= Vertex-count bounds for bounded scalar valence
{title2=$\tfrac12V\le L-1+\tfrac12n\le2V$}
For a connected <Feynman diagram> with external valence $n$ and cubic, quartic and sextic vertices, counting half-edges gives $2I+n=3V_3+4V_4+6V_6$. Using $L=I-(V_3+V_4+V_6)+1$ yields $L-1+n/2=V_3/2+V_4+2V_6$. Since each summand lies between half and twice its vertex count, the displayed bounds follow. More generally vertices of valence between $r_{\min}$ and $r_{\max}$ give $(r_{\min}/2-1)V\le L-1+n/2\le(r_{\max}/2-1)V$.
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