Thanks for your endeavor for my this question.And I think a good message deserve to bring to you.I receive a good solution here based on FindShortestTour.

source of the webpage...

It looks like we can get the message ID by copying the "Reply" hyperlink under the post. For example, for your post by copying I get

javascript:_19_addQuickReply('reply',%20'1077817');setQuickReplyTo(1077817);

where 1077817 obviously is the message ID. Then for generating the link we should just prepend http://community.wolfram.com/groups/-/m/t/:

http://community.wolfram.com/groups/-/m/t/1077817

and it works! No need to inspect the source. :)

Could I know how do you get that url(http://community.wolfram.com/groups/-/m/t/1075278) about a reply under a post?

Would FindShortestTour be a better choice, (if I have read this correctly) as an example

AbsoluteTiming[SeedRandom[1]; pts = RandomReal[100, {100, 2}]; 
 pair = Partition[pts[[Last[FindShortestTour[pts]]]], 2]; 
 Print[Graphics[{Line[pair], Red, PointSize[.015], Point[pts]}]]; 
 Total[Table[
   Sqrt[(pair[[q, 1, 1]] - pair[[q, 2, 1]])^2 + (pair[[q, 1, 2]] - 
      pair[[q, 2, 2]])^2], {q, 1, Length[pair]}]]]

or by rotating the points by 1 to get the other possible set of pairs

AbsoluteTiming[SeedRandom[1]; pts = RandomReal[100, {100, 2}]; 
 pair = Partition[pts[[Last[FindShortestTour[pts]]]], 2]; 
 pair = Partition[RotateLeft[Flatten[pair, 1]], 2]; 
 Print[Graphics[{Line[pair], Red, PointSize[.015], Point[pts]}]]; 
 Total[Table[
   Sqrt[(pair[[q, 1, 1]] - pair[[q, 2, 1]])^2 + (pair[[q, 1, 2]] - 
      pair[[q, 2, 2]])^2], {q, 1, Length[pair]}]]]

in comparison to this method

AbsoluteTiming[SeedRandom[1]; pts = RandomReal[100, {100, 2}]; 
 g = CompleteGraph[Length[pts]]; 
 verGraph = VertexReplace[g, Thread[VertexList[g] -> pts]]; 
 weighGraph = 
  Graph[verGraph, 
   EdgeWeight -> EuclideanDistance @@@ EdgeList[verGraph]]; 
 pair = Partition[FindHamiltonianPath[weighGraph], 2]; 
 Print[Graphics[{Line[pair], Red, PointSize[.015], Point[pts]}]]; 
 Total[Table[
   Sqrt[(pair[[q, 1, 1]] - pair[[q, 2, 1]])^2 + (pair[[q, 1, 2]] - 
      pair[[q, 2, 2]])^2], {q, 1, Length[pair]}]]]

The shortest tour option is also much faster, 3000 points is under a second on my machine.

What is this actually used for? I'm aware of connecting two sets of points, but i have not heard of this problem. How critical is it to get the global minimum versus sufficiently nice (practical) local minimum. I have a feeling this problem is very much akin to the travelling salesman problem, which for 50 or so points also already get tricky to ensure to get the global optimum. The one in mathematica switches to a 'good enough' algorithm for large number of points. The reference you cited in the opening post; do they really ensure it is a global minimum? the paper is hard to read (for me), as I'm not really into the field...

A better minimum as compared to the black one. I'm not sure if it is THE minimum. You can see that some pairs 'switched', especially clear in the top right with the 'triangle'.

I tried 100, 200 and 300 points, they all work fine. Good luck translating that huge code, the code is not hard, but without comments you have no idea what each subroutine accomplishes...

Using FindHamiltonianPath/FindShortestTour will not give the minimum, because you only consider half of the edges of the tour. But you CAN use it as start for my FindPairs function:

FindPairs[pts_List] := Module[{candidates, out, tmp, left, start},
  If[EvenQ[Length[pts]] \[And] Length[pts] >= 2,
   start = Most[FindShortestTour[pts][[2]]];
   TrySwitchAll[pts, Partition[start, 2]]
   ,
   Print["Even number of points >=2 needed!"]
   ]
  ]

As for the above example, there are clearly improvements that can be made to this solution:

SeedRandom[1]
pts = RandomReal[100, {100, 2}];
g = CompleteGraph[Length[pts]];
verGraph = VertexReplace[g, Thread[VertexList[g] -> pts]];
weighGraph = 
  Graph[verGraph, 
   EdgeWeight -> EuclideanDistance @@@ EdgeList[verGraph]];
pair = Partition[FindHamiltonianPath[weighGraph], 2];
pts = Join @@ pair;
conns = Partition[Range[Length[pts]], 2];
orig = Graphics[{Line[pts[[#]]] & /@ conns, Red, PointSize[.015], Point[pts]}];
newconn = TrySwitchAll[pts, conns];
new = Graphics[{Blue, Line[pts[[#]]] & /@ newconn}];
Show[{orig, new}]

You method, then improved using mine...

where the black are the old lines, and the blue ones the new lines.

Yep,I had realized it is very hard before arriving here. :) So I'm hope to implement that ready-made algorithm for min cost perfect matching in Mathemtica.But it is a pitty that I cannot read any C++ code.And as my try,we cannot find a pair even in 70 points if we use FindPairs,but that algorithm can make it in 1000 points within one second as my friend's test.

Brilliant try.Thanks very much.

But if I give some test pts for test,I found it not always give a right answer?Such as

SeedRandom[1]
pts = RandomReal[100, {8, 2}]

{{81.7389,11.142},{78.9526,18.7803},{24.1361,6.57388},{54.2247,23.1155},{39.6006,70.0474},{21.1826,74.8657},{42.2851,24.7495},{97.7172,82.5163}}

If we use a brute force method to find a definitely right answer:

First[pairs = 
  MinimalBy[Partition[#, 2] & /@ Permutations[pts], 
   Total[EuclideanDistance @@@ N[#]] &]]
First[Total[EuclideanDistance @@@ N[#]] & /@ pairs]

{{{81.7389,11.142},{78.9526,18.7803}},{{24.1361,6.57388},{42.2851,24.7495}},{{54.2247,23.1155},{97.7172,82.5163}},{{39.6006,70.0474},{21.1826,74.8657}}}

126.475(This is total distance)

But If we use your FindPairs

pts[[#]] & /@ FindPairs[pts]
Total[EuclideanDistance @@@ N[pts[[#]] & /@ FindPairs[pts]]]

{{{24.1361,6.57388},{54.2247,23.1155}},{{81.7389,11.142},{78.9526,18.7803}},{{39.6006,70.0474},{21.1826,74.8657}},{{42.2851,24.7495},{97.7172,82.5163}}}

141.565(This is total distance)