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Greedy Algorithm

- A greedy algorithm always makes the choice that

looks best at the moment - Key point Greed makes a locally optimal choice

in the hope that this choice will lead to a

globally optimal solution - Note Greedy algorithms do not always yield

optimal solutions, but for SOME problems they do

Greed

- When do we use greedy algorithms
- When we need a heuristic (e.g., hard problems

like the Traveling Salesman Problem) - When the problem itself is greedy
- Greedy Choice Property (CLRS 16.2)
- Optimal Substructure Property (shared with DP)

(CLRS 16.2) - Examples
- Minimum Spanning Tree (Kruskals algorithm)
- Optimal Prefix Codes (Huffmans algorithm)

Elements of the Greedy Algorithm

- Greedy-choice property A globally optimal

solution can be arrived at by making a locally

optimal (greedy) choice. - Must prove that a greedy choice at each step

yields a globally optimal solution - Optimal substructure property A problem

exhibits optimal substructure if an optimal

solution to the problem contains within it

optimal solutions to subproblems. This property

is a key ingredient of assessing the

applicability of greedy algorithm and dynamic

programming.

Greedy Algorithm Minimum Spanning Tree

- Kruskals minimum spanning tree algorithm
- INPUT edge-weighted graph G (V,E), with V

n - OUTPUT T a spanning tree of G (touches all

vertices, and therefore has n-1 edges) of minimum

cost ( total edge weight) - Will grow a set of edges T until it contains n-1

edges - Algorithm Start with T empty, then iteratively

add to T the smallest (minimum-cost) remaining

edge in the graph if the edge does not form a

cycle in T - Claim Greedy-MST is correct.
- Proof (by induction)
- Define a promising edge set E E to be any edge

set that is a subgraph of some MST. We will show

by induction that as T is constructed, T is

always promising.

Proof of Kruskals Algorithm

- Basis T 0, trivial.
- Induction Step T is promising by I.H., so it is

a subgraph of some MST, call it S. Let ei be the

smallest edge in E, s.t. Tei has no cycle,

eiT. - If eiS, were done.
- Suppose eiS, then S S ei has a unique

cycle containing ei, and all other arcs in cycle

ei (because S is an MST!) - Call the cycle C. Observe that C with ei cannot

be in T, because T ei is acyclic (because

Kruskal adds ei)

Proof of Kruskals Algorithm

- Then C must contains some edge ej s.t. ejS, and

we also know c(ej)c(ei). - Let S S ei \ ej
- S is an MST, so Tei is promising

Greedy Algorithm Huffman Codes

- Prefix codes
- one code per input symbol
- no code is a prefix of another
- Why prefix codes
- Easy decoding
- Since no codeword is a prefix of any other, the

codeword that begins an encoded file is

unambiguous - Identify the initial codeword, translate it back

to the original character, and repeat the

decoding process on the remainder of the encoded

file

Greedy Algorithm Huffman Codes

- Huffman coding
- Given frequencies with which with which source

symbols (e.g., A, B, C, , Z) appear in a message - Goal is to minimize the expected encoded message

length - Create tree (leaf) node for each symbol that

occurs with nonzero frequency - Node weights frequencies
- Find two nodes with smallest frequency
- Create a new node with these two nodes as

children, and with weight equal to the sum of the

weights of the two children - Continue until have a single tree

Greedy Algorithm Huffman Codes

- Example

Greedy Algorithm Huffman Codes

- Step 1 Step 2
- Step 3 Step 4

Greedy Algorithm Huffman Codes

- Step 5
- Step 6

Greedy Algorithm Huffman Codes

- Step 7
- Step 8

Greedy Algorithm Huffman Codes

- Step 9

Greedy Algorithm Huffman Codes

- Finally
- Note that the 0s (left branches) and 1s (right

branches) give the code words for each symbol

Proof That Huffmans Merge is Optimal

- Let T be an optimal prefix-code tree in which a,

b are siblings at deepest level, L(a) L(b) - Suppose that x, y are two other nodes that are

merged by the Huffman algorithm - x, y have lowest weights because Huffman chose

them - WLOG w(x) w(a), w(y) w(b) L(a) L(b)

L(x), L(y) - Swap a and x cost difference between T and new

T is - w(x)L(x) w(a)L(a) w(x)L(a) w(a)L(x)
- (w(a) w(x))(L(a) L(x)) // both factors

non-neg - 0
- Similar argument for b, y
- Huffman choice also optimal

Dynamic Programming

- Dynamic programming Divide problem into

overlapping subproblems recursively solve each

in the same way. - Similar to DQ, so whats the difference
- DQ partition the problem into independent

subproblems. - DP breaking it into overlapping subproblems, that

is, when subproblems share subproblems. - So DP saves work compared with DQ by solving

every subproblems just once ( when subproblems

are overlapping).

Elements of Dynamic Programming

- Optimal substructure A problem exhibits optimal

substructure if an optimal solution to the

problem contains within it optimal solutions to

subproblems. - Whenever a problem exhibits optimal

substructure, it is a good clue that DP might

apply.(a greedy method might apply also.) - Overlapping subproblems A recursive algorithm

for the problem solves the same subproblems over

and over, rather than always generating new

subproblems.

