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%u062c%u0645%u064a%u0639 %u0627%u0644%u062d%u0642%u0648%u0642 %u0645%u062d%u0641%u0648%u0638%u0629 %u0640 %u0627%u0625%u0644%u0639%u062a%u062f%u0627%u0621 %u0639%u0649%u0644 %u062d%u0642 %u0627%u0645%u0644%u0624%u0644%u0641 %u0628%u0627%u0644%u0646%u0633%u062e %u0623%u0648 %u0627%u0644%u0637%u0628%u0627%u0639%u0629 %u064a%u0639%u0631%u0636 %u0641%u0627%u0639%u0644%u0647 %u0644%u0644%u0645%u0633%u0627%u0626%u0644%u0629 %u0627%u0644%u0642%u0627%u0646%u0648%u0646%u064a%u062978-The 2-item subsets of {I2, I3, I5} are {I2, I3}, {I2, I5}, and {I3, I5}. {I3, I5} is not a member of L2, and so it is not frequent. Therefore, remove {I2, I3, I5} from C3. -The 2-item subsets of {I2, I4, I5} are {I2, I4}, {I2, I5}, and {I4, I5}. {I4, I5} is not a member of L2, and so it is not frequent. Therefore, remove {I2, I4, I5} from C3. (c) Therefore, C3 = {{I1, I2, I3}, {I1, I2, I5}} after pruning. How to Count Supports of Candidates? counting supports of candidates considered as a problem because of (a) the total number of candidates can be very huge. (b)one transaction may contain many candidates. To count support Candidate itemsets are stored in a hash-tree ,the Leaf node of hash-tree contains a list of itemsets and counts, and the Interior node contains a hash table .we use subset function to find all the candidates contained in a transaction 2- Improving the Efficiency of Apriori: We improve the apriori by : Reduce passes of transaction database scans ,shrink number of candidates ,facilitate support counting of candidates to overcome computational challenges (multiple scans of transaction database,huge number of candidates and tedious workload of support counting for candidates) a- Partition: Scan Database Only Twice: Any itemset that is potentially frequent in DB must be frequent in at least one of the partitions of the database, the first scan is partitionning database and finding local frequent patterns, the second scan is consolidating global frequent patterns. b-DHP(Direct hashing and pruning): Reduce the Number of Candidates: