# algorithm - How can I pair socks from a pile efficiently?

ID : 97

viewed : 223

### Top 5 Answer for algorithm - How can I pair socks from a pile efficiently?

97

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.

80

As the architecture of the human brain is completely different than a modern CPU, this question makes no practical sense.

Humans can win over CPU algorithms using the fact that "finding a matching pair" can be one operation for a set that isn't too big.

My algorithm:

``spread_all_socks_on_flat_surface(); while (socks_left_on_a_surface()) {      // Thanks to human visual SIMD, this is one, quick operation.      pair = notice_any_matching_pair();      remove_socks_pair_from_surface(pair); } ``

At least this is what I am using in real life, and I find it very efficient. The downside is it requires a flat surface, but it's usually abundant.

76

Case 1: All socks are identical (this is what I do in real life by the way).

Pick any two of them to make a pair. Constant time.

Case 2: There are a constant number of combinations (ownership, color, size, texture, etc.).

Use radix sort. This is only linear time since comparison is not required.

Case 3: The number of combinations is not known in advance (general case).

We have to do comparison to check whether two socks come in pair. Pick one of the `O(n log n)` comparison-based sorting algorithms.

However in real life when the number of socks is relatively small (constant), these theoretically optimal algorithms wouldn't work well. It might take even more time than sequential search, which theoretically requires quadratic time.

61

Non-algorithmic answer, yet "efficient" when I do it:

• step 2) go to Walmart and buy them by packets of 10 - n packet of white and m packets of black. No need for other colors in everyday's life.

Yet times to times, I have to do this again (lost socks, damaged socks, etc.), and I hate to discard perfectly good socks too often (and I wished they kept selling the same socks reference!), so I recently took a different approach.

Consider than if you draw only one sock for the second stack of socks, as you are doing, your odds of finding the matching sock in a naive search is quite low.

• So pick up five of them at random, and memorize their shape or their length.

Why five? Usually humans are good are remembering between five and seven different elements in the working memory - a bit like the human equivalent of a RPN stack - five is a safe default.

• Pick up one from the stack of 2n-5.

• Now look for a match (visual pattern matching - humans are good at that with a small stack) inside the five you drew, if you don't find one, then add that to your five.

• Keep randomly picking socks from the stack and compare to your 5+1 socks for a match. As your stack grows, it will reduce your performance but raise your odds. Much faster.

Feel free to write down the formula to calculate how many samples you have to draw for a 50% odds of a match. IIRC it's an hypergeometric law.

I do that every morning and rarely need more than three draws - but I have `n` similar pairs (around 10, give or take the lost ones) of `m` shaped white socks. Now you can estimate the size of my stack of stocks :-)

BTW, I found that the sum of the transaction costs of sorting all the socks every time I needed a pair were far less than doing it once and binding the socks. A just-in-time works better because then you don't have to bind the socks, and there's also a diminishing marginal return (that is, you keep looking for that two or three socks that when somewhere in the laundry and that you need to finish matching your socks and you lose time on that).

60

What I do is that I pick up the first sock and put it down (say, on the edge of the laundry bowl). Then I pick up another sock and check to see if it's the same as the first sock. If it is, I remove them both. If it's not, I put it down next to the first sock. Then I pick up the third sock and compare that to the first two (if they're still there). Etc.

This approach can be fairly easily be implemented in an array, assuming that "removing" socks is an option. Actually, you don't even need to "remove" socks. If you don't need sorting of the socks (see below), then you can just move them around and end up with an array that has all the socks arranged in pairs in the array.

Assuming that the only operation for socks is to compare for equality, this algorithm is basically still an n2 algorithm, though I don't know about the average case (never learned to calculate that).

Sorting, of course improves efficiency, especially in real life where you can easily "insert" a sock between two other socks. In computing the same could be achieved by a tree, but that's extra space. And, of course, we're back at NlogN (or a bit more, if there are several socks that are the same by sorting criteria, but not from the same pair).

Other than that, I cannot think of anything, but this method does seem to be pretty efficient in real life. :)