Is there an algorithm to find the lengths of all paths from a given vertice that can run in polynomial time?












2












$begingroup$


I have seen from a few sources (such as the CS Stack Exchange) that the problem of determining all the paths between two vertices $u$ and $v$ is NP hard. However, foR something that I am developing, I need to find only the number of paths of a given length whose first vertice is $v$. Is there an algorithm that can find this in Polynomial time ?



P.S. This is the first time I am posting on Stackexchange, so I apologize in case my question is too "rough" .










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  • $begingroup$
    I object to marking the question as "too broad". The question looks pretty specific.
    $endgroup$
    – lisyarus
    Dec 28 '18 at 10:31
















2












$begingroup$


I have seen from a few sources (such as the CS Stack Exchange) that the problem of determining all the paths between two vertices $u$ and $v$ is NP hard. However, foR something that I am developing, I need to find only the number of paths of a given length whose first vertice is $v$. Is there an algorithm that can find this in Polynomial time ?



P.S. This is the first time I am posting on Stackexchange, so I apologize in case my question is too "rough" .










share|cite|improve this question









$endgroup$












  • $begingroup$
    I object to marking the question as "too broad". The question looks pretty specific.
    $endgroup$
    – lisyarus
    Dec 28 '18 at 10:31














2












2








2





$begingroup$


I have seen from a few sources (such as the CS Stack Exchange) that the problem of determining all the paths between two vertices $u$ and $v$ is NP hard. However, foR something that I am developing, I need to find only the number of paths of a given length whose first vertice is $v$. Is there an algorithm that can find this in Polynomial time ?



P.S. This is the first time I am posting on Stackexchange, so I apologize in case my question is too "rough" .










share|cite|improve this question









$endgroup$




I have seen from a few sources (such as the CS Stack Exchange) that the problem of determining all the paths between two vertices $u$ and $v$ is NP hard. However, foR something that I am developing, I need to find only the number of paths of a given length whose first vertice is $v$. Is there an algorithm that can find this in Polynomial time ?



P.S. This is the first time I am posting on Stackexchange, so I apologize in case my question is too "rough" .







graph-theory algorithms computational-complexity






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asked Dec 28 '18 at 6:27









Aryaman GuptaAryaman Gupta

356




356












  • $begingroup$
    I object to marking the question as "too broad". The question looks pretty specific.
    $endgroup$
    – lisyarus
    Dec 28 '18 at 10:31


















  • $begingroup$
    I object to marking the question as "too broad". The question looks pretty specific.
    $endgroup$
    – lisyarus
    Dec 28 '18 at 10:31
















$begingroup$
I object to marking the question as "too broad". The question looks pretty specific.
$endgroup$
– lisyarus
Dec 28 '18 at 10:31




$begingroup$
I object to marking the question as "too broad". The question looks pretty specific.
$endgroup$
– lisyarus
Dec 28 '18 at 10:31










2 Answers
2






active

oldest

votes


















2












$begingroup$

If you only need the number of walks of a given length, you can do it in polynomial time (with respect to the number of vertices of the graph) using the incidence (or adjacency) matrix of the graph.



Indeed, assume $G= (V,E)$ is the graph in question, where $|V| = n$ and $A$ is the incidence matrix of $G$, i.e. $A$ is an $ntimes n$ matrix of ${0,1}$-s where $(i,j)$-th element is $1$ iff there is an edge from vertex $i$ to vertex $j$.



Fix an integer $k geq 1$, which is the length of the walk. It is a well-known fact that the $(i,j)$-th entry of the matrix $A^k$ shows the number of walks from $i$ to $j$ having length $k$ (the proof follows by a straightforward induction on $k$ and the definition of a matrix product, see wikipedia for instance).



Getting back to your original question: $A^k$ can be computed in $O(n^3 + nlog k)$ time using singular value decomposition (SVD). Here $n^3$ is the time to complete the SVD and $n log k$ is the time to raise all elements on the diagonal matrix of the SVD into power $k$. See here for further discussion.
Once you have $A^k$, then summing over all elements of the $v$-th row ($v$ was the vertex where you start the walk of length $k$) gives the number of walks from $v$ having length $k$.



