Lecture2.pdf

Competitive Programming-Lecture 2

Programming Club

IIT Kanpur

June 8, 2012

Programming Club (IIT Kanpur) Competitive Programming-Lecture 2 June 8, 2012 1 / 11

Divide and Conquer

Integer Multiplication

Multiply two numbers of x and y of n bits

Bitwise Multiplication: O(n2)

x=x1*2n/2+x0

y=y1*2n/2+y0

xy=x1*y1*2n+(x1*y0+y1*x0)*2

n/2+x0*y0

Compute 4 products:

T (n) = 4T (n/2) + O(n) = T (n) = O(n2)Compute three different products: x1*y1, x0*y0, (x1+x0).(y1+y0)

T(n)=3T(n/2)+O(n)T(n)=O(n1.59)

Divide and Conquer

x=x1*2n/2+x0

y=y1*2n/2+y0

xy=x1*y1*2n+(x1*y0+y1*x0)*2

n/2+x0*y0

Compute 4 products:

T(n)=3T(n/2)+O(n)T(n)=O(n1.59)

Divide and Conquer

x=x1*2n/2+x0

y=y1*2n/2+y0

xy=x1*y1*2n+(x1*y0+y1*x0)*2

n/2+x0*y0

Compute 4 products:

T(n)=3T(n/2)+O(n)T(n)=O(n1.59)

Divide and Conquer

x=x1*2n/2+x0

y=y1*2n/2+y0

xy=x1*y1*2n+(x1*y0+y1*x0)*2

n/2+x0*y0

Compute 4 products:

T(n)=3T(n/2)+O(n)T(n)=O(n1.59)

Divide and Conquer

x=x1*2n/2+x0

y=y1*2n/2+y0

xy=x1*y1*2n+(x1*y0+y1*x0)*2

n/2+x0*y0

Compute 4 products:

T(n)=3T(n/2)+O(n)T(n)=O(n1.59)

Dynamic Programming

DP solves problems by combining the solutions to overlappingsubproblems

Solves each subproblem just once and stores its result, therebyavoiding recomputing

Dynamic Programming

Traversing a matrix

Problem: Given a matrix with the cost of visiting a node and theconstraint that from (i,j), you can go to (i+1,j), (i+1,j-1) and(i+1,j+1)

Dynamic Programming

Coin Change

Problem: Given coins of certain denomination and infinite coins ofeach denominations, find the minimum number of coins such thatthey sum to N rupees

Dynamic Programming

Let the coins denomination be c1, c2,. . . , ck . Let C[i] be theminimum number of coins to exchange i rupees

Optimal Substructure

C[i]=1+minj{C[i-cj ]}

Dynamic Programming

Coin Change

Problem: Given coins of certain denomination and infinite coins ofeach denominations, find the minimum number of coins such thatthey sum to N rupees

Dynamic Programming

Let the coins denomination be c1, c2,. . . , ck . Let C[i] be theminimum number of coins to exchange i rupees

Optimal Substructure

C[i]=1+minj{C[i-cj ]}

Dynamic Programming

Longest Common Subsequence

Problem: Given two sequences, find the longest subsequence commonto both

opt(i,j) = length of subsequence in X[0.1. . . i] and Y[0.1. . . j]

opt(i,j) = 0 if i=0 or j=0

opt(i,j) = opt(i-1,j-1)+1 if xi=yj i , j > 0

opt(i,j) = Max[opt(i,j-1), opt(i-i,j)] where i , j > 0 xi 6= yj

Dynamic Programming

Matrix Chain Multiplication

Problem: To find an optimal bracketing so as to minimize the totalnumber of multiplication steps

opt(i,j): number of multiplications between matrices Ai and A jSubstructure

opt(i,j) = Min{opt(i,k)+opt(k+1,j)+mi .mk+1.mj+1}where k=i+1 to j-1

Base Case: opt(i,i)=0

Dynamic Programming

Matrix Chain Multiplication

Problem: To find an optimal bracketing so as to minimize the totalnumber of multiplication steps

opt(i,j): number of multiplications between matrices Ai and A jSubstructure

opt(i,j) = Min{opt(i,k)+opt(k+1,j)+mi .mk+1.mj+1}where k=i+1 to j-1

Base Case: opt(i,i)=0

Dynamic Programming

Longest Non-Decreasing Subsequence

Problem: Given an array, find the subsequence which isnon-decreasign and of maximum length

L[i]: Length of maximum subsequence ending at index i

Substructure

L[i]=1+Max{L[j] st. i > j and A[i ] A[j ]}

userSticky Note

Dynamic Programming

Longest Non-Decreasing Subsequence

Problem: Given an array, find the subsequence which isnon-decreasign and of maximum length

L[i]: Length of maximum subsequence ending at index i

Substructure

L[i]=1+Max{L[j] st. i > j and A[i ] A[j ]}

Dynamic Programming

Subsets of an array

Problem: Given 2 integers a,b and a set S, calculate #X such thata < X < b and X is not divisible by all elements of S

Dynamic Programming

Task Scheduling

Problem: To find an optimum scheduling of overlapping tasks withdifferent starting and ending time : tasks have no weight

Sort the events in the order of finishing time

Greedy Approach

Keep picking tasks compatible to previous one in successionThe greedy strategy stays ahead

Dynamic Approach

Substructure

opt(j)=Max{opt(p(j)), opt(j-1)}// where p(j) is the next compatible event with j

Dynamic Programming

Task Scheduling

Greedy Approach

Dynamic Approach

Substructure

Dynamic Programming

Task Scheduling

Greedy Approach

Dynamic Approach

Substructure

Dynamic Programming

Integer (0/1) Knapsack Problem

Problem: Given n items of sizes si and values vi and a knapsack ofcapacity C, find the maximum value of items that can be filled in theknapsack

M(i,j) = optimal value for filling a capacity j knapsack with somesubset of items 1. . . i

Substructure

M(i,j)=Max{M(i-1,j),M(i-1,j-si )+vi}

Dynamic Programming

Integer (0/1) Knapsack Problem

Problem: Given n items of sizes si and values vi and a knapsack ofcapacity C, find the maximum value of items that can be filled in theknapsack

M(i,j) = optimal value for filling a capacity j knapsack with somesubset of items 1. . . i

Substructure

M(i,j)=Max{M(i-1,j),M(i-1,j-si )+vi}

Divide and ConquerDynamic Programming

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