By Ian Anderson

Discrete arithmetic has now verified its position in such a lot undergraduate arithmetic classes. This textbook offers a concise, readable and obtainable advent to a few themes during this zone, resembling enumeration, graph concept, Latin squares and designs. it truly is aimed toward second-year undergraduate arithmetic scholars, and gives them with a few of the simple thoughts, rules and effects. It comprises many labored examples, and every bankruptcy ends with various workouts, with tricks or suggestions supplied for many of them. in addition to together with typical subject matters reminiscent of binomial coefficients, recurrence, the inclusion-exclusion precept, bushes, Hamiltonian and Eulerian graphs, Latin squares and finite projective planes, the textual content additionally comprises fabric at the ménage challenge, magic squares, Catalan and Stirling numbers, and event schedules.

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**Extra resources for A First Course in Discrete Mathematics**

**Sample text**

5) 24 Discrete Math ematic s Th en t he asserti on th at an = K 1an + K 2 /3 n is certainly true for n = 1, 2. We now pr oceed by ind uction . Assum e th e assertio n is t ru e for all n ::; k . Th en ak+1 = Aa k + Bak _ 1 = A(K1a k + K 2 /3 k) + B (K1a k- 1 + K 2 /3k-l ) = K 1a k- 1 (A a + B ) + K 2 /3 k-I (A/3 + B ) so the result follows. Th en th e assertion that an = (K 3 + n K 4)a n is certainly true for n Assume it is true for all n ::; k. Then = 1, 2. ak+1 = Aa k + Bak _1 = A(K 3 + kK4 )a k + B(K3 + (k - I )K4 )a k- 1 = K 3a k- I(A a + B ) + K 4 a k- I( A ka + B (k - 1)) = K3a k+J + K 4a k- 1(2k - a 2 (k - 1)) =K 3a k + 1 + K 4(k + l )a k+ 1 , as required.

Note also that Po = 1 37 2. Recurrence The num bers Pn are t he Catalan numbers, usually denoted by C«. 14) The sequence (Cn)n~O begins 1,1 ,2 , 5,14 ,42,139 ,4 29, .... 15) As remarked earlier, th e Catalan numbers a ppea r in many situations. One immediate interpret ation, obtained by replacing Rand U by 0 and 1 resp ectively, is: C n = number of binary sequences of length 2n containing exactly n Os and n Is , such th at at each stage in th e sequence t he number of I s up to that point never exceeds th e number of Os.

T{ + tk + e+ k - 1 = te + tk + n - 1. 12) 1). 2, first solve t he homogeneous recurrence am = 2a m - l . The solution is clearly an A2 " for some constant A . 12). ) We then requi re Bn2 n + C = 2B (n - 1)2n - 1 + 2C + 2n - 1 Le. 2" = = 1 to obtain So t ake B C A = - 1,giving + 2n finally an - an = 2fl(n - 1) Thus t2~ = 1 + 2m (m - 1). On putting n 1+ C . 2n + n 2n + 1. Bu t al l , so + 1. = 2m , we get t« = 1 + n(log2 n - 1), so th e merg esort method has com plexity O(n log n) , an improvement on the O(n 2 ) of bubblesort.