1. Machine Learning and Inductive Inference Hendrik Blockeel 2001-2002
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11. Example: given molecules that are active against some disease, find out what is common in them; this is probably the reason for their activity.
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21. Overview of design choices type of training experience games against self games against expert table of good moves determine type of target function determine representation determine learning algorithm … … … … ready! Board Board Move linear function of 6 features … gradient descent
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52. + G S Current versionspace contains all rectangles covering S and covered by G, e.g. h = <2-5,2-3> h S = {<3-5,2-3>} G = {<1-6, 1-3>}
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56. Equivalence between inductive and deductive systems inductive system training examples new instance deductive system training examples new instance inductive bias result (by proof) result (by inductive leap)
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71. Clustering tree abundance(Tubifex sp.,5) ? T = 0.357111 pH = -0.496808 cond = 1.23151 O2 = -1.09279 O2sat = -1.04837 CO2 = 0.893152 hard = 0.988909 NO2 = 0.54731 NO3 = 0.426773 NH4 = 1.11263 PO4 = 0.875459 Cl = 0.86275 SiO2 = 0.997237 KMnO4 = 1.29711 K2Cr2O7 = 0.97025 BOD = 0.67012 abundance(Sphaerotilus natans,5) ? yes no T = 0.0129737 pH = -0.536434 cond = 0.914569 O2 = -0.810187 O2sat = -0.848571 CO2 = 0.443103 hard = 0.806137 NO2 = 0.4151 NO3 = -0.0847706 NH4 = 0.536927 PO4 = 0.442398 Cl = 0.668979 SiO2 = 0.291415 KMnO4 = 1.08462 K2Cr2O7 = 0.850733 BOD = 0.651707 yes no abundance( ...) <- "standardized" values (how many standard deviations above mean)
88. accuracy on training data accuracy on unseen data size of tree accuracy effect of pruning
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91. Rules from trees: example Outlook Humidity Wind No Yes No Yes Yes Sunny Overcast Rainy High Normal Strong Weak if Outlook = Sunny and Humidity = High then No if Outlook = Sunny and Humidity = Normal then Yes …
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93. Pruning rules: example A false true B true true false true false if A=true then true if A=false and B=true then true if A=false and B=false then false Tree representing A B Rules represent A ( A B) A B
98. Why concave functions? E E 1 E 2 p p 2 p 1 Assume node with size n , entropy E and proportion of positives p is split into 2 nodes with n 1 , E 1 , p 1 and n 2 , E 2 p 2 . We have p = (n 1 /n)p 1 + (n 2 /n) p 2 and the new average entropy E’ = (n 1 /n)E 1 +(n 2 /n)E 2 is therefore found by linear interpolation between ( p 1 ,E 1 ) and ( p 2 ,E 2 ) at p . Gain = difference in height between ( p, E ) and ( p,E’ ). (n 1 /n)E 1 +(n 2 /n)E 2 Gain
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100. Generic TDIDT algorithm function TDIDT( E : set of examples) returns tree; T' := grow_tree( E ); T := prune ( T' ); return T ; function grow_tree( E : set of examples) returns tree; T := generate_tests ( E ); t := best_test ( T , E ); P := partition induced on E by t ; if stop_criterion ( E , P ) then return leaf( info ( E )) else for all E j in P: t j := grow_tree( E j ); return node( t , {( j,t j )};
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105. Clustering tree abundance(Tubifex sp.,5) ? T = 0.357111 pH = -0.496808 cond = 1.23151 O2 = -1.09279 O2sat = -1.04837 CO2 = 0.893152 hard = 0.988909 NO2 = 0.54731 NO3 = 0.426773 NH4 = 1.11263 PO4 = 0.875459 Cl = 0.86275 SiO2 = 0.997237 KMnO4 = 1.29711 K2Cr2O7 = 0.97025 BOD = 0.67012 abundance(Sphaerotilus natans,5) ? yes no T = 0.0129737 pH = -0.536434 cond = 0.914569 O2 = -0.810187 O2sat = -0.848571 CO2 = 0.443103 hard = 0.806137 NO2 = 0.4151 NO3 = -0.0847706 NH4 = 0.536927 PO4 = 0.442398 Cl = 0.668979 SiO2 = 0.291415 KMnO4 = 1.08462 K2Cr2O7 = 0.850733 BOD = 0.651707 yes no abundance( ...) <- "standardized" values (how many standard deviations above mean)
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