Mapping two integers to one, in a unique and deterministic way

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Imagine two positive integers A and B. I want to combine these two into a single integer C.

There can be no other integers D and E which combine to C. So combining them with the addition operator doesn't work. Eg 30 + 10 = 40 = 40 + 0 = 39 + 1 Neither does concatination work. Eg "31" + "2" = 312 = "3" + "12"

This combination operation should also be deterministic (always yield the same result with the same inputs) and should always yield an integer on either the positive or the negative side of integers.

19 Answers

You're looking for a bijective NxN -> N mapping. These are used for e.g. dovetailing. Have a look at this PDF for an introduction to so-called pairing functions. Wikipedia introduces a specific pairing function, namely the Cantor pairing function:

pi(k1, k2) = 1/2(k1 + k2)(k1 + k2 + 1) + k2

Three remarks:

  • As others have made clear, if you plan to implement a pairing function, you may soon find you need arbitrarily large integers (bignums).
  • If you don't want to make a distinction between the pairs (a, b) and (b, a), then sort a and b before applying the pairing function.
  • Actually I lied. You are looking for a bijective ZxZ -> N mapping. Cantor's function only works on non-negative numbers. This is not a problem however, because it's easy to define a bijection f : Z -> N, like so:
    • f(n) = n * 2 if n >= 0
    • f(n) = -n * 2 - 1 if n < 0

If A and B can be expressed with 2 bytes, you can combine them on 4 bytes. Put A on the most significant half and B on the least significant half.

In C language this gives (assuming sizeof(short)=2 and sizeof(int)=4):

int combine(short A, short B)
{
    return A<<16 | B;
}

short getA(int C)
{
    return C>>16;
}

short getB(int C)
{
    return C & 0xFFFF;
}

Is this even possible?
You are combining two integers. They both have the range -2,147,483,648 to 2,147,483,647 but you will only take the positives. That makes 2147483647^2 = 4,61169E+18 combinations. Since each combination has to be unique AND result in an integer, you'll need some kind of magical integer that can contain this amount of numbers.

Or is my logic flawed?

Let number a be the first, b the second. Let p be the a+1-th prime number, q be the b+1-th prime number

Then, the result is pq, if a<b, or 2pq if a>b. If a=b, let it be p^2.

The standard mathematical way for positive integers is to use the uniqueness of prime factorization.

f( x, y ) -> 2^x * 3^y

The downside is that the image tends to span quite a large range of integers so when it comes to expressing the mapping in a computer algorithm you may have issues with choosing an appropriate type for the result.

You could modify this to deal with negative x and y by encoding a flags with powers of 5 and 7 terms.

e.g.

f( x, y ) -> 2^|x| * 3^|y| * 5^(x<0) * 7^(y<0)

Although Stephan202's answer is the only truly general one, for integers in a bounded range you can do better. For example, if your range is 0..10,000, then you can do:

#define RANGE_MIN 0
#define RANGE_MAX 10000

unsigned int merge(unsigned int x, unsigned int y)
{
    return (x * (RANGE_MAX - RANGE_MIN + 1)) + y;
}

void split(unsigned int v, unsigned int &x, unsigned int &y)
{
    x = RANGE_MIN + (v / (RANGE_MAX - RANGE_MIN + 1));
    y = RANGE_MIN + (v % (RANGE_MAX - RANGE_MIN + 1));
}

Results can fit in a single integer for a range up to the square root of the integer type's cardinality. This packs slightly more efficiently than Stephan202's more general method. It is also considerably simpler to decode; requiring no square roots, for starters :)

f(a, b) = s(a+b) + a, where s(n) = n*(n+1)/2

  • This is a function -- it is deterministic.
  • It is also injective -- f maps different values for different (a,b) pairs. You can prove this using the fact: s(a+b+1)-s(a+b) = a+b+1 < a.
  • It returns quite small values -- good if your are going to use it for array indexing, as the array does not have to be big.
  • It is cache-friendly -- if two (a, b) pairs are close to each other, then f maps numbers to them which are close to each other (compared to other methods).

I did not understand what You mean by:

should always yield an integer on either the positive or the negative side of integers

How can I write (greater than), (less than) characters in this forum?

Check this: http://en.wikipedia.org/wiki/Pigeonhole_principle. If A, B and C are of same type, it cannot be done. If A and B are 16-bit integers, and C is 32-bit, then you can simply use shifting.

The very nature of hashing algorithms is that they cannot provide a unique hash for each different input.

It isn't that tough to construct a mapping:

   1  2  3  4  5  use this mapping if (a,b) != (b,a)
1  0  1  3  6 10
2  2  4  7 11 16
3  5  8 12 17 23
4  9 13 18 24 31
5 14 19 25 32 40

   1  2  3  4  5 use this mapping if (a,b) == (b,a) (mirror)
1  0  1  2  4  6
2  1  3  5  7 10
3  2  5  8 11 14
4  4  8 11 15 19
5  6 10 14 19 24


    0  1 -1  2 -2 use this if you need negative/positive
 0  0  1  2  4  6
 1  1  3  5  7 10
-1  2  5  8 11 14
 2  4  8 11 15 19
-2  6 10 14 19 24

Figuring out how to get the value for an arbitrary a,b is a little more difficult.

