Summation of 'for loop' with conditional? Announcing the arrival of Valued Associate #679:...

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Summation of 'for loop' with conditional?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Proof for a summation-procedure using the matrix of Eulerian numbers?Solving Summation ExpressionsHelp me understand this algorithm problem.Finding a recurrence for a sumSimplifying $sum_{i=1}^{n-2}i(n-1-i)$Write the following for loop as a double summationSummation with arithmetic seriesCalculating part of a summation function as a constant?Is summation the equivalent of a for loop? (Programming)Conditional summation with or without complex roots












1












$begingroup$


I am trying to convert instances of nested 'for' loops into a summation expression. The current code fragment I have is:



for i = 1 to n:

    for j = 1 to n:

        if (i*j >= n):

            for k = 1 to n:

                sum++

        endif


Basically, the 'if' conditional is confusing me. I know that the loops prior will be called n^2 times, but the third loop is only called when $i*j >= n$. How would I write the third summation to account for this, and then evaluate the overall loop's time complexity?






summation







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is often context dependent. If, say, you wanted to iterate over the entries $a_{ij}$ of an $n times n$ matrix, then the first two loops in your code sample could be adequately expressed as $$sum_{ij geq n} a_{ij}$$. Perhaps if you provided more context about why you have this problem we will be able to provide more helpful answers.
    $endgroup$
    – Brian
    Mar 26 at 1:18










  • $begingroup$
    You want to convert the nested loops into a summation, but then you say you want to evaluate the time complexity of your code, which does not require you to convert the loops into a summation, and, in fact, does not require you to find what value sum has after the loops.
    $endgroup$
    – Fabio Somenzi
    Mar 26 at 3:15
















1












$begingroup$


I am trying to convert instances of nested 'for' loops into a summation expression. The current code fragment I have is:



for i = 1 to n:

    for j = 1 to n:

        if (i*j >= n):

            for k = 1 to n:

                sum++

        endif


Basically, the 'if' conditional is confusing me. I know that the loops prior will be called n^2 times, but the third loop is only called when $i*j >= n$. How would I write the third summation to account for this, and then evaluate the overall loop's time complexity?






summation







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is often context dependent. If, say, you wanted to iterate over the entries $a_{ij}$ of an $n times n$ matrix, then the first two loops in your code sample could be adequately expressed as $$sum_{ij geq n} a_{ij}$$. Perhaps if you provided more context about why you have this problem we will be able to provide more helpful answers.
    $endgroup$
    – Brian
    Mar 26 at 1:18










  • $begingroup$
    You want to convert the nested loops into a summation, but then you say you want to evaluate the time complexity of your code, which does not require you to convert the loops into a summation, and, in fact, does not require you to find what value sum has after the loops.
    $endgroup$
    – Fabio Somenzi
    Mar 26 at 3:15














1












1








1


1



$begingroup$


I am trying to convert instances of nested 'for' loops into a summation expression. The current code fragment I have is:



for i = 1 to n:

    for j = 1 to n:

        if (i*j >= n):

            for k = 1 to n:

                sum++

        endif


Basically, the 'if' conditional is confusing me. I know that the loops prior will be called n^2 times, but the third loop is only called when $i*j >= n$. How would I write the third summation to account for this, and then evaluate the overall loop's time complexity?






summation







share|cite|improve this question









$endgroup$




I am trying to convert instances of nested 'for' loops into a summation expression. The current code fragment I have is:



for i = 1 to n:

    for j = 1 to n:

        if (i*j >= n):

            for k = 1 to n:

                sum++

        endif


Basically, the 'if' conditional is confusing me. I know that the loops prior will be called n^2 times, but the third loop is only called when $i*j >= n$. How would I write the third summation to account for this, and then evaluate the overall loop's time complexity?






summation




summation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 26 at 0:59









bpryanbpryan

83




83












  • $begingroup$
    This is often context dependent. If, say, you wanted to iterate over the entries $a_{ij}$ of an $n times n$ matrix, then the first two loops in your code sample could be adequately expressed as $$sum_{ij geq n} a_{ij}$$. Perhaps if you provided more context about why you have this problem we will be able to provide more helpful answers.
    $endgroup$
    – Brian
    Mar 26 at 1:18










