Week-6 (Matrix Chain Order / LCS)

CE100 Algorithms and Programming II

Week-6 (Matrix Chain Order / LCS)

Spring Semester, 2021-2022

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Matrix Chain Order / Longest Common Subsequence

Outline

• Elements of Dynamic Programming
• Optimal Substructure
• Overlapping Subproblems

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• Recursive Matrix Chain Order Memoization
• Top-Down Approach
• RMC
• MemoizedMatrixChain
  • LookupC
• Dynamic Programming vs Memoization Summary

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• Dynamic Programming
• Problem-2 : Longest Common Subsequence
  • Definitions
  • LCS Problem
  • Notations
  • Optimal Substructure of LCS
  • Proof Case-1
  • Proof Case-2
  • Proof Case-3

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• A recursive solution to subproblems (inefficient)
• Computing the length of and LCS
• LCS Data Structure for DP
• Bottom-Up Computation
• Constructing and LCS
• PRINT-LCS
• Back-pointer space optimization for LCS length

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• Most Common Dynamic Programming Interview Questions

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Elements of Dynamic Programming

• When should we look for a DP solution to an optimization problem?
• Two key ingredients for the problem
• Optimal substructure
• Overlapping subproblems

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DP Hallmark #1

• Optimal Substructure
• A problem exhibits optimal substructure
  • if an optimal solution to a problem contains within it optimal solutions to subproblems
• Example: matrix-chain-multiplication
  • Optimal parenthesization of \(A_1 A_2 \dots A_n\) that splits the product between \(A_k\) and \(A_{k+1}\), contains within it optimal soln’s to the problems of parenthesizing \(A_1A_2 \dots A_k\) and \(A_{k+1} A_{k+2} \dots A_n\)

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Optimal Substructure

• Finding a suitable space of subproblems
• Iterate on subproblem instances
• Example: matrix-chain-multiplication
  • Iterate and look at the structure of optimal soln’s to subproblems, sub-subproblems, and so forth
  • Discover that all subproblems consists of subchains of \(\langle A_1, A_2, \dots , A_n \rangle\)
  • Thus, the set of chains of the form \(\langle A_i,A_{i+1}, \dots , A_j \rangle\) for \(1 \leq i \leq j \leq n\)
  • Makes a natural and reasonable space of subproblems

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DP Hallmark #2

• Overlapping Subproblems
• Total number of distinct subproblems should be polynomial in the input size
• When a recursive algorithm revisits the same problem over and over again,
  • We say that the optimization problem has overlapping subproblems

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Overlapping Subproblems

• DP algorithms typically take advantage of overlapping subproblems
• by solving each problem once
• then storing the solutions in a table
  • where it can be looked up when needed
• using constant time per lookup

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Overlapping Subproblems

• Recursive matrix-chain order

\[ \begin{align*} & \text{RMC}(p, i, j) \{ \\[5 pt] & \quad \text{if} \ i = j \ \text{then} \\ & \qquad \text{return} \ 0 \\[5 pt] & \quad m[i, j] \leftarrow \infty \\[5 pt] & \quad \text{for} \ k \leftarrow i \text{to} \ j - 1 \ \text{do} \\ & \qquad q \leftarrow \text{RMC}(p, i, k) + \text{RMC}(p, k+1, j) + p_{i-1} p_k p_j \\[5 pt] & \quad if \ q < m[i, j] \ \text{then} \\ & \qquad m[i, j] \leftarrow q \\[5 pt] & \quad \text{return} \ m[i, j] \ \} \end{align*} \]

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Direct Recursion: Inefficient!

• Recursion tree for \(RMC(p,1,4)\)
• Nodes are labeled with \(i\) and \(j\) values

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Running Time of RMC

\(T(1) \geq 1\) \(T(n) \geq 1 + \sum \limits_{k=1}^{n-1} (T(k)+T(n-k)+1)\ \text{for} \ n>1\)

• For \(i =1,2, \dots ,n\) each term \(T(i)\) appears twice
• Once as \(T(k)\), and once as \(T(n-k)\)
• Collect \(n-1,\) \(1\)’s in the summation together with the front \(1\)

\[ \begin{align*} T(n) \geq 2 \sum \limits_{i=1}^{n-1}T(i)+n \end{align*} \]

• Prove that \(T(n)= \Omega(2n)\) using the substitution method

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Running Time of RMC: Prove that \(T(n)= \Omega(2n)\)

• Try to show that \(T(n) \geq 2^{n-1}\) (by substitution)
• Base case: \(T(1) \geq 1 = 2^0 = 2^{1-1}\) for \(n=1\)
• Ind. Hyp.:

\[ \begin{align*} T(i) & \geq 2^{i-1} \ \text{for all} \ i=1, 2, \dots, n-1 \ \text{and} \ n \geq 2 \\ T(n) & \geq 2 \sum \limits_{i=1}^{n-1}2^{i-1} + n \\[15 pt] & = 2 \sum \limits_{i=1}^{n-1} 2^{i-1} + n \\ & = 2(2^{n-1}-1) + n \\ & = 2^{n-1} + (2^{n-1} - 2 + n) \\ & \Rightarrow T(n) \geq 2^{n-1} \ \text{ Q.E.D.} \end{align*} \]

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Running Time of RMC: \(T(n) \geq 2^{n-1}\)

• Whenever
• a recursion tree for the natural recursive solution to a problem contains the same subproblem repeatedly
• the total number of different subproblems is small
  • it is a good idea to see if \(DP (Dynamic \ Programming)\) can be applied

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Memoization

• Offers the efficiency of the usual \(DP\) approach while maintaining top-down strategy
• Idea is to memoize the natural, but inefficient, recursive algorithm

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Memoized Recursive Algorithm

• Maintains an entry in a table for the soln to each subproblem
• Each table entry contains a special value to indicate that the entry has yet to be filled in
• When the subproblem is first encountered its solution is computed and then stored in the table
• Each subsequent time that the subproblem encountered the value stored in the table is simply looked up and returned

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Memoized Recursive Matrix-chain Order

• Shaded subtrees are looked-up rather than recomputing

\[ \begin{align*} \begin{aligned} & \text{MemoizedMatrixChain(p)} \\ & \quad n \leftarrow length[p] - 1 \\ & \quad \text{for} \ i \leftarrow 1 \ \text{to} \ n \ \text{do} \\ & \qquad \text{for} \ j \leftarrow 1 \ \text{to} \ n \ \text{do} \\ & \qquad \quad m[i, j] \leftarrow \infty \\ & \quad \text{return} \ \text{LookupC}(p, 1, n) \Longrightarrow \end{aligned} \begin{aligned} & \Longrightarrow \text{LookupC}(p, i, j) \\ & \quad \text{if} \ m[i, j] = \infty \ \text{then} \\ & \qquad \text{if} \ i = j \ \text{then} \\ & \qquad \quad m[i, j] \leftarrow 0 \\ & \qquad \text{else} \\ & \qquad \quad \text{for} \ k \leftarrow i \ \text{to} \ j-1 \ \text{do} \\ & \qquad \quad \quad q \leftarrow \text{LookupC}(p, i, k) + \text{LookupC}(p, k+1, j) + p_{i-1} p_k p_j \\ & \qquad \quad \quad \text{if} \ q < m[i, j] \ \text{then} \\ & \qquad \quad \quad \quad m[i, j] \leftarrow q \\ & \quad \text{return} \ m[i, j] \end{aligned} \end{align*} \]

