Yesterday I was pairing the socks from the clean laundry, and figured out the way I was doing it is not very efficient. I was doing a naive search — picking one sock and “iterating” the pile in order to find its pair. This requires iterating over n/2 * n/4 = n2/8 socks on average.
As a computer scientist I was thinking what I could do? Sorting (according to size/color/…) of course came into mind to achieve an O(NlogN) solution.
Hashing or other not-in-place solutions are not an option, because I am not able to duplicate my socks (though it could be nice if I could).
So, the question is basically:
Given a pile of
n pairs of socks, containing
2n elements (assume each sock has exactly one matching pair), what is the best way to pair them up efficiently with up to logarithmic extra space? (I believe I can remember that amount of info if needed.)
I will appreciate an answer that addresses the following aspects:
- A general theoretical solution for a huge number of socks.
- The actual number of socks is not that large, I don’t believe me and my spouse have more than 30 pairs. (And it is fairly easy to distinguish between my socks and hers, can this be utilized as well?)
- Is it equivalent to the element distinctness problem?
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).
- For each color of socks, form a pile. Iterate over all socks in your input basket and distribute them onto the color piles.
- Iterate over each pile and distribute it by some other metric (e.g. pattern) into a second set of piles
- 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 pattern. Why a rectangle? Because we need O(1) random-access to piles. (A 3D cuboidwould also work, but that is not very practical.)
What about parallelism? Can multiple humans match the socks faster?
- The simplest parallization 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!).
- It scales nearly indefinitely if each worker has its own set of piles. Workers can then take from the input basket big chunks of socks (very little contention as they are doing it rarely) and they do not need to sync 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.