Dynamic programming approach for finding perfect square subsequence












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Given an array of numbers $a_i$ with length $n$, such that $0<a_i<30$, how can we find the number of subsequences that the product of numbers is perfect square.



I have tried to find a dynamic programming approach but had no success so far.






algorithms







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  • $begingroup$
    Use linear algebra instead.
    $endgroup$
    – Yuval Filmus
    16 hours ago
















0












$begingroup$


Given an array of numbers $a_i$ with length $n$, such that $0<a_i<30$, how can we find the number of subsequences that the product of numbers is perfect square.



I have tried to find a dynamic programming approach but had no success so far.






algorithms







share|cite|improve this question







New contributor




besmel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Use linear algebra instead.
    $endgroup$
    – Yuval Filmus
    16 hours ago














0












0








0





$begingroup$


Given an array of numbers $a_i$ with length $n$, such that $0<a_i<30$, how can we find the number of subsequences that the product of numbers is perfect square.



I have tried to find a dynamic programming approach but had no success so far.






algorithms







share|cite|improve this question







New contributor




besmel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Given an array of numbers $a_i$ with length $n$, such that $0<a_i<30$, how can we find the number of subsequences that the product of numbers is perfect square.



I have tried to find a dynamic programming approach but had no success so far.






algorithms




algorithms






share|cite|improve this question







New contributor




besmel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




besmel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 17 hours ago









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besmel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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besmel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






besmel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Use linear algebra instead.
    $endgroup$
    – Yuval Filmus
    16 hours ago


















  • $begingroup$
    Use linear algebra instead.
    $endgroup$
    – Yuval Filmus
    16 hours ago
















$begingroup$
Use linear algebra instead.
$endgroup$
– Yuval Filmus
16 hours ago




$begingroup$
Use linear algebra instead.
$endgroup$
– Yuval Filmus
16 hours ago










1 Answer
1






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2












$begingroup$

Let us assume first that by subsequence you mean non-contiguous subsequence.



There are 10 primes in the range $1,ldots,29$:
$$ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. $$
You can represent each number in the range $1,ldots,29$ as a vector of length 10 of exponents. A set of such vectors correspond to numbers whose product is a perfect square iff they sum to a vector whose entries are all even. Alternatively, if we think of the exponents modulo 2, then a set of such vectors correspond to numbers whose product is a perfect square iff they sum to zero.



This suggests the following algorithm. Construct a matrix whose rows correspond to the vectors outlined above (modulo 2), and determine its rank. From the rank you can compute the size of the kernel, which is the quantity you’re after.




Suppose now that you’re after contiguous subsequences.



Let $b_i = a_1 cdot a_2 cdots a_i$. The subsequence $a_i,ldots,a_j$ multiplies to a perfect square iff $b_j/b_{i-1}$ is a perfect square, which happens if the vectors corresponding to $b_{i-1}$ and $b_j$ are equal.



This suggests the following algorithm. Compute the vectors corresponding to each $b_i$ (do this by computing the vectors for $a_i$ and summing them as you go), sort them (you can represent them as 10-bit integers in an arbitrary way), and divide them into runs of length $c_1,ldots,c_ell$. The answer is then $sum_i binom{c_i}{2}$.






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    1 Answer
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    1 Answer
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    active

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    2












    $begingroup$

    Let us assume first that by subsequence you mean non-contiguous subsequence.



    There are 10 primes in the range $1,ldots,29$:
    $$ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. $$
    You can represent each number in the range $1,ldots,29$ as a vector of length 10 of exponents. A set of such vectors correspond to numbers whose product is a perfect square iff they sum to a vector whose entries are all even. Alternatively, if we think of the exponents modulo 2, then a set of such vectors correspond to numbers whose product is a perfect square iff they sum to zero.



    This suggests the following algorithm. Construct a matrix whose rows correspond to the vectors outlined above (modulo 2), and determine its rank. From the rank you can compute the size of the kernel, which is the quantity you’re after.




    Suppose now that you’re after contiguous subsequences.



    Let $b_i = a_1 cdot a_2 cdots a_i$. The subsequence $a_i,ldots,a_j$ multiplies to a perfect square iff $b_j/b_{i-1}$ is a perfect square, which happens if the vectors corresponding to $b_{i-1}$ and $b_j$ are equal.