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Memoized Recursive Algorithm

• The approach assumes that
• The set of all possible subproblem parameters are known
• The relation between the table positions and subproblems is established
• Another approach is to memoize
• by using hashing with subproblem parameters as key

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Dynamic Programming vs Memoization Summary (1)

• Matrix-chain multiplication can be solved in \(O(n^3)\) time
• by either a top-down memoized recursive algorithm
• or a bottom-up dynamic programming algorithm
• Both methods exploit the overlapping subproblems property
• There are only \(\Theta(n^2)\) different subproblems in total
• Both methods compute the soln to each problem once
• Without memoization the natural recursive algorithm runs in exponential time since subproblems are solved repeatedly

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Dynamic Programming vs Memoization Summary (2)

• In general practice
• If all subproblems must be solved at once
  • a bottom-up DP algorithm always outperforms a top-down memoized algorithm by a constant factor
• because, bottom-up DP algorithm
  • Has no overhead for recursion
  • Less overhead for maintaining the table
• DP: Regular pattern of table accesses can be exploited to reduce the time and/or space requirements even further
• Memoized: If some problems need not be solved at all, it has the advantage of avoiding solutions to those subproblems

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Problem 3: Longest Common Subsequence

Definitions

• A subsequence of a given sequence is just the given sequence with some elements (possibly none) left out

• Example:

• \(X = \langle A, B, C, B, D, A, B \rangle\)

• \(Z = \langle B, C, D, B \rangle\)
  • \(Z\) is a subsequence of \(X\)

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Problem 3: Longest Common Subsequence

Definitions

• Formal definition: Given a sequence \(X = \langle x_1, x_2, \dots , x_m \rangle\), sequence \(Z = \langle z_1, z_2, \dots, z_k \rangle\) is a subsequence of \(X\)

• if \(\exists\) a strictly increasing sequence \(\langle i_1, i_2,\dots, i_k \rangle\) of indices of \(X\) such that \(x_{i_j} = z_j\) for all \(j = 1, 2, …, k\), where \(1 \leq k \leq m\)

• Example: \(Z = \langle B,C,D,B \rangle\) is a subsequence of \(X = \langle A,B,C,B,D,A,B \rangle\) with the index sequence \(\langle i_1, i_2, i_3, i_4 \rangle = \langle 2, 3, 5, 7 \rangle\)

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Problem 3: Longest Common Subsequence

Definitions

• If \(Z\) is a subsequence of both \(X\) and \(Y\), we denote \(Z\) as a common subsequence of \(X\) and \(Y\).

• Example:

\[ \begin{align*} X &= \langle A,B^*,C^*,B,D,A^*,B \rangle \\ Y &= \langle B^*,D,C^*,A^*,B,A \rangle \end{align*} \]

• \(Z_1 = \langle B^*, C^*, A^* \rangle\) is a common subsequence (of length 3) of \(X\) and \(Y\).
• Two longest common subsequence (LCSs) of \(X\) and \(Y\)?
• \(Z2 = \langle B, C, B, A \rangle\) of length \(4\)
• \(Z3 = \langle B, D, A, B \rangle\) of length \(4\)
  • The optimal solution value = 4

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Longest Common Subsequence (LCS) Problem

• LCS problem: Given two sequences
• \(X = \langle x_1, x_2, \dots, x_m \rangle\) and
• \(Y = \langle y_1, y_2, \dots , y_n \rangle\), find the LCS of \(X \& Y\)
• Brute force approach:
• Enumerate all subsequences of \(X\)
• Check if each subsequence is also a subsequence of \(Y\)
• Keep track of the LCS
• What is the complexity?
• There are \(2^m\) subsequences of \(X\)
  • Exponential runtime

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Notation

• Notation: Let \(X_i\) denote the \(i^{th}\) prefix of \(X\)
• i.e. \(X_i = \langle x_1, x_2, \dots, x_i \rangle\)
• Example:

\[ \begin{align*} X &= \langle A, B, C, B, D, A, B \rangle \\[10 pt] X_4 &= \langle A, B, C, B \rangle \\ X_0 &= \langle \rangle \end{align*} \]

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Optimal Substructure of an LCS

• Let \(X = <x1, x2, …, xm>\) and \(Y = \langle y_1, y_2, \dots, y_n \rangle\) are given
• Let \(Z = \langle z_1, z_2, \dots, z_k \rangle\) be an LCS of \(X\) and \(Y\)

• Question 1: If \(x_m = y_n\), how to define the optimal substructure?
• We must have \(z_k = x_m = y_n\) and
• \(Z_{k-1} = \text{LCS}(X_{m-1}, Y_{n-1})\)

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Optimal Substructure of an LCS

• Let \(X = <x1, x2, …, xm>\) and \(Y = \langle y_1, y_2, \dots, y_n \rangle\) are given
• Let \(Z = \langle z_1, z_2, \dots, z_k \rangle\) be an LCS of \(X\) and \(Y\)

• Question 2: If \(x_m \neq y_n \ \text{and} \ z_k \neq x_m\), how to define the optimal substructure?
• We must have \(Z = \text{LCS}(X_{m-1}, Y)\)

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Optimal Substructure of an LCS

• Let \(X = <x1, x2, …, xm>\) and \(Y = \langle y_1, y_2, \dots, y_n \rangle\) are given
• Let \(Z = \langle z_1, z_2, \dots, z_k \rangle\) be an LCS of \(X\) and \(Y\)

• Question 3: If \(x_m \neq y_n \ \text{and} \ z_k \neq y_n\), how to define the optimal substructure?
• We must have \(Z = \text{LCS}(X, Y_{n-1})\)

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Theorem: Optimal Substructure of an LCS

• Let \(X = \langle x_1, x_2, \dots, x_m \rangle\) and Y = are given
• Let \(Z = \langle z_1, z_2, \dots, z_k \rangle\) be an LCS of \(X\) and \(Y\)
• Theorem: Optimal substructure of an LCS:
• If \(x_m = y_n\)
  • then \(z_k = x_m = y_n\) and \(Z_{k-1}\) is an LCS of \(X_{m-1}\) and \(Y_{n-1}\)
• If \(x_m \neq y_n\) and \(z_k \neq x_m\)
  • then \(Z\) is an LCS of \(X_{m-1}\) and \(Y\)
• If \(x_m \neq y_n\) and \(z_k \neq y_n\)
  • then \(Z\) is an LCS of \(X\) and \(Y_{n-1}\)

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Optimal Substructure Theorem (case 1)

• If \(x_m = y_n\) then \(z_k = x_m = y_n\) and \(Z_{k-1}\) is an LCS of \(X_{m-1}\) and \(Y_{n-1}\)

