[SOLVED] CS代考程序代写 Bioinformatics algorithm Lecture 7: Dynamic Programming II

Lecture 7: Dynamic Programming II
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Changes to the unit
More resources
– Recorded tutorials
– Online office hours (Tue 4-6, Fri 2-4). Contact me beforehand on Slack. Maybe available at other times as well
Assignments
– Scale back in both difficulty and workload
Final Exam
– Online, 2 hours long on ProctorU
– Some knowledge-based questions, some algorithm design problems, easier than assignments
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How do I get good at problem solving?
– Work on lots of problems, especially the tutorial ones
– Present your solution during tutorial or on Ed to get
feedback
– Feel free to ask about solution to other problems, such as those from Jeff Erickson’s textbook
– Approaches:
– Make simplifying assumptions, solve special
cases
– Reuse computation (esp. useful for D&C) – Reformulate problem
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General techniques in this course
– Greedy algorithms [Lecture 2]
– Divide & Conquer algorithms [Lecture 3]
– Dynamic programming algorithms [Lectures 4 and today] – Network flow algorithms [Lectures 6 and 7]
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Algorithmic Paradigms
– Greedy. Build up a solution incrementally, myopically optimizing some local criterion.
– Divide-and-conquer. Break up a problem into two sub- problems, solve each sub-problem independently, and combine solution to sub-problems to form solution to original problem.
– Dynamic programming. Break up a problem into a series of overlapping sub-problems, and build up solutions to larger and larger sub-problems.
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Main Idea: Dynamic Programming
Recurrence equation relating optimal solution in terms optimal solutions to smaller subproblems
Given optimal solutions to smaller subproblems, how to construct solution to original problem?
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Key steps: Dynamic programming
1. Define subproblems 2. Find recurrences
3. Solve the base cases
4. Transform recurrence into an efficient algorithm
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Recap: Weighted Interval Scheduling
– Job j starts at sj, finishes at fj, and has weight vj.
– Two jobs compatible if they don’t overlap.
– Goal: find maximum weight subset of mutually compatible jobs.
b
a
c
d
e
f
g
h
0 1 2 3 4 5 6 7 8 9 10 11 The University of Sydney
Time
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Recap: Weighted Interval Scheduling
Notation. Label jobs by finishing time: f1 £ f2 £ . . . £ fn .
Def. p(j) = largest index i M[j-1])
print j
Find-Solution(p(j))
else
Find-Solution(j-1)
}
# of recursive calls £ n Þ O(n). The University of Sydney
picked job j
16
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Recap: Knapsack Problem
– Knapsackproblem.
– Givennobjectsanda”knapsack.”
– Item i weighs wi > 0 kilograms and has value vi > 0. – KnapsackhascapacityofWkilograms.
– Goal: fill knapsack so as to maximize total value.
W = 11
Item
Value
Weight
1
1
1
2
6
2
3
18
5
4
22
6
5
28
7
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Recap: Dynamic Programming – Step 1 Step 1: Define subproblems
OPT(i, w) = max profit with subset of items 1, …, i with weight limit w.
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Recap: Dynamic Programming – Step 2 Step 2: Find recurrences
– Case 1: OPT does not select item i.
• OPT selects best of { 1, 2, …, i-1 } using weight limit w
– Case 2: OPT selects item i.
• newweightlimit=w–wi
• OPT selects best of { 1, 2, …, i–1 } using this new weight limit
OPT(i,w) = max { vi + OPT (i-1,w-wi), OPT(i-1,w) }
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Recap: Dynamic Programming – Step 2 Step 2: Find recurrences
– Case 1: OPT does not select item i.
• OPT selects best of { 1, 2, …, i-1 } using weight limit w
– Case 2: OPT selects item i.
