1 Example 1: Axisaligned rectangles


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1 COS 511: Theoretcal Machne Learnng Lecturer: Rob Schapre Lecture # 6 Scrbe: Aaron Schld February 21, 2013 Last class, we dscussed an analogue for Occam s Razor for nfnte hypothess spaces that, n conjuncton wth VCdmenson, reduced the problem of fndng a good PAClearnng algorthm to the problem of computng the VCdmenson of a gven hypothess space. Recall that VCdmeson s defned usng the noton of a shattered set,.e. a subset S of the doman such that Π H (S 2 S. In ths lecture, we compute the VCdmenson of several hypothess spaces by computng the maxmum sze of a shattered set. 1 Example 1: Axsalgned rectangles Not all sets of four ponts are shattered. For example the followng arrangement s mpossble:  Fgure 1: An mpossble assgnment of / to the data, as all rectangles that contan the outer three ponts (marked must also contan the one pont. However, ths s not suffcent to conclude that the VCdmenson s at most three. Note that the followng set does shatter: Fgure 2: A set of four ponts that shatters, as there s an axsalgned rectangle that contans any gven subset of the ponts but contans no others. Therefore, the VCdmenson s at least four. In fact, t s exactly four. Consder any set of fve dstnct ponts {v 1, v 2, v 3, v 4, v 5 } R 2. Consder a rectangle that contans the ponts wth maxmum xcoordnate, mnmum xcoordnate, maxmum ycoordnate, and mnmum ycoordnate. These ponts may not be dstnct. However, there are at most four such ponts. Call ths set of ponts S {v 1, v 2, v 3, v 4, v 5 }. Any axsalgned rectangle that
2 contans S must also contan all of the ponts v 1, v 2, v 3, v 4, and v 5. There s at least one v that s not n S, but stll must be n the rectangle. Therefore, the labelng that labels all vertces n S wth and v wth cannot be consstent wth any axsalgned rectangle. Ths means that there s no shattered set of sze 5, snce all possble labelngs of a shattered set must be realzed by some concept. By a smlar argument, we can show that the VCdmenson of axsalgned rectangles n R n s 2n. By generalzng the approach for provng that the VCdmenson of the postve half nterval learnng problem s 1, one can show that the VCdmenson of n 1 dmensonal hyperplanes n R n that pass through the orgn s n. Ths concepts are nequaltes of the form w x > 0 for any fxed w R n and varable x R n. In ths case, concepts label ponts wth f they are one sde of a hyperplane and otherwse. 2 Other remarks on VCdmenson In the cases mentoned prevously, note that the VCdmenson s smlar to the number of parameters needed to specfy any partcular concept. In the case of axsalgned rectangles, for example, they are equal snce rectangles requre a left boundary, a rght boundary, a top boundary, and a bottom boundary. Unfortunately, ths smlarty does not always hold, although t often does. There are some hypothess spaces wth nfnte VCdmenson that can be specfed wth one parameter. Note that f H s fnte, the VCdmenson s at most log 2 H, as at least 2 r dstnct hypotheses must exst to shatter a set of sze r. For a hypothess space wth nfnte VCdmenson, there s a set of sze m that s shattered for any m > 0. Therefore, Π H (m 2 m, whch we mentoned last class as an ndcaton of a class that s hard to learn. In the next secton, we wll show that all classes wth bounded VCdmenson d have Π H (m O(m d, completng the descrpton of PAClearnablty by VCdmenson. 3 Sauer s Lemma Recall that ( n k n! (n k!k! f 0 k n and ( n k 0 f k < 0 or k > n. k and n are ntegers and n s nonnegatve for our purposes. Note that ( n k O(n k when k s regarded as a postve constant. We wll show the followng lemma, whch mmedately mples the desred result: Lemma 3.1 (Sauer s Lemma. Let H be a hypothess wth fnte VCdmenson d. Then, Π H (m d ( m : Φ d (m Proof. We wll prove ths by nducton on m d. There are two base cases: Case 1 (m 0. There s only one possble assgnment of and to the empty set,.e. Π H (m 1 here. Note that Φ d (0 ( 0 0 ( ( 0 d 1, as desred. 2
