chapter 5 fundamental algorithm design techniques
DESCRIPTION
Chapter 5 Fundamental Algorithm Design Techniques. Overview. The Greedy Method (5.1) Divide and Conquer (5.2) Dynamic Programming (5.3). Greedy Outline. The Greedy Method Technique (§5.1) Fractional Knapsack Problem (§5.1.1) Task Scheduling (§5.1.2) - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 5
Fundamental Algorithm
Design Techniques
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Overview The Greedy Method (5.1) Divide and Conquer (5.2) Dynamic Programming (5.3)
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Greedy Outline
The Greedy Method Technique (§5.1) Fractional Knapsack Problem (§5.1.1) Task Scheduling (§5.1.2) Minimum Spanning Trees (§7.3) [future lecture]
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A New Vending Machine? Coke machine in the
lounge behind the ulab—can we hack it?
Being efficient computer scientists, we want to return the smallest possible number of coins in change
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The Greedy Method
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The Greedy Method
The greedy method is a general algorithm design paradigm, built on the following elements:
configurations: different choices, collections, or values to find
objective function: a score assigned to configurations, which we want to either maximize or minimize
It works best when applied to problems with the greedy-choice property:
a globally-optimal solution can always be found by a series of local improvements from a starting configuration.
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Making Change Problem: A dollar amount to reach and a collection
of coin amounts to use to get there. Configuration: A dollar amount yet to return to a
customer plus the coins already returned Objective function: Minimize number of coins
returned. Greedy solution: Always return the largest coin you
can Example 1: Coins are valued $.32, $.08, $.01
Has the greedy-choice property, since no amount over $.32 can be made with a minimum number of coins by omitting a $.32 coin (similarly for amounts over $.08, but under $.32).
Example 2: Coins are valued $.30, $.20, $.05, $.01 Does not have greedy-choice property, since $.40 is
best made with two $.20’s, but the greedy solution will pick three coins (which ones?)
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How to rob a bank…
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The Fractional Knapsack Problem Given: A set S of n items, with each item i having
bi - a positive benefit wi - a positive weight
Goal: Choose items with maximum total benefit but with weight at most W.
If we are allowed to take fractional amounts, then this is the fractional knapsack problem.
In this case, we let xi denote the amount we take of item i
Objective: maximize
Constraint:
Si
iii wxb )/(
Si
i Wx
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Example
Given: A set S of n items, with each item i having bi - a positive benefit wi - a positive weight
Goal: Choose items with maximum total benefit but with weight at most W.
Weight:Benefit:
1 2 3 4 5
4 ml 8 ml 2 ml 6 ml 1 ml
$12 $32 $40 $30 $50
Items:
Value: 3($ per ml)
4 20 5 5010 ml
Solution:• 1 ml of 5• 2 ml of 3• 6 ml of 4• 1 ml of 2
“knapsack”
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The Fractional Knapsack Algorithm
Greedy choice: Keep taking item with highest value (benefit to weight ratio)
Since Run time: O(n log n). Why?
Correctness: Suppose there is a better solution
there is an item i with higher value than a chosen item j, but xi<wi, xj>0 and vi<vj
If we substitute some i with j, we get a better solution
How much of i: min{wi-xi, xj} Thus, there is no better
solution than the greedy one
Algorithm fractionalKnapsack(S, W)Input: set S of items w/
benefit bi and weight wi; max. weight W Output: amount xi of each item i to maximize benefit w/ weight
at most Wfor each item i in S
xi 0vi bi / wi {value}
w 0{total weight}
while w < W remove item i w/ highest
vi
xi min{wi , W - w}
w w + min{wi , W - w}
Si
iiiSi
iii xwbwxb )/()/(
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Task Scheduling Given: a set T of n tasks, each having:
A start time, si
A finish time, fi (where si < fi) Goal: Perform all the tasks using a minimum
number of “machines.”
1 98765432
Machine 1
Machine 3
Machine 2
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Task Scheduling Algorithm
Greedy choice: consider tasks by their start time and use as few machines as possible with this order.
