MIT6_042JF10_rec16_sol

MIT6_042JF10_rec16_sol - 6.042/18.062J Mathematics for...

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Unformatted text preview: 6.042/18.062J Mathematics for Computer Science November 5, 2010 Tom Leighton and Marten van Dijk Notes for Recitation 16 1 Combinatorial Proof A combinatorial proof is an argument that establishes an algebraic fact by relying on counting principles. Many such proofs follow the same basic outline: 1. Define a set S . 2. Show that | S | = n by counting one way. 3. Show that | S | = m by counting another way. 4. Conclude that n = m . Consider the following theorem: Theorem. n k + i k + n + 1 = k k + 1 i =0 We can prove it with a combinatorial approach: Proof. We give a combinatorial proof. Let S be the set of all binary sequences with exactly n zeroes and k + 1 ones. On the one hand, we know from a previous recitation that the number of such sequences is equal to k + k n +1 . On the other hand, the number of zeroes i to the left of the rightmost one ranges from to n . For a fixed value of i , there are k + i possible choices for the sequence of bits before k n k + i the rightmost one. If we sum over all possible i , we find that | S | = i =0 k . Equating these two expressions for | S | proves the theorem. | | Recitation 16 2 2 Triangles Let T = { X 1 ,...,X t } be a set whose elements X i are themselves sets such that each X i has size 3 and is { 1 , 2 ,...,n } . We call the elements of T triangles. Suppose that for all edges E { 1 , 2 ,...,n } with | E | = 2 there are exactly triangles X T with E X . For example, if we might have the triangles depicted in the following diagram, which has = 2, n = 4, and t = 4: 1 2 3 4 In this example, each edge appears in exactly two of the following triangles: { 1 , 2 , 3 } , { 1 , 2 , 4 } , { 1 , 3 , 4 } , { 2 , 3 , 4 } Prove n ( n 1) = 3 t 2 by counting the set C = { ( E,X ) : X T,E X, | E | = 2 } in two different ways. Solution. We give a combinatorial proof. Let C be { ( E,X ) : X T,E X, E = 2 } . On the one hand, there are n sets E { 1 ,...,n } of size E = 2. For each such E there 2 n | | n ( n 1) are exactly triangles X T with E X . So, | C | = = ....
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MIT6_042JF10_rec16_sol - 6.042/18.062J Mathematics for...

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