Download Dynamic Programming-Algorithms and Data Representation-Lecture Slides and more Slides Data Representation and Algorithm Design in PDF only on Docsity!
1
Chapter 6
Dynamic Programming
2
Algorithmic Paradigms
Greedy. Build up a solution incrementally, myopically optimizing some
local criterion.
Divide-and-conquer. Break up a problem into 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.
3
Dynamic Programming History
Bellman. [1950s] Pioneered the systematic study of dynamic programming.
Etymology.
!! Dynamic programming = planning over time.
!! Secretary of Defense was hostile to mathematical research.
!! Bellman sought an impressive name to avoid confrontation.
Reference: Bellman, R. E. Eye of the Hurricane, An Autobiography.
"it's impossible to use dynamic in a pejorative sense"
"something not even a Congressman could object to"
4
Dynamic Programming Applications
Areas.
!! Bioinformatics.
!! Control theory.
!! Information theory.
!! Operations research.
!! Computer science: theory, graphics, AI, compilers, systems, ….
Some famous dynamic programming algorithms.
!! Unix diff for comparing two files.
!! Viterbi for hidden Markov models.
!! Smith-Waterman for genetic sequence alignment.
!! Bellman-Ford for shortest path routing in networks.
!! Cocke-Kasami-Younger for parsing context free grammars.
6.1 Weighted Interval Scheduling
6
Weighted Interval Scheduling
Weighted interval scheduling problem.
!! Job j starts at s
j
, finishes at f
j
, and has weight or value v
j
!! Two jobs compatible if they don't overlap.
!! Goal: find maximum weight subset of mutually compatible jobs.
Time
f
g
h
e
a
b
c
d
0 1 2 3 4 5 6 7 8 9 10
7
Unweighted Interval Scheduling Review
Recall. Greedy algorithm works if all weights are 1.
!! Consider jobs in ascending order of finish time.
!! Add job to subset if it is compatible with previously chosen jobs.
Observation. Greedy algorithm can fail spectacularly if arbitrary
weights are allowed.
Time
0 1 2 3 4 5 6 7 8 9 10 11
b
a
weight = 999
weight = 1
8
Weighted Interval Scheduling
Notation. Label jobs by finishing time: f
1
! f
2
!...! f
n
Def. p(j) = largest index i < j such that job i is compatible with j.
Ex: p(8) = 5, p(7) = 3, p(2) = 0.
Time
0 1 2 3 4 5 6 7 8 9 10 11
6
7
8
4
3
1
2
5
13
Weighted Interval Scheduling: Running Time
Claim. Memoized version of algorithm takes O(n log n) time.
!! Sort by finish time: O(n log n).
!! Computing p(#) : O(n log n) via sorting by start time.
!! M-Compute-Opt(j): each invocation takes O(1) time and either
-! (i) returns an existing value M[j] -! (ii) fills in one new entry M[j] and makes two recursive calls
!! Progress measure $ = # nonempty entries of M[].
-! initially $ = 0, throughout $! n. -! (ii) increases $ by 1 " at most 2n recursive calls.
!! Overall running time of M-Compute-Opt(n) is O(n).!
Remark. O(n) if jobs are pre-sorted by start and finish times.
14
Weighted Interval Scheduling: Finding a Solution
Q. Dynamic programming algorithms computes optimal value.
What if we want the solution itself?
A. Do some post-processing.
!! # of recursive calls! n " O(n).
Run M-Compute-Opt(n)
Run Find-Solution(n)
Find-Solution(j) {
if (j = 0)
output nothing
**else if (v j
print j
Find-Solution(p(j))
else
Find-Solution(j-1)
15
Weighted Interval Scheduling: Bottom-Up
Bottom-up dynamic programming. Unwind recursion.
Input: n, s 1
,…,s n ,
f 1
,…,f n ,
v 1
,…,v n
Sort jobs by finish times so that f 1
! f 2
! ...! f n
Compute p(1), p(2), …, p(n)
Iterative-Compute-Opt {
M[0] = 0
for j = 1 to n
M[j] = max(v j
+ M[p(j)], M[j-1])
6.3 Segmented Least Squares
17
Segmented Least Squares
Least squares.
!! Foundational problem in statistic and numerical analysis.
!! Given n points in the plane: (x
1
, y
1
), (x
2
, y
2
) ,... , (x
n
, y
n
!! Find a line y = ax + b that minimizes the sum of the squared error:
Solution. Calculus " min error is achieved when
SSE = ( y
i
" ax
i
" b )
2
i = 1
n
a =
n x
i
y
i
" ( x
i
i
( y
i
i
i
n x
i
2
" ( x
i
2
i
i
, b =
y
i
" a x
ii
i
n
x
y
18
Segmented Least Squares
Segmented least squares.
!! Points lie roughly on a sequence of several line segments.
