analysis & design of algorithms (csce 321)

Prof. Amr Goneid, AUC 1

Analysis & Design of Analysis & Design of AlgorithmsAlgorithms(CSCE 321)(CSCE 321)

Prof. Amr GoneidDepartment of Computer Science, AUC

Part 10. Dynamic Programming


Dynamic ProgrammingDynamic Programming



Introduction What is Dynamic Programming? How To Devise a Dynamic Programming

Approach The Sum of Subset Problem The Knapsack Problem Minimum Cost Path Coin Change Problem Optimal BST DP Algorithms in Graph Problems Comparison with Greedy and D&Q Methods


1. Introduction1. Introduction

We have demonstrated that

Sometimes, the divide and conquerapproach seems appropriate but failsto produce an efficient algorithm.

One of the reasons is that D&Q produces overlapping subproblems.


IntroductionIntroduction

Solution: Buy speed using space Store previous instances to compute current instance instead of dividing the large problem into two (or more)

smaller problems and solving those problems (as we did in the divide and conquer approach), we start with the simplest possible problems.

We solve them (usually trivially) and save these results. These results are then used to solve slightly larger problems which are, in turn, saved and used to solve larger problems again.

This method is called Dynamic Programming


2. 2. What is Dynamic ProgrammingWhat is Dynamic Programming

An algorithm design method used when the solution is a result of a sequence of decisions (e.g. Knapsack, Optimal Search Trees, Shortest Path, .. etc).

Makes decisions one at a time. Never make an erroneous decision.

Solves a sub-problem by making use of previously stored solutions for all other sub-problems.



Invented by American mathematician Richard Bellman in the 1950s to solve optimization problems“Programming” here means “planning”


When is Dynamic ProgrammingWhen is Dynamic Programming

Two main properties of a problem that suggest

that the given problem can be solved using

Dynamic programming.

Overlapping Subproblems

Optimal Substructure


Overlapping SubproblemsOverlapping Subproblems

Like Divide and Conquer, Dynamic Programming

combines solutions to subproblems. Dynamic Programming is mainly used when solutions of same subproblems are needed again and again.

Examples are computing the Fibonacci Sequence, Binomial Coefficients, etc.


Optimal Substructure: Principle of Optimal Substructure: Principle of OptimalityOptimality

Dynamic programming uses the Principle of Optimality to avoid non-optimal decision sequences.

For an optimal sequence of decisions, the remaining decisions must constitute an optimal sequence.

Example: Shortest PathFind Shortest path from vertex (i) to vertex (j)


Principle of OptimalityPrinciple of Optimality

Let k be an intermediate vertex on a shortest i-to-j path i , a , b, … , k , l , m , … , j . The path i , a , b, … , k must be shortest i-to-k , and the path k , l , m , … , j must be shortest k-to-j

i

a b

k

l

m

j


3. How To Devise a Dynamic 3. How To Devise a Dynamic Programming ApproachProgramming Approach Given a problem that is solvable by a Divide &

Conquer method Prepare a table to store results of sub-problems Replace base case by filling the start of the table Replace recursive calls by table lookups Devise for-loops to fill the table with sub-problem

solutions instead of returning values Solution is at the end of the table Notice that previous table locations also contain valid

(optimal) sub-problem solutions


Example(1): Fibonacci SequenceExample(1): Fibonacci Sequence

The Fibonacci graph is not a tree, indicating an Overlapping Subproblem.

Optimal Substructue:If F(n-2) and F(n-1) are optimal, thenF(n) = F(n-2) + F(n-1) is optimal


Fibonacci SequenceFibonacci Sequence

Dynamic Programming Solution:Buy speed with space, a table

F(n)Store previous instances to

compute current instance



Fib(n):

if (n < 2) return 1;

else

return

Fib(n-1) + Fib(n-2)

Table F[n]

F[0] = F[1] = 1;

if ( n >= 2)

for i = 2 to n

F[i] =

F[i-1] + F[i-2];

return F[n];i-2 i-1 i



Dynamic Programming Solution:Space Complexity is O(n) Time Complexity is T(n) = O(n)


