Sept 16, 2013: Persistent homology III

Transcription

1 MATH:7450 (22M:305) Topics in Topology: Scien:fic and Engineering Applica:ons of Algebraic Topology Sept 16, 2013: Persistent homology III Fall 2013 course offered through the University of Iowa Division of Con:nuing Educa:on Isabel K. Darcy, Department of Mathema:cs Applied Mathema:cal and Computa:onal Sciences, University of Iowa htp://

2 htp://homepage.math.uiowa.edu/~idarcy/at/schedule.html

3 htp://link.springer.com/ar:cle/ %2fs y

4 Compu:ng Persistent Homology by Afra Zomorodian, Gunnar Carlsson 1 C 1 à C 0 htp://link.springer.com/ar:cle/ %2fs y

5 B 0 = image of 1 = column space of M 1 = < tc + td, tb + c, ta + tb> Z 1 = kernel of 1 = null space of M 1 = < ad + cd + t(bc) + t(ab), ac + t 2 bc + t 2 ab >

6 1 0 C 1 à C 0 à 0 H 0 = Z 0 /B 0 = (kernel of 0 )/ (image of 1 ) = null space of M 0 column space of M 1 = < a, b, c, d : tc + td, tb + c, ta + tb >

7 B 0 = image of 1 = column space of M 1 = < tc + td, tb + c, ta + tb> Z 1 = kernel of 1 = null space of M 1 = < ad + cd + t(bc) + t(ab), ac + t 2 bc + t 2 ab >

8 1 1 C 2 à C 1 à C 0 H 1 = Z 1 /B 1 = (kernel of 1 )/ (image of 2 ) = null space of M 1 column space of M 2 = < ad + cd + t(bc) + t(ab), ac + t 2 bc + t 2 ab :??????? > Let z 1 = ad + cd + t(bc) + t(ab), z 2 = ac + t 2 bc + t 2 ab H 1 = Z 1 /B 1 = <z 1, z 2 :?? >.

9 Compu:ng Persistent Homology by Afra Zomorodian, Gunnar Carlsson 2 C 2 à C 1 M 2 = htp://link.springer.com/ar:cle/ %2fs y

10 Long method for determining column space of M2

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14 k+1 k C k+1 à C k à C k- 1, m k = # of k- simplices htp://link.springer.com/ar:cle/ %2fs y

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17 Short method for determining column space of M2

18 1 1 C 2 à C 1 à C 0 H 1 = Z 1 /B 1 = (kernel of 1 )/ (image of 2 ) = null space of M 1 column space of M 2 = < ad + cd + t(bc) + t(ab), ac + t 2 bc + t 2 ab :??????? > Let z 1 = ad + cd + t(bc) + t(ab), z 2 = ac + t 2 bc + t 2 ab H 1 = Z 1 /B 1 = <z 1, z 2 :?? >.

19 image of 2 = kernel of 2 =

20 image of 2 = < t z 2, t3 z 1 + t 2 z 2 > kernel of 2 = 0

21 2 1 C 2 à C 1 à C 0 H 1 = Z 1 /B 1 = (kernel of 1 )/ (image of 2 ) = null space of M 1 column space of M 2 where = <z 1, z 2 : t z 2, t 3 z 1 + t 2 z 2 > z 1 = ad + cd + t(bc) + t(ab), z 2 = ac + t 2 bc + t 2 ab

22 3 2 C 3 à C 2 à C 1 H 2 = Z 2 /B 2 = (kernel of 2 )/ (image of 3 ) = null space of M 2 column space of M 3 = 0

23 <z 1, z 2 : t z 2, t 3 z 1 + t 2 z 2 > where z 1 = ad + cd + t(bc) + t(ab), z 2 = ac + t 2 bc + t 2 ab i, p i i+p H 1 = Z 1 / ( B 1 Z 1 ) Note deg z 1 = 2, deg z 2 = 3 i, p i i+p U H 1 = Z 1 / ( B 1 Z 1 ) = 0 for i < 2 U i i