!! Given n points in the plane (x
1
, y
1
), (x
2
, y
2
) ,... , (x
n
, y
n
) with
!! x
1
< x
2
< ... < x
n
, find a sequence of lines that minimizes f(x).
Q. What's a reasonable choice for f(x) to balance accuracy and
parsimony?
x
y
goodness of fit
number of lines
19
Segmented Least Squares
Segmented least squares.
!! Points lie roughly on a sequence of several line segments.
!! Given n points in the plane (x
1
, y
1
), (x
2
, y
2
) ,... , (x
n
, y
n
) with
!! x
1
< x
2
< ... < x
n
, 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.
x
y
20
Dynamic Programming: Multiway Choice
Notation.
!! OPT(j) = minimum cost for points p
1
, p
i+
,... , p
j
!! e(i, j) = minimum sum of squares for points p
i
, p
i+
,... , p
j
To compute OPT(j):
!! Last segment uses points p
i
, p
i+
,... , p
j
for some i.
!! Cost = e(i, j) + c + OPT(i-1).
OPT ( j ) =
0 if j = 0
min
1 " i " j
e ( i , j ) + c + OPT ( i # 1 )
otherwise
25
Dynamic Programming: Adding a New Variable
Def. OPT(i, w) = max profit subset of items 1, …, i with weight limit w.
!! 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.
-! new weight limit = w – w i -! OPT selects best of { 1, 2, …, i–1 } using this new weight limit
OPT ( i , w ) =
0 if i = 0
OPT ( i " 1 , w ) if w
i
> w
max OPT ( i " 1 , w ), v
i
+ OPT ( i " 1 , w " w
i
otherwise
26
Input: n, W, w 1
,…,w N,
v 1
,…,v N
for w = 0 to W
M[0, w] = 0
for i = 1 to n
for w = 1 to W
if (w i
> w)
M[i, w] = M[i-1, w]
else
M[i, w] = max {M[i-1, w], v i
+ M[i-1, w-w i
]}
return M[n, W]
Knapsack Problem: Bottom-Up
Knapsack. Fill up an n-by-W array.
27
Knapsack Algorithm
n + 1
Value
Weight
Item
W + 1
W = 11
OPT: { 4, 3 }
value = 22 + 18 = 40
28
Knapsack Problem: Running Time
Running time. &(n W).
!! Not polynomial in input size!
!! "Pseudo-polynomial."
!! Decision version of Knapsack is NP-complete. [Chapter 8]
Knapsack approximation algorithm. There exists a poly-time algorithm
that produces a feasible solution that has value within 0.01% of
optimum. [Section 11.8]
6.5 RNA Secondary Structure
30
RNA Secondary Structure
RNA. String B = b
1
b
2
…b
n
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
A G
A
G
G C
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
Ex: GUCGAUUGAGCGAAUGUAACAACGUGGCUACGGCGAGA
complementary base pairs: A-U, C-G
31
RNA Secondary Structure
Secondary structure. A set of pairs S = { (b
i
, b
j
) } 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 (b
i
, b
j
) ' S, then i < j - 4.
!! [Non-crossing.] If (b
i
, b
j
) and (b
k
, b
l
) are two pairs in S, then we
cannot have i < k < j < l.
Free energy. Usual hypothesis is that an RNA molecule will form the
secondary structure with the optimum total free energy.
Goal. Given an RNA molecule B = b
1
b
2
…b
n
, find a secondary structure S
that maximizes the number of base pairs.
approximate by number of base pairs
32
RNA Secondary Structure: Examples
Examples.
C
G G
C
A
G
U
U
U A
A U G U G G C C A U
G G
C
A
G
U
U A
A U G G G C A U
C
G G
C
A
U
G
U
U A
A G U U G G C C A U
ok sharp turn crossing
G
G
! 4
base pair
6.6 Sequence Alignment
38
String Similarity
How similar are two strings?
!! ocurrance
!! occurrence
o c u r r a n c e
o c c u r r e n c e
-
o c u r r n c e
o c c u r r n c e
- - a
e -
o c u r r a n c e
o c c u r r e n c e
-
6 mismatches, 1 gap
1 mismatch, 1 gap
0 mismatches, 3 gaps
39
Applications.
!! Basis for Unix diff.
!! Speech recognition.
!! Computational biology.
Edit distance. [Levenshtein 1966, Needleman-Wunsch 1970]
!! Gap penalty ); mismatch penalty *
pq
!! Cost = sum of gap and mismatch penalties.
CA
C G A C C T A C C T
C T G A C T A C A T
T G A C C T A C C T
C T G A C T A C A T
T -
C
C
C
TC
GT
AG
CA
-
Edit Distance
40
Goal: Given two strings X = x
1
x
2
... x
m
and Y = y
1
y
2
... y
n
find
alignment of minimum cost.
Def. An alignment M is a set of ordered pairs x
i
-y
j
such that each item
occurs in at most one pair and no crossings.
