Find 2 numbers in an unsorted array equal to a given sum

Here's a summary of Dimitris Andreou's link.

Remember sum of i-th powers, where i=1,2,..,k. This reduces the problem to solving the system of equations

a₁ + a₂ + ... + a_k = b₁

a₁² + a₂² + ... + a_k² = b₂

...

a₁^k + a₂^k + ... + a_k^k = b_k

Using Newton's identities, knowing b_i allows to compute

c₁ = a₁ + a₂ + ... a_k

c₂ = a₁a₂ + a₁a₃ + ... + a_k-1a_k

...

c_k = a₁a₂ ... a_k

If you expand the polynomial (x-a₁)...(x-a_k) the coefficients will be exactly c₁, ..., c_k - see Viète's formulas. Since every polynomial factors uniquely (ring of polynomials is an Euclidean domain), this means a_i are uniquely determined, up to permutation.

This ends a proof that remembering powers is enough to recover the numbers. For constant k, this is a good approach.

However, when k is varying, the direct approach of computing c₁,...,c_k is prohibitely expensive, since e.g. c_k is the product of all missing numbers, magnitude n!/(n-k)!. To overcome this, perform computations in Z_q field, where q is a prime such that n <= q < 2n - it exists by Bertrand's postulate. The proof doesn't need to be changed, since the formulas still hold, and factorization of polynomials is still unique. You also need an algorithm for factorization over finite fields, for example the one by Berlekamp or Cantor-Zassenhaus.

High level pseudocode for constant k:

• Compute i-th powers of given numbers
• Subtract to get sums of i-th powers of unknown numbers. Call the sums b_i.
• Use Newton's identities to compute coefficients from b_i; call them c_i. Basically, c₁ = b₁; c₂ = (c₁b₁ - b₂)/2; see Wikipedia for exact formulas
• Factor the polynomial x^k-c₁x^k-1 + ... + c_k.
• The roots of the polynomial are the needed numbers a₁, ..., a_k.

For varying k, find a prime n <= q < 2n using e.g. Miller-Rabin, and perform the steps with all numbers reduced modulo q.

EDIT: The previous version of this answer stated that instead of Z_q, where q is prime, it is possible to use a finite field of characteristic 2 (q=2^(log n)). This is not the case, since Newton's formulas require division by numbers up to k.

Sorting solutions have been proposed, but sorting is a little too much: We don't need order; we just need equality groups.

So hashing would be enough (and faster).

1. For each color of socks, form a pile. Iterate over all socks in your input basket and distribute them onto the color piles.
2. Iterate over each pile and distribute it by some other metric (e.g. pattern) into the second set of piles
3. Recursively apply this scheme until you have distributed all socks onto very small piles that you can visually process immediately

This kind of recursive hash partitioning is actually being done by SQL Server when it needs to hash join or hash aggregate over huge data sets. It distributes its build input stream into many partitions which are independent. This scheme scales to arbitrary amounts of data and multiple CPUs linearly.

You don't need recursive partitioning if you can find a distribution key (hash key) that provides enough buckets that each bucket is small enough to be processed very quickly. Unfortunately, I don't think socks have such a property.

If each sock had an integer called "PairID" one could easily distribute them into 10 buckets according to PairID % 10 (the last digit).

The best real-world partitioning I can think of is creating a rectangle of piles: one dimension is color, the other is the pattern. Why a rectangle? Because we need O(1) random-access to piles. (A 3D cuboid would also work, but that is not very practical.)

Update:

What about parallelism? Can multiple humans match the socks faster?

1. The simplest parallelization strategy is to have multiple workers take from the input basket and put the socks onto the piles. This only scales up so much - imagine 100 people fighting over 10 piles. The synchronization costs (manifesting themselves as hand-collisions and human communication) destroy efficiency and speed-up (see the Universal Scalability Law!). Is this prone to deadlocks? No, because each worker only needs to access one pile at a time. With just one "lock" there cannot be a deadlock. Livelocks might be possible depending on how the humans coordinate access to piles. They might just use random backoff like network cards do that on a physical level to determine what card can exclusively access the network wire. If it works for NICs, it should work for humans as well.
2. It scales nearly indefinitely if each worker has its own set of piles. Workers can then take big chunks of socks from the input basket (very little contention as they are doing it rarely) and they do not need to synchronise when distributing the socks at all (because they have thread-local piles). At the end, all workers need to union their pile-sets. I believe that can be done in O(log (worker count * piles per worker)) if the workers form an aggregation tree.

What about the element distinctness problem? As the article states, the element distinctness problem can be solved in O(N). This is the same for the socks problem (also O(N), if you need only one distribution step (I proposed multiple steps only because humans are bad at calculations - one step is enough if you distribute on md5(color, length, pattern, ...), i.e. a perfect hash of all attributes)).

Clearly, one cannot go faster than O(N), so we have reached the optimal lower bound.

Although the outputs are not exactly the same (in one case, just a boolean. In the other case, the pairs of socks), the asymptotic complexities are the same.