24 <z 1, z 2 : tz 2, t 3 z 1 + t 2 z 2 > where z 1 = ad + cd + t(bc) + t(ab), z 2 = ac + t 2 bc + t 2 ab i, p i i+p H 1 = Z 1 / ( B 1 Z 1 ) deg z 1 = 2, deg z 2 = 3, deg tz 2 = 4, deg t 3 z 1 + t 2 z 2 = 5 U i 2, p 2 2+p 2 H 1 = Z 1 / ( B 1 Z 1 U ) = Z/2Z for p =

25 <z 1, z 2 : tz 2, t 3 z 1 + t 2 z 2 > where z 1 = ad + cd + t(bc) + t(ab), z 2 = ac + t 2 bc + t 2 ab i, p i i+p H 1 = Z 1 / ( B 1 Z 1 ) deg z 1 = 2, deg z 2 = 3, deg tz 2 = 4, deg t 3 z 1 + t 2 z 2 = 5 U i 2, p 2 2+p 2 H 1 = Z 1 / ( B 1 Z 1 U ) = Z/2Z for p = 0, 1, 2

26 <z 1, z 2 : tz 2, t 3 z 1 + t 2 z 2 > where z 1 = ad + cd + t(bc) + t(ab), z 2 = ac + t 2 bc + t 2 ab i, p i i+p H 1 = Z 1 / ( B 1 Z 1 ) deg z 1 = 2, deg z 2 = 3, deg tz 2 = 4, deg t 3 z 1 + t 2 z 2 = 5 2, p 2 2+p 2 H 1 = Z 1 / ( B 1 Z 1 ) = 0 for p = U U i

27 <z 1, z 2 : tz 2, t 3 z 1 + t 2 z 2 > where z 1 = ad + cd + t(bc) + t(ab), z 2 = ac + t 2 bc + t 2 ab i, p i i+p H 1 = Z 1 / ( B 1 Z 1 ) deg z 1 = 2, deg z 2 = 3, deg tz 2 = 4, deg t 3 z 1 + t 2 z 2 = 5 2, p 2 2+p 2 H 1 = Z 1 / ( B 1 Z 1 ) = 0 for p = 3 U U i

28 <z 1, z 2 : tz 2, t 3 z 1 + t 2 z 2 > where z 1 = ad + cd + t(bc) + t(ab), z 2 = ac + t 2 bc + t 2 ab i, p i i+p H 1 = Z 1 / ( B 1 Z 1 ) deg z 1 = 2, deg z 2 = 3, deg tz 2 = 4, deg t 3 z 1 + t 2 z 2 = 5 3, p 3 3+p 3 U H 1 = Z 1 / ( B 1 Z 1 ) =??? U i

29 <z 1, z 2 : tz 2, t 3 z 1 + t 2 z 2 > where z 1 = ad + cd + t(bc) + t(ab), z 2 = ac + t 2 bc + t 2 ab i, p i i+p H 1 = Z 1 / ( B 1 Z 1 ) deg z 1 = 2, deg z 2 = 3, deg tz 2 = 4, deg t 3 z 1 + t 2 z 2 = 5 4, H 1 = Z 1 / ( B 1 Z 1 ) = U U i

30 <z 1, z 2 : tz 2, t 3 z 1 + t 2 z 2 > where z 1 = ad + cd + t(bc) + t(ab), z 2 = ac + t 2 bc + t 2 ab i, p i i+p H 1 = Z 1 / ( B 1 Z 1 ) deg z 1 = 2, deg z 2 = 3, deg tz 2 = 4, deg t 3 z 1 + t 2 z 2 = 5 4, H 1 = Z 1 / ( B 1 Z 1 ) = U U i

31 <z 1, z 2 : tz 2, t 3 z 1 + t 2 z 2 > where z 1 = ad + cd + t(bc) + t(ab), z 2 = ac + t 2 bc + t 2 ab i, p i i+p H 1 = Z 1 / ( B 1 Z 1 ) deg z 1 = 2, deg z 2 = 3, deg tz 2 = 4, deg t 3 z 1 + t 2 z 2 = 5 5, H 1 = Z 1 / ( B 1 Z 1 ) = U U i

