How exactly does wheel factorization work and what is it used for?
up vote
1
down vote
favorite
I would like to learn how to use wheel factorization but am having trouble understanding it. I tried reading the wikipedia article but found it confusing (even the talk page says it's a mess). What exactly is it and how is it used? To my understanding it eliminates some (but not all) composite numbers in a list up to a certain number. So in this sense it's a technique that can be used to speed up existing factorization algorithms? It seems to almost be the same as the sieve of Eratosthenes except it starts with a small list of known prime numbers?
If someone could please give the general procedure and a simple example that would be much appreciated.
algorithms prime-factorization sieve-theory
add a comment |
up vote
1
down vote
favorite
I would like to learn how to use wheel factorization but am having trouble understanding it. I tried reading the wikipedia article but found it confusing (even the talk page says it's a mess). What exactly is it and how is it used? To my understanding it eliminates some (but not all) composite numbers in a list up to a certain number. So in this sense it's a technique that can be used to speed up existing factorization algorithms? It seems to almost be the same as the sieve of Eratosthenes except it starts with a small list of known prime numbers?
If someone could please give the general procedure and a simple example that would be much appreciated.
algorithms prime-factorization sieve-theory
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I would like to learn how to use wheel factorization but am having trouble understanding it. I tried reading the wikipedia article but found it confusing (even the talk page says it's a mess). What exactly is it and how is it used? To my understanding it eliminates some (but not all) composite numbers in a list up to a certain number. So in this sense it's a technique that can be used to speed up existing factorization algorithms? It seems to almost be the same as the sieve of Eratosthenes except it starts with a small list of known prime numbers?
If someone could please give the general procedure and a simple example that would be much appreciated.
algorithms prime-factorization sieve-theory
I would like to learn how to use wheel factorization but am having trouble understanding it. I tried reading the wikipedia article but found it confusing (even the talk page says it's a mess). What exactly is it and how is it used? To my understanding it eliminates some (but not all) composite numbers in a list up to a certain number. So in this sense it's a technique that can be used to speed up existing factorization algorithms? It seems to almost be the same as the sieve of Eratosthenes except it starts with a small list of known prime numbers?
If someone could please give the general procedure and a simple example that would be much appreciated.
algorithms prime-factorization sieve-theory
algorithms prime-factorization sieve-theory
asked Nov 26 at 7:07
northerner
1203
1203
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
Since trial division is mostly useless for factoring large numbers, and using a prime number sieve for factoring is just a minor refinement of trial division, you shouldn't think of it as a factorization algorithm. Instead, this is a prime-generating algorithm: the goal is to generate the list of primes in the set $[n] := {1, 2, 3, dots, n}$ as quickly as possible.
We are trying to improve on the sieve of Eratosthenes in efficiency, which does $Theta(n cdot log log n)$ arithmetical operations on elements of $[n]$.
Wheel factorization does this by using the fact that for the first few primes, the sieve we're constructing is periodic, and there's no point in extending the periodic pattern all the way to $n$. Instead, we only generate the list of numbers not divisible by the first $k$ primes $p_1, p_2, dots, p_k$ only up to their product $p_1 p_2 dotsm p_k$. That is, we:
- Begin by generating the list of numbers in ${1,2,3,4,5,6}$ not divisible by $2$ or $3$: it is ${1,5}$.
- Extend this to the list of numbers in ${1,2,3,dots,30}$ not divisible by $2$, $3$, or $5$: it is ${1,7,11,13,17,19,23,29}$.
- Extend this to the list of numbers in ${1,2,3,dots, 210}$ not divisible by $2$, $3$, $5$, or $7$, and so on.
For each extension step, if the set we've generated is $S$ and the next prime we're adding is $p$, then the next set consists of $p$ translated copies of $S$, with $p cdot S$ removed. For example, if $S = {1,5}$ and $p=5$, then we repeat $S$ $5$ times (to get ${1,5} cup {7,11} cup {13,17} cup {19, 23} cup {25, 29}$) and remove $5cdot S = {5,25}$. By the way, $p$ is also easy to find: it is the element of $S$ after $1$.
