Why aren't primality tests easily linear in time complexity?
Why don't we consider them as linear? I don't understand. You just have to check for factorization up to sqrt of n. So it's even faster than linear.
I assume it's not linear only if we compare the number of operations relative to the input in terms of binary representation. But why would we do so?
It seems to me wrong. The growth in calculation should be calculated compared to the number itself. Why do we compare it to the binary representation?
algorithm-analysis time-complexity runtime-analysis
edited 1 hour ago
David Richerby
65.9k15100190
asked 5 hours ago
bilanush
223
|
show 3 more comments
Why don't we consider them as linear? I don't understand. You just have to check for factorization up to sqrt of n. So it's even faster than linear.
I assume it's not linear only if we compare the number of operations relative to the input in terms of binary representation. But why would we do so?
It seems to me wrong. The growth in calculation should be calculated compared to the number itself. Why do we compare it to the binary representation?
algorithm-analysis time-complexity runtime-analysis
edited 1 hour ago
David Richerby
65.9k15100190
asked 5 hours ago
bilanush
223
You're assuming that a single divisibility check is a constant-time operation, which is definitely not true if you let the numbers get arbitrarily large. In fact it's more than O(n).
– hobbs
1 hour ago
So this is the answer. Not that we care about the binaric representation . You are basically giving a different reasoning is that correct?
– bilanush
1 hour ago
No, just pointing out an additional problem.
– hobbs
1 hour ago
Ok,. It doesn't even say that it takes more than linear for testing devisability only for division itself.
– bilanush
1 hour ago
Also, I would be happy if anyone answers why we care about the binaric representation.
– bilanush
1 hour ago
|
show 3 more comments
Why don't we consider them as linear? I don't understand. You just have to check for factorization up to sqrt of n. So it's even faster than linear.
I assume it's not linear only if we compare the number of operations relative to the input in terms of binary representation. But why would we do so?
It seems to me wrong. The growth in calculation should be calculated compared to the number itself. Why do we compare it to the binary representation?
algorithm-analysis time-complexity runtime-analysis
edited 1 hour ago
David Richerby
65.9k15100190
asked 5 hours ago
bilanush
223
Why don't we consider them as linear? I don't understand. You just have to check for factorization up to sqrt of n. So it's even faster than linear.
I assume it's not linear only if we compare the number of operations relative to the input in terms of binary representation. But why would we do so?
It seems to me wrong. The growth in calculation should be calculated compared to the number itself. Why do we compare it to the binary representation?
algorithm-analysis time-complexity runtime-analysis
algorithm-analysis time-complexity runtime-analysis
edited 1 hour ago
David Richerby
65.9k15100190
asked 5 hours ago
bilanush
223
edited 1 hour ago
David Richerby
65.9k15100190
asked 5 hours ago
bilanush
223
edited 1 hour ago
David Richerby
65.9k15100190
edited 1 hour ago
David Richerby
65.9k15100190
edited 1 hour ago
David Richerby
65.9k15100190
65.9k15100190
asked 5 hours ago
bilanush
223
asked 5 hours ago
bilanush
223
asked 5 hours ago
bilanush
223
223
You're assuming that a single divisibility check is a constant-time operation, which is definitely not true if you let the numbers get arbitrarily large. In fact it's more than O(n).
– hobbs
1 hour ago
So this is the answer. Not that we care about the binaric representation . You are basically giving a different reasoning is that correct?
– bilanush
1 hour ago
No, just pointing out an additional problem.
– hobbs
1 hour ago
Ok,. It doesn't even say that it takes more than linear for testing devisability only for division itself.
– bilanush
1 hour ago
Also, I would be happy if anyone answers why we care about the binaric representation.
– bilanush
1 hour ago
|
show 3 more comments
You're assuming that a single divisibility check is a constant-time operation, which is definitely not true if you let the numbers get arbitrarily large. In fact it's more than O(n).
