Make the Stretchy Snakes Kiss A stretchy snake looks something like this:
&lt;||=|||:)~

Each separate sequence of vertical bars (|) in a stretchy snake, known as a stretchy portion, is individually extendable to twice its width, and is drawn with alternating slashes (/,\) once extended.
The particular snake above has two such stretchy portions, giving it four possible poses:
&lt;||=|||:)~

&lt;/\/\=|||:)~

&lt;||=/\/\/\:)~

&lt;/\/\=/\/\/\:)~

The general form of a stretchy snake in its least stretched pose is defined by this regex:
&lt;(\|+=)*\|+:\)~

Which can be stated in words as:

&lt;, followed by any number of sequences of |'s joined with = signs, followed by :)~.

So &lt;|:)~ and &lt;||:)~ and &lt;|=|:)~ and &lt;|=|=||=|||||=||:)~ are stretchy snakes, but &lt;=:)~ and &lt;=|:)~ and &lt;||=:)~ and &lt;|==||:)~ are not.
Stretchy snakes can also face left instead of right, e.g. ~(:|||=||&gt;. The forms are the same, just mirrored.
Challenge
Write a program that takes in a single line string of two stretchy snakes facing each other, with some number of spaces in between. Both snakes will be in their least stretched pose (all vertical bars, no slashes). The string will start with the tail of the right-facing snake and end with the tail of the left-facing snake (you may optionally assume there's also a trailing newline).
For example, here's a possible input with five spaces between the snakes:
&lt;|=||:)~.....~(:||||&gt;

I'm using periods (.) instead of actual space characters for clarity.
Zero spaces between snakes is also valid input:
&lt;|=||:)~~(:||||&gt;

We say the snakes are kissing when their tongues are touching like this.
Your program needs to extend some combination of the stretchy portions of both of the snakes such that the snakes have the fewest number of spaces possible between them (without overlapping), i.e. such that the snakes are as close to kissing as possible.
Both the snakes' tails are fixed but their heads and bodies can move - right for the right-facing snake, left for the left-facing snake - according to what stretchy portions have been extended.
The output of your program is the single line string (plus optional trailing newline) that shows the snakes as close to kissing as possible, with alternating slashes drawn in place of vertical bars for stretchy portions that have been extended.

For example, the output for &lt;|=||:)~.....~(:||||&gt; (from above) would be:
&lt;/\=||:)~~(:/\/\/\/\&gt;

This is the only solution here because with any other combination of the stretchy portions extended, the snakes would either overlap or be farther away from kissing.

If there are multiple solutions possible, the output may be any one of them.
For example, if the input were
&lt;|=||:)~.....~(:|||=|&gt;

the output could be
&lt;|=/\/\:)~~(:/\/\/\=|&gt;

or
&lt;/\=||:)~~(:/\/\/\=/\&gt;


Remember that it won't always be possible to make the snakes kiss, but you still need to get them as close as possible.
For example, if the input were
&lt;||=||||:)~...~(:||&gt;

the output could be
&lt;/\/\=||||:)~.~(:||&gt;

or
&lt;||=||||:)~.~(:/\/\&gt;


If the snakes are already kissing, the output will be the same as the input. e.g.
&lt;|=||:)~~(:||||&gt;

In general, the output will be the same as the input if the extension of any stretchy portion would make the snakes overlap. e.g.
&lt;|||=|||:)~..~(:||||=|||||=||||||&gt;


Notes

Takes input from stdin or the command line as usual, or write a function that takes a string. Print or return the output.
You can use periods (.) in the input and output in place of spaces ( ) if you prefer.
It's only important that slashes alternate within the sequence of vertical bars they replaced. Their ordering in the snake at large or whether a forward or backward slash comes first doesn't matter.
Stretchy portions cannot extend partway - it's exactly double or no extension at all.

Scoring
This is code-golf. The shortest submission in bytes wins. Tiebreaker is earlier answer.
Retina, 209 107 99 97 92 bytes

.(?=(.+)(?&lt;=(?=&lt;((\|+|(?&lt;-5&gt;\|(?=\1())?)+)[^|]+)+$(?(5)!)).+( )+\S+))\4
/ \
+` (.*~|~.*) 
$1


For counting purposes, each line goes in a separate file, but you can run the code from a single file with the -s flag.

Bringing the best features of .NET regex and Retina together: balancing groups, arbitrary length lookbehinds and repeated regex substitution.

