 Kauffman bracket breadth bound

A state with $k$ B smoothings has at most $s_A+k$ circles, so its largest A exponent is at most $c(D)+2s_A-2$. The dual argument bounds the smallest exponent below by $-c(D)-2s_B+2$. For a connected link diagram, the Turaev surface inequality bounds the resulting breadth by $4c(D)$. For a connected reduced alternating diagram, adequacy prevents cancellation at both extremes and planar Euler characteristic gives equality. A disconnected projection needs the additional term in the Jones polynomial breadth bound for a disconnected diagram.