BTreeMerger: Fix bugs with common subtree detection.

Merge methods (i.e., set operations) sometimes did not correctly handle duplicate keys at the boundaries of common subtrees.

This is extremely hard to reproduce using conventional means.

Affected methods:
- BTree.distinctUnion
- BTree.subtracting
- BTree.bagSubtracting
- BTree.symmetricDifference
- BTree.bagSymmetricDifference
- BTree.intersection
- BTree.bagIntersection

- BTreeMerger.copyCommonElementsFromSecond
- BTreeMerger.copyMatchingNumberOfCommonElementsFromSecond
- BTreeMerger.skipCommonElements
- BTreeMerger.skipMatchingNumberOfCommonElements

This fixes issue #20.

Sources/BTreeComparisons.swift

@@ -92,11 +92,11 @@ extension BTree {
                     return false
                 }
                 while a.key < b.key {
-                    a.nextPart(until: b.key, inclusive: false)
+                    a.nextPart(until: .excluding(b.key))
                     if a.isAtEnd { break outer }
                 }
                 while b.key < a.key {
-                    b.nextPart(until: a.key, inclusive: false)
+                    b.nextPart(until: .excluding(a.key))
                     if b.isAtEnd { break outer }
                 }
             }
@@ -164,10 +164,10 @@ extension BTree {
                     }
                     let key = a.key
                     repeat {
-                        a.nextPart(until: key, inclusive: true)
+                        a.nextPart(until: .including(key))
                     } while !a.isAtEnd && a.key == key
                     repeat {
-                        b.nextPart(until: key, inclusive: true)
+                        b.nextPart(until: .including(key))
                     } while !b.isAtEnd && b.key == key
                     if a.isAtEnd {
                         knownStrict = knownStrict || !b.isAtEnd
@@ -177,7 +177,7 @@ extension BTree {
                 }
                 while b.key < a.key {
                     knownStrict = true
-                    b.nextPart(until: a.key, inclusive: false)
+                    b.nextPart(until: .excluding(a.key))
                     if b.isAtEnd { return false }
                 }
             }