How to solve system of linear congruences with the same modulo?
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I have to write program which is solving linear congruences withe the same modulo. I have system of congruences like that(only 2 unknowns x and y):
$$begin{cases}
a_1x+b_1y equiv c_1pmod n \
a_2x+b_2y equiv c_2pmod n \
end{cases}
$$
I'm using Cramer's rule now and it works when the modulus is prime, but when modulus is not prime, my program behaves incorrectly.
modular-arithmetic matrix-congruences
asked Nov 22 at 12:13
Jacek Kurek
1
add a comment |
up vote
0
down vote
favorite
I have to write program which is solving linear congruences withe the same modulo. I have system of congruences like that(only 2 unknowns x and y):
$$begin{cases}
a_1x+b_1y equiv c_1pmod n \
a_2x+b_2y equiv c_2pmod n \
end{cases}
$$
I'm using Cramer's rule now and it works when the modulus is prime, but when modulus is not prime, my program behaves incorrectly.
modular-arithmetic matrix-congruences
asked Nov 22 at 12:13
Jacek Kurek
1
Search on Hermite(-Smith) normal form for the general ideas.
– Bill Dubuque
Nov 22 at 19:20
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have to write program which is solving linear congruences withe the same modulo. I have system of congruences like that(only 2 unknowns x and y):
$$begin{cases}
a_1x+b_1y equiv c_1pmod n \
a_2x+b_2y equiv c_2pmod n \
end{cases}
$$
I'm using Cramer's rule now and it works when the modulus is prime, but when modulus is not prime, my program behaves incorrectly.
modular-arithmetic matrix-congruences
asked Nov 22 at 12:13
Jacek Kurek
1
I have to write program which is solving linear congruences withe the same modulo. I have system of congruences like that(only 2 unknowns x and y):
$$begin{cases}
a_1x+b_1y equiv c_1pmod n \
a_2x+b_2y equiv c_2pmod n \
end{cases}
$$
I'm using Cramer's rule now and it works when the modulus is prime, but when modulus is not prime, my program behaves incorrectly.
modular-arithmetic matrix-congruences
modular-arithmetic matrix-congruences
asked Nov 22 at 12:13
Jacek Kurek
1
asked Nov 22 at 12:13
Jacek Kurek
1
asked Nov 22 at 12:13
Jacek Kurek
1
asked Nov 22 at 12:13
Jacek Kurek
1
asked Nov 22 at 12:13
Jacek Kurek
1
1
Search on Hermite(-Smith) normal form for the general ideas.
– Bill Dubuque
Nov 22 at 19:20
add a comment |
Search on Hermite(-Smith) normal form for the general ideas.
– Bill Dubuque
Nov 22 at 19:20
Search on Hermite(-Smith) normal form for the general ideas.
– Bill Dubuque
Nov 22 at 19:20
Search on Hermite(-Smith) normal form for the general ideas.
– Bill Dubuque
Nov 22 at 19:20
add a comment |
1 Answer
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This is not as simple as it may look. For example, there may be multiple solutions for even a single equation ($2xequiv 0 pmod 4$ has 2 solutions$pmod 4$, for example). This breaks the nice theory one knows from the reals. That it works for primes $p$ is due to the fact that the the numbers you are working with also form a field, just like the reals.
It seems you have to go the hard way and
- factor $n$ into prime powers: $n=p_1^{alpha_1}p_2^{alpha_2}ldots$,
- you need to solve the problem independently for each prime power $p_i^{alpha_i}$ (see next steps) and at the end combine the results using the Chineese reminder theorem.
- for each prime power $p_i^{alpha_i}$, first solve it$pmod {p_i}$. This will work with what you have it seems. This gives you for all variables either a result$pmod {p_i}$ or no information (if the coefficinet before the variable was a multiple of $p_i$ in all equations). Introduce new variables ($x_i'$) for the ones you got results for ($x_i$) in the form of $x_i=p_ix_i' + r_i$.
- reformulate your equations using the new $x_i'$ variables and the older ones that did not yield information in step 3. You will see that now all coefficients before variables contain the factor $p_i$. Now it's time to consider$pmod {p_i^2}$
- You can divide all your equations by $p_i$. If the constant term is not divisible by $p_i$, the system has no solution. You must also divide the modulus by $p_i$, this is an equivalent operation! ($paequiv pb pmod {p^2}$ means the same as $aequiv b pmod {p}$.