    This suggests the following algorithm. Compute the vectors corresponding to each $b_i$ (do this by computing the vectors for $a_i$ and summing them as you go), sort them (you can represent them as 10-bit integers in an arbitrary way), and divide them into runs of length $c_1,ldots,c_ell$. The answer is then $sum_i binom{c_i}{2}$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Let us assume first that by subsequence you mean non-contiguous subsequence.



      There are 10 primes in the range $1,ldots,29$:
      $$ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. $$
      You can represent each number in the range $1,ldots,29$ as a vector of length 10 of exponents. A set of such vectors correspond to numbers whose product is a perfect square iff they sum to a vector whose entries are all even. Alternatively, if we think of the exponents modulo 2, then a set of such vectors correspond to numbers whose product is a perfect square iff they sum to zero.



      This suggests the following algorithm. Construct a matrix whose rows correspond to the vectors outlined above (modulo 2), and determine its rank. From the rank you can compute the size of the kernel, which is the quantity you’re after.




      Suppose now that you’re after contiguous subsequences.



      Let $b_i = a_1 cdot a_2 cdots a_i$. The subsequence $a_i,ldots,a_j$ multiplies to a perfect square iff $b_j/b_{i-1}$ is a perfect square, which happens if the vectors corresponding to $b_{i-1}$ and $b_j$ are equal.



      This suggests the following algorithm. Compute the vectors corresponding to each $b_i$ (do this by computing the vectors for $a_i$ and summing them as you go), sort them (you can represent them as 10-bit integers in an arbitrary way), and divide them into runs of length $c_1,ldots,c_ell$. The answer is then $sum_i binom{c_i}{2}$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Let us assume first that by subsequence you mean non-contiguous subsequence.



        There are 10 primes in the range $1,ldots,29$:
        $$ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. $$
        You can represent each number in the range $1,ldots,29$ as a vector of length 10 of exponents. A set of such vectors correspond to numbers whose product is a perfect square iff they sum to a vector whose entries are all even. Alternatively, if we think of the exponents modulo 2, then a set of such vectors correspond to numbers whose product is a perfect square iff they sum to zero.



        This suggests the following algorithm. Construct a matrix whose rows correspond to the vectors outlined above (modulo 2), and determine its rank. From the rank you can compute the size of the kernel, which is the quantity you’re after.




        Suppose now that you’re after contiguous subsequences.



        Let $b_i = a_1 cdot a_2 cdots a_i$. The subsequence $a_i,ldots,a_j$ multiplies to a perfect square iff $b_j/b_{i-1}$ is a perfect square, which happens if the vectors corresponding to $b_{i-1}$ and $b_j$ are equal.



        This suggests the following algorithm. Compute the vectors corresponding to each $b_i$ (do this by computing the vectors for $a_i$ and summing them as you go), sort them (you can represent them as 10-bit integers in an arbitrary way), and divide them into runs of length $c_1,ldots,c_ell$. The answer is then $sum_i binom{c_i}{2}$.






        share|cite|improve this answer









        $endgroup$



        Let us assume first that by subsequence you mean non-contiguous subsequence.



        There are 10 primes in the range $1,ldots,29$:
        $$ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. $$
        You can represent each number in the range $1,ldots,29$ as a vector of length 10 of exponents. A set of such vectors correspond to numbers whose product is a perfect square iff they sum to a vector whose entries are all even. Alternatively, if we think of the exponents modulo 2, then a set of such vectors correspond to numbers whose product is a perfect square iff they sum to zero.



        This suggests the following algorithm. Construct a matrix whose rows correspond to the vectors outlined above (modulo 2), and determine its rank. From the rank you can compute the size of the kernel, which is the quantity you’re after.




        Suppose now that you’re after contiguous subsequences.



        Let $b_i = a_1 cdot a_2 cdots a_i$. The subsequence $a_i,ldots,a_j$ multiplies to a perfect square iff $b_j/b_{i-1}$ is a perfect square, which happens if the vectors corresponding to $b_{i-1}$ and $b_j$ are equal.



        This suggests the following algorithm. Compute the vectors corresponding to each $b_i$ (do this by computing the vectors for $a_i$ and summing them as you go), sort them (you can represent them as 10-bit integers in an arbitrary way), and divide them into runs of length $c_1,ldots,c_ell$. The answer is then $sum_i binom{c_i}{2}$.







        share|cite|improve this answer












        share|cite|improve this answer



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        answered 15 hours ago









        Yuval FilmusYuval Filmus

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