Dynamic Programming and Applications.ppt

Topic Overview
• Overview of Serial Dynamic Programming
• Serial Monadic DP Formulations
• Nonserial Monadic DP Formulations
• Serial Polyadic DP Formulations
• Nonserial Polyadic DP Formulations

Overview of Serial Dynamic Programming
• Dynamic programming (DP) is used to solve a wide
variety of discrete optimization problems such as
scheduling, string-editing, packaging, and inventory
management.
• Break problems into subproblems and combine their
solutions into solutions to larger problems.
• In contrast to divide-and-conquer, there may be
relationships across subproblems.

Dynamic Programming: Example
• Consider the problem of finding a shortest path between
a pair of vertices in an acyclic graph.
• An edge connecting node i to node j has cost c(i,j).
• The graph contains n nodes numbered 0,1,…, n-1, and
has an edge from node i to node j only if i < j. Node 0 is
source and node n-1 is the destination.
• Let f(x) be the cost of the shortest path from node 0 to
node x.

• A graph for which the shortest path between nodes 0
and 4 is to be computed.

Dynamic Programming
• The solution to a DP problem is typically expressed as a
minimum (or maximum) of possible alternate solutions.
• If r represents the cost of a solution composed of
subproblems x1, x2,…, xl, then r can be written as
Here, g is the composition function.
• If the optimal solution to each problem is determined by
composing optimal solutions to the subproblems and
selecting the minimum (or maximum), the formulation is
said to be a DP formulation.

The computation and composition of subproblem solutions to
solve problem f(x8).

Dynamic Programming
• The recursive DP equation is also called the functional
equation or optimization equation.
• In the equation for the shortest path problem the
composition function is f(j) + c(j,x). This contains a single
recursive term (f(j)). Such a formulation is called
monadic.
• If the RHS has multiple recursive terms, the DP
formulation is called polyadic.

Dynamic Programming
• The dependencies between subproblems can be
expressed as a graph.
• If the graph can be levelized (i.e., solutions to problems
at a level depend only on solutions to problems at the
previous level), the formulation is called serial, else it is
called non-serial.
• Based on these two criteria, we can classify DP
formulations into four categories - serial-monadic, serial-
polyadic, non-serial-monadic, non-serial-polyadic.
• This classification is useful since it identifies concurrency
and dependencies that guide parallel formulations.

Serial Monadic DP Formulations
• It is difficult to derive canonical parallel formulations for
the entire class of formulations.
• For this reason, we select two representative examples,
the shortest-path problem for a multistage graph and the
0/1 knapsack problem.
• We derive parallel formulations for these problems and
identify common principles guiding design within the
class.

Shortest-Path Problem
• Special class of shortest path problem where the graph
is a weighted multistage graph of r + 1 levels.
• Each level is assumed to have n levels and every node
at level i is connected to every node at level i + 1.
• Levels zero and r contain only one node, the source and
destination nodes, respectively.
• The objective of this problem is to find the shortest path
from S to R.

An example of a serial monadic DP formulation for finding
the shortest path in a graph whose nodes can be
organized into levels.

• The ith
node at level l in the graph is labeled vi
l
and the
cost of an edge connecting vi
l
to node vj
l+1
is labeled ci
l
,j.
• The cost of reaching the goal node R from any node vi
l
is
represented by Ci
l
.
• If there are n nodes at level l, the vector
[C0
l
, C1
l,…,
Cn
l
-1]T
is referred to as Cl
. Note that
C0 = [C0
0
].
• We have Ci
l
= min {(ci
l
,j + Cj
l+1
) | j is a node at level l + 1}

• Since all nodes vj
r-1
have only one edge connecting them
to the goal node R at level r, the cost Cj
r-1
is equal to cj
r
,
-
R
1
.
• We have:
Notice that this problem is serial and monadic.

• The cost of reaching the goal node R from any node at
level l is (0 < l < r – 1) is

• We can express the solution to the problem as a
modified sequence of matrix-vector products.
• Replacing the addition operation by minimization and the
multiplication operation by addition, the preceding set of
equations becomes:
where Cl
and Cl+1
are n x 1 vectors representing the cost
of reaching the goal node from each node at levels l and
l + 1.

• Matrix Ml,l+1 is an n x n matrix in which entry (i, j) stores
the cost of the edge connecting node i at level l to node j
at level l + 1.
• The shortest path problem has been formulated as a
sequence of r matrix-vector products.

