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\begin { proof} %[Proof of \Cref{lem:lin-sys}]
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The proof consists of two parts. First we need to show that a vector $ \vct { b } $ satisfying the linear system exists and further can be computed in $ O ( m ) $ time. Second we need to show that $ \numocc { G } { \tri } , \numocc { G } { \threedis } $ can indeed be computed in time $ O ( 1 ) $ .
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The lemma claims that for $ \vct { M } =
\begin { pmatrix}
1 - 3p & -(3\prob ^ 2 - \prob ^ 3)\\
10(3\prob ^ 2 - \prob ^ 3) & 10(3\prob ^ 2 - \prob ^ 3)
\end { pmatrix} $ , $ \vct { x} =
\begin { pmatrix}
\numocc { G} { \tri } ]\\
\numocc { G} { \threedis }
\end { pmatrix} $
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satisfies the system $ \vct { M } \cdot \vct { x } = \vct { b } $ .
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To prove the first step, we use \Cref { lem:qE3-exp} to derive the following equality (dropping the superscript and referring to $ G ^ { ( 1 ) } $ as $ G $ ):
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\begin { align}
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\numocc { G} { \ed } \prob ^ 2 & + 6\numocc { G} { \twopath } \prob ^ 3 + 6\numocc { G} { \twodis } \prob ^ 4 + 6\numocc { G} { \tri } \prob ^ 3 + 6\numocc { G} { \oneint } \prob ^ 4 \nonumber \\
& + 6\numocc { G} { \threepath } \prob ^ 4 + 6\numocc { G} { \twopathdis } \prob ^ 5 + 6\numocc { G} { \threedis } \prob ^ 6 = \rpoly _ { G} ^ 3(\prob ,\ldots , \prob )\label { eq:lem-qE3-exp} \\
\numocc { G} { \tri } & +\numocc { G} { \threepath } \prob +\numocc { G} { \twopathdis } \prob ^ 2+\numocc { G} { \threedis } \prob ^ 3\nonumber \\
& = \frac { \rpoly _ { G} ^ 3(\prob ,\ldots , \prob )} { 6\prob ^ 3} - \frac { \numocc { G} { \ed } } { 6\prob } - \numocc { G} { \twopath } -\numocc { G} { \twodis } \prob -\numocc { G} { \oneint } \prob \label { eq:b1-alg-1} \\
\numocc { G} { \tri } (1-3p) & - \numocc { G} { \threedis } (3\prob ^ 2 -\prob ^ 3) = \nonumber \\
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\frac { \rpoly _ { G} ^ 3(\prob ,\ldots , \prob )} { 6\prob ^ 3} & - \frac { \numocc { G} { \ed } } { 6\prob } - \numocc { G} { \twopath } -\numocc { G} { \twodis } \prob -\numocc { G} { \oneint } \prob \nonumber \\
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& -\left [\numocc{G}{\threepath}\prob+3\numocc{G}{\tri}\prob\right] -\left [\numocc{G}{\twopathdis}\prob^2+3\numocc{G}{\threedis}\prob^2\right] \label { eq:b1-alg-2}
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\end { align}
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\Cref { eq:lem-qE3-exp} is the result of \Cref { lem:qE3-exp} . We obtain the remaining equations through standard algebraic manipulations.
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Note that the LHS of \Cref { eq:b1-alg-2} is obtained using \cref { eq:2pd-3d} and \cref { eq:3p-3tri} and is indeed the product $ \vct { M } [ 1 ] \cdot \vct { x } [ 1 ] $ . Further note that this product is equal to the RHS of \Cref { eq:b1-alg-2} , where every term is computable in $ O ( m ) $ time (by equations (\ref { eq:1e} )-(\ref { eq:3p-3tri} )). We set $ \vct { b } [ 1 ] $ to the RHS of \Cref { eq:b1-alg-2} .
