0 1 sum= sum= sum= sum= sum= sum= sum=64

Transcription

1 Biomial Coefficiets I how may ways ca we choose elemets from a elemet set? There are choices for the first elemet, - for the secod,..., dow to - + for the th, yieldig *(-)*...*(-+). So there are 4*3=2 ways to choose 2 elemets from 4. Cosider {a,b,c,d}. {a,b} {a,c} {a,d} {b,c} {b,d} {c,d} Problem : The above formula couts (a,b) ad (b,a). Must divide by!. I geeral, r r(r )...(r + ), iteger 0 = ( )... 0, iteger < 0 for arbitrary real (eve complex) r.! = itegers 0!( )! Note that 0! = = 0 Let's loo at some values (Pascal's Triagle) sum= sum=2 2 2 sum= sum= sum= sum= sum=64 Note first 3 colums: 0 = = 2 ( ) = (triag) 2 Note symmetry i each row. = iteger 0, iteger Symmetry Argumet:-To choose of objects, we must choose which - to omit.! - same if replaced by -!( )! Sometimes we wat to add or get rid of a factor of. Note that

2 ! =!( )! = ( )! ( )!(( ) ( ))!, so r = r r iteger 0 Absorptio/extractio From Pascal's triagle, each elemet = elemet up + elemet up-&-left r = r + r iteger Combiatorial argumet: To choose elemets from, mar oe of the r -it's ot icluded i the (choose from the rest), or r -it's icluded i the (choose - from the rest) Applyig this several times (eepig costat gap betwee top & bottom): always expad smaller lower idex 5 3 = = = = Last term & all subsequet terms are 0. I geeral, (above =3, r=) + r+ r+ = iteger Combiatorial iterpretatio: 2

3 Cosider all paths from (0 0) -> ( r+). There are +r+ steps from + r+ {right, up}, ad ways to choose right from them. All paths hit the top lie somewhere. Call it, 0. For each value of, there are paths first hittig the top lie at. Repeat the above expasio but, istead of eepig a costat gap betwee top & bottom, expad larger lower idex. 5 3 = = = = = Note that 3 = 0, yieldig: + = m + itegers m, 0 m 0 Combiatorial iterpretatio: To choose m+ ticets from a set of + ticets umbered 0.., there are ways to do this whe largest ticet is m umber (choose the remaiig m from 0.. -) We ofte wat to maipulate Products of biomial coefficiets : r * m m = r * r itegers m, m Combiatorial iterpretatio: left side - give r people, choose m to play a game, the choose of the m to play offese right side - give r people, choose to play offese, the from the remaiig r- choose m- to play defese r m r! m! = m m!(r m)!!(m )! (cacel m!'s) r! = (multiply by (r- )!)!(m )!(r m)! r! (r )! =!(r )! (m )!(r m)! = r r m Whece biomial coefficiets? r+ 3

4 (x +y )0 = x 0y 0 (x +y ) = x y 0 + x 0y (x +y )2 = x 2y 0 + 2x y + x 0y 2... What is coefficiet of x y - i (x +y )? I how may ways ca we choose x 's & - y 's (multiplicatio beig commutative, we ca gather ad accumulate them). Surprise aswer: r x y r = (x + y) r itegers r 0 or x/y < Biomial Theorem Note: if x=y= & r= >0 itegral, the =2. -Also ote the sums of the rows of Pascal's triagle. -Also ote there are 2 -bit words, for 0, there are to choose the -bits i a -bit word with -bits. -To compute + x, -<x<. Cosider x=.04,.04 = ( ) 2 = ! 3! = = Bouds o biomial coefficiets : For, ( )...( + ) = = ( ) From Stirlig's approximatio,! 2π, more accurately, e! = 2π + e O! 2π e e32] 2π e! 2π e + 2 we get that! e, so ways 6 4

5 ( )...( + ) =!! e e Ex : Cosider Direct Chaiig Hashig. Assume there are m bucets (lists), ad elemets distributed amog the lists. What is E[A']? E[A']= * Pr{h(ey) has items} 0 Pr{ st items go to h(ey)}= m m Pr{last - items somewhere besides h(ey)}= m Pr{h(ey) has exactly items}= m m m m (Note: for =0, ; for =, ) m m E[A']= m m m 0 Remove ad we ca use the biomial theorem, but replacig it with m or meas we ca move it out of the summatio. Absorptio/Extractio: = = E[A']= m = m m m 0 m m 0 2 problems: -egative term i bottom of -sum of expoets - Replace by + E[A']= m = + m m m m m (pull out (sice 0 m ) = m = 0) = m m 0 m m m m m 5

6 (biomial theorem) = m m + m = m m Variace computed i H.W.#4-9 Ex : Uiform probig hashig, Goet 3.3.2, pg. 48 E[A'] = +E[ {collisios} ] = + * Pr{ ' C = } It's easy to compute Pr{C'>} Pr{C' }= m =α 0 Pr{C' 2}= m m Pr{C' i }= 0 i m for i (0 for i > ) Theorem : If r.v. X taes values from {0,,2,...}, the E[X] = ipr{x = i} = ipr{x [ i} Pr{X i+ } ] = Pr{X i} 0 i 0 i i If r.v. X taes values from {0,,2,...,}, the E[X] = i Pr{X = i} = i[ Pr{X i} Pr{X i + } ] = Pr{X i} 0 i 0 i i Proof : Each Pr{X i } added i i times & subtracted out (i-) times.ó ' i E[A'] = + Pr{ C } = + 0 i m i = + m m... + m +! (m )! ( )! m!! (m )! m! ( )! Isight : Get biomials out of this, i.e., multiply st umerator & 2 d deomiator by (m- )! m!(m )! (m )! E[A'] = + m! ( )!(m )! m m m m m What idetity helps us here? = + m m + 0 m m m = m m m 0 So E[A'] = + m m m m 0 m Expadig the sum, m m = + m = m + m m m m m + 0 6

7 So E[A'] = + m + m m m + m = + (m + )! m (m + )!! m! (m )!! (m + )! m!(m + )! = + m! (m + )(m )!! m (m + )(m )!! = + m = + m + = m + + m + = m m + m + = m α + m For large m, So E[A'] = (same as radom probig) α Ex : Biomial lower boud o worst case umber of comparisos for mergig. Give two sorted lists of legths a ad b, show that uder biary decisio tree model the umber of comparisos ecessary to merge them ito oe sorted list is lg a + b a Hit : You may proceed by -otig that ay algorithm whose basic computatio is pairwise compariso may be modeled as a biary tree, -establishig a lower boud o the umber of outputs of the algorithm (leaves of the biary decisio tree), -fidig the miimum height of a biary tree which has at least a certai umber of leaves. The output of ay merge algorithm is a list x,...,xm, where m =a+b, ad a + b the list of a's is a sublist of x,...,xm of legth a. There are ways a to choose this sublist, so the algorithm must be able to distiguish a + b betwee outputs. The correspodig decisio tree must have at a a + b a + b least leaves. Ay biary tree of leaves must have have a a height lg a + b a 7