Run time: O(n log n). Why? Correctness: Suppose there is a
better schedule. We can use k-1 machines The algorithm uses k Let i be first task scheduled
on machine k Machine i must conflict with
k-1 other tasks But that means there is no
non-conflicting schedule using k-1 machines
Algorithm taskSchedule(T)Input: set T of tasks w/ start
time si and finish time fi
Output: non-conflicting schedule with minimum number of machines
m 0{no. of machines}
while T is not empty
remove task i w/ smallest si
if there’s a machine j for i then
schedule i on machine j
else m m + 1
schedule i on machine m
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Example Given: a set T of n tasks, each having:
A start time, si
A finish time, fi (where si < fi) [1,4], [1,3], [2,5], [3,7], [4,7], [6,9], [7,8] (ordered by
start) Goal: Perform all tasks on min. number of machines
1 98765432
Machine 1
Machine 3
Machine 2
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Divide-and-Conquer
7 2 9 4 2 4 7 9
7 2 2 7 9 4 4 9
7 7 2 2 9 9 4 4
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Outline and Reading Divide-and-conquer paradigm (§5.2) Review Merge-sort (§4.1.1) Recurrence Equations (§5.2.1)
Iterative substitution Recursion trees Guess-and-test The master method
Integer Multiplication (§5.2.2)
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Divide-and-Conquer
Divide-and conquer is a general algorithm design paradigm:
Divide: divide the input data S in two or more disjoint subsets S1, S2, …
Recur: solve the subproblems recursively
Conquer: combine the solutions for S1, S2, …, into a solution for S
The base case for the recursion are subproblems of constant size
Analysis can be done using recurrence equations
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Binary Search I have a number X between c1 and c2… Divide and conquer algorithm? Analysis:
Called a Recurrence Equation How to solve?
2if)2/(
2if )(
ncnT
nbnT
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Solving the Binary Search RecurrenceT(n) = T(n/2) + cT(n/2) = T(n/4) + cT(n/4) = T(n/8) + c …-----------T(n) = T(n/2) + c
= [T(n/4) + c] + c= T(n/22) + 2 c= T(n/23) + 3 c[telescope]= T(n/2k) + k c
done when k=lg n; then T(n/2k)=T(1) is base case
= b + c lg n
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Merge Sort Recurrence Equation Analysis Recurrence Equation?
We can analyze by finding a closed form solution.
Method: expand and telescope
2if)2/(2
2if )(
nbnnT
nbnT
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Solving the Merge Sort recurrenceT(n) = 2T(n/2) + b nT(n/2) = 2T(n/4) + b n/2T(n/4) = 2T(n/8) + b n/4 …-----------T(n) = 2T(n/2) + b n
= 2[2T(n/4) + b n/2] + b n= 22T(n/22) + 2 b n= 23T(n/23) + 3 b n[telescope]= 2kT(n/2k) + k b n
done when k=lg n; then T(n/2k)=T(1) is base case
= b n + b n lg n
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Min & Max Given a set of n items, find the min and max.
How many comparisons needed? Simple alg: n-1 comparisons to find min, n-1
for max Recursive algorithm:void minmax(int i, int j, int& min0, int& max0) {
if (j<=i+1) [handle base case; return] m=(i+j+1)/2; minmax(i,m-1,min1,max1); minmax(m,j,min2,max2); min0 = min(min1,min2); max0 = max(max1,max2);}
Recurrence:T(n)=2T(n/2) + 2; T(2)=1
Solution? 1.5 n - 2
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Solving the Min&Max RecurrenceT(n) = 2T(n/2) + 2; T(2)=1T(n/2) = 2T(n/4) + 2T(n/4) = 2T(n/8) + 2 …-----------T(n) = 2T(n/2) + 2
= 2[2T(n/4) + 2] + 2= 22T(n/22) + 22 + 2= 23T(n/23) + 23 + 22 + 2[telescope]= 2kT(n/2k) + = 2kT(n/2k) + 2k+1 - 2
done when k=(lg n) - 1; then T(n/2k)=T(2) is base case= n/2 + n – 2 = 1.5n - 2
k
i
i
1
2
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Hanoi runtime?void hanoi(n, from, to, spare { if (n > 0) { hanoi(n-1,from,spare,to); cout << from << “ – “ << to << endl; hanoi(n-1,spare,to,from); }}
T(n) = 2T(n-1) + cT(0) = b
Look it up: T(n) = O(2n)
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Master Method Another way to solve recurrences: “Look it
up” The Master Theorem solves many of them:If T(n)=c, n<d, and T(n)=aT(n/b)+f(n), n≥d, and for small constants e>0, k>=0, and d<1 then
If f(n) is O( ) then T(n) is Q( )
If f(n) is Q( ) then T(n) is Q( )
If f(n) is W( ) and a f(n/b) ≤ df(n), for n ≤ d, then T(n) is Q(f(n))
abnlog
abnlog
abnlog
nn kab 1log log nn kab loglog
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09-08-04
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More Divide and Conquer Examples pow(x,n) log(n,base) Point-in-convex-polygon Fractals
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09-08-04
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09-08-04
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http://www.3villagecsd.k12.ny.us/wmhs/Departments/Math/OBrien/fibonacci2.html
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http://www.religiousforums.com/forum/material-world/80450-fibonacci-sequence.html
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Creating Fractals Koch Snowflake Fractal Star Sierpinski Gasket Your own…
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Integer Multiplication Algorithm: Multiply two n-bit integers I and J.
