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Unformatted text preview: Math 55: Sample Final Exam, 9 December 2009 3: (a) B is countable: in fact, a 11 correspondence between B and the natural numbers is provided by the function f ( b b 1 b 2 ...b n ) = b + 2 1 b 1 + ... + 2 n b n . In other words, B is the set of binary representations of the natural numbers. (b) F is uncountable, by the diagonal process. Suppose F were countable or finite, and arrange the elements of F into a sequence f ,f 1 ,f 2 ,... . Define g : N → { , 1 } by g ( n ) = 1 − f n ( n ). Then g ( n ) negationslash = f n ( n ) for any n , so g negationslash∈ F . This contradicts the assumption that we have listed all the elements of F , so F cannot be countable or finite. (c) Since a program is a finite bit string, the number of C programs is countable by (a). But functions are uncountable by (b), so C programs cannot evaluate all of them. 4: (a) Subtracting the number of edges of C n from the number of edges of the complete graph K n gives  ¯ E  = n ( n − 1) / 2 − n = n ( n − 3) / 2. (b) Since n ( n − 3) / 2 negationslash = n for n negationslash = 5, C n and ¯ C n have different numbers of edges and hence cannot be isomorphic. For n = 5, they are isomorphic: the map 1 → 1, 2 → 4, 3 → 2, 4 → 5, 5 → 3 preserves edges. (c) By symmetry and the handshaking theorem, all vertices of ¯ C n have the same degree deg( v ) = 2 · n ( n − 3) / 2 n = ( n − 3), which is even iff n is odd. By Euler’s theorem, ¯ C n has an Euler circuit iff n is odd. 5: (a) A partial order is a relation which is reflexive, antisymmetric and transitive. R is a partial order because for any a , b and c ∈ Z n , a  a , a  b ∧ b  a → a = b and a  b ∧ b  c → a  c . (b) The matrix M has M ij = 1 whenever i  j , so summing the entries in column j counts exactly d ( j ) divisors i of j ∈ Z n ....
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 Fall '08
 STRAIN
 Math, Natural Numbers, Equivalence relation, partial order, topologically sorted order, ﬁnite bit string

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