– hobbs
1 hour ago
So this is the answer. Not that we care about the binaric representation . You are basically giving a different reasoning is that correct?
– bilanush
1 hour ago
No, just pointing out an additional problem.
– hobbs
1 hour ago
Ok,. It doesn't even say that it takes more than linear for testing devisability only for division itself.
– bilanush
1 hour ago
Also, I would be happy if anyone answers why we care about the binaric representation.
– bilanush
1 hour ago
You're assuming that a single divisibility check is a constant-time operation, which is definitely not true if you let the numbers get arbitrarily large. In fact it's more than O(n).
– hobbs
1 hour ago
You're assuming that a single divisibility check is a constant-time operation, which is definitely not true if you let the numbers get arbitrarily large. In fact it's more than O(n).
– hobbs
1 hour ago
So this is the answer. Not that we care about the binaric representation . You are basically giving a different reasoning is that correct?
– bilanush
1 hour ago
So this is the answer. Not that we care about the binaric representation . You are basically giving a different reasoning is that correct?
– bilanush
1 hour ago
No, just pointing out an additional problem.
– hobbs
1 hour ago
No, just pointing out an additional problem.
– hobbs
1 hour ago
Ok,. It doesn't even say that it takes more than linear for testing devisability only for division itself.
– bilanush
1 hour ago
Ok,. It doesn't even say that it takes more than linear for testing devisability only for division itself.
– bilanush
1 hour ago
Also, I would be happy if anyone answers why we care about the binaric representation.
– bilanush
1 hour ago
Also, I would be happy if anyone answers why we care about the binaric representation.
– bilanush
1 hour ago
|
show 3 more comments
2 Answers
2
active
oldest
votes
Simple. When you give the number one trillion as input to your algorithm, do you give it as 1'000'000'000'000, or as a terabyte large string of ones?
And by all means, you are free to choose whichever representation you feel comfortable with.
We analyze the runtime as a function of the size of the input, not as the magnitude of the number represented by the input were the input to be a number.
answered 4 hours ago
Pål GD
5,8871939
Sorry. But this is exactly what I don't understand. Why do we care about the binary representation ? It just looks like a manipulation of looking at it's binaric input and asking about the complexity of it. Why don't you look at the number itself which is what we truly care about. The whole point of complexity is to assess how fast the time grows as we move to greater numbers. So the most logical way of looking at it is by assessing how fast for example grow when we move from 10 to 100 . So it's 2^4 compared to 2^7 the growth is exactly linearly proportional to the input number
– bilanush
2 hours ago
We as humans deal with the number itself. Not with it's binaric shape. Why on Earth would it matter the complexity compared to binaric input? We are only trying to figure out how complicated the algorithm gets as we go to greater numbers. It doesn't matter at all the way computer choose to write it. Buttom line we should care only about the number of operations the computer does, and as well as the input. But of course, we should care about the magnitude of the number because this is what we are having in our heads , what is the rate in which the complexity grows as we go to great number
– bilanush
2 hours ago
2
@bilanush no, we as humans are incapable of doing math on "the number itself". Usually we use the base-10 representation. Which is the same size as the base-2 representation, within a constant factor.
– hobbs
1 hour ago
I agree. But think about it. Why should we care about the size of the input????? This is a very cosmetic factor here. What we really care about is how fast the complexity grows as we bring great numbers. What's the point of the need size of input? This tells me nothing. When I am talking about primes what I really care about is the rate of change of going from 1000 to 100000. I really do care about the magnitude of the number. Why should I care about the binaric input? Or even the base 10 representation for that matter.
– bilanush
1 hour ago
In complexity we are always talking about the magnitude of the number. Which is what we are dealing with. We are always interested in knowing how complex would it become as I move to greater numbers .
– bilanush
1 hour ago
|
show 1 more comment
We use binary simply because it is the base in which computers operate and it is irrelevant which base is used to analyze algorithm complexity.
