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 = n^{2}/8 socks on average.
As a computer scientist I was thinking what I could do? Sorting (according to size/color/…) of course came to mind to achieve an O(NlogN) solution.
Hashing or other notinplace 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 my spouse and I have more than 30 pairs. (And it is fairly easy to distinguish between my socks and hers; can this be used 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 the 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 realworld 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) randomaccess 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?
 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 handcollisions and human communication) destroy efficiency and speedup (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.
 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 threadlocal piles). At the end, all workers need to union their pilesets. 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.
Answer：
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.
Answer：
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)
comparisonbased 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.
Answer：
Nonalgorithmic answer, yet “efficient” when I do it:

step 1) discard all your existing socks

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.
Algorithmic answer:
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 2n5.

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 justintime 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).
Answer：
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 n^{2} 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. 🙂
Answer：
This is asking the wrong question. The right question to ask is, why am I spending time sorting socks? How much does it cost on yearly basis, when you value your free time for X monetary units of your choice?
And more often than not, this is not just any free time, it’s morning free time, which you could be spending in bed, or sipping your coffee, or leaving a bit early and not being caught in the traffic.
It’s often good to take a step back, and think a way around the problem.
And there is a way!
Find a sock you like. Take all relevant features into account: colour in different lighting conditions, overall quality and durability, comfort in different climatic conditions, and odour absorption. Also important is, they should not lose elasticity in storage, so natural fabrics are good, and they should be available in a plastic wrapping.
It’s better if there’s no difference between left and right foot socks, but it’s not critical. If socks are leftright symmetrical, finding a pair is O(1) operation, and sorting the socks is approximate O(M) operation, where M is the number of places in your house, which you have littered with socks, ideally some small constant number.
If you chose a fancy pair with different left and right sock, doing a full bucket sort to left and right foot buckets take O(N+M), where N is the number of socks and M is same as above. Somebody else can give the formula for average iterations of finding the first pair, but worst case for finding a pair with blind search is N/2+1, which becomes astronomically unlikely case for reasonable N. This can be sped up by using advanced image recognition algorithms and heuristics, when scanning the pile of unsorted socks with Mk1 Eyeball.
So, an algorithm for achieving O(1) sock pairing efficiency (assuming symmetrical sock) is:

You need to estimate how many pairs of socks you will need for the rest of your life, or perhaps until you retire and move to warmer climates with no need to wear socks ever again. If you are young, you could also estimate how long it takes before we’ll all have socksorting robots in our homes, and the whole problem becomes irrelevant.

You need to find out how you can order your selected sock in bulk, and how much it costs, and do they deliver.

Order the socks!

Get rid of your old socks.
An alternative step 3 would involve comparing costs of buying the same amount of perhaps cheaper socks a few pairs at a time over the years and adding the cost of sorting socks, but take my word for it: buying in bulk is cheaper! Also, socks in storage increase in value at the rate of stock price inflation, which is more than you would get on many investments. Then again there is also storage cost, but socks really do not take much space on the top shelf of a closet.
Problem solved. So, just get new socks, throw/donate your old ones away, and live happily ever after knowing you are saving money and time every day for the rest of your life.
Answer：
The theoretical limit is O(n) because you need to touch each sock (unless some are already paired somehow).
You can achieve O(n) with radix sort. You just need to pick some attributes for the buckets.
 First you can choose (hers, mine) – split them into 2 piles,
 then use colors (can have any order for the colors, e.g. alphabetically by color name) – split them into piles by color (remember to keep the initial order from step 1 for all socks in the same pile),
 then length of the sock,
 then texture,
….
If you can pick a limited number of attributes, but enough attributes that can uniquely identify each pair, you should be done in O(k * n), which is O(n) if we can consider k is limited.
Answer：
As a practical solution:
 Quickly make piles of easily distinguishable socks. (Say by color)
 Quicksort every pile and use the length of the sock for comparison. As a human you can make a fairly quick decision which sock to use to partition that avoids worst case. (You can see multiple socks in parallel, use that to your advantage!)
 Stop sorting piles when they reached a threshold at which you are comfortable to find spot pairs and unpairable socks instantly
If you have 1000 socks, with 8 colors and an average distribution, you can make 4 piles of each 125 socks in c*n time. With a threshold of 5 socks you can sort every pile in 6 runs. (Counting 2 seconds to throw a sock on the right pile it will take you little under 4 hours.)
