Generalized%20Parallel%20Prefix%20Computation

1
Generalized Parallel Prefix Computation

• GPC
• Given
• f(1),f(2) ,..., f(n) associative
  operation defined.
• y(1) ,..., y(n) linear ordering lt
  defined.
• Objective Compute D(1) ,..., D(n), where
• D(m) f(j1)f(j2) ... f(jk), j1 lt j2 lt ...
  lt jk and
• j1 , j2 ,..., jk is the sequences of indices
  such that
• ji lt m and y(ji) lt y(m) for i1,2,..,.k
• Example Range searching problem
• Q ((m,y(m)), m1,...., n
• lt is defined on numbers
• Query G consists of two intervals (-inf, m and
  (-inf, y(m), for every m from 1 to n

2
Generalized Parallel Prefix Computation

• GPC
• Given
• f(1),f(2) ,..., f(n) associative
  operation defined.
• y(1) ,..., y(n) linear ordering lt
  defined.
• Objective
• Compute D(1) ,..., D(n), where
• D(m) f(j1)f(j2) ... f(jk), j1 lt j2 lt ...
  lt jk and
• j1 , j2 ,..., jk is the sequences of indices
  such that
• ji lt m and y(ji) lt y(m) for i1,2,..,.k
• Example Range searching problem
• Q ((m,y(m)), m1,...., n
• lt is defined on numbers
• Query G consists of two intervals (-?, m and (-
  ?, y(m), for every m from 1 to n

3
Lower Bound of GPC

• If we can do GPC, then we can do sorting.
• Idea Let z(1), z(2),..., z(n) all distinct.
• (i) f(j) 1 for 1lt j ltn.
• (ii) y(j) z(j), for 1lti ltn.
• (iii) Compute D(m)
• (iv) y(j) z(n-j1), for 1lti ltn.
• (v) Compute D(m)
• (vi) D(m) D(m) of elements in Z smaller
  than z(m)
• Example
• Z4,5,3,7,1,6
• D(m) 0,1,0,3,0,4
• D(m) 2,2,1,2,0,0
• rank(m) 2,3,1,5,0,4

4
GPC Computation on PRAM

• D(m,S) D(m) restricted on a sequence of indices
  S. That is,
• D(m,S) f(j1)f(j2) ... f(jk), where ji?S
  and
• ji satisfies the conditions earlier (ji lt m)
• Y(S) the sequence of elements y(j), j?S in
  sorted order.
• B(m,S) The position of y(m) in Y(S)
• J(m,S) j1, j2, ... jr be the subsequence of S
  satisfying y(ji)lt y(m)
• For convenience, m is in J(m,S).
• E(m,S) f(j1)f(j2) ... f(jr).

y(i)
m
E(m,S)
D(m,S)
i
5
GPC Algorithm

• Initially, S1,...,n
• Partition S into two parts, L, and R
• Apply algorithm recursively to L and R
• gt Y(L), Y(R), D(l,L), D(r,R), E(l,L), E(r,R),
  B(l,L), B(r,R), for all l in L and r in R.
• Compute Y(S) by merge Y(L) and Y(R).
• Compute the rank B(m,S) in Y(S)
• for each r in R,
• gr point in L with the largest y-value such
  that y(gr) lt y(r),
• B(gr,L) B(r,S) - B(r,R) gt can find B(r,S)
  (How to find gr ?)
• for each l in L.
• gl The point in R with the largest y-value
  such that y(gl) lt y(l),
• B(gl,L) B(l,S) - B(l,L) gt can find B(l,S)

y(i)
L
R
y(r)
i
6
GPC Algorithm cont

• Compute D and E as follows
• D(l,S) D(l,L)
• D(r,S) E(gr,L) D(r,R)
• E(l,S) E(l,L) E(gl,R)
• E(r,S) E(gr,L) E(r,R)

y(i)
R
L
4
D(6,S) E(2,L) D(6,R)
f(1)f(2)f(3)f(5)
7
6
y(r)
2
1
D(r,R)
8
5
3
i
7
Complexity

• Similar to tree Computation
• Depth of recursion log2 n
• Merging L and R into S
• points of L, R sorted in y value
• Points in S should be also sorted in y value
• Then computing gr is trivial
• How to merge L and R in constant time?

8
Pipelined Merging of Two sorted list in a
constant time(Coles Algorithm)

• Leaves contain the value
• Internal nodes merge at each time by updating the
  values
• Lv the sequence of values of descendants of v
• Qv(j) At time j, a sorted sequence v has.
• An increasing subsequence of Lv
• When Qv(j) Lv, then node v is complete.
• All leaf nodes are complete.
• At step j1, if vs parent is not complete at
  j-th step, it sends Rv(j) and Qv(j) to its
  parent.
• Qv(j) merge Rw(j) and Rz(j), where w and z are
  children of v
• How to compute R?
• If w is not complete at j-1 step, Rw(j)
  consists of every 4-th elements of Qw(j-1).
• If w is complete after j step,
• (i) Rw(j1) consists of every 4-th elements of
  Qw(j)
• (ii) Rw(j2) consists of every 2nd elements of
  Qw(j)
• (iii) Rw(j3) Qw(j)
• If w and z becomes complete at the j-th step,
  then v becomes complete at j3 step
• gt total complexity 3logn
• How to merge Rw(j) and Rz(j) in constant time?

9
Merging two samples in constant time

• Two sequences S and T.
• Predecessor of x in S the largest element T
  smaller than x.
• Example S1,3,4,9, T2,5,6,7
• pred(3) 2, pred(4) 2, pred(5) 4.
• If each element of S and T know the position of
  its pred in T and S,
• gt S and T can be merged in constant time using
  S T PEs.
• How to find the pred of Rw(j) and Rz(j) ? gt
  Inductively.
• 1. Rw(j-1) and Rz(j-1) know their predecessors,
  and two sequence merged to Qv(j-1) .
• 2. each element in Rw(j-1) finds its pred in
  Qw(j-1) in constant time
• and its pred in Rw(j) in constant time.
• Note that no more than 4 elements of Rw(j-1)
  have the same pred in Rw(j)
• Each element in Rw(j) finds its pred in
  Rw(j-1)
• 3. Same for Rz.
• 4. With these pred knowledge, Rw(j) can determine
  their pred in Rz(j) in cons time.