Recent content by lemonthree

Let me edit that! I feel like I could be using better values instead. I changed the $(k-1)$ to $k$ instead, in turn changing $diam(G)$ to $k$ as I think it would make things easier. Let me attempt to find the maximum value of $apd(G)$, please correct me if I'm wrong. We want the sum of all...

Hi, I have attempted and I am stuck at this part. Fix $ k$. Consider G, with a path $v_{1},v_{2},...v_{k-1}$. Attach each node $u_{1},u_{2},...u_{n-(k-1)} $to $v_{1}$. This means we have attached $(n-k+1) $nodes to $v_{1}$. $diam(G) = d(u_{1},v_{k-1}) = k-1$. So the graph looks like this...

Ah, that makes sense. Any 2 vertices will definitely have a path connecting them, just that they do not have to be "directly" connected to each other?

I need to prove the above statement. I have a very strong gut feeling that the above equation is not true, and so I need to find a case where the graph diameter is greater than the average pairwise distance. First off, I would like to clarify about the average pairwise distance, which is given...

Alright, noted, thank you very much for the comments! I never thought that $x<y=u<v$ could be seen in that way too :)

Thank you! I understand it now. Since $K_{1}$ is an isolated vertex, there will be no edges anyway. And so we just look at $K_{n-k+1}$ which leads us to C(n-k+1,2).

Thank you for the hint. I think it's cool that you mentioned about SameGame too, I played that game when I was younger and never really thought that you could link it to this topic. In any case, I followed as per your comment and got this: Let $n_{q} \ge n_{p} \gt 1$ $K_{n_{q}} + K_{n_{p}} \lt...

Hello! I am having trouble solving the right part of the inequality. For left part of the inequality $n-k \le m$, here’s how I did it Let $ n = v_{1} + v_{2} + v_{3}...+v_{k}$, the sum of vertices of each component in G least number of edges = $(v_{1}-1) + (v_{2}-1) + (v_{3}-1)...+(v_{k}-1)$...

$\left | h + cos(h) \right | \leq C \left | h \right | $can be written as $\left |h + (1-\frac{h^2}{2!}...) \right | \leq C \left | h \right |$ $\frac{\left |h + (1-\frac{h^2}{2!}...) \right |}{ \left | h \right | } \leq C$ The $\frac{1}{\left | h \right |}$ part in the left hand side...

$\left | h + cos(h) \right | \leq C \left | h \right |$ And by taylor series, I know that $ h + cos(h) = h + (1-\frac{h^2}{2!}...) $ So solving for the equation, as h tends to 0, lhs tends to infinity while rhs is a constant. This is a contradiction, so h + cos h is not O(h)

Let $$ \mid h \mid $$< 1. Which of the following functions are O(h)? Explain. $$ -4h $$ $$h+h^2 $$ $$ \mid h \mid ^{0.5} $$ $$h + cos (h) $$ Based on my notes, f(h) = O(h) only if $$ \mid f \mid $$ ≤ C $$ \mid h \mid $$, where C is a constant independent of h. I can only solve for the...

Let S on X be {(a,b), (a,c), (a,a), (b,a), (b,b), (c,a), (c,c)}. S is reflexive and symmetric, and S∘S= {(a,a), (a,b), (a,c), (b,b), (b,a), (b,c), (c,c), (c,a), (c,b)} So S has 7 elements while S∘S has 9 elements, so S is reflexive and symmetric is not true.

I selected the 2 options S is reflexive and symmetric S is reflexive and transitive and I'm wrong...I don't understand why o_O

Thank you so much @Country Boy :D I finally understood it! I could see it better with a diagram, so that's what I did for testing symmetry and transitivity. Just as in the reflexivity case, S∘S is not S. So that means the first 3 options are out. Again, looking at the diagram, once S is...

Which of the following are true? Select all options. Assume that f:A→B and g:B→C. If f and g are injective, then so is g∘f If f and g are surjective, then so is g∘f If f and g are bijective, then so is g∘f If g∘f is bijective, then so are both f and g If g∘f is...