/* this is similar for rules with nonexistence graph elements*/
private void FindPositiveStartElementAvoidNegatives(option location)
{
#region Case #1: Location found! No empty slots left in the location
/* this is the only way to properly exist the recursive loop. */
if (!L.nodes.Any(n => ((ruleNode)n).MustExist && location.findLMappedNode(n) == null) &&
!L.arcs.Any(a => ((ruleArc)a).MustExist && location.findLMappedArc(a) == null) &&
!L.hyperarcs.Any(n => ((ruleHyperarc)n).MustExist && location.findLMappedHyperarc(n) == null))
{
/* as a recursive function, we first check how the recognition process terminates. If all nodes,
* hyperarcs and arcs within location have been filled with references to elements in the host,
* then we've found a location...well maybe. More details are described in the LocationFound function. */
if (!FinalRuleChecks(location) && !FinalRuleCheckRelaxed(location))
{
return;
}
Boolean resultNegativeNotFulfilled;
lock (AllNegativeElementsFound)
{
AllNegativeElementsFound = false;
negativeRelaxation = location.Relaxations.copy();
findNegativeStartElement(location);
resultNegativeNotFulfilled = !(bool)AllNegativeElementsFound;
}
if (resultNegativeNotFulfilled)
{
var locCopy = location.copy();
locCopy.Relaxations = negativeRelaxation;
lock (options) { options.Add(locCopy); }
}
return;
}
#endregion
#region Case #2: build off of a hyperarc found so far - by looking for unfulfilled nodes
/* the quickest approach to finding a new element in the LHS to host subgraph matching is to build
* directly off of elements found so far. This is because we don't need to check amongst ALL elements in the
* host (as is the case in the last three cases below). In this case we start with any hyperarcs
* that have already been matched to one in the host, and see if it connects to any nodes that
* have yet to be matched. */
var startHyperArc = (ruleHyperarc)L.hyperarcs.FirstOrDefault(ha => ((ruleHyperarc)ha).MustExist &&
((location.findLMappedHyperarc(ha) != null) &&
(ha.nodes.Any(n => ((ruleNode)n).MustExist &&
(location.findLMappedNode(n) == null)))));
if (startHyperArc != null)
{
var hostHyperArc = location.findLMappedHyperarc(startHyperArc);
var newLNode = (ruleNode)startHyperArc.nodes.FirstOrDefault(n => ((ruleNode)n).MustExist &&
(location.findLMappedNode(n) == null));
foreach (var n in hostHyperArc.nodes.Where(n => !location.nodes.Contains(n)))
{
checkNodeAvoidNegatives(location.copy(), newLNode, n);
}
return;
}
#endregion
#region Case #3: build off of a node found so far - by looking for unfulfilled arcs
/* as stated above, the quickest approach is to build from elements that have already been found.
* Therefore, we see if there are any nodes already matched to a node in L that has an arc in L
* that has yet to be matched with a host arc. This is more efficient than the last 3 cases
* because they look through the entire host, which is potentially large. */
var startNode = (ruleNode)L.nodes.FirstOrDefault(n => ((ruleNode)n).MustExist &&
((location.findLMappedNode(n) != null) &&
(n.arcs.Any(a =>
(((a is ruleHyperarc) && ((ruleHyperarc)a).MustExist) ||
((a is ruleArc) && ((ruleArc)a).MustExist)) &&
(location.findLMappedElement(a) == null)))));
/* is there a node already matched (which would only occur if your recursed to get here) that has an
* unrecognized arc attaced to it. If yes, try all possible arcs in the host with the one that needs
* to be fulfilled in L. */
if (startNode != null)
{
var newLArc = startNode.arcs.FirstOrDefault(a =>
(((a is ruleHyperarc) && ((ruleHyperarc)a).MustExist) ||
((a is ruleArc) && ((ruleArc)a).MustExist)) &&
(location.findLMappedElement(a) == null));
if (newLArc is ruleHyperarc)
{
checkHyperArcAvoidNegatives(location, startNode, location.findLMappedNode(startNode), (ruleHyperarc)newLArc);
}
else if (newLArc is ruleArc)
{
checkArcAvoidNegatives(location, startNode, location.findLMappedNode(startNode), (ruleArc)newLArc);
}
return;
}
#endregion
#region Case 4: Check entire host for a matching hyperarc
/* if the above cases didn't match we try to match a hyperarc in the L to any in the host. Since the
* prior three cases have conditions which require some non-nulls in the location, this is likely where the
* process will start when invoked from line 79 of recognize above. Hyperarcs are most efficient to start from
* since there are likely fewer hyperarcs in the host than nodes, or arcs. */
startHyperArc = (ruleHyperarc)L.hyperarcs.FirstOrDefault(ha => ((ruleHyperarc)ha).MustExist &&
(location.findLMappedHyperarc(ha) == null));
if (startHyperArc != null)
{
if (_in_parallel_)
{
Parallel.ForEach(host.hyperarcs, hostHyperArc =>
{
if (!location.hyperarcs.Contains(hostHyperArc))
{
checkHyperArcAvoidNegatives(location.copy(), startHyperArc, hostHyperArc);
}
});
}
else
{
foreach (var hostHyperArc in
host.hyperarcs.Where(hostHyperArc => !location.hyperarcs.Contains(hostHyperArc)))
{
checkHyperArcAvoidNegatives(location.copy(), startHyperArc, hostHyperArc);
}
}
return;
}
#endregion
#region Case 5: Check entire host for a matching node
/* If no other hyperarcs to recognize look to a unlocated node. If one gets here then none of the above
* three conditions were met (obviously) but this also implies that there are multiple components in the