What you suggest is impossible. You will always have collisions.

In order to map two objects to another single set, the mapped set must have a minimum size of the number of combinations expected:

Assuming a 32-bit integer, you have 2147483647 positive integers. Choosing two of these where order doesn't matter and with repetition yields 2305843008139952128 combinations. This does not fit nicely in the set of 32-bit integers.

You can, however fit this mapping in 61 bits. Using a 64-bit integer is probably easiest. Set the high word to the smaller integer and the low word to the larger one.

Say you have a 32 bit integer, why not just move A into the first 16 bit half and B into the other?

def vec_pack(vec):
    return vec[0] + vec[1] * 65536;


def vec_unpack(number):
    return [number % 65536, number // 65536];

Other than this being as space efficient as possible and cheap to compute, a really cool side effect is that you can do vector math on the packed number.

a = vec_pack([2,4])
b = vec_pack([1,2])

print(vec_unpack(a+b)) # [3, 6] Vector addition
print(vec_unpack(a-b)) # [1, 2] Vector subtraction
print(vec_unpack(a*2)) # [4, 8] Scalar multiplication

We can encode two numbers into one in O(1) space and O(N) time. Suppose you want to encode numbers in the range 0-9 into one, eg. 5 and 6. How to do it? Simple,

  5*10 + 6 = 56. 
   
    5 can be obtained by doing 56/10 
    6 can be obtained by doing 56%10.

Even for two digit integer let's say 56 and 45, 56*100 + 45 = 5645. We can again obtain individual numbers by doing 5645/100 and 5645%100

But for an array of size n, eg. a = {4,0,2,1,3}, let's say we want to encode 3 and 4, so:

 3 * 5 + 4 = 19               OR         3 + 5 * 4 = 23
 3 :- 19 / 5 = 3                         3 :- 23 % 5 = 3
 4 :- 19 % 5 = 4                         4 :- 23 / 5 = 4

Upon generalising it, we get

    x * n + y     OR       x + n * y

But we also need to take care of the value we changed; so it ends up as

    (x%n)*n + y  OR x + n*(y%n)

You can obtain each number individually by dividing and finding mod of the resultant number.

Given positive integers A and B, let D = number of digits A has, and E=number of digits B has The result can be a concatenation of D, 0, E, 0, A, and B.

Example: A = 300, B = 12. D = 3, E=2 result = 302030012. This takes advantage of the fact that the only number that starts with 0, is 0,

Pro: Easy to encode, easy to decode, human readable, significant digits can be compared first, potential for compare without calculation, simple error checking.

Cons: Size of results is an issue. But that's ok, why are we storing unbounded integers in a computer anyways.

If you want more control such as allocate X bits for the first number and Y bits for the second number, you can use this code:

class NumsCombiner
{

    int num_a_bits_size;
    int num_b_bits_size;

    int BitsExtract(int number, int k, int p)
    {
        return (((1 << k) - 1) & (number >> (p - 1)));
    }

public:
    NumsCombiner(int num_a_bits_size, int num_b_bits_size)
    {
        this->num_a_bits_size = num_a_bits_size;
        this->num_b_bits_size = num_b_bits_size;
    }

    int StoreAB(int num_a, int num_b)
    {
        return (num_b << num_a_bits_size) | num_a;
    }

    int GetNumA(int bnum)
    {
        return BitsExtract(bnum, num_a_bits_size, 1);
    }

    int GetNumB(int bnum)
    {
        return BitsExtract(bnum, num_b_bits_size, num_a_bits_size + 1);
    }
};

I use 32 bits in total. The idea here is that if you want for example that first number will be up to 10 bits and second number will be up to 12 bits, you can do this:

NumsCombiner nums_mapper(10/*bits for first number*/, 12/*bits for second number*/);

Now you can store in num_a the maximum number that is 2^10 - 1 = 1023 and in num_b naximum value of 2^12 - 1 = 4095.

To set value for num A and num B:

int bnum = nums_mapper.StoreAB(10/*value for a*/, 12 /*value from b*/);

Now bnum is all of the bits (32 bits in total. You can modify the code to use 64 bits) To get num a:

int a = nums_mapper.GetNumA(bnum);

To get num b:

int b = nums_mapper.GetNumB(bnum);

EDIT: bnum can be stored inside the class. I did not did it because my own needs I shared the code and hope that it will be helpful.

Thanks for source: https://www.geeksforgeeks.org/extract-k-bits-given-position-number/ for function to extract bits and thanks also to mouviciel answer in this post. Using these to sources I could figure out more advanced solution

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