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Algorithm for the Aggregate Function SUM

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Algorithm for the Aggregate Function SUM

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  1. Algorithm for the Aggregate Function SUM Algorithm 4.1 Evaluating sum () with P-tree. total = 0.00; For i = 0 to n {total = total + 2i * RootCount (Pi);} Return total SUM function: Sum function can total a field of numerical values. P4,3P4,2P4,1 P4,0 0 1 1 1 0 0 0 0 1 0 1 1 1 1 10 5 6 7 11 9 3 1 0 0 0 1 1 0 1 0 1 1 1 0 1 • For example, if we want to know the total number of products which were sold out in relation S, the procedure is showed on left {3} {3} {5} {5} 51 23 * + 22 * + 21 * + 20 * =

  2. Algorithm of Aggregates AVERAGE, MAX Algorithm 4.2 Evaluating max () with P-tree. max = 0.00; c = 0; /*Pc is set all 1s*/ For i=n to 0 { c=Count (Pc AND Pi); If (c >= 1) Pc = Pc AND Pi; max = max + 2i } Return max; Average function: Average function will show the average value in a field. It can be calculated from function COUNT and SUM. Average () = Sum ()/Count (). Steps IFPos Bits P4,3P4,2P4,1 P4,0 1. Pc = P4,3 RootCount (Pc) = 3 >=1 0 1 1 1 0 0 0 0 1 0 1 1 1 1 10 5 6 7 11 9 3 1 0 0 0 1 1 0 1 0 1 1 1 0 1 {1} {0} 2. RootCount (Pc ANDP4,2) = 0 <1 Pc = Pc AND P’4,2 3. RootCount (Pc ANDP4,1 ) = 2 >=1 Pc = Pc AND P4,1 {1} 4. RootCount (Pc ANDP4,0 ) = 1 >=1 {1} {1} {1} {1} 11 23 * + 22 * + 21 * + 20 * = {0}

  3. Algorithm of Aggregate Function MIN Algorithm 4.3. Evaluating Min () with P-tree. min=0.00; c = 0; /*Pc is set all 1s*/ For i=n to 0 { c=RootCount (Pc AND NOT (Pi)); If (c >= 1) Pc=Pc AND NOT (Pi); Else min = min + 2i; } Return min; Steps IFPos Bits P4,3P4,2P4,1 P4,0 1. Pc = P’4,3 RootCount (Pc) = 4 > =1 0 1 1 1 0 0 0 0 1 0 1 1 1 1 10 5 6 7 11 9 3 1 0 0 0 1 1 0 1 0 1 1 1 0 1 {0} 2. RootCount (Pc ANDP’4,2) = 1 >=1 Pc = Pc AND P’4,2 {0} 3. RootCount (Pc ANDP’4,1 ) = 0 <1 Pc = Pc AND P4,1 {1} 4. RootCount (Pc ANDP’4,0 ) = 0 <1 {1} {0} {0} {1} {1} 23 * + 22 * + 21 * + 20 * = 3

  4. Algorithms of Aggregate Function MEDIAN, RANK, TOP-K Algorithm 4.4. Evaluating Median () with P-tree median=0.00; pos=N/2; for rank pos=K; c = 0; /*Pc is set all 1s for single attribute*/ For i=n to 0 { c=RootCount (Pc&Pi); If (c >= pos) median=median+2i; Pc=Pc&Pi; Else pos=pos-c; Pc=Pc &NOT(Pi);} Return median; Rank (K) function returns the value that is the kth largest value in a field. Steps IFPos Bits P4,3P4,2P4,1 P4,0 1. Pc = P4,3 RootCount (Pc) = 3 < 4 Pc = P’4,3 pos = 4 – 3 = 1 0 1 1 1 0 0 0 0 1 0 1 1 1 1 10 5 6 7 11 9 3 1 0 0 0 1 1 0 1 0 1 1 1 0 1 {0} 2. RootCount (Pc ANDP4,2) = 3 >=1 Pc = Pc AND P4,2 {1} 3. RootCount (Pc ANDP4,1 ) = 2 >=1 Pc = Pc AND P4,1 {1} 4. RootCount (Pc ANDP4,0 ) = 1 >=1 {1} 7 23 * + 22 * + 21 * + 20 * = {0} {1} {1} {1} Top-k: (largest k values): Find rank k value Vk. Find all tuples  Vk using EINRING.