32 H 0 = < a, b, c, d : tc + td, tb + c, ta + tb> H 1 = <z 1, z 2 : t z 2, t 3 z 1 + t 2 z 2 > [ ) [ ) [ ) [ ) [ z 1 = ad + cd + t(bc) + t(ab), z 2 = ac + t 2 bc + t 2 ab

33 htp://comptop.stanford.edu/programs/ Tausz, Andrew; Vejdemo- Johansson, Mikael; Adams, Henry

34 Download: htp://code.google.com/p/javaplex/

35 Download: htp://code.google.com/p/javaplex/

36 Some useful linux/mac/matlab commands: tar - xvvf foo.tar tar - xvvzf foo.tar.gz cd Directory cd../ cd pwd ls ls lrt extract foo.tar extract gzipped foo.tar.gz change directory go up one directory go to main directory print working directory list directory content list long format in reverse order :me

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38 >> version - java ans = Java 1.6.0_17- b04 ************ javaplex requires version number 1.5 or higher. >> cd AT/matlab_examples >> load_javaplex Confirm that javaplex is working properly: >> api.plex4.createexplicitsimplexstream() ans = edu.stanford.math.plex4.streams.impl.explicitsimplexstream@513fd4

39 create an empty explicit simplex stream: >> stream = api.plex4.createexplicitsimplexstream(); add simplicies: >> stream.addvertex(0); >> stream.addvertex(1); >> stream.addvertex(2); >> stream.addelement([0, 1]); >> stream.addelement([0, 2]); >> stream.addelement([1, 2]); >> stream.finalizestream();

40 Create filtered complex: >> stream = api.plex4.createexplicitsimplexstream(); >> stream.addvertex(1, 0); >> stream.addvertex(2, 0); >> stream.addvertex(3, 0); >> stream.addvertex(4, 0); >> stream.addvertex(5, 1); >> stream.addelement([1, 2], 0); >> stream.addelement([2, 3], 0); >> stream.addelement([3, 4], 0); >> stream.addelement([4, 1], 0); >> stream.addelement([3, 5], 2); >> stream.addelement([4, 5], 3); >> stream.addelement([3, 4, 5], 7); >> stream.finalizestream();

41 Determine if you have created a simplicial complex: >> stream.validateverbose() ans = 1 >> stream.addelement([1, 4, 5], 0); >> stream.validateverbose() Filtra:on index of face [4,5] exceeds that of element [1,4,5] (3 > 0) Stream does not contain face [1,5] of element [1,4,5] ans = 0

42 Create 2- dimensional sphere, S 2 >> dimension = 2; >> stream = api.plex4.createexplicitsimplexstream(); >> stream.addelement(0:(dimension + 1)); >> stream.ensureallfaces(); >> stream.removeelementifpresent(0:(dimension + 1)); >> stream.finalizestream();

43 Determine H i for i < 3 with Z 2 coefficients: >> persistence = api.plex4.getmodularsimplicialalgorithm(3, 2); >> intervals = persistence.computeintervals(stream) intervals = Dimension: 1 [3.0, 7.0) [0.0, infinity) Dimension: 0 [1.0, 2.0) [0.0, infinity)

44 compute a representa:ve cycle for each barcode: >> intervals = persistence.computeannotatedintervals(stream) intervals = Dimension: 1 [3.0, 7.0): [4,5] + [3,4] + - [3,5] [0.0, infinity): [1,4] + [2,3] + [1,2] + [3,4] Dimension: 0 [1.0, 2.0): - [3] + [5] [0.0, infinity): [1]

45 To enter a CSV file into Matlab, you can use the following command: >> M = csvread( filename,row,col); reads data from the file star:ng at the specified row and column. The row and column arguments are zero based, so that row = 0 and col = 0 specify the first value in the file. From: htp://