For the number of paths (instead of walks), see the other answer.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Just to note: even the most straightforward way is polynomial-time, with single matrix product in $O(n^3)$, thus $O(k n^3)$ for $A^k$, or $O(log_2 k cdot n^3)$ if using exponentiation by squaring. SVD, though, may suffer from precision issues.
    $endgroup$
    – lisyarus
    Dec 28 '18 at 10:28












  • $begingroup$
    @lisyarus, thanks for your comment. Well, of course the straightforward matrix product is still polynomial in $n$ (assuming $k$ is fixed). I though of SVD only to remove the factor coming from $k$ from the polynomial on $n$, but I suppose there should be better approaches here.
    $endgroup$
    – Hayk
    Dec 28 '18 at 10:39










  • $begingroup$
    @Hayk Does this calculate the number of paths between $i$ and $j$ though, or the number of walks between $i$ and $j$. A walk may have cycles and it may even backtrack on an edge.
    $endgroup$
    – Mike
    Dec 28 '18 at 19:16












  • $begingroup$
    In fact the above answer is incorrect: If it were true you could use the above to count the number of paths between $i$ and $j$: Simply take the sum $sum_k A^k_{ij}$.. However, finding the number of such paths is NP-Hard.
    $endgroup$
    – Mike
    Dec 28 '18 at 19:24








  • 1




    $begingroup$
    @Mike, fair enough, it was an oversight from my part: "walk" vs "path". Indeed, my answer computes the number of walks rather than the number of paths, there's no restriction on $A^k$ to take into account distinct vertices only. The answer is now edited accordingly.
    $endgroup$
    – Hayk
    Dec 28 '18 at 20:02



















1












$begingroup$

No, there is no such algorithm, unless P=NP. More precisely: If determining the number of path between two vertices $u$ and $v$ is NP-Hard, then so is calculating the number of paths starting at any vertex $v$.



Indeed, let $G$ be a graph, and $u$ and $v$ any two vertices in $G$. Then write as $n_G(v)$ the number of paths starting at $v$ and write as $n_G(v,u)$ the number of paths starting at $v$ and ending at $u$. Next, let $G'$ be the graph formed from $G$ by adding an extra vertex $z$ and attaching that extra vertex to precisely $u$. Then
$n_G(v)+ n_G(v,u) = n_{G'}(v)$. If we can calculate $n_G(v)$ and $n_{G'}(v)$ in polynomial time then we can calculate $n_{G}(v,u)$ in polynomial time.



And if we can calculate the number $^kn_G(v)$ of paths of length $k$ starting at $v$ then we can certainly calculate $n_G(v)$; indeed take the sum $sum_k$
$^kn_G(v)$ to get $n_G(v)$.



[Computing the total number of walks of a given length $k$ between two vertices is easy to do in polynomial time however; the above answer shows how to do this.]






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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

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    2












    $begingroup$

    If you only need the number of walks of a given length, you can do it in polynomial time (with respect to the number of vertices of the graph) using the incidence (or adjacency) matrix of the graph.



    Indeed, assume $G= (V,E)$ is the graph in question, where $|V| = n$ and $A$ is the incidence matrix of $G$, i.e. $A$ is an $ntimes n$ matrix of ${0,1}$-s where $(i,j)$-th element is $1$ iff there is an edge from vertex $i$ to vertex $j$.



    Fix an integer $k geq 1$, which is the length of the walk. It is a well-known fact that the $(i,j)$-th entry of the matrix $A^k$ shows the number of walks from $i$ to $j$ having length $k$ (the proof follows by a straightforward induction on $k$ and the definition of a matrix product, see wikipedia for instance).



    Getting back to your original question: $A^k$ can be computed in $O(n^3 + nlog k)$ time using singular value decomposition (SVD). Here $n^3$ is the time to complete the SVD and $n log k$ is the time to raise all elements on the diagonal matrix of the SVD into power $k$. See here for further discussion.
    Once you have $A^k$, then summing over all elements of the $v$-th row ($v$ was the vertex where you start the walk of length $k$) gives the number of walks from $v$ having length $k$.



    For the number of paths (instead of walks), see the other answer.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Just to note: even the most straightforward way is polynomial-time, with single matrix product in $O(n^3)$, thus $O(k n^3)$ for $A^k$, or $O(log_2 k cdot n^3)$ if using exponentiation by squaring. SVD, though, may suffer from precision issues.
      $endgroup$
      – lisyarus
      Dec 28 '18 at 10:28












    • $begingroup$
      @lisyarus, thanks for your comment. Well, of course the straightforward matrix product is still polynomial in $n$ (assuming $k$ is fixed). I though of SVD only to remove the factor coming from $k$ from the polynomial on $n$, but I suppose there should be better approaches here.
      $endgroup$
      – Hayk
      Dec 28 '18 at 10:39










