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[组合] 自避行走 步数期望

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hbghlyj Posted 2023-5-27 18:34 |Read mode
Last edited by hbghlyj 2023-5-28 14:16$(0,0)\cdots(n,n)$方形网格从左下角沿网格走(顶点不能重复), 走每个相邻顶点的概率相同, 直到无法继续走(所有相邻顶点都被走过)时结束
结束时走过的步数为$f(n)$, 求$f(n)$的期望?

当$n=1$时

总可以走3步,$f(1)=3$.

当$n=2$时目前只知道$5<f(2)<8$
可能走8步



可能走7步

可能走6步

可能走5步

可能走4步

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 Author| hbghlyj Posted 2023-5-28 04:08
Last edited by hbghlyj 2023-5-28 14:18$f(2)=$?

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 Author| hbghlyj Posted 2023-5-28 21:11

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