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PageRank 顶点Rank值总和收敛讨论

Pagerank算法在迭代过程中顶点Rank值的总和是可以收敛到等于顶点数的,这一性质与Pagerank算法每个顶点的rank初值无关。证明如下

假设 d i d_i di为第i轮的rank值总和,可得
d 2 d_2 d2 = d 1 ∗ 0.8 + 0.2 ∗ n ; d_1*0.8 + 0.2*n; d10.8+0.2n;
d 3 d_3 d3 = d 2 ∗ 0.8 + 0.2 ∗ n ; d_2*0.8 + 0.2*n; d20.8+0.2n;
d 4 d_4 d4 = d 3 ∗ 0.8 + 0.2 ∗ n ; d_3*0.8 + 0.2*n; d30.8+0.2n;

d n d_n dn = d n − 1 ∗ 0.8 + 0.2 ∗ n ; d_{n-1}*0.8 + 0.2*n; dn10.8+0.2n;

d 2 d_2 d2代入 d 3 d_3 d3可得

d 3 d_3 d3 = ( d 1 ∗ 0.8 + 0.2 ∗ n ) ∗ 0.8 + 0.2 ∗ n (d_1*0.8 + 0.2*n)*0.8 + 0.2*n (d10.8+0.2n)0.8+0.2n

d 3 d_3 d3代入 d 4 d_4 d4可得

d 4 d_4 d4 = ( ( d 1 ∗ 0.8 + 0.2 ∗ n ) ∗ 0.8 + 0.2 ∗ n ) ∗ 0.8 + 0.2 ∗ n ((d_1*0.8 + 0.2*n)*0.8 + 0.2*n ) * 0.8 +0.2 * n ((d10.8+0.2n)0.8+0.2n)0.8+0.2n

展开得

d 4 d_4 d4 = d 1 ∗ 0. 8 3 + 0.2 ∗ n ∗ ( 1 + 0.8 + 0. 8 2 ) d_1*0.8^3 + 0.2 * n * (1 + 0.8 + 0.8^2) d10.83+0.2n(1+0.8+0.82)

进而递推得

d n d_n dn = d n − 1 ∗ 0. 8 n − 1 + 0.2 ∗ n ∗ ( 1 + 0.8 + 0. 8 2 . . . + 0. 8 n − 2 ) d_{n-1}*0.8^{n-1} + 0.2 * n * (1 + 0.8 + 0.8^2... + 0.8^{n-2}) dn10.8n1+0.2n(1+0.8+0.82...+0.8n2)

1 + 0.8 + 0. 8 2 . . . + 0. 8 n − 2 1 + 0.8 + 0.8^2... + 0.8^{n-2} 1+0.8+0.82...+0.8n2为等比数列 ,求和为 ( 1 − 0. 8 n − 2 ) / 0.2 (1-0.8^{n-2})/0.2 (10.8n2)/0.2

带入后可得原式

d n d_n dn = d n − 1 ∗ 0. 8 n − 1 + n ∗ ( 1 − 0. 8 n − 2 ) d_{n-1}*0.8^{n-1} + n * (1-0.8^{n-2}) dn10.8n1+n(10.8n2)

对于上式,当n等于40附近, 0. 8 n − 1 , 0. 8 n − 2 0.8^{n-1} , 0.8^{n-2} 0.8n1,0.8n2 均接近0.0001,原式可近似为

d n d_n dn = $ n $,其中n为图数据顶点数

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