If you have just 60 socks, 3 colors and 2 sort of socks (yours / your wife’s) you can sort every pile of 10 socks in 1 runs (Again threshold = 5). (Counting 2 seconds it will take you 2 min).
The initial bucket sorting will speed up your process, because it divides your n socks into k buckets in c*n
time so than you will only have to do c*n*log(k)
work. (Not taking into account the threshold). So all in all you do about n*c*(1 + log(k))
work, where c is the time to throw a sock on a pile.
This approach will be favourable compared to any c*x*n + O(1)
method roughly as long as log(k) < x  1
.
In computer science this can be helpful:
We have a collection of n things, an order on them (length) and also an equivalence relation (extra information, for example the color of socks). The equivalence relation allows us to make a partition of the original collection, and in every equivalence class our order is still maintained. The mapping of a thing to it’s equivalence class can be done in O(1), so only O(n) is needed to assign each item to a class. Now we have used our extra information and can proceed in any manner to sort every class. The advantage is that the data sets are already significantly smaller.
The method can also be nested, if we have multiple equivalence relations > make colour piles, than within every pile partition on texture, than sort on length. Any equivalence relation that creates a partition with more than 2 elements that have about even size will bring a speed improvement over sorting (provided we can directly assign a sock to its pile), and the sorting can happen very quickly on smaller data sets.
Answer：
You are trying to solve the wrong problem.
Solution 1: Each time you put dirty socks in your laundry basket, tie them in a little knot. That way you will not have to do any sorting after the washing. Think of it like registering an index in a Mongo database. A little work ahead for some CPU savings in the future.
Solution 2: If it’s winter, you don’t have to wear matching socks. We are programmers. Nobody needs to know, as long as it works.
Solution 3: Spread the work. You want to perform such a complex CPU process asynchronously, without blocking the UI. Take that pile of socks and stuff them in a bag. Only look for a pair when you need it. That way the amount of work it takes is much less noticeable.
Hope this helps!
Answer：
This question is actually deeply philosophical. At heart it’s about whether the power of people to solve problems (the “wetware” of our brains) is equivalent to what can be accomplished by algorithms.
An obvious algorithm for sock sorting is:
Let N be the set of socks that are still unpaired, initially empty
for each sock s taken from the dryer
if s matches a sock t in N
remove t from N, bundle s and t together, and throw them in the basket
else
add s to N
Now the computer science in this problem is all about the steps
 “if s pairs with a sock t in N”. How quickly can we “remember” what we’ve seen so far?
 “remove t from N” and “add s to N”. How expensive is keeping track of what we’ve seen so far?
Human beings will use various strategies to effect these. Human memory is associative, something like a hash table where feature sets of stored values are paired with the corresponding values themselves. For example, the concept of “red car” maps to all the red cars a person is capable of remembering. Someone with a perfect memory has a perfect mapping. Most people are imperfect in this regard (and most others). The associative map has a limited capacity. Mappings may bleep out of existence under various circumstances (one beer too many), be recorded in error (“I though her name was Betty, not Nettie”), or never be overwritten even though we observe that the truth has changed (“dad’s car” evokes “orange Firebird” when we actually knew he’d traded that in for the red Camaro).
In the case of socks, perfect recall means looking at a sock s
always produces the memory of its sibling t
, including enough information (where it is on the ironing board) to locate t
in constant time. A person with photographic memory accomplishes both 1 and 2 in constant time without fail.
Someone with less than perfect memory might use a few commonsense equivalence classes based on features within his capability to track: size (papa, mama, baby), color (greenish, redish, etc.), pattern (argyle, plain, etc.), style (footie, kneehigh, etc.). So the ironing board would be divided into sections for the categories. This usually allows the category to be located in constant time by memory, but then a linear search through the category “bucket” is needed.
Someone with no memory or imagination at all (sorry) will just keep the socks in one pile and do a linear search of the whole pile.
A neat freak might use numeric labels for pairs as someone suggested. This opens the door to a total ordering, which allows the human to use exactly the same algorithms we might with a CPU: binary search, trees, hashes, etc.
