CS 336. of n 1 objects with order unimportant but repetition allowed.

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1 CS 336. The importat issue is the logic you used to arrive at your aswer.. Use extra paper to determie your solutios the eatly trascribe them oto these sheets. 3. Do ot submit the scratch sheets. However, all of the logic ecessary to obtai the solutio should be o these sheets. 4. Commet o all logical flaws ad omissios ad eclose the commets i boxes. [0] Let A ad B be o-empty sets with cardialities m ad, respectively. How may fuctios from A to B are ot oe-to-oe? (You may assume m.) For each of the m elemets of A there are choices for the value of the fuctio m so there are fuctios from A to B. To cout the oe-to-oe fuctios, we otice that there are There are choices for the value of the fuctio for the first elemet of A, choices for the value of the fuctio for the first elemet of A,, ad m+ choices for the value of the fuctio for the m th elemet of A.! Thus, there are oe-to-oe fuctios from A to B. There are ( m)! m! fuctios from A to B that are ot oe-to-oe. ( m)!. Cosider choosig objects from a set { a a },,..., of objects with order uimportat but repetitio allowed. a. [5] How may such selectios have exactly oe a? We eed to determie how may ways we have of choosig objects from a set of the set { a,..., a } of objects with order uimportat but repetitio allowed. This is equivalet to placig balls i bis ad there are ( ) + ( ) + 3 = ways of doig it. a

2 [5] How may such selectios have at least a? Sice the total umber of ways of choosig objects from a set { a a },,..., a of + objects with order uimportat but repetitio allowed is ad there ( ) + + are = ways of choosig objects from the set { a,..., a } of objects with order uimportat but repetitio allowed, there are + + such selectios havig at least oe a. Alteratively, you could see this as choosig a oce the choosig objects from the set + { a, a,..., a }. This way you get which equals a. [0] Usig a combiatorial argumet, prove that for : 3 = = 0 We see to determie how may strigs of legth there are cosistig of elemets of { abc,,} allowig repetitio. Sice there are three choices for each of the positios there are 3 such strigs. Alteratively, let deote the umber of positios i the strig occupied by a or b. The value of varies from 0 to. For a fixed value of, there are ways to select these positios ad the optios for each of the positios, thus for the fixed value of ad = 0 overall. This must equal3.

3 b. [0] Usig a combiatorial argumet, prove that for m,, p : m+ + p + p m+ + p m+ m = p. Give a set S of cardiality m+ + p, cosider how may partitios there are of S ito disjoit subsets AB,, ad C of cardialities m,, ad p, respectively. For the left side of the equality, we cout this by first selectig the m elemets for m+ + p the subset A i m ways ad the selectig the elemets for the subset B from the remaider i + p ways. The remaiig elemets form the subset C. For the right side of the equality, we cout this by first selectig the p m+ + p elemets for the subset C i p ways ad the selectig the elemets m+ for the subset B from the remaider i ways. The remaiig elemets form the subset A. The two must be equal so m+ + p + p m+ + p m+ m = p.

4 4. a. [5] Suppose objects are beig chose from a set { a a },,..., a of objects with order uimportat ad repetitio ot allowed. Suppose all such selectios are equally liely. What is the probability that a selectio cotais exactly two of { a, a, a? (You may assume + 3. ) There are equally liely selectios of objects from objects with order uimportat ad repetitio ot allowed. If exactly two of { a, a, a are chose 3 there are ways to select the two from { a, a, a ad the ways to choose the remaiig from the 3 elemets { a4, a5,..., a }. So there are 3 selectios cotaiig exactly two of { a, a, a ad the probability of 3 such a selectio is b. [5] Now suppose objects are beig chose from a set { a a },,..., a of objects with order importat ad repetitio still ot allowed. Suppose all such selectios are equally liely. What is the probability that a selectio cotais exactly two of { a, a, a?! There are equally liely selectios of objects from objects with order ( )! importat ad repetitio ot allowed. If exactly two of { a, a, a are chose there 3 are ways to select the two from { a, a, a ad the ways to choose the remaiig from the 3 elemets { a4, a5,..., a }. The there are! 3 ways to permute the elemets selected. Thus, there are! selectios cotaiig exactly two of { a, a, a ad the probability of such a selectio is 3!. (Alteratively, we could recogize that this is the same as part a.! ( )! sice, if orser is imposed, there are! ways of permutig the selectios with exactly two of { a, a, a but also! ways of permutig all of the selectios. )