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Optimal Substructure Theorem (case 2)

• If \(x_m \neq y_n\) and \(z_k \neq x_m\) then \(Z\) is an LCS of \(X_{m-1}\) and \(Y\)

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Optimal Substructure Theorem (case 3)

• If \(x_m \neq y_n\) and \(z_k \neq y_n\) then \(Z\) is an LCS of \(X\) and \(Y_{n-1}\)

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Proof of Optimal Substructure Theorem (case 1)

• If \(x_m = y_n\) then \(z_k = x_m = y_n\) and \(Z_{k-1}\) is an LCS of \(X_{m-1}\) and \(Y_{n-1}\)
• Proof: If \(z_k \neq x_m = y_n\) then
• we can append \(x_m = y_n\) to \(Z\) to obtain a common subsequence of length \(k+1 \Longrightarrow\) contradiction
• Thus, we must have \(z_k = x_m = y_n\)
• Hence, the prefix \(Z_{k-1}\) is a length-(\(k-1\)) CS of \(X_{m-1}\) and \(Y_{n-1}\)
• We have to show that \(Z_{k-1}\) is in fact an LCS of \(X_{m-1}\) and \(Y_{n-1}\)
• Proof by contradiction:
• Assume that \(\exists\) a CS \(W\) of \(X_{m-1}\) and \(Y_{n-1}\) with \(|W| = k\)
• Then appending \(x_m = y_n\) to \(W\) produces a CS of length \(k+1\)

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Proof of Optimal Substructure Theorem (case 2)

• If \(x_m \neq y_n\) and \(z_k \neq x_m\) then \(Z\) is an LCS of \(X_{m-1}\) and \(Y\)
• Proof : If \(z_k \neq x_m\) then \(Z\) is a CS of \(X_{m-1}\) and \(Y_n\)
• We have to show that \(Z\) is in fact an LCS of \(X_{m-1}\) and \(Y_n\)
• (Proof by contradiction)
• Assume that \(\exists\) a CS \(W\) of \(X_{m-1}\) and \(Y_n\) with \(|W| > k\)
• Then \(W\) would also be a CS of \(X\) and \(Y\)
• Contradiction to the assumption that
  • \(Z\) is an LCS of \(X\) and \(Y\) with \(|Z| = k\)
• Case 3: Dual of the proof for (case 2)

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A Recursive Solution to Subproblems

• Theorem implies that there are one or two subproblems to examine
• if \(x_m = y_n\) then
• we must solve the subproblem of finding an LCS of \(X_{m-1} \& Y_{n-1}\)
• appending \(x_m = y_n\) to this LCS yields an LCS of \(X \& Y\)
• else
• we must solve two subproblems
  • finding an LCS of \(X_{m-1} \& Y\)
  • finding an LCS of \(X \& Y_{n-1}\)
• longer of these two LCS s is an LCS of \(X \& Y\)
• endif

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Recursive Algorithm (Inefficient)

\[ \begin{align*} & \text{LCS}(X, Y) \ \{ \\ & \quad m \leftarrow length[X] \\ & \quad n \leftarrow length[Y] \\ & \quad \text{if} \ x_m = y_n \ \text{then} \\ & \qquad Z \leftarrow \text{LCS}(X_{m-1}, Y_{n-1}) \triangleright \text{solve one subproblem} \\ & \qquad \text{return} \ \langle Z, x_m = y_n \rangle \triangleright \text{append} \ x_m = y_n \ \text{to} \ Z \\ & \quad else \\ & \qquad Z^{'} \leftarrow \text{LCS}(X_{m-1}, Y) \triangleright \text{solve two subproblems} \\ & \qquad Z^{''} \leftarrow \text{LCS}(X, Y_{n-1}) \\ & \qquad \text{return longer of} \ Z^{'} \ \text{and} \ Z^{''} \\ & \} \end{align*} \]

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A Recursive Solution

• \(c[i, j]:\) length of an LCS of \(X_i\) and \(Y_j\)

\[ \begin{align*} c[i,j] = \begin{cases} & 0 & \text{if}& \ i=0 \ \text{or} \ j=0 \\ & c[i-1,j-1]+1 & \text{if}& \ i,j>0 \ \text{and} \ x_i=y_j \\ & \text{max}\{c[i,j-1],c[i-1,j]\} & \text{if}& \ i,j>0 \ \text{and} \ x_i \neq y_j \\ \end{cases} \end{align*} \]

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Computing the Length of an LCS

• We can easily write an exponential-time recursive algorithm based on the given recurrence. \(\Longrightarrow\) Inefficient!
• How many distinct subproblems to solve?
• \(\Theta(mn)\)
• Overlapping subproblems property: Many subproblems share the same sub-subproblems.
• e.g. Finding an LCS to \(X_{m-1} \& Y\) and an LCS to \(X \& Y_{n-1}\)
• has the sub-subproblem of finding an LCS to \(X_{m-1} \& Y_{n-1}\)
• Therefore, we can use dynamic programming.

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Data Structures

• Let:
• \(c[i, j]:\) length of an LCS of \(X_i\) and \(Y_j\)
• \(b[i, j]:\) direction towards the table entry corresponding to the optimal subproblem solution chosen when computing \(c[i, j]\).
• Used to simplify the construction of an optimal solution at the end.
• Maintain the following tables:
• \(c[0 \dots m, 0 \dots n]\)
• \(b[1 \dots m, 1 \dots n]\)

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Bottom-up Computation

• Reminder:

\[ \begin{align*} c[i,j] = \begin{cases} & 0 & \text{if}& \ i=0 \ \text{or} \ j=0 \\ & c[i-1,j-1]+1 & \text{if}& \ i,j>0 \ \text{and} \ x_i=y_j \\ & \text{max}\{c[i,j-1],c[i-1,j]\} & \text{if}& \ i,j>0 \ \text{and} \ x_i \neq y_j \\ \end{cases} \end{align*} \]

• How to choose the order in which we process \(c[i, j]\) values?
• The values for \(c[i-1, j-1]\), \(c[i, j-1]\), and \(c[i-1,j]\) must be computed before computing \(c[i, j]\).