Once $p_1 p_2 dotsm p_k > n$, we no longer take repeated copies of $S$, and just remove $p cdot S$ from $S$ to extend. We stop, as with the sieve of Eratosthenes, when $p_k > sqrt n$. At this point, $S$ contains all the primes larger than $p_k$; the primes smaller than $p_k$ are the ones we used along the way, which we keep track of separately.
According to this paper, this only requires $Theta(frac{n}{log log n})$ arithmetical operations on elements of $[n]$, if implemented carefully.
I guess my question is much more specific. What I don't get is how does one efficiently come up with the list given a base? From examples I've seen, given primes {2,3,5} they some how know the next possible prime can be calculated by adding from {4, 2, 4, 2, 4, 6, 2} periodically. Where do these numbers come from and how can it be extended for different bases?
– northerner
Nov 27 at 9:11
"Extend this to the list of numbers in {1,2,3,…,30} not divisible by 2, 3, or 5: it is {1,7,11,13,17,19,23,29}." how? I mean if you're just using trial division then what's the difference of wheel factorization?
– northerner
Nov 27 at 9:18
See the very next paragraph. We are not using trial division: we are adding $2cdot3$ to $S$ $5$ times, and then removing $5 cdot S$.
– Misha Lavrov
Nov 27 at 14:37
I know you said the goal of wheel factorization is to generate a list of prime numbers, but can/do people use it to factorize a number by generating a list up to the square root and then using trial division? Or using it to make a sieve?
– northerner
Dec 9 at 1:23
I'm not sure what you mean by a sieve if it's not a list of prime numbers. Factoring by wheel factorization plus trial division is plausible, but I expect that at the point where naive trial division is not practical, you should be switching to fancier factorization algorithms that don't use trial division at all.
– Misha Lavrov
Dec 9 at 1:46
add a comment |
up vote
0
down vote
There are many useful context about "Wheel factorization" in Web. Consider for example:
https://primes.utm.edu/glossary/page.php?sort=WheelFactorization
https://www.revolvy.com/page/Wheel-factorization
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013969%2fhow-exactly-does-wheel-factorization-work-and-what-is-it-used-for%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Since trial division is mostly useless for factoring large numbers, and using a prime number sieve for factoring is just a minor refinement of trial division, you shouldn't think of it as a factorization algorithm. Instead, this is a prime-generating algorithm: the goal is to generate the list of primes in the set $[n] := {1, 2, 3, dots, n}$ as quickly as possible.
We are trying to improve on the sieve of Eratosthenes in efficiency, which does $Theta(n cdot log log n)$ arithmetical operations on elements of $[n]$.
Wheel factorization does this by using the fact that for the first few primes, the sieve we're constructing is periodic, and there's no point in extending the periodic pattern all the way to $n$. Instead, we only generate the list of numbers not divisible by the first $k$ primes $p_1, p_2, dots, p_k$ only up to their product $p_1 p_2 dotsm p_k$. That is, we:
- Begin by generating the list of numbers in ${1,2,3,4,5,6}$ not divisible by $2$ or $3$: it is ${1,5}$.
- Extend this to the list of numbers in ${1,2,3,dots,30}$ not divisible by $2$, $3$, or $5$: it is ${1,7,11,13,17,19,23,29}$.
- Extend this to the list of numbers in ${1,2,3,dots, 210}$ not divisible by $2$, $3$, $5$, or $7$, and so on.
For each extension step, if the set we've generated is $S$ and the next prime we're adding is $p$, then the next set consists of $p$ translated copies of $S$, with $p cdot S$ removed. For example, if $S = {1,5}$ and $p=5$, then we repeat $S$ $5$ times (to get ${1,5} cup {7,11} cup {13,17} cup {19, 23} cup {25, 29}$) and remove $5cdot S = {5,25}$. By the way, $p$ is also easy to find: it is the element of $S$ after $1$.
