7 -2
Fibonacci Sequence
Fibonacci sequence: 0 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , …
Fi = i if i 1
Fi = Fi-1 + Fi-2 if i 2 Solved by a recursive program:
Much replicated computation is done. It should be solved by a simple loop.
f2
f4 f3
f1
f3
f2 f1
f1 f0
f2
f1 f0
f1
f5
f0
7 -3
Dynamic Programming Dynamic Programming is an
algorithm design method that can be used when the solution to a problem may be viewed as the result of a sequence of decisions.
7 -4
The Shortest Path
To find a shortest path in a multi-stage graph
Apply the greedy method: the shortest path from S to T: 1 + 2 + 5 = 8.
S A B T
3
4
5
2 7
1
5 6
7 -5
The Shortest Path in Multi-stage Graphs
e.g.
The greedy method cannot be applied to this case: (S, A, D, T) 1 + 4 + 18 = 23.
The real shortest path is: (S, C, F, T) 5 + 2 + 2 = 9.
S T132
B E
9
A D4
C F2
1
5
11
5
16
18
2
7 -6
Dynamic Programming Approach
Dynamic programming approach (forward approach):
d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C, T)}
S T2
B
A
C
1
5d(C, T)
d(B, T)
d(A, T)
A
T
4
E
D
11d(E, T)
d(D, T) d(A,T) = min{4+ d(D,T), 11+d(E,T)} = min{4+18, 11+13} = 22.
S T132
B E
9
A D4
C F2
1
5
11
5
16
18
2
7 -7
d(B, T) = min{9+d(D, T), 5+d(E, T), 16+d(F, T)} = min{9+18, 5+13, 16+2} = 18.
d(C, T) = min{ 2+d(F, T) } = 2+2 = 4 d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C, T)} = min{1+22, 2+18, 5+4} = 9. The above way of reasoning is called backward reasoning.
S T132
B E
9
A D4
C F2
1
5
11
5
16
18
2
7 -8
Backward Approach (Forward Reasoning)
d(S, A) = 1d(S, B) = 2d(S, C) = 5
d(S,D)=min{d(S, A)+d(A, D),d(S, B)+d(B, D)} = min{ 1+4, 2+9 } = 5 d(S,E)=min{d(S, A)+d(A, E),d(S, B)+d(B, E)} = min{ 1+11, 2+5 } = 7 d(S,F)=min{d(S, A)+d(A, F),d(S, B)+d(B, F)} = min{ 2+16, 5+2 } = 7
S T132
B E
9
A D4
C F2
1
5
11
5
16
18
2
7 -9
d(S,T) = min{d(S, D)+d(D, T),d(S,E)+ d(E,T), d(S, F)+d(F, T)}
= min{ 5+18, 7+13, 7+2 } = 9
S T132
B E
9
A D4
C F2
1
5
11
5
16
18
2
7 -10
Principle of Optimality Principle of optimality: Suppose that in solving
a problem, we have to make a sequence of decisions D1, D2, …, Dn. If this sequence is optimal, then the last k decisions, 1 k n must be optimal.
e.g. the shortest path problem
If i, i1, i2, …, j is a shortest path from i to j, then i1, i2, …, j must be a shortest path from i1 to j
In summary, if a problem can be described by a multi-stage graph, then it can be solved by dynamic programming.
7 -11
Forward approach and backward approach: If the recurrence relations are formulated using
the forward approach, the relations are solved backward, i.e., beginning with the last decision.
If the relations are formulated using the backward approach, they are solved forwards.
To solve a problem by using dynamic programming: Find out the recurrence relations. Represent the problem by a multi-stage graph.
Dynamic Programming
7 -12
The Resource Allocation Problem
m resources, n projects p(i, j): the profit obtained when j
resources are allocated to project i. Maximize the total profit.
Resource Project
1
2
3
1 2 8 9 2 5 6 7 3 4 4 4 4 2 4 5
7 -13
The Multi-Stage Graph Solution
The resource allocation problem can be described as a multi-stage graph.
(i, j): i resources allocated to projects 1, 2, …, je.g. node H = (3, 2): 3 resources allocated to projects 1, 2.