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Bottom-up Computation

\[ \begin{align*} c[i,j] = \begin{cases} & 0 & \text{if}& \ i=0 \ \text{or} \ j=0 \\ & c[i-1,j-1]+1 & \text{if}& \ i,j>0 \ \text{and} \ x_i=y_j \\ & \text{max}\{c[i,j-1],c[i-1,j]\} & \text{if}& \ i,j>0 \ \text{and} \ x_i \neq y_j \\ \end{cases} \end{align*} \]

Need to process: \(c[i, j]\) after computing: \(c[i-1, j-1]\), \(c[i, j-1]\),
\(c[i-1,j]\)

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Bottom-up Computation

\[ \begin{align*} c[i,j] = \begin{cases} & 0 & \text{if}& \ i=0 \ \text{or} \ j=0 \\ & c[i-1,j-1]+1 & \text{if}& \ i,j>0 \ \text{and} \ x_i=y_j \\ & \text{max}\{c[i,j-1],c[i-1,j]\} & \text{if}& \ i,j>0 \ \text{and} \ x_i \neq y_j \\ \end{cases} \end{align*} \]

\[ \Downarrow \]

\[ \begin{align*} & \text{for} \ i \leftarrow 1 \ \text{to} \ m \\ & \quad \text{for} \ j \leftarrow 1 \ \text{to} \ n \\ & \qquad \dots \\ & \qquad \dots \\ & \qquad c[i, j] = \cdots \end{align*} \]

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Computing the Length of an LCS

\[ \begin{align*} \frac{\text{Total Runtime} = \Theta(mn)}{\text{Total Space} = \Theta(mn)} \begin{cases} & LCS-LENGTH(X,Y) \\ & \quad m \leftarrow length[X]; n \leftarrow length[Y] \\ & \quad \text{for} \ i \leftarrow 0 \ \text{to} \ m \ \text{do} \ c[i, 0] \leftarrow 0 \\ & \quad \text{for} \ j \leftarrow 0 \ \text{to} \ n \ \text{do} \ c[0, j] \leftarrow 0 \\ & \quad \text{for} \ i \leftarrow 1 \ \text{to} \ m \ \text{do} \\ & \qquad \text{for} \ j \leftarrow 1 \ \text{to} \ n \ \text{do} \\ & \qquad \quad \text{if} \ x_i = y_j \ \text{then} \\ & \qquad \quad \quad c[i, j] \leftarrow c[i-1, j-1]+1 \\ & \qquad \quad \quad b[i, j] \leftarrow " \nwarrow " \\ & \qquad \quad \text{else if} \ c[i - 1, j] \geq c[i, j-1] \\ & \qquad \quad \quad c[i, j] \leftarrow c[i-1, j] \\ & \qquad \quad \quad b[i, j] \leftarrow "\uparrow " \\ & \qquad \quad \text{else} \\ & \qquad \quad \quad c[i, j] \leftarrow c[i, j-1] \\ & \qquad \quad \quad b[i, j] \leftarrow " \leftarrow " \\ \end{cases} \end{align*} \]

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Computing the Length of an LCS-1

Operation of LCS-LENGTH on the sequences $$ \begin{align*} X &= \langle \overset{1}{A}, \overset{2}{B}, \overset{3}{C}, \overset{4}{B}, \overset{5}{D}, \overset{6}{A}, \overset{7}{B} \rangle \ Y &= \langle \overset{1}{B}, \overset{2}{D}, \overset{3}{C}, \overset{4}{A}, \overset{5}{B}, \overset{6}{A} \rangle \end{align*} $$

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Computing the Length of an LCS-2

Operation of LCS-LENGTH on the sequences $$ \begin{align*} X &= \langle \overset{1}{A}, \overset{2}{B}, \overset{3}{C}, \overset{4}{B}, \overset{5}{D}, \overset{6}{A}, \overset{7}{B} \rangle \ Y &= \langle \overset{1}{B}, \overset{2}{D}, \overset{3}{C}, \overset{4}{A}, \overset{5}{B}, \overset{6}{A} \rangle \end{align*} $$

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Computing the Length of an LCS-3

Operation of LCS-LENGTH on the sequences $$ \begin{align*} X &= \langle \overset{1}{A}, \overset{2}{B}, \overset{3}{C}, \overset{4}{B}, \overset{5}{D}, \overset{6}{A}, \overset{7}{B} \rangle \ Y &= \langle \overset{1}{B}, \overset{2}{D}, \overset{3}{C}, \overset{4}{A}, \overset{5}{B}, \overset{6}{A} \rangle \end{align*} $$

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Computing the Length of an LCS-4

Operation of LCS-LENGTH on the sequences $$ \begin{align*} X &= \langle \overset{1}{A}, \overset{2}{B}, \overset{3}{C}, \overset{4}{B}, \overset{5}{D}, \overset{6}{A}, \overset{7}{B} \rangle \ Y &= \langle \overset{1}{B}, \overset{2}{D}, \overset{3}{C}, \overset{4}{A}, \overset{5}{B}, \overset{6}{A} \rangle \end{align*} $$

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Computing the Length of an LCS-5

Operation of LCS-LENGTH on the sequences $$ \begin{align*} X &= \langle \overset{1}{A}, \overset{2}{B}, \overset{3}{C}, \overset{4}{B}, \overset{5}{D}, \overset{6}{A}, \overset{7}{B} \rangle \ Y &= \langle \overset{1}{B}, \overset{2}{D}, \overset{3}{C}, \overset{4}{A}, \overset{5}{B}, \overset{6}{A} \rangle \end{align*} $$

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Computing the Length of an LCS-6

Operation of LCS-LENGTH on the sequences $$ \begin{align*} X &= \langle \overset{1}{A}, \overset{2}{B}, \overset{3}{C}, \overset{4}{B}, \overset{5}{D}, \overset{6}{A}, \overset{7}{B} \rangle \ Y &= \langle \overset{1}{B}, \overset{2}{D}, \overset{3}{C}, \overset{4}{A}, \overset{5}{B}, \overset{6}{A} \rangle \end{align*} $$

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Computing the Length of an LCS-7

Operation of LCS-LENGTH on the sequences $$ \begin{align*} X &= \langle \overset{1}{A}, \overset{2}{B}, \overset{3}{C}, \overset{4}{B}, \overset{5}{D}, \overset{6}{A}, \overset{7}{B} \rangle \ Y &= \langle \overset{1}{B}, \overset{2}{D}, \overset{3}{C}, \overset{4}{A}, \overset{5}{B}, \overset{6}{A} \rangle \end{align*} $$

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Computing the Length of an LCS-8

Operation of LCS-LENGTH on the sequences $$ \begin{align*} X &= \langle \overset{1}{A}, \overset{2}{B}, \overset{3}{C}, \overset{4}{B}, \overset{5}{D}, \overset{6}{A}, \overset{7}{B} \rangle \ Y &= \langle \overset{1}{B}, \overset{2}{D}, \overset{3}{C}, \overset{4}{A}, \overset{5}{B}, \overset{6}{A} \rangle \end{align*} $$

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Computing the Length of an LCS-9

Operation of LCS-LENGTH on the sequences $$ \begin{align*} X &= \langle \overset{1}{A}, \overset{2}{B}, \overset{3}{C}, \overset{4}{B}, \overset{5}{D}, \overset{6}{A}, \overset{7}{B} \rangle \ Y &= \langle \overset{1}{B}, \overset{2}{D}, \overset{3}{C}, \overset{4}{A}, \overset{5}{B}, \overset{6}{A} \rangle \end{align*} $$

---

Computing the Length of an LCS-10

Operation of LCS-LENGTH on the sequences $$ \begin{align*} X &= \langle \overset{1}{A}, \overset{2}{B}, \overset{3}{C}, \overset{4}{B}, \overset{5}{D}, \overset{6}{A}, \overset{7}{B} \rangle \ Y &= \langle \overset{1}{B}, \overset{2}{D}, \overset{3}{C}, \overset{4}{A}, \overset{5}{B}, \overset{6}{A} \rangle \end{align*} $$

