Dynamic Programming

Dynamic Programming

• Dynamic Programming is a general algorithm design technique

• It breaks up a problem into a series of overlapping sub-problems ( sub-problem whose results can be reused several times)

• Main idea:- set up a recurrence relating a solution to a larger instance to

solutions of some smaller instances- solve smaller instances once- record solutions in a table - extract solution to the initial instance from that table

Discussed topics

•Assembly-line scheduling

•Partition of Data Sequence

•Route Segmentation and Classification for GPS data

•Longest Common Subsequence (LCS)

Assembly-line scheduling

)]1[,]1[min(][ ,11,22,111 jjj atjfajfjf

Recursive equation:

where f1[j] and f2[j] are the accumulated time cost to reach station S1,j and S2,j

a1,j is the time cost for station S1,j

t2,j-1 is the time cost to change station from S2,j-1 to S1,j

Partition of Data Sequence

Partition of the sequence X into to K non-

overlapping groups with given cost functions

f(xi, xj) so that the total value of the cost

function is minimal:

min),(...),(),( 111 322 Niiii xxfxxfxxf

K

Partition of Data SequenceG(k,n) is cost function for optimal partition of n points into k non-overlapping groups:

k

jjj

iiifnkG

j 11

}{1,min),(

.0;,)1,1(min),( NnnjfjkGnkGnjk

Recursive equation:

Examples: Image QuantizationInput

Quantized

Here, cost function is:

2( , ) ( ) ( )

where ( )

i k j

i k j

f i j n k k q

q n k k

Square Error

Quantize value

Examples: Polygonal Approximation5004 points are approximated by 78 points.Given approximated points M, errors are minimized.Given error ε, number of approximated points are minimized.

Here, cost function is:

2or

( , ) max

{ }

( , )

i j i k j k

i j ki k j

f x x d

f x x d

Examples: Route segmentationSki

0 200 400 600 800 10000

5

10

15

20

25

time

spee

d

estimated segment result

0 1000 2000 3000 40000

1

2

3

4

5

6

time

spee

d

estimated segment result

0 1000 2000 3000 4000 5000 60000

2

4

6

8

10

time

spee

d

estimated segment result

Running and Jogging

Non-movingDivide the routes into several segments byspeed consistencyHere, cost function is:

( , ) ( ) ( )k i k j j if i j v t t

Speed variance Time duration

Frequent Moving type dependency of 1st order HMM

Examples: Route classification

0 10 20 30 40

0.2

0.4

0.6

0.8

1

Speed(km/h)

Pro

bab

ility

BikeRun

WalkStop

Car

Determine the moving type only by speed will cause mis-classification.

0 1000 2000 3000 4000 5000 60000

2

4

6

8

10

12

14

time

speed

estimated segment result

1st order HMM, maximize:

0 1000 2000 3000 4000 5000 60000

2

4

6

8

10

12

14

time

speed

estimated segment result

i 11

( | X , )M

i ii

f P m m

mi : moving type of segment i Xi : feature vector (e.g. speed)

Solve by dynamic programming similar with the Assembly-line scheduling problem

Examples: Route classification (cont.)

Longest Common Subsequence (LCS)

Find a maximum length common subsequence between two sequences.

For instance, Sequence 1: president Sequence 2: providence Its LCS is priden.

president

providence

How to compute LCS?

Let A=a1a2…am and B=b1b2…bn .

len(i, j): the length of an LCS between a1a2…ai and b1b2…bj

With proper initializations, len(i, j) can be computed as follows.

,

. and 0, if)),1(),1,(max(

and 0, if1)1,1(

,0or 0 if0

),(

ji

ji

bajijilenjilen

bajijilen

ji

jilen

i j 0 1 p

2 r

3 o

4 v

5 i

6 d

7 e

8 n

9 c

10 e

0 0 0 0 0 0 0 0 0 0 0 0

1 p 2

0 1 1 1 1 1 1 1 1 1 1

2 r 0 1 2 2 2 2 2 2 2 2 2

3 e 0 1 2 2 2 2 2 3 3 3 3

4 s 0 1 2 2 2 2 2 3 3 3 3

5 i 0 1 2 2 2 3 3 3 3 3 3

6 d 0 1 2 2 2 3 4 4 4 4 4

7 e 0 1 2 2 2 3 4 5 5 5 5

8 n 0 1 2 2 2 3 4 5 6 6 6

9 t 0 1 2 2 2 3 4 5 6 6 6

Running time and memory: O(mn) and O(mn).

i j 0 1 p

2 r

3 o

4 v

5 i

6 d

7 e

8 n

9 c

10 e

0 0 0 0 0 0 0 0 0 0 0 0

1 p 2

0 1 1 1 1 1 1 1 1 1 1

2 r 0 1 2 2 2 2 2 2 2 2 2

3 e 0 1 2 2 2 2 2 3 3 3 3

4 s 0 1 2 2 2 2 2 3 3 3 3

5 i 0 1 2 2 2 3 3 3 3 3 3

6 d 0 1 2 2 2 3 4 4 4 4 4

7 e 0 1 2 2 2 3 4 5 5 5 5

8 n 0 1 2 2 2 3 4 5 6 6 6

9 t 0 1 2 2 2 3 4 5 6 6 6

Output: priden

Time Series Matching using Longest Common Subsequence

0 20 40 60 80 100 120-4

-2

0

2Minimum Bounding Envelope (MBE) for LCSS

0 20 40 60 80 100 120

Point Correspondense, Similarity [=10, =0.3]

= 0.875

delta = time matching region (left & right)

epsilon = spatial matching region (up & down)

Recursive equation: 0 0, 0

( , ) ( 1, 1) 1 0, 0,| | , ( , )

max( ( 1, ), ( , 1))i j i j

i j

len i j len i j i j t t d p p

len i j len i j else

for 0, 0i j

Spatial Data Matching using Longest Common Subsequence

epsilon = spatial matching distance

Recursive equation:0 0, 0

( , ) ( 1, 1) 1 0, 0, ( , )

max( ( 1, ), ( , 1))i j

i j

len i j len i j i j d p p

len i j len i j else

Conclusion

Main idea:

- Divide into sub-problems

- Design cost function and recursive function

- Optimization (fill in the table)

- Backtracking