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How do I convert a simple algorithm to a mathematical notation?
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I am trying to write a simple algorithm given below in mathematical notation. I wrote the formula up to a certain point, but I have no idea of restrictions. For example, I used the NULL statement even though it was not "else". It also needs a "break" statement or equivalent.
var thr = 0.5;
var M = 100;
var sum = 0;
var T = 0;
for i 0 to G
sum = sum + i;
if sum/M >= thr then
T = i;
break;
return T;
My math notations; I defined "sum" in equation as follows.
$$sum = sum_j=0^ij$$
(Eq.1)$$T = sum_i=0^Gf=
begincases
i & fracsum_j=0^ijMgeq thr\
mathrmNULL & textotherwise
endcases$$
(Eq.2)$$T = sum_i=0,fneq mathrmNULL^Gf=
begincases
i & fracsum_j=0^ijMgeq thr\
mathrmNULL & textotherwise
endcases$$
Thank you very much for your help already.
summation conditional-expectation computational-mathematics
asked Jul 27 at 3:57
Abdullah Elen
1235
add a comment |Â
up vote
4
down vote
favorite
I am trying to write a simple algorithm given below in mathematical notation. I wrote the formula up to a certain point, but I have no idea of restrictions. For example, I used the NULL statement even though it was not "else". It also needs a "break" statement or equivalent.
var thr = 0.5;
var M = 100;
var sum = 0;
var T = 0;
for i 0 to G
sum = sum + i;
if sum/M >= thr then
T = i;
break;
return T;
My math notations; I defined "sum" in equation as follows.
$$sum = sum_j=0^ij$$
(Eq.1)$$T = sum_i=0^Gf=
begincases
i & fracsum_j=0^ijMgeq thr\
mathrmNULL & textotherwise
endcases$$
(Eq.2)$$T = sum_i=0,fneq mathrmNULL^Gf=
begincases
i & fracsum_j=0^ijMgeq thr\
mathrmNULL & textotherwise
endcases$$
Thank you very much for your help already.
summation conditional-expectation computational-mathematics
asked Jul 27 at 3:57
Abdullah Elen
1235
What is the value you wish to get out?
– Q the Platypus
Jul 27 at 4:02
I want to get the value of the "i" that provides the condition. This is actually the P-tile algorithm. I use histogram information for this, but I've assigned constant variables to simplify the syntax.
– Abdullah Elen
Jul 27 at 4:10
what is G ? $$
– mercio
Jul 27 at 7:37
G can be selected as a number greater than 10. That is, it's big enough to provide the "if" condition.
– Abdullah Elen
Jul 27 at 22:21
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am trying to write a simple algorithm given below in mathematical notation. I wrote the formula up to a certain point, but I have no idea of restrictions. For example, I used the NULL statement even though it was not "else". It also needs a "break" statement or equivalent.
var thr = 0.5;
var M = 100;
var sum = 0;
var T = 0;
for i 0 to G
sum = sum + i;
if sum/M >= thr then
T = i;
break;
return T;
My math notations; I defined "sum" in equation as follows.
$$sum = sum_j=0^ij$$
(Eq.1)$$T = sum_i=0^Gf=
begincases
i & fracsum_j=0^ijMgeq thr\
mathrmNULL & textotherwise
endcases$$
(Eq.2)$$T = sum_i=0,fneq mathrmNULL^Gf=
begincases
i & fracsum_j=0^ijMgeq thr\
mathrmNULL & textotherwise
endcases$$
Thank you very much for your help already.
summation conditional-expectation computational-mathematics
asked Jul 27 at 3:57
Abdullah Elen
1235
I am trying to write a simple algorithm given below in mathematical notation. I wrote the formula up to a certain point, but I have no idea of restrictions. For example, I used the NULL statement even though it was not "else". It also needs a "break" statement or equivalent.
var thr = 0.5;
var M = 100;
var sum = 0;
var T = 0;
for i 0 to G
sum = sum + i;
if sum/M >= thr then
T = i;
break;
return T;
My math notations; I defined "sum" in equation as follows.