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We follow the same process in deriving an equality for $ G ^ { ( 2 ) } $ . Replacing occurrences of $ G $ with $ G ^ { ( 2 ) } $ , we obtain an equation (below) of the form of \cref { eq:b1-alg-2} for $ G ^ { ( 2 ) } $ . Substituting identities from \cref { lem:3m-G2} and \Cref { lem:tri} we obtain
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\begin { align}
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0-\left (8\numocc { G} { \threedis } \right .& \left .+6\numocc { G} { \twopathdis } +4\numocc { G} { \oneint } +4\numocc { G} { \threepath } +2\numocc { G} { \tri } (3\prob ^ 2 -\prob ^ 3)\right )=\nonumber \\
& \frac { \rpoly _ { \graph { 2} } ^ 3(\prob ,\ldots , \prob )} { 6\prob ^ 3} - \frac { \numocc { \graph { 2} } { \ed } } { 6\prob } - \numocc { \graph { 2} } { \twopath } -\numocc { \graph { 2} } { \twodis } \prob -\numocc { \graph { 2} } { \oneint } \prob \nonumber \\
& -\left [\numocc{\graph{2}}{\twopathdis}\prob^2+3\numocc{\graph{2}}{\threedis}\prob^2\right] -\left [\numocc{\graph{2}}{\threepath}\prob + 3\numocc{\graph{2}}{\tri}\prob\right] \label { eq:b2-sub-lem} \\
(10\numocc { G} { \tri } & + 10{ G} { \threedis } )(3\prob ^ 2 -\prob ^ 3) = \nonumber \\
& \frac { \rpoly _ { \graph { 2} } ^ 3(\prob ,\ldots , \prob )} { 6\prob ^ 3} - \frac { \numocc { \graph { 2} } { \ed } } { 6\prob } - \numocc { \graph { 2} } { \twopath } -\numocc { \graph { 2} } { \twodis } \prob -\numocc { \graph { 2} } { \oneint } \prob \nonumber \\
& -\left [\numocc{\graph{2}}{\threepath}\prob+3\numocc{\graph{2}}{\tri}\prob\right] -\left [\numocc{\graph{2}}{\twopathdis}\prob^2-3\numocc{\graph{2}}{\threedis}\prob^2\right] \nonumber \\
& +\left (4\numocc { G} { \oneint } +\left [6\numocc{G}{\twopathdis}+18\numocc{G}{\threedis}\right] +\left [4\numocc{G}{\threepath}+12\numocc{G}{\tri}\right] \right )(3\prob ^ 2 - \prob ^ 3)\label { eq:b2-final}
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\end { align}
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The steps to obtaining \cref { eq:b2-final} are analogous to the derivation immediately preceding. As in the previous derivation, note that the LHS of \Cref { eq:b2-final} is the same as $ \vct { M } [ 2 ] \cdot \vct { x } [ 2 ] $ . The RHS of \Cref { eq:b2-final} has terms all computable (by equations (\ref { eq:1e} )-(\ref { eq:3p-3tri} )) in $ O ( m ) $ time. Setting $ \vct { b } [ 2 ] $ to the RHS then completes the proof of step 1.
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Note that if $ \vct { M } $ has full rank then one can compute $ \numocc { G } { \tri } $ and $ \numocc { G } { \threedis } $ in $ O ( 1 ) $ using Gaussian elimination.
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To show that $ \vct { M } $ indeed has full rank, we show derive in what follows that $ \dtrm { \vct { M } } \ne 0 $ for every $ \prob \in ( 0 , 1 ) $ .
$ \dtrm { \vct { M } } = $
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\begin { align}
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& \begin { vmatrix}
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1-3\prob & -(3\prob ^ 2 - \prob ^ 3)\\
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10(3\prob ^ 2 - \prob ^ 3) & 10(3\prob ^ 2 - \prob ^ 3)
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\end { vmatrix}
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= (1-3\prob )\cdot 10(3\prob ^ 2-\prob ^ 3) +10(3\prob ^ 2-\prob ^ 3)\cdot (3\prob ^ 2 - \prob ^ 3)\nonumber \\
& =10(3\prob ^ 2-\prob ^ 3)\cdot (1-3\prob +3\prob ^ 2-\prob ^ 3) = 10(3\prob ^ 2-\prob ^ 3)\cdot (-\prob ^ 3+3\prob ^ 2-3\prob + 1)\nonumber \\
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& =10\prob ^ 2(3 - \prob )\cdot (1-\prob )^ 3\label { eq:det-final}
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\end { align}
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From \Cref { eq:det-final} it can easily be seen that the roots of $ \dtrm { \vct { M } } $ are $ 0 , 1 , $ and $ 3 $ . Hence there are no roots in $ ( 0 , 1 ) $ and \Cref { lem:lin-sys} follows.
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\qed
\end { proof}
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