design and analysis of algorithms lecture dynamic programming alexander kolesnikov 17.10.205
DESCRIPTION
DEPARTMENT OF COMPUTER SCIENCE UNIVERSITY OF JOENSUU JOENSUU, FINLAND. Design and Analysis of Algorithms Lecture Dynamic programming Alexander Kolesnikov 17.10.205. List of sample problems. Shortest path in trellis graph; Optimal allocation of constrained resource; - PowerPoint PPT PresentationTRANSCRIPT
Design and Analysis of Algorithms
Lecture
Dynamic programmingAlexander Kolesnikov
17.10.205
DEPARTMENT OF COMPUTER SCIENCE UNIVERSITY OF JOENSUU JOENSUU, FINLAND
List of sample problemsList of sample problems
• Shortest path in trellis graph;Shortest path in trellis graph;
• Optimal allocation of constrained resource; Optimal allocation of constrained resource;
• Optimal sequence partition (k-link shortest Optimal sequence partition (k-link shortest
path).path).
• to be continued ...
Shortest path in trellis graphShortest path in trellis graph
Stagecoach problemStagecoach problem
A traveler wishes to minimize the length of a journey from town A A to JJ.
Greedy algorithmGreedy algorithm
The length of the route A-B-F-I-J: 2+4+3+4=13.
Can we find shorter route?
Exhaustive search: try all Exhaustive search: try all
Route A-D-F-I-J: 3+1+3+4=11
The total number of routes to be tested: 3321=18
Can we avoid exhaustive search?
Shortest path construction: 1st stage (B)
?
Shortest path construction: 1st stage
S(A,B)=2
S(A,C)=4
S(A,D)=3
Shortest path construction: 2nd stage (E)
1. (A..B)-E: 2+7=9
2. (A..C)-E: 4+3=7
3. (A..D)-E: 3+4=7
----------------------
(A..C)-E: 7
?
Shortest path construction: 2nd stage (E)
1. (A..B)-E: 2+7=9
2. (A..C)-E: 4+3=7 *)
3. (A..D)-E: 3+4=7
----------------------
(A..C)-E: 7
Shortest path construction: 2nd stage (F)
1. (A..B)-F: 2+4=6
2. (A..C)-F: 4+2=6
3. (A..D)-F: 3+1=4 *)
----------------------
(A..C)-F: 4
?
Shortest path construction: 2nd stage (F)
1. (A..B)-F: 2+4=6
2. (A..C)-F: 4+2=6
3. (A..D)-F: 3+1=4 *)
----------------------
(A..D)-F: 4
Shortest path construction: 2nd stage (G)
1. (A..B)-G: 2+6=8 *)
2. (A..C)-G: 4+6=10
3. (A..D)-G: 3+5=8
----------------------
(A..B)-G: 8?
Shortest path construction: 2nd stage (G)
1. (A..B)-G: 2+6=8 *)
2. (A..C)-G: 4+6=10
3. (A..D)-G: 3+5=8
----------------------
(A..B)-G: 8
Shortest path construction: 3rd stage (H)
1. (A..E)-H: 7+1=8 *)
2. (A..F)-H: 4+6=10
3. (A..G)-H: 5+3=8
----------------------
(A..E)-H: 5
Shortest path construction: 3rd stage (H)
1. (A..E)-H: 7+1=8 *)
2. (A..F)-H: 4+6=10
3. (A..G)-H: 5+3=8
----------------------
(A..E)-H: 5
Shortest path construction: 3rd stage (I)
1. (A..E)-I: 7+4=11
2. (A..F)-I: 4+3=7 *)
3. (A..G)-I: 5+3=8
----------------------
(A..F)-I: 7
Shortest path construction: 3rd stage (I)
1. (A..E)-I: 7+4=11
2. (A..F)-I: 4+3=7 *)
3. (A..G)-I: 5+3=8
----------------------
(A..F)-I: 7
Shortest path construction: 4th stage (J)
1. (A..H)-J: 8+3=11 *)
2. (A..I) -J: 7+4=11
----------------------
(A..H)-J: 11
Shortest path construction: 4th stage (J)
1. (A..H)-J: 8+3=11 *)
2. (A..I) -J: 7+4=11
----------------------
(A..H)-J: 11
Backtrack the shortest path
11
8
7
8
4
7
3
4
2
0
The shortest path
11
8
7
8
4
7
3
4
2
0
Route A-C-E-H-J: 4+3+1+3=11
Trellis graphTrellis graph
1 2 3 4 K-1 K
),(),,1( jkikf ),(),,1( jkikf
Trellis graphTrellis graph
• Distance (weight) from point i1 at stage (j1) to point i2
at stage j:
min),(),,1(1
1
K
jjj ijijfL
•The total value of cost function:
Principle of optimality of BellmanPrinciple of optimality of Bellman
An optimal path has the property that whatever
the initial
conditions and control variables (choices) over
some initial period, the control (or decision
variables) chosen over the remaining period must
be optimal for the remaining problem, with the
state resulting from the early decisions taken to
be the initial condition.
