Dynamic Programming Part One

1
Dynamic ProgrammingPart One

• HKOI Training Team 2004

2
Recurrences

• Function defined in terms of itself
• M(X) X M(X-1)
• Fn2 Fn1 Fn
• Kn1 maxKn-1100, Kn
• One or more base cases are needed
• M(0) 1
• F0 1, F1 1
• K10 100, K11 177

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Evaluating recurrences

• Combination
• C(n, r) C(n-1, r) C(n-1, r-1)
• C(n, 0) C(n, n) 1
• Algorithm
• function C(n, r)
• if (r0) or (rn)
• return 1
• else
• return C(n-1, r) C(n-1, r-1)

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Slow...
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A table...
C(5, 2)
C(4, 2) C(4, 1)
C(3, 2) C(3, 1) C(3, 0)
C(2, 2) C(2, 1) C(2, 0)
C(1, 1) C(1, 0)
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Redundant calculations
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Fast...
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Dynamic programming

• Abbreviation DP / DyP
• Not particularly related to programming
• As in Linear Programming!
• An approach (paradigm) for solving certain kinds
  of problems
• Not a particular algorithm
• Usually applied on recurrences

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Principles of DP

• Evaluating recurrences may sometimes be slow
  (exponential time)
• Accelerate by avoiding redundant calculations!
• Memorize previously calculated values
• Usually applied to problems with
• a recurrence relation
• overlapping subproblems
• solutions exhibiting optimal substructure

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Why DP?

• Reduce runtime
• In the previous computation of C(n, k), how many
  times is the function invoked?
• Slow algorithm 2C(n,k) 1 exponential in n
• Fast algorithm O(nk) polynomial in n
• Tradeoff memory
• In the C(n, k) problem, the memory requirement is
  O(n) using pure recursion O(nk) is required for
  memorization
• In fact, the O(nk) bound can be improved

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Classification of problems

• Induction
• Evaluation of a certain function is the main
  concern
• Examples
• The N-th Fibonacci number
• Number of different binary trees with N nodes
• Optimization
• Maximization or minimization of certain function
  is the objective
• Examples
• Shortest paths
• Activity scheduling (maximize no. of activities)

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Triangle (IOI 94)

• Given a triangle with N levels like the one on
  the left, find a path with maximum sum from the
  top to the bottom
• Only the sum is required

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Triangle (analysis)

• Exhaustion?
• How many paths are there in total?
• Greedy?
• It doesnt work. Why?
• Graph problem?
• Possible, but not simple enough

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Triangle (formulation)

• Let Aij denote the number in the i-th row and
  j-th column, (i, j)
• Let Fij denote the sum of the numbers on a
  path with max sum from (1, 1) to (i, j)
• Answer maximum of FN1, FN2, , FNN

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Triangle (formulation)

• Base case F11 A11
• Progress (i gt 1, 1 lt j lt i)
• Fi1 Fi-11Ai1
• Fii Fi-1i-1Aii
• Fij maxFi-1j-1,Fi-1jAij

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Triangle (order of calc.)

• Fi depends on Fi-1
• Compute F row by row, from top to bottom

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Triangle (algorithm)

• Algorithm
• F11 ? A11
• for i ? 2 to N do
• Fi1 Fi-11Ai1
• Fii Fi-1i-1Aii
• for j ? 2 to i-1 do
• Fij ? maxFij,Fij-1
• Aij
• answer ? maxFN1, , FNN

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Triangle (complexity)

• Number of array entries to be computed about
  N(N-1)/2
• Time for computing one entry O(1)
• Thus the total time complexity is O(N2)
• Memory complexity O(N2)
• This can be reduced to O(N) since Fi only
  depends on Fi-1
• We can discard some previous entries after
  starting on a new row

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Longest Common Subsequence

• Given two strings A and B of length n and m
  respectively, find their longest common
  subsequence (NOT substring)
• Example
• A aabcaabcaadyyyefg
• B cdfehgjaefazadxex
• LCS caaade
• Explanation
• A a a b c a a b c a a d y y y e f g
• B c d f e h g j a e f a z a d x e x

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LCS (analysis)

