06DynamicProgramming in algorithms CSE.pdf

Lecture slides by Kevin Wayne
Copyright © 2005 Pearson-Addison Wesley
http://www.cs.princeton.edu/~wayne/kleinberg-tardos
Last updated on 2/10/21 2:39 PM
6. DYNAMIC PROGRAMMING I
‣ weighted interval scheduling
‣ segmented least squares
‣ knapsack problem
‣ RNA secondary structure

Algorithmic paradigms
Greed. Process the input in some order, myopically making irrevocable
decisions.
Divide-and-conquer. Break up a problem into independent subproblems;
solve each subproblem; combine solutions to subproblems to form solution
to original problem.
Dynamic programming. Break up a problem into a series of overlapping
subproblems; combine solutions to smaller subproblems to form solution
to large subproblem.
2
fancy name for
caching intermediate results
in a table for later reuse

Dynamic programming history
Bellman. Pioneered the systematic study of dynamic programming in 1950s.
Etymology.
・Dynamic programming = planning over time.
・Secretary of Defense had pathological fear of mathematical research.
・Bellman sought a “dynamic” adjective to avoid conflict.
3
THE THEORY OF DYNAMIC PROGRAMMING
RICHARD BELLMAN
1. Introduction. Before turning to a discussion of some representa-
tive problems which will permit us to exhibit various mathematical
features of the theory, let us present a brief survey of the funda-
mental concepts, hopes, and aspirations of dynamic programming.
To begin with, the theory was created to treat the mathematical
problems arising from the study of various multi-stage decision
processes, which may roughly be described in the following way: We
have a physical system whose state at any time / is determined by a
set of quantities which we call state parameters, or state variables.
At certain times, which may be prescribed in advance, or which may
be determined by the process itself, we are called upon to make de-
cisions which will affect the state of the system. These decisions are
equivalent to transformations of the state variables, the choice of a
decision being identical with the choice of a transformation. The out-
come of the preceding decisions is to be used to guide the choice of
future ones, with the purpose of the whole process that of maximizing
some function of the parameters describing the final state.
Examples of processes fitting this loose description are furnished
by virtually every phase of modern life, from the planning of indus-
trial production lines to the scheduling of patients at a medical
clinic ; from the determination of long-term investment programs for
universities to the determination of a replacement policy for ma-
chinery in factories; from the programming of training policies for
skilled and unskilled labor to the choice of optimal purchasing and in-
ventory policies for department stores and military establishments.
It is abundantly clear from the very brief description of possible
applications that the problems arising from the study of these
processes are problems of the future as well as of the immediate
present.

Dynamic programming applications
Application areas.
・Computer science: AI, compilers, systems, graphics, theory, ….
・Operations research.
・Information theory.
・Control theory.
・Bioinformatics.
Some famous dynamic programming algorithms.
・Avidan–Shamir for seam carving.
・Unix diff for comparing two files.
・Viterbi for hidden Markov models.
・De Boor for evaluating spline curves.
・Bellman–Ford–Moore for shortest path.
・Knuth–Plass for word wrapping text in .
・Cocke–Kasami–Younger for parsing context-free grammars.
・Needleman–Wunsch/Smith–Waterman for sequence alignment.
4
T1X
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SECTIONS 6.1–6.2

Weighted interval scheduling
・Job j starts at sj, finishes at fj, and has weight wj > 0.
・Two jobs are compatible if they don’t overlap.
・Goal: find max-weight subset of mutually compatible jobs.
7
time
f
g
h
e
a
b
c
d
0 1 2 3 4 5 6 7 8 9 10 11
sj fj
wj

Earliest-finish-time first algorithm
Earliest finish-time first.
・Consider jobs in ascending order of finish time.
・Add job to subset if it is compatible with previously chosen jobs.
Recall. Greedy algorithm is correct if all weights are 1.
Observation. Greedy algorithm fails spectacularly for weighted version.
8
weight = 999
weight = 1
time
0 1 2 3 4 5 6 7 8 9 10 11
b
a
h
weight = 1

Weighted interval scheduling
Convention. Jobs are in ascending order of finish time: f1 ≤ f2 ≤ . . . ≤ fn .
Def. p(j) = largest index i < j such that job i is compatible with j.
Ex. p(8) = 1, p(7) = 3, p(2) = 0.
9
time
0 1 2 3 4 5 6 7 8 9 10 11
6
7
8
4
3
1
2
5
i is rightmost interval
that ends before j begins