Example(2): Counting CombinationsExample(2): Counting Combinations

Overlapping Subproblem.5,3

4,2 4,3

3,1 3,2 3,3

2,1 2,2

1,0 1,1

3,2

2,1 2,2

1,0 1,1

10 with

, 1

1

1

)n,n(comb),n(comb

nmm

n

m

n

m

n)m,n(comb


Counting CombinationsCounting Combinations

Optimal SubstructureThe value of Comb(n, m) can be recursively calculated using following standard formula for Binomial Coefficients:

Comb(n, m) = Comb(n-1, m-1) + Comb(n-1, m)

Comb(n, 0) = Comb(n, n) = 1



Dynamic Programming Solution:Buy speed with space, Pascal’s

Triangle. Use a table T[0..n, 0..m].Store previous instances to

compute current instance



comb(n,m):if ((m == 0) || (m == n)) return1; else return

comb (n − 1, m − 1) + comb (n − 1, m);

Table T[n,m]for (i = 0 to n − m) T[i, 0] = 1;for (i = 0 to m) T[i, i] = 1;for (j = 1 to m) for (i = j + 1 to n − m + j) T[i, j] = T[i − 1, j − 1] + T[i − 1, j];return T[n, m];

i-1, j-1 i-1 , j

i , j



Dynamic Programming Solution:Space Complexity is O(nm)Time Complexity is T(n) = O(nm)


ExerciseExerciseConsider the following function:

Consider the number of arithmetic operations used tobe T(n): Show that a direct recursive algorithm would give

an exponential complexity. Explain how, by not re-computing the same F(i)

value twice, one can obtain an algorithm with T(n) = O(n2)

Give an algorithm for this problem that only uses O(n) arithmetic operations.

1

1

21011n

i

)(F)(Fwithnfor)i(F)i(F)n(F


4. The Sum of Subset Problem4. The Sum of Subset Problem

Given a set of positive integers W = {w1,w2...wn} The problem: is there a subset of W that sums exactly to m?

i.e, is SumSub (w,n,m) true?

Example:W = { 11 , 13 , 27 , 7} , m = 31A possible subset that sums exactly to 31 is {11 , 13 , 7}

Hence, SumSub (w,4,31) is true

w1 w2 .... wn m


The Sum of Subset ProblemThe Sum of Subset Problem

Consider the partial problem SumSub (w,i,j)

SumSub (w, i, j) is true if: wi is not needed, {w1,..,wi-1} has a subset that sums to (j), i.e.,

SumSub (w, i-1, j) is true, OR wi is needed to fill the rest of (j), i.e., {w1,..,wi-1} has a subset that

sums to (j-wi)

If there are no elements, i.e. (i = 0) then SumSub (w, 0, j) is true if (j = 0) and false otherwise

w1 w2 .... wi j


Divide & Conquer ApproachDivide & Conquer Approach

Algorithm:bool SumSub (w, i, j)

{

if (i == 0) return (j == 0);

else

if (SumSub (w, i-1, j)) return true;

else if ((j - wi) >= 0)

return SumSub (w, i-1, j - wi);

else return false;

}


Dynamic Programming ApproachDynamic Programming Approach

Use a tabel t[i,j], i = 0.. n, j = 0..m Base case:

set t[0,0] = true and t[0,j] = false for (j != 0) Recursive calls are constructed as follows:

loop on i = 1 to n

loop on j = 1 to m test on SumSub (w, i-1, j) is replaced by t[i,j] = t[i-1,j] return SumSub (w, i-1, j - wi) is replaced by

t[i,j] = t[i-1,j] OR t[i-1,j – wi]


Dynamic Programming AlgorithmDynamic Programming Algorithm

bool SumSub (w , n , m)

{

1. t[0,0] = true; for (j = 1 to m) t[0,j] = false;

2. for (i = 1 to n)

3. for (j = 0 to m)

4. t[i,j] = t[i-1,j];

5. if ((j-wi) >= 0)

6. t[i,j] = t[i-1,j] || t[i-1, j – wi];

7. return t[n,m];

}

i-1,

j - wi

i-1,

j

i , j



Analysis:

1. costs O(1) + O(m)

2. costs O(n)

3. costs O(m)

4. costs O(1)

5. costs O(1)

6. costs O(1)

7. costs O(1)

Hence, space complexity is O(nm)

Time complexity is T(n) = O(m) + O(nm) = O(nm)


5. The (0/1) Knapsack Problem5. The (0/1) Knapsack Problem

Given n indivisible objects with positive integer weights W = {w1,w2...wn} and positive integer values V = {v1,v2...vn} and a knapsack of size (m)

Find the highest valued subset of objects with total weight at most (m)

i = 1 i = 2 i = n

mw1

p1

w2

p2

wn

pn

……..


