Solutions to the Final Exam

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1 CS/Math 24: Intro to Discrete Math 5//2 Instructor: Dieter van Mekebeek Soutions to the Fina Exam Probem Let D be the set of a peope. From the definition of R we see that (x, y) R if and ony if x is a bioogica chid of y. Then (x, z) R R if and ony if there is some y such that x is the bioogica chid of y (i.e., if (x, y) R ), and y is the bioogica mother of z. In that case x and z have the same mother, so they are sibings. Since everybody has a bioogica mother, every x D is actuay reated to itsef, too. Just et y be the bioogica mother of x and note that (x, y) R as we as (y, x) R. The reation is symmetric. In fact, any reation of the form R R is symmetric (see part (b)). Finay, the reation is transitive. Any pair of sibings has the same bioogica mother, so if x and z are sibings, and z and w are aso sibings, it foows that x and w are sibings too. More formay, say (x, z), (z, w) R R. Since the reation R is injective and our hypothesis impies that x and z have the same mother, as we as z and w have the same mother, we have (y, x), (y, z), (y, w) R. Then (x, y) R and (y, w) R, and we see that (x, w) R R. The equivaence casses are the sets of peope who have the same mother. That is, we have a cass C y = {x (y, x) R} for every mother y D. The reation R R is not aways refexive. Consider the empty reation, and pick any x D. Since R is empty, there is no y D such that (x, y) R, which means that x cannot be reated to itsef. The reation R R is aways symmetric. Suppose (x, z) R R. That means that there is some y D such that (x, y) R and (y, z) R. But then (z, y) R and (y, x) R, so (z, x) R R. The reation R R is not necessariy transitive. For exampe, suppose we have D = {a, b, c, d, e} and R = {(b, a), (b, c), (d, c), (d, e)}. Then note (a, b) R, so (a, c) R R. Aso, (c, d) R, so (c, e) R R. But (a, e) are not reated by R R because in R, a is ony reated to b, and (b, e) / R. Probem 2 Let v be a vertex of odd degree and consider the connected component C = (V, E) that it beongs to. We know that for any graph, v V deg(v) = 2 E, and we derived as a coroary that every graph has an even number of vertices of odd degree. Since v is a vertex of odd degree in C, there are at east 2 vertices of odd degree in C. Let u v be another vertex of odd degree in C. Since C is connected, there is a path from v to u.

2 Probem 3 The first probem appears in the appication of Euer s formua. This ony hods when the graph is connected, but the graph for this probem is not connected. Second, it is not true that a simpe cyces in the graph have ength 4. Going around the boundary of the 4-vertex subgraph that ooks ike K 4 minus one edge yieds a simpe cyce of ength 4. Thus, the caim that every face has 3 edges on its border is incorrect. There is another reason why the atter caim is incorrect, as the fact that the border of a face forms a simpe cyce does not hod for the outer face of a disconnected graph. Indeed, in this case the outer face of the graph has edges on its border (a edges except for the diagona one in the four-vertex subgraph), whereas the argest simpe cyce has ength 4. Finay, the impication V 3 E = 2 E > 3 V is incorrect. For exampe, if V = 3 and E = 3, this impication does not hod. A graph with V = E = 3 is for exampe the graph K 3. Probem 4 For the graph on 4 vertices without edges, we can pick any coor for each vertex. Since there are k coors, the number of coorings is k 4. We have k choices for the coor of vertex. Given that coor, we have k choices for the coor of vertex 2, then k 2 choices for the coor of vertex 3, and finay k 3 choices for the coor of vertex 4. The generaized product rue appies and tes us that the tota number of coorings is k(k )(k 2)(k 3) = k!/(k 4)!. Part c Since vertex 5 is connected to a vertices, it is quite convenient to pick its coor first. We have k options for this coor. After the vertex 5 is coored, we can treat the subgraph with vertices, 4, 5 as K 3, where one of its vertices, 5, is aready coored. Now we have k options for the coor of vertex, and since that vertex is connected to both and 5, we get k 2 options for the coor of 4. Simiary, we get k options for the coor of 2 and k 2 options for the coor of 3. It foows by the generaized product rue that we have k(k ) 2 (k 2) 2 coorings. Part d First et s focus on the subgraph with vertices, 2, 4. This graph is K 3, so there are k(k )(k 2) ways to coor its vertices. Now vertex 3 must have a coor that differs from coors of 2 and 4. Those two coors are different, so we have k 2 choices for the coor of 3. Hence, the tota number of coorings is k(k )(k 2) 2. 2

