// Copyright 2012 Georg-August-Universität Göttingen, Germany
//
// Licensed under the Apache License, Version 2.0 (the "License");
// you may not use this file except in compliance with the License.
// You may obtain a copy of the License at
//
// http://www.apache.org/licenses/LICENSE-2.0
//
// Unless required by applicable law or agreed to in writing, software
// distributed under the License is distributed on an "AS IS" BASIS,
// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
// See the License for the specific language governing permissions and
// limitations under the License.
package de.ugoe.cs.autoquest.tasktrees.nodeequality;
import java.util.List;
import de.ugoe.cs.autoquest.tasktrees.treeifc.IEventTask;
import de.ugoe.cs.autoquest.tasktrees.treeifc.IIteration;
import de.ugoe.cs.autoquest.tasktrees.treeifc.ISelection;
import de.ugoe.cs.autoquest.tasktrees.treeifc.ISequence;
import de.ugoe.cs.autoquest.tasktrees.treeifc.ITaskTreeNode;
/**
*
* This class is capable of comparing Iterations. Iterations equal at distinct levels
* in distinct situations. The following table shows the results of the comparison for the
* specific situations (the parameters are commutative). In any other situation, the comparison
* returns NodeEquality.UNEQUAL
:
*
*
*
*
* iteration 1 |
* iteration 2 |
* comparison result |
*
*
* any iteration |
* any iteration with a child that is lexically equal to the child of iteration 1 |
* NodeEquality.LEXICALLY_EQUAL |
*
*
* any iteration |
* any iteration with a child that is syntactically equal to the child of iteration 1 |
* NodeEquality.SYNTACTICALLY_EQUAL |
*
*
* any iteration |
* any iteration with a child that is semantically equal to the child of iteration 1 |
* NodeEquality.SEMANTICALLY_EQUAL |
*
*
* an iteration with a selection of syntactically equal children |
* an iteration with a child that is syntactically equal to the children of the child
* selection of iteration 1 |
* NodeEquality.SYNTACTICALLY_EQUAL |
*
*
* an iteration with a selection of syntactically equal children |
* an iteration with a selection of syntactically equal children that are all syntactically
* equal to the selection of children of iteration 1 |
* NodeEquality.SYNTACTICALLY_EQUAL |
*
*
* an iteration with a selection of semantically equal children |
* an iteration with a child that is semantically equal to the children of the child
* selection of iteration 1 |
* NodeEquality.SEMANTICALLY_EQUAL |
*
*
* an iteration with a selection of semantically equal children |
* an iteration with a selection of semantically equal children that are all semantically
* equal to the selection of children of iteration 1 |
* NodeEquality.SEMANTICALLY_EQUAL |
*
*
*
* @version $Revision: $ $Date: 19.02.2012$
* @author 2012, last modified by $Author: patrick$
*/
public class IterationComparisonRule implements NodeComparisonRule {
/** the rule manager for internally comparing task tree nodes */
private NodeEqualityRuleManager mRuleManager;
/**
*
* simple constructor to provide the rule with the node equality rule manager to be able
* to perform comparisons of the children of provided task tree nodes
*
*
* @param ruleManager the rule manager for comparing task tree nodes
*/
IterationComparisonRule(NodeEqualityRuleManager ruleManager) {
super();
mRuleManager = ruleManager;
}
/* (non-Javadoc)
* @see NodeComparisonRule#isApplicable(ITaskTreeNode, ITaskTreeNode)
*/
@Override
public boolean isApplicable(ITaskTreeNode node1, ITaskTreeNode node2) {
return (node1 instanceof IIteration) && (node2 instanceof IIteration);
}
/* (non-Javadoc)
* @see NodeComparisonRule#areLexicallyEqual(ITaskTreeNode, ITaskTreeNode)
*/
@Override
public boolean areLexicallyEqual(ITaskTreeNode node1, ITaskTreeNode node2) {
List children1 = node1.getChildren();
List children2 = node2.getChildren();
if (children1.size() == children2.size()) {
if (children1.size() == 0) {
return true;
}
else {
ITaskTreeNode child1 = children1.get(0);
ITaskTreeNode child2 = children2.get(0);
// iterations may have 3 different structures.
// 1. they have one child, which is the iterated one
// 2. they have a sequence of children, which is iterated
// 3. they have a selection of different iterated variants (usually the variants
// are semantically equal)
// check if the type of children match. If not, return false. If they match,
// use the equality manager to perform further comparisons
if (((child1 instanceof ISelection) && (child2 instanceof ISelection)) ||
((child1 instanceof ISequence) && (child2 instanceof ISequence)) ||
((child1 instanceof IEventTask) && (child2 instanceof IEventTask)))
{
return getNodeEquality
(child1, child2).isAtLeast(NodeEquality.LEXICALLY_EQUAL);
}
}
}
return false;
}
/* (non-Javadoc)
* @see NodeComparisonRule#areSyntacticallyEqual(ITaskTreeNode, ITaskTreeNode)
*/
@Override
public boolean areSyntacticallyEqual(ITaskTreeNode node1, ITaskTreeNode node2) {
List children1 = node1.getChildren();
List children2 = node2.getChildren();
if (children1.size() == children2.size()) {
if (children1.size() == 0) {
return true;
}
else {
ITaskTreeNode child1 = children1.get(0);
ITaskTreeNode child2 = children2.get(0);
// iterations may have 3 different structures.
