Guichard-Combinatorics-Graph-Theory · Problem 5.1.13

Graph Theory

Mathematics — Probability · graph-theory-basics

Problem

Prove the ``if'' part of theorem 5.1.3, as follows: The proof is by induction on s=∑i=1ndis = \sum_{i=1}^n d_i. This is easy to see if s=2s = 2, so suppose s>2s > 2. Without loss of generality we may suppose that dn>0d_n > 0. Let tt be the least integer such that dt>dt+1d_t > d_{t+1}, or t=n−1t = n - 1 if there is no such integer. Let dt′=dt−1d'_t = d_t - 1, dn′=dn−1d'_n = d_n - 1, and di′=did'_i = d_i for all other ii. Note that d1′≥d2′≥⋯dn′d'_1 \geq d'_2 \geq \cdots d'_n. We want to show that the sequence {di′}\{d'_i\} satisfies the condition of the theorem, that is, that for all k∈{1,2,…,n}k \in \{1, 2, \ldots, n\}, ∑i=1kdi′≤k(k−1)+∑i=k+1nmin⁡(di′,k).\sum_{i=1}^k d'_i \leq k(k-1) + \sum_{i=k+1}^n \min(d'_i, k). There are five cases: • k≥tk \geq t • k<tk < t, dk<kd_k < k • k<tk < t, dk=kd_k = k • k<tk < t, dn>kd_n > k • k<tk < t, dk>kd_k > k, dn≤kd_n \leq k By the induction hypothesis, there is a simple graph with degree sequence {di′}\{d'_i\}. Finally, show that there is a graph with degree sequence {di}\{d_i\}.

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