Once $p_1 p_2 dotsm p_k > n$, we no longer take repeated copies of $S$, and just remove $p cdot S$ from $S$ to extend. We stop, as with the sieve of Eratosthenes, when $p_k > sqrt n$. At this point, $S$ contains all the primes larger than $p_k$; the primes smaller than $p_k$ are the ones we used along the way, which we keep track of separately.
According to this paper, this only requires $Theta(frac{n}{log log n})$ arithmetical operations on elements of $[n]$, if implemented carefully.
I guess my question is much more specific. What I don't get is how does one efficiently come up with the list given a base? From examples I've seen, given primes {2,3,5} they some how know the next possible prime can be calculated by adding from {4, 2, 4, 2, 4, 6, 2} periodically. Where do these numbers come from and how can it be extended for different bases?
– northerner
Nov 27 at 9:11
"Extend this to the list of numbers in {1,2,3,…,30} not divisible by 2, 3, or 5: it is {1,7,11,13,17,19,23,29}." how? I mean if you're just using trial division then what's the difference of wheel factorization?
– northerner
Nov 27 at 9:18
See the very next paragraph. We are not using trial division: we are adding $2cdot3$ to $S$ $5$ times, and then removing $5 cdot S$.
– Misha Lavrov
Nov 27 at 14:37
I know you said the goal of wheel factorization is to generate a list of prime numbers, but can/do people use it to factorize a number by generating a list up to the square root and then using trial division? Or using it to make a sieve?
– northerner
Dec 9 at 1:23
I'm not sure what you mean by a sieve if it's not a list of prime numbers. Factoring by wheel factorization plus trial division is plausible, but I expect that at the point where naive trial division is not practical, you should be switching to fancier factorization algorithms that don't use trial division at all.
– Misha Lavrov
Dec 9 at 1:46
add a comment |
up vote
2
down vote
Since trial division is mostly useless for factoring large numbers, and using a prime number sieve for factoring is just a minor refinement of trial division, you shouldn't think of it as a factorization algorithm. Instead, this is a prime-generating algorithm: the goal is to generate the list of primes in the set $[n] := {1, 2, 3, dots, n}$ as quickly as possible.
We are trying to improve on the sieve of Eratosthenes in efficiency, which does $Theta(n cdot log log n)$ arithmetical operations on elements of $[n]$.
Wheel factorization does this by using the fact that for the first few primes, the sieve we're constructing is periodic, and there's no point in extending the periodic pattern all the way to $n$. Instead, we only generate the list of numbers not divisible by the first $k$ primes $p_1, p_2, dots, p_k$ only up to their product $p_1 p_2 dotsm p_k$. That is, we:
- Begin by generating the list of numbers in ${1,2,3,4,5,6}$ not divisible by $2$ or $3$: it is ${1,5}$.
- Extend this to the list of numbers in ${1,2,3,dots,30}$ not divisible by $2$, $3$, or $5$: it is ${1,7,11,13,17,19,23,29}$.
- Extend this to the list of numbers in ${1,2,3,dots, 210}$ not divisible by $2$, $3$, $5$, or $7$, and so on.
For each extension step, if the set we've generated is $S$ and the next prime we're adding is $p$, then the next set consists of $p$ translated copies of $S$, with $p cdot S$ removed. For example, if $S = {1,5}$ and $p=5$, then we repeat $S$ $5$ times (to get ${1,5} cup {7,11} cup {13,17} cup {19, 23} cup {25, 29}$) and remove $5cdot S = {5,25}$. By the way, $p$ is also easy to find: it is the element of $S$ after $1$.
Once $p_1 p_2 dotsm p_k > n$, we no longer take repeated copies of $S$, and just remove $p cdot S$ from $S$ to extend. We stop, as with the sieve of Eratosthenes, when $p_k > sqrt n$. At this point, $S$ contains all the primes larger than $p_k$; the primes smaller than $p_k$ are the ones we used along the way, which we keep track of separately.
According to this paper, this only requires $Theta(frac{n}{log log n})$ arithmetical operations on elements of $[n]$, if implemented carefully.
I guess my question is much more specific. What I don't get is how does one efficiently come up with the list given a base? From examples I've seen, given primes {2,3,5} they some how know the next possible prime can be calculated by adding from {4, 2, 4, 2, 4, 6, 2} periodically. Where do these numbers come from and how can it be extended for different bases?
