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Unformatted text preview: 6.045J/18.400J: Automata, Computability and Complexity Prof. Nancy Lynch Homework 1
Due: February 12, 2007 Elena Grigorescu Problem 1: Construct truth tables for each of the following formulas. Also, for each pair of formulas, state which of the following holds: They are equivalent, They are not equivalent, but one implies the other (make sure to state which is which), or Neither of the above: (a) (b) (c) (d) Solution 1: f f t t f f t t f f t t $" (a) f t f t f t f t f t f t '" f f f t t f t t t t f t t f t f
11 2 01 (b) (c) f (d) f t t f t f t f t t f t t f f t f f t t f f t '" % $# " )543 0)( '" % $" ( '" % &$# " ! t t t t t t t t f f t t ) Comparison: By reading the truth tables we note that only (a) and (b) are equivalent, both (c) and (d) imply each of (a) and (d), while (c) and (d) are incomparable. Problem 2: Fix any finite set and let the power set of be denoted by . Let be the relation between elements of such that if and only if there is a bijection between and .
8 (a) Show that is an equivalence relation. that is reflexive and symmetric but not transitive. that is reflexive and transitive but not symmetric. that is symmetric and transitive but not reflexive.
8 GFD8 E C (b) Define a relation (c) Define a relation (d) Define a relation Solution 2: There are many correct answers to this problem. Here is one example for each part. (a) To show that is an equivalence relation, we must show that it is reflexive, symmetric, and transitive. First, is reflexive, because a bijection can be formed by mapping to itself. Next, is symmetric, because we can simply invert each element mapping. By the onetoone property of the bijection, we know that each the element of has a single corresponding element in and by the onto property, we know that all of will be defined under this mapping. Finally, is transitive, because given and , we consider the corresponding pairs and for and define a new mapping by the set all elements . (b) Let be the relation between elements of such that if and only if (1) there is a bijection between and , and (2) and differ in at most one element. The relation is reflexive: a set does not differ from itself. The relation is also symmetric: if differs from in at most one element; then obviously differs from in at most one element. However, this relation is not transitive. Suppose , , and . Now, is true (i.e., there is a bijection and they only differ in one element) and likewise is true, however is not true! and differ in two elements. (c) Suppose we define an ordering on the elements in the finite power set . Let be the relation between elements of such that if and only if (1) there is a bijection between and , and (2) it is possible to list the elements of and in such a way that each is the corresponding . Since the operator is both reflexive and transitive, so is . However, is not symmetric. For example, if we let and , the obviously but there is no valid ordering of . Thus, is also not symmetric.
9 6 7 A uca S X A H 8 e1P9 H 18 6 s Yi 7 x A 9 yuR S X A H V8 s i 9 x 9 8 V8 t H A A C wI9 8 W C w44A 8 A C wP9 8 A 9 6 7 s Y g p i 9 A x W A 9 s cg i 9 W x A C D8 s 9 bg 9 i x y9 8 vu8 t C W C 8 P9 C 18 A X ed U X `ca W U h9 X ba U X `YR 8 A 8 W @[email protected] S X d g A S X a g 9 S X 8 W 8 V4A 9 TS A R 8 9 A 8 VI9 A s c X d g X ccrqR U T9 pi 12 9 8 6 7 6 A [email protected] 8 8 GPI8 E H 8 GPI8 E Q 6 7 6 A (d) Let be the relation between elements of such that if and only if (1) there is a bijection between and , and (2) a specific element is contained in . First, is symmetric, because obviously if then . is also transitive, since and implies that . However, we see that is not reflexive, because the relation only holds when . By the definition of reflexivity ( ), for a relation to be reflexive, it must hold for all elements of , not just those containing . Problem 3: Proof practice Part (a). Sipser, Problem 0.10: Find the error in the following proof that . Consider the equation . Multiply both sides by to obtain . Subtract from . Now factor each side, , and both sides to get divide each side by , to get . Finally, let and , which shows that . Part (b). Prove that there exists a natural number such that, for every natural number , there exist natural numbers such that .
z y F2Ff(ofw U z y { z y x x 9 4A W fg9 S D S & 6 S g A Q 8 eVI9 x s p 9 Q 8 cVP9 A u9 S 1 W @A 6 7 S & 8 g h A A fe9 9 S d 6 7 Q I8 Q P8 8 TIP8 t Q A u9 9 f1 S Q 8 Part (c). Let function be defined recursively as follows: @n~ x Solution 3: TFa x R a R 1. (a) We cannot divide by , since .
p Base case: . Here we see that . Inductive Hypothesis: Assume that for a natural number there exist natural numbers such that . Inductive Step: Given the inductive hypothesis, we wish to show that there exist integers such that . Notice that and . Thus, to "add one" to , we simply "add 7" and "subtract 6" as follows: . Thus, we see that integers and do, in fact, exist.
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j x v hx R TFd x k 4p v 4u v k u Tx a v 5IqT(p k R u x v @Fx k d u k a v k q j R k p u @x p a g ctR v k q x (p g cd k a '@ x u j a v qu k R x q Fp 2. (b) By inspection, one can deduce that statement holds for any .
s p . We prove by induction on that the p ~ g c`p ~ `p j j w p }~ p vb}e`p ~ g ~ g Prove that for every choose "), which is defined as
~ , is equal to the binomial coefficient (" p GT ~ s p for
g q j H a a j R a ix x oa a R a 5VR k j l a R Yix R H p Tn~ x a v k R u VqT(p x j x nR ~ eg q rp R j p w k a TmVR x a k ~ eg a g ctR j p g g H a j w j a R ix x H a @dR x a ~ g e`p a j j !H R w g p G2nc}c`}h ~ g ~ g p R q p $gp s x C ~ j p j j p k dp k &p }~ x ~ j x k ~ j k 1~ j p j ~ }~ p k ~ j p p }~ x ~ eg k rp w `p k D~ j ~ `p p ~ k k ~ k 1~ }~ x `p k 1~ j p `p p [email protected] ~ w 5k ~ g e`p c p TT& ~ 3. (c) Proof by Induction on . Inductive Step: We want to show that for any . We know by the definition of that For we apply the inductive hypothesis on the right handside: ~ g e`p j 2d ~ C wix z z ~ cg k np ~ eg w w C x k np w c p Base case: Let . Then by the definition. Inductive Hypothesis: Assume that for some , it is true that for any . q C C x x g w Tx cg w x p p Finally, for we have that C z C z z C z C x x ~ cg k Vp w now conclude that for any k g2u p ~ c C z z x 14
~ eg k dp w . We can , . k p VTn~ x k dp x ~ g e`p w ...
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This note was uploaded on 08/12/2009 for the course CS 420 taught by Professor Nancylynch during the Spring '07 term at New Mexico Junior College.
 Spring '07
 NancyLynch

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