- Now you are back where you started, you can solve the system with (possibly) some new variables $pmod {p}$ and continue as in steps 3-5. Each time you get information about your variables modulo a higer power of $p_i$, and at the end you get them for what you need, $pmod {p_i^{alpha_i}}$
Again, this is probably at least a few days work, not a few hours. Maybe somebody else has a better idea, though.
answered Nov 22 at 15:27
Ingix
3,204145
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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up vote
0
down vote
This is not as simple as it may look. For example, there may be multiple solutions for even a single equation ($2xequiv 0 pmod 4$ has 2 solutions$pmod 4$, for example). This breaks the nice theory one knows from the reals. That it works for primes $p$ is due to the fact that the the numbers you are working with also form a field, just like the reals.
It seems you have to go the hard way and
- factor $n$ into prime powers: $n=p_1^{alpha_1}p_2^{alpha_2}ldots$,
- you need to solve the problem independently for each prime power $p_i^{alpha_i}$ (see next steps) and at the end combine the results using the Chineese reminder theorem.
- for each prime power $p_i^{alpha_i}$, first solve it$pmod {p_i}$. This will work with what you have it seems. This gives you for all variables either a result$pmod {p_i}$ or no information (if the coefficinet before the variable was a multiple of $p_i$ in all equations). Introduce new variables ($x_i'$) for the ones you got results for ($x_i$) in the form of $x_i=p_ix_i' + r_i$.
- reformulate your equations using the new $x_i'$ variables and the older ones that did not yield information in step 3. You will see that now all coefficients before variables contain the factor $p_i$. Now it's time to consider$pmod {p_i^2}$
- You can divide all your equations by $p_i$. If the constant term is not divisible by $p_i$, the system has no solution. You must also divide the modulus by $p_i$, this is an equivalent operation! ($paequiv pb pmod {p^2}$ means the same as $aequiv b pmod {p}$.
- Now you are back where you started, you can solve the system with (possibly) some new variables $pmod {p}$ and continue as in steps 3-5. Each time you get information about your variables modulo a higer power of $p_i$, and at the end you get them for what you need, $pmod {p_i^{alpha_i}}$
Again, this is probably at least a few days work, not a few hours. Maybe somebody else has a better idea, though.
answered Nov 22 at 15:27
Ingix
3,204145
add a comment |
up vote
0
down vote
This is not as simple as it may look. For example, there may be multiple solutions for even a single equation ($2xequiv 0 pmod 4$ has 2 solutions$pmod 4$, for example). This breaks the nice theory one knows from the reals. That it works for primes $p$ is due to the fact that the the numbers you are working with also form a field, just like the reals.
It seems you have to go the hard way and
- factor $n$ into prime powers: $n=p_1^{alpha_1}p_2^{alpha_2}ldots$,
- you need to solve the problem independently for each prime power $p_i^{alpha_i}$ (see next steps) and at the end combine the results using the Chineese reminder theorem.
- for each prime power $p_i^{alpha_i}$, first solve it$pmod {p_i}$. This will work with what you have it seems. This gives you for all variables either a result$pmod {p_i}$ or no information (if the coefficinet before the variable was a multiple of $p_i$ in all equations). Introduce new variables ($x_i'$) for the ones you got results for ($x_i$) in the form of $x_i=p_ix_i' + r_i$.
- reformulate your equations using the new $x_i'$ variables and the older ones that did not yield information in step 3. You will see that now all coefficients before variables contain the factor $p_i$. Now it's time to consider$pmod {p_i^2}$
- You can divide all your equations by $p_i$. If the constant term is not divisible by $p_i$, the system has no solution. You must also divide the modulus by $p_i$, this is an equivalent operation! ($paequiv pb pmod {p^2}$ means the same as $aequiv b pmod {p}$.
