1 / 31

Improve Run Generation

Strategies to Improve Disk Memory Efficiency, Reduce Runs, and Increase CPU Utilization with Quick Sort and Internal Sorting Schemes. Includes a new buffer strategy for steady-state operations.

djodi
Télécharger la présentation

Improve Run Generation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. DISK MEMORY DISK Improve Run Generation • Overlap input,output, and internal CPU work. • Reduce the number of runs (equivalently, increase average run length).

  2. 6 2 8 5 11 10 4 1 9 7 3 4 2 3 5 1 6 10 11 9 7 8 Internal Quick Sort Use 6 as the pivot (median of 3). Input first, middle, and last blocks first. In-place partitioning. Input blocks from the ends toward the middle. Sort left and right groups recursively. Can begin output as soon as left most block is ready.

  3. B1 B2 B3 Alternative Internal Sort Scheme Partition into 3 buffers. DISK DISK

  4. Read from disk Write to disk Run generation Steady State Operation • Synchronization is done when the active input buffer gets empty (the active output buffer will be full at this time).

  5. Use 2 input and 2 output buffers. Rest of memory is used for a min loser tree. Loser Tree Input 0 Output 0 Output 1 Input 1 DISK MEMORY DISK New Strategy

  6. Read from disk Write to disk Run generation Steady State Operation • Synchronization is done when the active input buffer gets empty (the active output buffer will be full at this time).

  7. O0 O1 I1 I0 Initialize 3 4 8 4 3 6 8 1 5 7 3 2 6 9 4 5 2 5 8 Fill From Disk

  8. O0 O1 I1 I0 Initialize 3 6 1 4 8 5 7 4 3 6 8 1 5 7 3 2 6 9 4 5 2 5 8 Fill From Disk

  9. O0 O1 I1 I0 Initialize 1 3 6 3 2 4 8 5 7 6 9 4 3 6 8 1 5 7 3 2 6 9 4 5 2 5 8 Fill From Disk

  10. O0 O1 I1 I0 Initialize 1 3 2 6 3 4 2 4 8 5 7 6 9 5 8 4 3 6 8 1 5 7 3 2 6 9 4 5 2 5 8 Fill From Disk

  11. O0 O1 I1 I0 Initialize 1 3 2 6 3 4 5 4 8 5 7 6 9 5 8 4 3 6 8 1 5 7 3 2 6 9 4 5 2 5 8 Fill From Disk

  12. O0 O1 I1 I0 Initialize 2 3 2 6 3 4 5 4 8 5 7 6 9 5 8 4 3 6 8 1 5 7 3 2 6 9 4 5 2 5 8 Fill From Disk

  13. 2 O0 O1 3 2 6 3 4 5 4 8 5 7 6 9 5 8 4 3 6 8 1 5 7 3 2 6 9 4 5 2 5 8 I1 I0 Fill From Tree Generate Run 1 3 5 4 Fill From Disk

  14. 3 O0 O1 I1 I0 Fill From Tree 1 Generate Run 1 2 3 2 6 3 4 5 4 8 5 7 6 9 5 8 3 4 3 6 8 5 7 3 2 6 9 4 5 2 5 8 3 5 4 Fill From Disk

  15. 4 5 O0 O1 I1 I0 Fill From Tree 1 2 Generate Run 1 3 3 2 6 3 4 5 4 8 5 7 6 9 5 8 3 5 4 3 6 8 5 7 3 6 9 4 5 2 5 8 3 5 4 Fill From Disk

  16. 5 4 O0 O1 I1 I0 Fill From Tree 1 2 Generate Run 1 3 2 3 2 6 3 4 5 4 8 5 7 6 9 5 8 3 5 4 4 3 6 8 5 7 3 6 9 4 5 5 8 3 5 4 Interchange Role Of Buffers Fill From Disk

  17. 4 5 O0 O1 I1 I0 Interchange Role Of Buffers Write To Disk Fill From Tree 1 2 3 2 3 2 6 3 4 5 4 8 5 7 6 9 5 8 3 5 4 4 3 6 8 5 7 3 6 9 4 5 5 8 1 9 2 Fill From Disk