Def. The pair x
i
-y
j
and x
i'
-y
j'
cross if i < i', but j > j'.
Ex: CTACCG vs. TACATG.
Sol: M = x
2
-y
1
, x
3
-y
2
, x
4
-y
3
, x
5
-y
4
, x
6
-y
6
Sequence Alignment
cost( M ) = " x i y j
( x i , y j ) # M
mismatch
i : x i unmatched
j : y j unmatched
gap
C T A C C -
- T A C A T
G
G
y 1
y 2
y 3
y 4
y 5
y 6
x 2 x 3 x 4 x 5
x 1 x 6
41
Sequence Alignment: Problem Structure
Def. OPT(i, j) = min cost of aligning strings x
1
x
2
... x
i
and y
1
y
2
... y
j
!! Case 1: OPT matches x
i
-y
j
-! pay mismatch for x i
-y
j
+ min cost of aligning two strings
x
1
x
2
... x
i-
and y
1
y
2
... y
j-
!! Case 2a: OPT leaves x
i
unmatched.
-! pay gap for x i
and min cost of aligning x
1
x
2
... x
i-
and y
1
y
2
... y
j
!! Case 2b: OPT leaves y
j
unmatched.
-! pay gap for y j
and min cost of aligning x
1
x
2
... x
i
and y
1
y
2
... y
j-
OPT ( i , j ) =
j & if i = 0
min
' x i y j
& + OPT ( i ( 1 , j )
& + OPT ( i , j ( 1 )
otherwise
i & if j = 0
42
Sequence Alignment: Algorithm
Analysis. &(mn) time and space.
English words or sentences: m, n! 10.
Computational biology: m = n = 100,000. 10 billions ops OK, but 10GB array?
Sequence-Alignment(m, n, x 1
x 2
...x m
, y 1
y 2
...y n
for i = 0 to m
M[i, 0] = i )
for j = 0 to n
M[0, j] = j )
for i = 1 to m
for j = 1 to n
M[i, j] = min( * [x i,
y j
] + M[i-1, j-1],
) + M[i-1, j],
) + M[i, j-1])
return M[m, n]
6.7 Sequence Alignment in Linear Space
44
Sequence Alignment: Linear Space
Q. Can we avoid using quadratic space?
Easy. Optimal value in O(m + n) space and O(mn) time.
!! Compute OPT(i, •) from OPT(i-1, •).
!! No longer a simple way to recover alignment itself.
Theorem. [Hirschberg 1975] Optimal alignment in O(m + n) space and
O(mn) time.
!! Clever combination of divide-and-conquer and dynamic programming.
!! Inspired by idea of Savitch from complexity theory.
49
Observation 1. The cost of the shortest path that uses (i, j) is
f(i, j) + g(i, j).
Sequence Alignment: Linear Space
i-j
m-n
x 1
x 2
y 1
x 3
y 2
y 3
y 4
y 5
y 6
0-
50
Observation 2. let q be an index that minimizes f(q, n/2) + g(q, n/2).
Then, the shortest path from (0, 0) to (m, n) uses (q, n/2).
Sequence Alignment: Linear Space
i-j
m-n
x 1
x 2
y 1
x 3
y 2
y 3
y 4
y 5
y 6
0-
n / 2
q
51
Divide: find index q that minimizes f(q, n/2) + g(q, n/2) using DP.
!! Align x
q
and y
n/
Conquer: recursively compute optimal alignment in each piece.
Sequence Alignment: Linear Space
i-j x 1
x 2
y 1
x 3
y 2
y 3
y 4
y 5
y 6
0-
q
n / 2
m-n
52
Theorem. Let T(m, n) = max running time of algorithm on strings of
length at most m and n. T(m, n) = O(mn log n).
Remark. Analysis is not tight because two sub-problems are of size
(q, n/2) and (m - q, n/2). In next slide, we save log n factor.
Sequence Alignment: Running Time Analysis Warmup
T ( m , n ) " 2 T ( m , n / 2 ) + O ( mn ) # T ( m , n ) = O ( mn log n )
53
Theorem. Let T(m, n) = max running time of algorithm on strings of
length m and n. T(m, n) = O(mn).
Pf. (by induction on n)
!! O(mn) time to compute f( •, n/2) and g ( •, n/2) and find index q.
!! T(q, n/2) + T(m - q, n/2) time for two recursive calls.
!! Choose constant c so that:
!! Base cases: m = 2 or n = 2.
!! Inductive hypothesis: T(m, n)! 2cmn.
Sequence Alignment: Running Time Analysis
cmn
cqn cmn cqn cmn
cqn cm qn cmn
Tmn Tqn Tm qn cmn
T ( m , 2 ) " cm
T ( 2 , n ) " cn
T ( m , n ) " cmn + T ( q , n / 2 ) + T ( m # q , n / 2 )