– northerner
Nov 27 at 9:11
"Extend this to the list of numbers in {1,2,3,…,30} not divisible by 2, 3, or 5: it is {1,7,11,13,17,19,23,29}." how? I mean if you're just using trial division then what's the difference of wheel factorization?
– northerner
Nov 27 at 9:18
See the very next paragraph. We are not using trial division: we are adding $2cdot3$ to $S$ $5$ times, and then removing $5 cdot S$.
– Misha Lavrov
Nov 27 at 14:37
I know you said the goal of wheel factorization is to generate a list of prime numbers, but can/do people use it to factorize a number by generating a list up to the square root and then using trial division? Or using it to make a sieve?
– northerner
Dec 9 at 1:23
I'm not sure what you mean by a sieve if it's not a list of prime numbers. Factoring by wheel factorization plus trial division is plausible, but I expect that at the point where naive trial division is not practical, you should be switching to fancier factorization algorithms that don't use trial division at all.
– Misha Lavrov
Dec 9 at 1:46
add a comment |
up vote
2
down vote
up vote
2
down vote
Since trial division is mostly useless for factoring large numbers, and using a prime number sieve for factoring is just a minor refinement of trial division, you shouldn't think of it as a factorization algorithm. Instead, this is a prime-generating algorithm: the goal is to generate the list of primes in the set $[n] := {1, 2, 3, dots, n}$ as quickly as possible.
We are trying to improve on the sieve of Eratosthenes in efficiency, which does $Theta(n cdot log log n)$ arithmetical operations on elements of $[n]$.
Wheel factorization does this by using the fact that for the first few primes, the sieve we're constructing is periodic, and there's no point in extending the periodic pattern all the way to $n$. Instead, we only generate the list of numbers not divisible by the first $k$ primes $p_1, p_2, dots, p_k$ only up to their product $p_1 p_2 dotsm p_k$. That is, we:
- Begin by generating the list of numbers in ${1,2,3,4,5,6}$ not divisible by $2$ or $3$: it is ${1,5}$.
- Extend this to the list of numbers in ${1,2,3,dots,30}$ not divisible by $2$, $3$, or $5$: it is ${1,7,11,13,17,19,23,29}$.
- Extend this to the list of numbers in ${1,2,3,dots, 210}$ not divisible by $2$, $3$, $5$, or $7$, and so on.
For each extension step, if the set we've generated is $S$ and the next prime we're adding is $p$, then the next set consists of $p$ translated copies of $S$, with $p cdot S$ removed. For example, if $S = {1,5}$ and $p=5$, then we repeat $S$ $5$ times (to get ${1,5} cup {7,11} cup {13,17} cup {19, 23} cup {25, 29}$) and remove $5cdot S = {5,25}$. By the way, $p$ is also easy to find: it is the element of $S$ after $1$.
Once $p_1 p_2 dotsm p_k > n$, we no longer take repeated copies of $S$, and just remove $p cdot S$ from $S$ to extend. We stop, as with the sieve of Eratosthenes, when $p_k > sqrt n$. At this point, $S$ contains all the primes larger than $p_k$; the primes smaller than $p_k$ are the ones we used along the way, which we keep track of separately.
According to this paper, this only requires $Theta(frac{n}{log log n})$ arithmetical operations on elements of $[n]$, if implemented carefully.
Since trial division is mostly useless for factoring large numbers, and using a prime number sieve for factoring is just a minor refinement of trial division, you shouldn't think of it as a factorization algorithm. Instead, this is a prime-generating algorithm: the goal is to generate the list of primes in the set $[n] := {1, 2, 3, dots, n}$ as quickly as possible.
We are trying to improve on the sieve of Eratosthenes in efficiency, which does $Theta(n cdot log log n)$ arithmetical operations on elements of $[n]$.