    • $begingroup$
      @Hayk Does this calculate the number of paths between $i$ and $j$ though, or the number of walks between $i$ and $j$. A walk may have cycles and it may even backtrack on an edge.
      $endgroup$
      – Mike
      Dec 28 '18 at 19:16












    • $begingroup$
      In fact the above answer is incorrect: If it were true you could use the above to count the number of paths between $i$ and $j$: Simply take the sum $sum_k A^k_{ij}$.. However, finding the number of such paths is NP-Hard.
      $endgroup$
      – Mike
      Dec 28 '18 at 19:24








    • 1




      $begingroup$
      @Mike, fair enough, it was an oversight from my part: "walk" vs "path". Indeed, my answer computes the number of walks rather than the number of paths, there's no restriction on $A^k$ to take into account distinct vertices only. The answer is now edited accordingly.
      $endgroup$
      – Hayk
      Dec 28 '18 at 20:02
















    2












    $begingroup$

    If you only need the number of walks of a given length, you can do it in polynomial time (with respect to the number of vertices of the graph) using the incidence (or adjacency) matrix of the graph.



    Indeed, assume $G= (V,E)$ is the graph in question, where $|V| = n$ and $A$ is the incidence matrix of $G$, i.e. $A$ is an $ntimes n$ matrix of ${0,1}$-s where $(i,j)$-th element is $1$ iff there is an edge from vertex $i$ to vertex $j$.



    Fix an integer $k geq 1$, which is the length of the walk. It is a well-known fact that the $(i,j)$-th entry of the matrix $A^k$ shows the number of walks from $i$ to $j$ having length $k$ (the proof follows by a straightforward induction on $k$ and the definition of a matrix product, see wikipedia for instance).



    Getting back to your original question: $A^k$ can be computed in $O(n^3 + nlog k)$ time using singular value decomposition (SVD). Here $n^3$ is the time to complete the SVD and $n log k$ is the time to raise all elements on the diagonal matrix of the SVD into power $k$. See here for further discussion.
    Once you have $A^k$, then summing over all elements of the $v$-th row ($v$ was the vertex where you start the walk of length $k$) gives the number of walks from $v$ having length $k$.



    For the number of paths (instead of walks), see the other answer.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Just to note: even the most straightforward way is polynomial-time, with single matrix product in $O(n^3)$, thus $O(k n^3)$ for $A^k$, or $O(log_2 k cdot n^3)$ if using exponentiation by squaring. SVD, though, may suffer from precision issues.
      $endgroup$
      – lisyarus
      Dec 28 '18 at 10:28












    • $begingroup$
      @lisyarus, thanks for your comment. Well, of course the straightforward matrix product is still polynomial in $n$ (assuming $k$ is fixed). I though of SVD only to remove the factor coming from $k$ from the polynomial on $n$, but I suppose there should be better approaches here.
      $endgroup$
      – Hayk
      Dec 28 '18 at 10:39










    • $begingroup$
      @Hayk Does this calculate the number of paths between $i$ and $j$ though, or the number of walks between $i$ and $j$. A walk may have cycles and it may even backtrack on an edge.
      $endgroup$
      – Mike
      Dec 28 '18 at 19:16












    • $begingroup$
      In fact the above answer is incorrect: If it were true you could use the above to count the number of paths between $i$ and $j$: Simply take the sum $sum_k A^k_{ij}$.. However, finding the number of such paths is NP-Hard.
      $endgroup$
      – Mike
      Dec 28 '18 at 19:24








    • 1




      $begingroup$
      @Mike, fair enough, it was an oversight from my part: "walk" vs "path". Indeed, my answer computes the number of walks rather than the number of paths, there's no restriction on $A^k$ to take into account distinct vertices only. The answer is now edited accordingly.
      $endgroup$
      – Hayk
      Dec 28 '18 at 20:02














    2












    2








    2





    $begingroup$

    If you only need the number of walks of a given length, you can do it in polynomial time (with respect to the number of vertices of the graph) using the incidence (or adjacency) matrix of the graph.



    Indeed, assume $G= (V,E)$ is the graph in question, where $|V| = n$ and $A$ is the incidence matrix of $G$, i.e. $A$ is an $ntimes n$ matrix of ${0,1}$-s where $(i,j)$-th element is $1$ iff there is an edge from vertex $i$ to vertex $j$.