So the “best” algorithm depends on the qualities of the wetware/hardware/software that is running it and our willingness to “cheat” by imposing a total order on pairs. Certainly a “best” metaalgorithm is to hire the worlds best socksorter: a person or machine that can aquire and quickly store a huge set N of sock attribute sets in a 11 associative memory with constant time lookup, insert, and delete. Both people and machines like this can be procured. If you have one, you can pair all the socks in O(N) time for N pairs, which is optimal. The total order tags allow you to use standard hashing to get the same result with either a human or hardware computer.
Answer：
Cost: Moving socks > high, finding/search socks in line > small
What we want to do is reduce the number of moves, and compensate with the number of searches. Also, we can utilize the multithreded environment of the Homo Sapiens to hold more things in the descision cache.
X = Yours, Y = Your spouses
From pile A of all socks:
Pick two socks, place corresponding X sock in X line, and Y sock in Y line at next available position.
Do until A is empty.
For each line X and Y

Pick the first sock in line, search along the line until it finds the corresponding sock.

Put into the corresponding finished line of socks.
 Optional While you are searching the line and and the current sock you are looking at is identical to the previous, do step 2 for these socks.
Optionally to step one, you pick up two sock from that line instead of two, as the caching memory is large enough we can quickly identify if either sock matches the current one on the line you are observing. If you are fortunate enough to have three arms, you could possibly parse three socks at the same time given that the memory of the subject is large enough.
Do until both X and Y is empty.
Done
However, as this have simillar complexity as selection sort, the time taken is far less due to the speeds of I/O(moving socks) and search(searching the line for a sock).
Answer：
Here’s an Omega(n log n) lower bound in comparison based model. (The only valid operation is comparing two socks.)
Suppose that you know that your 2n socks are arranged this way:
p_{1} p_{2} p_{3} … p_{n} p_{f(1)} p_{f(2)} … p_{f(n)}
where f is an unknown permutation of the set {1,2,…,n}. Knowing this cannot make the problem harder. There are n! possible outputs (matchings between first and second half), which means you need log(n!) = Omega(n log n) comparisons. This is obtainable by sorting.
Since you are interested in connections to element distinctness problem: proving the Omega(n log n) bound for element distinctness is harder, because the output is binary yes/no. Here, the output has to be a matching and the number of possible outputs suffices to get a decent bound. However, there’s a variant connected to element distinctness. Suppose you are given 2n socks and wonder if they can be uniquely paired. You can get a reduction from ED by sending (a_{1}, a_{2}, …, a_{n}) to (a_{1}, a_{1}, a_{2}, a_{2}, …, a_{n}, a_{n}). (Parenthetically, the proof of hardness of ED is very interesting, via topology.)
I think that there should be an Omega(n^{2}) bound for the original problem if you allow equality tests only. My intuition is: Consider a graph where you add an edge after a test, and argue that if the graph is not dense the output is not uniquely determined.
Answer：
This is how I actually do it, for p pairs of socks (n = 2p individual socks):
 Grab a sock at random from the pile.
 For the first sock, or if all previouslychosen socks have been paired, simply place the sock into the first “slot” of an “array” of unpaired socks in front of you.
 If you have one or more selected unpaired socks, check your current sock against all the unpaired socks in the array.
 It is possible to separate socks into general classes or types (white/black, ankle/crew, athletic/dress) when building your array, and “drilldown” to only compare likeforlike.
 If you find an acceptable match, put both socks together and remove them from the array.
 If you do not, put the current sock into the first open slot in the array.
 Repeat with every sock.
The worstcase scenario of this scheme is that every pair of socks is different enough that it must be matched exactly, and that the first n/2 socks you pick are all different. This is your O(n^{2}) scenario, and it’s extremely unlikely. If the number of unique types of sock t is less than the number of pairs p = n/2, and the socks in each type are alike enough (usually in wearrelated terms) that any sock of that type can be paired with any other, then as I inferred above, the maximum number of socks you will ever have to compare to is t, after which the next one you pull will match one of the unpaired socks. This scenario is much more likely in the average sock drawer than the worstcase, and reduces the worstcase complexity to O(n*t) where usually t << n.
