# An independent set in a graph is a set of mutually non-adjacent vertices in the graph

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Why was Roosevelt unhappy with the Treaty of Portsmouth? The Treaty of Portsmouth: After lengthy negotiations that started on the sixth day of August and concluded on the 30th day of the same month, the Treaty of Portsmouth marked the beginning of peace between Japan and Russia for a long time. The war lasted for nearly a year, from 1904 and concluded in 1905.
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How long did the Treaty of Portsmouth take to discuss? Teddy Roosevelt: Teddy Roosevelt, also known as Theodore Roosevelt Jr., was an accomplished American statesman, historian, and conservationist. He was also the twenty-sixth President of the United States and served in this capacity from 1901 to 1909.

a) Compute a0, bo, a1, b1 Answer; We know that the number of vertices in La is 3. So, by definition, a0=3. Now for a1: if an independent set does not include (n- 1, 0), then it cannot have (n, 0) as an endpoint. The only way to avoid having (n- 1, 0) is not to have any vertex of L. as an endpoint. So, a1 =3- 1.

For b1: we need to count the number of graphs on 3 vertices which have 1 vertex adjacent to (n, 0) and 2 vertices adjacent to (n- 1, 0). The order of the resulting graph is at most 3 because it has 3 vertices. And this graph must contain an edge between each vertex with (n, 0) and a vertex with (n-1,0) by definition. So this gives a sum of terms each on the order of 4(=l+l+?+l).

Thus b1 =4·3- 2·2 = 12. So total number of independent sets in Le is =12. Similarly, for all r, assume that the number of vertices in Lr is s. Then a0+a1+b1+c1 =s-2=s+1. b0+b1=s-2 +s = s- 1 + 1 = 3·s – 9 =4·(4294967296) So for b0: if an independent set does not include (n, 0) then it cannot have (n, 0) as an endpoint. So we need to count the number of graphs on 3 vertices which have 2 vertices adjacent to (n, 0). So we need to count the number of graphs on 3 vertices which have 1 vertex adjacent to (n, 0). The order of the resulting graph is at most 3 because it has 3 vertices. And this graph must contain an edge between each vertex with (n, 0) and a vertex with (n-1,0) by definition. So this gives a sum of terms each on the order of 4(=l+l+?+l). Thus b0=4·3- 2·1 = 11. So total number of independent sets in Lr is =11.

(b) Give, with justification, recurrence relations that express a nplus b nplus1 , and bat, in terms of an and bn (Here, each of an+1 and but is expressed using both a,, and b,.) Answer; a) We know that a0+a1+b0+b1 =s-2=s+1. b0+b1 =s-2 +s = s- 1 + 1 =3·s -9=4·(4294967296). So b0=4·3- 2·1=11. So for an: if an independent set does not include (n, 0), then it cannot have (n, 0) as an endpoint. So this gives a sum of terms each on the order of 4(=l+l+?). Thus b0=4·3- 2.

(c) Prove, by induction on n, that for all n 2 0, On S v2(v2+ 1)” and S (V2+1)”. We just need to count the number of independent sets in each case. So we need to count the number of graphs on 3 vertices which have 2 vertices adjacent to (n+ 1, 0) and a vertex with (n,0). The order of the resulting graph is at most 3 because it has 3 vertices. And this graph must contain an edge between each vertex with (n+ 1,0) and a vertex with (n,0). So this gives a sum of terms each on the order of 4(=l+l+?). Thus v2+v2+ 2 =4·3 +4·3 +4·3 = 102. So for all n: if an independent set does not include (n, 0), then it cannot have (n, 0) as an endpoint. So this gives a sum of terms each on the order of 4(=l+l+?). Now b0=4·3- 2. So for all n 2 0: if an independent set does not include (n, 0), then it cannot have (n, 0) as an endpoint. So this gives a sum of terms each on the order of 4(=l+l+?).

Thus b0=4·3- 2. SUMMARY: An lndependent set in a graph is a set of mutually non-adjacent vertices in the graph. So, no edge can have both its endpoints in an independent set. In this problem, we will count independent sets in ladder graphs. A ladder graph L is a graph obtained from the points and lines formed by a row of n squares. More formally: . its vertices correspond to ordered pairs (0, 0), (0, 1), (1, 0), (1, 1), … (d) Hence give a good closed-form upper bound for c. (i.e., an upper-bound expression in terms of n, not a recurrence relation). Your bound should be significantly better than 3″ (which is the upper bound you get by ignoring horizontal edges and only taking vertical edges into account). Answer; ( c = \frac{9}{(n+1)^{2}}. )

Notation: a typical counterexample of independent sets in total could be (0,0), (1,1) and (3,3). We can have the counterexample either as n=2 or n=3. In the first case we have c=6, in the latter it is 9. If we interchange endpoints of the second vertex (1, 1) and endpoints of the last vertex (0, 3), we can find different counts: 6 or 9 depending on which set we want. This means that all the three sets are adjacent, so there is no independence at all. M= 3·2n is the total number of spanning trees of the graph. As we saw above, M·3·2(n+1)- 5·3(n+1)- 5·3(n-1) = 12M and therefore formula_66 is just smaller by a factor of formula_67 So we may conclude iN.B.: Now you can show that there are more independent sets in ladder graphs with even orders than odd orders (see section on independence number).