    Fix an integer $k geq 1$, which is the length of the walk. It is a well-known fact that the $(i,j)$-th entry of the matrix $A^k$ shows the number of walks from $i$ to $j$ having length $k$ (the proof follows by a straightforward induction on $k$ and the definition of a matrix product, see wikipedia for instance).



    Getting back to your original question: $A^k$ can be computed in $O(n^3 + nlog k)$ time using singular value decomposition (SVD). Here $n^3$ is the time to complete the SVD and $n log k$ is the time to raise all elements on the diagonal matrix of the SVD into power $k$. See here for further discussion.
    Once you have $A^k$, then summing over all elements of the $v$-th row ($v$ was the vertex where you start the walk of length $k$) gives the number of walks from $v$ having length $k$.



    For the number of paths (instead of walks), see the other answer.






    share|cite|improve this answer











    $endgroup$



    If you only need the number of walks of a given length, you can do it in polynomial time (with respect to the number of vertices of the graph) using the incidence (or adjacency) matrix of the graph.



    Indeed, assume $G= (V,E)$ is the graph in question, where $|V| = n$ and $A$ is the incidence matrix of $G$, i.e. $A$ is an $ntimes n$ matrix of ${0,1}$-s where $(i,j)$-th element is $1$ iff there is an edge from vertex $i$ to vertex $j$.



    Fix an integer $k geq 1$, which is the length of the walk. It is a well-known fact that the $(i,j)$-th entry of the matrix $A^k$ shows the number of walks from $i$ to $j$ having length $k$ (the proof follows by a straightforward induction on $k$ and the definition of a matrix product, see wikipedia for instance).



    Getting back to your original question: $A^k$ can be computed in $O(n^3 + nlog k)$ time using singular value decomposition (SVD). Here $n^3$ is the time to complete the SVD and $n log k$ is the time to raise all elements on the diagonal matrix of the SVD into power $k$. See here for further discussion.
    Once you have $A^k$, then summing over all elements of the $v$-th row ($v$ was the vertex where you start the walk of length $k$) gives the number of walks from $v$ having length $k$.



    For the number of paths (instead of walks), see the other answer.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 28 '18 at 19:58

























    answered Dec 28 '18 at 10:07









    HaykHayk

    2,6371214




    2,6371214








    • 1




      $begingroup$
      Just to note: even the most straightforward way is polynomial-time, with single matrix product in $O(n^3)$, thus $O(k n^3)$ for $A^k$, or $O(log_2 k cdot n^3)$ if using exponentiation by squaring. SVD, though, may suffer from precision issues.
      $endgroup$
      – lisyarus
      Dec 28 '18 at 10:28












    • $begingroup$
      @lisyarus, thanks for your comment. Well, of course the straightforward matrix product is still polynomial in $n$ (assuming $k$ is fixed). I though of SVD only to remove the factor coming from $k$ from the polynomial on $n$, but I suppose there should be better approaches here.
      $endgroup$
      – Hayk
      Dec 28 '18 at 10:39










    • $begingroup$
      @Hayk Does this calculate the number of paths between $i$ and $j$ though, or the number of walks between $i$ and $j$. A walk may have cycles and it may even backtrack on an edge.
      $endgroup$
      – Mike
      Dec 28 '18 at 19:16












    • $begingroup$
      In fact the above answer is incorrect: If it were true you could use the above to count the number of paths between $i$ and $j$: Simply take the sum $sum_k A^k_{ij}$.. However, finding the number of such paths is NP-Hard.
      $endgroup$
      – Mike
      Dec 28 '18 at 19:24








    • 1




      $begingroup$
      @Mike, fair enough, it was an oversight from my part: "walk" vs "path". Indeed, my answer computes the number of walks rather than the number of paths, there's no restriction on $A^k$ to take into account distinct vertices only. The answer is now edited accordingly.
      $endgroup$
      – Hayk
      Dec 28 '18 at 20:02














    • 1




      $begingroup$
      Just to note: even the most straightforward way is polynomial-time, with single matrix product in $O(n^3)$, thus $O(k n^3)$ for $A^k$, or $O(log_2 k cdot n^3)$ if using exponentiation by squaring. SVD, though, may suffer from precision issues.
      $endgroup$
      – lisyarus
      Dec 28 '18 at 10:28












    • $begingroup$
      @lisyarus, thanks for your comment. Well, of course the straightforward matrix product is still polynomial in $n$ (assuming $k$ is fixed). I though of SVD only to remove the factor coming from $k$ from the polynomial on $n$, but I suppose there should be better approaches here.
      $endgroup$
      – Hayk
      Dec 28 '18 at 10:39