- Now you are back where you started, you can solve the system with (possibly) some new variables $pmod {p}$ and continue as in steps 3-5. Each time you get information about your variables modulo a higer power of $p_i$, and at the end you get them for what you need, $pmod {p_i^{alpha_i}}$
Again, this is probably at least a few days work, not a few hours. Maybe somebody else has a better idea, though.
answered Nov 22 at 15:27
Ingix
3,204145
add a comment |
up vote
0
down vote
up vote
0
down vote
This is not as simple as it may look. For example, there may be multiple solutions for even a single equation ($2xequiv 0 pmod 4$ has 2 solutions$pmod 4$, for example). This breaks the nice theory one knows from the reals. That it works for primes $p$ is due to the fact that the the numbers you are working with also form a field, just like the reals.
It seems you have to go the hard way and
- factor $n$ into prime powers: $n=p_1^{alpha_1}p_2^{alpha_2}ldots$,
- you need to solve the problem independently for each prime power $p_i^{alpha_i}$ (see next steps) and at the end combine the results using the Chineese reminder theorem.
- for each prime power $p_i^{alpha_i}$, first solve it$pmod {p_i}$. This will work with what you have it seems. This gives you for all variables either a result$pmod {p_i}$ or no information (if the coefficinet before the variable was a multiple of $p_i$ in all equations). Introduce new variables ($x_i'$) for the ones you got results for ($x_i$) in the form of $x_i=p_ix_i' + r_i$.
- reformulate your equations using the new $x_i'$ variables and the older ones that did not yield information in step 3. You will see that now all coefficients before variables contain the factor $p_i$. Now it's time to consider$pmod {p_i^2}$
- You can divide all your equations by $p_i$. If the constant term is not divisible by $p_i$, the system has no solution. You must also divide the modulus by $p_i$, this is an equivalent operation! ($paequiv pb pmod {p^2}$ means the same as $aequiv b pmod {p}$.
- Now you are back where you started, you can solve the system with (possibly) some new variables $pmod {p}$ and continue as in steps 3-5. Each time you get information about your variables modulo a higer power of $p_i$, and at the end you get them for what you need, $pmod {p_i^{alpha_i}}$
Again, this is probably at least a few days work, not a few hours. Maybe somebody else has a better idea, though.
answered Nov 22 at 15:27
Ingix
3,204145
This is not as simple as it may look. For example, there may be multiple solutions for even a single equation ($2xequiv 0 pmod 4$ has 2 solutions$pmod 4$, for example). This breaks the nice theory one knows from the reals. That it works for primes $p$ is due to the fact that the the numbers you are working with also form a field, just like the reals.
It seems you have to go the hard way and
- factor $n$ into prime powers: $n=p_1^{alpha_1}p_2^{alpha_2}ldots$,
- you need to solve the problem independently for each prime power $p_i^{alpha_i}$ (see next steps) and at the end combine the results using the Chineese reminder theorem.
- for each prime power $p_i^{alpha_i}$, first solve it$pmod {p_i}$. This will work with what you have it seems. This gives you for all variables either a result$pmod {p_i}$ or no information (if the coefficinet before the variable was a multiple of $p_i$ in all equations). Introduce new variables ($x_i'$) for the ones you got results for ($x_i$) in the form of $x_i=p_ix_i' + r_i$.
- reformulate your equations using the new $x_i'$ variables and the older ones that did not yield information in step 3. You will see that now all coefficients before variables contain the factor $p_i$. Now it's time to consider$pmod {p_i^2}$
- You can divide all your equations by $p_i$. If the constant term is not divisible by $p_i$, the system has no solution. You must also divide the modulus by $p_i$, this is an equivalent operation! ($paequiv pb pmod {p^2}$ means the same as $aequiv b pmod {p}$.
- Now you are back where you started, you can solve the system with (possibly) some new variables $pmod {p}$ and continue as in steps 3-5. Each time you get information about your variables modulo a higer power of $p_i$, and at the end you get them for what you need, $pmod {p_i^{alpha_i}}$
Again, this is probably at least a few days work, not a few hours. Maybe somebody else has a better idea, though.
answered Nov 22 at 15:27
Ingix
3,204145
answered Nov 22 at 15:27
Ingix
3,204145
answered Nov 22 at 15:27
Ingix
3,204145
answered Nov 22 at 15:27
Ingix
3,204145
3,204145
add a comment |
add a comment |
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Search on Hermite(-Smith) normal form for the general ideas.
– Bill Dubuque
Nov 22 at 19:20