[1113] | 1 | // Copyright 2012 Georg-August-Universität Göttingen, Germany |
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| 2 | // |
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| 3 | // Licensed under the Apache License, Version 2.0 (the "License"); |
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| 4 | // you may not use this file except in compliance with the License. |
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| 5 | // You may obtain a copy of the License at |
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| 6 | // |
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| 7 | // http://www.apache.org/licenses/LICENSE-2.0 |
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| 8 | // |
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| 9 | // Unless required by applicable law or agreed to in writing, software |
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| 10 | // distributed under the License is distributed on an "AS IS" BASIS, |
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| 11 | // WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. |
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| 12 | // See the License for the specific language governing permissions and |
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| 13 | // limitations under the License. |
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| 14 | |
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[922] | 15 | package de.ugoe.cs.autoquest.tasktrees.temporalrelation; |
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[439] | 16 | |
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| 17 | import java.util.ArrayList; |
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| 18 | import java.util.List; |
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| 19 | |
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[922] | 20 | import de.ugoe.cs.autoquest.tasktrees.nodeequality.NodeEquality; |
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| 21 | import de.ugoe.cs.autoquest.tasktrees.nodeequality.NodeEqualityRuleManager; |
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| 22 | import de.ugoe.cs.autoquest.tasktrees.treeifc.IEventTask; |
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| 23 | import de.ugoe.cs.autoquest.tasktrees.treeifc.IIteration; |
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| 24 | import de.ugoe.cs.autoquest.tasktrees.treeifc.ISelection; |
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| 25 | import de.ugoe.cs.autoquest.tasktrees.treeifc.ISequence; |
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| 26 | import de.ugoe.cs.autoquest.tasktrees.treeifc.ITaskTreeBuilder; |
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| 27 | import de.ugoe.cs.autoquest.tasktrees.treeifc.ITaskTreeNode; |
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| 28 | import de.ugoe.cs.autoquest.tasktrees.treeifc.ITaskTreeNodeFactory; |
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[439] | 29 | |
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| 30 | /** |
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[813] | 31 | * <p> |
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| 32 | * iterations in a list of nodes are equal subsequences following each other directly. The |
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| 33 | * subsequences can be of any length depending on the type of equality they need to have. If the |
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| 34 | * subsequences have to be lexically equal, then they have to have the same length if they only |
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| 35 | * contain event tasks. As an example entering text can be done through appropriate keystrokes or |
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| 36 | * through pasting the text. As a result, two syntactically different sequences are semantically |
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| 37 | * equal. If both follow each other, then they are an iteration of semantically equal children. |
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| 38 | * But they are not lexically equal. |
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| 39 | * </p> |
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| 40 | * <p> |
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| 41 | * This class determines equal subsequences following each other. It is provided with a minimal node |
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| 42 | * equality the equal nodes should have. Through this, it is possible to find e.g. lexically |
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| 43 | * equal subsequence through a first application of this rule and semantically equal children to |
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| 44 | * a later application of this rule. This is used by the {@link TemporalRelationshipRuleManager} |
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| 45 | * which instantiates this rule three times, each with a different minimal equality. |
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| 46 | * </p> |
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| 47 | * <p> |
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| 48 | * The equal subsequences are determined through trial and error. This algorithm has a high effort |
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| 49 | * as it tries in the worst case all possible combinations of sub lists in all possible parts of |
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| 50 | * the list of children of a provided parent node. The steps for each trial are. |
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| 51 | * <ul> |
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| 52 | * <li>for all possible subparts of the children of the provided parent |
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| 53 | * <ul> |
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| 54 | * <li>for all possible first sublists in the subpart |
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| 55 | * <ul> |
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| 56 | * <li>for all succeeding next sublists in this part</li> |
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| 57 | * <ul> |
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| 58 | * <li>check if this sublist is equal to all previously identified sublist in this part</li> |
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| 59 | * </ul> |
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| 60 | * </ul> |
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| 61 | * <li> |
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| 62 | * if a combination of sublists is found in this subpart which are all equal to each other |
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| 63 | * at the provided minimal equality level, an iteration in this subpart was found. |
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| 64 | * </li> |
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| 65 | * <ul> |
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| 66 | * <li>merge the identified equal sublists to an iteration</li> |
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| 67 | * </ul> |
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| 68 | * </ul> |
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| 69 | * </ul> |
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| 70 | * The algorithm tries to optimize if all children are event tasks and if the sublists shall be |
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| 71 | * lexically equal. In this case, the sublist all have to have the same length. The trial and |