Dynamic programming: binary choice
Def. OPT(j) = max weight of any subset of mutually compatible jobs for
subproblem consisting only of jobs 1, 2, ..., j.
Goal. OPT(n) = max weight of any subset of mutually compatible jobs.
Case 1. OPT(j) does not select job j.
・Must be an optimal solution to problem consisting of remaining
jobs 1, 2, ..., j – 1.
Case 2. OPT(j) selects job j.
・Collect profit wj.
・Can’t use incompatible jobs { p(j) + 1, p(j) + 2, ..., j – 1 }.
・Must include optimal solution to problem consisting of remaining
compatible jobs 1, 2, ..., p(j).
Bellman equation.
10
optimal substructure property
(proof via exchange argument)
OPT(j) =
0 B7 j = 0
max { OPT(j 1), wj + OPT(p(j)) } B7 j > 0
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Weighted interval scheduling: brute force
11
BRUTE-FORCE (n, s1, …, sn, f1, …, fn, w1, …, wn)
_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Sort jobs by finish time and renumber so that f1 ≤ f2 ≤ … ≤ fn.
Compute p[1], p[2], …, p[n] via binary search.
RETURN COMPUTE-OPT(n).
COMPUTE-OPT( j )
_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
IF (j = 0)
RETURN 0.
ELSE
RETURN max {COMPUTE-OPT( j – 1), wj + COMPUTE-OPT(p[ j ]) }.

What is running time of COMPUTE-OPT(n) in the worst case?
A. Θ(n log n)
B. Θ(n2
)
C. Θ(1.618n
)
D. Θ(2n
)
12
Dynamic programming: quiz 1
COMPUTE-OPT( j )
_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
IF (j = 0)
RETURN 0.
ELSE
RETURN max {COMPUTE-OPT( j – 1), wj + COMPUTE-OPT(p[ j ]) }.
T(n) =
(1) B7 n = 1
2T(n 1) + (1) B7 n > 1
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Weighted interval scheduling: brute force
Observation. Recursive algorithm is spectacularly slow because of
overlapping subproblems ⇒ exponential-time algorithm.
Ex. Number of recursive calls for family of “layered” instances grows like
Fibonacci sequence.
13
3
4
5
1
2
p(1) = 0, p(j) = j-2
4 3
3 2 2 1
2 1
1 0
1 0 1 0
recursion tree
5

Weighted interval scheduling: memoization
Top-down dynamic programming (memoization).
・Cache result of subproblem j in M[j].
・Use M[j] to avoid solving subproblem j more than once.
14
M-COMPUTE-OPT( j )
_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
IF (M[j] is uninitialized)
M[j] ← max { M-COMPUTE-OPT (j – 1), wj + M-COMPUTE-OPT(p[j]) }.
RETURN M[j].
TOP-DOWN(n, s1, …, sn, f1, …, fn, w1, …, wn)
_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Compute p[1], p[2], …, p[n] via binary search.
M[0] ← 0.
RETURN M-COMPUTE-OPT(n).
global array

Weighted interval scheduling: running time
Claim. Memoized version of algorithm takes O(n log n) time.
Pf.
・Sort by finish time: O(n log n) via mergesort.
・Compute p[j] for each j : O(n log n) via binary search.
・M-COMPUTE-OPT(j): each invocation takes O(1) time and either
- (1) returns an initialized value M[j]
- (2) initializes M[j] and makes two recursive calls
・Progress measure Φ = # initialized entries among M[1.. n].
- initially Φ = 0; throughout Φ ≤ n.
- (2) increases Φ by 1 ⇒ ≤ 2n recursive calls.
・Overall running time of M-COMPUTE-OPT(n) is O(n). ▪
15

Weighted interval scheduling: finding a solution
Q. DP algorithm computes optimal value. How to find optimal solution?
A. Make a second pass by calling FIND-SOLUTION(n).
Analysis. # of recursive calls ≤ n ⇒ O(n).
17
FIND-SOLUTION(j)
_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
____________________________________________________________________________________________________________
IF (j = 0)
RETURN ∅.
ELSE IF (wj + M[p[j]] > M[ j – 1])
RETURN { j } ∪ FIND-SOLUTION(p[j]).
ELSE
RETURN FIND-SOLUTION( j – 1).
M[j] = max { M[j – 1], wj + M[p[j]] }.

