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8. External Sorting

8. External Sorting. Suppose that a file is so large that the whole file cannot be accommodated in the internal memory of a computer. What shall we do? Need to use EXTERNAL STORAGE DEVICE !!! External Sorting - Disk Sort - Tape Sort

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8. External Sorting

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  1. 8. External Sorting Suppose that a file is so large that the whole file cannot be accommodated in the internal memory of a computer. What shall we do? Need to use EXTERNALSTORAGE DEVICE !!! External Sorting - Disk Sort - Tape Sort What is a major difference between two external sorts?

  2. Sorting with Disk k - way merging “mergesort” merge internal sort . . . . . . . . . . . .

  3. Example 4500 records 250 records/block available memory = 3 blocks Def’n : A segment of a file is said to be a run if all the records in the segment are sorted. 1 2 3 4 5 6 I 1 3 5 D1 …… 2 4 6 D2 ……

  4. 3 D1D2 …… 6 n D3D4 : the size of a run

  5. 1 3 5 7 Run size 2 4 6 8 1 3 5 7 2 4 6 8 3 12 34 56 78 6 1256 3478 12 12345678 24 How many passes? 1 + log2r (r # of initial runs)

  6. k-way merging … … …… … …… logkr ………………………………………………. …… # of passes 1+logkr # of I/O operations? O(nlogkr) better than 2-way merging !!!

  7. How about # of comparisons? Is k-way merging always better than 2-way merging?

  8. Replacement Selection … … …… … …… ………………………………………………. …… # of passes 1+logkr  #(P) #(P)  k r r run size 

  9. # of comparisons(k-way merge) 16 38 30 25 50 16 110 20 15 20 20 25 15 11 120 18 10 9 20 15 8 9 90 17 10 9 20 15 8 9 90 17 8 9 10 11 12 13 14 15 9 15 8 17 4 5 6 7 9 8 2 3 1 8 8 8

  10. How many comparisons in a pass? nlog2k why? Total # of comparisons? (# of passes) (# of comparisons in a pass) = (logkr)(nlog2k) = (nlog2r) independent of k !!! #(c)  r 

  11. How to increase run size(initial run size) x1, x2, x3,…,xm, xm+1, xm+2, xm+3,…,x2m, x2m+1, x2m+2, x2m+3,… m keys m keys m keys r = # of runs =   Any improvement? Observation See p.94 in textbook !!! …...

  12. 4 11 11 2 5 4,2,32,12,18,24,91,11 (record size >> the size of pointer) why do we need this? 91 11 6 24 3 18 7 18

  13. A tree of losers 4 parent 2 loser 32 12 Updating pointers 18 ptr := winner.parent; 24 while ptr  nil do 91 if (ptr.loser.key < winner.key) then 11 interchange(ptr.loser, winner); end {if} ptr := ptr.parent; end {while} 11 91 24 18 winner

  14. Explain p.97-101, textbook !!! Exercise : In a complete 2-tree(T) with n leaf nodes, show that total # of nodes in T = 2n -1

  15. Performance Analysis (Average size of runs) m0  # of records in (real) memory. H. Seward (M.S. Thesis, MIT, 1954) gave a good reason to believe that a run contains more than 1.5m0 records (no proof) E. Friend (JACM, 3, (1966)) experiment  2m0 E. Moore (1961) Proved that 2m0 is the expected run length.

  16. Sketch of Moore’s Proof Snowplow falling snow 2m0 m0 uniform distribution  2m0

  17. Tape Sorting • Balanced k-way merging (similar to disk sorting) • Polyphase merging  • Cascade merging 

  18. Polyphase Merging (Motivation) • (R1, R2, …, R5000) • length (Ri)  20 bytes • Only 1000 records fitted in the internal memory at one time. ( 20k bytes) • 4 tapes available Balanced 2-way merge T1T2T3T4 R1,1000R1001,2000 R2001,3000R3001,4000   R4001,5000   R1,2000R2001,4000 R4001,5000 R1,4000R4001,5000     R1,5000  Total # of operations = 15000

  19. Tape 1 Tape 2 Tape 3 Tape 4 R1,1000R1001,2000R2001,3000  R3001,4000R4001,5000 (rewind) R3001,4000R4001,5000  R1,3000 R1,5000  • Total # of I/O operations 3000 + 5000 = 8000 Balanced Merge is not always best !!!

  20. What if only 3 tapes available? Tape 1 Tape 2 Tape 3 R1,1000 R1001,2000 R2001,3000 R3001,4000  R4001,5000 R1,2000 R2001,4000 R4001,5000 R1,2000 R2001,4000 R4001,5000 R1,4000 R4001,5000   R4001,5000 R1,4000 R1,5000  Total # of I/O Operations 5000 + 2000 + 5000 + 4000 + 5000 = 21,000 !!!