Wheel factorization does this by using the fact that for the first few primes, the sieve we're constructing is periodic, and there's no point in extending the periodic pattern all the way to $n$. Instead, we only generate the list of numbers not divisible by the first $k$ primes $p_1, p_2, dots, p_k$ only up to their product $p_1 p_2 dotsm p_k$. That is, we:
- Begin by generating the list of numbers in ${1,2,3,4,5,6}$ not divisible by $2$ or $3$: it is ${1,5}$.
- Extend this to the list of numbers in ${1,2,3,dots,30}$ not divisible by $2$, $3$, or $5$: it is ${1,7,11,13,17,19,23,29}$.
- Extend this to the list of numbers in ${1,2,3,dots, 210}$ not divisible by $2$, $3$, $5$, or $7$, and so on.
For each extension step, if the set we've generated is $S$ and the next prime we're adding is $p$, then the next set consists of $p$ translated copies of $S$, with $p cdot S$ removed. For example, if $S = {1,5}$ and $p=5$, then we repeat $S$ $5$ times (to get ${1,5} cup {7,11} cup {13,17} cup {19, 23} cup {25, 29}$) and remove $5cdot S = {5,25}$. By the way, $p$ is also easy to find: it is the element of $S$ after $1$.
Once $p_1 p_2 dotsm p_k > n$, we no longer take repeated copies of $S$, and just remove $p cdot S$ from $S$ to extend. We stop, as with the sieve of Eratosthenes, when $p_k > sqrt n$. At this point, $S$ contains all the primes larger than $p_k$; the primes smaller than $p_k$ are the ones we used along the way, which we keep track of separately.
According to this paper, this only requires $Theta(frac{n}{log log n})$ arithmetical operations on elements of $[n]$, if implemented carefully.
edited Nov 26 at 8:11
answered Nov 26 at 8:03
Misha Lavrov
42.9k555101
42.9k555101
I guess my question is much more specific. What I don't get is how does one efficiently come up with the list given a base? From examples I've seen, given primes {2,3,5} they some how know the next possible prime can be calculated by adding from {4, 2, 4, 2, 4, 6, 2} periodically. Where do these numbers come from and how can it be extended for different bases?
– northerner
Nov 27 at 9:11
"Extend this to the list of numbers in {1,2,3,…,30} not divisible by 2, 3, or 5: it is {1,7,11,13,17,19,23,29}." how? I mean if you're just using trial division then what's the difference of wheel factorization?
– northerner
Nov 27 at 9:18
See the very next paragraph. We are not using trial division: we are adding $2cdot3$ to $S$ $5$ times, and then removing $5 cdot S$.
– Misha Lavrov
Nov 27 at 14:37
I know you said the goal of wheel factorization is to generate a list of prime numbers, but can/do people use it to factorize a number by generating a list up to the square root and then using trial division? Or using it to make a sieve?
– northerner
Dec 9 at 1:23
I'm not sure what you mean by a sieve if it's not a list of prime numbers. Factoring by wheel factorization plus trial division is plausible, but I expect that at the point where naive trial division is not practical, you should be switching to fancier factorization algorithms that don't use trial division at all.
– Misha Lavrov
Dec 9 at 1:46
add a comment |
I guess my question is much more specific. What I don't get is how does one efficiently come up with the list given a base? From examples I've seen, given primes {2,3,5} they some how know the next possible prime can be calculated by adding from {4, 2, 4, 2, 4, 6, 2} periodically. Where do these numbers come from and how can it be extended for different bases?
– northerner
Nov 27 at 9:11
"Extend this to the list of numbers in {1,2,3,…,30} not divisible by 2, 3, or 5: it is {1,7,11,13,17,19,23,29}." how? I mean if you're just using trial division then what's the difference of wheel factorization?
– northerner
Nov 27 at 9:18
See the very next paragraph. We are not using trial division: we are adding $2cdot3$ to $S$ $5$ times, and then removing $5 cdot S$.
– Misha Lavrov
Nov 27 at 14:37
I know you said the goal of wheel factorization is to generate a list of prime numbers, but can/do people use it to factorize a number by generating a list up to the square root and then using trial division? Or using it to make a sieve?
