More search utils

python-minimalistic/K23.py

@@ -10,3 +10,9 @@
 K23Binomials=[monomial(nvars, b[0]).divide(monomial(nvars, b[1])) for b in K23Edges]
 
 degreeTwo = listMonomials (2, nvars)
+degreeThree = listMonomials (3, nvars)
+degreeFour = listMonomials (4, nvars)
+
+oneMonomial = degreeTwo[122];
+
+enumerateConnectedComponent (oneMonomial, K23Binomials)

python-minimalistic/searching.py

@@ -36,16 +36,16 @@ def neighbors (mons, binomials):
                 pass
     return unique (result, idfun=lambda m: str(m.exponents))
 
+def isPresent (l, m):
+    """Check if a monomial m is present in a list l"""
+    for li in l:
+        if li.isSame(m): return True
+    return False
 
 def enumerateConnectedComponent (m, binomials):
     """ Starting from a monomial m, perform a breadth first exploration of the
     connected component"""
     print ("Starting breadth-first exploration of connected component")
-    def isPresent (l, m):
-        """Check if m is already in l"""
-        for li in l:
-            if li.isSame(m): return True
-        return False
     result = [m]
     # This will store previously unseen elements in the next neighborhood
     newKids = [m] 
@@ -62,3 +62,29 @@ def isPresent (l, m):
         else: print ("Done :)")
     return result
 
+def inSameComponent (m1, m2, binomials):
+    """
+    Check if two monomials m1, m2 can be connected via the given binomials
+
+    Use breadth-first like in enumeration of a connected component
+    """
+    print ("Starting breadth-first exploration of connected component")
+    # This will store all seen monomials
+    known = [m1]
+    # This will store previously unseen elements in the next neighborhood
+    newKids = [m1] 
+    while len (newKids) > 0:
+        newneighbors = neighbors (newKids, binomials)
+        # Sort the new neighbors into newKids and oldKids
+        newKids = []
+        for n in newneighbors:
+            if n.isSame (m2):
+                return True
+            if not isPresent (known, n):
+                newKids.append(n)
+                known.append(n)
+        if newKids != []:
+            print ("Next iteration will run with " + str(len(newKids)) + " new neighbours.")
+        else: print ("Done :)")
+    return False
+