3 Case 2 (d 0. Not even a sngle pont can be shattered n ths stuaton. Therefore, on any gven pont, all hypotheses have the same value. Therefore, there s only one possble hypothess and Π H (m 1. Ths agrees wth Φ, as Φ 0 (m ( m 0 1. Now, we wll prove the nducton step. For ths, we wll need Pascal s Identty, whch states that ( ( ( n n n 1 k k 1 k 1 for all ntegers n and k wth n 0. Consder a hypothess space H wth VCdmenson d and a set of m examples S : {x 1, x 2,..., x m }. Let T : {x 1, x 2,..., x m 1 }. Form two hypothess spaces H 1 and H 2 on T as follows (an example s n Fgure 3. Let H 1 be the set of restrctons of hypotheses from H to T. Let h T denote the restrcton of h to T for h H,.e. the functon h T : T {, } such that h T (x h(x for all x T. An element ρ on T s added to H 2 f and only f there are two dstnct hypotheses h 1, h 2 H such that h 1 T h 2 T ρ. Note that Π H (S Π H1 (T Π H2 (T. What are the VCdmensons of H 1 and H 2? Frst, note that the VCdmenson of H 1 s at most d, as any shatterng set of sze d 1 n T s also a subset of S that s shattered by the elements of H, contradctng the fact that the VCdmenson of H s d. Suppose that there s a set of sze d n T that s shattered by H 2. Snce every hypothess n H 2 s the restrcton of two dfferent hypotheses n H, x m can be added to the shattered set of sze d n T to obtan a set shattered by H of sze d 1. Ths s a contradcton, so the VCdmenson of H 2 s at most d 1. By the nductve hypothess, Π H1 (m 1 Φ d (m 1. Smlarly, Π H2 (m 1 Φ d 1 (m 1. Combnng these two nequaltes shows that Π H (m Φ d (m 1 Φ d 1 (m 1 ( d ( m 1 d 1 ( m 1 j j0 ( m 1 d 1 (( m 1 0 ( m d 1 ( m 0 1 Φ d (m completng the nductve step. ( m 1 1 Often, the polynomal Φ d (m s hard to work wth. Instead, we often use the followng result: Lemma 3.2. Φ d (m (em/d d when m d 1. Proof. m d 1 mples that d m 1. Therefore, snce d n the summand, 3
4 H x 1 x 2 x 3 x 4 x H 1 x 1 x 2 x 3 x H 2 x 1 x 2 x 3 x Fgure 3: The constructon of H 1 and H 2 ( d m d d ( m d ( d ( m m ( 1 d m m e d Multplyng on both sdes by (m/d d on both sdes gves the desred result. Pluggng ths result nto the examples bound proven last class shows that ( ( 1 err(h O d ln m m d ln 1 δ We can also wrte ths n terms of the number of examples requred to learn: ( 1 m O (ln 1/δ d ln 1/ɛ ɛ Note that the number of examples requred to learn scales lnearly wth the VCdmenson. 4 Lower bounds on learnng The bound proven n the prevous secton shows that the VCdmenson of a hypothess space yelds an upper bound on the number of examples needed to learn. Lower bounds on the requred number of examples also exst. If the VCdmenson of a hypothess space s d, there s a shattered set of sze d. Intutvely, any hypothess learned from a subset of sze at most d 1 cannot predct the value of the last element wth probablty better than 1/2. Ths suggests that at least Ω(d examples are requred to learn. In future classes, we wll prove the followng Theorem 4.1. For all learnng algorthms A, there s a concept c C and a dstrbuton D such that f A s gven m d/2 examples labeled by c and dstrbuted accordng to D, then Pr[err(h A > 1/8] 1 8 4
5 One can try to prove ths as follows. Choose a unform dstrbuton D on examples {z 1,..., z d } and run A on m d/2 examples. Call ths set of examples S. Label the elements of S arbtrarly wth and . Suppose that c C s selected to be consstent wth all of the labels on S and c(x h A (x for all x / S. err D (h A 1 2 snce c agrees wth h A on at most (d/2/2 1/2 of the probablty mass of the doman, whch means that there s no PAClearnng algorthm on d/2 examples. Ths proof s flawed, as c needs to be chosen before the examples. We wll dscuss a correct proof n future classes. 5
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