$$sum = sum_j=0^ij$$
(Eq.1)$$T = sum_i=0^Gf=
begincases
i & fracsum_j=0^ijMgeq thr\
mathrmNULL & textotherwise
endcases$$
(Eq.2)$$T = sum_i=0,fneq mathrmNULL^Gf=
begincases
i & fracsum_j=0^ijMgeq thr\
mathrmNULL & textotherwise
endcases$$
Thank you very much for your help already.
summation conditional-expectation computational-mathematics
asked Jul 27 at 3:57
Abdullah Elen
1235
edited Jul 27 at 4:56
asked Jul 27 at 3:57
Abdullah Elen
1235
asked Jul 27 at 3:57
Abdullah Elen
1235
asked Jul 27 at 3:57
Abdullah Elen
1235
1235
What is the value you wish to get out?
– Q the Platypus
Jul 27 at 4:02
I want to get the value of the "i" that provides the condition. This is actually the P-tile algorithm. I use histogram information for this, but I've assigned constant variables to simplify the syntax.
– Abdullah Elen
Jul 27 at 4:10
what is G ? $$
– mercio
Jul 27 at 7:37
G can be selected as a number greater than 10. That is, it's big enough to provide the "if" condition.
– Abdullah Elen
Jul 27 at 22:21
add a comment |Â
What is the value you wish to get out?
– Q the Platypus
Jul 27 at 4:02
I want to get the value of the "i" that provides the condition. This is actually the P-tile algorithm. I use histogram information for this, but I've assigned constant variables to simplify the syntax.
– Abdullah Elen
Jul 27 at 4:10
what is G ? $$
– mercio
Jul 27 at 7:37
G can be selected as a number greater than 10. That is, it's big enough to provide the "if" condition.
– Abdullah Elen
Jul 27 at 22:21
What is the value you wish to get out?
– Q the Platypus
Jul 27 at 4:02
What is the value you wish to get out?
– Q the Platypus
Jul 27 at 4:02
I want to get the value of the "i" that provides the condition. This is actually the P-tile algorithm. I use histogram information for this, but I've assigned constant variables to simplify the syntax.
– Abdullah Elen
Jul 27 at 4:10
I want to get the value of the "i" that provides the condition. This is actually the P-tile algorithm. I use histogram information for this, but I've assigned constant variables to simplify the syntax.
– Abdullah Elen
Jul 27 at 4:10
what is G ? $$
– mercio
Jul 27 at 7:37
what is G ? $$
– mercio
Jul 27 at 7:37
G can be selected as a number greater than 10. That is, it's big enough to provide the "if" condition.
– Abdullah Elen
Jul 27 at 22:21
G can be selected as a number greater than 10. That is, it's big enough to provide the "if" condition.
– Abdullah Elen
Jul 27 at 22:21
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
The idea here is that $T$ is equal to $i$, where $i$ is the first index in the loop for which $verb|sum|,/M ge verb|thr|$. This can be written in a formula as follows:
$$
T = begincases0 &tfrac1Mtextstylesum_j=0^Gj<verb|thr|\minimid iin 0,1,dots,G,tfrac1Mtextstylesum_j=0^ijge verb & textOtherwise
endcases
$$
So you take the set of $i$ for which the loop would have stopped by the $i^textth$ iteration, and declare $T$ to be the smallest (minimum) of this set.
answered Jul 27 at 4:24
Mike Earnest
14.9k11644
Thank you very much, Mike.
– Abdullah Elen
Jul 27 at 4:37
1
@AbdullahELEN By the way, see my edit, I originally forgot about what would happen if the loop ended before the sum got large enough.
