How to calculate mod inverse












0












$begingroup$


Given a number set of integers $mathbb{Z}$, how do I find the inverse of a given number?



I am trying to test an algorithm to extract the $k$ and $x$ values from the Elgamal Signature algorithm given that $k$ is repeated.



What I have is
$k$ congruent to $(m_1 - m_2)times(s_1 - s_2)^{-1} mod p - 1$
given $k$ is used twice.



I am not sure how to calculate the mod inverse though?
_
Is the above formula the same thing as $((m_1 - m_2) mod p -1 times (s_1 - s_2)^{-1} mod p -1) mod p -1$



I am not sure if it is any different since I am doing a mod inverse.



PS. I am a programmer, not a mathematician so please elaborate.






number-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Use the Extended Euclidean Algorithm, e.g. see here
    $endgroup$
    – Bill Dubuque
    Nov 19 '18 at 20:36










  • $begingroup$
    I know how to find a mod inverse. But if I have a number A*B^-1 mod p-1 is that equivalent to A mod p-1 * B mod p-1 mod p-1. That is what I found online but I wasn't sure.
    $endgroup$
    – User
    Nov 19 '18 at 20:39










  • $begingroup$
    $ab$ is invertible $iff a,b$ are invertible $iff a,b,$ are coprime to the modulus. When so we have $(ab)^{-1}equiv b^{-1}a^{-1},$ by $ b^{-1}a^{-1} (ab) equiv b^{-1}(a^{-1}a)bequiv b^{-1}b equiv 1 $ (inverses are always unique)
    $endgroup$
    – Bill Dubuque
    Nov 19 '18 at 20:58












  • $begingroup$
    So what if I have a number * an inverse mod p -1. How would I break that down?
    $endgroup$
    – User
    Nov 19 '18 at 21:01










  • $begingroup$
    Calculate the inverse then modular_multiply the two as you would any pair of (modular) integers - using the mod prodcut rule
    $endgroup$
    – Bill Dubuque
    Nov 19 '18 at 21:05


















0












$begingroup$


Given a number set of integers $mathbb{Z}$, how do I find the inverse of a given number?



I am trying to test an algorithm to extract the $k$ and $x$ values from the Elgamal Signature algorithm given that $k$ is repeated.



What I have is
$k$ congruent to $(m_1 - m_2)times(s_1 - s_2)^{-1} mod p - 1$
given $k$ is used twice.



I am not sure how to calculate the mod inverse though?
_
Is the above formula the same thing as $((m_1 - m_2) mod p -1 times (s_1 - s_2)^{-1} mod p -1) mod p -1$



I am not sure if it is any different since I am doing a mod inverse.



PS. I am a programmer, not a mathematician so please elaborate.






number-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Use the Extended Euclidean Algorithm, e.g. see here
    $endgroup$
    – Bill Dubuque
    Nov 19 '18 at 20:36










  • $begingroup$
    I know how to find a mod inverse. But if I have a number A*B^-1 mod p-1 is that equivalent to A mod p-1 * B mod p-1 mod p-1. That is what I found online but I wasn't sure.
    $endgroup$
    – User
    Nov 19 '18 at 20:39










  • $begingroup$
    $ab$ is invertible $iff a,b$ are invertible $iff a,b,$ are coprime to the modulus. When so we have $(ab)^{-1}equiv b^{-1}a^{-1},$ by $ b^{-1}a^{-1} (ab) equiv b^{-1}(a^{-1}a)bequiv b^{-1}b equiv 1 $ (inverses are always unique)
    $endgroup$
    – Bill Dubuque
    Nov 19 '18 at 20:58












  • $begingroup$
    So what if I have a number * an inverse mod p -1. How would I break that down?
    $endgroup$
    – User
    Nov 19 '18 at 21:01










  • $begingroup$
    Calculate the inverse then modular_multiply the two as you would any pair of (modular) integers - using the mod prodcut rule
    $endgroup$
    – Bill Dubuque
    Nov 19 '18 at 21:05
















0












0








0


1



$begingroup$


Given a number set of integers $mathbb{Z}$, how do I find the inverse of a given number?



I am trying to test an algorithm to extract the $k$ and $x$ values from the Elgamal Signature algorithm given that $k$ is repeated.



What I have is
$k$ congruent to $(m_1 - m_2)times(s_1 - s_2)^{-1} mod p - 1$
given $k$ is used twice.



I am not sure how to calculate the mod inverse though?
_
Is the above formula the same thing as $((m_1 - m_2) mod p -1 times (s_1 - s_2)^{-1} mod p -1) mod p -1$



I am not sure if it is any different since I am doing a mod inverse.