Answer：
Real world approach:
As rapidly as possible, remove socks from the unsorted pile one at a time and place in piles in front of you. The piles should be arranged somewhat spaceefficiently, with all socks pointing the same direction; the number of piles is limited by the distance you can easily reach. The selection of a pile on which to put a sock should be — as rapidly as possible — by putting a sock on a pile of apparently like socks; the occasional type I (putting a sock on a pile it doesn’t belong to) or type II (putting a sock in its own pile when there’s an existing pile of like socks) error can be tolerated — the most important consideration is speed. Once all the socks are in piles, rapidly go through the multisock piles creating pairs and removing them (these are heading for the drawer). If there are nonmatching socks in the pile, repile them to their best (within the asfastaspossible constraint) pile. When all the multisock piles have been processed, match up remaining pairable socks that weren’t paired due to type II errors. Whoosh, you’re done — and I have a lot of socks and don’t wash them until a large fraction are dirty. Another practical note: I flip the top of one of a pair of socks down over the other, taking advantage of their elastic properties, so they stay together while being transported to the drawer and while in the drawer.
Answer：
From your question it is clear you don’t have much actual experience with laundry :). You need an algorithm that works well with a small number of nonpairable socks.
The answers till now don’t make good use of our human pattern recognition capabilities. The game of Set provides a clue of how to do this well: put all socks in a twodimensional space so you can both recognize them well and easily reach them with your hands. This limits you to an area of about 120 * 80 cm or so. From there select the pairs you recognize and remove them. Put extra socks in the free space and repeat. If you wash for people with easily recognizable socks (small kids come to mind), you can do a radix sort by selecting those socks first. This algorithm works well only when the number of single socks is low
Answer：
Pick up a first sock and place it on a table. Now pick another sock; if it matches the first picked, place it on top of the first. If not, place it on the table a small distance from the first. Pick a third sock; if it matches either of the previous two, place it on top of them or else place it a small distance from the third. Repeat until you have picked up all the socks.
Answer：
In order to say how efficient it is to pair socks from a pile, we have to define the machine first, because the pairing isn’t done whether by a turing nor by a random access machine, which are normally used as the basis for an algorithmic analysis.
The machine
The machine is an abstraction of a the real world element called human being. It is able to read from the environment via a pair of eyes. And our machine model is able to manipulate the environment by using 2 arms. Logical and arithmetic operations are calculated using our brain (hopefully ;)).
We also have to consider the intrinsic runtime of the atomic operations that can be carried out with these instruments. Due to physical constraints, operations which are carried out by an arm or eye have non constant time complexity. This is because we can’t move an endlessly large pile of socks with an arm nor can an eye see the top sock on an endlessly large pile of socks.
However mechanical physics give us some goodies as well. We are not limited to move at most one sock with an arm. We can move a whole couple of them at once.
So depending on the previous analysis following operations should be used in descending order:
 logical and arithmetic operations
 environmental reads
 environmental modifications
We can also make use of the fact that people only have a very limited amount of socks. So an environmental modification can involve all socks in the pile.
The algorithm
So here is my suggestion:
 Spread all socks in the pile over the floor.
 Find a pair by looking at the socks on the floor.
 Repeat from 2 until no pair can be made.
 Repeat from 1 until there are no socks on the floor.
Operation 4 is necessary, because when spreading socks over the floor some socks may hide others. Here is the analysis of the algorithm:
The analysis
The algorithm terminates with high probability. This is due to the fact that one is unable to find pairs of socks in step number 2.
For the following runtime analysis of pairing n
pairs of socks, we suppose that at least half of the 2n
socks aren’t hidden after step 1. So in the average case we can find n/2
pairs. This means that the loop is step 4 is executed O(log n)
times. Step 2 is executed O(n^2)
times. So we can conclude:
 The algorithm involves
O(ln n + n)
environmental modifications (step 1O(ln n)
plus picking every pair of sock from the floor)  The algorithm involves
O(n^2)
environmental reads from step 2  The algorithm involves
O(n^2)
logical and arithmetic operations for comparing a sock with another in step 2
So we have a total runtime complexity of O(r*n^2 + w*(ln n + n))
where r
and w
are the factors for environmental read and environmental write operations respectively for a reasonable amount of socks. The cost of the logical and arithmetical operations are omitted, because we suppose that it takes a constant amount of logical and arithmetical operations to decide whether 2 socks belong to the same pair. This may not be feasible in every scenario.
