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发表于 2014-4-14 10:39:01
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本帖最后由 sunwukong 于 2014-4-14 10:40 编辑
(参考“屠龙刀的概率升级问题”http://bbs.emath.ac.cn/forum.php?mod=viewthread&tid=3911)
设\(k\)月时年龄为\(i\)月的兔子有 \(x_{k,i}\) 个 (\(1<=i<=c-1\))
记\(x_{k,c}=x_{k,1}+x_{k,2}+x_{k,3}+…+x_{k,c-1}\)为\(k\)月时的兔子总数
那么
\(x_{1,1}=1\),\(x_{1,2}=x_{1,3}=…=x_{1,c-1}=0\),\(x_{1,c}=1\)
\(x_{k+1,1}=x_{k,a}+x_{k,a+1}+x_{k,a+2}+…+x_{k,b-1}\)
\(x_{k+1,2}=x_{k,1}\)
\(x_{k+1,3}=x_{k,2}\)
…
\(x_{k+1,c-1}=x_{k,c-2}\)
\(x_{k+1,c}=x_{k,a}+x_{k,a+1}+x_{k,a+2}+…+x_{k,b-1}-x_{k,c-1}+x_{k,c}\)
设矩阵\[A=[(0,0,0…,0,1,1,…,1,0,0,…,0,0),(1,0,0,…,0),(0,1,0,0,…,0),(0,0,1,0,…,0),…,(0,0,0,…,0,1,0,0),(0,0,0…,0,1,1,…,1,0,0,…,0,-1,1)]\]
其中,
\(A\)的第\(1\)行是:\(a-1\)个\(0\),跟着\(b-a\)个\(1\),再跟着\(c-b+1\)个\(0\),
\(A\)的第\(i\)行是:第\(i-1\)个分量为\(1\),其余分量为\(0\)的行向量(\(2<=i<=c-1\))
\(A\)的第\(c\)行是:前\(c-2\)个分量与第\(1\)行的前\(c-2\)个分量一致,后\(2\)个分量是\(-1\),\(1\)
列向量
\(X_1=[(1,0,0…,0,1)]'\)
(头尾为\(1\),其余为\(0\))
则
\(X_n=A^{n-1}*X_1\)
\(X_n\)的最后一个分量就是第\(n\)月的兔子总数 |
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