PS. I am a programmer, not a mathematician so please elaborate.






number-theory







share|cite|improve this question











$endgroup$




Given a number set of integers $mathbb{Z}$, how do I find the inverse of a given number?



I am trying to test an algorithm to extract the $k$ and $x$ values from the Elgamal Signature algorithm given that $k$ is repeated.



What I have is
$k$ congruent to $(m_1 - m_2)times(s_1 - s_2)^{-1} mod p - 1$
given $k$ is used twice.



I am not sure how to calculate the mod inverse though?
_
Is the above formula the same thing as $((m_1 - m_2) mod p -1 times (s_1 - s_2)^{-1} mod p -1) mod p -1$



I am not sure if it is any different since I am doing a mod inverse.



PS. I am a programmer, not a mathematician so please elaborate.






number-theory




number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 '18 at 21:02









Mason

1,9651530




1,9651530










asked Nov 19 '18 at 20:29









UserUser

11




11








  • 2




    $begingroup$
    Use the Extended Euclidean Algorithm, e.g. see here
    $endgroup$
    – Bill Dubuque
    Nov 19 '18 at 20:36










  • $begingroup$
    I know how to find a mod inverse. But if I have a number A*B^-1 mod p-1 is that equivalent to A mod p-1 * B mod p-1 mod p-1. That is what I found online but I wasn't sure.
    $endgroup$
    – User
    Nov 19 '18 at 20:39










  • $begingroup$
    $ab$ is invertible $iff a,b$ are invertible $iff a,b,$ are coprime to the modulus. When so we have $(ab)^{-1}equiv b^{-1}a^{-1},$ by $ b^{-1}a^{-1} (ab) equiv b^{-1}(a^{-1}a)bequiv b^{-1}b equiv 1 $ (inverses are always unique)
    $endgroup$
    – Bill Dubuque
    Nov 19 '18 at 20:58












  • $begingroup$
    So what if I have a number * an inverse mod p -1. How would I break that down?
    $endgroup$
    – User
    Nov 19 '18 at 21:01










  • $begingroup$
    Calculate the inverse then modular_multiply the two as you would any pair of (modular) integers - using the mod prodcut rule
    $endgroup$
    – Bill Dubuque
    Nov 19 '18 at 21:05
















  • 2




    $begingroup$
    Use the Extended Euclidean Algorithm, e.g. see here
    $endgroup$
    – Bill Dubuque
    Nov 19 '18 at 20:36










  • $begingroup$
    I know how to find a mod inverse. But if I have a number A*B^-1 mod p-1 is that equivalent to A mod p-1 * B mod p-1 mod p-1. That is what I found online but I wasn't sure.
    $endgroup$
    – User
    Nov 19 '18 at 20:39










  • $begingroup$
    $ab$ is invertible $iff a,b$ are invertible $iff a,b,$ are coprime to the modulus. When so we have $(ab)^{-1}equiv b^{-1}a^{-1},$ by $ b^{-1}a^{-1} (ab) equiv b^{-1}(a^{-1}a)bequiv b^{-1}b equiv 1 $ (inverses are always unique)
    $endgroup$
    – Bill Dubuque
    Nov 19 '18 at 20:58












  • $begingroup$
    So what if I have a number * an inverse mod p -1. How would I break that down?
    $endgroup$
    – User
    Nov 19 '18 at 21:01










  • $begingroup$
    Calculate the inverse then modular_multiply the two as you would any pair of (modular) integers - using the mod prodcut rule
    $endgroup$
    – Bill Dubuque
    Nov 19 '18 at 21:05










2




2




$begingroup$
Use the Extended Euclidean Algorithm, e.g. see here
$endgroup$
– Bill Dubuque
Nov 19 '18 at 20:36




$begingroup$
Use the Extended Euclidean Algorithm, e.g. see here
$endgroup$
– Bill Dubuque
Nov 19 '18 at 20:36












$begingroup$
I know how to find a mod inverse. But if I have a number A*B^-1 mod p-1 is that equivalent to A mod p-1 * B mod p-1 mod p-1. That is what I found online but I wasn't sure.
$endgroup$
– User
Nov 19 '18 at 20:39




$begingroup$
I know how to find a mod inverse. But if I have a number A*B^-1 mod p-1 is that equivalent to A mod p-1 * B mod p-1 mod p-1. That is what I found online but I wasn't sure.
$endgroup$
– User
Nov 19 '18 at 20:39












$begingroup$
$ab$ is invertible $iff a,b$ are invertible $iff a,b,$ are coprime to the modulus. When so we have $(ab)^{-1}equiv b^{-1}a^{-1},$ by $ b^{-1}a^{-1} (ab) equiv b^{-1}(a^{-1}a)bequiv b^{-1}b equiv 1 $ (inverses are always unique)
$endgroup$
– Bill Dubuque
Nov 19 '18 at 20:58






