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4 4 3 3 3 2 3 4 4 4 4 3 4 3 4 4 3 4

E 31 E 32 E 33. E 3. S. 4 4 3 3 3 2 3 4 4 4 4 3 4 3 4 4 3 4. 1 1 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 1.

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4 4 3 3 3 2 3 4 4 4 4 3 4 3 4 4 3 4

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  1. E31 E32 E33 E3 S 4 4 3 3 3 2 3 4 4 4 4 3 4 3 4 4 3 4 1 1 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 1 0 0 1 1 1 1 1 0 0 0 0 1 0 1 0 0 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 1 0 0 1 0 E 8 2 1 7E3pt (purity template) 4 1 0 3 0 0 1 1 0 1 0 0 2 2 1 2 18 0 2 3 2 1 13021 02 C41 1 0 0 1 1 1 S5 0 1 1 0 1 1 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 1 1 1 0 0 1 1 1 0 0 0 0 0 0 0 1 1 0 0 1 1 1 1 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 2 1 4 E31 2 0 1 0 0 1 1 0 1 0 0 0 2 1 1 10 01 3 3 2 1 021 2 5 0 0 3 E32 2 1 0 2 2 0 0 1 8 230 01 13 0 C 4 0 0 3 E33 2 1 0 1 2 0 0 1 7 230 1 13 0 4 3 C4 purity tree 2 1 7 1 E31NZP1 [] 1111 0110 00 10 1 00 0 11 0111 0110 00.00 1100 00.11 00 1 11.11 0 1 S5 NZP1 [] 11 00 0 11 00 1 10 10 0.0 01 0.1 10 C4 NZP1 [] 11 01 0 10 11 1 11 11 1.1 1 1 S5C4NZP1 [] 1111 0000 00 1010 0000 01 1111 0000 10 1000 1000 11 1100 1100 00.00 0011 00.10 1100 01.00 0011 01.01 0 1 01.10 1100 01.11 1 0 11.01 1 1 11.11 0 0 E32 NZP1 [] 1001 0000 00 10 1 00 0 11 1001 1000 00.00 0011 00.11 11 0 11.11 1 0 E33NZP1 [] 1001 0000 00 10 1 00 0 11 1001 1000 00.00 0011 00.11 01 0 11.11 1 0

  2. Query: Find the GPA of married students in classes that require permits. 1. Form the selection vectors, S5 = 0110 1100 and C4 = 1100 111 on the dimensions and “cross” them. (one for each dimension - if there’s no selection on a given dimension, use the pure1 P-tree for it - I.e. select everything) 3. And with E31 , multiply the rc by 4 4. And with E32 , multiply the rc by 2 5. And with E33 , multiply the rc by 1 6. Add values 7. Calculate S5xC4^E3pt 8. Divide sum by rc(S5xC4^E3pt) E31 E32 E33 E3pt 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 1 0 0 1 1 1 1 1 0 0 0 0 1 0 1 0 0 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 1 0 0 1 0 S5 NZP1 [] 11 00 0 11 00 1 10 10 0.0 01 0.1 10 S5C4 NZP1 [] 1111 0000 00 1010 0000 01 1111 0000 10 1000 1000 11 1100 1100 00.00 0011 00.10 1100 01.00 0011 01.01 0 1 01.10 1100 01.11 1 1 0 0 0 0 0 0 0 1 1 0 0 1 1 1 1 1 0 0 1 1 1 0 0 0 0 0 0 0 1 1 0 0 1 1 1 1 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 C4 NZP1 [] 11 01 0 10 11 1 11 11 1.1 1 1 3. 4. 5. 7. E31NZP1 [] 1111 0110 00 10 1 00 0 11 0111 0110 00.00 1100 00.11 00 1 11.11 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 E32 NZP1 [] 1001 0000 00 10 1 00 0 11 1001 1000 00.00 0011 00.11 11 0 11.11 1 0 E33NZP1 [] 1001 0000 00 10 1 00 0 11 1001 1000 00.00 0011 00.11 01 0 11.11 1 0 6. 2*22 + 4*21 + 4*20 = 20 8. 6 8. 20/6 = 3.33

  3. Query: What is the GPA of married grad students taking courses in rooms holding less than 45 students? 1. Form S= S5 ^ SV(S.y > 4) = 0110 1100 ^ 0000 1111 = 0000 1100 and C=SV(C.cp < 45) = 1000 001: 2. Cross the P-tree forms of those vectors (one for each dimension) 3. And with E31, mult rc by 4; with E32, mult rc by 2; with E33 , mult rc by 1 4. Add values 5. Calculate rc(S5xC4^E3pt) 6. Divide E31 E32 E33 E3pt 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 1 0 0 1 1 1 1 1 0 0 0 0 1 0 1 0 0 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6. 2*22 + 0*21 + 0*20 = 8 7. 2 8. 8/2 = 4

  4. Query: Change the grad of the married student in FM (Finite Math) from B to A. (I don’t remember the name or s-# or c-# and therefore hope there is only one such student!!) 1. Form S = S5 = 0110 1100 and C=SV(C.nm = FM) = 0100 000: 2. Cross the P-tree forms of those vectors (one for each dimension) 3. Calculate S5xC4^E3pt 4.Lock the record and change the grade. What is the best structure for CC? If bfr=2 and the file is clustered on s (allowing only 2 students per page therefore using ( DIV2(s), c ) as RIDs) one can use the OR-rollup of the QV to get the pg-RV (in ptree form). E31 E32 E33 E3pt 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 1 0 0 1 1 1 1 1 0 0 0 0 1 0 1 0 0 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 WRITE-ROLL null Rv0 NZP1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 RV (tuple) NZP1 [] 1000 0000 00 1000 0000 00.11 0001 Rvi-2NZP1 POST Rvi-1NZP1 4. It is student with qid=00.00.11 and therefore coordinates, (001, 001) or (1,1). That is enrollment of s=1 and c=1. Use the green Ptree to request access to this enrollment record, By ANDing this ptree into the write-ROLL ptree, if you get a 1, wait, if you get a 0 then OR it into the ROLL and you have the lock. RViNZP1 [] 1000 0000 00 1000 0000 00.11 0001

  5. Use the ROLL as the log?? That is, retain all entries in the ROLL (both read and write) back to the last checkpoint. • Thesis topic: Develop this approach and prove it effective for all types of checkpointing algorithms, all UNDO/REDO scenarios, etc. How much improvement does it offer? • Note that finding the lcv (last committed value) should be easy PG NZP1 [] 10 00 0 10 10 val 5. When access is available, use rolled-up RV (to S only) determine exactly which pages need to be accessed (pg=0) WRITE-ROLL null Rv0 NZP1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 RV (tuple) NZP1 [] 1000 0000 00 1000 0000 00.11 0001 Rvi-2NZP1 POST Rvi-1NZP1 4. It is student with qid=00.00.11 and therefore coordinates, (001, 001) or (1,1). That is enrollment of s=1 and c=1. Use the green Ptree to request access to this enrollment record, By ANDing this ptree into the write-ROLL ptree, if you get a 1, wait, if you get a 0 then OR it into the ROLL and you have the lock. RViNZP1 [] 1000 0000 00 1000 0000 00.11 0001

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