1 | package de.ugoe.cs.quest.tasktrees.nodeequality; |
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2 | |
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3 | import de.ugoe.cs.quest.tasktrees.treeifc.IIteration; |
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4 | import de.ugoe.cs.quest.tasktrees.treeifc.ITaskTreeNode; |
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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 an iteration with an |
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9 | * iteration. This is needed, because iterations may iterate exactly that node. In this |
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10 | * case, the iteration would be equal to that node if it was executed exactly once. The rule |
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11 | * returns lexically equal, it the child of the iteration is lexically equal to the node |
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12 | * or if the child of the iteration is a selection and this selections contains a lexically equal |
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13 | * node. The same applies for syntactical and semantical equality. |
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14 | * </p> |
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15 | |
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16 | * @author Patrick Harms |
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17 | */ |
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18 | public class NodeAndIterationComparisonRule implements NodeComparisonRule { |
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19 | |
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20 | /** the rule manager for internally comparing task tree nodes */ |
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21 | private NodeEqualityRuleManager mRuleManager; |
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22 | |
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23 | /** |
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24 | * <p> |
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25 | * simple constructor to provide the rule with the node equality rule manager to be able |
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26 | * to perform comparisons of the children of provided task tree nodes |
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27 | * </p> |
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28 | * |
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29 | * @param ruleManager the rule manager for comparing task tree nodes |
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30 | */ |
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31 | NodeAndIterationComparisonRule(NodeEqualityRuleManager ruleManager) { |
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32 | super(); |
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33 | mRuleManager = ruleManager; |
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34 | } |
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35 | |
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36 | /* |
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37 | * (non-Javadoc) |
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38 | * |
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39 | * @see de.ugoe.cs.tasktree.nodeequality.NodeEqualityRule#apply(TaskTreeNode, TaskTreeNode) |
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40 | */ |
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41 | @Override |
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42 | public NodeEquality compare(ITaskTreeNode node1, ITaskTreeNode node2) { |
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43 | IIteration iteration = null; |
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44 | ITaskTreeNode node = null; |
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45 | |
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46 | if (node1 instanceof IIteration) { |
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47 | if (node2 instanceof IIteration) { |
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48 | // the rule is not responsible for two iterations |
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49 | return null; |
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50 | } |
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51 | |
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52 | iteration = (IIteration) node1; |
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53 | node = node2; |
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54 | } |
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55 | else if (node2 instanceof IIteration) { |
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56 | if (node1 instanceof IIteration) { |
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57 | // the rule is not responsible for two iterations |
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58 | return null; |
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59 | } |
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60 | |
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61 | iteration = (IIteration) node2; |
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62 | node = node1; |
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63 | } |
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64 | else { |
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65 | return null; |
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66 | } |
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67 | |
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68 | // now, that we found the iteration and the node, lets compare the child of the iteration |
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69 | // with the node. |
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70 | if (iteration.getChildren().size() < 1) { |
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71 | return null; |
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72 | } |
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73 | |
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74 | NodeEquality nodeEquality = mRuleManager.applyRules(iteration.getChildren().get(0), node); |
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75 | |
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76 | // although the subtask may be identical to the node, we can not return identical, as |
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77 | // the iteration is not identical to the node, but at most lexically equal |
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78 | if (nodeEquality == NodeEquality.IDENTICAL) { |
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79 | return NodeEquality.LEXICALLY_EQUAL; |
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80 | } |
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81 | else { |
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82 | return nodeEquality; |
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83 | } |
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84 | |
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85 | } |
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86 | } |
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