Weighted interval scheduling: bottom-up dynamic programming
Bottom-up dynamic programming. Unwind recursion.
Running time. The bottom-up version takes O(n log n) time.
18
BOTTOM-UP(n, s1, …, sn, f1, …, fn, w1, …, wn)
_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Compute p[1], p[2], …, p[n].
M[0] ← 0.
FOR j = 1 TO n
M[j] ← max { M[ j – 1], wj + M[p[j]] }.
_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
previously computed values

MAXIMUM SUBARRAY PROBLEM
Goal. Given an array x of n integer (positive or negative),
find a contiguous subarray whose sum is maximum.
Applications. Computer vision, data mining,
genomic sequence analysis, technical job interviews, ….
19
12 5 −1 31 −61 59 26 −53 58 97 −93 −23 84 −15 6
187

MAXIMUM SUBARRAY PROBLEM
Goal. Given an array x of n integer (positive or negative),
find a contiguous subarray whose sum is maximum.
Brute-force algorithm.
・For each i and j : computer a[i] + a[i+1] + … + a[j].
・Takes Θ(n3
) time.
Apply “cumulative sum” trick.
・Precompute cumulative sums: S[i] = a[0] + a[1] + … + a[i].
・Now a[i] + a[i+1] + … + a[j] = S[j] − S[i−1].
・Improves running time Θ(n2
).
20
12 5 −1 31 −61 59 26 −53 58 97 −93 −23 84 −15 6
187
i j

KADANE’S ALGORITHM
Def. OPT(i) = max sum of any subarray of x whose rightmost
index is i.
Goal.
Bellman equation.
Running time. O(n).
21
take only
element i
take element i
together with best subarray
ending at index i − 1
OPT(i) =
x1 B7 i = 1
max { xi, xi + OPT(i 1) } B7 i > 1
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max
i
OPT(i)
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Goal. Given an n-by-n matrix A, find a rectangle whose sum is maximum.
Applications. Databases, image processing, maximum likelihood
estimation, technical job interviews, …
A =
2 5 0 5 2 2 3
4 3 1 3 2 1 1
5 6 3 5 1 4 2
1 1 3 1 4 1 1
3 3 2 0 3 3 2
2 1 2 1 1 3 1
2 4 0 1 0 3 1
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MAXIMUM RECTANGLE PROBLEM
22
13

j jʹ
Assumption. Suppose you knew the left and right column indices j and jʹ.
An O(n3
) algorithm.
・Precompute cumulative row sums .
・For each j < jʹ :
– define array x using row-sum differences:
– run Kadane’s algorithm in array x
Open problem. O(n3−ε
) for any constant ε > 0.
A =
2 5 0 5 2 2 3
4 3 1 3 2 1 1
5 6 3 5 1 4 2
1 1 3 1 4 1 1
3 3 2 0 3 3 2
2 1 2 1 1 3 1
2 4 0 1 0 3 1
BENTLEY’S ALGORITHM
23
Sij =
j
k = 1
Aik
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solve maximum
subarray problem
in this array
x =
7
4
3
6
5
0
1
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0 − 5 − 2
xi = Sij Sij
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SECTION 6.3

Least squares
Least squares. Foundational problem in statistics.
・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:
Solution. Calculus ⇒ min error is achieved when
x
y
25
a =
n
!
i xiyi − (
!
i xi)(
!
i yi)
n
!
i x2
i − (
!
i xi)2
, b =
!
i yi − a
!
i xi
n
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<latexit sha1_base64="osgmybCFlDRwfvDA5+Vdeo/74To=">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</latexit>
<latexit sha1_base64="uSR6MIDaEDPJwkbPgs/ygO2X/DQ=">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</latexit>
SSE =
n
!
i=1
(yi − axi − b)2
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Segmented least squares
Segmented least squares.
・Points lie roughly on a sequence of several line segments.
・Given n points in the plane: (x1, y1), (x2, y2) , …, (xn, yn) with
x1 < x2 < ... < xn, find a sequence of lines that minimizes f (x).
Q. What is a reasonable choice for f (x) to balance accuracy and parsimony?
x
y
26
goodness of fit number of lines

Segmented least squares
Segmented least squares.
・Points lie roughly on a sequence of several line segments.
・Given n points in the plane: (x1, y1), (x2, y2) , …, (xn, yn) with
x1 < x2 < ... < xn, find a sequence of lines that minimizes f (x).
Goal. Minimize f (x) = E + c L for some constant c > 0, where
・E = sum of the sums of the squared errors in each segment.
・L = number of lines.
x
y
27