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| 72 | * error reduces to a minimum of possible sublists. |
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| 73 | * </p> |
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[439] | 74 | * |
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[813] | 75 | * @author Patrick Harms |
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[439] | 76 | */ |
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[1107] | 77 | class DefaultIterationDetectionRule implements TemporalRelationshipRule { |
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| 78 | |
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| 79 | /** |
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| 80 | * <p> |
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| 81 | * the maximum length for iterated sequences |
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| 82 | * </p> |
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| 83 | */ |
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[1045] | 84 | private static final int MAX_LENGTH_OF_ITERATED_SEQUENCE = 50; |
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[1044] | 85 | |
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[813] | 86 | /** |
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| 87 | * <p> |
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[1107] | 88 | * the task tree node factory to be used for creating substructures for the temporal |
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| 89 | * relationships identified during rule |
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| 90 | * </p> |
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| 91 | */ |
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| 92 | private ITaskTreeNodeFactory taskTreeNodeFactory; |
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| 93 | /** |
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| 94 | * <p> |
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| 95 | * the task tree builder to be used for creating substructures for the temporal relationships |
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| 96 | * identified during rule application |
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| 97 | * </p> |
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| 98 | */ |
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| 99 | private ITaskTreeBuilder taskTreeBuilder; |
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| 100 | |
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| 101 | /** |
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| 102 | * <p> |
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[813] | 103 | * the node equality manager needed for comparing task tree nodes with each other |
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| 104 | * </p> |
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| 105 | */ |
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[557] | 106 | private NodeEqualityRuleManager nodeEqualityRuleManager; |
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[439] | 107 | |
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[813] | 108 | /** |
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| 109 | * <p> |
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| 110 | * the minimal node equality two identified sublists need to have to consider them as equal |
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| 111 | * and to create an iteration for |
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| 112 | * </p> |
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| 113 | */ |
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[805] | 114 | private NodeEquality minimalNodeEquality; |
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| 115 | |
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[557] | 116 | /** |
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[813] | 117 | * <p> |
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| 118 | * instantiates the rule and initializes it with a node equality rule manager and the minimal |
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| 119 | * node equality identified sublist must have to consider them as iterated. |
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| 120 | * </p> |
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[557] | 121 | */ |
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[805] | 122 | DefaultIterationDetectionRule(NodeEqualityRuleManager nodeEqualityRuleManager, |
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[1107] | 123 | NodeEquality minimalNodeEquality, |
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| 124 | ITaskTreeNodeFactory taskTreeNodeFactory, |
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| 125 | ITaskTreeBuilder taskTreeBuilder) |
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[805] | 126 | { |
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[557] | 127 | this.nodeEqualityRuleManager = nodeEqualityRuleManager; |
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[805] | 128 | this.minimalNodeEquality = minimalNodeEquality; |
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[1107] | 129 | this.taskTreeNodeFactory = taskTreeNodeFactory; |
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| 130 | this.taskTreeBuilder = taskTreeBuilder; |
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[557] | 131 | } |
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[439] | 132 | |
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[557] | 133 | /* |
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| 134 | * (non-Javadoc) |
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| 135 | * |
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[1107] | 136 | * @see de.ugoe.cs.tasktree.temporalrelation.TemporalRelationshipRule#apply(TaskTreeNode, |
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| 137 | * boolean) |
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[557] | 138 | */ |
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| 139 | @Override |
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[1107] | 140 | public RuleApplicationResult apply(ITaskTreeNode parent, boolean finalize) { |
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[557] | 141 | if (!(parent instanceof ISequence)) { |
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| 142 | return null; |
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| 143 | } |
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| 144 | |
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[813] | 145 | if (!finalize) { |
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| 146 | // the rule is always feasible as iterations may occur at any time |
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| 147 | RuleApplicationResult result = new RuleApplicationResult(); |
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| 148 | result.setRuleApplicationStatus(RuleApplicationStatus.RULE_APPLICATION_FEASIBLE); |
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| 149 | return result; |
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| 150 | } |
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| 151 | |
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[557] | 152 | // parent must already have at least 2 children |
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| 153 | if ((parent.getChildren() == null) || (parent.getChildren().size() < 2)) { |
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| 154 | return null; |
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| 155 | } |
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[813] | 156 | |