* LHS, and we are now jumping to a new one with this. This is potentially time intensive if there are
* a lot of nodes in the host. We allow for the possibility that this recognition can be done in parallel. */
startNode = (ruleNode)L.nodes.FirstOrDefault(n => ((ruleNode)n).MustExist && (location.findLMappedNode(n) == null));
if (startNode != null)
{
if (_in_parallel_)
{
Parallel.ForEach(host.nodes, hostNode =>
{
if (!location.nodes.Contains(hostNode))
{
checkNodeAvoidNegatives(location.copy(), startNode, hostNode);
}
});
}
else
{
foreach (var hostNode in host.nodes
.Where(hostNode => !location.nodes.Contains(hostNode)))
{
checkNodeAvoidNegatives(location.copy(), startNode, hostNode);
}
}
return;
}
#endregion
#region Case 6: Check entire host for a matching arc
var looseArc = (ruleArc)L.arcs.FirstOrDefault(a => ((ruleArc)a).MustExist && (location.findLMappedArc(a) == null));
/* the only way one can get here is if there are one or more arcs NOT connected to any nodes
* in L - a floating arc, dangling on both sides, like an eyelash. */
if (looseArc != null)
{
if (_in_parallel_)
{
Parallel.ForEach(host.arcs, hostArc =>
{
if ((!location.arcs.Contains(hostArc)) && (!location.nodes.Contains(hostArc.From)) &&
(!location.nodes.Contains(hostArc.To)) &&
(arcMatches(looseArc, hostArc) || arcMatchRelaxed(looseArc, hostArc, location)))
{
var newLocation = location.copy();
newLocation.arcs[L.arcs.IndexOf(looseArc)] = hostArc;
FindPositiveStartElementAvoidNegatives(newLocation);
}
});
}
else
{
foreach (var hostArc in host.arcs)
{
if ((!location.arcs.Contains(hostArc)) && (!location.nodes.Contains(hostArc.From)) &&
(!location.nodes.Contains(hostArc.To)) &&
(arcMatches(looseArc, hostArc) || arcMatchRelaxed(looseArc, hostArc, location)))
{
var newLocation = location.copy();
newLocation.arcs[L.arcs.IndexOf(looseArc)] = hostArc;
FindPositiveStartElementAvoidNegatives(newLocation);
}
}
}
}
#endregion
}
private void findNegativeStartElement(option location)
{
if ((bool)AllNegativeElementsFound)
{
return; /* another sub-branch found a match to the negative elements.
* There's no point in finding more than one, so this statement
* aborts the search down this branch. */
}
#region Case #1: Location found! No empty slots left in the location
/* this is the only way to properly exit the recursive loop. */
if (!location.nodes.Contains(null) && !location.arcs.Contains(null) && !location.hyperarcs.Contains(null))
{
if (FinalRuleChecks(location))
{
if (!InvalidateWithRelaxation(location))
{
AllNegativeElementsFound = true;
}
}
return;
}
#endregion
#region Case #2: build off of a hyperarc found so far - by looking for unfulfilled nodes
/* the quickest approach to finding a new element in the LHS to host subgraph matching is to build
* directly off of elements found so far. This is because we don't need to check amongst ALL elements in the
* host (as is the case in the last three cases below). In this case we start with any hyperarcs
* that have already been matched to one in the host, and see if it connects to any nodes that
* have yet to be matched. */
var startHyperArc = (ruleHyperarc)L.hyperarcs.FirstOrDefault(ha => ((location.findLMappedHyperarc(ha) != null) &&
(ha.nodes.Any(n => (location.findLMappedNode(n) == null)))));
if (startHyperArc != null)
{
var hostHyperArc = location.findLMappedHyperarc(startHyperArc);
var newLNode = (ruleNode)startHyperArc.nodes.FirstOrDefault(n => (location.findLMappedNode(n) == null));
foreach (var n in hostHyperArc.nodes.Where(n => !location.nodes.Contains(n)))
{
checkForNegativeNode(location.copy(), newLNode, n);
if ((bool)AllNegativeElementsFound)
{
return;
}
}
return;
}
#endregion
#region Case #2.5: build off of a partially matched arc
/* unlike the other renditions of this function (findNewStartElement,
* findPositiveStartElementAvoidNegatives) this has a situation in which an arc has only been
* partially matched but the connected nodes were not touched because they were negative elements.*/
var startArc = (ruleArc)L.arcs.FirstOrDefault(a => (location.findLMappedArc(a) != null) &&
a.To != null && location.findLMappedNode(a.To) == null);
if (startArc != null)
{
var hostArc = location.findLMappedArc(startArc);
if ((hostArc.To != null) && !location.nodes.Contains(hostArc.To))
{
checkForNegativeNode(location.copy(), (ruleNode)startArc.To, hostArc.To);
}
else if (!startArc.directionIsEqual && hostArc.From != null &&
!location.nodes.Contains(hostArc.From))
{
checkForNegativeNode(location.copy(), (ruleNode)startArc.To, hostArc.From);
}
return;
}
startArc = (ruleArc)L.arcs.FirstOrDefault(a => (location.findLMappedArc(a) != null) &&
a.From != null && location.findLMappedNode(a.From) == null);
if (startArc != null)
{
var hostArc = location.findLMappedArc(startArc);
if ((hostArc.From != null) && !location.nodes.Contains(hostArc.From))
{
checkForNegativeNode(location.copy(), (ruleNode)startArc.From, hostArc.From);
}
else if (!startArc.directionIsEqual && hostArc.To != null &&
!location.nodes.Contains(hostArc.To))
{
checkForNegativeNode(location.copy(), (ruleNode)startArc.From, hostArc.To);
}
return;
}
#endregion
#region Case #3: build off of a node found so far - by looking for unfulfilled arcs
/* as stated above, the quickest approach is to build from elements that have already been found.