Dynamic programmingDynamic programming
• Initialization:
),(),,1()(min)( nkikfiGnG ki
k • Recursive eqution:
• Cost function:
k
jjjjjk ipipfnG
111 )(),(min)(
0)(0 nG
ComplexityComplexity
• Exhaustive search: O(nK)
• Dynamic programming algorithm: O(Kn2)
where K is the number of stages, n is the number of points in a stage
Optimal allocation Optimal allocation of constrained resourceof constrained resource
Problem formulationProblem formulation• N units of a resource;
• This resource must be distributed among K activities;
• Functions fk(x) - profit for allocated resource;
• Allocate N units of resource to K activities with given return functions so that the total profit is maximal:
K
jjj nfP
1
max)(
K
jj Nn
1
subject to: 0jn
Dynamic programming formulationDynamic programming formulation
k
jjj
nnnk nfnG
j 1...1
max)(
.0;)(max)( 10
NnmnfmGnG kknm
k
NnnG 0;0)(0
Initialization:
Recursive equation:
Optimal value function:
)(NGF K
Allocate 3 mln euros into four projectsAllocate 3 mln euros into four projects
x Proj #1 Proj #2 Proj #3 Proj #40 0 0 0 01 2 5 4 22 8 6 4 43 9 7 4 5
Profit fk(x), K=3, N=3.
4
1
max)(j
jj nfP
4
1
3j
jn
Trellis graphTrellis graph
SolutionSolution
#1: 2; f1(2)=8#2: 1; f2(1)=5#3: 0; f3(0)=0#4: 0; f4(0)=0---------------------- N=3; G4(3)=13
*)
Search in the state space
K
N00
n
k
j
GGkk(n)(n)
GGk-1k-1(j)(j)
GGKK(N)(N)
Start state
fk(n-j)
Gk(n) = max{Gk(0) + fk(n), Gk(1) + fk(n-1), . . . Gk(j-1)+ fk(j),
. . . Gk(n) + fk(0)}
k-1 *
Ak(n)=jopt
1
Optimal partition of Optimal partition of data sequencedata sequence
Problem formulationProblem formulation
Given a sequence of data X={x1, x2, …,xN}
Do partition of the sequence X into to K groups with
given cost functions f(xi,xj) so that the total value of
the cost function is minimal:
K
jii jj
xxf1
min),(1
min),(...),(),(12111
Niiii xxfxxfxxfK
Partition into groups: Example• Data: x 0= < x1 <... < xj < ... < xN
• Partition indices: i0= 0 < i1 <... < ij < ... < iM =N. • Groups:
• • • • •• • • •• •• • • •• ... ]( ]( ](#1 #2 #3
(x0=) x1 x2 x3 x4 x5 x6 x7x8 x9 x10 x11 x12 x13 x14 xN
(i0=0) i1=4 i2=10 iK =N=15
15,1010,44,0 fffP
K=3
Problem formulationProblem formulation
k
jjjjk niiifnG
11 .;,)(
Recursive equation:
Cost function:
.0;,)(min)( 11
NnnjfjGnG knjk
k
.0;0)(0 NnnG
Initialization:
Search in the state space
K
N10
b
n
k
j
State space
Gk (n)
Gk-1(j)
GGKK(N)(N)
Start state
f(j,n)
Gk(n) = min{Gk(k) + f(k, n], Gk(k) + f(k+1,n], . . . Gk(j-1) + f(j, n]*),
. . . Gk(n-1) + f(n, n]}
k-1 *
Ak(n)=jopt
Scheme of the DP algorithm// Initialization
FOR n = 1 TO N DO GG11(n)(n)= f(1,n]// Minimum searchFOR k = 2 TO K DO FOR n = k TO N DO
dmin FOR j= k-1 TO n-1 DO c GGk-1k-1(j)(j) + f(j,n]
IF(c < cmin)
cmin c;
jmin j ENDIF ENDFOR GGk k (n)(n) dmin
AAk k (n)(n) jmin
ENDFORENDFOR
Complexity: O(KN2)Complexity: O(KN2)
Backtrack in the state space
K
N10
b
nj
State space
AK(N)
Start state
S(M+1)= NFOR m = K+1 TO 2 DO S(m1) = A(S(m), m))P = GGKK(N)(N)
S(M+1)= NFOR m = K+1 TO 2 DO S(m1) = A(S(m), m))P = GGKK(N)(N)
N=22, K=8: S={22,18,14,12,9,6,4,3,1}
(x0,x3], (x3,x4], (x4,x6], (x6,x9], (x9,x12],
(x12,x14], (x14,x18], (x18,x22]