Essentially, the long regex encodes a valid solution and the regex engine's backtracker finds one of the optimal ones for me.

Explanation

First, let's consider how we can find a valid solution (not necessarily producing the correct output) with a regex. We can use .NET's balancing groups to help us count the stretchy parts. Consider the following simpler regex:

\S+( )+.+(?&lt;=(?(1)!)^([^|]+(\|+|(?&lt;-1&gt;\|)*))+&gt;)


We can disect that.

\S+( )+.+


This matches the entire string, pushing one capture onto the group 1 stack for each space in the input. We will use that stack to ensure that the stretchy parts exactly fill the space captured into those groups.

Next is a lookbehind. The catch is that lookbehinds are matched from right to left in .NET (so that's how you should read them). This gives us the opportunity to traverse the string a second time, figuring out whether there's a subset of stretchy part that sums to the number of matched spaces. Going through the lookbehind from right-to-left:

&gt;


This is just ensure that we're actually starting from the end of the string (the snake's tail).

(
  [^|]+
  (
    \|+
  |
    (?&lt;-1&gt;\|)+
  )
)+


For each stretchy part, this either just matches the entire part without doing anything (\|+), or it matches the entire part while popping captures off stack 1 ((?&lt;-1&gt;\|)*). Having this alternation ensures that we can only fully extend a stretchy part or leave it unchanged, and not get stuff like |/\|. Then we move on to the next stretchy part with [^|]+.

(?(1)!)^


Finally, we ensure that we've traverse the entire string (both snakes), and that stack 1 is completely empty. I.e. that we've found a subset of stretchy part that exactly sums to the number of spaces we've captured earlier.

The backtracker will go back and forth through the string trying all combinations of unchanged and extended parts until the subset sum problem is solved. If such a subset doesn't exist, the lookbehind will fail. This will cause the backtracker to go back to the \S+( )+.+ part and try capturing one space less with ( )+ (which will just be cover by .+ instead). Due to the greediness of + we therefore try filling as many spaces as possible.

You can check the validity of this approach with this slightly modified substitution:

\S+(( )+).+(?&lt;=(?(2)!)^([^|]+(\|+|(?&lt;-2&gt;\|)*))+&gt;)
=$1=


Which will give you a string =spaces= with exactly the number of spaces that can be filled with the given snakes.

I've had to add some more trickery to actually expand the correct |s. Basically, I want to replace all |s which were matched using the (?&lt;-1&gt;\|)+ branch. The idea is to match an individual character, put the solver in a lookaround and set a flag if the match happens to be inside that branch. If that flag wasn't set we invalidate the match at the end to avoid replacement of other characters.

To do this, we use a bunch of nested lookarounds. Again, .NET's variable-length lookbehinds are matched from right to left, so if we nest lookaheads and lookbehinds, we can let the regex engine traverse the string several times. For golfing reasons, the solver is reversed in my actual solution (starting at the end, picking up the spaces from right to left, and then solving the subset sum from left to right), but otherwise the structure of the solver is exactly the same. Let's dissect the full regex:

.(?=(.+)...)


We match a single character, then capture the remainder of the string and move the cursor to the end of the string. We will use this group 1 later to check in the solver whether we're at the position of the match.

(?&lt;=....+( )+\S+)


This is like the first part of the simple solver above, except that we pick up the spaces from right to left. The backtracking of the number of spaces works exactly the same as before, except that we're using group 5 now.

(?=&lt;((\|+|(?&lt;-5&gt;\|(?=\1())?)+)[^|]+)+$(?(5)!))


This is the same as before, except we're going from left to right, and whenever we match a | in the expanding branch, we check if it's the one being matched right with

(?=\1())?


This is an optional lookahead. It tries to match group 1 again (which, here, is only possible if we're right after the character being matched), and if we do, we capture an empty string into group 4, which indicates that we did find the current character in one of the expanded bits. If \1 doesn't match, 4 won't capture anything, and the ? ensures that the failing lookahead won't affect the solver at all.

Finally, after all the solving is done, we just check with \4 whether this lookahead was used. If so, we want to replace the current character with /\.

One difficulty remains: removing the right amount of spaces. The shortest way to do this I've found so far is to insert a space along with the /\ and then get rid of one space between the tongues for each of those marker spaces in a separate step:

+` (.*~|~.*) 
$1