Mixing
Any good way to calculate $frac {alpha ^ n - 1 } {alpha - 1} pmod{c}$
4
$begingroup$
I tried by multiplying modular inverse of denominator to the numerator and then taking modulo $c$, but there are problems when the inverse does not exist.
So is there a good way to solve this problem.
Constraints
$$ 1 le alpha le 1e9 $$
$c$ is a prime
$$ 1 le n le 1e9 $$
number-theory
edited Jan 4 at 21:45
rtybase
10.7k21533
asked Jan 3 at 16:59
satvik choudharysatvik choudhary
215
$endgroup$
add a comment |
4
$begingroup$
I tried by multiplying modular inverse of denominator to the numerator and then taking modulo $c$, but there are problems when the inverse does not exist.
So is there a good way to solve this problem.
Constraints
$$ 1 le alpha le 1e9 $$
$c$ is a prime
$$ 1 le n le 1e9 $$
number-theory
edited Jan 4 at 21:45
rtybase
10.7k21533
asked Jan 3 at 16:59
satvik choudharysatvik choudhary
215
$endgroup$
1
$begingroup$
If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
$endgroup$
– Wojowu
Jan 3 at 17:04
add a comment |
4
4
4
$begingroup$
I tried by multiplying modular inverse of denominator to the numerator and then taking modulo $c$, but there are problems when the inverse does not exist.
So is there a good way to solve this problem.
Constraints
$$ 1 le alpha le 1e9 $$
$c$ is a prime
$$ 1 le n le 1e9 $$
number-theory
edited Jan 4 at 21:45
rtybase
10.7k21533
asked Jan 3 at 16:59
satvik choudharysatvik choudhary
215
$endgroup$
I tried by multiplying modular inverse of denominator to the numerator and then taking modulo $c$, but there are problems when the inverse does not exist.
So is there a good way to solve this problem.
Constraints
$$ 1 le alpha le 1e9 $$
$c$ is a prime
$$ 1 le n le 1e9 $$
number-theory
number-theory
edited Jan 4 at 21:45
rtybase
10.7k21533
asked Jan 3 at 16:59
satvik choudharysatvik choudhary
215
edited Jan 4 at 21:45
rtybase
10.7k21533
asked Jan 3 at 16:59
satvik choudharysatvik choudhary
215
edited Jan 4 at 21:45
rtybase
10.7k21533
edited Jan 4 at 21:45
rtybase
10.7k21533
edited Jan 4 at 21:45
rtybase
10.7k21533
10.7k21533
asked Jan 3 at 16:59
satvik choudharysatvik choudhary
215
asked Jan 3 at 16:59
satvik choudharysatvik choudhary
215
asked Jan 3 at 16:59
satvik choudharysatvik choudhary
215
215
1
$begingroup$
If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
$endgroup$
– Wojowu
Jan 3 at 17:04
add a comment |
1
$begingroup$
If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
$endgroup$
– Wojowu
Jan 3 at 17:04
1
1
$begingroup$
If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
$endgroup$
– Wojowu
Jan 3 at 17:04
$begingroup$
If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
$endgroup$
– Wojowu
Jan 3 at 17:04
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
Set $S_0:=1$ and then recursively $S_k:=alpha S_{k-1}+1 pmod c$ for all $k=1,dotsc,n-1$. The last value $S_{n-1}$ is what you seek.
answered Jan 3 at 18:21
W-t-PW-t-P
64559
$endgroup$
$begingroup$
Its too slow to be just linearly calculated with n ~ 1e9
$endgroup$
– satvik choudhary
Jan 4 at 10:19
$begingroup$
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
$endgroup$
– satvik choudhary
Jan 4 at 10:29
$begingroup$
Sure, as long as running time is an issue. (From your question I understood that you are mostly struggling with the situation where the denominator is not invertible.)