Divide step: Split I and J into high-order and low-order bits
We can then define I*J by multiplying the parts and adding:
So, T(n) = 4T(n/2) + n, which implies T(n) is O(n2). But that is no better than the algorithm we learned in
grade school.
ln
h
ln
h
JJJ
III
2/
2/
2
2
lln
hln
lhn
hh
ln
hln
h
JIJIJIJI
JJIIJI
2/2/
2/2/
222
)2(*)2(*
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An Improved Integer Multiplication Algorithm Algorithm: Multiply two n-bit integers I
and J. Divide step: Split I and J into high-order and low-order
bits
Observe that there is a different way to multiply parts:
So, T(n) = 3T(n/2) + n, which implies T(n) is O(nlog2
3) [by master theorem].
Thus, T(n) is O(n1.585).
ln
h
ln
h
JJJ
III
2/
2/
2
2
lln
hllhn
hh
lln
llhhhlhhlllhn
hh
lln
llhhhllhn
hh
JIJIJIJI
JIJIJIJIJIJIJIJI
JIJIJIJJIIJIJI
2/
2/
2/
2)(2
2])[(2
2]))([(2*
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Practice Analyzing Recursive Functions1. int fun1(int n) { if (n==1) return 50; return fun1(n/2) * fun1(n/2) + 97; }
2. int fun2(int n) { if (n==1) return 1; int sum=0; for (int i=1; i<=n; i++) sum += fun(n-1); return sum; }
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Dynamic Programming
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Outline and Reading Fibonacci Making Change Matrix Chain-Product (§5.3.1) The General Technique (§5.3.2) 0-1 Knapsack Problem (§5.3.3)
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Opinions?public class Fib {
public static BigInteger fibonacci(long n) { if (n <= 2) return new BigInteger("1"); return fibonacci(n-1).add(fibonacci(n-2)); }
public static void main( String args[] ) { long n = Integer.parseInt(args[0]); System.out.println("Fib(" + n + ") = " +
fibonacci(n).toString()); }}
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Bottom up…public class Fib2 { static BigInteger[] fibs;
public static BigInteger fibonacci(int n) { BigInteger result; fibs[1]=new BigInteger("1"); fibs[2]=new BigInteger("1"); for (int i=3; i<=n; i++) fibs[i] = fibs[i-1].add(fibs[i-2]); return fibs[n]; }
public static void main( String args[] ) { int n = Integer.parseInt(args[0]); fibs = new BigInteger[n+1]; System.out.println("Fib(" + n + ") = " +
fibonacci(n).toString()); }}
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Memoize…public class Fib3 { static BigInteger[] fibs;
public static BigInteger fibonacci(int n) { long result; if (fibs[n] != null) return fibs[n]; else { fibs[n] = fibonacci(n-1).add(fibonacci(n-2)); return fibs[n]; } }
public static void main( String args[] ) { int n = Integer.parseInt(args[0]); fibs = new BigInteger [n+1]; fibs[1] = new BigInteger("1");
fibs[2] = new BigInteger("1"); System.out.println("Fib("+n+") = " + fibonacci(n));} }
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Dynamic Programming Applies to problems that can be solved
recursively, but the subproblems overlap
(Typically optimization problems) Method:
Write definition of value you want to compute
Express solution in terms of sub-problem solutions
Compute bottom-up or “memoize”
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Making Change Being good Computer Scientists, we
want to have a General Solution for making change in our vending machine.
Suppose that our coin values were 40, 30, 5, and 1 cent? How would we make change using the smallest number of coins?
