pset5soln - EECS 310, Fall 2011 Instructor: Nicole...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
EECS 310, Fall 2011 Instructor: Nicole Immorlica Problem Set #5 Due: October 27, 2011 1. (25 points) Let ( s 1 ,s 2 ,...,s n ) be an arbitrarily distributed sequence of the numbers 1 , 2 ,...,n - 1 ,n (i.e., a permutation). For instance, for n = 5, one arbitrary sequence could be (5 , 3 , 4 , 2 , 1). Define the graph G = ( V,E ) as follows: 1. V = { v 1 ,v 2 ,...,v n } 2. { v i ,v j } ∈ E if either: j = i + 1, for 1 i n - 1 i = s k , and j = s k +1 for 1 k n - 1 (a) Prove that this graph is 4-colorable for any ( s 1 ,s 2 ,...,s n ). 1 (b) Suppose ( s 1 ,s 2 ,...,s n ) = (1 ,a 1 , 3 ,a 2 , 5 ,a 3 ,... ) where a 1 ,a 2 .... is an arbitrary dis- tributed sequence of the even numbers in 1 ,...,n . Prove that the resulting graph is 2-colorable. Solution: (a) One important observation for this problem is that each of the above edge conditions form a line graph. For the first condition, states that if we label all the nodes 1. ..n, then we will add edges (1 , 2) , (2 , 3) , (3 , 4) ,... ( n - 1 ,n ). From previous assignments, we know this graph is bipartite and is thus 2-colorable. Denote the set of edges coming from this condition as E 1 . The second condition will also give edges that will look like a line graph but instead of having the standard ordering of 1 , 2 ,...n , we will have some different ordering defined by the permutation. For example, if the permutation we have gives us 3 , 5 , 2 , 1 , 4 , 6, then we will have edges (3 , 5) , (5 , 2) , (2 , 1) , (1 , 4) , (4 , 6). You can convince yourself that this is a line graph by drawing it out. Denote the set of edges coming from this condition as E 2 . Now let G 1 = ( V,E 1 ) be the graph where I have nodes 1 ...n but the edge set are only the edges that come from the first condition (so (1 , 2) , (2 , 3) ,... ( n - 1 ,n )). Let G 2 = ( V,E 2 ) have the same node set but only edges that come from the second edge condition. We will eventually form the original graph G = ( V,E 1 E 2 ) by taking the union of these two edge sets. Since both G 1 and G 2 are line graphs, then they are both 2-colorable. Color G 1 with colors black and white and color G 2 with colors red and blue. Now we note that each node i in G is colored either black or white in G 1 and colored either red or blue in G 2 . Then we can color G with 4 colors by using 1 Hint: Use the fact that a line graph is 2-colorable.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
pairs of colors as our colors. That is the 4 possible colors for graph G will be ( black,blue ) , ( black,red ) , ( white,blue ) , ( white,red ). So if a particular node i is colored white in G 1 and red in G 2 , the node i will have color ( white,red ) in G. Claim: In G , there does not exist any edge e = ( u,v ) such that nodes u and v have the same color in G . Proof: By contradiction. Assume there exists some edge
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

pset5soln - EECS 310, Fall 2011 Instructor: Nicole...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online