3 Probem 5 We construct an NFA for this anguage using two NFAs for simper anguages: (i) the anguage of strings that have exacty two ones, and (ii) the anguage of strings that have an even number of zeros. To construct an NFA that accepts strings containing exacty two ones, we with the DFA in Figure a. The states s, s and s 2 indicate that, and 2 ones have been read so far, respectivey. After reading a third one, we go to the garbage state s 3 and stay there for the rest of the computation. Now the key observation here is that since we are constructing an NFA, we can omit a transition from s 2 to the garbage state. If our NFA reads a third, it has nowhere to go from s 2, and, therefore, rejects right away, which is correct behavior. We constructed an automaton accepting strings that have an odd number of ones in ecture, and the construction of an automaton that accepts strings with an even number of zeros ooks amost the same. We show it in Figure b. Finay, we add an initia state s init. From there, we have transitions into the NFAs described in the previous two paragraphs. In particuar, we copy the transitions from states s and s e so that, s s s 2 s 3 (a) DFA M for strings with exacty two ones. s e s o (b) DFA M 2 for strings with an even number of zeros. s e s o s init s s s 2 s e s o s e s 2e s 3e s o s 2o s 3o (c) NFA for part (a) Figure : Automata for probem 5. (d) DFA for part (b) 3

4 the same transitions can be made from s init. We have seen an idea ike this in the constructions of NFAs for the concatenation of two anguages and for the Keene cosure of a anguage. The fina automaton is in Figure c. Since the NFA can correcty guess whether the string has two ones or an even number of zeros, it can correcty choose whether to go to a state of the modified machine M or to a state of M 2 after reading the first input symbo. On the other hand, if a string is not in the anguage, it does not matter whose machine s state our automaton goes to after reading the first input symbo because both M and M 2 are going to reject that string. Our anguage is reay a union of two anguages. We can use the construction from cass to design a DFA for this anguage. The resut is in Figure d. Part c We saw a reguar expression for a string containing an even number of occurrences of some symbo in ecture as we as on the homework. For an even number of occurrences of in a binary string, we use the reguar expression ( ). A string with exacty two ones may with some number of zeros, then contain a one, then have some more zeros, another one, and end with another sequence of zeros. One reguar expression for such a string is. To get a reguar expression for our anguage, we just take the union of the two reguar expressions we constructed earier, and obtain Probem 6 ( ). We are going to count the set S of binary strings of ength n with an even number of ones in two different ways. For the rest of this probem, we assume our strings are binary without saying so. We can pick out of n positions in a string of ength n for k {,..., n/2 }. Put ones in those positions, and put zeros everywhere ese. This process yieds a string of ength n with an even number of ones. We can pick positions for the ones in ) ways, and we can pick ones for k {,..., n/2 }. Since the set of strings with 2i ones and the set of strings with 2j ones are disjoint whenever i j, we can use the sum rue to count the number of strings with an even number of ones using n/2 ) k=. Now et T be the set of strings of ength n, and et p : {, } n {, } be the parity function. That is, p(x) = if x contains an odd number of ones, and is otherwise. Let f : T S be defined by f(x) = xp(x), i.e., x concatenated with the parity of x. This function is we-defined. If x has an odd number of ones, p(x) =, so the number of ones in xp(x) is one more than the number of ones in x, which is even. If x has an even number of ones, p(x) =, so concatenating x and p(x) produces a string with an even number of ones. Observe that f is tota since we defined it for every string in T. Aso, f is injective. Suppose f(x) = f(y). This means that the first n bits of f(x) and f(y) are the same, and those bits are x and y, respectivey, so x = y. Finay, f is surjective. Let z be a string of ength n with an even number of ones, and et x be the first n bits of z. If the ast bit of z is, x has an odd 4

5 number of ones, so p(x) = and xp(x) = z. If the ast bit of z is, x has an even number of ones, p(x) =, and xp(x) = z. Thus, f is a bijection from the set of strings of ength n to the set of strings of ength n with an even number of ones. Since there are 2 n strings of ength n, there are 2 n strings of ength n with an even number of ones by the bijection rue. Since both of our approaches counted the number of eements in the set S, we have 2 n = S = n/2 ) k=, which is what we wanted. The binomia theorem states that (x + y) n = = x y n. () Pugging in (x, y) = (, ) and (x, y) = (, ), respectivey, into () yieds the foowing two equations: Now we can add (2) and (3) to get = = = = ( n + ( ) n ) = ) n (2) ) ( ) n (3) k= ( + ( ) ) n. (4) k Since + ( ) is 2 if is even and if is odd, and since the argest term with an even vaue of in (4) is 2 n/2, we can rewrite (4) as n/2 k= 2, and divide by 2 to get 2 n = n/2 k=. 5