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Restricted Boltzmann Machines (RBM)
- 1. A friendly introduction to Restricted Boltzmann Machines (RBM) Luis Serrano
- 2. The mystery
- 5. Beto Mystery
- 9. Beto Cameron Mystery Aisha Don’t like himUgh, Beto… Aisha and Cameron? Nope
- 10. Beto Cameron Mystery Aisha Aisha? Beto?Who are they? No clue who they are.
- 11. Beto Cameron Solution Aisha Descartes Euler WOOF YEAH!DOGS!!! I LOVE CATS!
- 12. Beto Cameron Weights Aisha Descartes Euler 4 -2 2 2 -2 -4 1 1 1 12 Hidden Layer Visible Layer
- 13. Restricted Boltzmann Machine (RBM) 4 -2 2 2 -2 -4 1 1 1 12 Hidden Layer Visible Layer
- 14. Scores
- 15. Beto Cameron Scores Aisha Descartes Euler 4 -2 2 2 -2 -4 1 1 1 12 Participants Score B CA D E4 -2 2 2 -2 -4 1 1 1 12
- 16. Beto Cameron Scores Aisha Descartes Euler 4 -2 2 2 -2 -4 1 1 1 12 Participants Score B CA D E 4 -222 -2 -4 1 1 112 6
- 21. Scenario Score None 0 A 1 B 1 C 1 D 2 E 1 AB 2 AC 2 AD 5 AE 0 BC 2 BD -2 BE 7 CD 5 CE 0 DE 3 ABC 3 ABD 1 ABE 6 ACD 8 ACE -1 ADE 4 BCD 1 BCE 6 BDE 4 CDE 4 ABCD 4 ABCE 5 ABDE 5 ACDE 5 BCDE 5 ABCDE 6 Beto Cameron Scores Aisha Descartes Euler 4 -2 2 2 -2 -4 1 1 1 12
- 22. Scenario Score None 0 A 1 B 1 C 1 D 2 E 1 AB 2 AC 2 AD 5 AE 0 BC 2 BD -2 BE 7 CD 5 CE 0 DE 3 ABC 3 ABD 1 ABE 6 ACD 8 ACE -1 ADE 4 BCD 1 BCE 6 BDE 4 CDE 4 ABCD 4 ABCE 5 ABDE 5 ACDE 5 BCDE 5 ABCDE 6 Beto Cameron Scores Aisha Descartes Euler 4 -2 2 2 -2 -4 1 1 1 12
- 23. Restricted Boltzmann Machine (RBM) 4 -2 2 2 -2 -4 1 1 1 12 Hidden Layer Visible Layer E = − ∑ i bivi − ∑ i aihi − ∑ i,j Wijvihj Energy = -Score
- 24. Probabilities
- 25. Scores to probabilities Sum = 6 Score Probability 3 1/2 2 1/3 1 1/6 Sum = 1
- 26. Scores to probabilities Sum = 0 Score Probability 1 1/0 0 0/0 -1 -1/0
- 27. Scores to probabilities Sum = 4.086 Score escore Normalize 1 e1 = 2.718 0.665 0 e0 = 1 0.245 -1 e-1 = 0.368 0.09 Sum = 1
- 28. Beto CameronAisha Descartes Euler 4 -2 2 2 -2 -4 1 1 1 12 Scenario Score eScore Probability None 0 1 0 A 1 2.72 0 B 1 2.72 0 C 1 2.72 0 D 2 7.38 0 E 1 2.72 0 AB 2 7.38 0 AC 2 7.38 0 AD 5 148.41 0.02 AE 0 2.72 0 BC 2 7.38 0 BD -2 0.14 0 BE 7 1096.63 0.17 CD 5 148.41 0.02 CE 0 1 0 DE 3 20.08 0 ABC 3 20.08 0 ABD 1 2.72 0 ABE 6 403.43 0.06 ACD 8 2980.96 0.45 ACE -1 0.37 0 ADE 4 54.6 0 BCD 1 2.72 0 BCE 6 403.43 0.06 BDE 4 54.6 0 CDE 4 54.6 0 ABCD 4 54.6 0.02 ABCE 5 148.41 0.02 ABDE 5 148.41 0.02 ACDE 5 148.41 0.02 BCDE 5 148.41 0.02 ABCDE 6 403.43 0.06
- 29. Energy to probability 4 -2 2 2 -2 -4 1 1 1 12 Hidden Layer Visible Layer E = − ∑ i bivi − ∑ i aihi − ∑ i,j Wijvihj p(v, h) = 1 Z e−E(v,h) Z = ∑ v,h e−E(v,h)
- 30. How to train an RBM? What exactly do we want?