  • $begingroup$
    You want to convert the nested loops into a summation, but then you say you want to evaluate the time complexity of your code, which does not require you to convert the loops into a summation, and, in fact, does not require you to find what value sum has after the loops.
    $endgroup$
    – Fabio Somenzi
    Mar 26 at 3:15


















  • $begingroup$
    This is often context dependent. If, say, you wanted to iterate over the entries $a_{ij}$ of an $n times n$ matrix, then the first two loops in your code sample could be adequately expressed as $$sum_{ij geq n} a_{ij}$$. Perhaps if you provided more context about why you have this problem we will be able to provide more helpful answers.
    $endgroup$
    – Brian
    Mar 26 at 1:18










  • $begingroup$
    You want to convert the nested loops into a summation, but then you say you want to evaluate the time complexity of your code, which does not require you to convert the loops into a summation, and, in fact, does not require you to find what value sum has after the loops.
    $endgroup$
    – Fabio Somenzi
    Mar 26 at 3:15
















$begingroup$
This is often context dependent. If, say, you wanted to iterate over the entries $a_{ij}$ of an $n times n$ matrix, then the first two loops in your code sample could be adequately expressed as $$sum_{ij geq n} a_{ij}$$. Perhaps if you provided more context about why you have this problem we will be able to provide more helpful answers.
$endgroup$
– Brian
Mar 26 at 1:18




$begingroup$
This is often context dependent. If, say, you wanted to iterate over the entries $a_{ij}$ of an $n times n$ matrix, then the first two loops in your code sample could be adequately expressed as $$sum_{ij geq n} a_{ij}$$. Perhaps if you provided more context about why you have this problem we will be able to provide more helpful answers.
$endgroup$
– Brian
Mar 26 at 1:18












$begingroup$
You want to convert the nested loops into a summation, but then you say you want to evaluate the time complexity of your code, which does not require you to convert the loops into a summation, and, in fact, does not require you to find what value sum has after the loops.
$endgroup$
– Fabio Somenzi
Mar 26 at 3:15




$begingroup$
You want to convert the nested loops into a summation, but then you say you want to evaluate the time complexity of your code, which does not require you to convert the loops into a summation, and, in fact, does not require you to find what value sum has after the loops.
$endgroup$
– Fabio Somenzi
Mar 26 at 3:15










3 Answers
3






active

oldest

votes


















1












$begingroup$

Hint:



The "for" cycle in $k$ is very easy to turn into something simpler...



As for the other parts, see if you can split the problem into easier steps. For example, what happens for $i=1$? And what happens for $i=2$? And $i=3$? And...






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$


    We obtain
    begin{align*}
    color{blue}{sum_{i=1}^n}&color{blue}{sum_{j=1}^n[i jgeq n]sum_{k=1}^n1}tag{1}\
    &=nsum_{i=1}^nsum_{j=1}^n[i jgeq n]\
    &=nleft(sum_{i=1}^nsum_{{j=leftlfloor n/i rightrfloor}atop{imid n}}^n1+sum_{i=1}^nsum_{{j=leftlfloor n/i rightrfloor+1}atop{inot mid n}}^n1right)tag{2}\
    &=nleft(sum_{i=1}^nsum_{j=leftlfloor n/i rightrfloor+1}^n1+sum_{{i=1}atop{imid n}}^n1right)\
    &=nleft(sum_{i=1}^nleft(n-leftlfloorfrac{n}{i}rightrfloorright)+tau(n)right)tag{3}\
    &=nleft(n^2-sum_{i=1}^nleftlfloorfrac{n}{i}rightrfloor+tau(n)right)\
    &,,color{blue}{=nleft(n^2-sum_{i=1}^{n-1}tau(i)right)}tag{4}\
    &,,color{blue}{sim n^3-n^2left(log n+2gamma-1right)+Oleft(n^{3/2}right)}tag{5}
    end{align*}




    Comment:




    • In (1) we use Iverson brackets to represent the condition $ijgeq n$.