– Mike Earnest
Jul 27 at 4:41
It's exactly what I wanted. Thank you again :)
– Abdullah Elen
Jul 27 at 4:47
I think we have a little problem. $T$ needs to be equal to $i$. As a result, the function must return $i$ value. Can i define it as follows? $displaystyle operatorname*argmin_i i$
– Abdullah Elen
Jul 29 at 1:27
add a comment |Â
up vote
1
down vote
You need to think about what the code is doing, not just translate the code. Wehn you write $$sum = sum_j=0^ii$$
there is a small error that the iteration variable is $j$ so you should write
$$sum = sum_j=0^ij$$
More importantly, $sum$ is changing through the computation, so it would be better to write $$sum(i) = sum_j=0^ij$$
then you should perform the sum and write
$$sum(i)=frac 12i(i+1)$$
Then $T$ is the $i$ for which $sum$ exceeds $Mcdot thr$. Some analysis will show that $$T=leftlceilfrac 12(-1+sqrt1+8cdot Mcdot thr) right rceil$$
which is much more useful for someone trying to understand your code.
This last shows a way to get the result without iteration as well.
answered Jul 27 at 4:13
Ross Millikan
275k21186351
First of all thank you for your quick response. Oh yes. I wrote i instead of j and am going to fix it immediately. However, the "sum" statement does not need to be stored in stages. You can see this in the algorithm. I could express this mathematically.
– Abdullah Elen
Jul 27 at 4:21
My point is that you are trying to find the $i$ where the sum exceeds $Mcdot thr$. That is the first thing you should say-it is the spec for the algorithm. Then you don't compute sum all the way up to $G$, which just needs to be some large number. You compute it in stages and compare at each stage. Whether you store it or not is not important, the important thing is the value as a function of $i$. Writing it as a sum shows how you do it, writing it as $frac 12i(i+1)$ shows what it is.
– Ross Millikan
Jul 27 at 4:29
Yes. If you look at "if conditional", you can see the "break" statement there. For this reason, the loop does not operate until G.
– Abdullah Elen
Jul 27 at 4:36
Yes, I understood that. You asked about expressing the algorithm mathematically but you continue to talk about how the code works. The important thing is to say what value of $T$ is returned based on the input parameters. Mike Earnest's expression is very close to your code. I think it is much less useful to a reader.
– Ross Millikan
Jul 27 at 4:47
You're right. Not everybody has coding knowledge. For this reason, I should have written with a clearer expression.
– Abdullah Elen
Jul 27 at 4:53
 |Â
show 1 more comment
up vote
1
down vote
Your code is iterating $i$ from $0$ to $G$ and cumulatively incrementing the $textsum$ by that iterator, breaking when $textsum/Mgeq textthr$ (or $textsumgeq Mtimestextthr$ when rounding is not an issue ). Only when breaking is $T$ set to the iterator, otherwise it is left at zero.
Since $sum_i=0^n i =tfracn,(n+1)2$ , and the values of $M$ and $textthr$ are hard coded constants: $100$ and $0.5$, that code should result in
$$beginalignT & =begincases 0 &:& G,(G+1)< 100\ mintin0..G: t(t+1)geqslant 100 &:&textotherwise endcases \ &= begincases 0 &:& G<10\ 10 &:& Ggeq 10endcasesendalign$$
answered Jul 27 at 5:03
Graham Kemp
80k43275
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The idea here is that $T$ is equal to $i$, where $i$ is the first index in the loop for which $verb|sum|,/M ge verb|thr|$. This can be written in a formula as follows:
$$
T = begincases0 &tfrac1Mtextstylesum_j=0^Gj<verb|thr|\minimid iin 0,1,dots,G,tfrac1Mtextstylesum_j=0^ijge verb & textOtherwise
endcases
$$
So you take the set of $i$ for which the loop would have stopped by the $i^textth$ iteration, and declare $T$ to be the smallest (minimum) of this set.
answered Jul 27 at 4:24
Mike Earnest
14.9k11644
Thank you very much, Mike.
– Abdullah Elen
Jul 27 at 4:37
1
@AbdullahELEN By the way, see my edit, I originally forgot about what would happen if the loop ended before the sum got large enough.