$endgroup$
– W-t-P
Jan 5 at 9:01
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
Set $S_0:=1$ and then recursively $S_k:=alpha S_{k-1}+1 pmod c$ for all $k=1,dotsc,n-1$. The last value $S_{n-1}$ is what you seek.
answered Jan 3 at 18:21
W-t-PW-t-P
64559
$endgroup$
$begingroup$
Its too slow to be just linearly calculated with n ~ 1e9
$endgroup$
– satvik choudhary
Jan 4 at 10:19
$begingroup$
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
$endgroup$
– satvik choudhary
Jan 4 at 10:29
$begingroup$
Sure, as long as running time is an issue. (From your question I understood that you are mostly struggling with the situation where the denominator is not invertible.)
$endgroup$
– W-t-P
Jan 5 at 9:01
add a comment |
2
$begingroup$
Set $S_0:=1$ and then recursively $S_k:=alpha S_{k-1}+1 pmod c$ for all $k=1,dotsc,n-1$. The last value $S_{n-1}$ is what you seek.
answered Jan 3 at 18:21
W-t-PW-t-P
64559
$endgroup$
$begingroup$
Its too slow to be just linearly calculated with n ~ 1e9
$endgroup$
– satvik choudhary
Jan 4 at 10:19
$begingroup$
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
$endgroup$
– satvik choudhary
Jan 4 at 10:29
$begingroup$
Sure, as long as running time is an issue. (From your question I understood that you are mostly struggling with the situation where the denominator is not invertible.)
$endgroup$
– W-t-P
Jan 5 at 9:01
add a comment |
2
2
2
$begingroup$
Set $S_0:=1$ and then recursively $S_k:=alpha S_{k-1}+1 pmod c$ for all $k=1,dotsc,n-1$. The last value $S_{n-1}$ is what you seek.
answered Jan 3 at 18:21
W-t-PW-t-P
64559
$endgroup$
Set $S_0:=1$ and then recursively $S_k:=alpha S_{k-1}+1 pmod c$ for all $k=1,dotsc,n-1$. The last value $S_{n-1}$ is what you seek.
answered Jan 3 at 18:21
W-t-PW-t-P
64559
answered Jan 3 at 18:21
W-t-PW-t-P
64559
answered Jan 3 at 18:21
W-t-PW-t-P
64559
answered Jan 3 at 18:21
W-t-PW-t-P
64559
64559
$begingroup$
Its too slow to be just linearly calculated with n ~ 1e9
$endgroup$
– satvik choudhary
Jan 4 at 10:19
$begingroup$
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
$endgroup$
– satvik choudhary
Jan 4 at 10:29
$begingroup$
Sure, as long as running time is an issue. (From your question I understood that you are mostly struggling with the situation where the denominator is not invertible.)
$endgroup$
– W-t-P
Jan 5 at 9:01
add a comment |
$begingroup$
Its too slow to be just linearly calculated with n ~ 1e9
$endgroup$
– satvik choudhary
Jan 4 at 10:19
$begingroup$
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
$endgroup$
– satvik choudhary
Jan 4 at 10:29
$begingroup$
Sure, as long as running time is an issue. (From your question I understood that you are mostly struggling with the situation where the denominator is not invertible.)
$endgroup$
– W-t-P
Jan 5 at 9:01
$begingroup$
Its too slow to be just linearly calculated with n ~ 1e9
$endgroup$
– satvik choudhary
Jan 4 at 10:19
$begingroup$
Its too slow to be just linearly calculated with n ~ 1e9
$endgroup$
– satvik choudhary
Jan 4 at 10:19
$begingroup$
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
$endgroup$
– satvik choudhary
Jan 4 at 10:29
$begingroup$
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
$endgroup$
– satvik choudhary
Jan 4 at 10:29
$begingroup$
Sure, as long as running time is an issue. (From your question I understood that you are mostly struggling with the situation where the denominator is not invertible.)
$endgroup$
– W-t-P
Jan 5 at 9:01
$begingroup$
Sure, as long as running time is an issue. (From your question I understood that you are mostly struggling with the situation where the denominator is not invertible.)
$endgroup$
– W-t-P
Jan 5 at 9:01
add a comment |
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1
$begingroup$
If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
$endgroup$
– Wojowu
Jan 3 at 17:04