Symmetric Bijective Algorithm for Integers

Question

I need an algorithm that can do a one-to-one mapping (ie. no collision) of a 32-bit signed integer onto another 32-bit signed integer.

My real concern is enough entr

Answer 1

Can you use a random generated lookup-table? As long as the random numbers in the table are unique, you get a bijective mapping. It's not symmetric, though.

One 16 GB lookup-table for all 32 bit values is probably not practical, but you could use two separate 16-bit lookup tables for the high-word and the low word.

PS: I think you can generate a symmetric bijective lookup table, if that's important. The algorithm would start with an empty LUT:

+----+        +----+
|  1 |   ->   |    |
+----+        +----+
|  2 |   ->   |    |
+----+        +----+
|  3 |   ->   |    |
+----+        +----+
|  4 |   ->   |    |
+----+        +----+

Pick the first element, assign it a random mapping. To make the mapping symmetric, assign the inverse, too:

+----+        +----+
|  1 |   ->   |  3 |
+----+        +----+
|  2 |   ->   |    |
+----+        +----+
|  3 |   ->   |  1 |
+----+        +----+
|  4 |   ->   |    |
+----+        +----+

Pick the next number, again assign a random mapping, but pick a number that's not been assigned yet. (i.e. in this case, don't pick 1 or 3). Repeat until the LUT is complete. This should generate a random bijective symmetric mapping.