    • $begingroup$
      @Hayk Does this calculate the number of paths between $i$ and $j$ though, or the number of walks between $i$ and $j$. A walk may have cycles and it may even backtrack on an edge.
      $endgroup$
      – Mike
      Dec 28 '18 at 19:16












    • $begingroup$
      In fact the above answer is incorrect: If it were true you could use the above to count the number of paths between $i$ and $j$: Simply take the sum $sum_k A^k_{ij}$.. However, finding the number of such paths is NP-Hard.
      $endgroup$
      – Mike
      Dec 28 '18 at 19:24








    • 1




      $begingroup$
      @Mike, fair enough, it was an oversight from my part: "walk" vs "path". Indeed, my answer computes the number of walks rather than the number of paths, there's no restriction on $A^k$ to take into account distinct vertices only. The answer is now edited accordingly.
      $endgroup$
      – Hayk
      Dec 28 '18 at 20:02








    1




    1




    $begingroup$
    Just to note: even the most straightforward way is polynomial-time, with single matrix product in $O(n^3)$, thus $O(k n^3)$ for $A^k$, or $O(log_2 k cdot n^3)$ if using exponentiation by squaring. SVD, though, may suffer from precision issues.
    $endgroup$
    – lisyarus
    Dec 28 '18 at 10:28






    $begingroup$
    Just to note: even the most straightforward way is polynomial-time, with single matrix product in $O(n^3)$, thus $O(k n^3)$ for $A^k$, or $O(log_2 k cdot n^3)$ if using exponentiation by squaring. SVD, though, may suffer from precision issues.
    $endgroup$
    – lisyarus
    Dec 28 '18 at 10:28














    $begingroup$
    @lisyarus, thanks for your comment. Well, of course the straightforward matrix product is still polynomial in $n$ (assuming $k$ is fixed). I though of SVD only to remove the factor coming from $k$ from the polynomial on $n$, but I suppose there should be better approaches here.
    $endgroup$
    – Hayk
    Dec 28 '18 at 10:39




    $begingroup$
    @lisyarus, thanks for your comment. Well, of course the straightforward matrix product is still polynomial in $n$ (assuming $k$ is fixed). I though of SVD only to remove the factor coming from $k$ from the polynomial on $n$, but I suppose there should be better approaches here.
    $endgroup$
    – Hayk
    Dec 28 '18 at 10:39












    $begingroup$
    @Hayk Does this calculate the number of paths between $i$ and $j$ though, or the number of walks between $i$ and $j$. A walk may have cycles and it may even backtrack on an edge.
    $endgroup$
    – Mike
    Dec 28 '18 at 19:16






    $begingroup$
    @Hayk Does this calculate the number of paths between $i$ and $j$ though, or the number of walks between $i$ and $j$. A walk may have cycles and it may even backtrack on an edge.
    $endgroup$
    – Mike
    Dec 28 '18 at 19:16














    $begingroup$
    In fact the above answer is incorrect: If it were true you could use the above to count the number of paths between $i$ and $j$: Simply take the sum $sum_k A^k_{ij}$.. However, finding the number of such paths is NP-Hard.
    $endgroup$
    – Mike
    Dec 28 '18 at 19:24






    $begingroup$
    In fact the above answer is incorrect: If it were true you could use the above to count the number of paths between $i$ and $j$: Simply take the sum $sum_k A^k_{ij}$.. However, finding the number of such paths is NP-Hard.
    $endgroup$
    – Mike
    Dec 28 '18 at 19:24






    1




    1




    $begingroup$
    @Mike, fair enough, it was an oversight from my part: "walk" vs "path". Indeed, my answer computes the number of walks rather than the number of paths, there's no restriction on $A^k$ to take into account distinct vertices only. The answer is now edited accordingly.
    $endgroup$
    – Hayk
    Dec 28 '18 at 20:02




    $begingroup$
    @Mike, fair enough, it was an oversight from my part: "walk" vs "path". Indeed, my answer computes the number of walks rather than the number of paths, there's no restriction on $A^k$ to take into account distinct vertices only. The answer is now edited accordingly.
    $endgroup$
    – Hayk
    Dec 28 '18 at 20:02











    1












    $begingroup$

    No, there is no such algorithm, unless P=NP. More precisely: If determining the number of path between two vertices $u$ and $v$ is NP-Hard, then so is calculating the number of paths starting at any vertex $v$.