    • In (2) we rewrite the expression using the floor function. Note the $pm 1$ technicality regarding the range of summation.


    • In (3) we simplify the inner sum of the left-hand term and use the $tau$-function which counts the number of divisors.


    • In (4) we use an identity stated in A006218 to get rid of the floor function.


    • In (5) we use an asymptotic expansion which is also stated in OEIS/A006218.







    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      At the basic level, for loops are also rewritable as conditionals. Anyways, on to the meat of the problem. The conditional can be removed if you replace 1 in the previous loop with ceil(n/i). So it can be rewritten as:



      for i = 1 to n:
      for j = ceil(n/i) to n:
      for k = 1 to n:
      sum++



      Then the inner 2 loops compress to:



      sum+=n*(n-ceil(n/i)+1)



      Which then gets called n times, but the sum can be wriiten as:



      $$nsum_{i=1}^{n}(n-lceilfrac{n}{i}rceil+1)$$



      As to the time complexity of the code as wriiten..., it'll take, okay I don't quite know that part.






      share|cite|improve this answer









      $endgroup$














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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        Hint:



        The "for" cycle in $k$ is very easy to turn into something simpler...



        As for the other parts, see if you can split the problem into easier steps. For example, what happens for $i=1$? And what happens for $i=2$? And $i=3$? And...






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Hint:



          The "for" cycle in $k$ is very easy to turn into something simpler...



          As for the other parts, see if you can split the problem into easier steps. For example, what happens for $i=1$? And what happens for $i=2$? And $i=3$? And...






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Hint:



            The "for" cycle in $k$ is very easy to turn into something simpler...



            As for the other parts, see if you can split the problem into easier steps. For example, what happens for $i=1$? And what happens for $i=2$? And $i=3$? And...






            share|cite|improve this answer









            $endgroup$



            Hint:



            The "for" cycle in $k$ is very easy to turn into something simpler...



            As for the other parts, see if you can split the problem into easier steps. For example, what happens for $i=1$? And what happens for $i=2$? And $i=3$? And...







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 26 at 1:18









            ErtxiemErtxiem

            941212




            941212























                0












                $begingroup$


                We obtain
                begin{align*}
                color{blue}{sum_{i=1}^n}&color{blue}{sum_{j=1}^n[i jgeq n]sum_{k=1}^n1}tag{1}\
                &=nsum_{i=1}^nsum_{j=1}^n[i jgeq n]\
                &=nleft(sum_{i=1}^nsum_{{j=leftlfloor n/i rightrfloor}atop{imid n}}^n1+sum_{i=1}^nsum_{{j=leftlfloor n/i rightrfloor+1}atop{inot mid n}}^n1right)tag{2}\
                &=nleft(sum_{i=1}^nsum_{j=leftlfloor n/i rightrfloor+1}^n1+sum_{{i=1}atop{imid n}}^n1right)\
                &=nleft(sum_{i=1}^nleft(n-leftlfloorfrac{n}{i}rightrfloorright)+tau(n)right)tag{3}\
                &=nleft(n^2-sum_{i=1}^nleftlfloorfrac{n}{i}rightrfloor+tau(n)right)\
                &,,color{blue}{=nleft(n^2-sum_{i=1}^{n-1}tau(i)right)}tag{4}\
                &,,color{blue}{sim n^3-n^2left(log n+2gamma-1right)+Oleft(n^{3/2}right)}tag{5}
                end{align*}




                Comment:




                • In (1) we use Iverson brackets to represent the condition $ijgeq n$.


                • In (2) we rewrite the expression using the floor function. Note the $pm 1$ technicality regarding the range of summation.


                • In (3) we simplify the inner sum of the left-hand term and use the $tau$-function which counts the number of divisors.


                • In (4) we use an identity stated in A006218 to get rid of the floor function.


                • In (5) we use an asymptotic expansion which is also stated in OEIS/A006218.