• Exhaustion
• Number of subsequences of A 2n
• Number of subsequences of B 2m
• Exponential time!
• Observation
• If a LCS of A and B is nonempty then it must have
  a last character (trivial)

A a a b c a a b c a a d y y y e f g B c d f e
h g j a e f a z a d x e x
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LCS (analysis)

• Observation
• Suppose the last LCS character corresponds to
  Ai and Bj
• Removing the last LCS character results in a LCS
  of A1..i-1 and B1..j-1

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LCS (formulation)

• Thus it is reasonable to define Fij as the
  length of any LCS of A1..i and B1..j
• Base cases
• Fi0 0 for all i
• F0j 0 for all j
• Progress (0 lt i lt n, 0 lt j lt m)
• Fij Fi-1j-1 1 if AiBj
• Fij maxFi-1j, Fij-1 else

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LCS (order of calc.)

• Fij depends on Fab, a i and b j
• Order from top to bottom, from left to right
• There are some other possible orders!

0 1 2 3 4
0
1
2
3
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LCS (complexity)

• Time complexity O(NM)
• Memory complexity O(NM)
• Again, this can be reduced to O(NM), but why and
  how?

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Coins

• Given an unlimited supply of 2, 3 and 5 coins,
  find the number of different ways to form a
  denomination of N
• Example
• N 10
• 2, 2, 2, 2, 2, 2, 2, 3, 3, 2, 3, 5,
• 5, 5
• Solve this induction problem by yourself

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Matrix Chain Multiplication

• A matrix is a rectangle of numbers
• The number of usual multiplications needed to
  multiply a nm matrix and a mr matrix is nmr
• Matrix multiplication is associative
• (AB)C A(BC)

A matrix of size 23
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Matrix Chain Multiplication

• Given a sequence of matrices A1, A2, , An, find
  an order of multiplications such that the total
  number of usual multiplications is minimized
• Example
• A 37, B 75, C 52
• (AB)C takes 375352 135 muls
• A(BC) takes 752372 112 muls

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MCM (analysis)

• How many different orders of multiplications are
  there?
• This can be solved by DP!
• Does greedy work?
• Lets have a look at the brackets
• (A((BC)((DE)F)))(G((HI)J))
• There must be one last multiplication
• (A((BC)((DE)F)))(G((HI)J))
• Every subexpression with more than one matrix has
  one last multiplication

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MCM (analysis)

• Optimal substructure
• If the last multiplication is (A1A2Ai)(Ai1AN)
  and the multiplication scheme is optimal, then
  the multiplication schemes for A1A2Ai and
  Ai1AN must also be optimal

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MCM (formulation)

• Let ri and ci be the number of rows and the
  number of columns of Ai respectively
• Let Fij be the minimum number of usual
  multiplications required to compute AiAi1Aj
• Base case Fii 0 for all i
• Progress for all i lt j
• Fij minFikFk1j
• rickcj

i k lt j
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MCM (order of calc.)

• F with fewer matrices first

j, i 1 2 3 4 5 6
1
2
3
4
5
6
first
last
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MCM (alternative)

• Let Gij be the minimum number of usual
  multiplications required to compute AiAi1Aij-1
• Base case Gi1 0 for all i
• Progress for all 1 lt j lt N-i
• Gij minGikGikj-k
• ririkcij-1
• Explain!!

1 k lt j
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MCM (alternative)

• Still, G with fewer matrices first

i, j 1 2 3 4 5 6
1
2
3
4
5
6
last
first
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MCM (algorithm)

• Bottom-up algorithm (algorithm 2)
• for i ? 1 to N do
• Gi1 ? 0
• for j ? 2 to N do
• for i ? 1 to N-j1
• Gij ? minGik
• Gikj-k
• ririkcij-1
• answer ? G1N

1 k lt j
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MCM algorithm)

• Top-down algorithm (algorithm 1)
• function MCM_DP(i, j)
• if Fij is not 8, return Fij
• m ? 8
• for k ? i to j-1 do
• m ? minm,
• MCM_DP(i, k)
• MCM_DP(k1, j)
• rickcj
• Fij ? m
• return Fij

