460 likes | 514 Views
Bitonic Sorting and Its Circuit Design. Kenneth E. Batcher Professor, Kent State University. http://www.cs.kent.edu/~batcher. “Sorting networks and their applications”, AFIPS Proc. of 1968 Spring Joint Computer Conference, Vol. 32, pp 307-314. Background. Sorting is fundamental
E N D
Kenneth E. Batcher Professor, Kent State University http://www.cs.kent.edu/~batcher “Sorting networks and their applications”, AFIPS Proc. of 1968 Spring Joint Computer Conference, Vol. 32, pp 307-314.
Background • Sorting is fundamental • Low bound of any sequential sorting algorithms is O(nlogn) • Can we improve the time complexity further? • Parallel algorithms • Circuit/Network Design • Parallel Computing Models
①Bitonic Sequence 双调序列 • sequence of elements {a0, a1, …, an-1} where either • (1) there exists an index, i, 0 i n-1, such that {a0, …, ai} is monotonically increasing, and {ai+1, …, an-1} is monotonically decreasing, • e.g. {1, 2, 4, 7, 6, 0} Or • (2) there exists a cyclic shift of indices so that (1) is satisfied • e.g. {8, 9, 2, 1, 0, 4} {0, 4, 8, 9, 2, 1}
Value of element { 3, 5, 7, 9, 8, 6, 4, 2 } ai a0 a1 a2 a3 a4 a5 a6 a7 Value of element { 8, 6, 4, 2, 3, 5, 7, 9} ai a0 a1 a2 a3 a4 a5 a6 a7 ①Bitonic Sequence : Examples
Value of element { 3, 5, 7, 9, 11, 13, 15, 17 } ai a0 a1 a2 a3 a4 a5 a6 a7 Value of element { 5, 3, 1, 2, 4, 6, 8, 7 } ai a0 a1 a2 a3 a4 a5 a6 a7 ①Bitonic Sequence : Examples
Value of element ai a0 a1... an/2-1 an/2 an/2 +1 … an-1 Bitonic Sort: basic idea • Consider a bitonic sequence S of size n where • the first half ( {a0, a1, …, an/2-1} ) is increasing, and the second half ( {an/2, an/2+1, …, an-1} ) is decreasing
ai a0 a1... an/2-1 an/2 an/2 +1 … an-1 Compare and exchange an/2 an/2-1 value value a0 an-1 ②“Bitonic Split” 双调分裂 • Pair-wise min-max comparison • s1 = {min(a0, an/2), min(a1, an/2+1), … , min(an/2-1, an-1)} • s2 = {max(a0, an/2), max(a1, an/2+1), … , max(an/2-1, an-1)} S2 S1
value • There exists • an element b in S1such that all elements before b is increasing and all elements after b is decreasing • an element c in S2 such that all elements before c is decreasing and all elements after c is increasing • S1 and S2 • Both S1 and S2 are bitonic sequences • Any elements in S1 < any elements in S2 (because b < c and b is the maximum value in S1 and c is the minimum value in S2) S2 c b S1
pair-wise min-max comparison e.g. { 2, 4, 6, 8, 7, 5, 3, 1} { 2, 4, 6, 8 7, 5, 3, 1 } => S1={2, 4, 3, 1} S2={7, 5, 6, 8} bitonic sequence of size 8 => 2 bitonic sequence of size 4 Compare and exchange
②Bitonic Split • The split is applicable to any bitonic sequence. • Need not to have the 1st half to be increasing/decreasing and the 2nd half to be decreasing/increasing: Bitonic Split 2 Bitonic(n/2) Bitonic(n)
Sorting a bitonic sequence • By using bitonic split recursively, INPUT:a bitonic sequence of size n • Phase 1: 2 bitonic sequence of size n/2 • Phase 2: 4 bitonic sequence of size n/4 • … • … • Phase (log n): n bitonic sequence of size 1 • a sorted sequence can be generated by concatenating the n bitonic sequence of size 1
③Bitonic Merge 双调合并 • sort a bitonic sequence using bitonic splits 1 2 3 4 5 6 7 8 9 10111213141516 length 16 8 4 2 Anything wrong with this slide?
Bitonic Merge Circuit : BM[16] What do you think of ?