Parallel Shortest-Path
• We can parallelize this algorithm using the parallel
algorithms for the matrix-vector product.
• Θ(n) processing elements can compute each vector Cl
in
time Θ(n) and solve the entire problem in time Θ(rn).
• In many instances of this problem, the matrix M may be
sparse. For such problems, it is highly desirable to use
sparse matrix techniques.

0/1 Knapsack Problem
• We are given a knapsack of capacity c and a set of n objects
numbered 1,2,…,n. Each object i has weight wi and profit pi.
• Let v = [v1, v2,…, vn] be a solution vector in which vi = 0 if object i is
not in the knapsack, and vi = 1 if it is in the knapsack.
• The goal is to find a subset of objects to put into the knapsack so
that
(that is, the objects fit into the knapsack) and
is maximized (that is, the profit is maximized).

• The naive method is to consider all 2n
possible subsets
of the n objects and choose the one that fits into the
knapsack and maximizes the profit.
• Let F[i,x] be the maximum profit for a knapsack of
capacity x using only objects {1,2,…,i}. The DP
formulation is:

• Construct a table F of size n x c in row-major order.
• Filling an entry in a row requires two entries from the
previous row: one from the same column and one from
the column offset by the weight of the object
corresponding to the row.
• Computing each entry takes constant time; the
sequential run time of this algorithm is Θ(nc).
• The formulation is serial-monadic.

Computing entries of table F for the 0/1 knapsack problem. The computation of
entry F[i,j] requires communication with processing elements containing
entries F[i-1,j] and F[i-1,j-wi].

• Using c processors in a PRAM, we can derive a simple
parallel algorithm that runs in O(n) time by partitioning
the columns across processors.
• In a distributed memory machine, in the jth
iteration, for
computing F[j,r] at processing element Pr-1, F[j-1,r] is
available locally but F[j-1,r-wj] must be fetched.
• The communication operation is a circular shift and the
time is given by (ts + tw) log c. The total time is therefore
tc + (ts + tw) log c.
• Across all n iterations (rows), the parallel time is O(n log
c). Note that this is not cost optimal.

• Using p-processing elements, each processing element
computes c/p elements of the table in each iteration.
• The corresponding shift operation takes time (2ts + twc/p),
since the data block may be partitioned across two
processors, but the total volume of data is c/p.
• The corresponding parallel time is n(tcc/p + 2ts + twc/p),
or O(nc/p) (which is cost-optimal).
• Note that there is an upper bound on the efficiency of
this formulation.

Nonserial Monadic DP Formulations: Longest-
Common-Subsequence
• Given a sequence A = <a1, a2,…, an>, a subsequence of
A can be formed by deleting some entries from A.
• Given two sequences A = <a1, a2,…, an> and B = <b1, b2,
…, bm>, find the longest sequence that is a subsequence
of both A and B.
• If A = <c,a,d,b,r,z> and B = <a,s,b,z>, the longest
common subsequence of A and B is <a,b,z>.

Longest-Common-Subsequence Problem
• Let F[i,j] denote the length of the longest common
subsequence of the first i elements of A and the first j
elements of B. The objective of the LCS problem is to
find F[n,m].
• We can write:

• The algorithm computes the two-dimensional F table in a
row- or column-major fashion. The complexity is Θ(nm).
• Treating nodes along a diagonal as belonging to one
level, each node depends on two subproblems at the
preceding level and one subproblem two levels prior.
• This DP formulation is nonserial monadic.

(a) Computing entries of table for the longest-common-
subsequence problem. Computation proceeds along the dotted
diagonal lines. (b) Mapping elements of the table to processing
elements.

Longest-Common-Subsequence: Example
• Consider the LCS of two amino-acid sequences H E A G A W G H E E and P A W H E A E. For the interested reader,
the names of the corresponding amino-acids are A: Alanine, E: Glutamic acid, G: Glycine, H: Histidine, P: Proline,
and W: Tryptophan.
• The F table for computing the LCS of the sequences. The LCS is A W H E E.

Parallel Longest-Common-Subsequence
• Table entries are computed in a diagonal sweep from the
top-left to the bottom-right corner.
• Using n processors in a PRAM, each entry in a diagonal
can be computed in constant time.
• For two sequences of length n, there are 2n-1 diagonals.
• The parallel run time is Θ(n) and the algorithm is cost-
optimal.