    Indeed, let $G$ be a graph, and $u$ and $v$ any two vertices in $G$. Then write as $n_G(v)$ the number of paths starting at $v$ and write as $n_G(v,u)$ the number of paths starting at $v$ and ending at $u$. Next, let $G'$ be the graph formed from $G$ by adding an extra vertex $z$ and attaching that extra vertex to precisely $u$. Then
    $n_G(v)+ n_G(v,u) = n_{G'}(v)$. If we can calculate $n_G(v)$ and $n_{G'}(v)$ in polynomial time then we can calculate $n_{G}(v,u)$ in polynomial time.



    And if we can calculate the number $^kn_G(v)$ of paths of length $k$ starting at $v$ then we can certainly calculate $n_G(v)$; indeed take the sum $sum_k$
    $^kn_G(v)$ to get $n_G(v)$.



    [Computing the total number of walks of a given length $k$ between two vertices is easy to do in polynomial time however; the above answer shows how to do this.]






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      No, there is no such algorithm, unless P=NP. More precisely: If determining the number of path between two vertices $u$ and $v$ is NP-Hard, then so is calculating the number of paths starting at any vertex $v$.



      Indeed, let $G$ be a graph, and $u$ and $v$ any two vertices in $G$. Then write as $n_G(v)$ the number of paths starting at $v$ and write as $n_G(v,u)$ the number of paths starting at $v$ and ending at $u$. Next, let $G'$ be the graph formed from $G$ by adding an extra vertex $z$ and attaching that extra vertex to precisely $u$. Then
      $n_G(v)+ n_G(v,u) = n_{G'}(v)$. If we can calculate $n_G(v)$ and $n_{G'}(v)$ in polynomial time then we can calculate $n_{G}(v,u)$ in polynomial time.



      And if we can calculate the number $^kn_G(v)$ of paths of length $k$ starting at $v$ then we can certainly calculate $n_G(v)$; indeed take the sum $sum_k$
      $^kn_G(v)$ to get $n_G(v)$.



      [Computing the total number of walks of a given length $k$ between two vertices is easy to do in polynomial time however; the above answer shows how to do this.]






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        No, there is no such algorithm, unless P=NP. More precisely: If determining the number of path between two vertices $u$ and $v$ is NP-Hard, then so is calculating the number of paths starting at any vertex $v$.



        Indeed, let $G$ be a graph, and $u$ and $v$ any two vertices in $G$. Then write as $n_G(v)$ the number of paths starting at $v$ and write as $n_G(v,u)$ the number of paths starting at $v$ and ending at $u$. Next, let $G'$ be the graph formed from $G$ by adding an extra vertex $z$ and attaching that extra vertex to precisely $u$. Then
        $n_G(v)+ n_G(v,u) = n_{G'}(v)$. If we can calculate $n_G(v)$ and $n_{G'}(v)$ in polynomial time then we can calculate $n_{G}(v,u)$ in polynomial time.



        And if we can calculate the number $^kn_G(v)$ of paths of length $k$ starting at $v$ then we can certainly calculate $n_G(v)$; indeed take the sum $sum_k$
        $^kn_G(v)$ to get $n_G(v)$.



        [Computing the total number of walks of a given length $k$ between two vertices is easy to do in polynomial time however; the above answer shows how to do this.]






        share|cite|improve this answer









        $endgroup$



        No, there is no such algorithm, unless P=NP. More precisely: If determining the number of path between two vertices $u$ and $v$ is NP-Hard, then so is calculating the number of paths starting at any vertex $v$.



        Indeed, let $G$ be a graph, and $u$ and $v$ any two vertices in $G$. Then write as $n_G(v)$ the number of paths starting at $v$ and write as $n_G(v,u)$ the number of paths starting at $v$ and ending at $u$. Next, let $G'$ be the graph formed from $G$ by adding an extra vertex $z$ and attaching that extra vertex to precisely $u$. Then
        $n_G(v)+ n_G(v,u) = n_{G'}(v)$. If we can calculate $n_G(v)$ and $n_{G'}(v)$ in polynomial time then we can calculate $n_{G}(v,u)$ in polynomial time.



        And if we can calculate the number $^kn_G(v)$ of paths of length $k$ starting at $v$ then we can certainly calculate $n_G(v)$; indeed take the sum $sum_k$
        $^kn_G(v)$ to get $n_G(v)$.



        [Computing the total number of walks of a given length $k$ between two vertices is easy to do in polynomial time however; the above answer shows how to do this.]







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 28 '18 at 19:46









        MikeMike

        4,396412




        4,396412






























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