The Decision InstanceThe Decision Instance

Assume that we have tried objects of type (1,2,..., i -1) to fill the sack up to a capacity (j) with a maximum profit of P(i-1,j)

If j wi then P(i-1, j - wi) is the maximum profit if we remove the equivalent weight wi of an object of type (i).

By trying to add object (i), we expect the maximum profit to change to P(i-1, j - wi) + vi


The Decision InstanceThe Decision Instance

If this change is better, we do it, otherwise we leave things as they were, i.e.,

P(i , j) = max { P(i-1, j) , P(i-1, j - wi) + vi } for j wi P(i , j) = P(i-1, j) for j < wi

The above instance can be solved for P(n , m) by initializing P(0,j) = 0 and successively computing P(1,j) , P(2,j) ....., P (n , j) for all 0 j m



Algorithm:int Knapsackr (int w[ ], int v[ ], int i, int j){

if (i == 0) return 0;else{

int a = Knapsackr (w,v,i-1,j);if ((j - w[i]) >= 0){ int b = Knapsackr (w,v,i-1,j-w[i]) + v[i];

return (b > a? b : a); }else return a;

}}



Analysis:

T(n) = no. of calls to Knapsackr (w, v, n, m): For n = 0, one main call, T(0) = 1 For n > 0, one main call plus two calls each with n-1 The recurrence relation is:

T(n) = 2T(n-1) + 1 for n > 0 with T(0) = 1 Hence T(n) = 2n+1 -1 = O(2n) = exponential time


Dynamic Programming ApproachDynamic Programming Approach The following approach will give the maximum

profit, but not the collection of objects that produced this profit

Initialize P(0 , j) = 0 for 0 j m Initialize P(i , 0) = 0 for 0 i n for each object i from 1 to n do

for a capacity j from 0 to m doP(i , j) = P(i-1 , j) if ( j >= wi) if (P(i-1 , j) < P(i-1, j - wi) + vi ) P(i , j) P(i-1 , j - wi) + vi


DP AlgorithmDP Algorithmint Knapsackdp (int w[ ], int v[ ], int n, int m){ int p[N][M];

for (int j = 0; j <= m; j++) p[0][j] = 0; for (int i= 0; i <= n; i++) p[i][0] = 0;for (int i = 1; i <= n; i++)

for (j = 0; j <= m; j++){ int a = p[i-1][j]; p[i][j] = a;

if ((j-w[i]) >= 0) { int b = p[i-1][j-w[i]]+v[i]; if (b > a) p[i][j] = b; }

}return p[n][m];

}Hence, space complexity is O(nm)Time complexity is T(n) = O(n) + O(m) + O(nm) = O(nm)


ExampleExample

Example: Knapsack capacity m = 5

item weight value ($)

1 2 12

2 1 10

3 3 20

4 2 15


ExampleExample

wi, vi

0 j-wi j m

0 0 0 0 0

i-1 0

i 0

n 0

P(i-1, j-wi) P(i-1, j)

P(i , j) Goal P(n,m)


ExampleExample

.


ExercisesExercises Modify the previous Knapsack algorithm so

that it could also list the objects contributing to the maximum profit.

Explain how to reduce the space complexity of the Knapsack problem to only O(m). You need only to find the maximum profit, not the actual collection of objects.


Exercise: Exercise: Longest Common Sequence ProblemLongest Common Sequence Problem Given two sequences A = {a1, . . . , an} and B = {b1, . . . , bm}.

Find the longest sequence that is a subsequence of both A and B. For example, if A = {aaadebcbac} and

B = {abcadebcbec}, then {adebcb} is subsequence of length 6 of both sequences.