Dynamic programming: multiway choice
Notation.
・OPT(j) = minimum cost for points p1, p2, …, pj.
・eij = SSE for for points pi, pi+1, …, pj.
To compute OPT(j):
・Last segment uses points pi, pi+1, …, pj for some i ≤ j.
・Cost = eij + c + OPT(i – 1).
Bellman equation.
28
OPT(j) =
0 B7 j = 0
min
1 i j
{ eij + c + OPT(i 1) } B7 j > 0
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Segmented least squares algorithm
29
SEGMENTED-LEAST-SQUARES(n, p1, …, pn, c)
__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
FOR j = 1 TO n
FOR i = 1 TO j
Compute the SSE eij for the points pi, pi+1, …, pj.
M[0] ← 0.
FOR j = 1 TO n
M[j] ← min 1 ≤ i ≤ j { eij + c + M[i – 1] }.
RETURN M[n].
__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
previously computed value

Segmented least squares analysis
Theorem. [Bellman 1961] DP algorithm solves the segmented least squares
problem in O(n3
) time and O(n2
) space.
Pf.
・Bottleneck = computing SSE eij for each i and j.
・O(n) to compute eij. ▪
Remark. Can be improved to O(n2
) time.
・For each i : precompute cumulative sums .
・Using cumulative sums, can compute eij in O(1) time.
30
aij =
n
!
k xkyk − (
!
k xk)(
!
k yk)
n
!
k x2
k − (
!
k xk)2
, bij =
!
k yk − aij
!
k xk
n
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<latexit sha1_base64="BChx1F3fNhGq6mUCFDX2bORLMoA=">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</latexit>
i
k=1
xk,
i
k=1
yk,
i
k=1
x2
k,
i
k=1
xkyk
<latexit sha1_base64="E9Bo6rLIFL0/cuHYqsa/2SHtfn0=">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</latexit>

SECTION 6.4

Knapsack problem
Goal. Pack knapsack so as to maximize total value of items taken.
・There are n items: item i provides value vi > 0 and weighs wi > 0.
・Value of a subset of items = sum of values of individual items.
・Knapsack has weight limit of W.
Ex. The subset { 1, 2, 5 } has value $35 (and weight 10).
Ex. The subset { 3, 4 } has value $40 (and weight 11).
Assumption. All values and weights are integral.
32
weights and values
can be arbitrary
positive integers
$28 7 kg
$22 6 kg
$6 2 kg
$1 1 kg
$18 5 kg
11 kg
Creative Commons Attribution-Share Alike 2.5
by Dake
i vi wi
1 $1 1 kg
2 $6 2 kg
3 $18 5 kg
4 $22 6 kg
5 $28 7 kg
knapsack instance
(weight limit W = 11)

Which algorithm solves knapsack problem?
A. Greedy-by-value: repeatedly add item with maximum vi.
B. Greedy-by-weight: repeatedly add item with minimum wi.
C. Greedy-by-ratio: repeatedly add item with maximum ratio vi / wi.
D. None of the above.
33
Creative Commons Attribution-Share Alike 2.5
by Dake
$28 7 kg
$22 6 kg
$6 2 kg
$1 1 kg
$18 5 kg
11 kg
i vi wi
1 $1 1 kg
2 $6 2 kg
3 $18 5 kg
4 $22 6 kg
5 $28 7 kg
knapsack instance
(weight limit W = 11)

Which subproblems?
A. OPT(w) = optimal value of knapsack problem with weight limit w.
B. OPT(i) = optimal value of knapsack problem with items 1, …, i.
C. OPT(i, w) = optimal value of knapsack problem with items 1, …, i
subject to weight limit w.
D. Any of the above.
34

Dynamic programming: two variables
Def. OPT(i, w) = optimal value of knapsack problem with items 1, …, i,
subject to weight limit w.
Goal. OPT(n, W).
Case 1. OPT(i, w) does not select item i.
・OPT(i, w) selects best of { 1, 2, …, i – 1 } subject to weight limit w.
Case 2. OPT(i, w) selects item i.
・Collect value vi.
・New weight limit = w – wi.
・OPT(i, w) selects best of { 1, 2, …, i – 1 } subject to new weight limit.
Bellman equation.
35
possibly because wi > w
OPT(i, w) =
0 B7 i = 0
OPT(i 1, w) B7 wi > w
max { OPT(i 1, w), vi + OPT(i 1, w wi) } Qi?2`rBb2
<latexit sha1_base64="yGc+dBk4QwRLdPr1J0EKEi2YjzI=">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</latexit>

