GPT 6 Astra Ultra attempt on Math-5446
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FORMAL RESTATEMENT
Erdős #127; reconstruction, 2026-09-23. Let $b(m)$ be the minimum, over graphs with $m$ edges, of the size of a maximum cut. Put \[ f(m)=b(m)-\left(\frac m2+\frac{\sqrt{8m+1}-1}{8}\right). \] Is $f(m)$ unbounded along some sequence? We reconstruct Alon's proof that $f(n^2/2)\gg\sqrt n$ for large even $n$.QUICK LITERATURE AND CONTEXT CHECK
The current page credits Alon's 1996 paper. We follow its chromatic number and large-clique dichotomy. We explicitly use the previously proved Edwards inequality $b(m)\ge m/2+(\sqrt{8m+1}-1)/8$. The proof below is a reconstruction of the known argument, not a new improvement. Integer parity in the complete-graph example is recorded separately.WORK
If a graph is $2s$-colorable, randomly divide its $2s$ independent color classes into two groups of size $s$. Each edge crosses with probability $s/(2s-1)$, so some cut has size at least \[ m/2+m/(4s-2). \tag{1} \] Fix $\epsilon=10^{-3}$, take large even $n$, and let $m=n^2/2$. If the chromatic number is below $n-\epsilon\sqrt n$, round it up to an even integer $2s\le n-\epsilon\sqrt n+1$. Formula $1$ gives \[ \operatorname{maxcut}(G)\ge m/2+ \frac{n^2}{4(n-\epsilon\sqrt n)} \ge m/2+n/4+(\epsilon/4)\sqrt n. \tag{2} \]Otherwise write $\chi(G)=n-k$, where $0\le k\le\epsilon\sqrt n$. The upper bound $\chi(G)\le n$ follows because a minimum proper coloring has an edge between every two color classes, while $m<\binom{n+1}{2}$. Take a vertex-critical subgraph $H$ with the same chromatic number. Every vertex has degree at least $n-k-1$; hence \[ |V(H)|\le n^2/(n-k-1)\le n+2\epsilon\sqrt n \] for large $n$. In a proper coloring of $H$ with $n-k$ colors there are at least $2(n-k)-|V(H)|\ge n-4\epsilon\sqrt n$ singleton classes. Every two are joined, again because merging nonadjacent color classes would reduce the chromatic number. Thus $G$ contains a clique $U$ of size $h=n-r$, with $0\le r\le4\epsilon\sqrt n$. Let $W=V(G)\setminus U$, $w=e(W)$, and $t=e(U,W)$. Direct subtraction gives \[ w+t=rh+n/2+(r^2-r)/2. \tag{3} \]
If $w\ge n/32$, Edwards' inequality gives a cut of $G[W]$ with at least $w/2+\sqrt{w/8}-1/8\ge w/2+\sqrt n/16-1/8$ edges. A balanced cut of the clique has at least $e(U)/2+(h-1)/4$ edges. Choose the better of the two relative orientations of these cuts; this retains at least $t/2$ crossing edges between $U$ and $W$. Altogether the cut has size at least \[ m/2+n/4+(1/16-\epsilon)\sqrt n-O(1). \tag{4} \]
Suppose instead $w<n/32$. In $3$, the remainder $R=n/2+(r^2-r)/2-w$ lies between $h/4$ and $3h/4$ for large $n$; this follows from $r\le4\epsilon\sqrt n$ and the chosen small constant. In particular $t/h=r+R/h$ is at distance at least $1/4$ from an integer. Write the degrees from $U$ into $W$ in increasing order $d_1\le\cdots\le d_h$, with mean $\bar d=t/h$. If $d_{\lfloor h/2\rfloor}\ge\bar d$, every degree in the upper half is at least $\bar d+1/4$. Otherwise all degrees in the lower half are at most $\bar d-1/4$. In either case the upper $\lceil h/2\rceil$ vertices have at least \[ t/2+\tfrac14\lfloor h/2\rfloor\ge t/2+n/12 \] neighbors in $W$, counted with multiplicity, for large $n$. Put all of $W$ on the other side from that upper half. The clique is cut as evenly as possible, and the resulting cut has at least \[ e(U)/2+(h-1)/4+t/2+n/12 =m/2+n/4+n/12-w/2-r/4-O(1). \] Since $w<n/32$ and $r=O(\sqrt n)$, this is larger than $m/2+n/4+c\sqrt n$ for a fixed $c>0$ and all large $n$. Together with $2$ and $4$, this proves \[ b(n^2/2)\ge n^2/4+n/4+c\sqrt n. \] The Edwards correction at $m=n^2/2$ is $(\sqrt{4n^2+1}-1)/8=n/4-1/8+O(1/n)$. Therefore $f(n^2/2)\ge c\sqrt n+O(1)\to\infty$ as required. $\square$