  21. Tape 1 Tape 2 Tape 3 R1,1000 R1001,2000 R2001,3000 R3001,4000  R4001,5000 R1,2000 R4001,5000 R2001,4000 (rewind) R1,2000; 4001,5000 (rewind) R1,5000   Total # of I/O Operations 4000 + 3000 + 5000 = 11,000 !!!

  22. Polyphase merge T1T2T3T4T5T6 131 130 128 124 116  115 114 112 18  516 17 16 14  98 58 13 12  174 94 54 11  332 172 92 52  651 331 171 91 51 1291      How to assign initial runs?

  23. Cascade Merge T1T2T3T4T5T6 155 150 141 129 115  140 135 126 114  515 Pass 1 126 121 112  414 515 114 19  312 414 515 15  29 312 414 515 ( 15 29 312 414 515) 155  24 37 49 510 155 144  33 45 56 Pass 2 155 144 123  42 53 155 144 123 92  51 (155 144 123 92 51 ) 154 143 122 91 551 153 142 121  501 551 Pass 3 152 141  411 501 551 151  291 411 501 551 ( 151 291 411 501 551) Pass 4 1901     

  24. Polyphase Merge T1T2T3T4T5T6 phase 1 131 130 128 124 116  2 115 114 112 18  516 3 17 16 14  98 58 4 13 12  174 94 54 Gilstad(1960) 5 11  332 172 92 52 6 651 331 171 91 51 71291      {{1,0,0,0,0},{1,1,1,1,1},{2,2,2,2,1},{4,4,4,3,2},{8,8,7,6,4}, {16,15,14,12,8},{31,30,28,24,16}} Perfect Fibonacci Distribution !!! What is the underlying rule?

  25. iaibicidiei 0 1 0 0 0 0 1 1 1 1 1 1 2 2 2 2 2 1 3 4 4 4 3 2 4 8 8 7 6 4 5 16 15 14 12 8 6 31 30 28 24 16

  26. (a0 + b0) (a0 + c0) (a0 + d0) (a0 + e0) a0 (a1 + b1) (a1 + c1) (a1 + d1) (a1 + e1) a1 (a2 + b2) (a2 + c2) (a2 + d2) (a2 + e2) a2 n an bn cn dn en n+1 an + bn an + cn an + dn an + en an an  bn  cn dn en

  27. iaibicidiei output 0 1 0 0 0 0 T6 1 1 1 1 1 1 T1 2 2 2 2 2 1 T2 3 4 4 4 3 2 T3 2 2 2 1 0 2 1 1 1 0 1 1 4 8 8 7 6 4 T4 5 16 15 14 12 8 T5 6 31 30 28 24 16 T6 7 61 59 55 47 31 T1T2T3T4T5

  28. n-1 an-1 bn-1 cn-1 dn-1 en-1 n an-1+bn-1 an-1+cn-1 an-1+dn-1 an-1+en-1 an-1 an bn cn dn en  en = an-1 dn = an-1 + en = an-1 + an-2 cn = an-1 + dn-1 = an-1 + (an-2 + en-2) = an-1 + an-2 + an-3 …………. en = an-1 dn = an-1 + an-2 cn = an-1 + an-2 + an-3 bn = an-1 + an-2 + an-3 + an-4 an = an-1 + an-2 + an-3 + an-4 + an-5 (a0 = 1, ai = 0, i = -1, -2, -3, -4)

  29. e = an-1 d = an-1 + an-2 c = an-1 + an-2 + an-3 b = an-1 + an-2 + an-3 + an-4 a = an-1 + an-2 + an-3 + an-4 + an-4

  30. i -4 -3 -2 -1 0 1 2 3 4 5 6 7 ai 0 0 0 0 1 1 2 4 8 16 31 61 1 bi 0 ci 0 di 0 ei 0

  31. 1 2 4 8 16 31 61 1 2 4 8 15 30 59 1 2 4 7 14 28 55 1 2 3 6 12 24 47 1 1 2 4 8 16 31

  32. ai = < 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 31, 61, …… >, i = -4, -3, -2, -1, 0, 1, 2,... “The kth order Fibonacci number” Fnk = Fn-1k + Fn-2k + …… + Fn-kk 0, 0  nk-2 Fnk= 1, n = k-1 e.g) The second order Fibonacci number 0 1 1 2 3 5 …… Fn2 = Fn-12 + Fn-22 0, if n = 0 Fn2 = 1, if n = 1 Fibonacci number !!! an = Fn+k-1k if k tapes(input) are used why?