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| 157 | |
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[1107] | 158 | SubSequences subSequences = getEqualSubsequences(parent); |
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[557] | 159 | |
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[813] | 160 | if (subSequences != null) { |
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| 161 | RuleApplicationResult result = new RuleApplicationResult(); |
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[557] | 162 | |
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[1107] | 163 | mergeEqualNodes(subSequences.equalVariants); |
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| 164 | IIteration newIteration = |
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| 165 | createIterationBasedOnIdentifiedVariants(subSequences, result); |
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[557] | 166 | |
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[813] | 167 | determineNewlyCreatedParentTasks(parent, newIteration, result); |
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| 168 | |
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| 169 | // remove iterated children |
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| 170 | for (int j = subSequences.start; j < subSequences.end; j++) { |
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[1107] | 171 | taskTreeBuilder.removeChild((ISequence) parent, subSequences.start); |
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[813] | 172 | } |
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[557] | 173 | |
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[813] | 174 | // add the new iteration instead |
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[1107] | 175 | taskTreeBuilder.addChild((ISequence) parent, subSequences.start, newIteration); |
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[557] | 176 | |
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[813] | 177 | result.setRuleApplicationStatus(RuleApplicationStatus.RULE_APPLICATION_FINISHED); |
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| 178 | return result; |
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[557] | 179 | } |
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[439] | 180 | |
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[557] | 181 | return null; |
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[439] | 182 | } |
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| 183 | |
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[557] | 184 | /** |
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[813] | 185 | * <p> |
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| 186 | * this method initiates the trial and error algorithm denoted in the description of this class. |
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| 187 | * Its main purpose is the selection of a subpart of all children in the parent node in which |
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| 188 | * equal sublists shall be searched. It is important, to always find the last iterations in a |
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| 189 | * part first. The reason for this are iterations of iterations. If we always found the first |
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| 190 | * iteration in a subpart first, then this may be an iteration of iterations. However, there |
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| 191 | * may be subsequent iterations to be included in this iteration. But these iterations are not |
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| 192 | * found yet, as they occur later in the sequence. Therefore, if we always find the last |
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| 193 | * iteration in a sequence first, iterations of iterations are identified, last. |
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| 194 | * </p> |
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[557] | 195 | * |
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[813] | 196 | * @param parent the parent node in which iterations of children shall be found |
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| 197 | * |
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| 198 | * @return the iterated subsequences identified in a specific part (contains the equal |
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| 199 | * subsequences as well as the start (inclusive) and end (exclusive) index of the |
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| 200 | * subpart in which the sequences were found) |
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[557] | 201 | */ |
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[1107] | 202 | private SubSequences getEqualSubsequences(ITaskTreeNode parent) { |
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[813] | 203 | SubSequences subSequences = null; |
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[557] | 204 | |
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[970] | 205 | // to find longer iterations first, start with long sequences |
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[813] | 206 | FIND_ITERATION: |
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| 207 | for (int end = parent.getChildren().size(); end > 0; end--) { |
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| 208 | for (int start = 0; start < end; start++) { |
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| 209 | boolean useEqualSublistLengths = equalSublistLengthsCanBeUsed(parent, start, end); |
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[557] | 210 | |
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[813] | 211 | subSequences = new SubSequences(); |
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| 212 | subSequences.start = start; |
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[557] | 213 | |
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[813] | 214 | boolean foundFurtherVariants = findFurtherVariants |
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[1107] | 215 | (subSequences, parent, start, end, useEqualSublistLengths); |
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[813] | 216 | |
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| 217 | if (foundFurtherVariants) { |
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| 218 | break FIND_ITERATION; |
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[557] | 219 | } |
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[813] | 220 | else { |
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| 221 | subSequences = null; |
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| 222 | } |
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[557] | 223 | } |
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[439] | 224 | } |
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[813] | 225 | |
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| 226 | return subSequences; |
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[439] | 227 | } |
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| 228 | |
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[557] | 229 | /** |
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[813] | 230 | * <p> |
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| 231 | * for optimization purposes, we check if the length of the sublists to be identified as |
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| 232 | * iterations has to be the same for any sublist. This only applies, if the minimum node |
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| 233 | * equality to be checked for is lexical equality. If the children of the parent are all event |