• newweightlimit=w–wi
• OPT selects best of { 1, 2, …, i–1 } using this new weight limit
If wi>w
OPT(i,w) = OPT (i-1,w)
otherwise
OPT(i,w) = max { vi + OPT (i-1,w-wi), OPT(i-1,w) }
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Recap: Dynamic Programming – Step 3 Step 3: Solve the base cases
OPT(0,w) = 0
ì0 if i=0 OPT(i,w)=ïíOPT(i-1,w) if i>0andwi >w
ïîmax{OPT(i-1,w),vi+OPT(i-1,w-wi)} otherwise
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Done…more or less

Knapsack Problem: Bottom-Up
– Knapsack. Fill up an (n+1)-by-(W+1) array. Space complexity: O(nW).
Input: n, w1,…,wN, v1,…,vN for w = 0 to W
M[0, w] = 0
for i = 1 to n
for w = 1 to W
if (wi > w)
M[i, w] = M[i-1, w]
else
M[i, w] = max {M[i-1, w], vi + M[i-1, w-wi ]} return M[n, W]
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n+1
W+1
0 1 2 3 4 5 6 7 8 9 10 11
Æ 000000000000 {1} 011111111111 {1,2} 016777777777
{1,2,3} 0 1 6 7 7 18 19 24 25 25 25 25
Knapsack Algorithm
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{1,2,3,4} {1,2,3,4,5}
0 1 6 7 7 18 22 24 28 29 29 40 0 1 6 7 7 18 22 28 29 34 34 40
OPT: {4,3}
value = 22 + 18 = 40
Item Value Weight
111 262 W=11 3 18 5 4 22 6
5 28 7

Longest Common Subsequence
Given two sequences X[1..n] and Y[1..m], find the longest subsequences X’ of X and Y’ of Y such that X’ = Y’.
Example 1 X = BANANAS Y = KATANA
LCS(X,Y) = AANA
Example 2 X = DYNAMIC
Y = PROGRAMMING
LCS(X,Y) = AMI The University of Sydney
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Longest Common Subsequence: Matching Definition
Given two sequences X[1..n] and Y[1..m], find the longest non- crossing matching between elements of X and elements of Y
Example 1 X = BANANAS Y = KATANA
LCS(X,Y) = AANA. Matching: X[2] – Y[2], X[4] – Y[4], X[5] – Y[5], X[6] – Y[6].
Example 2 X = DYNAMIC
Y = PROGRAMMING
LCS(X,Y) = AMI. Matching: X[4] – Y[6], X[5] – Y[7], X[6] – Y[8],
Note: if crossings allowed, then can add X[3] – Y[10]. The University of Sydney
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Longest Common Subsequence: Applications
– Measures similarity of two strings – Bioinformatics
– Merging in version control
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LCS Dynamic Programming: Step 1 Step 1: Define subproblems
OPT(i,j) = length of LCS(X[1..i], Y[1..j]). Subproblems: finding LCS of prefixes of X and Y
Example
X = BANANAS Y = KATANA
OPT(6, 4) = length of LCS(BANANA, KATA) = 2 The University of Sydney
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LCS Dynamic Programming: Step 2
Notations:
– OPT(i,j) = length of LCS(X[1..i], Y[1..j]). Step 2: Finding recurrences
– Case 1: X[i] ≠ Y[j].
– Leave X[i] unmatched and take LCS(X[1..i-1], Y[1..j]). OR – Leave Y[j] unmatched and take LCS(X[1..i], Y[1..j-1]).
– OPT(i,j)= max{OPT(i-1,j), OPT(i,j-1)}
Example
X[1..5] = BANAN Y[1..6] = KATANA
X[5] unmatched: LCS = LCS(X[1..4], Y[1..6]) = ANA
Y[6] unmatched: LCS = LCS(X[1..5], Y[1..5]) = AAN The University of Sydney
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LCS Dynamic Programming: Step 2
Notations:
– OPT(i,j) = length of LCS(X[1..i], Y[1..j]). Step 2: Finding recurrences
– Case 1: X[i] ≠ Y[j].
– Leave X[i] unmatched and take LCS(X[1..i-1], Y[1..j]). OR – Leave Y[j] unmatched and take LCS(X[1..i], Y[1..j-1]).