– Mike Earnest
Jul 27 at 4:41
It's exactly what I wanted. Thank you again :)
– Abdullah Elen
Jul 27 at 4:47
I think we have a little problem. $T$ needs to be equal to $i$. As a result, the function must return $i$ value. Can i define it as follows? $displaystyle operatorname*argmin_i i$
– Abdullah Elen
Jul 29 at 1:27
add a comment |Â
up vote
3
down vote
accepted
The idea here is that $T$ is equal to $i$, where $i$ is the first index in the loop for which $verb|sum|,/M ge verb|thr|$. This can be written in a formula as follows:
$$
T = begincases0 &tfrac1Mtextstylesum_j=0^Gj<verb|thr|\minimid iin 0,1,dots,G,tfrac1Mtextstylesum_j=0^ijge verb & textOtherwise
endcases
$$
So you take the set of $i$ for which the loop would have stopped by the $i^textth$ iteration, and declare $T$ to be the smallest (minimum) of this set.
answered Jul 27 at 4:24
Mike Earnest
14.9k11644
Thank you very much, Mike.
– Abdullah Elen
Jul 27 at 4:37
1
@AbdullahELEN By the way, see my edit, I originally forgot about what would happen if the loop ended before the sum got large enough.
– Mike Earnest
Jul 27 at 4:41
It's exactly what I wanted. Thank you again :)
– Abdullah Elen
Jul 27 at 4:47
I think we have a little problem. $T$ needs to be equal to $i$. As a result, the function must return $i$ value. Can i define it as follows? $displaystyle operatorname*argmin_i i$
– Abdullah Elen
Jul 29 at 1:27
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The idea here is that $T$ is equal to $i$, where $i$ is the first index in the loop for which $verb|sum|,/M ge verb|thr|$. This can be written in a formula as follows:
$$
T = begincases0 &tfrac1Mtextstylesum_j=0^Gj<verb|thr|\minimid iin 0,1,dots,G,tfrac1Mtextstylesum_j=0^ijge verb & textOtherwise
endcases
$$
So you take the set of $i$ for which the loop would have stopped by the $i^textth$ iteration, and declare $T$ to be the smallest (minimum) of this set.
answered Jul 27 at 4:24
Mike Earnest
14.9k11644
The idea here is that $T$ is equal to $i$, where $i$ is the first index in the loop for which $verb|sum|,/M ge verb|thr|$. This can be written in a formula as follows:
$$
T = begincases0 &tfrac1Mtextstylesum_j=0^Gj<verb|thr|\minimid iin 0,1,dots,G,tfrac1Mtextstylesum_j=0^ijge verb & textOtherwise
endcases
$$
So you take the set of $i$ for which the loop would have stopped by the $i^textth$ iteration, and declare $T$ to be the smallest (minimum) of this set.
answered Jul 27 at 4:24
Mike Earnest
14.9k11644
edited Jul 27 at 4:40
answered Jul 27 at 4:24
Mike Earnest
14.9k11644
answered Jul 27 at 4:24
Mike Earnest
14.9k11644
answered Jul 27 at 4:24
Mike Earnest
14.9k11644
14.9k11644
Thank you very much, Mike.
– Abdullah Elen
Jul 27 at 4:37
1
@AbdullahELEN By the way, see my edit, I originally forgot about what would happen if the loop ended before the sum got large enough.
– Mike Earnest
Jul 27 at 4:41
It's exactly what I wanted. Thank you again :)
– Abdullah Elen
Jul 27 at 4:47
I think we have a little problem. $T$ needs to be equal to $i$. As a result, the function must return $i$ value. Can i define it as follows? $displaystyle operatorname*argmin_i i$
– Abdullah Elen
Jul 29 at 1:27
add a comment |Â
Thank you very much, Mike.
– Abdullah Elen
Jul 27 at 4:37
1
@AbdullahELEN By the way, see my edit, I originally forgot about what would happen if the loop ended before the sum got large enough.
– Mike Earnest
Jul 27 at 4:41
It's exactly what I wanted. Thank you again :)
– Abdullah Elen
Jul 27 at 4:47
I think we have a little problem. $T$ needs to be equal to $i$. As a result, the function must return $i$ value. Can i define it as follows? $displaystyle operatorname*argmin_i i$
– Abdullah Elen
Jul 29 at 1:27
Thank you very much, Mike.
– Abdullah Elen
Jul 27 at 4:37
Thank you very much, Mike.
– Abdullah Elen
Jul 27 at 4:37
1
1
@AbdullahELEN By the way, see my edit, I originally forgot about what would happen if the loop ended before the sum got large enough.