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Making Change – DP solution Suppose we have coin values c1, c2, …, ck and we want to
make change for value V using the smallest number of coins?
What sub-problem solutions would give us solution to the whole problem?
Define NC(V) = smallest number of coins that will add up to V.
Base case? Run time?
Some kind of exponential, if you’re not careful… How can we implement this efficiently?
)(1)( 1 ini cVNCMINVNC
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Dynamic Programming: Two methods for efficiency Build a table bottom-up
Compute NC(1), NC(2), …, NC(V), storing sub-problem results in a table as you go up
Memoize Compute NC(V) recursively, but store new
sub-problem results when you first compute them
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Making change example
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
11
12
13
∞ ∞ ∞ ∞ ∞ ∞ 0 1 2 3 4 1 2 1 2 3
Suppose coin values are 1, 5, 7 and v=10
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Making change example
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
11
12
13
∞ ∞ ∞ ∞ ∞ ∞ 0 1 2 3 4 1 2 1 2 3
Suppose coin values are 1, 5, 7 and v=10
NC[10] = 1 + Min (NC[9], NC[5], NC[3]) Runtime?
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Forgot your hacksaw?
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The 0/1 Knapsack Problem
Given: A set S of n items, with each item i having bi - a positive benefit wi - a positive weight
Goal: Choose items with maximum total benefit but with weight at most W.
If we are not allowed to take fractional amounts, then this is the 0/1 knapsack problem.
In this case, we let T denote the set of items we take
Objective: maximize
Constraint:
Ti
ib
Ti
i Ww
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Given: A set S of n items, with each item i having bi - a positive benefit wi - a positive weight
Goal: Choose items with maximum total benefit but with weight at most W.
Example
Weight:Benefit:
1 2 3 4 5
4 in 2 in 2 in 6 in 2 in
$20 $3 $6 $25 $80
Items:
9 in
Solution:• 5 (2 in)• 3 (2 in)• 1 (4 in)
“knapsack”
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A 0/1 Knapsack Algorithm, First Attempt Sk: Set of items numbered 1 to k. Define B[k] = best selection from Sk. Problem: does not have subproblem optimality:
Consider S={(3,2),(5,4),(8,5),(4,3),(10,9)} weight-benefit pairs
Best for S4:
Best for S5:
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A 0/1 Knapsack Algorithm, Second Attempt Sk: Set of items numbered 1 to k. Define B[k,w] = best selection from Sk with
weight exactly equal to w Good news: this does have subproblem
optimality:
I.e., best subset of Sk with weight exactly w is either the best subset of Sk-1 w/ weight w or the best subset of Sk-1 w/ weight w-wk plus item k.
else}],1[],,1[max{
if],1[],[
kk
k
bwwkBwkB
wwwkBwkB
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(Weight,Value): (4,6) (3,8) (5,10) (8,13) (7,15)
B 1 2 3 4 5 6 7 8 9 10
11
12
13
14
15
16
17
18
19
20
S1
0 0 0 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6
S2
0 0 8 8 8 8 14
14
14
14
14
14
14
14
14
14
14
14
14
14
S3
0 0 8 8 10
10
14
S4
S5
else}],1[],,1[max{
if],1[],[
kk
k
bwwkBwkB
wwwkBwkB
How to tell which objects were used for each box?
Runtime?
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The 0/1 Knapsack Algorithm
Recall def. of B[k,w]:
Since B[k,w] is defined in terms of B[k-1,*], we can reuse the same array
Running time: O(nW). Not a polynomial-time
algorithm if W is large This is a pseudo-polynomial
time algorithm
Algorithm 01Knapsack(S, W):
Input: set S of items w/ benefit bi and
weight wi; max. weight WOutput: benefit of best subset with
weight at most W
for w 0 to W doB[w] 0
for k 1 to n do
for w W downto wk do
if B[w-wk]+bk > B[w] then
B[w] B[w-wk]+bk
else}],1[],,1[max{
if],1[],[
kk
k
bwwkBwkB
wwwkBwkB
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Matrix Chain Products
Matrix Chain Products Review: Matrix Multiplication.
C = A*B A is d × e and B is e × f
O(def ) time
A C
B
d d
f
e
f
e
i
j
i,j
1
0
],[*],[],[e
k
jkBkiAjiC
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Matrix Chain Products Matrix Chain Product:
Compute A=A0*A1*…*An-1
Ai is di × di+1 Problem: How to parenthesize?