---

Computing the Length of an LCS-11

Operation of LCS-LENGTH on the sequences $$ \begin{align*} X &= \langle \overset{1}{A}, \overset{2}{B}, \overset{3}{C}, \overset{4}{B}, \overset{5}{D}, \overset{6}{A}, \overset{7}{B} \rangle \ Y &= \langle \overset{1}{B}, \overset{2}{D}, \overset{3}{C}, \overset{4}{A}, \overset{5}{B}, \overset{6}{A} \rangle \end{align*} $$

---

Computing the Length of an LCS-12

Operation of LCS-LENGTH on the sequences $$ \begin{align*} X &= \langle \overset{1}{A}, \overset{2}{B}, \overset{3}{C}, \overset{4}{B}, \overset{5}{D}, \overset{6}{A}, \overset{7}{B} \rangle \ Y &= \langle \overset{1}{B}, \overset{2}{D}, \overset{3}{C}, \overset{4}{A}, \overset{5}{B}, \overset{6}{A} \rangle \end{align*} $$

---

Computing the Length of an LCS-13

Operation of LCS-LENGTH on the sequences $$ \begin{align*} X &= \langle \overset{1}{A}, \overset{2}{B}, \overset{3}{C}, \overset{4}{B}, \overset{5}{D}, \overset{6}{A}, \overset{7}{B} \rangle \ Y &= \langle \overset{1}{B}, \overset{2}{D}, \overset{3}{C}, \overset{4}{A}, \overset{5}{B}, \overset{6}{A} \rangle \end{align*} $$

• Running-time = \(O(mn)\) since each table entry takes \(O(1)\) time to compute

---

Computing the Length of an LCS-14

Operation of LCS-LENGTH on the sequences $$ \begin{align*} X &= \langle \overset{1}{A}, \overset{2}{B}, \overset{3}{C}, \overset{4}{B}, \overset{5}{D}, \overset{6}{A}, \overset{7}{B} \rangle \ Y &= \langle \overset{1}{B}, \overset{2}{D}, \overset{3}{C}, \overset{4}{A}, \overset{5}{B}, \overset{6}{A} \rangle \end{align*} $$

• Running-time = \(O(mn)\) since each table entry takes \(O(1)\) time to compute

• LCS of \(X \& Y = \langle B, C, B, A \rangle\)

---

Constructing an LCS

• The \(b\) table returned by LCS-LENGTH can be used to quickly construct an LCS of \(X \& Y\)
• Begin at \(b[m, n]\) and trace through the table following arrows
• Whenever you encounter a "\(\nwarrow\)" in entry \(b[i, j]\) it implies that \(x_i = y_j\) is an element of LCS
• The elements of LCS are encountered in reverse order

---

Constructing an LCS

• The recursive procedure \(\text{PRINT-LCS}\) prints out \(\text{LCS}\) in proper order
• This procedure takes \(O(m+n)\) time since at least one of \(i\) and \(j\) is decremented in each stage of the recursion

\[ \begin{align*} & \text{PRINT-LCS}(b, X, i, j) \\ & \quad \text{if} \ i = 0 \ \text{or} j = 0 \ \text{then} \\ & \quad \text{return} \\ & \quad \text{if} \ b[i, j] = " \nwarrow " \ \text{then} \\ & \qquad \text{PRINT-LCS}(b, X, i-1, j-1) \\ & \qquad \text{print} \ x_i \\ & \quad \text{else if} \ b[i, j] = " \uparrow " \ \text{then} \\ & \qquad \text{PRINT-LCS}(b, X, i-1, j) \\ & \quad \text{else} \\ & \qquad \text{PRINT-LCS}(b, X, i, j-1) \end{align*} \]

• The initial invocation: \(\text{PRINT-LCS}(b, X, length[X], length[Y])\)

---

Do we really need the b table (back-pointers)?

• Question: From which neighbor did we expand to the highlighted cell?
• Answer: Upper-left neighbor,because \(X[i] = Y[j]\).

---

Do we really need the b table (back-pointers)?

• Question: From which neighbor did we expand to the highlighted cell?
• Answer: Left neighbor, because \(X[i] \neq Y[j]\) and \(LCS[i, j-1] > LCS[i-1, j]\).

---

Do we really need the b table (back-pointers)?

• Question: From which neighbor did we expand to the highlighted cell?
• Answer: Upper neighbor,because \(X[i] \neq Y[j]\) and \(LCS[i, j-1] = LCS[i-1, j]\). (See pseudo-code to see how ties are handled.)

---

Improving the Space Requirements

• We can eliminate the b table altogether
• each \(c[i, j]\) entry depends only on \(3\) other \(c\) table entries: \(c[i-1, j-1]\), \(c[i-1, j]\) and \(c[i, j-1]\)
• Given the value of \(c[i, j]\):
• We can determine in \(O(1)\) time which of these \(3\) values was used to compute \(c[i, j]\) without inspecting table \(b\)
• We save \(\Theta(mn)\) space by this method
• However, space requirement is still \(\Theta(mn)\) since we need \(\Theta(mn)\) space for the \(c\) table anyway

---

What if we store the last 2 rows only?

• To compute \(c[i, j]\), we only need \(c[i-1, j-1]\), \(c[i-1, j]\),and \(c[i-1, j-1]\)
• So, we can store only the last two rows.

---

What if we store the last 2 rows only?

• To compute \(c[i, j]\), we only need \(c[i-1, j-1]\), \(c[i-1, j]\), and \(c[i-1, j-1]\)
• So, we can store only the last two rows.

---

What if we store the last 2 rows only?

• To compute \(c[i, j]\), we only need \(c[i-1, j-1]\), \(c[i-1, j]\), and \(c[i-1, j-1]\)

• So, we can store only the last two rows.

• This reduces space complexity from \(\Theta(mn)\) to \(\Theta(n)\).

• Is there a problem with this approach?

---

What if we store the last 2 rows only?

• Is there a problem with this approach?

• We cannot construct the optimal solution because we cannot backtrace anymore.

• This approach works if we only need the length of an LCS, not the actual LCS.

---

Problem 4 Optimal Binary Search Tree

---

Reminder: Binary Search Tree (BST)

---

Binary Search Tree Example

• Example: English-to-French translation
• Organize (English, French) word pairs in a BST
  • Keyword: English word
  • Satellite Data: French word
• We can search for an English word (node key) efficiently, and return the corresponding French word (satellite data).