Knapsack problem: bottom-up dynamic programming
36
KNAPSACK(n, W, w1, …, wn, v1, …, vn )
__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
FOR w = 0 TO W
M[0, w] ← 0.
FOR i = 1 TO n
FOR w = 0 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].
__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
previously computed values
OPT(i, w) =
0 B7 i = 0
max { OPT(i 1, w), vi + OPT(i 1, w wi) } Qi?2`rBb2

Knapsack problem: bottom-up dynamic programming demo
37
0 1 2 3 4 5 6 7 8 9 10 11
{ } 0 0 0 0 0 0 0 0 0 0 0 0
{ 1 } 0 1 1 1 1 1 1 1 1 1 1 1
{ 1, 2 } 0 1 6 7 7 7 7 7 7 7 7 7
{ 1, 2, 3 } 0 1 6 7 7 18 19 24 25 25 25 25
{ 1, 2, 3, 4 } 0 1 6 7 7 18 22 24 28 29 29 40
{ 1, 2, 3, 4, 5 } 0 1 6 7 7 18 22 28 29 34 35 40
weight limit w
subset
of items
1, …, i
OPT(i, w) = optimal value of knapsack problem with items 1, …, i, subject to weight limit w
OPT(i, w) =
0 B7 i = 0
max {OPT(i 1, w), vi + OPT(i 1, w wi} Qi?2`rBb2
<latexit sha1_base64="Z2vromkJx+wQ2wpWsn/gLtchm2Q=">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</latexit>
i vi wi
1 $1 1 kg
2 $6 2 kg
3 $18 5 kg
4 $22 6 kg
5 $28 7 kg

Knapsack problem: running time
Theorem. The DP algorithm solves the knapsack problem with n items
and maximum weight W in Θ(n W) time and Θ(n W) space.
Pf.
・Takes O(1) time per table entry.
・There are Θ(n W) table entries.
・After computing optimal values, can trace back to find solution:
OPT(i, w) takes item i iff M [i, w] > M [i – 1, w]. ▪
Remarks.
・Algorithm depends critically on assumption that weights are integral.
・Assumption that values are integral was not used.
38
weights are integers
between 1 and W

Does there exist a poly-time algorithm for the knapsack problem?
A. Yes, because the DP algorithm takes Θ(n W) time.
B. No, because Θ(n W) is not a polynomial function of the input size.
C. No, because the problem is NP-hard.
D. Unknown.
39
“pseudo-polynomial”
equivalent to P ≠ NP conjecture
because knapsack problem is NP-hard

COIN CHANGING
Problem. Given n coin denominations { d1, d2, …, dn } and a target value V,
find the fewest coins needed to make change for V (or report impossible).
Recall. Greedy cashier’s algorithm is optimal for U.S. coin denominations,
but not for arbitrary coin denominations.
Ex. { 1, 10, 21, 34, 70, 100, 350, 1295, 1500 }.
Optimal. 140¢ = 70 + 70.
40

Def. OPT(v) = min number of coins to make change for v.
Goal. OPT(V).
Multiway choice. To compute OPT(v),
・Select a coin of denomination ci for some i.
・Select fewest coins to make change for v – ci.
Bellman equation.
Running time. O(n V).
COIN CHANGING
41
<latexit sha1_base64="6B5tGCszYw5e5nluRWvumveoBFs=">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</latexit>
OPT(v) =
8
>
>
>
>
<
>
>
>
>
:
1 B7 v < 0
0 B7 v = 0
min
1in
{ 1 + OPT(v di) } B7 v > 0

SECTION 6.5

RNA 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.
G
U
C
A
G
A
A
G
C
G
A
U
G
A
U
U
A
G
A
C A
A
C
U
G
A
G
U
C
A
U
C
G
G
G
C
C
G
RNA secondary structure for GUCGAUUGAGCGAAUGUAACAACGUGGCUACGGCGAGA
43
base
base pair

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.
44
A C G U G G C C A U
S is not a secondary structure
(C-A is not a valid Watson-Crick pair)
B = ACGUGGCCCAU
S = { (b1, b10), (b2, b9), (b3, b8) }
base pair
in secondary structure
C
G G
C
A
G
C
U
U A