S T
6
0,1
1,1
2,1
3,1
0,2
1,2
2,2
3,2
0,3
1,3
2,3
3,3
A
7
6
44
4
B
C
D H
G
F
E I
J
K
L
0 5
8
9
0
0
0
0
5
5
5
0
0
0
0
4
4
4
42
2
0
7 -14
Find the longest path from S to T: (S, C, H, L, T), 8 + 5 + 0 + 0 = 13.
2 resources allocated to project 1.1 resource allocated to project 2.0 resource allocated to projects 3, 4.
7 -15
The Longest Common Subsequence (LCS)
problem A string: A = b a c a d A subsequence of A: deleting 0 or more
symbols from A (not necessarily consecutive).
e.g. ad, ac, bac, acad, bacad, bcd. Common subsequences of A = b a c a d and B = a c c b a d c b : ad, ac, bac, acad. The longest common subsequence (LCS) of
A and B: a c a d.
7 -16
The LCS Algorithm Let A = a1 a2 am and B = b1 b2 bn
Let Li,j denote the length of the longest common subsequence of a1 a2 ai and b1 b2 bj.
Li,j = Li-1,j-1 + 1 if ai=bj
max{ Li-1,j, Li,j-1 } if aibj
L0,0 = L0,j = Li,0 = 0 for 1im, 1jn.
7 -17
The dynamic programming approach for solving the LCS problem:
Time complexity: O(mn)
L1,1
L2,1
L3,1
L1,2 L1,3
L2,2
Lm,n
7 -18
Tracing Back in the LCS Algorithm
e.g. A = b a c a d, B = a c c b a d c b
After all Li,j’s have been found, we can trace back to find the longest common subsequence of A and B.
2
43
100000
0 0 0
111
1 1 11 1 1 1 1
22 2 2
222
2 2 2 222
33 3 3
33
4 4
0 0 0 0 0 0 0 00 0 0 0 0
bac
a
d
A
ac c ba d cbB
2
7 -19
The Two-Sequence Alignment Problem
Sequence alignment may be viewed as a method to measure the similarity of two sequences
Let A = a1 a2 am and B = b1 b2 bn A sequence alignment of A and B : a 2k
matrix M (k m, n ) of characters over {-}. e.g.
if A = a b c d and B = c b d.a possible alignment of them would be:
a b c - d - - c b d
7 -20
Let f(x, y) denote the score for aligning x with y. If x and y are the same, f(x, y)= 2.
If x and y are not the same, f(x, y)= 1.If x or y is “-“, f(x, y)= -1.
e.g.
A= a b c d
B= c b - dThe total score of the alignment is 1+ 2 – 1 + 2 = 4.
7 -21
Let Ai,j denote the optimal alignment score between a1 a2 ai and b1 b2 bj and, where 1 i m and 1 j n.
Then, Ai,j can be expressed as follows:
Ai,0 : a1 a2 ai are all aligned with “-”.
A0,j : b1 b2 bj are all aligned with “-”.
f(ai ,-): ai is aligned with “-”.
f(ai ,bj): ai is aligned with bj.
f(- ,bj): bj is aligned with “-”.
),(
),(
),(
max
),(
),(
0
1,
1,1
,1
,
,0
0,
0,0
jji
jiji
iji
ji
jj
ii
bfA
bafA
afA
A
bfjA
afiA
A
7 -22
The Ai,j’s for A = a b d a d and B = b a c d using the above recursive formula are listed below: ai
bj
a b d a d
0 1 2 3 4 5
0 0 -1 -2 -3 -4 -5
b 1 -1 1 1 0 -1 -2
a 2 -2 1 2 2 2 1
c 3 -3 0 2 3 3 3
d 4 -4 -1 1 4 4 5
7 -23
In the above table, we have also recorded how each Ai,j is obtained.An arrow from (ai,bj) to (ai-1,bj-1): ai matched with bj.
An arrow from (ai,bj) to (ai-1,bj): ai matched with “-”,
An arrow from (ai,bj) to (ai,bj-1): bj matched with “-”.
Based upon the arrows in the table, we can trace back and find the optimal alignment is as follows:
a b d a d
- b a c d
7 -24
Edit Distance Edit Distance is also used quite often to
measure the similarity between two
sequences. Transform A to B by three edit operations:
deletion of a character from A. insertion of a character into A. substitution of a character in A with another
character.
7 -25
For exampleLet A = GTAAHTY and B = TAHHYC
(1) Delete the first character G of A. GTAAHTY → TAAHTY(2) Substitute the third character of A by H.
