Is there a way to reverse the Chinese Remainder Theorem? What extra information do we need?
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Dear math stackexchange community,
Given a list of numbers < N, after generating the Chinese Remainder, is there a way to get back to the same list of numbers?
Example:
N = 100
List of numbers = [32, 11, 57, 12] (all below N)
For Chinese Remainder we need pairwise coprimes, so I use a list of primes < N. Because a list of primes is automatically pairwise coprimes...
Primes = [2, 3, 5, 7] = a
List of remainder = [32 % 2, 11 % 3, 57 % 5, 12 % 7] = [0, 2, 2, 5] = m
Given that we know [a] and [m] we can find the lowest Chinese Remainder: 152
We can double check this result:
- 152 % 2 = 0
- 152 % 3 = 2
- 152 % 5 = 2
- 152 % 7 = 5
So just by knowing the number 152, we know the sets of remainders for the original list, what extra information do I need to find the original list of numbers?
(current idea is to generate all possible sequences that fit the prerequisites of [m] and use a number for the index within that sequence...) But maybe I'm missing something, any advice or suggestions would be more than welcome!
number-theory prime-numbers modular-arithmetic information-theory chinese-remainder-theorem
asked Nov 21 at 12:20
Eli
1534
add a comment |
up vote
0
down vote
favorite
Dear math stackexchange community,
Given a list of numbers < N, after generating the Chinese Remainder, is there a way to get back to the same list of numbers?
Example:
N = 100
List of numbers = [32, 11, 57, 12] (all below N)
For Chinese Remainder we need pairwise coprimes, so I use a list of primes < N. Because a list of primes is automatically pairwise coprimes...
Primes = [2, 3, 5, 7] = a
List of remainder = [32 % 2, 11 % 3, 57 % 5, 12 % 7] = [0, 2, 2, 5] = m
Given that we know [a] and [m] we can find the lowest Chinese Remainder: 152
We can double check this result:
- 152 % 2 = 0
- 152 % 3 = 2
- 152 % 5 = 2
- 152 % 7 = 5
So just by knowing the number 152, we know the sets of remainders for the original list, what extra information do I need to find the original list of numbers?
(current idea is to generate all possible sequences that fit the prerequisites of [m] and use a number for the index within that sequence...) But maybe I'm missing something, any advice or suggestions would be more than welcome!
number-theory prime-numbers modular-arithmetic information-theory chinese-remainder-theorem
asked Nov 21 at 12:20
Eli
1534
You need all the residues for all the numbers. Then, you can recover the original numbers.
– Peter
Nov 21 at 19:05
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Dear math stackexchange community,
Given a list of numbers < N, after generating the Chinese Remainder, is there a way to get back to the same list of numbers?
Example:
N = 100
List of numbers = [32, 11, 57, 12] (all below N)
For Chinese Remainder we need pairwise coprimes, so I use a list of primes < N. Because a list of primes is automatically pairwise coprimes...
Primes = [2, 3, 5, 7] = a
List of remainder = [32 % 2, 11 % 3, 57 % 5, 12 % 7] = [0, 2, 2, 5] = m
Given that we know [a] and [m] we can find the lowest Chinese Remainder: 152
We can double check this result:
- 152 % 2 = 0
- 152 % 3 = 2
- 152 % 5 = 2
- 152 % 7 = 5
So just by knowing the number 152, we know the sets of remainders for the original list, what extra information do I need to find the original list of numbers?
(current idea is to generate all possible sequences that fit the prerequisites of [m] and use a number for the index within that sequence...) But maybe I'm missing something, any advice or suggestions would be more than welcome!
number-theory prime-numbers modular-arithmetic information-theory chinese-remainder-theorem
asked Nov 21 at 12:20
Eli
1534
Dear math stackexchange community,
Given a list of numbers < N, after generating the Chinese Remainder, is there a way to get back to the same list of numbers?
Example:
N = 100
List of numbers = [32, 11, 57, 12] (all below N)
For Chinese Remainder we need pairwise coprimes, so I use a list of primes < N. Because a list of primes is automatically pairwise coprimes...
Primes = [2, 3, 5, 7] = a
List of remainder = [32 % 2, 11 % 3, 57 % 5, 12 % 7] = [0, 2, 2, 5] = m
Given that we know [a] and [m] we can find the lowest Chinese Remainder: 152
We can double check this result:
- 152 % 2 = 0
- 152 % 3 = 2
- 152 % 5 = 2
- 152 % 7 = 5
So just by knowing the number 152, we know the sets of remainders for the original list, what extra information do I need to find the original list of numbers?
(current idea is to generate all possible sequences that fit the prerequisites of [m] and use a number for the index within that sequence...) But maybe I'm missing something, any advice or suggestions would be more than welcome!
number-theory prime-numbers modular-arithmetic information-theory chinese-remainder-theorem
number-theory prime-numbers modular-arithmetic information-theory chinese-remainder-theorem
asked Nov 21 at 12:20
Eli
1534
asked Nov 21 at 12:20
Eli
1534
asked Nov 21 at 12:20
Eli
1534
asked Nov 21 at 12:20
Eli
1534
asked Nov 21 at 12:20
Eli
1534
1534
You need all the residues for all the numbers. Then, you can recover the original numbers.
– Peter
Nov 21 at 19:05
add a comment |
You need all the residues for all the numbers. Then, you can recover the original numbers.
– Peter
Nov 21 at 19:05
You need all the residues for all the numbers. Then, you can recover the original numbers.
– Peter
Nov 21 at 19:05
You need all the residues for all the numbers. Then, you can recover the original numbers.
– Peter
Nov 21 at 19:05
add a comment |
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You need all the residues for all the numbers. Then, you can recover the original numbers.
– Peter
Nov 21 at 19:05