The number of digits of a number $n$ in bases $b_1$ and $b_2$ is about $log_{b_1}n$ and $log_{b_2}n$. Those two expressions relate as
$$log_{b_1}n = frac{1}{log_{b_2}b_1} cdot log_{b_2}n$$
So the algorithm will end up (for asymptotical complexity analysis concerns) in the same class no matter which base we use to represent input, because they differ only by a constant factor.
answered 22 mins ago
Sandro Lovnički
8761115
add a comment |
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2 Answers
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2 Answers
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Simple. When you give the number one trillion as input to your algorithm, do you give it as 1'000'000'000'000, or as a terabyte large string of ones?
And by all means, you are free to choose whichever representation you feel comfortable with.
We analyze the runtime as a function of the size of the input, not as the magnitude of the number represented by the input were the input to be a number.
answered 4 hours ago
Pål GD
5,8871939
Sorry. But this is exactly what I don't understand. Why do we care about the binary representation ? It just looks like a manipulation of looking at it's binaric input and asking about the complexity of it. Why don't you look at the number itself which is what we truly care about. The whole point of complexity is to assess how fast the time grows as we move to greater numbers. So the most logical way of looking at it is by assessing how fast for example grow when we move from 10 to 100 . So it's 2^4 compared to 2^7 the growth is exactly linearly proportional to the input number
– bilanush
2 hours ago
We as humans deal with the number itself. Not with it's binaric shape. Why on Earth would it matter the complexity compared to binaric input? We are only trying to figure out how complicated the algorithm gets as we go to greater numbers. It doesn't matter at all the way computer choose to write it. Buttom line we should care only about the number of operations the computer does, and as well as the input. But of course, we should care about the magnitude of the number because this is what we are having in our heads , what is the rate in which the complexity grows as we go to great number
– bilanush
2 hours ago
2
@bilanush no, we as humans are incapable of doing math on "the number itself". Usually we use the base-10 representation. Which is the same size as the base-2 representation, within a constant factor.
– hobbs
1 hour ago
I agree. But think about it. Why should we care about the size of the input????? This is a very cosmetic factor here. What we really care about is how fast the complexity grows as we bring great numbers. What's the point of the need size of input? This tells me nothing. When I am talking about primes what I really care about is the rate of change of going from 1000 to 100000. I really do care about the magnitude of the number. Why should I care about the binaric input? Or even the base 10 representation for that matter.
– bilanush
1 hour ago
In complexity we are always talking about the magnitude of the number. Which is what we are dealing with. We are always interested in knowing how complex would it become as I move to greater numbers .
– bilanush
1 hour ago
|
show 1 more comment
Simple. When you give the number one trillion as input to your algorithm, do you give it as 1'000'000'000'000, or as a terabyte large string of ones?
And by all means, you are free to choose whichever representation you feel comfortable with.
We analyze the runtime as a function of the size of the input, not as the magnitude of the number represented by the input were the input to be a number.
answered 4 hours ago
Pål GD
5,8871939
Sorry. But this is exactly what I don't understand. Why do we care about the binary representation ? It just looks like a manipulation of looking at it's binaric input and asking about the complexity of it. Why don't you look at the number itself which is what we truly care about. The whole point of complexity is to assess how fast the time grows as we move to greater numbers. So the most logical way of looking at it is by assessing how fast for example grow when we move from 10 to 100 . So it's 2^4 compared to 2^7 the growth is exactly linearly proportional to the input number
– bilanush
2 hours ago
We as humans deal with the number itself. Not with it's binaric shape. Why on Earth would it matter the complexity compared to binaric input? We are only trying to figure out how complicated the algorithm gets as we go to greater numbers. It doesn't matter at all the way computer choose to write it. Buttom line we should care only about the number of operations the computer does, and as well as the input. But of course, we should care about the magnitude of the number because this is what we are having in our heads , what is the rate in which the complexity grows as we go to great number
– bilanush
2 hours ago
2
@bilanush no, we as humans are incapable of doing math on "the number itself". Usually we use the base-10 representation. Which is the same size as the base-2 representation, within a constant factor.