  5. LOC 0 0 0 0 1 P1,4 P1,3 P1,2 P1.1 P1.0P’1,4 P’1,3 P’1,2 P’1.1 P1.0 PNY PNY 1 0 1 1 0 0 0 PMN 0 1 0 0 1 0 1 PCH 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1 1 0 1 1 0 0 0 0 0 0 0 0 1 0 1 0 1 1 0 0 0 Iceberg Query Operation Using P-trees SELECT Loc, Type, Sum (# Product) FROM Relation S GROUPBY Loc, Type HAVING Sum (# Product) >= 15 Step one: Build value P-trees for the 4 values, {Loc| New York, Minneapolis, Chicago}, of attribute Loc. Calculation of value P-tree PNY. Because binary value of New York is 00001, we will get (1). PNY = P’1,4 AND P’1,3 AND P’1,2 AND P’1,1 AND P1,0 (1)

  6. PNotebook PDesktop PPrinter PFAX 1 0 0 1 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 After getting all the value P-trees for each location, we calculate the total number of products sold in each place. We still use the value, New York, as our example. Sum(# product | New York) = 23 * RootCount (P4,3 AND PNY) + 22 * RootCount (P4,2 AND PNY) + 21 * RootCount (P4,1 AND PNY) + 20 * RootCount (P4,0 AND PNY) = 8 * 1 + 4 * 2 + 2 * 3 + 1 * 1 = 23 (2) Table shows total number of products sold out in each location. Because threshold is 15, eliminate Chicago. Step 2: Similarly, build P-trees for every value of attribute Type. Attribute Type has 4 values {Type | Notebook, desktop, Printer, Fax}. Figure shows P-tree of 4 values of attrib Type. Similarly we get the summary table for each value of attribute Type. According to the threshold, T equals 15, only value P-tree of notebook will be used in the future.

  7. Calculate total notebooks sold in New York by Sum(# Product | New York) = 23 * RootCount (P4,3 ANDPNY AND Notebook) + 22 * RootCount (P4,2 AND PNY AND Notebook) + 21 * RootCount (P4,1 AND PNY AND Notebook) + 20 * RootCount (P4,0 ANDPNY AND Notebook) = 8 * 1 + 4 * 1 + 2 * 2 + 1* 1 = 17 (3) PMN PNotebook PMN AND Notebook 0 0 0 0 1 0 0 0 1 0 0 1 0 1 1 0 0 1 1 0 0 AND = Iceberg Step three: We only generate candidate Loc and Type pairs for local store and Product type, which can pass T. By Performing And op on PNY with PNotebook, obtain P-treePNYAND Notebook By performing And operations on PMN with P Notebook, obtain value P-tree PMNAND Notebook PNY PNotebookPNY AND Notebook 1 0 1 1 0 0 0 1 0 0 1 1 0 0 1 0 0 1 0 0 0 = AND

  8. We calculate the total number of notebook sold out in Minneapolis by formula 4. Sum (# product | Minneapolis) = 23 * RootCount (P4,3 AND PMNAND Notbook) + 22 * RootCount (P4,2 AND PMN ANDNotbook) + 21 * RootCount (P4,1 AND PMNAND Notbook) + 20 * RootCount (P4,0 AND PMNAND Notbook) = 8 * 1 + 4 * 0 + 2 * 1 + 1 * 1 = 11 (4) Finally, we obtain the summary table 5. According to the threshold T=15, we can see that only group pair “New York And Notebook” pass our threshold T. From value P-tree PNYAND Notebook, we can see that tuple 1 and 4 are in the results of our iceberg query example. PNYAND Notebook 1 0 0 1 0 0 0