- 31. 0 0 0 0 0 0 0 0 0 00 A B C D E Scenario Score eScore Probability None 0 1 1/32 A 0 1 1/32 B 0 1 1/32 C 0 1 1/32 D 0 1 1/32 E 0 1 1/32 AB 0 1 1/32 AC 0 1 1/32 AD 0 1 1/32 AE 0 1 1/32 BC 0 1 1/32 BD 0 1 1/32 BE 0 1 1/32 CD 0 1 1/32 CE 0 1 1/32 DE 0 1 1/32 Scenario Score eScore Probability ABC 0 1 1/32 ABD 0 1 1/32 ABE 0 1 1/32 ACD 0 1 1/32 ACE 0 1 1/32 ADE 0 1 1/32 BCD 0 1 1/32 BCE 0 1 1/32 BDE 0 1 1/32 CDE 0 1 1/32 ABCD 0 1 1/32 ABCE 0 1 1/32 ABDE 0 1 1/32 ACDE 0 1 1/32 BCDE 0 1 1/32 ABCDE 0 1 1/32
- 32. None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE 0 0 0 0 0 0 0 0 0 00 A B C D E
- 38. How to train an RBM? Contrastive divergence
- 40. A, C, and no B None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE
- 41. A, C, and no B None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE
- 42. A, C, and no B None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE All scenarios
- 43. A, C, and no B None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE
- 44. B and no A,C None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE
- 45. B and no A,C None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE
- 46. B and no A,C None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE All scenarios
- 47. B and no A,C None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE All scenarios
- 49. Maximizing the probability of the data 4 -2 2 2 -2 -4 1 1 1 12 Hidden Layer Visible Layer arg max W 𝔼[log P(v)]Maximize ∂ ∂W log P(vn)Derivative: = 𝔼 [ ∂ ∂W − E(v, h)|v = vn] − 𝔼 [ ∂ ∂W − E(v, h) ] arg max W ∏ v∈V P(v) Find
- 50. The end? nope…
- 51. Small problem
- 52. There are way too many possibilities! Problem None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE 32 = 25
- 53. How many? … … 100 nodes 200 nodes 2300 configurations
- 55. Partition function is intractable 4 -2 2 2 -2 -4 1 1 1 12 Hidden Layer Visible Layer p(v, h) = 1 Z e−E(v,h) Z = ∑ v,h e−E(v,h) Intractable
- 56. Problem What can we do? None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE
- 63. A, C, and no B None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE
- 64. A, C, and no B None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE
- 65. A, C, and no B None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE
- 66. A, C, and no B None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE All scenarios
- 67. A, C, and no B None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE All scenarios
- 68. A, C, and no B None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE All scenarios
- 86. None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE 4 -2 2 2 -2 -4 1 1 1 12 A B C D E
- 87. How to increase (or decrease) the probability of a configuration?