Example B is 3 × 100 C is 100 × 5 D is 5 × 5 (B*C)*D takes 1500 + 75 = 1575 ops B*(C*D) takes 1500 + 2500 = 4000 ops
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An Enumeration Approach Matrix Chain-Product Algorithm:
Try all possible ways to parenthesize A=A0*A1*…*An-
1 Calculate number of ops for each one Pick the one that is best
Running time: The number of parethesizations is equal to the
number of binary trees with n nodes This is exponential! It is called the Catalan number, and it is almost 4n. This is a terrible algorithm!
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A Greedy Approach Idea #1: repeatedly select the product
that uses (up) the most operations. Counter-example:
A is 10 × 5 B is 5 × 10 C is 10 × 5 D is 5 × 10 Greedy idea #1 gives (A*B)*(C*D), which
takes 500+1000+500 = 2000 ops A*((B*C)*D) takes 500+250+250 = 1000 ops
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Another Greedy Approach Idea #2: repeatedly select the product that
uses the fewest operations. Counter-example:
A is 101 × 11 B is 11 × 9 C is 9 × 100 D is 100 × 99 Greedy idea #2 gives A*((B*C)*D)), which takes
109989+9900+108900=228789 ops (A*B)*(C*D) takes 9999+89991+89100=189090 ops
The greedy approach is not giving us the optimal value.
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A “Recursive” Approach Define subproblems:
Find the best parenthesization of Ai*Ai+1*…*Aj. Let Ni,j denote the number of operations done by this
subproblem. The optimal solution for the whole problem is N0,n-1.
Subproblem optimality: The optimal solution can be defined in terms of optimal subproblems
There has to be a final multiplication (root of the expression tree) for the optimal solution.
Say, the final multiply is at index i: (A0*…*Ai)*(Ai+1*…*An-1). Then the optimal solution N0,n-1 is the sum of two optimal
subproblems, N0,i and Ni+1,n-1 plus the time for the last multiply.
If the global optimum did not have these optimal subproblems, we could define an even better “optimal” solution.
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A Characterizing Equation
The global optimal has to be defined in terms of optimal subproblems, depending on where the final multiply is at.
Let us consider all possible places for that final multiply: Recall that Ai is a di × di+1 dimensional matrix. So, a characterizing equation for Ni,j is the following:
Note that subproblems are not independent--the subproblems overlap.
}{min 11,1,, jkijkki
jkiji dddNNN
}{min 11,1,, jkijkki
jkiji dddNNN
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A Dynamic Programming Algorithm
Since subproblems overlap, we don’t use recursion.
Instead, we construct optimal subproblems “bottom-up.”
Ni,i’s are easy, so start with them
Then do length 2,3,… subproblems, and so on.
Running time: O(n3)
Algorithm matrixChain(S):Input: sequence S of n matrices to be
multipliedOutput: number of operations in an optimal
paranethization of S
for i 1 to n-1 do
Ni,i 0 for b 1 to n-1 do
for i 0 to n-b-1 doj i+b
Ni,j +infinityfor k i to j-1 do
Ni,j min{Ni,j , Ni,k +Nk+1,j +di dk+1 dj+1}
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A Dynamic Programming Algorithm Visualization
The bottom-up construction fills in the N array by diagonals
Ni,j gets values from pervious entries in i-th row and j-th column
Filling in each entry in the N table takes O(n) time.
Total run time: O(n3) Getting actual
parenthesization can be done by remembering “k” for each N entry
answerN 0 1
0
1
2 …
n-1
…
n-1j
i
}{min 11,1,, jkijkki
jkiji dddNNN
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The General Dynamic Programming Technique
Applies to a problem that at first seems to require a lot of time (possibly exponential), provided we have:
Simple subproblems: the subproblems can be defined in terms of a few variables, such as j, k, l, m, and so on.
Subproblem optimality: the global optimum value can be defined in terms of optimal subproblems
Subproblem overlap: the subproblems are not independent, but instead they overlap (hence, should be constructed bottom-up).
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Parallel Decomposition Strategies How would you parallelize a divide-and-
conquer algorithm? Differences for dynamic programming
algorithms? Alternate approaches to parallelizing? Decomposition strategies:
Algorithm decomposition Data decomposition