1. Data Structures and Algorithms (AT70. 02) Comp. Sc. and Inf. Mgmt
Data Structures and Algorithms (AT70.02) Comp. Sc. and Inf. Mgmt. Asian Institute of Technology
Instructor: Prof. Sumanta Guha
Slide Sources: CLRS “Intro. To Algorithms” book website
(copyright McGraw Hill)
adapted and supplemented

2. CLRS “Intro. To Algorithms” Ch. 15: Dynamic Programming

3. What is the brute-force, i. e
What is the brute-force, i.e., exhaustive method, to solve the problem?
What is its running time?

4. fi[j] = fastest possible time from start through station Si,j
Fastest time to completion = f* = min{ f1[n] + x1, f2[n] + x2 }
f1[j] = e1 + a1, if j = 1
min( f1[j-1] + a1,j , f2[j-1] + t2,j-1 + a1,j ) if j > 1
Similar equation for f2[j].
What’s the problem with a recursive algorithm
based on these formulae?
Think of the Fibonacci numbers!

5. Idea is to fill out the tables of f1,
f2, l1 and l2. Linear time!
Ques: How does
PRINT-STATIONS work?

6. Straight-forward matrix multiplication: if A is p  q and B is q  r then
C is p  r and requires pqr scalar multiplications to compute.
Consider 3 matrices A1 (size 10  100), A2 (size 100  5) and A3 (size 5  50).
There are two ways to compute the product A1A2A3:
(A1A2)A3 = A1(A2A3)
Ques: Is the complexity in terms of the number scalar of multiplications the same?
Matrix-chain multiplication problem: Given a chain A1, A2, …, An of n matrices,
where matrix Ai has size pi-1  pi, fully parenthesize the product A1A2…An in a way
that minimizes the number of scalar multiplications.
Note: There are exponentially many ways to fully parenthesize the product so an
exhaustive solution is infeasible.

7. Let Ai..j for the product matrix AiAi+1…Aj, where i ≤ j.
Let m[i,j] be the minimum number of scalar multiplications to compute Ai..j.
The problem is to find m[1,n]. Now,
m[i,j] = if i = j
mini≤k<j { m[i,k] + m[k+1,j] + pi-1pkpj } if i < j
Ques: What would be the performance of a recursive algorithm based on
this formula?

8. Ex

9. Optimal Binary Search Trees
0.1x1
0.15x2
0.1x2
0.05x3
0.2x3
0.05x3
0.1x3
0.05x4
0.05x4
0.05x4
0.1x4
Cost of node in red: total cost 2.80
ki are keys (k1 < k2 < … < k5), di are dummy values
representing “space between” keys k-i and ki+1.
Check the expected
search cost of the
two trees!
←probability of ki
←probability of di

10. Finding a Recursive Solution through Dynamic Programming
Critical Observation: If an optimal BST T has a subtree T’ containing keys
ki, …, kj, then this must be the optimal BST for the subproblem with keys
ki, …, kj and dummy keys di-1, di, …, dj.
e[i, j] is the expected cost of searching an optimal BST containing the keys ki, …, kj. Ultimately, we wish to compute e[1, n].
For subtree with keys ki, …, kj denote
w(i, j) = ∑l=i..j pl + ∑l=i-1..j ql
If kr is the root of the optimal subtree containing ki, …, kj, we have
e[i, j] = pr + (e(i, r-1] + w(i, r-1)) + (e[r+1, j] + w(r+1, j))
which is equivalent to
e[i, j] = e[i, r-1] + e[r+1, j] + w(i, j)
because w(i, j) = w(i, r-1) + pr + w(r+1, j)
Final, recursive formula:
e[i, j] = qi if j = i -1
mini≤r≤j {e[i, r-1] + e[r+1, j] } + w(i, j) if i ≤ j
Must add because the depth
of each node in the left and
right subtrees increases by
one in the big tree.

13. Problems Read Sec. 15.3 Read Sec. 15.4: Longest common subsequencesEx Ex
Prob. 15-6
Prob. 15-7