  33. What if not perfect Fib. Dist’n? Use dummy runs !!! 5 input tapes and 53 initial runs. Level T1 T2 T3 T4 T5 1 1 1 1 1 1 5 2 2 2 2 2 1 9 1 1 1 1 0 3 4 4 4 3 2 17 2 2 2 1 1 4 8 8 7 6 4 33 4 4 3 3 2 5 16 15 14 12 8 65>53 (8 7 7 6 4) ……………………………… T1 T2 T3 T4 T5 (34) (35) (36) (37) (38) (39) (40) (41) (42) (43) (44) (45) (46) (47) (48) (49) (50) (51) (52) (53)            

  34. T1 T2 T3 T4 T5 T6  (2) (2) (2) (3) (3) 18 17 16 14 58 (2) (2) (2) (3) 55 53 not best but simple and good !!! For better one, see Knuth !!!

  35. Example (3 tapes) T1 T2 T3 (k)8 (k)5  (k)3 (2k)5  (3k)3 (2k)2 0, 1, 1, 2, 3, 5, 8 (5k)2 (3k)1  (5k)1 (8k)1  (13k)1 Runs on two input tapes (k) # of runs run size(k) # of pairs # of I/O’s 8,5 1,1 5 10 5,3 2,1 3 9 3,2 3,2 2 10 2,1 5,3 1 8 1,1 8,5 1 13 1 13 How many passes over the data?

  36. Total number  Fs for some s. of initial runs the sth Fibonacci number Fs Fs-1 Fs-2 T1 T2 T3 Fs-1 Fs-2  Fs-3 Fs-2 Fs-3 Fs-4 ………… See Fig. p.107, textbook !!! Total # of I/O operations =  # of passes =

  37. Lemma : [proof] (By induction on S) (s=2) LHS = RHS = (s=3) LHS = RHS = (s=k) Suppose that (s=k+1) Exercise !!! See page 106-107 in textbook !!!

  38. From the previous lemma, # of passes = Fs = r (1) why? . Golden Ratio !!! From (1) ,

  39. Theorem: Fs-1 Fs-2 Polyphase merge merge 3 tapes Fs = r = # of initial runs # of passes = 1.04 log2r

  40. APPROXIMATED BEHAVIOR OF POLYPHASE MERGE SORTING Tapes Phases Passes Pass/phase Growth percent ratio 3 2.078 lnS + 0.672 1.504 lnS + 0.992 72 1.6180340 4 1.641 lnS + 0.364 1.015 lnS + 0.965 62 1.8392868 5 1.524 lnS + 0.078 0.863 lnS + 0.921 57 1.9275620 6 1.479 lnS + 0.185 0.795 lnS + 0.864 54 1.9659482 7 1.460 lnS + 0.424 0.762 lnS + 0.797 52 1.9835828 8 1.451 lnS + 0.642 0.744 lnS + 0.723 51 1.9919642 9 1.447 lnS + 0.838 0.734 lnS + 0.646 51 1.9960312 10 1.445 lnS + 1.017 0.728 lnS + 0.568 50 1.9980295 20 1.443 lnS + 2.170 0.721 lnS– 0.030 50 1.9999981 APPROXIMATED BEHAVIOR OF CASCADE MERGE SORTING Tapes Phases Passes Growth ratio 3 2.078 lnS + 0.672 1.504 lnS + 0.992 1.6180840 4 1.235 lnS + 0.754 1.012 lnS + 0.820 2.2469796 5 0.946 lnS + 0.796 0.897 lnS + 0.800 2.8793852 6 0.796 lnS + 0.821 0.773 lnS + 0.808 3.5133371 7 0.703 lnS + 0.839 0.691 lnS + 0.822 4.1481149 8 0.639 lnS + 0.852 0.632 lnS + 0.834 4.7833861 9 0.592 lnS + 0.861 0.587 lnS + 0.845 5.4189757 10 0.555 lnS + 0.869 0.552 lnS + 0.854 6.0547828 20 0.397 lnS + 0.905 0.397 lnS + 0.901 12.4174426

  41. Cascade Merge Level aibicidiei 0 1 0 0 0 0 1 1 1 1 1 1 2 5 4 3 2 1 3 15 14 12 9 5 4 55 50 41 29 15 nanbncndnen n+1 an+bn+cnan+1bn+1cn+1dn+1 +dn+en -en -dn -cn -bn an+1an Perfect dist’n for detail see Knuth Vol III !!!

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