TAAHTY → TAHHTY(3) Delete the fifth character of A.
TAHHTY → TAHHY(4) Insert C after the last character of A.
TAHHY → TAHHYC = B
7 -26
Associate a cost with each operation. Let , and denote the costs of insertion, deletion and
substitution respectively. Ai,j : the edit distance between a1 a2 .. ai and b1 b2 ..bj.
It takes O(nm) time to find an optimal alignment.
otherwisemin
if
0
1,
1,1
,1
1,1
,
,0
0,
0,0
ji
ji
ji
jiji
ji
j
i
A
A
A
baA
A
jA
iA
A
7 -27
The RNA Maximum Base Pair Matching Problem
Ribonucleic acid (RNA) is a single strand of nucleotides (bases) adenine (A), guanine (G), cytosine (C) and uracil (U).
The sequence of the bases A, G, C and U is called the primary structure of an RNA.
The primary structure of an RNA can fold back on itself to form its secondary structure. G and C can form a base pair G≡C by a triple-
hydrogen bond. A and U can form a base pair A=U by a
double-hydrogen bond. G and U can form a base pair GU by a single
hydrogen bond.
7 -28
E.g. The primary structure of an RNA, R:
A–G–G–C–C–U–U–C–C–U Six possible secondary structures of RNA, R:
7 -29
An RNA sequence will be represented as a string of n characters R = r1r2 · · ·rn, where ri {A, C, G, U}.
A secondary structure of R is a set S of base pairs (ri, rj), where 1 i < j n. (1) j - i > t, where t is a small positive constant. Typically, t
= 3. (2) If (ri, rj) and (rk, rl) are two base pairs in S and i k,
then either(a) i = k and j = l, i.e., (ri, rj) and (rk, rl) are the same
base pair,(b) i < j < k < l, i.e., (ri, rj) precedes (rk, rl), or
(c) i < k < l < j, i.e., (ri, rj) includes (rk, rl). Pseudoknot: Two base pairs (ri, rj) and (rk, rl) ,if i < k < j < l Due to different hydrogen bonds, the energies of base pairs
are usually assigned different values. For example, the reasonable values for A≡U, G=C and G–U are -3, -2 and -1, respectively.
7 -30
Dynamic Programming Given an RNA sequence R = r1r2 · · ·rn, find a secondary
structure of RNA with the maximum number of base pairs. Let Si,j denote the secondary structure of the maximum
number of base pairs on the substring Ri,j = riri+1 · · · rj.
Denote the number of matched base pairs in Si,j by Mi,j .
If j - i 3, then ri and rj cannot be a base pair of Si,j and Mi,j = 0 .
Let WW = {(A, U), (U, A), (G, C), (C, G), (G, U), (U, G)}. ρ(ri, rj) : whether any two bases ri and rj can be a legal
base pair:
otherwise0
),( if1),(
WWrrrr
jiji
7 -31
To compute Mi,j, where j- i > 3, we consider the following cases from rj’s point of view. Case 1: In the optimal solution, rj is
not paired with any other base. In this case, find an optimal solution for riri+1 . . . rj-1 and Mi,j =Mi,j-1.
7 -32
Case 2: In the optimal solution, rj is paired with ri. In this case, find an optimal solution for ri+1ri+2 . . . rj-1 and Mi,j = 1 + Mi+1,j-1.
7 -33
Case 3: In the optimal solution, rj is paired with some rk, where i + 1 k j - 4. In this case, find the optimal solutions for riri+1 . . . rk-1 and rk+1rk+1 . . . rj-1 and Mi,j = 1 + Mi,k-i+ Mk+1,j-1.
7 -34
Find the k between i+1 and j+ 4 such that Mi,j
is the maximum, we have
Compute Mi,j by the following recursive formula
If j - i ≦ 3, then Mi,j = 0. If j - i > 3, then
1,11,41
, 1max jkki
jkiji MMM
),()1(max
),()1(max
1,11,41
1,1
1,
,
jkjkkijki
jkji
ji
ji
rrMM
rrM
M
M
7 -35
The following table illustrates the computation of Mi,j, where 1 i < j 10, for an RNA sequence R1,10 = A–G–G–C–C–U–U–C–C–U.
