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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

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-overlappinggroups with given cost functions f(xi, xj) sothat the totalvalue of the cost function is minimal:

Partition of Data Sequence

G(k,n) is cost function for optimal partition of n points into

k non-overlapping groups:

Recursive equation:

Examples: Polygonal Approximation

5004points are approximated by 78 points.

Givenapproximatedpoints M, errorsareminimized.

Givenerrorε, number of approximatedpointsareminimized.

Here, cost function is:

Examples: Route segmentation

Ski

Running and Jogging

Non-moving

Divide the routes into severalsegmentsby

speedconsistency

Here, cost function is:

Speedvariance

Time duration

Examples: Route classification

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

Frequent Moving type dependency of 1st order HMM

Examples: Route classification (cont.)

1st order HMM, maximize:

mi : moving type of segment i

Xi : feature vector (e.g. speed)

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

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…aiand b1b2…bj

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

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

Time Series Matching using Longest Common Subsequence

delta = time matching region (left & right)

epsilon = spatial matching region (up & down)

Recursive equation:

Spatial Data Matching using Longest Common Subsequence

epsilon = spatial matching distance

Recursive equation:

Conclusion

- Main idea:
- Divide into sub-problems
- Design cost function and recursive function
- Optimization (fill in the table)
- Backtracking

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