– Mike Earnest
Jul 27 at 4:41
@AbdullahELEN By the way, see my edit, I originally forgot about what would happen if the loop ended before the sum got large enough.
– Mike Earnest
Jul 27 at 4:41
It's exactly what I wanted. Thank you again :)
– Abdullah Elen
Jul 27 at 4:47
It's exactly what I wanted. Thank you again :)
– Abdullah Elen
Jul 27 at 4:47
I think we have a little problem. $T$ needs to be equal to $i$. As a result, the function must return $i$ value. Can i define it as follows? $displaystyle operatorname*argmin_i i$
– Abdullah Elen
Jul 29 at 1:27
I think we have a little problem. $T$ needs to be equal to $i$. As a result, the function must return $i$ value. Can i define it as follows? $displaystyle operatorname*argmin_i i$
– Abdullah Elen
Jul 29 at 1:27
add a comment |Â
up vote
1
down vote
You need to think about what the code is doing, not just translate the code. Wehn you write $$sum = sum_j=0^ii$$
there is a small error that the iteration variable is $j$ so you should write
$$sum = sum_j=0^ij$$
More importantly, $sum$ is changing through the computation, so it would be better to write $$sum(i) = sum_j=0^ij$$
then you should perform the sum and write
$$sum(i)=frac 12i(i+1)$$
Then $T$ is the $i$ for which $sum$ exceeds $Mcdot thr$. Some analysis will show that $$T=leftlceilfrac 12(-1+sqrt1+8cdot Mcdot thr) right rceil$$
which is much more useful for someone trying to understand your code.
This last shows a way to get the result without iteration as well.
answered Jul 27 at 4:13
Ross Millikan
275k21186351
First of all thank you for your quick response. Oh yes. I wrote i instead of j and am going to fix it immediately. However, the "sum" statement does not need to be stored in stages. You can see this in the algorithm. I could express this mathematically.
– Abdullah Elen
Jul 27 at 4:21
My point is that you are trying to find the $i$ where the sum exceeds $Mcdot thr$. That is the first thing you should say-it is the spec for the algorithm. Then you don't compute sum all the way up to $G$, which just needs to be some large number. You compute it in stages and compare at each stage. Whether you store it or not is not important, the important thing is the value as a function of $i$. Writing it as a sum shows how you do it, writing it as $frac 12i(i+1)$ shows what it is.
– Ross Millikan
Jul 27 at 4:29
Yes. If you look at "if conditional", you can see the "break" statement there. For this reason, the loop does not operate until G.
– Abdullah Elen
Jul 27 at 4:36
Yes, I understood that. You asked about expressing the algorithm mathematically but you continue to talk about how the code works. The important thing is to say what value of $T$ is returned based on the input parameters. Mike Earnest's expression is very close to your code. I think it is much less useful to a reader.
– Ross Millikan
Jul 27 at 4:47
You're right. Not everybody has coding knowledge. For this reason, I should have written with a clearer expression.
– Abdullah Elen
Jul 27 at 4:53
 |Â
show 1 more comment
up vote
1
down vote
You need to think about what the code is doing, not just translate the code. Wehn you write $$sum = sum_j=0^ii$$
there is a small error that the iteration variable is $j$ so you should write
$$sum = sum_j=0^ij$$
More importantly, $sum$ is changing through the computation, so it would be better to write $$sum(i) = sum_j=0^ij$$
then you should perform the sum and write
$$sum(i)=frac 12i(i+1)$$
Then $T$ is the $i$ for which $sum$ exceeds $Mcdot thr$. Some analysis will show that $$T=leftlceilfrac 12(-1+sqrt1+8cdot Mcdot thr) right rceil$$
which is much more useful for someone trying to understand your code.
This last shows a way to get the result without iteration as well.
answered Jul 27 at 4:13
Ross Millikan
275k21186351
First of all thank you for your quick response. Oh yes. I wrote i instead of j and am going to fix it immediately. However, the "sum" statement does not need to be stored in stages. You can see this in the algorithm. I could express this mathematically.
