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
$begingroup$
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
asked Mar 26 at 0:59
bpryanbpryan
83
$endgroup$
add a comment |
$begingroup$
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
asked Mar 26 at 0:59
bpryanbpryan
83
$endgroup$
$begingroup$
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 valuesumhas after the loops.
$endgroup$
– Fabio Somenzi
Mar 26 at 3:15
add a comment |
$begingroup$
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
asked Mar 26 at 0:59
bpryanbpryan
83
$endgroup$
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
asked Mar 26 at 0:59
bpryanbpryan
83
asked Mar 26 at 0:59
bpryanbpryan
83
asked Mar 26 at 0:59
bpryanbpryan
83
asked Mar 26 at 0:59
bpryanbpryan
83
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 valuesumhas after the loops.
$endgroup$
– Fabio Somenzi
Mar 26 at 3:15
add a comment |
$begingroup$
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 valuesumhas 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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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...
answered Mar 26 at 1:18
ErtxiemErtxiem
941212
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 26 at 12:58
Markus ScheuerMarkus Scheuer
64.7k460154
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 27 at 16:25
Roddy MacPheeRoddy MacPhee
927118
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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...
answered Mar 26 at 1:18
ErtxiemErtxiem
941212
$endgroup$
add a comment |
$begingroup$
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...
answered Mar 26 at 1:18
ErtxiemErtxiem
941212
$endgroup$
add a comment |
$begingroup$
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...
answered Mar 26 at 1:18
ErtxiemErtxiem
941212
$endgroup$
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...
answered Mar 26 at 1:18
ErtxiemErtxiem
941212
answered Mar 26 at 1:18
ErtxiemErtxiem
941212
answered Mar 26 at 1:18
ErtxiemErtxiem
941212
answered Mar 26 at 1:18
ErtxiemErtxiem
941212
941212
add a comment |
add a comment |
$begingroup$
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.
answered Mar 26 at 12:58
Markus ScheuerMarkus Scheuer
64.7k460154
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 26 at 12:58
Markus ScheuerMarkus Scheuer
64.7k460154
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 26 at 12:58
Markus ScheuerMarkus Scheuer
64.7k460154
$endgroup$
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.
answered Mar 26 at 12:58
Markus ScheuerMarkus Scheuer
64.7k460154
answered Mar 26 at 12:58
Markus ScheuerMarkus Scheuer
64.7k460154
answered Mar 26 at 12:58
Markus ScheuerMarkus Scheuer
64.7k460154
answered Mar 26 at 12:58
Markus ScheuerMarkus Scheuer
64.7k460154
64.7k460154
add a comment |
add a comment |
$begingroup$
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.
answered Mar 27 at 16:25
Roddy MacPheeRoddy MacPhee
927118
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 27 at 16:25
Roddy MacPheeRoddy MacPhee
927118
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 27 at 16:25
Roddy MacPheeRoddy MacPhee
927118
$endgroup$
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.
answered Mar 27 at 16:25
Roddy MacPheeRoddy MacPhee
927118
answered Mar 27 at 16:25
Roddy MacPheeRoddy MacPhee
927118
answered Mar 27 at 16:25
Roddy MacPheeRoddy MacPhee
927118
answered Mar 27 at 16:25
Roddy MacPheeRoddy MacPhee
927118
927118
add a comment |
add a comment |
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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.
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– Brian
Mar 26 at 1:18
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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
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– Fabio Somenzi
Mar 26 at 3:15