・[No sharp turns] The ends of each pair are separated by at least 4
intervening bases. If (bi, bj) ∈ S, then i < j – 4.
45
G G
C
A
G
U
U A
G
A U G G G C A U
(≤4 intervening bases between G and C)
G
B = AUGGGGCAU
S = { (b1, b9), (b2, b8), (b3, b7) }

・[Non-crossing] If (bi, bj) and (bk, bℓ) are two pairs in S, then we cannot
have i < k < j < ℓ.
46
A G U U G G C C A U
(G-C and U-A cross)
C
G G
C
A
U
G
U
U A
B = ACUUGGCCAU
S = { (b1, b10), (b2, b8), (b3, b9) }

47
A U G U G G C C A U
S is a secondary structure
(with 3 base pairs)
C
G G
C
A
G
U
U
U A
B = AUGUGGCCAU
S = { (b1, b10), (b2, b9), (b3, b8) }

Free-energy hypothesis. RNA molecule will form the secondary structure
with the minimum total free energy.
Goal. Given an RNA molecule B = b1b2…bn, find a secondary structure S
that maximizes the number of base pairs.
approximate by number of base pairs
(more base pairs ⇒ lower free energy)
48

Is the following a secondary structure?
A. Yes.
B. No, violates Watson–Crick condition.
C. No, violates no-sharp-turns condition.
D. No, violates no-crossing condition.
49
G
U
C
A
G
A
A
G
C
G
A
U
G
A
U
U
A
G
C G
U
G
C
C
G

Which subproblems?
A. OPT(j) = max number of base pairs in secondary structure
of the substring b1b2 … bj.
B. OPT(j) = max number of base pairs in secondary structure
of the substring bj bj+1 … bn.
C. Either A or B.
D. Neither A nor B.
50

RNA secondary structure: subproblems
First attempt. OPT(j) = maximum number of base pairs in a secondary
structure of the substring b1b2 … bj.
Goal. OPT(n).
Choice. Match bases bt and bj.
Difficulty. Results in two subproblems (but one of wrong form).
・Find secondary structure in b1b2 … bt–1.
・Find secondary structure in bt+1bt+2 … bj–1.
OPT(t–1)
need more subproblems
(first base no longer b1)
51
1 t j
match bases bt and bn
last base

Dynamic programming over intervals
Def. OPT(i, j) = maximum number of base pairs in a secondary structure
of the substring bi bi+1 … bj.
Case 1. If i ≥ j – 4.
・OPT(i, j) = 0 by no-sharp-turns condition.
Case 2. Base bj is not involved in a pair.
・OPT(i, j) = OPT(i, j – 1).
Case 3. Base bj pairs with bt for some i ≤ t < j – 4.
・Non-crossing condition decouples resulting two subproblems.
・OPT(i, j) = 1 + max t { OPT(i, t – 1) + OPT(t + 1, j – 1) }.
take max over t such that i ≤ t < j – 4 and
bt and bj are Watson–Crick complements
52
i t j
match bases bj and bt

In which order to compute OPT(i, j) ?
A. Increasing i, then j.
B. Increasing j, then i.
C. Either A or B.
D. Neither A nor B.
53
i t j
match bases bj and bt
OPT(i, j) depends upon OPT(i, t-1) and OPT(t+1, j-1)

Bottom-up dynamic programming over intervals
Q. In which order to solve the subproblems?
A. Do shortest intervals first—increasing order of ⎜j − i⎟.
Theorem. The DP algorithm solves the RNA secondary structure problem in
O(n3
) time and O(n2
) space.
54
6 7 8 9 10
4 0 0 0
3 0 0
2 0
1
i
j
order in which to solve subproblems
RNA-SECONDARY-STRUCTURE(n, b1, …, bn )
__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
FOR k = 5 TO n – 1
FOR i = 1 TO n – k
j ← i + k.
Compute M[i, j] using formula.
RETURN M[1, n].
__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
all needed values
are already computed

Dynamic programming summary
Outline.
・Define a collection of subproblems.
・Solution to original problem can be computed from subproblems.
・Natural ordering of subproblems from “smallest” to “largest” that
enables determining a solution to a subproblem from solutions to
smaller subproblems.
Techniques.
・Binary choice: weighted interval scheduling.
・Multiway choice: segmented least squares.
・Adding a new variable: knapsack problem.
・Intervals: RNA secondary structure.
Top-down vs. bottom-up dynamic programming. Opinions differ.
55
typically, only a polynomial
number of subproblems

06DynamicProgramming in algorithms CSE.pdf

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