Parallel Longest-Common-Subsequence
• Consider a (logical) linear array of processors. Processing
element Pi is responsible for the (i+1)th
column of the
table.
• To compute F[i,j], processing element Pj-1 may need either
F[i-1,j-1] or F[i,j-1] from the processing element to its left.
This communication takes time ts + tw.
• The computation takes constant time (tc).
• We have:
• Note that this formulation is cost-optimal, however, its
efficiency is upper-bounded by 0.5!
• Can you think of how to fix this?

Serial Polyadic DP Formulation: Floyd's All-
Pairs Shortest Path
• Given weighted graph G(V,E), Floyd's algorithm determines
the cost di,j of the shortest path between each pair of nodes in
V.
• Let di
k
,j be the minimum cost of a path from node i to node j,
using only nodes v0,v1,…,vk-1.
• We have:
• Each iteration requires time Θ(n2
) and the overall run time of
the sequential algorithm is Θ(n3
).

Serial Polyadic DP Formulation: Floyd's All-
Pairs Shortest Path
• A PRAM formulation of this algorithm uses n2
processors
in a logical 2D mesh. Processor Pi,j computes the value
of di
k
,j for k=1,2,…,n in constant time.
• The parallel runtime is Θ(n) and it is cost-optimal.
• The algorithm can easily be adapted to practical
architectures, as discussed in our treatment of Graph
Algorithms.

Nonserial Polyadic DP Formulation: Optimal Matrix-
Parenthesization Problem
• When multiplying a sequence of matrices, the order of
multiplication significantly impacts operation count.
• Let C[i,j] be the optimal cost of multiplying the matrices
Ai,…Aj.
• The chain of matrices can be expressed as a product of
two smaller chains, Ai,Ai+1,…,Ak and Ak+1,…,Aj.
• The chain Ai,Ai+1,…,Ak results in a matrix of dimensions ri-
1 x rk, and the chain Ak+1,…,Aj results in a matrix of
dimensions rk x rj.
• The cost of multiplying these two matrices is ri-1rkrj.

Optimal Matrix-Parenthesization Problem
• We have:

A nonserial polyadic DP formulation for finding an optimal matrix
parenthesization for a chain of four matrices. A square node represents
the optimal cost of multiplying a matrix chain. A circle node represents a
possible parenthesization.

• The goal of finding C[1,n] is accomplished in a bottom-up
fashion.
• Visualize this by thinking of filling in the C table
diagonally. Entries in diagonal l corresponds to the cost
of multiplying matrix chains of length l+1.
• The value of C[i,j] is computed as min{C[i,k] + C[k+1,j] +
ri-1rkrj}, where k can take values from i to j-1.
• Computing C[i,j] requires that we evaluate (j-i) terms and
select their minimum.
• The computation of each term takes time tc, and the
computation of C[i,j] takes time (j-i)tc. Each entry in
diagonal l can be computed in time ltc.

• The algorithm computes (n-1) chains of length two. This
takes time (n-1)tc; computing n-2 chains of length three
takes time (n-2)tc. In the final step, the algorithm
computes one chain of length n in time (n-1)tc.
• It follows that the serial time is Θ(n3
).

The diagonal order of computation for the optimal matrix-
parenthesization problem.

Parallel Optimal Matrix-Parenthesization
Problem
• Consider a logical ring of processors. In step l, each processor computes a
single element belonging to the lth
diagonal.
• On computing the assigned value of the element in table C, each processor
sends its value to all other processors using an all-to-all broadcast.
• The next value can then be computed locally.
• The total time required to compute the entries along diagonal l is ltc+tslog
n+tw(n-1).
• The corresponding parallel time is given by:

Parallel Optimal Matrix-Parenthesization
Problem
• When using p (<n) processors, each processor stores n/p nodes.
• The time taken for all-to-all broadcast of n/p words is
and the time to compute n/p entries of the table in the lth
diagonal is ltcn/p.
• This formulation can be improved to use up to n(n+1)/2 processors using
pipelining.

Discussion of Parallel Dynamic Programming
Algorithms
• By representing computation as a graph, we identify
three sources of parallelism: parallelism within nodes,
parallelism across nodes at a level, and pipelining nodes
across multiple levels. The first two are available in serial
formulations and the third one in non-serial formulations.
• Data locality is critical for performance. Different DP
formulations, by the very nature of the problem instance,
have different degrees of locality.

Dynamic Programming and Applications.ppt

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