By Nagpal, Radhika; Meyer, Albert R

**Read Online or Download 6.042J / 18.062J Mathematics for Computer Science (SMA 5512), Fall 2002 PDF**

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**Extra resources for 6.042J / 18.062J Mathematics for Computer Science (SMA 5512), Fall 2002**

**Sample text**

There is a party. Some people shake hands an even number of times and some shake an odd number of times. Show that an even number of people shake hands an odd number of times. We can represent the party by a graph. ) Each person is represented by a vertex. If two people shake hands, then there is an edge between the corresponding vertices. 2. In every graph, there are an even number of vertices of odd degree. Proof. Partitioning the vertices into those of even degree and those of odd degree, we know � � � d(v) = d(v) + d(v) v∈V d(v) is odd d(v) is even The value of the lefthand side of this equation is even, and the second summand on the righthand side is even since it is entirely a sum of even values.

Try t = 3. Has a chain of length 4. Try t = 4. Has no chain of length 5, but has an antichain of size 4 ≥ 10/4. Posets arise in all sorts of contexts, and when they do, Dilworth’s theorem can have interesting implications. 28. In any sequence of n different numbers, there is either an increasing subsequence of length √ √ greater than n or a decreasing subsequence of length at least n. 29. �6, 4, 7, 9, 1, 2, 5, 3, 8� has the decreasing sequence �6, 4, 1� and the increasing sequence �1, 2, 3, 8�.

The preceding algorithm colors a graph with at most p + 1 colors, where p is the maximum degree of any vertex. This theorem implies, for example, that a graph with thousands of vertices, each of degree 3, requires at most 4 colors. The proof is surprisingly easy: Proof. The proof is by induction. Let P (n) be the predicate that the preceding algorithm colors every n-vertex graph in which every vertex has degree at most p using at most p + 1 colors. In the base case, P (1), there is a single vertex with degree zero.