Let \(G\) be a graph. The cardinality of any largest independent set of vertices in \(G\) is called the independence number of \(G\) and is denoted by \(\alpha(G)\). Let \(a\) and \(b\) be integers with \(0 \leq a \leq b\). If \(a = b\), it is assumed that \(G\) be a connected graph, furthermore, \(a \geq \alpha(G)\), \(a/|V(G)| = 0 \pmod{2}\) if \(a\) is odd. We prove that every graph \(G\) has an \([a, b]\)-factor if its minimum degree is at least \((\frac{b+\alpha(G)a-\alpha(G)}{b})\lfloor \frac{b+\alpha(G)a}{2\alpha(G)} \rfloor -\frac{\alpha(G)}{b}(\lfloor \frac{b+\alpha(G)a}{2\alpha(G)}\rfloor )^2+ \theta\frac{\alpha(G)^2}{b}+\frac{a}{b}\alpha(G)\), where \(\theta = 0\) if \(a < b\), and \(\theta = 1\) if \(a = b\). This degree condition is sharp.
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