X fix metric to keep Z <- N order stable over key^

with just sum of abs, it was reverting the order for e.g.

	[2,3) [3,∞)  wrt  [-∞,1) [1,2)

where cost matrix was

	⎡∞ + 4  2⋅∞ + 2⎤
	⎢              ⎥
	⎣  2       ∞   ⎦

and 0-0 + 1-1 is the same as 0-1 + 1-0

with squares 2 in (2⋅∞)² becomes 4 and wins ∞²+∞².

This assert in T2/T1-T3/B0-B1-T-T/B2-B3 -> T2/T1-T/T-T-B2,3/B0-B1.

wcfs/internal/xbtree.py

@@ -262,7 +262,8 @@ def Restructure(ztree, newStructure):
     # - find solution to linear assignment problem A <- B with the cost given by D
     #   https://en.wikipedia.org/wiki/Assignment_problem
     #
-    # D(A,B) = |A.lo - B.lo| + |A.hi - B.hi|
+    #                       2                2
+    # D(A,B) = (A.lo - B.lo)  + (A.hi - B.hi)
 
     # we will modify nodes from new set:
     # - node.Z              will point to associated znode
@@ -292,16 +293,19 @@ def Restructure(ztree, newStructure):
                 if v == -inf: return -1E4
                 assert abs(v) < 1E3
                 return v
-            def d(x,y):
-                return abs(fin(x)-fin(y))
-            return d(a.range.klo, b.range.klo) + \
-                   d(a.range.khi, b.range.khi)
+            def d2(x,y):
+                #return abs(fin(x)-fin(y))
+                return (fin(x)-fin(y))**2
+            return d2(a.range.klo, b.range.klo) + \
+                   d2(a.range.khi, b.range.khi)
 
         # cost matrix
         C = np.zeros((len(RNv), len(RZv)))
         for j in range(len(RNv)):       # "workers"
             for i in range(len(RZv)):   # "jobs"
                 C[j,i] = D(RZv[i], RNv[j])
+        #print('\nCOST:')
+        #print(C)
 
         # find the assignments.
         # TODO try to avoid scipy dependency; we could also probably make