---

ASCII Table

---

Binary Search Tree Example

Suppose we know the frequency of each keyword in texts: $$ \underset{5\%}{\underline{begin}}, \underset{40\%}{\underline{do}}, \underset{8\%}{\underline{else}}, \underset{4\%}{\underline{end}}, \underset{10\%}{\underline{if}}, \underset{10\%}{\underline{then}}, \underset{23\%}{\underline{while}}, $$

---

Cost of a Binary Search Tree

Example: If we search for keyword "while", we need to access \(3\) nodes. So, \(23%\) of the queries will have cost of \(3\).

\[ \begin{align*} \text{Total Cost} &= \sum \limits_{i}^{}(\text{depth}(i)+1)\text{freq}(i) \\ &= 1 \times 0.04 + 2 \times 0.4 + \\ & 2 \times 0.1 + 3 \times 0.05 + \\ & 3 \times 0.08 + 3 \times 0.1 + \\ & 3 \times 0.23 \\ &= 2.42 \end{align*} \]

---

Cost of a Binary Search Tree

Example: If we search for keyword "while", we need to access \(3\) nodes. So, \(23%\) of the queries will have cost of \(3\).

\[ \begin{align*} \text{Total Cost} &= \sum \limits_{i}^{}(\text{depth}(i)+1)\text{freq}(i) \\ &= 1 \times 0.4 + 2 \times 0.05 + 2 \times 0.23 + \\ & 3 \times 0.1 + 4 \times 0.08 + \\ & 4 \times 0.1 + 5 \times 0.04 \\ &= 2.18 \end{align*} \]

• This is in fact an optimal BST.

---

Optimal Binary Search Tree Problem

• Given:
• A collection of \(n\) keys \(K_1 < K_2 < \dots K_n\) to be stored in a BST.
• The corresponding \(p_i\) values for \(1 \leq i \leq n\)
  • \(p_i\): probability of searching for key \(K_i\)
• Find:
• An optimal BST with minimum total cost:

\[ \begin{align*} \text{Total Cost} &= \sum \limits_{i}^{}(\text{depth}(i)+1)\text{freq}(i) \end{align*} \]

• Note: The BST will be static. Only search operations will be performed. No insert, no delete, etc.

---

Cost of a Binary Search Tree

• Lemma 1: Let \(Tij\) be a BST containing keys \(K_i < K_{i+1} < \dots < K_j\). Let \(T_L\) and \(T_R\) be the left and right subtrees of \(T\). Then we have:

\[ \begin{align*} \text{cost}(T_{ij})=\text{cost}(T_{L})+\text{cost}(T_{R})+\sum \limits_{h=i}^{j}p_h \end{align*} \]

Intuition: When we add the root node, the depth of each node in \(T_L\) and \(T_R\) increases by \(1\). So, the cost of node \(h\) increases by \(p_h\). In addition, the cost of root node \(r\) is \(p_r\). That’s why, we have the last term at the end of the formula above.

---

Optimal Substructure Property

• Lemma 2: Optimal substructure property
• Consider an optimal BST \(T_{ij}\) for keys \(K_i < K_{i+1} < \dots < K_j\)
• Let \(K_m\) be the key at the root of \(T_{ij}\)
• Then:
• \(T_{i,m-1}\) is an optimal BST for subproblem containing keys:
  • \(K_i < \dots < K_{m-1}\)
• \(T_{m+1,j}\) is an optimal BST for subproblem containing keys:
  • \(K_{m+1} < \dots < K_j\)

\[ \begin{align*} \text{cost}(T_{ij})=\text{cost}(T_{i,m-1})+\text{cost}(T_{m+1,j})+\sum \limits_{h=i}^{j}p_h \end{align*} \]

---

Recursive Formulation

• Note: We don’t know which root vertex leads to the minimum total cost. So, we need to try each vertex \(m\), and choose the one with minimum total cost.

• \(c[i, j]\): cost of an optimal BST \(T_{ij}\) for the subproblem \(K_i < \dots < K_j\)

\[ \begin{align*} & c[i,j] = \begin{cases} & 0 & \text{if} \ i>j \\ & \underset{i \leq r \leq j}{\text{min}}\{ c[i,r-1]+c[r+1,j]+P_{ij} \} & \text{otherwise} \\ \end{cases} \\ & \text{where} \ P_{ij}= \sum \limits_{h=i}^{j}p_h \end{align*} \]

---

Bottom-up computation

\[ \begin{align*} & c[i,j] = \begin{cases} & 0 & \text{if} \ i>j \\ & \underset{i \leq r \leq j}{\text{min}}\{ c[i,r-1]+c[r+1,j]+P_{ij} \} & \text{otherwise} \\ \end{cases} \end{align*} \]

• How to choose the order in which we process \(c[i, j]\) values?
• Before computing \(c[i, j]\), we have to make sure that the values for \(c[i, r-1]\) and \(c[r+1,j]\) have been computed for all \(r\).

---

Bottom-up computation

\[ \begin{align*} & c[i,j] = \begin{cases} & 0 & \text{if} \ i>j \\ & \underset{i \leq r \leq j}{\text{min}}\{ c[i,r-1]+c[r+1,j]+P_{ij} \} & \text{otherwise} \\ \end{cases} \end{align*} \]

• \(c[i,j]\) must be processed after \(c[i,r-1]\) and \(c[r+1,j]\)

---

Bottom-up computation

\[ \begin{align*} & c[i,j] = \begin{cases} & 0 & \text{if} \ i>j \\ & \underset{i \leq r \leq j}{\text{min}}\{ c[i,r-1]+c[r+1,j]+P_{ij} \} & \text{otherwise} \\ \end{cases} \end{align*} \]

• If the entries \(c[i,j]\) are computed in the shown order, then \(c[i,r-1]\) and \(c[r+1,j]\) values are guaranteed to be computed before \(c[i,j]\).

---

Computing the Optimal BST Cost

\[ \begin{align*} & \text{OPTIMAL-BST-COST} (p, n) \\ & \quad \text{for} \ i \leftarrow 1 \ \text{to} \ n \ \text{do} \\ & \qquad c[i, i-1] \leftarrow 0 \\ & \qquad c[i, i] \leftarrow p[i] \\ & \qquad R[i, j] \leftarrow i \\ & \quad PS[1] \leftarrow p[1] \Longleftarrow PS[i] \rightarrow \text{ prefix-sum } (i): \text{Sum of all} \ p[j] \ \text{values for} \ j \leq i \\ & \quad \text{for} \ i \leftarrow 2 \ \text{to} \ n \ \text{do} \\ & \qquad PS[i] \leftarrow p[i] + PS[i-1] \Longleftarrow \text{compute the prefix sum} \\ & \quad \text{for} \ d \leftarrow 1 \ \text{to} \ n−1 \ \text{do} \Longleftarrow \text{BSTs with} \ d+1 \ \text{consecutive keys} \\ & \qquad \text{for} \ i \leftarrow 1 \ \text{to} \ n – d \ \text{do} \\ & \qquad \quad j \leftarrow i + d \\ & \qquad \quad c[i, j] \leftarrow \infty \\ & \qquad \quad \text{for} \ r \leftarrow i \ \text{to} \ j \ \text{do} \\ & \qquad \qquad q \leftarrow min\{c[i,r-1] + c[r+1, j]\} + PS[j] – PS[i-1]\} \\ & \qquad \qquad \text{if} \ q < c[i, j] \ \text{then} \\ & \qquad \qquad \quad c[i, j] \leftarrow q \\ & \qquad \qquad \quad R[i, j] \leftarrow r \\ & \quad \text{return} \ c[1, n], R \end{align*} \]

---

Note on Prefix Sum

• We need \(P_{ij}\) values for each \(i, j (1 ≤ i ≤ n \ \text{and} \ 1 ≤ j ≤ n)\), where:

\[ \begin{align*} P_{ij} = \sum \limits_{h=i}^{j}p_h \end{align*} \]

• If we compute the summation directly for every \((i, j)\) pair, the runtime would be \(\Theta(n^3)\).