– Abdullah Elen
Jul 27 at 4:21
My point is that you are trying to find the $i$ where the sum exceeds $Mcdot thr$. That is the first thing you should say-it is the spec for the algorithm. Then you don't compute sum all the way up to $G$, which just needs to be some large number. You compute it in stages and compare at each stage. Whether you store it or not is not important, the important thing is the value as a function of $i$. Writing it as a sum shows how you do it, writing it as $frac 12i(i+1)$ shows what it is.
– Ross Millikan
Jul 27 at 4:29
Yes. If you look at "if conditional", you can see the "break" statement there. For this reason, the loop does not operate until G.
– Abdullah Elen
Jul 27 at 4:36
Yes, I understood that. You asked about expressing the algorithm mathematically but you continue to talk about how the code works. The important thing is to say what value of $T$ is returned based on the input parameters. Mike Earnest's expression is very close to your code. I think it is much less useful to a reader.
– Ross Millikan
Jul 27 at 4:47
You're right. Not everybody has coding knowledge. For this reason, I should have written with a clearer expression.
– Abdullah Elen
Jul 27 at 4:53
 |Â
show 1 more comment
up vote
1
down vote
up vote
1
down vote
You need to think about what the code is doing, not just translate the code. Wehn you write $$sum = sum_j=0^ii$$
there is a small error that the iteration variable is $j$ so you should write
$$sum = sum_j=0^ij$$
More importantly, $sum$ is changing through the computation, so it would be better to write $$sum(i) = sum_j=0^ij$$
then you should perform the sum and write
$$sum(i)=frac 12i(i+1)$$
Then $T$ is the $i$ for which $sum$ exceeds $Mcdot thr$. Some analysis will show that $$T=leftlceilfrac 12(-1+sqrt1+8cdot Mcdot thr) right rceil$$
which is much more useful for someone trying to understand your code.
This last shows a way to get the result without iteration as well.
answered Jul 27 at 4:13
Ross Millikan
275k21186351
You need to think about what the code is doing, not just translate the code. Wehn you write $$sum = sum_j=0^ii$$
there is a small error that the iteration variable is $j$ so you should write
$$sum = sum_j=0^ij$$
More importantly, $sum$ is changing through the computation, so it would be better to write $$sum(i) = sum_j=0^ij$$
then you should perform the sum and write
$$sum(i)=frac 12i(i+1)$$
Then $T$ is the $i$ for which $sum$ exceeds $Mcdot thr$. Some analysis will show that $$T=leftlceilfrac 12(-1+sqrt1+8cdot Mcdot thr) right rceil$$
which is much more useful for someone trying to understand your code.
This last shows a way to get the result without iteration as well.
answered Jul 27 at 4:13
Ross Millikan
275k21186351
answered Jul 27 at 4:13
Ross Millikan
275k21186351
answered Jul 27 at 4:13
Ross Millikan
275k21186351
answered Jul 27 at 4:13
Ross Millikan
275k21186351
275k21186351
First of all thank you for your quick response. Oh yes. I wrote i instead of j and am going to fix it immediately. However, the "sum" statement does not need to be stored in stages. You can see this in the algorithm. I could express this mathematically.
– Abdullah Elen
Jul 27 at 4:21
My point is that you are trying to find the $i$ where the sum exceeds $Mcdot thr$. That is the first thing you should say-it is the spec for the algorithm. Then you don't compute sum all the way up to $G$, which just needs to be some large number. You compute it in stages and compare at each stage. Whether you store it or not is not important, the important thing is the value as a function of $i$. Writing it as a sum shows how you do it, writing it as $frac 12i(i+1)$ shows what it is.
– Ross Millikan
Jul 27 at 4:29
Yes. If you look at "if conditional", you can see the "break" statement there. For this reason, the loop does not operate until G.
– Abdullah Elen
Jul 27 at 4:36
Yes, I understood that. You asked about expressing the algorithm mathematically but you continue to talk about how the code works. The important thing is to say what value of $T$ is returned based on the input parameters. Mike Earnest's expression is very close to your code. I think it is much less useful to a reader.
