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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                                Fil:Tokke komm.svg

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                                Skildring Skildring Tokke komm.svg Norsk nynorsk: Våpen for Tokke kommune i Telemark fylke, godkjent ved kongeleg resolusjon 6. februar 1987. Motiv: På gull grunn ein gåande svart bjørn. [1] Tokke ligg i eit gamalt bjørneområde og bjørnen er eit viktig innslag i fleire lokale historier. Bjørnen kan òg symbolisere naturen og dyrelivet i kommunen. English: Coat of arms of the Norwegian municipality of Tokke, Telemark. Dato 6. februar 1987 Kjelde Original by Stein Davidsen, based upon image from caplex.no. Opphavsperson Eige arbeid Vektorgrafikken vart laga med Inkscape. Løyve (Gjenbruk av denne fila) Public domain Public domain false false This official Norwegian coat of arms is ineligible for copyright, but its usage is restricted according to Norwegian penal law, independent of the copyright status of the depiction shown here. Dette biletet er sett saman av eit flagg, eit våpen, eit sigill eller eit anna offisielt insignium . Bruken av slike symbol er avgrensa i fleire land. Desse r
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