Give the recursive Divide & Conquer algorithm and the Dynamic programming algorithm together with their analyses

Hint:

Let L(i, j) be the length of the longest common subsequence of {a1, . . . , ai} and {b1, . . . , bj}. If ai = bj then

L(i, j) = L(i−1, j −1)+1. Otherwise, one can see that

L(i, j) = max (L(i, j − 1), L(i − 1, j)).


6. Minimum Cost Path6. Minimum Cost Path Given a cost matrix C[ ][ ] and a position (n, m) in

C[ ][ ], find the cost of minimum cost path to reach (n , m) from (0, 0).

Each cell of the matrix represents a cost to traverse through that cell. Total cost of a path to reach (n, m) is the sum of all the costs on that path (including both source and destination).

From a given cell (i, j), You can only traverse down to cell (i+1, j), , right to cell (i, j+1) and diagonally to cell (i+1, j+1) . Assume that all costs are positive integers


Minimum Cost PathMinimum Cost Path

Example: what is the minimum cost path to (2, 2)?

The path is (0, 0) –> (0, 1) –> (1, 2) –> (2, 2). The cost of the path is 8 (1 + 2 + 2 + 3).Optimal SubstructureMinimum cost to reach (n, m) is

“minimum of the 3 cells plus cost[n][m]“.

i.e.,

minCost(n, m) = min (minCost(n-1, m-1), minCost(n-1, m), minCost(n, m-1)) + C[n][m]


Minimum Cost Path (D&Q)Minimum Cost Path (D&Q) Overlapping Subproblems:The recursive definition suggests a D&Q approach with

overlapping subproblems:

int MinCost(int C[ ][M], int n, int m)

{

if (n < 0 || m < 0) return ∞;

else if (n == 0 && m == 0) return C[n][m];

else return C[n][m] + min( MinCost(C, n-1, m-1),

MinCost(C, n-1, m), MinCost(C, n, m-1) );

}

Anaysis: For m=n, T(n) = 3 T(n-1) + 3 for n > 0 with T(0) = 0

Hence T(n) = O(3n) , exponential complexity



In the Dynamic Programming(DP) algorithm, recomputations of same subproblems can be avoided by constructing a temporary array T[ ][ ] in a bottom up manner.int minCost(int C[ ][M], int n, int m)

{ int i, j; int T[N][M];

T[0][0] = C[0][0];

/* Initialize first column */

for (i = 1; i <= n; i++) T[i][0] = T[i-1][0] + C[i][0];

/* Initialize first row */

for (j = 1; j <= m; j++) T[0][j] = T[0][j-1] + C[0][j];

/* Construct rest of the array */

for (i = 1; i <= n; i++)

for (j = 1; j <= m; j++)

T[i][j] = min(T[i-1][j-1], T[i-1][j], T[i][j-1]) + C[i][j];

return T[n][m];

}

Space complexityis O(nm)Time complexity isT(n) = O(n)+O(m) +

O(nm) = O(nm)


7. Coin Change Problem7. Coin Change Problem We want to make change for N cents, and we have infinite

supply of each of S = {S1 , S2, …Sm} valued coins, how many ways can we make the change? (For simplicity's sake, the order does not matter.)

Mathematically, how many ways can we express N as

For example, for N = 4,S = {1,2,3}, there are four solutions: {1,1,1,1},{1,1,2},{2,2},{1,3}.

We are trying to count the number of distinct sets. Since order does not matter, we will impose that our solutions

(sets) are all sorted in non-decreasing order (Thus, we are

looking at sorted-set solutions: collections).

}....1{,0,1

mkxSxN kk

m

kk


Coin Change ProblemCoin Change Problem With S1 < S2 < …<Sm the number of possible sets C(N,m) is

Composed of: Those sets that contain at least 1 Sm, i.e. C(N-Sm, m)

Those sets that do not contain any Sm, i.e. C(N, m-1)

Hence, the solution can be represented by the recurrence relation:

C(N,m) = C(N, m-1) + C(N-Sm , m)

with the base cases:

C(N,m) = 1 for N = 0 C(N,m) = 0 for N < 0

C(N,m) = 0 for N 1 , M ≤ 0 Therefore, the problem has optimal substructure property as

the problem can be solved using solutions to subproblems. It also has the property of overlapping subproblems.