- 88. A B C ED RBM
- 89. A B C ED Increase probability of
- 90. A B C ED Increase probability of
- 91. Beto CameronAisha Descartes Euler 0 0 0 0 0 0 0 0 0 00 Increase probability of
- 92. Beto CameronAisha Descartes Euler 0 0 0 0 0 0 0 0 0 00 Increase probability of Learning rate = 0.1
- 93. Beto CameronAisha Descartes Euler 0 0.1 0 0 0.1 0 0.1 0 0.1 0.10 Increase probability of Learning rate = 0.1
- 94. Beto CameronAisha Descartes Euler 0 0.1 0 0 0.1 0 0.1 0 0.1 0.10 Decrease probability of
- 95. Beto CameronAisha Descartes Euler 0 0.1 0 0 0.1 0 0.1 0 0.1 0.10 Decrease probability of
- 96. Beto CameronAisha Descartes Euler 0 0.1 -0.1 0 0.1 0 0.1 0 0 0.1-0.1 Decrease probability of
- 97. Beto CameronAisha Descartes Euler 4 -2 2 2 -2 -4 1 1 1 12 None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE
- 98. Are we done now? Still no…
- 101. None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE A, C, and no B All other scenarios
- 102. None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE Picking random items from here is really hard! How to pick a random one with conditions How to pick a completely random one
- 103. How to pick a sample that agrees with our data point?
- 104. Gibbs Sampling
- 105. Independent sampling
- 106. BetoAisha Descartes 1 1 1 Euler Fernando -2 Gloria -1 Igor 0 2 1 Hypatia 3 Cameron
- 107. BetoAisha 1 1 1 Fernando -2 Gloria -1 Igor 0 Descartes Euler2 1 Hypatia 3 Cameron
- 108. Beto CameronAisha 1 1 1 Fernando Gloria Igor Hypatia -2 -1 0 3 2 -3 1 -2 P( ) =
- 109. Beto CameronAisha 1 1 1 Fernando Gloria Igor Hypatia -2 -1 0 3 2 -3 1 -2P( ) =
- 110. Beto CameronAisha 1 1 1 Fernando Gloria Igor Hypatia -2 -1 0 32-31-2P( ) = 1σ( ) = 0.73 σ(x) = 1 1 + e−x 1 1 0.73
- 111. Descartes Fernando Euler Hypatia -2 2 1 3 BetoAisha 1 1 1 Gloria Igor-1 0Cameron P( ) =
- 112. Descartes Fernando Euler Hypatia -2 2 1 3 -1-1 P( ) =
- 113. Descartes Fernando Euler Hypatia -2 2 1 3 -1-1P( ) =
- 114. Descartes Fernando Euler Hypatia -2 2 1 3 -1-1P( ) = σ( ) = 0.018 -4
- 115. How do we pick random samples?
- 116. None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE How to pick a random one with conditions How to pick a completely random one
- 117. BetoAisha Descartes Euler -0.5 -0.4 0.5 -1 1 0.8 0.5 0.3 10.9 -0.8 Cameron
- 120. 0.50.9-0.4 Aisha Descartes P( ) = -0.4 0.5 0.9 Cameron = 0.731σ( )
- 121. σ( = 0.731 )P( ) = Aisha Euler -1 1 -0.8 σ( = 0.31-0.8 )P( ) = Cameron
- 122. BetoAisha Descartes Euler Cameron P = 0.73 P = 0.31
- 123. Gibbs Sampling
- 124. How to pick a completely random sample?
- 125. None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE How to pick a random one with conditions How to pick a completely random one
- 126. Pick a random spot in the world
- 127. None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE Gibbs Sampling How to pick a totally random sample from this distribution
- 128. A B C ED A B C ED A B C ED A B C ED A B C ED A B C ED None AC BC DED A D BC DE None DD
- 129. A B C ED A B C ED A B C ED A B C ED None BC DE A B C ED AC D D BC DE None D A B C ED A D
- 130. A B C ED AC D Increase scores Decrease scores A B C ED A D
- 131. Summary
- 132. Data None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE ProbabilitiesRestricted Boltzmann Machine 4 -2 2 2 -2 -4 1 1 1 12 A B C D E
- 133. Generated Data None A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ABCD ABCE ABDE ACDE BCDE ABCDE ProbabilitiesRestricted Boltzmann Machine 4 -2 2 2 -2 -4 1 1 1 12 A B C D E
- 135. Thank you!
- 137. This presentation was done on Keynote
- 138. https://www.manning.com/books/grokking-machine-learning Discount code: serranoyt Grokking Machine Learning By Luis G. Serrano