  18. 4 5 O0 O1 I1 I0 Write To Disk Fill From Tree 1 2 Continue With Run 1 3 2 3 4 6 3 4 5 4 8 5 7 6 9 5 8 3 5 4 4 3 6 8 5 7 3 6 9 4 5 5 8 1 9 2 Fill From Disk

  19. 5 1 5 O0 O1 I1 I0 Write To Disk Fill From Tree 1 3 2 Continue With Run 1 4 2 3 4 6 3 4 5 4 8 5 7 6 9 5 8 5 4 4 3 6 8 5 7 3 6 9 4 5 5 8 1 1 9 2 Fill From Disk

  20. 5 1 5 O0 O1 7 9 5 I1 I0 Write To Disk Fill From Tree 1 3 2 3 Continue With Run 1 4 2 3 4 6 3 4 5 4 8 5 7 6 9 5 8 9 5 4 4 3 6 8 5 7 6 9 4 5 5 8 1 1 9 2 Fill From Disk

  21. 5 5 1 O0 O1 7 5 9 I1 I0 Write To Disk Fill From Tree 1 3 2 3 Continue With Run 1 4 2 3 3 4 6 3 4 5 4 8 5 7 6 9 5 8 2 9 5 4 4 6 8 5 7 6 9 4 5 5 8 1 1 9 2 Interchange Role Of Buffers Fill From Disk

  22. 5 1 5 O0 O1 7 5 9 I1 I0 Fill From Tree Interchange Role Of Buffers Write To Disk 3 3 4 3 3 4 6 3 4 5 4 8 5 7 6 9 5 8 2 9 5 4 4 6 8 5 7 6 9 4 5 5 8 1 6 1 3 Fill From Disk

  23. 5 2 1 5 O0 O1 7 5 9 I1 I0 Fill From Tree Write To Disk 3 3 Continue With Run 1 4 3 3 4 6 3 4 5 4 8 5 7 6 9 5 8 2 9 5 4 4 6 8 5 7 6 9 4 5 5 8 1 6 1 3 Fill From Disk

  24. 5 2 5 1 O0 O1 6 9 5 5 7 I1 I0 Fill From Tree Write To Disk 4 3 3 Continue With Run 1 4 3 3 4 6 3 4 5 4 8 5 7 6 9 5 8 6 2 9 5 4 6 8 5 7 6 9 4 5 5 8 1 6 1 3 Fill From Disk

  25. 5 1 2 1 5 9 O0 O1 4 5 9 7 5 5 6 I1 I0 Fill From Tree Write To Disk 4 3 4 3 Continue With Run 1 4 3 3 6 3 4 5 4 8 5 7 6 9 5 8 6 2 9 5 1 4 6 8 5 7 6 9 5 5 8 1 6 1 3 Fill From Disk

  26. Buffer Reduction • We may reduce the number of buffers to 3. • At any time, 1 us used to read into, 1 to write from, and the 3rd both feeds the loser tree and is filled from the tree. • The 3 physical buffers rotate through these three roles.

  27. RUN SIZE • Let k be number of external nodes in loser tree. • Run size >= k. • Sorted input => 1 run. • Reverse of sorted input => n/k runs. • Average run size is ~2k.

  28. Comparison • Memory capacity = m records. • Run size using fill memory, sort, and output run scheme = m. • Use loser tree scheme. • Assume block size is b records. • Need memory for 4 buffers (4b records). • Loser tree k = m – 4b. • Average run size = 2k = 2(m – 4b). • 2k >= m when m >= 8b.

  29. m 600 1000 5000 10000 k 200 600 4600 9600 2k 400 1200 9200 19200 Comparison • Assume b = 100.

  30. Comparison • Total internal processing time using fill memory, sort, and output run scheme = O((n/m) m log m) = O(n log m). • Total internal processing time using loser tree = O(n log k). • Loser tree scheme generates runs that differ in their lengths.

  31. 9 6 4 3 6 9 4 3 Merging Runs Of Different Length 22 22 13 7 15 7 Cost = 42 Cost = 44 Best merge order?

More Related