Homework

An important aspect of the course is the clear presentation of solutions, which involves careful thinking. Producing a solution involves two main steps: finding a proof and properly writing it. Even when a problem asks for a numerical value (instead of a proof), you still have to justify your answer.

 

You may work together on the homework, but each of you should turn in your own version, written in your own words. You may not copy someone else's paper, or allow another student to copy yours.

 

Homework 1: Due 1/21, Friday

1.1: 16, 19; 1.2: 2, 4; 1.4: 12, 23; 1.5: 2, 7ce, 22

 

Homework 2: Due 2/4, Friday

2.1: 38; 2.2: 7, 22; 2.3: 21; 2.4: 7; 3.1: 14, 25

Two additional problems (required):

1.      Prove or disprove that every Eulerian bipartite graph has an even number of edges.

2.      Find the number of trees on n labeled vertices (as a function of n). You do not need to simply your answer.

 

Homework 3: Due 2/21, Monday

  1. Prove that every tree T has a unique bipartition V(T) = A + B. In addition, if |A| ≥ |B|, then A has at least |A|- |B| + 1 leaves.
  2. Prove that following algorithm outputs a minimal spanning tree in G.

Consider the edges of G in the order of decreasing weights. Remove an edge if it is not a cut-edge. Output the remaining edges.

5.1: 4, 17; 5.2: 8; 6.1: 26; 6.2: 6

 

Homework 4: Due 3/11, Friday

  1. Determine all r,s such that Kr,s is planar. Explain why.
  2. Determine (with proofs) all n such that cube Qn is planar. Explain why.

10.1: 6, 19; 10.2: 7; 10.3: 18, 28

 

Homework 5: Due 4/1, Monday

  1. In class we showed that every tournament has a king. Let T be a tournament containing no vertex of indegree 0. (a) Prove that if x is a king in T, then T has another king from the vertices that beat x. (b) Use (a) to prove that T has at least three kings.

11.1: 18; 11.3: 8, 12.4: 9, 17, 27

 

Homework 6: Due 4/25, Monday

12.1: 7; 12.2: 3, 12.3: 10 (I replaced #14 by #10, which is easier) and 21.

Also: use Kuratowski’s Theorem to prove that Petersen graph is not planar.