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| 234 | * tasks, then sublists can only be lexically equal, if they all have the same length. |
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| 235 | * Therefore we check, if the minimal node equality is lexical equality. And if so, we also |
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| 236 | * check if all children of the parent in which an iteration shall be searched for are event |
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| 237 | * tasks. |
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| 238 | * </p> |
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[805] | 239 | * |
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[813] | 240 | * @param parent the parent node to search for iterations of its children |
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| 241 | * @param start the beginning of the subpart (inclusive) to be considered |
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| 242 | * @param end the end of the subpart (exclusive) to be considered |
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| 243 | * |
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| 244 | * @return true, if the sublists must have the same lengths, false else |
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[805] | 245 | */ |
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[813] | 246 | private boolean equalSublistLengthsCanBeUsed(ITaskTreeNode parent, int start, int end) { |
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| 247 | boolean equalLengthsCanBeUsed = minimalNodeEquality.isAtLeast(NodeEquality.LEXICALLY_EQUAL); |
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| 248 | |
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| 249 | if (equalLengthsCanBeUsed) { |
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| 250 | for (int i = start; i < end; i++) { |
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| 251 | if (!(parent.getChildren().get(i) instanceof IEventTask)) { |
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| 252 | equalLengthsCanBeUsed = false; |
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| 253 | break; |
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| 254 | } |
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| 255 | } |
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[439] | 256 | } |
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[557] | 257 | |
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[813] | 258 | return equalLengthsCanBeUsed; |
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| 259 | } |
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[557] | 260 | |
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[813] | 261 | /** |
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| 262 | * <p> |
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| 263 | * this method starts at a specific position in the list of children of the provided parent |
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| 264 | * and checks, if it finds a further sublist, that matches the already found sublists. If |
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| 265 | * the sublist lengths must be equal, it only searches for a sublist of the same length of the |
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| 266 | * already found sublists. The method calls itself if it identifies a further equal sublist but |
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| 267 | * if the end of the subpart of children is not yet reached. |
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| 268 | * </p> |
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| 269 | * |
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| 270 | * @param subSequences the sublist found so far against which equality of the next |
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| 271 | * sublist must be checked |
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| 272 | * @param parent the parent node of which the children are analyzed |
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| 273 | * @param start the starting index from which to start the next sublist to be |
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| 274 | * identified |
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| 275 | * @param end the end index (exclusive) of the current subpart of children |
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| 276 | * in which iterations are searched for |
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| 277 | * @param useEqualSublistLengths true if the sublists to be searched for all need to have the |
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| 278 | * same length |
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| 279 | * |
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| 280 | * @return true if a further equal variant was found, false else |
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| 281 | */ |
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| 282 | private boolean findFurtherVariants(SubSequences subSequences, |
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| 283 | ITaskTreeNode parent, |
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| 284 | int start, |
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| 285 | int end, |
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| 286 | boolean useEqualSublistLengths) |
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| 287 | { |
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| 288 | boolean foundFurtherVariants = (start == end) && (subSequences.equalVariants.size() > 1); |
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| 289 | |
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| 290 | int minChildCount = 1; |
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[1044] | 291 | int maxChildCount = Math.min(MAX_LENGTH_OF_ITERATED_SEQUENCE, end - start); |
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[813] | 292 | |
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| 293 | if (useEqualSublistLengths && (subSequences.equalVariants.size() > 0)) { |
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| 294 | minChildCount = subSequences.equalVariants.get(0).getChildren().size(); |
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| 295 | maxChildCount = Math.min(minChildCount, maxChildCount); |
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| 296 | } |
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| 297 | |
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| 298 | for (int childCount = minChildCount; childCount <= maxChildCount; childCount++) { |
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| 299 | if (useEqualSublistLengths && (((end - start) % childCount) != 0)) { |
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| 300 | continue; |
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| 301 | } |
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| 302 | |
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[1107] | 303 | ISequence furtherVariant = taskTreeNodeFactory.createNewSequence(); |
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[813] | 304 | |
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| 305 | for (int j = start; j < start + childCount; j++) { |
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[1107] | 306 | taskTreeBuilder.addChild(furtherVariant, parent.getChildren().get(j)); |
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[813] | 307 | } |
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| 308 | |
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| 309 | boolean allVariantsEqual = true; |
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| 310 | |