– OPT(i,j)= max{OPT(i-1,j), OPT(i,j-1)}
Example
X[1..5] = BANAN Y[1..6] = KATANA
X[5] unmatched: LCS = LCS(X[1..4], Y[1..6]) = ANA
Y[6] unmatched: LCS = LCS(X[1..5], Y[1..5]) = AAN
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LCS Dynamic Programming: Step 2
Notations:
– OPT(i,j) = length of LCS(X[1..i], Y[1..j]). Step 2: Finding recurrences
– Case 1: X[i] ≠ Y[j].
– Leave X[i] unmatched and take LCS(X[1..i-1], Y[1..j]). OR – Leave Y[j] unmatched and take LCS(X[1..i], Y[1..j-1]).
– OPT(i,j)= max{OPT(i-1,j), OPT(i,j-1)}
Example
X[1..3] = BAN Y[1..6] = KATANA
X[3] unmatched: LCS(X[1..2], Y[1..6]) = A
Y[6] unmatched: LCS(X[1..3], Y[1..5]) = AN The University of Sydney
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LCS Dynamic Programming: Step 2
Notations:
– OPT(i,j) = length of LCS(X[1..i], Y[1..j]). Step 2: Finding recurrences
– Case 1: X[i] ≠ Y[j].
– Leave X[i] unmatched and take LCS(X[1..i-1], Y[1..j]). OR – Leave Y[j] unmatched and take LCS(X[1..i], Y[1..j-1]).
– OPT(i,j)= max{OPT(i-1,j), OPT(i,j-1)}
Example
X[1..3] = BAN Y[1..6] = KATANA
X[3] unmatched: LCS(X[1..2], Y[1..6]) = A
Y[6] unmatched: LCS(X[1..3], Y[1..5]) = AN
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LCS Dynamic Programming: Step 2
Notations:
– OPT(i,j) = length of LCS(X[1..i], Y[1..j]). Step 2: Finding recurrences
– Case 2: X[i] = Y[j].
– Match X[i] to Y[j] and take LCS(X[1..i-1], Y[1..j-1]) + X[i] OR
– Leave X[i] unmatched or Y[j] unmatched
– OPT(i,j) = max{OPT(i-1, j-1) + 1, OPT(i-1, j), OPT(i, j-1)
Example
X[1..6] = BANANA Y[1..6] = KATANA
Match X[6] – Y[6]: LCS = LCS(X[1..5], Y[1..5]) + A = AANA
X[6] unmatched: LCS = LCS(X[1..5], Y[1..6]) = AAN
Y[6] unmatched: LCS = LCS(X[1..6], Y[1..5]) = AAN The University of Sydney
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LCS Dynamic Programming: Step 3 Step 3: Solving the base cases
OPT(i,0) = 0 for all i, OPT(0,j) = 0 for all j
8>0 if i = 0 or j = 0 <OPT(i,j)= max{OPT(i1,j1)+1,OPT(i,j1),OPT(i1,j)} ifX[i]=Y[j] >
max{OPT(i, j 1), OPT(i 1, j)} if X[i] 6= Y [j]
:
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LCS Dynamic Programming: Algorithm
INPUT: n, m, X[1..n], Y[1..m]
for j = 0 to m
M[0,j] = 0
for i = 0 to n
M[i,0] = 0
for i = 1 to n
for j = 1 to m
if (X[i] = Y[j])
M[i,j] = max(M[i-1,j-1] + 1, M[i-1,j], M[i,j-1])
else
M[i,j] = max(M[i-1,j], M[i,j-1])
return M[n,m]
8><>:0 if i = 0 or j = 0 OPT(i,j)= max{OPT(i1,j1)+1,OPT(i,j1),OPT(i1,j)} ifX[i]=Y[j] max{OPT(i, j 1), OPT(i 1, j)} if X[i] 6= Y [j]
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LCS Dynamic Programming: Analysis
INPUT: n, m, X[1..n], Y[1..m]
for j = 0 to m
M[0,j] = 0
for i = 0 to n
M[i,0] = 0
for i = 1 to n
for j = 1 to m
O(n + m)
O(nm)
if (X[i] = Y[j])
M[i,j] = max(M[i-1,j-1] + 1, M[i-1,j], M[i,j-1])
else
M[i,j] = max(M[i-1,j], M[i,j-1])
return M[n,m] Running time: O(nm)
Space: O(nm)
See 3927 Lecture 6 for Sequence Alignment
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6.5 RNA Secondary Structure
Dynamic programming over intervals
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RNA (Ribonucleic acid) Secondary Structure
– RNA. String B = b1b2…bn over alphabet { A, C, G, U }.