D&Q AlgorithmD&Q Algorithm int count( int S[ ], int m, int n )

{

// If n is 0 then there is 1 solution (do not include any coin)

if (n == 0) return 1;

// If n is less than 0 then no solution exists

if (n < 0) return 0;

// If there are no coins and n is greater than 0, then no solution

if (m <=0 && n >= 1) return 0;

// count is sum of solutions (i) including S[m-1] (ii) excluding

// S[m-1]

return count( S, m - 1, n ) + count( S, m, n-S[m-1] );

}

The algorithm has exponential complexity.


DP AlgorithmDP Algorithm int count( int S[ ], int m, int n ){ int i, j, x, y; int table[n+1][m]; // n+1 rows to include the case (n = 0) for (i=0; i < m; i++) table[0][i] = 1; // Fill for the case (n = 0) // Fill rest of the table enteries bottom up for (i = 1; i < n+1; i++) for (j = 0; j < m; j++) { x = (i-S[j] >= 0)? table[i - S[j]][j]: 0; // solutions including S[j] y = (j >= 1)? table[i][j-1]: 0; // solutions excluding S[j] table[i][j] = x + y; // total count } return table[n][m-1];}Space comlexity is O(nm)Time comlexity is O(m) + O(nm) = O(nm)


8. Optimal Binary Search Trees8. Optimal Binary Search Trees

Problem:

Given a set of keys K1 , K2 , … , Kn and their corresponding search frequencies P1 , P2 , … , Pn , find a binary search tree for the keys such that the total search cost is minimum.

Remark:The problem is similar to that of the optimal merge trees (Huffman Coding) but more difficult because now:- Keys can exist in internal nodes- The binary search tree condition (Left < Parent < Right) is imposed


(a) Example(a) Example

A Binary Search Tree of 5 words:

A Greedy Algorithm:

Insert words in the tree in order of decreasing frequency of search.

10 30 20 18 22Freq

(tot 100)

two if and am aWord


A Greedy BSTA Greedy BST Insert (if) , (a) , (and) , (am) , (two)

Total Search Cost (100 searches) = 1*30 + 2*22 + 2* 10 + 3*20 + 4*18 = 226

if

twoa

and

am

30

22 10

20

18

Level

1

2

4

3


Another way to Compute CostAnother way to Compute Cost

For a tree containing n keys (K 1 , K 2 , … , K n), the total cost C(tree) is:

Total Cost = C(tree) =

( P)All keys + C(left subtree) + C(right subtree)

For the previous example:

C(tree) = 100 + {60 + [0] + [38 + (18) + (0)]}+ {10} = 226

i

C (Left)1 .. (i-1)

C (right)(i+1).. n


An Optimal BSTAn Optimal BST A Dynamic Programming Algorithm leads to the

following BST:

Cost = 1*20 +2*22 + 2*30 + 3*18 + 3*10 = 208 =100 + {40 + [0] + [18]} + {40 + [0] + [10] } = = 208

and

if

two

a

am

20

2230

1810


(b) Dynamic Programming Method(b) Dynamic Programming Method

For n keys, we perform n-1 iterations 1 j n-1 In each iteration (j), we compute the best way to build

a sub-tree containing j+1 keys (K i , K i+1 , … , K i+j) for all possible BST combinations of such j+1 keys ( i.e. for 1 i n-j).

Each sub-tree is tried with one of the keys as the root and a minimum cost sub-tree is stored.

For a given iteration (j), we use previously stored values to determine the current best sub-tree.


Simple ExampleSimple Example For 3 keys (A < B < C), we perform 2

iterations j = 1 , j = 2 For j = 1, we build sub-trees using 2 keys.

These come from i = 1, (A-B), andi = 2, (B-C).

For each of these combinations, we compute the least cost sub-tree, i.e., the least cost of the two sub-trees (A*,B) and (A, B*) and the least cost of the two sub-trees (B*,C) and (B, C*) , where (*) denotes parent.


Simple Example (Cont.)Simple Example (Cont.)

For j = 2, we build trees using 3 keys. These come from i = 1, (A-C).

For this combination, we compute the least cost tree of the trees

(A*,(B-C)) , (A , B* , C) , ((A-B) , C*).

