Pseudo Programmer

Definition of Full Binary Tree      A Binary Tree is full if and only if every node has 0 or 2 children. Proof by Induction Inductive Hypothesis The full binary tree with $L$ leaves has $N = 2L - 1$ total nodes. Base Case We can easily think of two base cases. Note that the shape of the tree at each base case is unique (Draw by yourself). If L = 1, then N = 1. If L = 2, then N = 3. Inductive Step We want to show that the full binary tree with $L + 1$ leaves has $2(L + 1) - 1$ total leaves $\forall n \gt 2, n \in \mathbb{N}$. Suppose that $\exists n \gt 2, n \in \mathbb{N} $ the full binary tree with $L$ leaves has $2L - 1$ total leaves. $T$ : the full binary tree with $L$ leaves. $v$ : a leaf node of $T$ If we want to append a new node in $T$, two nodes must be inserted on the left and right side of some leaf node of $T$. Which implies that every node insertion increases $N$ by $2$ and $L$ by $1$. $2L - 1 + 2 = 2(L + 1) - 1$ Hence, t...

$G$ : undirected simple finite graph, $\bar{G}$ : complementary graph of $G$. self-complementary : $G \cong \bar{G}$ Between complementary graph, $$|E_{G}| + |E_{ \bar{G}}| = \frac{|V_{G}|(|V_{G}| - 1)}{2}$$should be satisfied. If $G \cong \bar{G}$, then $|E_{G}| = |E_{\bar{G}}|$. Hence, with substitution, $$ 2|E_{G}| = \frac{|V_{G}|(|V_{G}| - 1)}{2} \\[10pt] 4|E_{G}| = |V_{G}|(|V_{G}| - 1) \\[10pt] $$ If $|V_{G}|$ is even integer, then $|V_{G}| - 1$ is odd integer. Converse is also trivially true. $$ \therefore 4 \mid |V_{G}| \ \text{or} \ 4 \mid |V_{G}| - 1 $$

$G$ : undirected simple finite graph, $\bar{G}$ : complementary graph of $G$. $C_{n} (n \geq 3)$ : n-cycle graph, which is the graph itself is a cycle. To be $G$ self-complementary, $G \cong \bar{G}$ should be satisfied. Hence, $G$ and $\bar{G}$ must have same number of edges. Which implies that $|E_{G}| = |E_{ \bar{G}}| = \frac{1} {2} \cdot \frac{|V_{G}|(|V_{G}| - 1)}{2}$ (Note that $|V_{G}| = |V_{\bar{G}}|$ by definition of complementary graph). From the fact that $C_{n}$ is cycle graph, $|E_{C_{n}}| = |V_{C_{n}}| = n$ and $|E_{C_{n}}| = |E_{\bar{C_{n}}}|$, $$ |E_{C_{n}}|  + |E_{\bar{C_{n}}}| = \frac{|E_{C_{n}}|(|E_{C_{n}}| - 1)}{2} \\[10pt] 2n = n(n-1) \\[10pt] n(n - 5) = 0 \\[10pt] \therefore n = 0, 5$$ Since that $n \geq 3$, the only answer is $5$.