5 5. [0] Usig Defiitio but o cardiality theorems, prove that the set P = { ifiitely log strigs of 0s ad s with exactly two s} is ifiite. Cosider the fuctio f : P P defied by f () s = 0 s for ay strig s P(where deotes cocateatio). Notice that if s has exactly two s, the so will f (). s The fuctio f is oe-to-oe sice if st, P ad s t, the f () s = 0 s 0 t = f(). t Lastly, otice that sice o strig maps to < >, f maps P ito a proper subset of P. We coclude that P is ifiite. 6. [0] Let A be a coutably ifiite set, B be a ucoutably ifiite set oempty set ad C = A B. Is the C fiite, coutably ifiite, or ucoutably ifiite? Prove your assertio. The set C is ucoutably ifiite. Sice A is coutably ifiite, it is o-empty. Let a be ay elemet of A. Cosider the fuctio f : B C defied as f ( b) = ( a, b). The fuctio f is oe-to-oe sice if b, b B ad b b, the f ( b) = ( a, b) ( a, b ) = f( b ). By Theorem, C is ucoutably ifiite. 7. [0] Give that for ad α > 0,( + α) + α, prove that =Ο(3 ). 3 Let M = ad N =. For, ( ) = ( + ) +. Thus, for, 3. So the for N, = 3 = M [0] Prove that if f=ο ( g) ad f= ο( g), the ff= ο( gg). Sice f=ο ( g), there exist M ad N so that for N, f( ) M g( ). ε Sice f= ο( g), give ε > 0 the also > 0 ad there exists N so that for M ε N, f( ) g( ), thus for M ε max{ N, N}, f( ) f( ) M g( ) g( ) = ε g( ) g( ), so M f f = ο( g g ).

6 9. [0] Assumig x ad y are iteger variables, prove correct with respect to precoditio y is defied ad postcoditio x > 0 : if y > 0 the x := *y if x > 5 the x : = x-4 edif x := 4-y y := y- if y = -3 the x : = x-3 edif edif y is defied if y > 0 the y > 0 x := *y y > 0 x = y x > 0 if x > 5 the x > x : = x-4 ( x' > 5) ( x = x' 4) x > edif ( x > 0) ( x > ) x > 0 y 0 x := 4-y ( y 0) ( x = 4 y) x > 3 y := y- x > 3 if y = -3 the x > 3 x : = x-3 ( x' > 3) ( x = x' 3) x > 0 edif ( x > 3) ( x> 0) x > 0 edif ( x > 0) ( x > 0) x > 0

7 0. [0] Prove the followig code is partially correct with respect to precoditio true ad postcoditio is eve. (Assume,, ad i are iteger variables ad that ad i are defied at iput.) := 34 while i = do if i 7 the : = - := 4*-6 edif i := i+5 edwhile Be explicit about your loop ivariat: I = is eve true := 34 = 34 is eve while i = do ( is eve ) ( i ) is eve if i 7 the ( is eve ) ( i 7) is eve : = - ( ' is eve ) ( = ' ) is eve ( is eve ) ( i < 7) is eve := 4*-6 ( ' is eve ) ( = 4 ' 6) is eve edif ( is eve) ( is eve ) is eve i := i+5 ( is eve ) ( i' = i+ 5) is eve edwhile ( is eve ) ( i > ) is eve

8 . [0] Prove that the code below termiates. (Assume s ad i are iteger variables.): s := 0 i := while i = do s := s+i i := 4*i+ edwhile First we recogize that if the quatity 00,000-i becomes egative, the loop will termiate. We will show that that quatity strictly decreases but to that ed we eed to guaratee that the variable i stays positive. Cosider the ivariat i : s := 0 i := i while i = do i s := s+i i i := 4*i+ (' i ) ( i = 4' i + ) i = i' + 3 i' + > i' ( i ) (00000 i< i') i edwhile The quatity 00,000-i strictly decreases through the loop. Sice this is a iteger expressio, evetually 00,000-i becomes egative ad the loop termiates.. [0] Determie the weaest precoditio with respect to the postcoditio S = 0 for the followig (assume S, y, ad x are iteger variables ad y ad x are defied if x 0 the x := y S := x-y S := y+x edif wp( if x 0 the x := y; S := x-y S := y+x edif, S = 0) = (( x 0) wp( x := y; S := x-y, S = 0)) (( x = 0) wp( S := y+x, S = 0) ) = (( x 0) wp( x := y; x y = 0)) (( x = 0) ( y+ x = 0)) = (( x 0) ( y y = 0)) ( x = y = 0)) = ( x 0) ( x = y = 0))

9 3. [0] Determie the weaest precoditio with respect to the postcoditio x y for the followig (assume y ad x are iteger variables ad x is defied). Simplify your aswer so that there are NO logical operators. if x 3 the y:= if x = the y := 6 y := x+ edif edif We cosider the ier if-the- first: wp (if x = the y := 6 y := x+ edif, x y ) = (( x = ) wp( y := 6, x y)) (( x ) wp( y := x+, x y)) = (( x = ) ( x 6)) (( x ) ( x x + )) = (( x = ) ( x ) = true Now, lettig S deote if x = the y := 6 y := x+ edif, wp (if x 3 the y:= S edif, x y ) = (( x 3) wp( y :=, x y)) (( x < 3) wp( S, x y)) = (( x 3) ( x )) (( x < 3) true) = (( x 3) ( x < 3) = true So, wp (if x 3 the y:= if x = the y := 6 y := x+ edif edif, x y ) = true.