– hobbs
1 hour ago
I agree. But think about it. Why should we care about the size of the input????? This is a very cosmetic factor here. What we really care about is how fast the complexity grows as we bring great numbers. What's the point of the need size of input? This tells me nothing. When I am talking about primes what I really care about is the rate of change of going from 1000 to 100000. I really do care about the magnitude of the number. Why should I care about the binaric input? Or even the base 10 representation for that matter.
– bilanush
1 hour ago
In complexity we are always talking about the magnitude of the number. Which is what we are dealing with. We are always interested in knowing how complex would it become as I move to greater numbers .
– bilanush
1 hour ago
|
show 1 more comment
Simple. When you give the number one trillion as input to your algorithm, do you give it as 1'000'000'000'000, or as a terabyte large string of ones?
And by all means, you are free to choose whichever representation you feel comfortable with.
We analyze the runtime as a function of the size of the input, not as the magnitude of the number represented by the input were the input to be a number.
answered 4 hours ago
Pål GD
5,8871939
Simple. When you give the number one trillion as input to your algorithm, do you give it as 1'000'000'000'000, or as a terabyte large string of ones?
And by all means, you are free to choose whichever representation you feel comfortable with.
We analyze the runtime as a function of the size of the input, not as the magnitude of the number represented by the input were the input to be a number.
answered 4 hours ago
Pål GD
5,8871939
answered 4 hours ago
Pål GD
5,8871939
answered 4 hours ago
Pål GD
5,8871939
answered 4 hours ago
Pål GD
5,8871939
5,8871939
Sorry. But this is exactly what I don't understand. Why do we care about the binary representation ? It just looks like a manipulation of looking at it's binaric input and asking about the complexity of it. Why don't you look at the number itself which is what we truly care about. The whole point of complexity is to assess how fast the time grows as we move to greater numbers. So the most logical way of looking at it is by assessing how fast for example grow when we move from 10 to 100 . So it's 2^4 compared to 2^7 the growth is exactly linearly proportional to the input number
– bilanush
2 hours ago
We as humans deal with the number itself. Not with it's binaric shape. Why on Earth would it matter the complexity compared to binaric input? We are only trying to figure out how complicated the algorithm gets as we go to greater numbers. It doesn't matter at all the way computer choose to write it. Buttom line we should care only about the number of operations the computer does, and as well as the input. But of course, we should care about the magnitude of the number because this is what we are having in our heads , what is the rate in which the complexity grows as we go to great number
– bilanush
2 hours ago
2
@bilanush no, we as humans are incapable of doing math on "the number itself". Usually we use the base-10 representation. Which is the same size as the base-2 representation, within a constant factor.
– hobbs
1 hour ago
I agree. But think about it. Why should we care about the size of the input????? This is a very cosmetic factor here. What we really care about is how fast the complexity grows as we bring great numbers. What's the point of the need size of input? This tells me nothing. When I am talking about primes what I really care about is the rate of change of going from 1000 to 100000. I really do care about the magnitude of the number. Why should I care about the binaric input? Or even the base 10 representation for that matter.
– bilanush
1 hour ago
In complexity we are always talking about the magnitude of the number. Which is what we are dealing with. We are always interested in knowing how complex would it become as I move to greater numbers .