Dynamic Programming Matrix Chain Product

- Matrix-chain multiplication problem Give a chain

of n matrices A1, A2, , An to be multiplied,

how to get the product A1 A2 An. with minimum

number of scalar multiplications. - Because of the associative law of matrix

multiplication, there are many possible orderings

to calculate the product for the same matrix

chain - Only one way to multiply A1 A2
- Best way for triple Cost (A1 , A2) Cost((A1

A2) A3) or Cost (A2 , A3) Cost(A1 (A2 A3)).

Dynamic Programming Matrix Chain Product

- How do we build bottom-up
- From last example
- Best way for triple Cost (A1 , A2) Cost((A1

A2) A3) or Cost (A2 , A3) Cost(A1 (A2 A3)). - Save the best solutions for contiguous groups of

Ai. - Cost of ( ij )( j k) is ijk.
- E.g.,

Each of 310 entries requires 5 multiplies ( 4

adds)

Dynamic Programming Matrix Chain Product

- Cost of final multiplication
- A1 A2 A3 Ak-1 Ak An.
- Each of these subproblems can be solved optimally

just look in the table

Dynamic Programming Matrix Chain Product

- FORMULATION
- Table entries aij, 1 i j n, where aij

optimal solution min multiplications for - Ai Ai1 Aj-1 Ak
- We want aij to fill the table.
- Let dimensions be given by vector di, 1 i n1,

i.e., Ai is didi1

Dynamic Programming Matrix Chain Product

- Build Table
- Diagonal S contains aij with j - i S.
- S 0 aij 0, i1, 2, , n
- S 1 ai, i1 di di1di2, i1, 2, , n-1
- 1 S n ai, is (ai, k ak1, is

di dkdis ) - Example (Brassard/Bratley)
- 4 matrices, d (13, 5, 89, 3, 34)
- S 1 a12 5785
- a231335
- a34 9078

Dynamic Programming Matrix Chain Product

- S 2 a13 min(a11 a23 1353, a12

a3313893) 1530 - a24 min(a22 a34 58934, a23 a44

5334) 1845 - S 3 a14 min( k1 a11 a24 13534,
- k2 a12 a34 138934,
- k3 a13 a44 13334) 2856

(note max cost is 54201 multiplies) - Complexity Sgt0, choose among S choices for each

of n-S elements in diagonal, so runtime is (n3).

- Proof åi 1 to n i(n-i) åi 1 to n (ni

i2) n(n(n1)/2) (n(n1)(2n1)/6) (n3)

Dynamic Programming Longest Common Subsequence

- Longest Common Subsequence Give two strings a1

a2 am and b1 b2 bn, what is the largest

value P such that - For indices 1 i1 i2 ip m, and
- 1 j1 j2 jp n,
- We have aix bjx, for 1 x P
- Example
- So P 4, i 1, 2, 3, 5, j 3, 4, 5, 6

b a a b a c b a c b a a a

Dynamic Programming Longest Common Subsequence

- Let L(k, l) denote length of LCS for a1 a2 ak

and b1 b2 bl. - Then we have facts
- L(p, q) L(p-1, q-1).
- L(p, q) L(p-1, q-1) 1 if ap bq when ap

and bq are both in LCS. - L(p, q) L(p-1, q) when ap is not in LCS.
- L(p, q) L(p, q-1)
- when bq is not in LCS.

Dynamic Programming Longest Common Subsequence

- ALGORITHM
- for i 1 to m
- for j 1 to n
- if ai bj then L(i, j) L(i-1, j-1) 1
- else L(i, j) maxL(i, j-1),
- L(i-1, j)
- Time complexity (n2).

Dynamic Programming Knapsack

- The problem The knapsack problem is a particular

type of integer program with just one constraint

Each item that can go into the knapsack has a

size and a benefit. The knapsack has a certain

capacity. What should go into the knapsack so as

to maximize the total benefit - Hint Recall shortest path method.
- Define Fk(y) max (0kn)
- with (0yb)
- Then, what is Fk(y)
- Max value possible using only first k items when

weight limit is y.

Dynamic Programming Knapsack

- B.C.s 1. F0(y) 0 y no items chosen
- 2. Fk(0) 0 k weight limit 0
- 3. F1(y) y/w1v1
- Generally speaking
- Fk(y) maxFk-1(y), Fk(y-wk)vk
- Then we could build matrix use entries above,

here is an example

Dynamic Programming Knapsack

- Example k4, b10, ypounds, kitem

types allowed - v11 w12 v23 w23
- v35 w34 v49 w47
- Fk(y) maxFk-1(y), Fk(y-wk)vk
- Fk(y)
- nb table

Dynamic Programming Knapsack

- Note 12 max(11, 9 F4(3))max(11,93)12
- What is missing here (Like in SP, we know the

SPs cost, but we dont know SP itself) - So, we need another table
- i(k,j) max index such that item type j is used

in Fk(y), i.e., i(k,y)j xj1 xq0 qgtj - B.C.s i(1,y) 0 if F1(y) 0
- i(1,y) 0 if F1(y) 0
- General

Dynamic Programming Knapsack

- Trace Back if i(k,y) q, use item q once, check

i(k,y-q). - Example
- E.g. F4(10) 12. i(4,10)4 4th item used once

Dynamic Programming Knapsack

- i(4, 10 - w4) i(4,3)2 2nd item used once
- i(4, 3 w2) i(4,0)0 done
- Notice i(4,8)3 dont use most valuable item.

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