Lecture 5: Fusion Trees (ctd.) Johannes Fischer

Transcription

1 Lecture 5: Fusion Trees (ctd.) Johannes Fischer

2 Fusion Trees fusion tree pred/succ O(lg n / lg w) w.c. construction O(n) w.c. + SORT(n,w) space O(n) w.c. 2 2

3 Idea B-tree with branching factor b=w /5 b height lg w n search for q by topdown traversal b... identifies pred/succ b... x x2 xb q xi xi+ 3 3

4 Sketches important bits b0<...<br- for x0<...<xb-: levels with branching nodes in trie SKETCH(xi) = xi restricted to important bits SKETCH preserves order: SKETCH(xi) < SKETCH(xj) xi < xj problem: does not hold for query q: SKETCH(xi) < SKETCH(q) SKETCH(xi+) xi < q xi+ 4 4

5 Example b = 3 0 b0 = SKETCH( )

6 Right Insertion Point SKETCH(xi) SKETCH(q) SKETCH(xi+) y max{lcp(q,xi), LCP(q,xi+)} point where q deviates from x i/xi+ most significant bit of q XOR xi/xi+ subtree of y must contain pred or succ repeat computation in that subtree 6 6

7 Right Insertion Point y y xi/xi+ q q xi/xi+ e=y e=y

8 Summary at every node of the B-tree with keys x0<...<xb- and important bits b0<...<br- TODO compute SKETCH(q) in O() time find i such that SKETCH(xi) SKETCH(q) SKETCH(xi+) y max{msb(q XOR x i),msb(q XOR xi+)} e q w-...qy-0... or qw-...qy compute SKETCH(e) in O() time find i' such that SKETCH(xi') SKETCH(e) SKETCH(xi'+) then x i' or xi'+ is predecessor or successor of q 8 TODO 2 TODO 3 8

9 TODO : MSB in O() time possible, but complicated if not supported by CPU: D.E. Knuth: The Art of Computer Programming, Vol. 4A. Addison-Wesley, 20. note the use of 5 multiplications 9 9

10 MSB in O(lg w) time idea: binary search over x with ANDs determine if search interval empty x= mask= x AND mask= continue in left half 0 0

11 MSB in O(lg w) time idea: binary search over x with ANDs determine if search interval empty x= mask= x AND mask= =0 continue in right half

12 MSB in O(lg w) time idea: binary search over x with ANDs determine if search interval empty x= mask= x AND mask= 0 0 continue in left half etc. 2 2

13 MSB in O(lg w) time neat variation of this idea: function MSB(x): λ 0 for k=lg(w)- downto 0 z x 2 k if (z 0) λ λ + 2 k x z return λ possible: table lookup for small blocks 3 3

14 Implementation unsigned int x; // 32-bit value to find the log2 of const unsigned int b[] = {0x2, 0xC, 0xF0, 0xFF00, 0xFFFF0000}; const unsigned int S[] = {, 2, 4, 8, 6}; int i; register unsigned int l = 0; // result will go here for (i = 4; i >= 0; i--) // unroll for speed... { if (x & b[i]) { x >>= S[i]; l = S[i]; } } see 4 4

15 Implementation with Table Lookup unsigned int l = 0; // will be lg(x) register unsigned int t, tt; // temporaries if (tt = x >> 6)! l = (t = x >> 24)? 24 + LogTable256[t] : 6 + LogTable256[tt & 0xFF]; else! l = (t = x >> 8)? 8 + LogTable256[t] : LogTable256[x]; return l; const char LogTable256[256] =! {!! 0,0,,,2,2,2,2,3,3,3,3,3,3,3,3,!! 4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,!! 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,!! 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,!! 6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,!! 6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,!! 6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,!! 6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,!! 7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,!! 7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,!! 7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,!! 7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,!! 7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,!! 7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,!! 7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,!! 7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7 5! }; 5

16 TODO 2: Computing SKETCH(q) q= br-... b3 b2 b b difficult! SKETCH(q)= APPSKETCH(q)= O(w 4/5 ) 6

17 APPSKETCH(q) q= br-... b3 b2 b b O(w 4/5 ) APPSKETCH'(q)= br-+mr-... b+m b0+m0 right shift by b0+m0 APPSKETCH(q)=

18 APPSKETCH'(q) q= br-... b3 b2 b b = B= = q AND B = = M= = *M= = = + APPSKETCH'(q)= br-+mr b+m b0+m0 = + 8

19 Requirements on mi's + should contain bits bi of q at, positions bi + mj no overflow a) bi + mj = bk + ml i=k and j=l b) b0 + m0 < b + m <... < br- + mr- c) (br- + mr-) - (b0 + m0) = O(w 4/5 ) 9 9

20 Existence of mi's strong (a): distinctness modulo r 3 : bi + m'j bk + m'l (mod r 3 ) i=k and j=l proof by induction: - for m' j avoid m'l + (bk-bi) (mod r 3 ) - for all 0 l < j, 0 k,i < r jr 2 < r 3 - r3 possible choices m'j exists can make bi + mi lie in different blocks of length r 3 : mj = m'j + ( (w-bj+jr 3 )/r 3 r 3 ) 20 20

21 Existence of mi's Requirements satisfied: a) bi + mj distinct modulo r 3 distinct overall b) bi + mi lie in consecutive r-blocks b0 + m0 < b + m <... < br- + mr- c) b0 + m0 = w bi + mi bi- + mi- + r 3 r<w (br- + mr-) - (b0 + m0) w+r 4 = O(w 4/5 ) 2 2

22 TODO 3: Finding q's Insertion Point B-tree node v with x0<...<xb-: find i s.th. APPSKETCH(xi) APPSKETCH(q) APPSKETCH(xi+) b=o(w /5 ) sketches of size O(w 4/5 ) b sketches fused=packed in O() words separate by -bits: APPSKETCH(v)=... APPSKETCH(x0) APPSKETCH(x) APPSKETCH(xb-) 22 22

23 APPSKETCH(q) = Parallel Comparison APPSKETCH(q) iff xi q C = APPSKETCH(v)=... APPSKETCH(x0) APPSKETCH(x) APPSKETCH(xb-)... APPSKETCH(q)*C= 0 APPSKETCH(q) 0 APPSKETCH(q) 0 0 APPSKETCH(q) = D 0...?...??...??...? APPSKETCH(v)-D= = Δ E= Δ AND E = xi's ordered 0 pattern will be find MSB 23 23

24 Summary multiple keys per B-tree node in O() time broadword computing: subtraction: parallel comparison AND: masking multiplication: - create copies - bit permutation O(lg n / lg w): beat information theoretic barrier of O(n lg n) 24 24