– bilanush
1 hour ago
|
show 1 more comment
Sorry. But this is exactly what I don't understand. Why do we care about the binary representation ? It just looks like a manipulation of looking at it's binaric input and asking about the complexity of it. Why don't you look at the number itself which is what we truly care about. The whole point of complexity is to assess how fast the time grows as we move to greater numbers. So the most logical way of looking at it is by assessing how fast for example grow when we move from 10 to 100 . So it's 2^4 compared to 2^7 the growth is exactly linearly proportional to the input number
– bilanush
2 hours ago
We as humans deal with the number itself. Not with it's binaric shape. Why on Earth would it matter the complexity compared to binaric input? We are only trying to figure out how complicated the algorithm gets as we go to greater numbers. It doesn't matter at all the way computer choose to write it. Buttom line we should care only about the number of operations the computer does, and as well as the input. But of course, we should care about the magnitude of the number because this is what we are having in our heads , what is the rate in which the complexity grows as we go to great number
– bilanush
2 hours ago
2
@bilanush no, we as humans are incapable of doing math on "the number itself". Usually we use the base-10 representation. Which is the same size as the base-2 representation, within a constant factor.
– hobbs
1 hour ago
I agree. But think about it. Why should we care about the size of the input????? This is a very cosmetic factor here. What we really care about is how fast the complexity grows as we bring great numbers. What's the point of the need size of input? This tells me nothing. When I am talking about primes what I really care about is the rate of change of going from 1000 to 100000. I really do care about the magnitude of the number. Why should I care about the binaric input? Or even the base 10 representation for that matter.
– bilanush
1 hour ago
In complexity we are always talking about the magnitude of the number. Which is what we are dealing with. We are always interested in knowing how complex would it become as I move to greater numbers .
– bilanush
1 hour ago
Sorry. But this is exactly what I don't understand. Why do we care about the binary representation ? It just looks like a manipulation of looking at it's binaric input and asking about the complexity of it. Why don't you look at the number itself which is what we truly care about. The whole point of complexity is to assess how fast the time grows as we move to greater numbers. So the most logical way of looking at it is by assessing how fast for example grow when we move from 10 to 100 . So it's 2^4 compared to 2^7 the growth is exactly linearly proportional to the input number
– bilanush
2 hours ago
Sorry. But this is exactly what I don't understand. Why do we care about the binary representation ? It just looks like a manipulation of looking at it's binaric input and asking about the complexity of it. Why don't you look at the number itself which is what we truly care about. The whole point of complexity is to assess how fast the time grows as we move to greater numbers. So the most logical way of looking at it is by assessing how fast for example grow when we move from 10 to 100 . So it's 2^4 compared to 2^7 the growth is exactly linearly proportional to the input number
– bilanush
2 hours ago
We as humans deal with the number itself. Not with it's binaric shape. Why on Earth would it matter the complexity compared to binaric input? We are only trying to figure out how complicated the algorithm gets as we go to greater numbers. It doesn't matter at all the way computer choose to write it. Buttom line we should care only about the number of operations the computer does, and as well as the input. But of course, we should care about the magnitude of the number because this is what we are having in our heads , what is the rate in which the complexity grows as we go to great number
– bilanush
2 hours ago
We as humans deal with the number itself. Not with it's binaric shape. Why on Earth would it matter the complexity compared to binaric input? We are only trying to figure out how complicated the algorithm gets as we go to greater numbers. It doesn't matter at all the way computer choose to write it. Buttom line we should care only about the number of operations the computer does, and as well as the input. But of course, we should care about the magnitude of the number because this is what we are having in our heads , what is the rate in which the complexity grows as we go to great number
– bilanush
2 hours ago
2
2
@bilanush no, we as humans are incapable of doing math on "the number itself". Usually we use the base-10 representation. Which is the same size as the base-2 representation, within a constant factor.
– hobbs
1 hour ago
@bilanush no, we as humans are incapable of doing math on "the number itself". Usually we use the base-10 representation. Which is the same size as the base-2 representation, within a constant factor.
– hobbs
1 hour ago
I agree. But think about it. Why should we care about the size of the input????? This is a very cosmetic factor here. What we really care about is how fast the complexity grows as we bring great numbers. What's the point of the need size of input? This tells me nothing. When I am talking about primes what I really care about is the rate of change of going from 1000 to 100000. I really do care about the magnitude of the number. Why should I care about the binaric input? Or even the base 10 representation for that matter.