• Instead, we spend \(O(n)\) time in preprocessing to compute the prefix sum array PS. Then we can compute each \(P_{ij}\) in \(O(1)\) time using PS.

---

Note on Prefix Sum

• In preprocessing, compute for each \(i\):
• \(PS[i]\): the sum of \(p[j]\) values for \(1 \leq j \leq i\)
• Then, we can compute \(P_{ij}\) in \(O(1)\) time as follows:
• \(P_{ij} = PS[i] – PS[j-1]\)
• Example:

\[ \begin{align*} p &: \overset{1}{0.05} \ \overset{2}{0.02} \ \overset{3}{0.06} \ \overset{4}{0.07} \ \overset{5}{0.20} \ \overset{6}{0.05} \ \overset{7}{0.08} \ \overset{8}{0.02} \\ PS &: \overset{1}{0.05} \ \overset{2}{0.07} \ \overset{3}{0.13} \ \overset{4}{0.20} \ \overset{5}{0.40} \ \overset{6}{0.45} \ \overset{7}{0.53} \ \overset{8}{0.55} \\[10 pt] P_{27} &= PS[7] – PS[1] = 0.53 – 0.05 = 0.48 \\ P_{36} &= PS[6] – PS[2] = 0.45 – 0.07 = 0.38 \end{align*} \]

---

REVIEW

Overlapping Subproblems Property in Dynamic Programming

Dynamic Programming is an algorithmic paradigm that solves a given complex problem by breaking it into subproblems and stores the results of subproblems to avoid computing the same results again.

---

Overlapping Subproblems Property in Dynamic Programming

Following are the two main properties of a problem that suggests that the given problem can be solved using Dynamic programming.

1. Overlapping Subproblems
2. Optimal Substructure

---

Overlapping Subproblems

• Like Divide and Conquer, Dynamic Programming combines solutions to sub-problems.
• Dynamic Programming is mainly used when solutions of the same subproblems are needed again and again.
• In dynamic programming, computed solutions to subproblems are stored in a table so that these don’t have to be recomputed.
• So Dynamic Programming is not useful when there are no common (overlapping) subproblems because there is no point storing the solutions if they are not needed again.

---

Overlapping Subproblems

• For example, Binary Search doesn’t have common subproblems.
• If we take an example of following recursive program for Fibonacci Numbers, there are many subproblems that are solved again and again.

---

Simple Recursion

• \(f(n) = f(n-1) + f(n-2)\)
• C sample code:

#include <stdio.h>
// a simple recursive program to compute fibonacci numbers
int fib(int n)
{
    if (n <= 1)
        return n;
    else
        return fib(n-1) + fib(n-2);
}

int main()
{
    int n = 5;
    printf("Fibonacci number is %d ", fib(n));
    return 0;
}

---

Simple Recursion

• Output

Fibonacci number is 5

---

Simple Recursion

• \(f(n) = f(n-1) + f(n-2)\)

/* a simple recursive program for Fibonacci numbers */
public class Fibonacci {
    public static void main(String[] args) {
        int n = Integer.parseInt(args[0]);
        System.out.println(fib(n));
    }

    public static int fib(int n) {
        if (n <= 1)
            return n;

        return fib(n - 1) + fib(n - 2);
    }
}

---

Simple Recursion

• \(f(n) = f(n-1) + f(n-2)\)

public class Fibonacci {
    public static void Main(string[] args) {
        int n = int.Parse(args[0]);
        Console.WriteLine(fib(n));
    }

    public static int fib(int n) {
        if (n <= 1)
            return n;

        return fib(n - 1) + fib(n - 2);
    }
}

---

Recursion tree for execution of fib(5)

                         fib(5)
                     /             \
               fib(4)                fib(3)
             /      \                /     \
         fib(3)      fib(2)         fib(2)    fib(1)
        /     \        /    \       /    \
  fib(2)   fib(1)  fib(1) fib(0) fib(1) fib(0)
  /    \
fib(1) fib(0)

• We can see that the function fib(3) is being called 2 times.
• If we would have stored the value of fib(3), then instead of computing it again, we could have reused the old stored value.

---

Recursion tree for execution of fib(5)

There are following two different ways to store the values so that these values can be reused:

1. Memoization (Top Down)
2. Tabulation (Bottom Up)

---

Memoization (Top Down)

• The memoized program for a problem is similar to the recursive version with a small modification that looks into a lookup table before computing solutions.
• We initialize a lookup array with all initial values as NIL. Whenever we need the solution to a subproblem, we first look into the lookup table.
• If the precomputed value is there then we return that value, otherwise, we calculate the value and put the result in the lookup table so that it can be reused later.

---

Memoization (Top Down)

• Following is the memoized version for the nth Fibonacci Number.
• C++ Version:

/* C++ program for Memoized version
for nth Fibonacci number */
#include <bits/stdc++.h>
using namespace std;
#define NIL -1
#define MAX 100

int lookup[MAX];

---

Memoization (Top Down)

• C++ Version:

/* Function to initialize NIL
values in lookup table */
void _initialize()
{
    int i;
    for (i = 0; i < MAX; i++)
        lookup[i] = NIL;
}

---

Memoization (Top Down)

• C++ Version:

/* function for nth Fibonacci number */
int fib(int n)
{
    if (lookup[n] == NIL) {
        if (n <= 1)
            lookup[n] = n;
        else
            lookup[n] = fib(n - 1) + fib(n - 2);
    }

    return lookup[n];
}

---

Memoization (Top Down)

• C++ Version:

// Driver code
int main()
{
    int n = 40;
    _initialize();
    cout << "Fibonacci number is " << fib(n);
    return 0;
}

---

Memoization (Top Down)

• Java Version:

/* Java program for Memoized version */
public class Fibonacci {
    final int MAX = 100;
    final int NIL = -1;

    int lookup[] = new int[MAX];

    /* Function to initialize NIL values in lookup table */
    void _initialize()
    {
        for (int i = 0; i < MAX; i++)
            lookup[i] = NIL;
    }

---

Memoization (Top Down)

• Java Version:

    /* function for nth Fibonacci number */
    int fib(int n)
    {
        if (lookup[n] == NIL) {
            if (n <= 1)
                lookup[n] = n;
            else
                lookup[n] = fib(n - 1) + fib(n - 2);
        }
        return lookup[n];
    }

---

Memoization (Top Down)

• Java Version:

    public static void main(String[] args)
    {
        Fibonacci f = new Fibonacci();
        int n = 40;
        f._initialize();
        System.out.println("Fibonacci number is"
                        + " " + f.fib(n));
    }
}