Maximum number of base pairs in S1,10 is 3 since M1,10 = 3.
i\j 1 2 3 4 5 6 7 8 9 101 0 0 0 0 0 1 2 2 2 32 - 0 0 0 0 1 1 2 2 23 - - 0 0 0 0 1 1 1 14 - - - 0 0 0 0 0 0 05 - - - - 0 0 0 0 0 06 - - - - - 0 0 0 0 07 - -- - - - - 0 0 0 08 - - - - - - - 0 0 09 - - - - - - - - 0 0
10 - - - - - - - - - 0
7 -36
Algorithm to Compute M1,n
Input: An RNA sequence R = r1r2 · · · rn.
Output: Find a secondary structure of RNA with the
maximum number of base pairs. Step 1: /* Computation of ρ(ri, rj) function for 1 i < j n
*/ WW = {(A, U), (U, A), (G, C), (C, G), (G, U), (U, G)};for i = 1 to n do for j = i to n do if (ri, rj) WW then ρ(ri, rj) = 1; else ρ(ri, rj) = 0;
end forend for
7 -37
Step 2: /* Initialization of Mi,j for j - i 3 */for i = 1 to n do for j = i to i + 3 do if j ≦ n then Mi,j = 0; end forend forStep 3: /* Calculation of Mi,j for j - i > 3 */for h = 4 to n - 1 do for i = 1 to n - h do j = i + h; case1 = Mi,j-1; case2 = (1+Mi+1,j-1) × ρ(ri, rj ); case3 = Mi,j = max{case1, case2, case3}; end forend for
Time-complexity : O(n3). ),()1(max 1,11,41
jkjkkijki
rrMM
7 -38
0/1 Knapsack Problem n objects , weights W1, W2, ,Wn
profits P1, P2, ,Pn
capacity M maximize
subject to M xi = 0 or 1, 1in e.g.
ni
iixP1
ni
iixW1
i Wi Pi 1 10 40 2 3 20 3 5 30
M=10
7 -39
The Multi-Stage Graph Solution
The 0/1 knapsack problem can be described by a multi-stage graph.
S T
0
1 0
10
00
01
100
010
011
000
001
0
0
0
0
00
40
020
0
30
0
0
30
x1=1
x1=0
x2=0
x2=1
x2=0
x3=0
x3=1
x3=0
x3=1
x3=0
7 -40
The Dynamic Programming Approach
The longest path represents the optimal solution:
x1=0, x2=1, x3=1
= 20+30 = 50 Let fi(Q) be the value of an optimal solution
to objects 1,2,3,…,i with capacity Q. fi(Q) = max{ fi-1(Q), fi-1(Q-Wi)+Pi }
The optimal solution is fn(M).
iixP
7 -41
Optimal Binary Search Trees
e.g. binary search trees for 3, 7, 9, 12;
3
7
12
9
(a) (b)
9
3
7
12
12
3
7
9
(c)
12
3
7
9
(d)
7 -42
Optimal Binary Search Trees
n identifiers : a1 <a2 <a3 <…< an
Pi, 1in : the probability that ai is searched.
Qi, 0in : the probability that x is searched
where ai < x < ai+1 (a0=-, an+1=).111
n
ii
n
ii QP
7 -43
Identifiers : 4, 5, 8, 10, 11, 12, 14
Internal node: successful search, Pi
External node: unsuccessful search, Qi
10
14
E 7
5
11
12E 4
4
E 0 E 1
8
E 2 E 3
E 5 E 6
The expected cost of a binary tree:
The level of the root: 1
n
0nii
n
1nii 1))(level(EQ)level(aP
7 -44
The Dynamic Programming Approach
Let C(i, j) denote the cost of an optimal binary search tree containing ai,…,aj .
The cost of the optimal binary search tree with ak as its root:
ak
a1...ak-1 ak+1...an
P 1 ...P k-1
Q 0...Q k-1
P k+1 ...P n
Q k ...Q n
C(1,k-1) C(k+1,n)
n1,kCQPQ1k1,CQPQPminn)C(1,n
1kmmmk
1k
1mmm0k
nk1
7 -45
j
immm1-i
jki
j
1kmmmk
1k
immm1-ik
jki
QPQj1,kC1ki,Cmin
j1,kCQPQ
1ki,CQPQPminj)C(i,
General Formula
ak
ai...ak-1 ak+1...aj
Pi...Pk-1
Qi-1...Qk-1
Pk+1...Pj
Qk...Qj
C(i,k-1) C(k+1,j)