– bilanush
1 hour ago
I agree. But think about it. Why should we care about the size of the input????? This is a very cosmetic factor here. What we really care about is how fast the complexity grows as we bring great numbers. What's the point of the need size of input? This tells me nothing. When I am talking about primes what I really care about is the rate of change of going from 1000 to 100000. I really do care about the magnitude of the number. Why should I care about the binaric input? Or even the base 10 representation for that matter.
– bilanush
1 hour ago
In complexity we are always talking about the magnitude of the number. Which is what we are dealing with. We are always interested in knowing how complex would it become as I move to greater numbers .
– bilanush
1 hour ago
In complexity we are always talking about the magnitude of the number. Which is what we are dealing with. We are always interested in knowing how complex would it become as I move to greater numbers .
– bilanush
1 hour ago
|
show 1 more comment
We use binary simply because it is the base in which computers operate and it is irrelevant which base is used to analyze algorithm complexity.
The number of digits of a number $n$ in bases $b_1$ and $b_2$ is about $log_{b_1}n$ and $log_{b_2}n$. Those two expressions relate as
$$log_{b_1}n = frac{1}{log_{b_2}b_1} cdot log_{b_2}n$$
So the algorithm will end up (for asymptotical complexity analysis concerns) in the same class no matter which base we use to represent input, because they differ only by a constant factor.
answered 22 mins ago
Sandro Lovnički
8761115
add a comment |
We use binary simply because it is the base in which computers operate and it is irrelevant which base is used to analyze algorithm complexity.
The number of digits of a number $n$ in bases $b_1$ and $b_2$ is about $log_{b_1}n$ and $log_{b_2}n$. Those two expressions relate as
$$log_{b_1}n = frac{1}{log_{b_2}b_1} cdot log_{b_2}n$$
So the algorithm will end up (for asymptotical complexity analysis concerns) in the same class no matter which base we use to represent input, because they differ only by a constant factor.
answered 22 mins ago
Sandro Lovnički
8761115
add a comment |
We use binary simply because it is the base in which computers operate and it is irrelevant which base is used to analyze algorithm complexity.
The number of digits of a number $n$ in bases $b_1$ and $b_2$ is about $log_{b_1}n$ and $log_{b_2}n$. Those two expressions relate as
$$log_{b_1}n = frac{1}{log_{b_2}b_1} cdot log_{b_2}n$$
So the algorithm will end up (for asymptotical complexity analysis concerns) in the same class no matter which base we use to represent input, because they differ only by a constant factor.
answered 22 mins ago
Sandro Lovnički
8761115
We use binary simply because it is the base in which computers operate and it is irrelevant which base is used to analyze algorithm complexity.
The number of digits of a number $n$ in bases $b_1$ and $b_2$ is about $log_{b_1}n$ and $log_{b_2}n$. Those two expressions relate as
$$log_{b_1}n = frac{1}{log_{b_2}b_1} cdot log_{b_2}n$$
So the algorithm will end up (for asymptotical complexity analysis concerns) in the same class no matter which base we use to represent input, because they differ only by a constant factor.
answered 22 mins ago
Sandro Lovnički
8761115
answered 22 mins ago
Sandro Lovnički
8761115
answered 22 mins ago
Sandro Lovnički
8761115
answered 22 mins ago
Sandro Lovnički
8761115
8761115
add a comment |
add a comment |
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You're assuming that a single divisibility check is a constant-time operation, which is definitely not true if you let the numbers get arbitrarily large. In fact it's more than O(n).
– hobbs
1 hour ago
So this is the answer. Not that we care about the binaric representation . You are basically giving a different reasoning is that correct?
– bilanush
1 hour ago
No, just pointing out an additional problem.
– hobbs
1 hour ago
Ok,. It doesn't even say that it takes more than linear for testing devisability only for division itself.
– bilanush
1 hour ago
Also, I would be happy if anyone answers why we care about the binaric representation.
– bilanush
1 hour ago