– Ross Millikan
Jul 27 at 4:47
You're right. Not everybody has coding knowledge. For this reason, I should have written with a clearer expression.
– Abdullah Elen
Jul 27 at 4:53
 |Â
show 1 more comment
First of all thank you for your quick response. Oh yes. I wrote i instead of j and am going to fix it immediately. However, the "sum" statement does not need to be stored in stages. You can see this in the algorithm. I could express this mathematically.
– Abdullah Elen
Jul 27 at 4:21
My point is that you are trying to find the $i$ where the sum exceeds $Mcdot thr$. That is the first thing you should say-it is the spec for the algorithm. Then you don't compute sum all the way up to $G$, which just needs to be some large number. You compute it in stages and compare at each stage. Whether you store it or not is not important, the important thing is the value as a function of $i$. Writing it as a sum shows how you do it, writing it as $frac 12i(i+1)$ shows what it is.
– Ross Millikan
Jul 27 at 4:29
Yes. If you look at "if conditional", you can see the "break" statement there. For this reason, the loop does not operate until G.
– Abdullah Elen
Jul 27 at 4:36
Yes, I understood that. You asked about expressing the algorithm mathematically but you continue to talk about how the code works. The important thing is to say what value of $T$ is returned based on the input parameters. Mike Earnest's expression is very close to your code. I think it is much less useful to a reader.
– Ross Millikan
Jul 27 at 4:47
You're right. Not everybody has coding knowledge. For this reason, I should have written with a clearer expression.
– Abdullah Elen
Jul 27 at 4:53
First of all thank you for your quick response. Oh yes. I wrote i instead of j and am going to fix it immediately. However, the "sum" statement does not need to be stored in stages. You can see this in the algorithm. I could express this mathematically.
– Abdullah Elen
Jul 27 at 4:21
First of all thank you for your quick response. Oh yes. I wrote i instead of j and am going to fix it immediately. However, the "sum" statement does not need to be stored in stages. You can see this in the algorithm. I could express this mathematically.
– Abdullah Elen
Jul 27 at 4:21
My point is that you are trying to find the $i$ where the sum exceeds $Mcdot thr$. That is the first thing you should say-it is the spec for the algorithm. Then you don't compute sum all the way up to $G$, which just needs to be some large number. You compute it in stages and compare at each stage. Whether you store it or not is not important, the important thing is the value as a function of $i$. Writing it as a sum shows how you do it, writing it as $frac 12i(i+1)$ shows what it is.
– Ross Millikan
Jul 27 at 4:29
My point is that you are trying to find the $i$ where the sum exceeds $Mcdot thr$. That is the first thing you should say-it is the spec for the algorithm. Then you don't compute sum all the way up to $G$, which just needs to be some large number. You compute it in stages and compare at each stage. Whether you store it or not is not important, the important thing is the value as a function of $i$. Writing it as a sum shows how you do it, writing it as $frac 12i(i+1)$ shows what it is.
– Ross Millikan
Jul 27 at 4:29
Yes. If you look at "if conditional", you can see the "break" statement there. For this reason, the loop does not operate until G.
– Abdullah Elen
Jul 27 at 4:36
Yes. If you look at "if conditional", you can see the "break" statement there. For this reason, the loop does not operate until G.
– Abdullah Elen
Jul 27 at 4:36
Yes, I understood that. You asked about expressing the algorithm mathematically but you continue to talk about how the code works. The important thing is to say what value of $T$ is returned based on the input parameters. Mike Earnest's expression is very close to your code. I think it is much less useful to a reader.
– Ross Millikan
Jul 27 at 4:47
Yes, I understood that. You asked about expressing the algorithm mathematically but you continue to talk about how the code works. The important thing is to say what value of $T$ is returned based on the input parameters. Mike Earnest's expression is very close to your code. I think it is much less useful to a reader.
– Ross Millikan
Jul 27 at 4:47
You're right. Not everybody has coding knowledge. For this reason, I should have written with a clearer expression.
– Abdullah Elen
Jul 27 at 4:53
You're right. Not everybody has coding knowledge. For this reason, I should have written with a clearer expression.