* Therefore, we see if there are any nodes already matched to a node in L that has an arc in L
* that has yet to be matched with a host arc. This is more efficient than the last 3 cases
* because they look through the entire host, which is potentially large. */
var startNode = (ruleNode)L.nodes.FirstOrDefault(n => ((location.findLMappedNode(n) != null) &&
(n.arcs.Any(a => (location.findLMappedElement(a) == null)))));
/* is there a node already matched (which would only occur if your recursed to get here) that has an
* unrecognized arc attaced to it. If yes, try all possible arcs in the host with the one that needs
* to be fulfilled in L. */
if (startNode != null)
{
var newLArc = startNode.arcs.FirstOrDefault(a => (location.findLMappedElement(a) == null));
if (newLArc is ruleHyperarc)
{
checkForNegativeHyperArc(location, startNode, location.findLMappedNode(startNode), (ruleHyperarc)newLArc);
}
else if (newLArc is ruleArc)
{
checkForNegativeArc(location, startNode, location.findLMappedNode(startNode), (ruleArc)newLArc);
}
return;
}
#endregion
#region Case 4: Check entire host for a matching hyperarc
/* if the above cases didn't match we try to match a hyperarc in the L to any in the host. Since the
* prior three cases have conditions which require some non-nulls in the location, this is likely where the
* process will start when invoked from line 87 of recognize above. Hyperarcs are most efficient to start from
* since there are likely fewer hyperarcs in the host than nodes, or arcs. */
startHyperArc = (ruleHyperarc)L.hyperarcs.FirstOrDefault(ha => (location.findLMappedHyperarc(ha) == null));
if (startHyperArc != null)
{
foreach (var hostHyperArc in
host.hyperarcs.Where(hostHyperArc => !location.hyperarcs.Contains(hostHyperArc)))
{
checkForNegativeHyperArc(location.copy(), startHyperArc, hostHyperArc);
if ((bool)AllNegativeElementsFound)
{
return;
}
}
return;
}
#endregion
#region Case 5: Check entire host for a matching node
/* If no other hyperarcs can be recognized, then look to a unlocated node. If one gets here then none of the above
* three conditions were met (obviously) but this also implies that there are multiple components in the
* LHS, and we are now jumping to a new one with this. This is potentially time intensive if there are
* a lot of nodes in the host. We allow for the possibility that this recognition can be done in parallel. */
startNode = (ruleNode)L.nodes.FirstOrDefault(n => (location.findLMappedNode(n) == null));
if (startNode != null)
{
foreach (var hostNode in host.nodes.Where(hostNode => !location.nodes.Contains(hostNode)))
{
checkForNegativeNode(location.copy(), startNode, hostNode);
if ((bool)AllNegativeElementsFound)
{
return;
}
}
return;
}
#endregion
#region Case 6: Check entire host for a matching arc
var looseArc = (ruleArc)L.arcs.FirstOrDefault(a => (location.findLMappedArc(a) == null));
/* the only way one can get here is if there are one or more arcs NOT connected to any nodes
* in L - a floating arc, dangling on both sides, like an eyelash. */
if (looseArc != null)
{
foreach (var hostArc in host.arcs)
{
if (!location.arcs.Contains(hostArc) && !location.nodes.Contains(hostArc.From) &&
!location.nodes.Contains(hostArc.To) && arcMatches(looseArc, hostArc))
{
var newLocation = location.copy();
newLocation.arcs[L.arcs.IndexOf(looseArc)] = hostArc;
findNegativeStartElement(newLocation);
}
if ((bool)AllNegativeElementsFound)
{
return;
}
}
}
#endregion
}