// 1. they have one child, which is the iterated one
// 2. they have a sequence of children, which is iterated
// 3. they have a selection of different iterated variants (usually the variants
// are semantically equal)
// ignore the type of the children but check them for equality.
return getNodeEquality(child1, child2).isAtLeast(NodeEquality.SYNTACTICALLY_EQUAL);
}
}
return false;
}
/* (non-Javadoc)
* @see NodeComparisonRule#areSemanticallyEqual(ITaskTreeNode, ITaskTreeNode)
*/
@Override
public boolean areSemanticallyEqual(ITaskTreeNode node1, ITaskTreeNode node2) {
return compare(node1, node2).isAtLeast(NodeEquality.SEMANTICALLY_EQUAL);
}
/* (non-Javadoc)
* @see NodeComparisonRule#compare(ITaskTreeNode, ITaskTreeNode)
*/
@Override
public NodeEquality compare(ITaskTreeNode node1, ITaskTreeNode node2) {
List children1 = node1.getChildren();
List children2 = node2.getChildren();
// if both iterations do not have children, they are equal although this doesn't make sense
if ((children1.size() == 0) && (children2.size() == 0)) {
return NodeEquality.LEXICALLY_EQUAL;
}
else if ((children1.size() == 0) || (children2.size() == 0)) {
return NodeEquality.UNEQUAL;
}
ITaskTreeNode child1 = children1.get(0);
ITaskTreeNode child2 = children2.get(0);
// iterations may have 3 different structures.
// 1. they have one child, which is the iterated one
// 2. they have a sequence of children, which is iterated
// 3. they have a selection of different iterated variants (usually the variants are
// semantically equal)
//
// the permutations of the three variants in combination must be checked
// check if both nodes are the same variants of iterations and if their children are equal.
// This condition matches, if both iterations are the same variants of iteration. I.e. three
// combinations of the permutation are handled herewith.
NodeEquality nodeEquality = getNodeEquality(child1, child2);
if (nodeEquality != null) {
return nodeEquality;
}
// compare one iteration with a single node as a child and another one with a selection of
// semantically equal nodes
return selectionChildrenSemanticallyEqualNode(child1, child2);
// all other combinations (i.e. sequence with single child and sequence with selection)
// can not match
}
/**
* TODO update comment
*/
private NodeEquality getNodeEquality(ITaskTreeNode child1, ITaskTreeNode child2) {
NodeEquality nodeEquality = callRuleManager(child1, child2, null);
if (nodeEquality.isAtLeast(NodeEquality.SEMANTICALLY_EQUAL)) {
// prevent, that identical is returned, because the iterations itself are not identical
// although the iterated tasks are
if (nodeEquality == NodeEquality.IDENTICAL) {
return NodeEquality.LEXICALLY_EQUAL;
}
else {
return nodeEquality;
}
}
return NodeEquality.UNEQUAL;
}
/**
*
* compares two task tree nodes. One of them must be a selection, the other one can be any task
* tree node. The method returns a node equality that is not NodeEquality.UNEQUAL
* if the other node is at least semantically equal to the children of the selection. It
* returns more concrete equalities, if the equality between the other node and the children
* of the selection is more concrete.
*
*
* @param taskTreeNode the first task tree node to compare
* @param taskTreeNode2 the second task tree node to compare
*
* @return as described
*/
private NodeEquality selectionChildrenSemanticallyEqualNode(ITaskTreeNode taskTreeNode,
ITaskTreeNode taskTreeNode2)
{
ISelection selection = null;
ITaskTreeNode node = null;
if (taskTreeNode instanceof ISelection) {
selection = (ISelection) taskTreeNode;
node = taskTreeNode2;
}
else if (taskTreeNode2 instanceof ISelection) {
selection = (ISelection) taskTreeNode2;
node = taskTreeNode;
}
else {
return NodeEquality.UNEQUAL;
}
// Iterations, where one has a selection and the other one not can at most be syntactically
// equal but not identical
NodeEquality commonDenominatorForAllComparisons = NodeEquality.SYNTACTICALLY_EQUAL;
for (ITaskTreeNode child : selection.getChildren()) {
NodeEquality nodeEquality =
callRuleManager(node, child, commonDenominatorForAllComparisons);
if ((nodeEquality == null) || (nodeEquality == NodeEquality.UNEQUAL))
{
return NodeEquality.UNEQUAL;
}
commonDenominatorForAllComparisons =
commonDenominatorForAllComparisons.getCommonDenominator(nodeEquality);
}
return commonDenominatorForAllComparisons;
}
/**
*
* TODO: comment
*
*
* @param child1
* @param child2
* @param requiredEqualityLevel
* @return
*/
private NodeEquality callRuleManager(ITaskTreeNode child1,
ITaskTreeNode child2,
NodeEquality requiredEqualityLevel)
{
if (requiredEqualityLevel == null) {
return mRuleManager.compare(child1, child2);
}
else if (mRuleManager.areAtLeastEqual(child1, child2, requiredEqualityLevel)) {
return requiredEqualityLevel;
}
else {
return NodeEquality.UNEQUAL;
}
}
}