Real Modulo Numbers in the Interval [0, 1), and Multiplication of The 2019 Stack Overflow...
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Real Modulo Numbers in the Interval [0, 1), and Multiplication of
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$begingroup$
Let's say we have a real number, A, in the interval [0, 1).
If we add another real number to it, it "Wraps" around back to zero.
So, for example:
Lets say: A = 5/13
If we multiply A by 2, we get: 10/13
If we multiple A by 3, we get: 2/13 (not 15/13)
If we multiple A by 4, we get 7/13 (not 20/13)
And so on...
My question is:
If given two such numbers, A and B, how can we find the smallest number, x, such that:
Ax = B
For example:
If A = 5/13 and B = 8/13, what do we need to multiply A by, to get B.
A solution for real numbers is preferable. However, if a solution does not exist for real numbers generally, then a solution for rational numbers should be sufficient.
Note that this problem is relevant to an algorithm that I'm writing. And I'd like to get it finished soon.
If I get my algorithm to work, and someone here provides a good solution, and that person (or people) have their name and contact details on their profile page, then I'm happy to give them a reference.
number-theory modular-arithmetic
asked Mar 21 at 6:24
Abs SpurdleAbs Spurdle
113
$endgroup$
add a comment |
$begingroup$
Let's say we have a real number, A, in the interval [0, 1).
If we add another real number to it, it "Wraps" around back to zero.
So, for example:
Lets say: A = 5/13
If we multiply A by 2, we get: 10/13
If we multiple A by 3, we get: 2/13 (not 15/13)
If we multiple A by 4, we get 7/13 (not 20/13)
And so on...
My question is:
If given two such numbers, A and B, how can we find the smallest number, x, such that:
Ax = B
For example:
If A = 5/13 and B = 8/13, what do we need to multiply A by, to get B.
A solution for real numbers is preferable. However, if a solution does not exist for real numbers generally, then a solution for rational numbers should be sufficient.
Note that this problem is relevant to an algorithm that I'm writing. And I'd like to get it finished soon.
If I get my algorithm to work, and someone here provides a good solution, and that person (or people) have their name and contact details on their profile page, then I'm happy to give them a reference.
number-theory modular-arithmetic
asked Mar 21 at 6:24
Abs SpurdleAbs Spurdle
113
$endgroup$
$begingroup$
My first thought would be to see if the (extended) Euclidean algorithm would be of any use: en.wikipedia.org/wiki/Modular_multiplicative_inverse
$endgroup$
– Matti P.
Mar 21 at 6:29
$begingroup$
For rational numbers $a/n$ and $b/n$, this is the same as solving the congruence $ax equiv b pmod n$. Solutions will not always exist (e.g., if $a = 2/4, b=1/4$).
$endgroup$
– FredH
Mar 21 at 9:07
$begingroup$
I assumed that there was a solution with O(1), my bad. I had a look at the algorithm you suggested, and if I understand it correctly, it should work for the example above, however, it's not as efficient as I would like.
$endgroup$
– Abs Spurdle
Mar 21 at 21:28
add a comment |
$begingroup$
Let's say we have a real number, A, in the interval [0, 1).
If we add another real number to it, it "Wraps" around back to zero.
So, for example:
Lets say: A = 5/13
If we multiply A by 2, we get: 10/13
If we multiple A by 3, we get: 2/13 (not 15/13)
If we multiple A by 4, we get 7/13 (not 20/13)
And so on...
My question is:
If given two such numbers, A and B, how can we find the smallest number, x, such that:
Ax = B
For example:
If A = 5/13 and B = 8/13, what do we need to multiply A by, to get B.
A solution for real numbers is preferable. However, if a solution does not exist for real numbers generally, then a solution for rational numbers should be sufficient.
Note that this problem is relevant to an algorithm that I'm writing. And I'd like to get it finished soon.
If I get my algorithm to work, and someone here provides a good solution, and that person (or people) have their name and contact details on their profile page, then I'm happy to give them a reference.
number-theory modular-arithmetic
asked Mar 21 at 6:24
Abs SpurdleAbs Spurdle
113
$endgroup$
Let's say we have a real number, A, in the interval [0, 1).
If we add another real number to it, it "Wraps" around back to zero.
So, for example:
Lets say: A = 5/13
If we multiply A by 2, we get: 10/13
If we multiple A by 3, we get: 2/13 (not 15/13)
If we multiple A by 4, we get 7/13 (not 20/13)
And so on...
My question is:
If given two such numbers, A and B, how can we find the smallest number, x, such that:
Ax = B
For example:
If A = 5/13 and B = 8/13, what do we need to multiply A by, to get B.