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| 311 | for (ITaskTreeNode equalVariant : subSequences.equalVariants) { |
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| 312 | NodeEquality nodeEquality = |
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| 313 | nodeEqualityRuleManager.applyRules(equalVariant, furtherVariant); |
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| 314 | |
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[805] | 315 | if (!nodeEquality.isAtLeast(minimalNodeEquality)) { |
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[813] | 316 | allVariantsEqual = false; |
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| 317 | break; |
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[557] | 318 | } |
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[813] | 319 | } |
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| 320 | |
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| 321 | if (allVariantsEqual) { |
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| 322 | |
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| 323 | // we found a further variant. Add it to the list of variants and try to find |
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| 324 | // further variants. Ignore, if none is available |
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| 325 | int index = subSequences.equalVariants.size(); |
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| 326 | subSequences.equalVariants.add(index, furtherVariant); |
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| 327 | |
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| 328 | foundFurtherVariants = findFurtherVariants |
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[1107] | 329 | (subSequences, parent, start + childCount, end, useEqualSublistLengths); |
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[813] | 330 | |
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| 331 | if (foundFurtherVariants) { |
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| 332 | subSequences.end = end; |
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| 333 | break; |
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[805] | 334 | } |
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[813] | 335 | else { |
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| 336 | subSequences.equalVariants.remove(index); |
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| 337 | } |
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[557] | 338 | } |
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[813] | 339 | } |
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| 340 | |
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| 341 | return foundFurtherVariants; |
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| 342 | } |
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[557] | 343 | |
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[813] | 344 | /** |
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| 345 | * <p> |
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| 346 | * this method merges task tree nodes in a list, if they can be merged. for this, it tries |
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| 347 | * to merge every node with every other node in the provided list using the |
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| 348 | * {@link #mergeEqualTasks(ITaskTreeNode, ITaskTreeNode, ITaskTreeBuilder, ITaskTreeNodeFactory)} |
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| 349 | * method. If a merge is possible, it removes the merged nodes from the list and adds the |
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| 350 | * merge result. |
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| 351 | * </p> |
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| 352 | * |
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| 353 | * @param nodes the list of nodes to be merged |
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| 354 | */ |
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[1107] | 355 | private void mergeEqualNodes(List<ITaskTreeNode> nodes) { |
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[813] | 356 | int index1 = 0; |
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| 357 | int index2 = 0; |
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| 358 | ITaskTreeNode variant1; |
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| 359 | ITaskTreeNode variant2; |
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| 360 | |
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| 361 | while (index1 < nodes.size()) { |
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| 362 | variant1 = nodes.get(index1); |
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| 363 | index2 = index1 + 1; |
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| 364 | |
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| 365 | while (index2 < nodes.size()) { |
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| 366 | variant2 = nodes.get(index2); |
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[1107] | 367 | ITaskTreeNode mergedChild = mergeEqualTasks(variant1, variant2); |
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[813] | 368 | |
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| 369 | if (mergedChild != null) { |
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| 370 | // if we merged something start from the beginning to perform the next merge |
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| 371 | nodes.remove(index2); |
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| 372 | nodes.remove(index1); |
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| 373 | nodes.add(index1, mergedChild); |
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| 374 | index1 = -1; |
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| 375 | break; |
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[557] | 376 | } |
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| 377 | else { |
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[813] | 378 | index2++; |
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[557] | 379 | } |
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| 380 | } |
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[813] | 381 | |
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| 382 | index1++; |
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| 383 | } |
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| 384 | } |
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[557] | 385 | |
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[813] | 386 | /** |
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| 387 | * <p> |
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| 388 | * this method merges two equal tasks with each other if possible. If the tasks are lexically |
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| 389 | * equal, the first of them is returned as merge result. If both tasks are of the same |
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| 390 | * temporal relationship type, the appropriate merge method is called to merge them. If one |
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| 391 | * of the nodes is a selection, the other one is added as a variant of this selection. |
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| 392 | * (However, if both nodes are selections, they are merged using the appropriate merge method.) |
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| 393 | * If merging is not possible, then a selection of both provided nodes is created and |
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| 394 | * returned as merge result. |