– Secondary structure. RNA is single-stranded so it tends to loop
back and form base pairs with itself. This structure is essential
for understanding behavior of molecule.
CA AA AU
GC CGUAA G
Ex: GUCGAUUGAGCGAAUGUAACAACGUGGCUACGGCGAGA UGAUUA
G
ACGCU CGCGAGC
G
G AU
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complementary base pairs: A-U, C-G G
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RNA Secondary Structure
– Secondary structure. A set of pairs S = { (bi, bj) } that satisfy: – [Watson-Crick.] S is a matching and each pair in S is a Watson-Crick
complement: A-U, U-A, C-G, or G-C.
– [No sharp turns.] The ends of each pair are separated by at least 4
intervening bases. If (bi, bj) Î S, then i < j – 4.– [Non-crossing.] If (bi, bj) and (bk, bl) are two pairs in S, then we cannothave i < k < j < l.– Free energy. Usual hypothesis is that an RNA molecule will form the secondary structure with the optimum total free energy.approximated by number of base pairs– Goal. Given an RNA molecule B = b1b2…bn, find a secondary structure S that maximizes the number of base pairs. The University of SydneyPage 3939 RNA Secondary Structure: ExamplesGGGGG CUGGCU CGCGCUAUAUAG UAUAUAbase pairAUGUGGCCAU AUGGGGCAU AGUUGGCCAU £4ok sharp turn crossing The University of SydneyPage 40 RNA Secondary Structure: Subproblems– First attempt (Step 1). OPT(j) = maximum number of base pairs in a secondary structure of the substring b1b2…bj.1match bt and bitj– Difficulty (in Step 2). Results in two sub-problems. – Finding secondary structure in: b1b2…bt-1.– Finding secondary structure in: bt+1bt+2…bj-1.OPT(t-1)The University of Sydney need more sub-problems Page 41 Dynamic Programming Over Intervals – Step 1 Step 1: Define subproblemsOPT(i, j) = maximum number of base pairs in a secondary structure of the substringbibi+1…bj.The University of SydneyPage 42 Dynamic Programming Over Intervals – Step 2 Notation. OPT(i, j) = maximum number of base pairs in asecondary structure of the substring bibi+1…bj. Step 2: Find recurrencesThe University of SydneyPage 43ijCase 1. Base bj is not involved in a pair. OPT(i, j) = OPT(i, j-1) Dynamic Programming Over Intervals – Step 2 Notation. OPT(i, j) = maximum number of base pairs in asecondary structure of the substring bibi+1…bj. Step 2: Find recurrences The University of SydneyPage 44imatch bt and bitjCase2. Basebjpairswithbt forsomei£t M[w] + cvw) {
M[v] ¬ M[w] + cvw
successor[v] ¬ w }
} }
If no M[w] value changed in iteration i, stop.
}
}
Analysis: Q(mn) time, Q(n) working space. The University of Sydney Page 67

Shortest Paths: Practical Improvements
– Practical improvements
– Maintain only one array M[v] = shortest v-t path that we have
found so far.
– No need to check edges of the form (v, w) unless M[w] changed in previous iteration.
– Theorem: Throughout the algorithm, M[v] is length of some v-t path, and after i rounds of updates, the value M[v] is no larger than the length of shortest v-t path using £ i edges.
– Overall impact
– Working space: O(n).
– Total space (including input): O(m+n)
– Running time: O(mn) worst case, but substantially faster in practice.
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6.3 Segmented Least Squares
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Segmented Least Squares
– Least squares.
– Foundational problem in statistic and numerical analysis.