– northerner
Dec 9 at 1:23
I'm not sure what you mean by a sieve if it's not a list of prime numbers. Factoring by wheel factorization plus trial division is plausible, but I expect that at the point where naive trial division is not practical, you should be switching to fancier factorization algorithms that don't use trial division at all.
– Misha Lavrov
Dec 9 at 1:46
I guess my question is much more specific. What I don't get is how does one efficiently come up with the list given a base? From examples I've seen, given primes {2,3,5} they some how know the next possible prime can be calculated by adding from {4, 2, 4, 2, 4, 6, 2} periodically. Where do these numbers come from and how can it be extended for different bases?
– northerner
Nov 27 at 9:11
I guess my question is much more specific. What I don't get is how does one efficiently come up with the list given a base? From examples I've seen, given primes {2,3,5} they some how know the next possible prime can be calculated by adding from {4, 2, 4, 2, 4, 6, 2} periodically. Where do these numbers come from and how can it be extended for different bases?
– northerner
Nov 27 at 9:11
"Extend this to the list of numbers in {1,2,3,…,30} not divisible by 2, 3, or 5: it is {1,7,11,13,17,19,23,29}." how? I mean if you're just using trial division then what's the difference of wheel factorization?
– northerner
Nov 27 at 9:18
"Extend this to the list of numbers in {1,2,3,…,30} not divisible by 2, 3, or 5: it is {1,7,11,13,17,19,23,29}." how? I mean if you're just using trial division then what's the difference of wheel factorization?
– northerner
Nov 27 at 9:18
See the very next paragraph. We are not using trial division: we are adding $2cdot3$ to $S$ $5$ times, and then removing $5 cdot S$.
– Misha Lavrov
Nov 27 at 14:37
See the very next paragraph. We are not using trial division: we are adding $2cdot3$ to $S$ $5$ times, and then removing $5 cdot S$.
– Misha Lavrov
Nov 27 at 14:37
I know you said the goal of wheel factorization is to generate a list of prime numbers, but can/do people use it to factorize a number by generating a list up to the square root and then using trial division? Or using it to make a sieve?
– northerner
Dec 9 at 1:23
I know you said the goal of wheel factorization is to generate a list of prime numbers, but can/do people use it to factorize a number by generating a list up to the square root and then using trial division? Or using it to make a sieve?
– northerner
Dec 9 at 1:23
I'm not sure what you mean by a sieve if it's not a list of prime numbers. Factoring by wheel factorization plus trial division is plausible, but I expect that at the point where naive trial division is not practical, you should be switching to fancier factorization algorithms that don't use trial division at all.
– Misha Lavrov
Dec 9 at 1:46
I'm not sure what you mean by a sieve if it's not a list of prime numbers. Factoring by wheel factorization plus trial division is plausible, but I expect that at the point where naive trial division is not practical, you should be switching to fancier factorization algorithms that don't use trial division at all.
– Misha Lavrov
Dec 9 at 1:46
add a comment |
up vote
0
down vote
There are many useful context about "Wheel factorization" in Web. Consider for example:
https://primes.utm.edu/glossary/page.php?sort=WheelFactorization
https://www.revolvy.com/page/Wheel-factorization
add a comment |
up vote
0
down vote
There are many useful context about "Wheel factorization" in Web. Consider for example:
https://primes.utm.edu/glossary/page.php?sort=WheelFactorization
https://www.revolvy.com/page/Wheel-factorization
add a comment |
up vote
0
down vote
up vote
0
down vote
There are many useful context about "Wheel factorization" in Web. Consider for example:
https://primes.utm.edu/glossary/page.php?sort=WheelFactorization
https://www.revolvy.com/page/Wheel-factorization
There are many useful context about "Wheel factorization" in Web. Consider for example:
https://primes.utm.edu/glossary/page.php?sort=WheelFactorization
https://www.revolvy.com/page/Wheel-factorization
edited Nov 26 at 8:21
answered Nov 26 at 8:16
Mahmood Dadkhah
113
113
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013969%2fhow-exactly-does-wheel-factorization-work-and-what-is-it-used-for%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown