101) | 0010) != 0001 -->> no match!
i:5 -->> ((0011 ^ 0010) | 0000) != 0000 -->> no match!