– Given n points in the plane: (x1, y1), (x2, y2) , . . . , (xn, yn).
– Find a line y = ax + b that minimizes the sum of the squared error:
y
SSE = ån (yi -axi -b)2 i =1
x
– Solution. Calculus Þ min error is achieved when
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a=nåixiyi -(åixi)(åi yi), b=åi yi -aåixi
n å x 2 – (å x )2 n ii ii
O(n)

Segmented Least Squares
– Segmented least squares.
– Points lie roughly on a sequence of several line segments.
– Givennpointsintheplane(x1,y1),(x2,y2),…,(xn,yn)with – x1 < x2 < … < xn, find a sequence of lines that minimizes f(x).– Question. What’s a reasonable choice for f(x) to balance accuracy and complexity?number of linesy The University of Sydneyx Page 71 Segmented Least Squares– Segmented least squares.– Points lie roughly on a sequence of several line segments.– Givennpointsintheplane(x1,y1),(x2,y2),…,(xn,yn)withx1 < x2 < … < xn, find a sequence of lines that minimizes:• the sum of the sums of the squared errors E in each segment • the number of lines L– Tradeoff function: E + c·L, for some constant c > 0.
y
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x Page 72

Segmented Least Squares
– Segmented least squares.
– Points lie roughly on a sequence of several line segments.
– Givennpointsintheplane(x1,y1),(x2,y2),…,(xn,yn)with
x1 < x2 < … < xn, find a partition into segments that minimizes: • the sum of the sums of the squared errors E in each segment • the number of segments L– Tradeoff function: E + c·L, for some constant c > 0.
y
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x Page 73

Dynamic Programming: Multiway Choice – Step 1 Step 1: Define subproblems
OPT(j) = minimum cost for points p1, p2 , . . . , pj.
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Dynamic Programming: Multiway Choice – Step 2
Notations:
– OPT(j)=minimumcostforpointsp1,p2 ,…,pj.
– e(i,j) =minimumsumofsquaresforpointspi,pi+1 ,…,pj.
Step 2: Finding recurrences
– Lastsegmentusespointspi,pi+1 ,…,pj forsomei. – Cost = e(i, j) + c + OPT(i-1).
OPT(j) = min { e(i,j) + c + OPT (i-1) } 1≤i≤j
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Dynamic Programming: Multiway Choice – Step 3 Step 3: Solving the base cases
OPT(0) = 0
OPT(j) = (0 if j = 0 min1ije(i,j)+c+OPT(i1) ifj>0
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Segmented Least Squares: Algorithm
O(n2) iterations
O(n) iterations
INPUT: n, (p1,…,pn), c
Segmented-Least-Squares() {
M[0] = 0
for j = 1 to n
for i = 1 to j
compute the least square error eij for the segment pi,…, pj
O(n)
O(n)
for j = 1 to n
M[j]=min1 £ i £ j(eij +c+M[i-1])
return M[n] }
OPT(j) = (0 if j = 0 min1ije(i,j)+c+OPT(i1) ifj>0
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Segmented Least Squares: Algorithm
2 O(n )
iterations
O(n) iterations
for j = 1 to n fori=1toj
compute the least square error eij for the segment pi,…, pj
for j = 1 to n
M[j]=min1 £ i £ j(eij +c+M[i-1])
O(n)
O(n)
INPUT: n, (p1,…,pn), c
Segmented-Least-Squares() {
M[0] = 0
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return M[n] }
Running time: O(n3) Space: O(n2)

Dynamic Programming Summary I
3 steps:
1. Definesubproblems
2. Findrecurrences
3. Solvethebasecases
4. Transformrecurrenceintoanefficientalgorithm [usually bottom-up]
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Dynamic Programming Summary II
– 1D dynamic programming
– Weighted interval scheduling
– Segmented Least Squares
– Maximum-sum contiguous subarray – Longest increasing subsequence
– 2D dynamic programming – Knapsack
– Shortest path
– Longest common subsequence
– Dynamic programming over intervals – RNA Secondary Structure
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