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| 395 | * </p> |
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| 396 | * |
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| 397 | * @param node1 the first task to be merged |
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| 398 | * @param node2 the second task to be merged |
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| 399 | * |
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| 400 | * @return the result of the merge |
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| 401 | */ |
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[1107] | 402 | private ITaskTreeNode mergeEqualTasks(ITaskTreeNode node1, ITaskTreeNode node2) { |
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[813] | 403 | ITaskTreeNode mergeResult = null; |
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| 404 | |
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[818] | 405 | if ((node1 instanceof ISequence) && (node2 instanceof ISequence)) { |
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[1107] | 406 | mergeResult = mergeEqualSequences((ISequence) node1, (ISequence) node2); |
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[818] | 407 | } |
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| 408 | else if ((node1 instanceof ISelection) && (node2 instanceof ISelection)) { |
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[1107] | 409 | mergeResult = mergeEqualSelections((ISelection) node1, (ISelection) node2); |
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[818] | 410 | } |
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| 411 | else if ((node1 instanceof IIteration) && (node2 instanceof IIteration)) { |
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[1107] | 412 | mergeResult = mergeEqualIterations((IIteration) node1, (IIteration) node2); |
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[818] | 413 | } |
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| 414 | else if (node1 instanceof ISelection) { |
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[1107] | 415 | taskTreeBuilder.addChild((ISelection) node1, node2); |
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[813] | 416 | mergeResult = node1; |
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| 417 | } |
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[818] | 418 | else if (node2 instanceof ISelection) { |
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[1107] | 419 | taskTreeBuilder.addChild((ISelection) node2, node1); |
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[818] | 420 | mergeResult = node2; |
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| 421 | } |
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| 422 | else if (node1 instanceof IIteration) { |
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[1107] | 423 | mergeResult = mergeEqualTasks(((IIteration) node1).getChildren().get(0), node2); |
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[818] | 424 | |
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| 425 | if (mergeResult != null) { |
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[1107] | 426 | IIteration iteration = taskTreeNodeFactory.createNewIteration(); |
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| 427 | taskTreeBuilder.setChild(iteration, mergeResult); |
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[818] | 428 | mergeResult = iteration; |
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[557] | 429 | } |
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[818] | 430 | } |
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| 431 | else if (node2 instanceof IIteration) { |
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[1107] | 432 | mergeResult = mergeEqualTasks(((IIteration) node2).getChildren().get(0), node1); |
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[818] | 433 | |
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| 434 | if (mergeResult != null) { |
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[1107] | 435 | IIteration iteration = taskTreeNodeFactory.createNewIteration(); |
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| 436 | taskTreeBuilder.setChild(iteration, mergeResult); |
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[818] | 437 | mergeResult = iteration; |
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[813] | 438 | } |
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[818] | 439 | } |
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| 440 | else { |
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| 441 | NodeEquality nodeEquality = nodeEqualityRuleManager.applyRules(node1, node2); |
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| 442 | |
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| 443 | if (nodeEquality.isAtLeast(NodeEquality.LEXICALLY_EQUAL)) { |
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[813] | 444 | mergeResult = node1; |
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| 445 | } |
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[439] | 446 | } |
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[818] | 447 | |
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[813] | 448 | if (mergeResult == null) { |
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[1107] | 449 | mergeResult = taskTreeNodeFactory.createNewSelection(); |
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| 450 | taskTreeBuilder.addChild((ISelection) mergeResult, node1); |
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| 451 | taskTreeBuilder.addChild((ISelection) mergeResult, node2); |
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[813] | 452 | } |
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| 453 | |
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| 454 | return mergeResult; |
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[439] | 455 | } |
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| 456 | |
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[805] | 457 | /** |
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| 458 | * <p> |
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[813] | 459 | * merges equal sequences. This is done through trying to merge each node of sequence 1 with |
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| 460 | * the node in sequence 2 being located at the same position. If not all children can be merged |
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| 461 | * or if the sequences have different lengths, null is returned to indicate, that merging is |
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| 462 | * not possible. For merging children, the |
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| 463 | * {@link #mergeEqualTasks(ITaskTreeNode, ITaskTreeNode, ITaskTreeBuilder, ITaskTreeNodeFactory)} |
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| 464 | * method is called. |
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[805] | 465 | * </p> |
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| 466 | * |
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[813] | 467 | * @param sequence1 the first sequence to be merged |
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| 468 | * @param sequence2 the second sequence to be merged |
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| 469 | * |
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| 470 | * @return the result of the merge or null if merging was not possible |
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[805] | 471 | */ |
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[1107] | 472 | private ISequence mergeEqualSequences(ISequence sequence1, ISequence sequence2) { |