$begingroup$
$ab$ is invertible $iff a,b$ are invertible $iff a,b,$ are coprime to the modulus. When so we have $(ab)^{-1}equiv b^{-1}a^{-1},$ by $ b^{-1}a^{-1} (ab) equiv b^{-1}(a^{-1}a)bequiv b^{-1}b equiv 1 $ (inverses are always unique)
$endgroup$
– Bill Dubuque
Nov 19 '18 at 20:58














$begingroup$
So what if I have a number * an inverse mod p -1. How would I break that down?
$endgroup$
– User
Nov 19 '18 at 21:01




$begingroup$
So what if I have a number * an inverse mod p -1. How would I break that down?
$endgroup$
– User
Nov 19 '18 at 21:01












$begingroup$
Calculate the inverse then modular_multiply the two as you would any pair of (modular) integers - using the mod prodcut rule
$endgroup$
– Bill Dubuque
Nov 19 '18 at 21:05






$begingroup$
Calculate the inverse then modular_multiply the two as you would any pair of (modular) integers - using the mod prodcut rule
$endgroup$
– Bill Dubuque
Nov 19 '18 at 21:05












1 Answer
1






active

oldest

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0












$begingroup$

Yes, the two formulas you wrote in the question give the same output.



More generally, as Bill Dubuque points out in the comments, you can usually just take mods at each step, instead of doing the whole computation and then modding at the end. However, exponentiation is a notable exception; you can reduce the base but generally not the exponent
$$ a^k bmod n quad=quad (abmod n)^k bmod n qquadneqqquad (abmod n)^{(k bmod n)}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This answer exists primarily to remove this question from the Unanswered list; please upvote (or give Best Answer) to complete the process.
    $endgroup$
    – aleph_two
    Dec 22 '18 at 5:06













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Yes, the two formulas you wrote in the question give the same output.



More generally, as Bill Dubuque points out in the comments, you can usually just take mods at each step, instead of doing the whole computation and then modding at the end. However, exponentiation is a notable exception; you can reduce the base but generally not the exponent
$$ a^k bmod n quad=quad (abmod n)^k bmod n qquadneqqquad (abmod n)^{(k bmod n)}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This answer exists primarily to remove this question from the Unanswered list; please upvote (or give Best Answer) to complete the process.
    $endgroup$
    – aleph_two
    Dec 22 '18 at 5:06


















0












$begingroup$

Yes, the two formulas you wrote in the question give the same output.



More generally, as Bill Dubuque points out in the comments, you can usually just take mods at each step, instead of doing the whole computation and then modding at the end. However, exponentiation is a notable exception; you can reduce the base but generally not the exponent
$$ a^k bmod n quad=quad (abmod n)^k bmod n qquadneqqquad (abmod n)^{(k bmod n)}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This answer exists primarily to remove this question from the Unanswered list; please upvote (or give Best Answer) to complete the process.
    $endgroup$
    – aleph_two
    Dec 22 '18 at 5:06
















0












0








0





$begingroup$

Yes, the two formulas you wrote in the question give the same output.



More generally, as Bill Dubuque points out in the comments, you can usually just take mods at each step, instead of doing the whole computation and then modding at the end. However, exponentiation is a notable exception; you can reduce the base but generally not the exponent
$$ a^k bmod n quad=quad (abmod n)^k bmod n qquadneqqquad (abmod n)^{(k bmod n)}.$$






share|cite|improve this answer











$endgroup$



Yes, the two formulas you wrote in the question give the same output.



More generally, as Bill Dubuque points out in the comments, you can usually just take mods at each step, instead of doing the whole computation and then modding at the end. However, exponentiation is a notable exception; you can reduce the base but generally not the exponent
$$ a^k bmod n quad=quad (abmod n)^k bmod n qquadneqqquad (abmod n)^{(k bmod n)}.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 22 '18 at 5:52


























community wiki





2 revs
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  • $begingroup$
    This answer exists primarily to remove this question from the Unanswered list; please upvote (or give Best Answer) to complete the process.
    $endgroup$
    – aleph_two
    Dec 22 '18 at 5:06




















  • $begingroup$
    This answer exists primarily to remove this question from the Unanswered list; please upvote (or give Best Answer) to complete the process.
    $endgroup$
    – aleph_two
    Dec 22 '18 at 5:06


















$begingroup$
This answer exists primarily to remove this question from the Unanswered list; please upvote (or give Best Answer) to complete the process.
$endgroup$
– aleph_two
Dec 22 '18 at 5:06






$begingroup$
This answer exists primarily to remove this question from the Unanswered list; please upvote (or give Best Answer) to complete the process.
$endgroup$
– aleph_two
Dec 22 '18 at 5:06




















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