Image: PageRanks-Example

Description: Numeric examples of PageRank values in a small graph with a damping factor of 0.85. The exact solution is:

R=[PR(A)PR(B)⋮PR(K)]=1579662461⋅(1900220122282280019877222022657320468864002265732093728409372840937284093728409372840)≈(0,032781490,384400950,342910290,039087090,080885690,039087090,016169480,016169480,016169480,016169480,01616948){\displaystyle \mathbf {R} ={\begin{bmatrix}PR(A)\\PR(B)\\\vdots \\PR(K)\end{bmatrix}}={\dfrac {1}{579662461}}\cdot {\begin{pmatrix}19002201\\222822800\\198772220\\22657320\\46886400\\22657320\\9372840\\9372840\\9372840\\9372840\\9372840\end{pmatrix}}\approx {\begin{pmatrix}0,03278149\\0,38440095\\0,34291029\\0,03908709\\0,08088569\\0,03908709\\0,01616948\\0,01616948\\0,01616948\\0,01616948\\0,01616948\end{pmatrix}}}And here’s the solution for any arbitrary damping factor d:

Rd=[PRd(A)PRd(B)⋮PRd(K)]=(1(d+1)⋅(7d4+28d2+12d−132))⋅[(d+1)(7d3+6d+12)(d−1)−9d3−18d2−46d−12−11d4−12d−12−18d3−32d22(d+1)(7d2+2d+6)(d−1)12(d+1)(4d2−3d−1)2(d+1)(7d2+2d+6)(d−1)−2(d+1)(−6+d2)(d−1)−2(d+1)(−6+d2)(d−1)−2(d+1)(−6+d2)(d−1)−2(d+1)(−6+d2)(d−1)−2(d+1)(−6+d2)(d−1)]{\displaystyle \mathbf {R_{d}} ={\begin{bmatrix}PR_{d}(A)\\PR_{d}(B)\\\vdots \\PR_{d}(K)\end{bmatrix}}=\left({\dfrac {1}{(d+1)\cdot (7\,d^{4}+28\,d^{2}+12\,d-132)}}\right)\cdot {\begin{bmatrix}\left({d}+1\right)\left(7\,{d}^{3}+6\,{d}+12\right)\left({d}-1\right)\\-9\,{d}^{3}-18\,{d}^{2}-46\,{d}-12\\-11\,{d}^{4}-12\,{d}-12-18\,{d}^{3}-32\,{d}^{2}\\2\,\left({d}+1\right)\left(7\,{d}^{2}+2\,{d}+6\right)\left({d}-1\right)\\12\,\left({d}+1\right)\left(4\,{d}^{2}-3\,{d}-1\right)\\2\,\left({d}+1\right)\left(7\,{d}^{2}+2\,{d}+6\right)\left({d}-1\right)\\-2\,\left({d}+1\right)\left(-6+{d}^{2}\right)\left({d}-1\right)\\-2\,\left({d}+1\right)\left(-6+{d}^{2}\right)\left({d}-1\right)\\-2\,\left({d}+1\right)\left(-6+{d}^{2}\right)\left({d}-1\right)\\-2\,\left({d}+1\right)\left(-6+{d}^{2}\right)\left({d}-1\right)\\-2\,\left({d}+1\right)\left(-6+{d}^{2}\right)\left({d}-1\right)\end{bmatrix}}}
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