Answer：
List<Sock> UnSearchedSocks = getAllSocks();
List<Sock> UnMatchedSocks = new list<Sock>();
List<PairOfSocks> PairedSocks = new list<PairOfSocks>();
foreach (Sock newSock in UnsearchedSocks)
{
Sock MatchedSock = null;
foreach(Sock UnmatchedSock in UnmatchedSocks)
{
if (UnmatchedSock.isPairOf(newSock))
{
MatchedSock = UnmatchedSock;
break;
}
}
if (MatchedSock != null)
{
UnmatchedSocks.remove(MatchedSock);
PairedSocks.Add(new PairOfSocks(MatchedSock, NewSock));
}
else
{
UnmatchedSocks.Add(NewSock);
}
}
Answer：
I came out with another solution which would not promise fewer operations, neither less time consumption, but it should be tried to see if it can be a goodenough heuristic to provide less time consumption in huge series of sock pairing.
Preconditions:
There is no guarantee that there are the same socks. If they are of the same color it doesn’t mean they have the same size or pattern. Socks are randomly shuffled. There can be odd number of socks (some are missing, we don’t know how many). Prepare to remember a variable “index” and set it to 0.
The result will have one or two piles: 1. “matched” and 2. “missing”
Heuristic:
 Find most distinctive sock.
 Find its match.
 If there is no match, put it on the “missing” pile.
 Repeat from 1. until there are no more most distinctive socks.
 If there are less then 6 socks, go to 11.
 Pair blindly all socks to its neighbor (do not pack it)
 Find all matched pairs, pack it and move packed pairs to “matched” pile;
If there were no new matches – increment “index” by 1  If “index” is greater then 2 (this could be value dependent on sock
number because with greater number of socks there are less chance to
pair them blindly) go to 11  Shuffle the rest
 Go to 1
 Forget “index”
 Pick a sock
 Find its pair
 If there is no pair for the sock, move it to the “missing” pile
 If match found pair it, pack pair and move it to the “matched” pile
 If there are still more then one socks go to 12
 If there is just one left go to 14
 Smile satisfied 🙂
Also, there could be added check for damaged socks also, as if the removal of those. It could be inserted between 2 and 3, and between 13 and 14.
I’m looking forward to hear about any experiences or corrections.
Answer：
When I sort socks, I do an approximate radix sort, dropping socks near other socks of the same colour/pattern type. Except in the case when I can see an exact match at/near the location I’m about to drop the sock I extract the pair at that point.
Almost all the other algorithms (including the top scoring answer by usr) sort, then remove pairs. I find that, as a human, it is better to minimize the number of socks being considered at one time.
I do this by:
 Picking a distinctive sock (whatever catches my eye first in the pile).
 Starting a radix sort from that conceptual location by pulling socks from the pile based on similarity to that one.
 Place the new sock near into the current pile, with a distance based on how different it is. If you find yourself putting the sock on top of another because it is identical, form the pair there, and remove them. This means that future comparisons take less effort to find the correct place.
This takes advantage of the human ability to fuzzymatch in O(1) time, which is somewhat equivalent to the establishment of a hashmap on a computing device.
By pulling the distinctive socks first, you leave space to “zoom” in on the features which are less distinctive, to begin with.
After eliminating the fluro coloured, the socks with stripes, and the three pairs of long socks, you might end up with mostly white socks roughly sorted by how worn they are.
At some point, the differences between socks are small enough that other people won’t notice the difference, and any further matching effort is not needed.
Answer：
Whenever you pick up a sock, put it in one place. Then the next sock you pick up, if it doesn’t match the first sock, set it beside the first one. If it does, there’s a pair. This way it doesn’t really matter how many combinations there are, and there are only two possibilities for each sock you pick up — either it has a match that’s already in your array of socks, or it doesn’t, which means you add it to a place in the array.
This also means that you will almost certainly never have all your socks in the array, because socks will get removed as they’re matched.
Answer：
Consider a hashtable of size ‘N’.
If we assume normal distribution, then the estimated number of ‘insertions’ to have atleast one sock mapped to one bucket is NlogN (ie, all buckets are full)
I had derived this as a part of another puzzle,but I would be happy to be proven wrong.
Here’s my blog article on the same
Let ‘N’ correspond to an approximate upperbound on the number of number of unique colors/pattern of socks that you have.
Once you have a collision(a.k.a : a match) simply remove that pair of socks.
Repeat the same experiment with the next batch of NlogN socks.
The beauty of it is that you could be making NlogN parallel comparisons(collisionresolution) because of the way the human mind works. 🙂
Answer：
Socks, whether real ones or some analogous data structure, would be supplied in pairs.