This is done using previously computed least cost sub-trees.


Simple Example (continued)Simple Example (continued) j = 1

i = 1, (A-B) i = 2 , (B-C)

k = 1 k = 2

k = 2 k = 3

j = 2 i = 1, (A-C)k = 1

k = 2

k = 3

A

B

B

A

B

C

C

B

A

B

A C

C

B-C

A-B

min

min

min = ? min = ?

min = ?


(c) Example Revisited: Optimal BST for 5 keys(c) Example Revisited: Optimal BST for 5 keys

Iterations 1..2I 1 2 3 4 5

1 Key a am and if two

j = 0 p 22 18 20 30 10

2 keys a..am am..and and..if if..two

j = 1 sum(p) 40 38 50 40

k = I 18 20 30 10

k = I + j 22 18 20 30

min 58 56 70 50

3 Keys a..and am..if and..two

j = 2 sum(p) 60 68 60

k = I 56 70 50

42 48 30

k = I + j 58 56 70

min 102 116 90


Optimal BST for 5 keysOptimal BST for 5 keys Iterations 3 .. n-1

4 Keys a..if am..two

j = 3 sum(p) 90 78

k = I 116 90

92 68

88 66

k = I + j 102 116

min 178 144

5 Keys a..two

j = 4 sum(p) 100

k = I 144

112

108

112

k = I + j 178

min 208


Construction of The TreeConstruction of The Tree Min BST at j = 4, i = 1, k = 3

cost = 100 + 58 + 50 = 208

(a .. am)min at j = 1, i = 1, k = 1

cost = 58 (if .. two)min at j = 1, i = 4, k = 1

cost = 50

Final min BST

cost = 1*20 + 2*22 + 2*30 +

3*18 + 3*10 = 208

and

if

two

a

am

if

two

a

and

am

a..am If..twomin min


(d) Complexity Analysis(d) Complexity Analysis Skeletal Algorithm:

for j = 1 to n-1 do { // sub-tree has j+1 nodes

for i = 1 to n-j do { // for each of the n-j sub-tree combinations

for k = i to i+j do { find cost of each of the j+1 configurations and determine minimum cost }

} }

T(n) = 1 j n-1 ( j + 1) (n – j) = O(n3) ,

S(n) = O(n2)


ExerciseExercise Find the optimal binary search tree for the

following words with the associated frequencies:a (18) , and (22) , I (19) , it (20) , or (21)

Answer: Min cost = 20 + 2*43 + 3*37 = 217 it

orand

Ia

20

22 21

18 19


99. Dynamic Programming Algorithms . Dynamic Programming Algorithms for Graph Problemsfor Graph ProblemsVarious optimization graph problems have been solved using Dynamic Programming algorithms. Examples are:

Dijkstra's algorithm solves the single-source shortest path problem for a graph with nonnegative edge path costsFloyd–Warshall algorithm for finding all pairs shortest paths in a weighted graph (with positive or negative edge weights) and also for finding transitive closureThe Bellman–Ford algorithm computes single-source shortest paths in a weighted digraph for graphs with negative edge weights.

These will be discussed later under “Graph Algorithms”.


1010. Comparison with Greedy and . Comparison with Greedy and Divide & Conquer MethodsDivide & Conquer Methods Greedy vs. DP :

Both are optimization techniques, building solutions from a collection of choices of individual elements.

The greedy method computes its solution by making its choices in a serial forward fashion, never looking back or revising previous choices.

DP computes its solution bottom up by synthesizing them from smaller subsolutions, and by trying many possibilities and choices before it arrives at the optimal set of choices.

There is no a priori test by which one can tell if the Greedy method will lead to an optimal solution.

By contrast, there is a test for DP, called The Principle of Optimality


Comparison with Greedy and Divide & Comparison with Greedy and Divide & Conquer MethodsConquer Methods D&Q vs. DP:

Both techniques split their input into parts, find subsolutions to the parts, and synthesize larger solutions from smaller ones.

D&Q splits its input at pre-specified deterministic points (e.g., always in the middle)

DP splits its input at every possible split point rather than at pre-specified points. After trying all split points, it determines which split point is optimal.

analysis & design of algorithms (csce 321)

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