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[813] | 473 | ISequence mergeResult = null; |
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| 474 | |
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| 475 | if (sequence1.getChildren().size() == sequence2.getChildren().size()) { |
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[1107] | 476 | mergeResult = taskTreeNodeFactory.createNewSequence(); |
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[813] | 477 | |
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| 478 | for (int i = 0; i < sequence1.getChildren().size(); i++) { |
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| 479 | ITaskTreeNode mergedNode = mergeEqualTasks |
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[1107] | 480 | (sequence1.getChildren().get(i), sequence2.getChildren().get(i)); |
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[813] | 481 | |
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| 482 | if (mergedNode != null) { |
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[1107] | 483 | taskTreeBuilder.addChild(mergeResult, mergedNode); |
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[813] | 484 | } |
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| 485 | else { |
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| 486 | mergeResult = null; |
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| 487 | break; |
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| 488 | } |
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[805] | 489 | } |
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[813] | 490 | } |
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| 491 | |
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| 492 | return mergeResult; |
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| 493 | } |
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[805] | 494 | |
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[813] | 495 | /** |
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| 496 | * <p> |
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[970] | 497 | * merges equal selections. This is done by adding those children of the second selection to |
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| 498 | * the first selection that can not be merged with any of the children of the first selection. |
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| 499 | * If a merge is possible and this merge is not a simple selection of the merged children, |
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| 500 | * then the merged child replaces the child of the first selection which was merged. |
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[813] | 501 | * </p> |
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| 502 | * |
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| 503 | * @param selection1 the first selection to be merged |
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| 504 | * @param selection2 the second selection to be merged |
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| 505 | * |
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[970] | 506 | * @return the result of the merge which is never null |
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[813] | 507 | */ |
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[1107] | 508 | private ITaskTreeNode mergeEqualSelections(ISelection selection1, ISelection selection2) { |
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[970] | 509 | ISelection mergeResult = selection1; |
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[813] | 510 | |
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[970] | 511 | ITaskTreeNode childToMerge = null; |
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| 512 | ITaskTreeNode mergedChild = null; |
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| 513 | |
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| 514 | // check for each child of selection 2 if it is a duplicate of one of the children |
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| 515 | // if selection 1. If not, add it as further child to the merge result, else skip it. |
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[813] | 516 | for (int i = 0; i < selection2.getChildren().size(); i++) { |
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[970] | 517 | childToMerge = selection2.getChildren().get(i); |
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| 518 | for (int j = 0; j < selection1.getChildren().size(); j++) { |
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[1107] | 519 | mergedChild = mergeEqualTasks(selection1.getChildren().get(j), childToMerge); |
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[970] | 520 | |
---|
| 521 | // a merge must not be a selection, except it is one of the children. Otherwise |
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| 522 | // no real merge was done. |
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| 523 | if ((mergedChild != null) && |
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| 524 | ((!(mergedChild instanceof ISelection)) || |
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| 525 | (selection1.getChildren().get(j) == mergedChild) || |
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| 526 | (childToMerge == mergedChild))) |
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| 527 | { |
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| 528 | // we found a real merge. So replace the original child in selection 1 with |
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| 529 | // the merged child |
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[1107] | 530 | taskTreeBuilder.removeChild(selection1, selection1.getChildren().get(j)); |
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| 531 | taskTreeBuilder.addChild(selection1, mergedChild); |
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[970] | 532 | mergedChild = null; |
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| 533 | childToMerge = null; |
---|
| 534 | break; |
---|
| 535 | } |
---|
| 536 | } |
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| 537 | |
---|
| 538 | if (childToMerge != null) { |
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[1107] | 539 | taskTreeBuilder.addChild(selection1, childToMerge); |
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[970] | 540 | } |
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[813] | 541 | } |
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| 542 | |
---|
| 543 | return mergeResult; |
---|
| 544 | } |
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[805] | 545 | |
---|
[813] | 546 | /** |
---|
| 547 | * <p> |
---|
| 548 | * merges equal iterations. This is done through merging the children of both iterations. If |
---|
| 549 | * this is possible, a resulting iteration with the merge result of the children as its own |
---|
| 550 | * child is returned. Otherwise null is returned to indicate that merging was not possible. |
---|
| 551 | * </p> |
---|
| 552 | * |
---|
| 553 | * @param selection1 the first iteration to be merged |
---|
| 554 | * @param selection2 the second iteration to be merged |
---|
| 555 | * |
---|
| 556 | * @return the result of the merge or null if merging is not possible |
---|
| 557 | */ |
---|
[1107] | 558 | private ITaskTreeNode mergeEqualIterations(IIteration iteration1, IIteration iteration2) { |
---|
[813] | 559 | ITaskTreeNode mergedChild = mergeEqualTasks |
---|
[1107] | 560 | (iteration1.getChildren().get(0), iteration2.getChildren().get(0)); |
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[813] | 561 | |
---|
| 562 | IIteration mergeResult = null; |
---|
| 563 | |
---|
| 564 | if (mergedChild != null) { |
---|