The simplest answer is prior to allowing the pair to be separated, a single data structure for the pair should have been initialized that contained a pointer to the left and right sock, thus enabling socks to be referred to directly or via their pair. A sock may also be extended to contain a pointer to its partner.
This solves any computational pairing problem by removing it with a layer of abstraction.
Applying the same idea to the practical problem of pairing socks, the apparent answer is: don’t allow your socks to ever be unpaired. Socks are provided as a pair, put in the drawer as a pair (perhaps by balling them together), worn as a pair. But the point where unpairing is possible is in the washer, so all that’s required is a physical mechanism that allows the socks to stay together and be washed efficiently.
There are two physical possibilities:
For a ‘pair’ object that keeps a pointer to each sock we could have a cloth bag that we use to keep the socks together. This seems like massive overhead.
But for each sock to keep a reference to the other, there is a neat solution: a popper (or a ‘snap button’ if you’re American), such as these:
http://www.aliexpress.com/compare/compareinvisiblesnapbuttons.html
Then all you do is snap your socks together right after you take them off and put them in your washing basket, and again you’ve removed the problem of needing to pair your socks with a physical abstraction of the ‘pair’ concept.
Answer：
I have taken simple steps to reduce my effort into a process taking O(1) time.
By reducing my inputs to one of two types of socks (white socks for recreation, black socks for work), I only need to determine which of two socks I have in hand. (Technically, since they are never washed together, I have reduced the process to O(0) time)
Some upfront effort is required to find desirable socks, and to purchase in sufficient quantity as to eliminate need for your existing socks. As I’d done this before my need for black socks, my effort was minimal, but mileage may vary.
Such an upfront effort has been seen many times in very popular and effective code. Examples include #DEFINE’ing pi to several decimals (other examples exist, but that’s the one that comes to mind right now).
Answer：
If the “move” operation is fairly expensive, and the “compare” operation is cheap, and you need to move the whole set anyway, into a buffer where search is much faster than in original storage… just integrate sorting into the obligatory move.
I found integrating the process of sorting into hanging to dry makes it a breeze. I need to pick up each sock anyway, and hang it (move) and it costs me about nothing to hang it in a specific place on the strings. Now just not to force search of the whole buffer (the strings) I choose to place socks by color/shade. Darker left, brighter right, more colorful front etc. Now before I hang each sock, I look in its “right vicinity” if a matching one is there already – this limits “scan” to 23 other socks – and if it is, I hang the other one right next to it. Then I roll them into pairs while removing from the strings, when dry.
Now this may not seem all that different from “forming piles by color” suggested by top answers but first, by not picking discrete piles but ranges, I have no problem classifying whether “purple” goes to “red” or “blue” pile; it just goes between. And then by integrating two operations (hang to dry and sort) the overhead of sorting while hanging is like 10% of what separate sorting would be.
Answer：
I’ve finished pairing my socks just right now, and I found that the best way to do it is the following:
 Choose one of the socks and put it away (create a ‘bucket’ for that pair)
 If the next one is the pair of the previous one, then put it to the existing bucket, otherwise create a new one.
In the worst case it means that you will have n/2 different buckets, and you will have n2 determinations about that which bucket contains the pair of the current sock. Obviously, this algorithm works well if you have just a few pairs; I did it with 12 pairs.
It is not so scientific, but it works well:)
Answer：
My solution does not exactly correspond to your requirements, as it formally requires O(n)
“extra” space. However, considering my conditions it is very efficient in my practical application. Thus I think it should be interesting.
Combine with Other Task
The special condition in my case is that I don’t use drying machine, just hang my cloths on an ordinary cloth dryer. Hanging cloths requires O(n)
operations (by the way, I always consider bin packing problem here) and the problem by its nature requires the linear “extra” space. When I take a new sock from the bucket I to try hang it next to its pair if the pair is already hung. If its a sock from a new pair I leave some space next to it.
Oracle Machine is Better 😉
It obviously requires some extra work to check if there is the matching sock already hanging somewhere and it would render solution O(n^2)
with coefficient about 1/2
for a computer. But in this case the “human factor” is actually an advantage — I usually can very quickly (almost O(1)
) identify the matching sock if it was already hung (probably some imperceptible inbrain caching is involved) — consider it a kind of limited “oracle” as in Oracle Machine 😉 We, the humans have these advantages over digital machines in some cases 😉
Have it Almost O(n)
!