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MCM (algorithm)

• Top-down algorithm (main)
• initialize everything in F to 8
• for i ? 1 to N do
• Fii ? 0
• answer ? MCM_DP(1, N)

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MCM (complexity)

• Number of array entries to be computed about
  N(N-1)/2
• Number of references to other entries when
  computing one single entry
• It varies
• On the average, about N/2
• Thus the total time complexity is O(N3)

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Fishing

• There are N fish ponds and you are going to spend
  M minutes on fishing
• Given the time-reward relationship of each pond,
  determine the time you should spend at each pond
  in order to get the biggest reward

time/pond 1 2 3
1 minute 0 fish 2 fish 1 fish
2 minutes 3 fish 2 fish 2 fish
3 minutes 3 fish 4 fish 4 fish
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Fishing (example)

• For example, if N3, M3 and the relationships
  are given in the previous slide, then the optimal
  schedule is
• Pond 1 2 minutes
• Pond 2 1 minute
• Pond 3 0 minute
• Reward 5 fish

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Fishing (analysis)

• You can think of yourself visiting ponds 1, 2, 3,
  , N in order
• Why?
• Suppose in an optimal schedule you spend K
  minutes on fishing at pond 1
• So you have M-K minutes to spend at the remaining
  N-1 ponds
• The problem is reduced
• But how can I know what is K?
• You dont know, so try all possible values!

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Fishing (formulation)

• Let Fij be the maximum reward you can get by
  spending j minutes at the first i ponds
• Base cases 0 i, j N
• Fi0 0
• F0j 0
• Progress 1 i N, 1 j M
• Fij maxFi-1kRij-k

0 k j
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Brackets

• A balanced-bracket expression (BBE) is defined as
  follows
• The empty string is a BBE
• If X and Y are BBEs then (X) and XY BBEs
• Nothing else is a BBE
• For example, (), (()), ()(()())(()) are BBEs
  while ((), )(, (()()))()() are not BBEs
• Find the number of different BBEs with exactly N
  pairs of brackets

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Brackets (analysis)

• Obviously this is an induction problem
• Listing out all BBEs with N pairs of brackets is
  not a good idea
• What makes a bracket expression not a BBE?
• the numbers of opening and closing brackets do
  not match
• the number of closing brackets exceeds the number
  of opening brackets at some point during a scan
  from left to right

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Brackets (analysis)

• Clearly only the second rule matters in this
  problem
• Intuitively it is easy to construct a BBE from
  left to right
• We call a bracket expression a BBE-prefix (BP) if
  the number of closing brackets never exceeds the
  number of opening brackets during a scan from
  left to right

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Brackets (formulation)

• Let Fij be the number of BPs with i opening
  brackets and j closing brackets (thus of length
  ij)
• Base cases
• F00 1
• Fij 0 for all 0 i lt j N
• Progress (0 lt j i N)
• Fij Fi-1jFij-1

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Polygon Triangulation

• Given an N-sided convex polygon A, find a
  triangulation scheme with minimum total cut length

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Polygon (analysis)

• Every edge of A belongs to exactly one triangle
  resulting from the triangulation

• We get two (or one) smaller polygons after
  deleting a triangle

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Polygon (analysis)

• The order of cutting does not matter
• Optimal substructure
• If the cutting scheme for A is optimal, then the
  cutting schemes for B and C must also be optimal

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Polygon (formulation)

• Take this problem as an exercise
• A small hint similar to Matrix Chain
  Multiplication
• Nothing more

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Summary

• Remember, DP is just a technique, not a
  particular algorithm
• The problems we have discussed are quite
  straightforward that you should be able to know
  they can be solved by DP with little inspection
• The DP problems in NOI and IOI are much harder
• They are well disguised
• Looking at a wide variety of DP problems seems to
  be the only way to master DP

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Looking forward

• Hopefully the trainer responsible for Advanced DP
  II will talk about DP on non-rectangular
  structures, such as trees and graphs
• The problems discussed will be more complicated
  than those you have just seen, so please be
  prepared

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The end
The last thing
LETS HAVE LUNCH!!!