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Re: The good will hunting problem solved

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MerelyOfModels is not online. MerelyOfModels
Joined: 23 Dec 2014
Total Posts: 516
08 Jun 2015 06:27 PM

Solution to 1). The adjacency matrix L is L = 0 & 1 & 0 & 1 \\ 1 & 0 & 2 & 1 \\ 0 & 2 & 0 & 0 \\ 1 & 1 & 0 & 0
Solution to 2). [L2]ij is by definition of the matrix product the sum Li 1 L1 j + Li 2 L2 j +... + Li n Ln j. Each term Li 1 L1 j is not 0 if and only if there is at least one path of length 2 going from i to j passing through k. Therefore [L2]ij is the number of paths of length 2 leading from node i to j. Similarly, [Ln]ij is the number of paths of length n going from i to j. The answer is L^3 = [2 & 7 & 2 & 3 \\ 7 & 2 & 12 & 7 \\ 2 & 12 & 0 & 2 \\ 3 & 7 & 2 & 2]
Solution to 3). The \sum_{n} x^n = (1-x)^{-1} for the summation of a geometric series holds also for matrices: f(z) = \sum_{n=0}^{\infty} [L^n]_{ij} z^n = [\sum_{n=0}^{\infty} L^n z^n)]_{ij} = [(1-Lz)^{-1}]_{ij}. Cramer's rule for the inverse of a matrix is A-1 = det(Adj(A)ij)/det(A) leads to det( Adj(1-z L)ij)/det(1-z L) which can also be written as det( Adj(L-z)ij)/det(L-z).
Solution to 4). Especially, when i=1 and j=3, we get det( Adj(A)13 = 2 z2 + 2 z3 and det(L-z) = 1-7 z2 - 2 z3+4 z4. The result can be written as (2 z2+2z3)/(1-7 z2 - 2 z3+4 z4) which simplifies to 2z^2/(1-z-6z^2+4z^3).
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Indasaur is not online. Indasaur
Joined: 15 Apr 2012
Total Posts: 1560
08 Jun 2015 06:29 PM
cool dud
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