---

Memoization (Top Down)

• C# Version:

// C# program for Memoized versionof nth Fibonacci number
using System;

class FiboCalcMemoized {

    static int MAX = 100;
    static int NIL = -1;
    static int[] lookup = new int[MAX];

    /* Function to initialize NIL
    values in lookup table */
    static void initialize()
    {
        for (int i = 0; i < MAX; i++)
            lookup[i] = NIL;
    }

---

Memoization (Top Down)

• C# Version:

    /* function for nth Fibonacci number */
    static int fib(int n)
    {
        if (lookup[n] == NIL) {
            if (n <= 1)
                lookup[n] = n;
            else
                lookup[n] = fib(n - 1) + fib(n - 2);
        }
        return lookup[n];
    }

---

Memoization (Top Down)

• C# Version:

    // Driver code
    public static void Main()
    {

        int n = 40;
        initialize();
        Console.Write("Fibonacci number is"
                    + " " + fib(n));
    }
}

---

Tabulation (Bottom Up)

• The tabulated program for a given problem builds a table in bottom-up fashion and returns the last entry from the table.
• For example, for the same Fibonacci number,
• we first calculate fib(0) then fib(1) then fib(2) then fib(3), and so on. So literally, we are building the solutions of subproblems bottom-up.

---

Tabulation (Bottom Up)

• C++ Version:

/* C program for Tabulated version */
#include <stdio.h>
int fib(int n)
{
    int f[n + 1];
    int i;
    f[0] = 0;
    f[1] = 1;
    for (i = 2; i <= n; i++)
        f[i] = f[i - 1] + f[i - 2];

    return f[n];
}

---

Tabulation (Bottom Up)

• C++ Version:

...
int main()
{
    int n = 9;
    printf("Fibonacci number is %d ", fib(n));
    return 0;
}

Output:

Fibonacci number is 34 

---

Tabulation (Bottom Up)

• Java Version:

/* Java program for Tabulated version */
public class Fibonacci {
  public static void main(String[] args)
  {
    int n = 9;
    System.out.println("Fibonacci number is " + fib(n));
  }

---

Tabulation (Bottom Up)

• Java Version:

  /* Function to calculate nth Fibonacci number */
  static int fib(int n)
  {
    int f[] = new int[n + 1];
    f[0] = 0;
    f[1] = 1;
    for (int i = 2; i <= n; i++)
      f[i] = f[i - 1] + f[i - 2];

    return f[n];
  }
}

---

Tabulation (Bottom Up)

• C# Version:

// C# program for Tabulated version
using System;

class Fibonacci {
    static int fib(int n)
    {
        int[] f = new int[n + 1];
        f[0] = 0;
        f[1] = 1;
        for (int i = 2; i <= n; i++)
            f[i] = f[i - 1] + f[i - 2];
        return f[n];
    }

    public static void Main()
    {
        int n = 9;
        Console.Write("Fibonacci number is"
                    + " " + fib(n));
    }
}

---

• Both Tabulated and Memoized store the solutions of subproblems.
• In Memoized version, the table is filled on demand while in the Tabulated version, starting from the first entry, all entries are filled one by one.
• Unlike the Tabulated version, all entries of the lookup table are not necessarily filled in Memoized version.

---

• To see the optimization achieved by Memoized and Tabulated solutions over the basic Recursive solution, see the time taken by following runs for calculating the 40^th Fibonacci number:

• Recursive Solution:

• https://ide.geeksforgeeks.org/vHt6ly

• Memoized Solution:

• https://ide.geeksforgeeks.org/Z94jYR

• Tabulated Solution:

• https://ide.geeksforgeeks.org/12C5bP

---

Optimal Substructure Property in Dynamic Programming

• A given problems has Optimal Substructure Property if optimal solution of the given problem can be obtained by using optimal solutions of its subproblems.

• For example, the Shortest Path problem has following optimal substructure property:

• If a node x lies in the shortest path from a source node u to destination node v then the shortest path from u to v is combination of shortest path from u to x and shortest path from x to v. The standard All Pair Shortest Path algorithm like Floyd–Warshall and Single Source Shortest path algorithm for negative weight edges like Bellman–Ford are typical examples of Dynamic Programming.

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Optimal Substructure Property in Dynamic Programming

• On the other hand, the Longest Path problem doesn’t have the Optimal Substructure property. Here by Longest Path we mean longest simple path (path without cycle) between two nodes

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Optimal Substructure Property in Dynamic Programming

• There are two longest paths from q to t: q→r→t and q→s→t. Unlike shortest paths, these longest paths do not have the optimal substructure property. For example, the longest path q→r→t is not a combination of longest path from q to r and longest path from r to t, because the longest path from q to r is q→s→t→r and the longest path from r to t is r→q→s→t.

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Most Common Dynamic Programming Interview Questions

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Problem-1: Longest Increasing Subsequence

• Problem-1: Longest Increasing Subsequence

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Problem-1: Longest Increasing Subsequence

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Problem-2: Edit Distance

• Problem-2: Edit Distance

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Problem-2: Edit Distance (Recursive)

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Problem-2: Edit Distance (DP)

https://www.coursera.org/learn/dna-sequencing

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Problem-2: Edit Distance (DP)

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Problem-2: Edit Distance (Other)

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Problem-3: Partition a set into two subsets such that the difference of subset sums is minimum

• Problem-3: Partition a set into two subsets such that the difference of subset sums is minimum

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Problem-4: Count number of ways to cover a distance

• Problem-4: Count number of ways to cover a distance

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Problem-5: Find the longest path in a matrix with given constraints

• Problem-5: Find the longest path in a matrix with given constraints

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Problem-6: Subset Sum Problem

• Problem-6: Subset Sum Problem

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Problem-7: Optimal Strategy for a Game

• Problem-7: Optimal Strategy for a Game

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Problem-8: 0-1 Knapsack Problem

• Problem-8: 0-1 Knapsack Problem

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Problem-9: Boolean Parenthesization Problem

• Problem-9: Boolean Parenthesization Problem

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Problem-10: Shortest Common Supersequence

• Problem-10: Shortest Common Supersequence

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Problem-11: Partition Problem

• Problem-11: Partition Problem

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Problem-12: Cutting a Rod

• Problem-12: Cutting a Rod

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Problem-13: Coin Change

• Problem-13: Coin Change

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Problem-14: Word Break Problem

• Problem-14: Word Break Problem

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Problem-15: Maximum Product Cutting

• Problem-15: Maximum Product Cutting

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Problem-16: Dice Throw

• Problem-16: Dice Throw

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Problem-16: Dice Throw

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Problem-17: Box Stacking Problem

• Problem-17: Box Stacking Problem

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Problem-18: Egg Dropping Puzzle

• Problem-18: Egg Dropping Puzzle

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References

• Introduction to Algorithms, Third Edition | The MIT Press

• CLRS

• Bilkent CS473 Course Notes (new)

• Bilkent CS473 Course Notes (old)

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Son Güncelleme: 13 Mayıs 2022