                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$


                  We obtain
                  begin{align*}
                  color{blue}{sum_{i=1}^n}&color{blue}{sum_{j=1}^n[i jgeq n]sum_{k=1}^n1}tag{1}\
                  &=nsum_{i=1}^nsum_{j=1}^n[i jgeq n]\
                  &=nleft(sum_{i=1}^nsum_{{j=leftlfloor n/i rightrfloor}atop{imid n}}^n1+sum_{i=1}^nsum_{{j=leftlfloor n/i rightrfloor+1}atop{inot mid n}}^n1right)tag{2}\
                  &=nleft(sum_{i=1}^nsum_{j=leftlfloor n/i rightrfloor+1}^n1+sum_{{i=1}atop{imid n}}^n1right)\
                  &=nleft(sum_{i=1}^nleft(n-leftlfloorfrac{n}{i}rightrfloorright)+tau(n)right)tag{3}\
                  &=nleft(n^2-sum_{i=1}^nleftlfloorfrac{n}{i}rightrfloor+tau(n)right)\
                  &,,color{blue}{=nleft(n^2-sum_{i=1}^{n-1}tau(i)right)}tag{4}\
                  &,,color{blue}{sim n^3-n^2left(log n+2gamma-1right)+Oleft(n^{3/2}right)}tag{5}
                  end{align*}




                  Comment:




                  • In (1) we use Iverson brackets to represent the condition $ijgeq n$.


                  • In (2) we rewrite the expression using the floor function. Note the $pm 1$ technicality regarding the range of summation.


                  • In (3) we simplify the inner sum of the left-hand term and use the $tau$-function which counts the number of divisors.


                  • In (4) we use an identity stated in A006218 to get rid of the floor function.


                  • In (5) we use an asymptotic expansion which is also stated in OEIS/A006218.







                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$


                    We obtain
                    begin{align*}
                    color{blue}{sum_{i=1}^n}&color{blue}{sum_{j=1}^n[i jgeq n]sum_{k=1}^n1}tag{1}\
                    &=nsum_{i=1}^nsum_{j=1}^n[i jgeq n]\
                    &=nleft(sum_{i=1}^nsum_{{j=leftlfloor n/i rightrfloor}atop{imid n}}^n1+sum_{i=1}^nsum_{{j=leftlfloor n/i rightrfloor+1}atop{inot mid n}}^n1right)tag{2}\
                    &=nleft(sum_{i=1}^nsum_{j=leftlfloor n/i rightrfloor+1}^n1+sum_{{i=1}atop{imid n}}^n1right)\
                    &=nleft(sum_{i=1}^nleft(n-leftlfloorfrac{n}{i}rightrfloorright)+tau(n)right)tag{3}\
                    &=nleft(n^2-sum_{i=1}^nleftlfloorfrac{n}{i}rightrfloor+tau(n)right)\
                    &,,color{blue}{=nleft(n^2-sum_{i=1}^{n-1}tau(i)right)}tag{4}\
                    &,,color{blue}{sim n^3-n^2left(log n+2gamma-1right)+Oleft(n^{3/2}right)}tag{5}
                    end{align*}




                    Comment:




                    • In (1) we use Iverson brackets to represent the condition $ijgeq n$.


                    • In (2) we rewrite the expression using the floor function. Note the $pm 1$ technicality regarding the range of summation.


                    • In (3) we simplify the inner sum of the left-hand term and use the $tau$-function which counts the number of divisors.


                    • In (4) we use an identity stated in A006218 to get rid of the floor function.


                    • In (5) we use an asymptotic expansion which is also stated in OEIS/A006218.







                    share|cite|improve this answer









                    $endgroup$




                    We obtain
                    begin{align*}
                    color{blue}{sum_{i=1}^n}&color{blue}{sum_{j=1}^n[i jgeq n]sum_{k=1}^n1}tag{1}\
                    &=nsum_{i=1}^nsum_{j=1}^n[i jgeq n]\
                    &=nleft(sum_{i=1}^nsum_{{j=leftlfloor n/i rightrfloor}atop{imid n}}^n1+sum_{i=1}^nsum_{{j=leftlfloor n/i rightrfloor+1}atop{inot mid n}}^n1right)tag{2}\
                    &=nleft(sum_{i=1}^nsum_{j=leftlfloor n/i rightrfloor+1}^n1+sum_{{i=1}atop{imid n}}^n1right)\
                    &=nleft(sum_{i=1}^nleft(n-leftlfloorfrac{n}{i}rightrfloorright)+tau(n)right)tag{3}\
                    &=nleft(n^2-sum_{i=1}^nleftlfloorfrac{n}{i}rightrfloor+tau(n)right)\
                    &,,color{blue}{=nleft(n^2-sum_{i=1}^{n-1}tau(i)right)}tag{4}\
                    &,,color{blue}{sim n^3-n^2left(log n+2gamma-1right)+Oleft(n^{3/2}right)}tag{5}
                    end{align*}




