Help me sort my socks!

Pyth, 25 bytes

hf!u #-R.-F{BhMs_BMGGT)./

Test suite


hf!u #-R.-F{BhMs_BMGGT)./
                       ./    Form all partitions (implicitly) of the input.
 f                           Filter the permutations on
   u                 T)      Run the following function on the partition
                             until it reaches a fixed point:
                _BMG         Bifurcate the lists on reversal
               s             Concatenate
             hM              Take the first element of each list. 
                             These elements are all the ones on the ends of lists.
           {B                Bifurcate on deduplication
        .-F                  Bagwise subtraction.
                             Only elements repeated in ends of lists remain.
      -R            G        Remove these elements from each list.
   ' #'                      Filter out empty lists.
  !                          Negate. Only an empty list as fixed point succeeds.
h                            Output the first successful partition.

JavaScript (ES6), 329

Not an easy task for a language that has no combinatorics builtins.

Probably sligthly more golfable.

Note: all partition are at least of size 2, as a partition with a single element is always less useful.

Example: [1] [2 3 4] // can take 1 or 2 or 4  
Better: [1 2] [3 4] // can take 3 too  
a=>{G=(v,i,u=v)=>{if(i--){for(;r[i]=--u;)if(G(u,i))return 1;}else for(w=[...r,n=l].map((x,i)=>a.slice(z,z=x-~i),z=0),y=w.join`;`;>[0,1].map(q=>(x=b[q*=~-b.length])&&(t[x]?([c,p]=t[x],n-=2,p?c.pop():c.shift(),q?b.pop():b.shift()):t[x]=[b,q])),c=0,t=[]),c;)if(!n)return 1};for(l=a.length,r=[],k=0;!G(l-k-1,k);k++);return y}

Explanation in parts

(it's overly verbose, but I found it tough to explain - eventually skip to "put it all together")

A recursive function to enumerate all possible splits of an array

// v: array length
// i number of splits
// fill the global array r that must exists
    // the current split position are in r, ready to use
    // for instance...
    parts = [...r,a.length].map(x=>a.slice(z,z=x),z=0)
    console.log(r, parts)

G(4, 2)

// output in console (firebug)
[2, 3] [["A", "B"], ["C"], ["D"]]
[1, 3] [["A"], ["B", "C"], ["D"]]
[1, 2] [["A"], ["B"], ["C", "D"]]

Now, I need partitions of size 2 or more, so I must use this function with sligtly different parameters. The parameter v is "array size - number of desired partitions - 1". Then I must build the partitions in a slightly different way.

// Same call (4,2), same r, but the array b is of size 7
part = [...r,b.length].map((x,i)=>
          b.slice(z,z=x+i+1) // add 1 more element to each partition
// output in console (firebug) 
[2, 3] [["A", "B", "C"], ["D", "E"], ["F", "G"]]
[1, 3] [["A", "B"], ["C", "D", "E"], ["F", "G"]]
[1, 2] [["A", "B"], ["C", "D"], ["E", "F", "G"]]

So, I can enumerate the list of partitions for no split, 1 split, 2 splits and so on. When I find a working partition I will stop and output the result found.

To check, scan the partition list, note the values at start and end of each, if found a repated value then remove it. Repeat until nothing can be removed, at last: if all pairs were removed then this partition is good.

t = []; // array to note the repeated values
// t[x] == [
//           subarray holding value x, 
//           position of value x (I care zero or nonzero)
//         ]
n = a.length // counter start, must reach 0
// remember part just in case, because this check will destroy it 
result = part.join(';') // a string representation for return value
  // in the golfed code there is a forr loop
  // all this body is inside the for condition
  c = 0; // init c to a falsy, if a pair is found c becomes truthy
  part.forEach(b=> // b: array, current partition
    [0,1].forEach(q=> ( // exec for 0 (start), 1 (end)
      q *= b.length-1, // now q is the correct index
      x = b[q]) // x is the value at start or end
      x && ( // b could be empty, check that x is not 'undefined'
        t[x] ? // is there a value in t at position x?
           ( // yes, remove the pair
             n-=2, // pair found, decrement counter
             [c, p] = t[x], // get stored array and position
             p ? c.pop() : c.shift(), // remove from c at start or end
             q ? b.pop() : b.shift()  // remove twin value from b
           : // no, remember the value in t
             t[x] = [b, q]
    )) // end [0,1].forEach
  ) // end part.forEach
while (c) // repeat until nothing can be removed
if(!n) return 1 // wow, result found (in 'result' variable)

Then, the missing part is just a loop caling the G function increasing the partition count. The loop exit when a result is found.

Put it all together

    if (i--)
      for(; r[i]=--u; )
        if (G(u,i)) 
          return 1;
      w = [...r,n=l].map((x,i)=>a.slice(z, z = x-~i), z = 0);
      y = w.join`;`;
      for(; // almost all the for body is inside the condition>
                :t[x]=[b,q])) // end [0,1].map
          ,c=0,t=[] // init variables for
        ),c; // the loop condition is on c
        if(!n)return 1 // this is the for body
  for(l = a.length, r = [], k = 0; !G(l-k-1, k); k++);
  return y


F=a=>{G=(v,i,u=v)=>{if(i--){for(;r[i]=--u;)if(G(u,i))return 1;}else for(w=[...r,n=l].map((x,i)=>a.slice(z,z=x-~i),z=0),y=w.join`;`;>[0,1].map(q=>(x=b[q*=~-b.length])&&(t[x]?([c,p]=t[x],n-=2,p?c.pop():c.shift(),q?b.pop():b.shift()):t[x]=[b,q])),c=0,t=[]),c;)if(!n)return 1};for(l=a.length,r=[],k=0;!G(l-k-1,k);k++);return y}



TestData.forEach(t=>console.log(t+' -> '+F(t)))

function RandomTest() {
  var l=I.value*2
  var a=[...Array(l)].map((_,i)=>1+i/2|0),i)=>a[a[i]=a[j=0|i+Math.random()*(a.length-i)],j]=v) // shuffle
  Q.textContent=a+''+'\n\n'+F(a).replace(/;/g, ';\n') // better readability
Base test
<pre id=O></pre>
Random test. Number of pairs: <input id=I value=15><button onclick="RandomTest()">-></button>
<pre id=Q></pre>


Code Golf