1 | package de.ugoe.cs.quest.tasktrees.nodeequality; |
---|
2 | |
---|
3 | import de.ugoe.cs.quest.tasktrees.treeifc.ISelection; |
---|
4 | import de.ugoe.cs.quest.tasktrees.treeifc.ITaskTreeNode; |
---|
5 | |
---|
6 | /** |
---|
7 | * <p> |
---|
8 | * this node comparison rule is capable of comparing selections. If both selections do not have |
---|
9 | * children, they are treated as lexically equal. If they have the same number of children other |
---|
10 | * than 0 and all these children are lexically equal, then the selections are lexically equal. |
---|
11 | * They are syntactically equal, if each child of both selections is syntactically equal to any |
---|
12 | * other child. The rule can not compare semantical equality if the nodes are not at least |
---|
13 | * syntactically equal and returns null, if it can not decide this. |
---|
14 | * </p> |
---|
15 | * |
---|
16 | * @version $Revision: $ $Date: 19.02.2012$ |
---|
17 | * @author 2012, last modified by $Author: patrick$ |
---|
18 | */ |
---|
19 | public class SelectionComparisonRule implements NodeComparisonRule { |
---|
20 | |
---|
21 | /** the rule manager for internally comparing task tree nodes */ |
---|
22 | private NodeEqualityRuleManager mRuleManager; |
---|
23 | |
---|
24 | /** |
---|
25 | * <p> |
---|
26 | * simple constructor to provide the rule with the node equality rule manager to be able |
---|
27 | * to perform comparisons of the children of provided task tree nodes |
---|
28 | * </p> |
---|
29 | * |
---|
30 | * @param ruleManager the rule manager for comparing task tree nodes |
---|
31 | */ |
---|
32 | SelectionComparisonRule(NodeEqualityRuleManager ruleManager) { |
---|
33 | super(); |
---|
34 | mRuleManager = ruleManager; |
---|
35 | } |
---|
36 | |
---|
37 | /* |
---|
38 | * (non-Javadoc) |
---|
39 | * |
---|
40 | * @see de.ugoe.cs.tasktree.nodeequality.NodeEqualityRule#apply(TaskTreeNode, TaskTreeNode) |
---|
41 | */ |
---|
42 | @Override |
---|
43 | public NodeEquality compare(ITaskTreeNode node1, ITaskTreeNode node2) { |
---|
44 | if ((!(node1 instanceof ISelection)) || (!(node2 instanceof ISelection))) { |
---|
45 | return null; |
---|
46 | } |
---|
47 | |
---|
48 | // if both sequences do not have children, they are equal although this doesn't make sense |
---|
49 | if ((node1.getChildren().size() == 0) && (node2.getChildren().size() == 0)) { |
---|
50 | return NodeEquality.LEXICALLY_EQUAL; |
---|
51 | } |
---|
52 | |
---|
53 | // Selections are syntactically equal, if they have children, which are all syntactically |
---|
54 | // equal. |
---|
55 | // They are lexically equals, if they have the same number and order of lexically equal |
---|
56 | // children |
---|
57 | boolean lexicallyEqual = node1.getChildren().size() == node2.getChildren().size(); |
---|
58 | |
---|
59 | for (int i = 0; i < node1.getChildren().size(); i++) { |
---|
60 | ITaskTreeNode child1 = node1.getChildren().get(i); |
---|
61 | boolean foundLexicallyEqualChild = false; |
---|
62 | |
---|
63 | for (int j = 0; j < node2.getChildren().size(); j++) { |
---|
64 | ITaskTreeNode child2 = node2.getChildren().get(j); |
---|
65 | |
---|
66 | NodeEquality nodeEquality = mRuleManager.applyRules(child1, child2); |
---|
67 | |
---|
68 | if (!nodeEquality.isAtLeast(NodeEquality.SYNTACTICALLY_EQUAL)) { |
---|
69 | return null; |
---|
70 | } |
---|
71 | else if (nodeEquality.isAtLeast(NodeEquality.LEXICALLY_EQUAL)) { |
---|
72 | foundLexicallyEqualChild = true; |
---|
73 | } |
---|
74 | } |
---|
75 | |
---|
76 | // if we compare two children at the same position and if they are lexically equal |
---|
77 | // then it can be further expected, that the selections are lexically equal |
---|
78 | lexicallyEqual &= foundLexicallyEqualChild; |
---|
79 | } |
---|
80 | |
---|
81 | if (lexicallyEqual) { |
---|
82 | return NodeEquality.LEXICALLY_EQUAL; |
---|
83 | } |
---|
84 | else { |
---|
85 | return NodeEquality.SYNTACTICALLY_EQUAL; |
---|
86 | } |
---|
87 | } |
---|
88 | } |
---|