[922] | 1 | package de.ugoe.cs.autoquest.tasktrees.nodeequality; |
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[816] | 2 | |
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[922] | 3 | import de.ugoe.cs.autoquest.tasktrees.treeifc.ISelection; |
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| 4 | import de.ugoe.cs.autoquest.tasktrees.treeifc.ITaskTreeNode; |
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[816] | 5 | |
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| 6 | /** |
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| 7 | * <p> |
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| 8 | * This class is capable of comparing any task tree node which is not a selection with a |
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| 9 | * selection. This is needed, because selections may contain exactly that node. Therefore, if |
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| 10 | * this node is selected out of a selection the selection is equal to the node itself. |
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| 11 | * The rule returns lexically equal, it the selection contains a lexically equal node. The same |
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| 12 | * applies for syntactical and semantical equality. |
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| 13 | * </p> |
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| 14 | |
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| 15 | * @author Patrick Harms |
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| 16 | */ |
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| 17 | public class NodeAndSelectionComparisonRule implements NodeComparisonRule { |
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| 18 | |
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| 19 | /** the rule manager for internally comparing task tree nodes */ |
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| 20 | private NodeEqualityRuleManager mRuleManager; |
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| 21 | |
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| 22 | /** |
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| 23 | * <p> |
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| 24 | * simple constructor to provide the rule with the node equality rule manager to be able |
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| 25 | * to perform comparisons of the children of provided task tree nodes |
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| 26 | * </p> |
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| 27 | * |
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| 28 | * @param ruleManager the rule manager for comparing task tree nodes |
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| 29 | */ |
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| 30 | NodeAndSelectionComparisonRule(NodeEqualityRuleManager ruleManager) { |
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| 31 | super(); |
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| 32 | mRuleManager = ruleManager; |
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| 33 | } |
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| 34 | |
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| 35 | /* |
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| 36 | * (non-Javadoc) |
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| 37 | * |
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| 38 | * @see de.ugoe.cs.tasktree.nodeequality.NodeEqualityRule#apply(TaskTreeNode, TaskTreeNode) |
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| 39 | */ |
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| 40 | @Override |
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| 41 | public NodeEquality compare(ITaskTreeNode node1, ITaskTreeNode node2) { |
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| 42 | ISelection selection = null; |
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| 43 | ITaskTreeNode node = null; |
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| 44 | |
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| 45 | if (node1 instanceof ISelection) { |
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| 46 | if (node2 instanceof ISelection) { |
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| 47 | // the rule is not responsible for two iterations |
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| 48 | return null; |
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| 49 | } |
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| 50 | |
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| 51 | selection = (ISelection) node1; |
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| 52 | node = node2; |
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| 53 | } |
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| 54 | else if (node2 instanceof ISelection) { |
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| 55 | if (node1 instanceof ISelection) { |
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| 56 | // the rule is not responsible for two iterations |
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| 57 | return null; |
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| 58 | } |
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| 59 | |
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| 60 | selection = (ISelection) node2; |
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| 61 | node = node1; |
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| 62 | } |
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| 63 | else { |
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| 64 | return null; |
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| 65 | } |
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| 66 | |
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| 67 | // now, that we found the iteration and the node, lets compare the child of the iteration |
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| 68 | // with the node. |
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| 69 | if (selection.getChildren().size() < 1) { |
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| 70 | return null; |
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| 71 | } |
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| 72 | |
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| 73 | NodeEquality mostConcreteNodeEquality = null; |
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| 74 | |
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| 75 | for (ITaskTreeNode child : selection.getChildren()) { |
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| 76 | NodeEquality nodeEquality = mRuleManager.applyRules(child, node); |
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| 77 | |
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| 78 | if (nodeEquality != NodeEquality.UNEQUAL) { |
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| 79 | if (mostConcreteNodeEquality == null) { |
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| 80 | mostConcreteNodeEquality = nodeEquality; |
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| 81 | } |
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| 82 | else { |
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| 83 | mostConcreteNodeEquality = |
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| 84 | mostConcreteNodeEquality.getCommonDenominator(nodeEquality); |
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| 85 | } |
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| 86 | } |
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| 87 | } |
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| 88 | |
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| 89 | // although the subtask may be identical to the node, we can not return identical, as |
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| 90 | // the selection is not identical to the node, but at most lexically equal |
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| 91 | if (mostConcreteNodeEquality == NodeEquality.IDENTICAL) { |
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| 92 | return NodeEquality.LEXICALLY_EQUAL; |
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| 93 | } |
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| 94 | else { |
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| 95 | return mostConcreteNodeEquality; |
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| 96 | } |
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| 97 | |
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| 98 | } |
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| 99 | } |
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