                    Comment:




                    • In (1) we use Iverson brackets to represent the condition $ijgeq n$.


                    • In (2) we rewrite the expression using the floor function. Note the $pm 1$ technicality regarding the range of summation.


                    • In (3) we simplify the inner sum of the left-hand term and use the $tau$-function which counts the number of divisors.


                    • In (4) we use an identity stated in A006218 to get rid of the floor function.


                    • In (5) we use an asymptotic expansion which is also stated in OEIS/A006218.








                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 26 at 12:58









                    Markus ScheuerMarkus Scheuer

                    64.7k460154




                    64.7k460154























                        0












                        $begingroup$

                        At the basic level, for loops are also rewritable as conditionals. Anyways, on to the meat of the problem. The conditional can be removed if you replace 1 in the previous loop with ceil(n/i). So it can be rewritten as:



                        for i = 1 to n:
                        for j = ceil(n/i) to n:
                        for k = 1 to n:
                        sum++



                        Then the inner 2 loops compress to:



                        sum+=n*(n-ceil(n/i)+1)



                        Which then gets called n times, but the sum can be wriiten as:



                        $$nsum_{i=1}^{n}(n-lceilfrac{n}{i}rceil+1)$$



                        As to the time complexity of the code as wriiten..., it'll take, okay I don't quite know that part.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          At the basic level, for loops are also rewritable as conditionals. Anyways, on to the meat of the problem. The conditional can be removed if you replace 1 in the previous loop with ceil(n/i). So it can be rewritten as:



                          for i = 1 to n:
                          for j = ceil(n/i) to n:
                          for k = 1 to n:
                          sum++



                          Then the inner 2 loops compress to:



                          sum+=n*(n-ceil(n/i)+1)



                          Which then gets called n times, but the sum can be wriiten as:



                          $$nsum_{i=1}^{n}(n-lceilfrac{n}{i}rceil+1)$$



                          As to the time complexity of the code as wriiten..., it'll take, okay I don't quite know that part.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            At the basic level, for loops are also rewritable as conditionals. Anyways, on to the meat of the problem. The conditional can be removed if you replace 1 in the previous loop with ceil(n/i). So it can be rewritten as:



                            for i = 1 to n:
                            for j = ceil(n/i) to n:
                            for k = 1 to n:
                            sum++



                            Then the inner 2 loops compress to:



                            sum+=n*(n-ceil(n/i)+1)



                            Which then gets called n times, but the sum can be wriiten as:



                            $$nsum_{i=1}^{n}(n-lceilfrac{n}{i}rceil+1)$$



                            As to the time complexity of the code as wriiten..., it'll take, okay I don't quite know that part.






                            share|cite|improve this answer









                            $endgroup$



                            At the basic level, for loops are also rewritable as conditionals. Anyways, on to the meat of the problem. The conditional can be removed if you replace 1 in the previous loop with ceil(n/i). So it can be rewritten as:



                            for i = 1 to n:
                            for j = ceil(n/i) to n:
                            for k = 1 to n:
                            sum++



                            Then the inner 2 loops compress to:



                            sum+=n*(n-ceil(n/i)+1)



                            Which then gets called n times, but the sum can be wriiten as:



                            $$nsum_{i=1}^{n}(n-lceilfrac{n}{i}rceil+1)$$



                            As to the time complexity of the code as wriiten..., it'll take, okay I don't quite know that part.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 27 at 16:25









                            Roddy MacPheeRoddy MacPhee

                            927118




                            927118






























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