[1107] | 565 | mergeResult = taskTreeNodeFactory.createNewIteration(); |
---|
| 566 | taskTreeBuilder.setChild(mergeResult, mergedChild); |
---|
[805] | 567 | } |
---|
[813] | 568 | |
---|
| 569 | return mergeResult; |
---|
| 570 | } |
---|
[805] | 571 | |
---|
| 572 | /** |
---|
| 573 | * <p> |
---|
[813] | 574 | * this is a convenience method to create an iteration based on the identified and already |
---|
| 575 | * merged iterated subsequences. This method creates the simplest iteration possible. As an |
---|
| 576 | * example, if always the same task tree node is iterated, it becomes the child of the |
---|
| 577 | * iteration. If a sequence of tasks is iterated, this sequence becomes the child of the |
---|
| 578 | * iteration. It several equal sublists or nodes which are not lexically equal are iterated |
---|
| 579 | * they become a selection which in turn become the child of the iteration. |
---|
[805] | 580 | * </p> |
---|
| 581 | * |
---|
[813] | 582 | * @param subsequences the identified and already merged equal subsequences |
---|
| 583 | * |
---|
| 584 | * @return the resulting iteration |
---|
[805] | 585 | */ |
---|
[813] | 586 | private IIteration createIterationBasedOnIdentifiedVariants(SubSequences subsequences, |
---|
[805] | 587 | RuleApplicationResult result) |
---|
| 588 | { |
---|
[1107] | 589 | IIteration newIteration = taskTreeNodeFactory.createNewIteration(); |
---|
[805] | 590 | result.addNewlyCreatedParentNode(newIteration); |
---|
| 591 | |
---|
[813] | 592 | if (subsequences.equalVariants.size() == 1) { |
---|
[805] | 593 | // all children are the same. Create an iteration of this child |
---|
[813] | 594 | if (subsequences.equalVariants.get(0).getChildren().size() == 1) { |
---|
| 595 | // there is only one equal variant and this has only one child. So create an |
---|
| 596 | // iteration of this child |
---|
[1107] | 597 | taskTreeBuilder.setChild |
---|
[813] | 598 | (newIteration, subsequences.equalVariants.get(0).getChildren().get(0)); |
---|
[805] | 599 | } |
---|
| 600 | else { |
---|
[813] | 601 | // there was an iteration of one equal sequence |
---|
[1107] | 602 | taskTreeBuilder.setChild(newIteration, subsequences.equalVariants.get(0)); |
---|
[813] | 603 | result.addNewlyCreatedParentNode(subsequences.equalVariants.get(0)); |
---|
[805] | 604 | } |
---|
| 605 | } |
---|
| 606 | else { |
---|
[813] | 607 | // there are distinct variants of equal subsequences or children --> create an |
---|
| 608 | // iterated selection |
---|
[1107] | 609 | ISelection selection = taskTreeNodeFactory.createNewSelection(); |
---|
[805] | 610 | result.addNewlyCreatedParentNode(selection); |
---|
| 611 | |
---|
[813] | 612 | for (ITaskTreeNode variant : subsequences.equalVariants) { |
---|
| 613 | if (variant.getChildren().size() == 1) { |
---|
[1107] | 614 | taskTreeBuilder.addChild(selection, variant.getChildren().get(0)); |
---|
[805] | 615 | } |
---|
| 616 | else { |
---|
[1107] | 617 | taskTreeBuilder.addChild(selection, variant); |
---|
[813] | 618 | result.addNewlyCreatedParentNode(variant); |
---|
[805] | 619 | } |
---|
| 620 | } |
---|
| 621 | |
---|
[1107] | 622 | taskTreeBuilder.setChild(newIteration, selection); |
---|
[805] | 623 | } |
---|
| 624 | |
---|
| 625 | return newIteration; |
---|
| 626 | } |
---|
| 627 | |
---|
[813] | 628 | /** |
---|
| 629 | * <p> |
---|
| 630 | * as the method has to denote all newly created parent nodes this method identifies them by |
---|
| 631 | * comparing the existing subtree with the newly created iteration. Only those parent nodes |
---|
| 632 | * in the new iteration, which are not already found in the existing sub tree are denoted as |
---|
| 633 | * newly created. We do this in this way, as during the iteration detection algorithm, many |
---|
| 634 | * parent nodes are created, which may be discarded later. It is easier to identify the |
---|
| 635 | * remaining newly created parent nodes through this way than to integrate it into the |
---|
| 636 | * algorithm. |
---|
| 637 | * </p> |
---|
| 638 | * |
---|
| 639 | * @param existingSubTree the existing subtree |
---|
| 640 | * @param newSubTree the identified iteration |
---|
| 641 | * @param result the rule application result into which the newly created parent nodes |
---|
| 642 | * shall be stored. |
---|
| 643 | */ |
---|
| 644 | private void determineNewlyCreatedParentTasks(ITaskTreeNode existingSubTree, |
---|
| 645 | ITaskTreeNode newSubTree, |
---|
| 646 | RuleApplicationResult result) |
---|
| 647 | { |
---|
| 648 | List<ITaskTreeNode> existingParentNodes = getParentNodes(existingSubTree); |
---|
| 649 | List<ITaskTreeNode> newParentNodes = getParentNodes(newSubTree); |
---|
| 650 | |
---|
| 651 | boolean foundNode; |
---|
| 652 | for (ITaskTreeNode newParentNode : newParentNodes) { |
---|
| 653 | foundNode = false; |
---|
| 654 | for (ITaskTreeNode existingParentNode : existingParentNodes) { |
---|
| 655 | // It is sufficient to compare the references. The algorithm reuses nodes as they |
---|
| 656 | // are. So any node existing in the new structure that is also in the old structure |
---|
| 657 | // was unchanged an therefore does not need to be handled as a newly created one. |
---|
| 658 | // but every node in the new structure that is not included in the old structure |
---|
| 659 | // must be treated as a newly created one. |
---|
| 660 | if (newParentNode == existingParentNode) { |
---|
| 661 | foundNode = true; |
---|
| 662 | break; |
---|
| 663 | } |
---|
| 664 | } |
---|
| 665 | |
---|
| 666 | if (!foundNode) { |
---|
| 667 | result.addNewlyCreatedParentNode(newParentNode); |
---|
| 668 | } |
---|
| 669 | } |
---|
| 670 | |
---|
| 671 | } |
---|
| 672 | |
---|
| 673 | /** |
---|
| 674 | * <p> |
---|
| 675 | * convenience method to determine all parent nodes existing in a subtree |
---|
| 676 | * </p> |
---|
| 677 | * |
---|
| 678 | * @param subtree the subtree to search for parent nodes in |
---|
| 679 | * |
---|
| 680 | * @return a list of parent nodes existing in the subtree |
---|
| 681 | */ |
---|
| 682 | private List<ITaskTreeNode> getParentNodes(ITaskTreeNode subtree) { |
---|
| 683 | List<ITaskTreeNode> parentNodes = new ArrayList<ITaskTreeNode>(); |
---|
| 684 | |
---|
| 685 | if (subtree.getChildren().size() > 0) { |
---|
| 686 | parentNodes.add(subtree); |
---|
| 687 | |
---|
| 688 | for (ITaskTreeNode child : subtree.getChildren()) { |
---|
| 689 | parentNodes.addAll(getParentNodes(child)); |
---|
| 690 | } |
---|
| 691 | } |
---|
| 692 | |
---|
| 693 | return parentNodes; |
---|
| 694 | } |
---|
| 695 | |
---|
| 696 | /** |
---|
| 697 | * <p> |
---|
| 698 | * used to have a container for equal sublists identified in a sub part of the children of |
---|
| 699 | * a parent node. |
---|
| 700 | * </p> |
---|
| 701 | * |
---|
| 702 | * @author Patrick Harms |
---|
| 703 | */ |
---|
[815] | 704 | private static class SubSequences { |
---|
[813] | 705 | |
---|
| 706 | /** |
---|
| 707 | * <p> |
---|
| 708 | * the beginning of the subpart of the children of the parent node in which the sublists |
---|
| 709 | * are found (inclusive) |
---|
| 710 | * </p> |
---|
| 711 | */ |
---|
| 712 | public int start; |
---|
| 713 | |
---|
| 714 | /** |
---|
| 715 | * <p> |
---|
| 716 | * the end of the subpart of the children of the parent node in which the sublists |
---|
| 717 | * are found (exclusive) |
---|
| 718 | * </p> |
---|
| 719 | */ |
---|
| 720 | public int end; |
---|
| 721 | |
---|
| 722 | /** |
---|
| 723 | * <p> |
---|
| 724 | * the equal sublists found in the subpart of the children of the parent node |
---|
| 725 | * </p> |
---|
| 726 | */ |
---|
| 727 | List<ITaskTreeNode> equalVariants = new ArrayList<ITaskTreeNode>(); |
---|
| 728 | |
---|
| 729 | } |
---|
| 730 | |
---|
[439] | 731 | } |
---|