Thus connecting the problem of pairing socks with the problem of hanging cloths I get O(n)
“extra space” for free, and have a solution that is about O(n)
in time, requires just a little more work than simple hanging cloths and allows to immediately access complete pair of socks even in a very bad Monday morning… 😉
Answer：
I hope I can contribute something new to this problem. I noticed that all of the answers neglect the fact that there are two points where you can perform preprocessing, without slowing down your overall laundry performance.
Also, we don’t need to assume a large number of socks, even for large families. Socks are taken out of the drawer and are worn, and then they are tossed in a place (maybe a bin) where they stay before being laundered. While I wouldn’t call said bin a LIFOStack, I’d say it is safe to assume that
 people toss both of their socks roughly in the same area of the
bin,  the bin is not randomized at any point, and therefore
 any subset taken from the top of this bin generally contains both
socks of a pair.
Since all washing machines I know about are limited in size (regardless of how many socks you have to wash), and the actual randomizing occurs in the washing machine, no matter how many socks we have, we always have small subsets which contain almost no singletons.
Our two preprocessing stages are “putting the socks on the clothesline” and “Taking the socks from the clothesline”, which we have to do, in order to get socks which are not only clean but also dry. As with washing machines, clotheslines are finite, and I assume that we have the whole part of the line where we put our socks in sight.
Here’s the algorithm for put_socks_on_line():
while (socks left in basket) {
take_sock();
if (cluster of similar socks is present) {
Add sock to cluster (if possible, next to the matching pair)
} else {
Hang it somewhere on the line, this is now a new cluster of similarlooking socks.
Leave enough space around this sock to add other socks later on
}
}
Don’t waste your time moving socks around or looking for the best match, this all should be done in O(n), which we would also need for just putting them on the line unsorted.
The socks aren’t paired yet, we only have several similarity clusters on the line. It’s helpful that we have a limited set of socks here, as this helps us to create “good” clusters (for example, if there are only black socks in the set of socks, clustering by colours would not be the way to go)
Here’s the algorithm for take_socks_from_line():
while(socks left on line) {
take_next_sock();
if (matching pair visible on line or in basket) {
Take it as well, pair 'em and put 'em away
} else {
put the sock in the basket
}
I should point out that in order to improve the speed of the remaining steps, it is wise not to randomly pick the next sock, but to sequentially take sock after sock from each cluster.
Both preprocessing steps don’t take more time than just putting the socks on the line or in the basket, which we have to do no matter what, so this should greatly enhance the laundry performance.
After this, it’s easy to do the hash partitioning algorithm. Usually, about 75% of the socks are already paired, leaving me with a very small subset of socks, and this subset is already (somewhat) clustered (I don’t introduce much entropy into my basket after the preprocessing steps). Another thing is that the remaining clusters tend to be small enough to be handled at once, so it is possible to take a whole cluster out of the basket.
Here’s the algorithm for sort_remaining_clusters():
while(clusters present in basket) {
Take out the cluster and spread it
Process it immediately
Leave remaining socks where they are
}
After that, there are only a few socks left. This is where I introduce previously unpaired socks into the system and process the remaining socks without any special algorithm – the remaining socks are very few and can be processed visually very fast.
For all remaining socks, I assume that their counterparts are still unwashed and put them away for the next iteration. If you register a growth of unpaired socks over time (a “sock leak”), you should check your bin – it might get randomized (do you have cats which sleep in there?)
I know that these algorithms take a lot of assumptions: a bin which acts as some sort of LIFO stack, a limited, normal washing machine, and a limited, normal clothesline – but this still works with very large numbers of socks.
About parallelism:
As long as you toss both socks into the same bin, you can easily parallelize all of those steps.
Answer：
Create a hash table which will be used for unmatched socks, using the pattern as the hash. Iterate over the socks one by one. If the sock has a pattern match in the hash table, take the sock out of the table and make a pair. If the sock does not have a match, put it into the table.
Answer：
The problem of sorting your n pairs of socks is O(n). Before you throw them in the laundry basket, you thread the left one to the right one. On taking them out, you cut the thread and put each pair into your drawer – 2 operations on n pairs, so O(n).
Now the next question is simply whether you do your own laundry and your wife does hers. That is a problem likely in an entirely different domain of problems. 🙂
Tags: algorithmalgorithm