– Abdullah Elen
Jul 27 at 4:53
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up vote
1
down vote
Your code is iterating $i$ from $0$ to $G$ and cumulatively incrementing the $textsum$ by that iterator, breaking when $textsum/Mgeq textthr$ (or $textsumgeq Mtimestextthr$ when rounding is not an issue ). Only when breaking is $T$ set to the iterator, otherwise it is left at zero.
Since $sum_i=0^n i =tfracn,(n+1)2$ , and the values of $M$ and $textthr$ are hard coded constants: $100$ and $0.5$, that code should result in
$$beginalignT & =begincases 0 &:& G,(G+1)< 100\ mintin0..G: t(t+1)geqslant 100 &:&textotherwise endcases \ &= begincases 0 &:& G<10\ 10 &:& Ggeq 10endcasesendalign$$
answered Jul 27 at 5:03
Graham Kemp
80k43275
add a comment |Â
up vote
1
down vote
Your code is iterating $i$ from $0$ to $G$ and cumulatively incrementing the $textsum$ by that iterator, breaking when $textsum/Mgeq textthr$ (or $textsumgeq Mtimestextthr$ when rounding is not an issue ). Only when breaking is $T$ set to the iterator, otherwise it is left at zero.
Since $sum_i=0^n i =tfracn,(n+1)2$ , and the values of $M$ and $textthr$ are hard coded constants: $100$ and $0.5$, that code should result in
$$beginalignT & =begincases 0 &:& G,(G+1)< 100\ mintin0..G: t(t+1)geqslant 100 &:&textotherwise endcases \ &= begincases 0 &:& G<10\ 10 &:& Ggeq 10endcasesendalign$$
answered Jul 27 at 5:03
Graham Kemp
80k43275
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your code is iterating $i$ from $0$ to $G$ and cumulatively incrementing the $textsum$ by that iterator, breaking when $textsum/Mgeq textthr$ (or $textsumgeq Mtimestextthr$ when rounding is not an issue ). Only when breaking is $T$ set to the iterator, otherwise it is left at zero.
Since $sum_i=0^n i =tfracn,(n+1)2$ , and the values of $M$ and $textthr$ are hard coded constants: $100$ and $0.5$, that code should result in
$$beginalignT & =begincases 0 &:& G,(G+1)< 100\ mintin0..G: t(t+1)geqslant 100 &:&textotherwise endcases \ &= begincases 0 &:& G<10\ 10 &:& Ggeq 10endcasesendalign$$
answered Jul 27 at 5:03
Graham Kemp
80k43275
Your code is iterating $i$ from $0$ to $G$ and cumulatively incrementing the $textsum$ by that iterator, breaking when $textsum/Mgeq textthr$ (or $textsumgeq Mtimestextthr$ when rounding is not an issue ). Only when breaking is $T$ set to the iterator, otherwise it is left at zero.
Since $sum_i=0^n i =tfracn,(n+1)2$ , and the values of $M$ and $textthr$ are hard coded constants: $100$ and $0.5$, that code should result in
$$beginalignT & =begincases 0 &:& G,(G+1)< 100\ mintin0..G: t(t+1)geqslant 100 &:&textotherwise endcases \ &= begincases 0 &:& G<10\ 10 &:& Ggeq 10endcasesendalign$$
answered Jul 27 at 5:03
Graham Kemp
80k43275
answered Jul 27 at 5:03
Graham Kemp
80k43275
answered Jul 27 at 5:03
Graham Kemp
80k43275
answered Jul 27 at 5:03
Graham Kemp
80k43275
80k43275
add a comment |Â
add a comment |Â
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What is the value you wish to get out?
– Q the Platypus
Jul 27 at 4:02
I want to get the value of the "i" that provides the condition. This is actually the P-tile algorithm. I use histogram information for this, but I've assigned constant variables to simplify the syntax.
– Abdullah Elen
Jul 27 at 4:10
what is G ? $$
– mercio
Jul 27 at 7:37
G can be selected as a number greater than 10. That is, it's big enough to provide the "if" condition.
– Abdullah Elen
Jul 27 at 22:21