Questions ? • How can we convert an unsorted sequence to a bitonic sequence ? (then, by using bitonic split recursively, a sorted sequence can be formed).
Turn an unsorted sequence into a bitonic sequence: ③Bitonic Merge (BM) Operation 1 2 3 4 5 6 7 8 9 10111213141516 length 4 8 16 At every phase, sort a bitonic sequence of size 2, 4, 8, 16 into a monotonically increasing or decreased sequence
④Bitonic Sort 1 2 3 4 5 6 7 8 9 10111213141516 length 4 8 16
Sort (any ordered of) sequence • Using bitonic merge repeatedly • Definition: • BM[n]: increasing bitonic merge of size n • bitonic merge : sort a bitonic sequence of size n into a monotonically increasing sequence • BM[n]: decreasing bitonic merge of size n • bitonic merge that sort a bitonic sequence of size n into a monotonically decreasing sequence
Steps: • Divide the sequence into a group of 2 • any sequence of size 2 is a bitonic sequence: either the increasing part is of size 2 and the decreasing part is of size 0, or vice versa • Using BM[2] on a group to form an increasing sequence, and BM[2] on the adjacent group to form an decreasing sequence • Concatenate the two group to form a bitonic sequence of size 4
Steps: • Repeat the above steps on other groups • Repeat the above steps recursively, until a bitonic sequence of size n is formed • Using bitonic merge again to turn the bitonic sequence into a sorted sequence
Bitonic Sorting Circuit: BS(18) • BM[n]: increasing bitonic merge of size n • BM[n]: decreasing bitonic merge of size n
Sort (any ordered of) sequence • Hence, • n unsorted numbers • n/2 group of 2-number bitonic sequence • n/4 group of 4-number bitonic sequence • … • 1 group of n-number bitonic sequence • a sorted sequence
⑤Complexity of Bitonic Sort • Parallel bitonic sort with n processor • The last stage of an n-element bitonic sorting need to merge n-element, and has a depth of log(n) • Other stages perform a complete sort of n/2 elements • Depth, d(n) = d(n/2) + log(n) • d(n) = 1 + 2 + 4 + … + log(n) = (log2n) • Complexity: T(n) = (log2n)
⑤Complexity of Bitonic Sort • Parallel sorting with a block of elements per processor • sort the local block of elements first (using any sorting algorithm such as quicksort, bitonic sort) • sort the elements among processors using parallel bitonic sort • T(n) = T(local_sort) + T(comparisons) +T(communication) • Only computation time is considered here (you need to consider all communication time also)
⑥Concluding Remarks • Bitonic Sorting: Common Sense • Regression to Computer Science • One of 10 Most Important Papers • Parallel Algorithm: Ascend/Descend • Another example: Prefix sum • Network Model:
Bitonic Sorting Network Hypercube connections! Try to Write Bitonic Sorting algorithm on hypercube.
PRAM Model … P1 P2 P3 Pn Memory Access time from any processor to any memory unit is equal It is impossible in practice So it is an ideal model for parallel computing
PRAM Model Program for Sum= a(1)+a(2)+…+a(N) for i = 1 to log N for j= 1 to n/ 2i parallel doa(j) = a(j)+ a(N/ 2i + j) endpar endfor endfor Finally a(1) is the sum
Hypercube Model Suppose node N(i) holds element a(i), where i is the value of node index x1x2…xn for i = 1 to n for j= i to n parallel do N(00…0 (xj=0) xj+1…xn) N(00…0 (xj=1) xj+1…xn); a(00…0 (xj=0) xj+1…xn)= a(00…0 (xj=0) xj+1…xn) + a(00…0 (xj=1) xj+1…xn) endpar endfor endfor Finally node 00…0 holds the sum
Hypercube Model Suppose node 000 holds element a(0) and 111holds element a(7) a(4) a(5) a(0) a(1) a(0)+a(4) a(1)+a(5) a(6) a(7) a(3) a(2) a(2)+a(6) a(3)+a(7) a(0)+a(4) +a(2)+a(6) a(0)+a(4) +a(2)+a(6) +a(1)+a(5) +a(3)+a(7) a(1)+a(5) +a(3)+a(7)