A solution for real numbers is preferable. However, if a solution does not exist for real numbers generally, then a solution for rational numbers should be sufficient.
Note that this problem is relevant to an algorithm that I'm writing. And I'd like to get it finished soon.
If I get my algorithm to work, and someone here provides a good solution, and that person (or people) have their name and contact details on their profile page, then I'm happy to give them a reference.
number-theory modular-arithmetic
number-theory modular-arithmetic
asked Mar 21 at 6:24
Abs SpurdleAbs Spurdle
113
asked Mar 21 at 6:24
Abs SpurdleAbs Spurdle
113
edited Mar 21 at 20:34
Abs Spurdle
asked Mar 21 at 6:24
Abs SpurdleAbs Spurdle
113
asked Mar 21 at 6:24
Abs SpurdleAbs Spurdle
113
asked Mar 21 at 6:24
Abs SpurdleAbs Spurdle
113
113
$begingroup$
My first thought would be to see if the (extended) Euclidean algorithm would be of any use: en.wikipedia.org/wiki/Modular_multiplicative_inverse
$endgroup$
– Matti P.
Mar 21 at 6:29
$begingroup$
For rational numbers $a/n$ and $b/n$, this is the same as solving the congruence $ax equiv b pmod n$. Solutions will not always exist (e.g., if $a = 2/4, b=1/4$).
$endgroup$
– FredH
Mar 21 at 9:07
$begingroup$
I assumed that there was a solution with O(1), my bad. I had a look at the algorithm you suggested, and if I understand it correctly, it should work for the example above, however, it's not as efficient as I would like.
$endgroup$
– Abs Spurdle
Mar 21 at 21:28
add a comment |
$begingroup$
My first thought would be to see if the (extended) Euclidean algorithm would be of any use: en.wikipedia.org/wiki/Modular_multiplicative_inverse
$endgroup$
– Matti P.
Mar 21 at 6:29
$begingroup$
For rational numbers $a/n$ and $b/n$, this is the same as solving the congruence $ax equiv b pmod n$. Solutions will not always exist (e.g., if $a = 2/4, b=1/4$).
$endgroup$
– FredH
Mar 21 at 9:07
$begingroup$
I assumed that there was a solution with O(1), my bad. I had a look at the algorithm you suggested, and if I understand it correctly, it should work for the example above, however, it's not as efficient as I would like.
$endgroup$
– Abs Spurdle
Mar 21 at 21:28
$begingroup$
My first thought would be to see if the (extended) Euclidean algorithm would be of any use: en.wikipedia.org/wiki/Modular_multiplicative_inverse
$endgroup$
– Matti P.
Mar 21 at 6:29
$begingroup$
My first thought would be to see if the (extended) Euclidean algorithm would be of any use: en.wikipedia.org/wiki/Modular_multiplicative_inverse
$endgroup$
– Matti P.
Mar 21 at 6:29
$begingroup$
For rational numbers $a/n$ and $b/n$, this is the same as solving the congruence $ax equiv b pmod n$. Solutions will not always exist (e.g., if $a = 2/4, b=1/4$).
$endgroup$
– FredH
Mar 21 at 9:07
$begingroup$
For rational numbers $a/n$ and $b/n$, this is the same as solving the congruence $ax equiv b pmod n$. Solutions will not always exist (e.g., if $a = 2/4, b=1/4$).
$endgroup$
– FredH
Mar 21 at 9:07
$begingroup$
I assumed that there was a solution with O(1), my bad. I had a look at the algorithm you suggested, and if I understand it correctly, it should work for the example above, however, it's not as efficient as I would like.
$endgroup$
– Abs Spurdle
Mar 21 at 21:28
$begingroup$
I assumed that there was a solution with O(1), my bad. I had a look at the algorithm you suggested, and if I understand it correctly, it should work for the example above, however, it's not as efficient as I would like.
$endgroup$
– Abs Spurdle
Mar 21 at 21:28
add a comment |
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$begingroup$
My first thought would be to see if the (extended) Euclidean algorithm would be of any use: en.wikipedia.org/wiki/Modular_multiplicative_inverse
$endgroup$
– Matti P.
Mar 21 at 6:29
$begingroup$
For rational numbers $a/n$ and $b/n$, this is the same as solving the congruence $ax equiv b pmod n$. Solutions will not always exist (e.g., if $a = 2/4, b=1/4$).
$endgroup$
– FredH
Mar 21 at 9:07
$begingroup$
I assumed that there was a solution with O(1), my bad. I had a look at the algorithm you suggested, and if I understand it correctly, it should work for the example above, however, it's not as efficient as I would like.
$endgroup$
– Abs Spurdle
Mar 21 at 21:28