In number theory, the Chinese remainder theorem states that if one knows the remainders of the Euclidean division of an integer n by several integers, then one can determine uniquely the remainder of the division of n by the product of these integers, under the condition that the divisors are pairwise coprime.
The earliest known statement of the theorem is by the Chinese mathematician Sun-tzu in the Sun-tzu Suan-ching in the 3rd century AD.
The Chinese remainder theorem is widely used for computing with large integers, as it allows replacing a computation for which one knows a bound on the size of the result by several similar computations on small integers.
There are certain things whose number is unknown. If we count them by threes, we have two left over; by fives, we have three left over; and by sevens, two are left over. How many things are there?
Sun-tzu's work contains neither a proof nor a full algorithm. What amounts to an algorithm for solving this problem was described by Aryabhata (6th century). Special cases of the Chinese remainder theorem were also known to Brahmagupta (7th century), and appear in Fibonacci's Liber Abaci (1202). The result was later generalized with a complete solution called Ta-yan-shu () in Ch'in Chiu-shao's 1247 Mathematical Treatise in Nine Sections (?, Shu-shu Chiu-chang) which was translated into English in early 19th century by British missionary Alexander Wylie.
The notion of congruences was first introduced and used by Gauss in his Disquisitiones Arithmeticae of 1801. Gauss illustrates the Chinese remainder theorem on a problem involving calendars, namely, "to find the years that have a certain period number with respect to the solar and lunar cycle and the Roman indiction." Gauss introduces a procedure for solving the problem that had already been used by Euler but was in fact an ancient method that had appeared several times.
The Chinese remainder theorem asserts that if the ni are pairwise coprime, and if a1, ..., ak are integers such that 0 ai < ni for every i, then there is one and only one integer x, such that 0 x < N and the remainder of the Euclidean division of x by ni is ai for every i.
This may be restated as follows in term of congruences: If the ni are pairwise coprime, and if a1, ..., ak are any integers, then there exist integers x such that
and any two solutions, say x1 and x2, are congruent modulo N, that is, x1 ? x2 (mod N).
In abstract algebra, the theorem is often restated as: if the ni are pairwise coprime, the map
between the ring of integers modulo N and the direct product of the rings of integers modulo the ni. This means that for doing a sequence of arithmetic operations in one may do the same computation independently in each and then get the result by applying the isomorphism (from the right to the left). This may be much faster than the direct computation if N and the number of operations are large. This is widely used, under the name multi-modular computation, for linear algebra over the integers or the rational numbers.
The existence and the uniqueness of the solution may be proven independently. However, the first proof of existence, given below, uses this uniqueness.
Suppose that x and y are both solutions to all the congruences. As x and y give the same remainder, when divided by ni, their difference x - y is a multiple of each ni. As the ni are pairwise coprime, their product N divides also x - y, and thus x and y are congruent modulo N. If x and y are supposed to be non negative and less than N (as in the first statement of the theorem), then their difference may be a multiple of N only if x = y.
maps congruence classes modulo N to sequences of congruence classes modulo ni. The proof of uniqueness shows that this map is injective. As the domain and the codomain of this map have the same number of elements, the map is also surjective, which proves the existence of the solution.
This proof is very simple but does not provide any direct way for computing a solution. Moreover, it cannot be generalized to other situations where the following proof can.
Existence may be established by an explicit construction of x. This construction may be split into two steps, firstly by solving the problem in the case of two moduli, and the second one by extending this solution to the general case by induction on the number of moduli.
We want to solve the system:
where and are coprime.
Bézout's identity asserts the existence of two integers and such that
The integers and may be computed by the extended Euclidean algorithm.
A solution is given by
implying that The second congruence is proved similarly, by exchanging the subscripts 1 and 2.
Consider a sequence of congruence equations:
where the are pairwise coprime. The two first equations have a solution provided by the method of the previous section. The set of the solutions of these two first equations is the set of all solutions of the equation
As the other are coprime with this reduces solving the initial problem of k equations to a similar problem with equations. Iterating the process, one gets eventually the solutions of the initial problem.
For constructing a solution, it is not necessary to make an induction on the number of moduli. However, such a direct construction involves more computation with large numbers, which makes it less efficient and less used. Nevertheless, Lagrange interpolation is a special case of this construction, applied to polynomials instead of integers.
Let be the product of all moduli but one. As the are pairwise coprime, and are coprime. Thus Bézout's identity applies, and there exist integers and such that
A solution of the system of congruences is
In fact, as is a multiple of for we have
Consider a system of congruences:
where the are pairwise coprime, and let In this section several methods are described for computing the unique solution for , such that and these methods are applied on the example:
It is easy to check whether a value of x is a solution: it suffices to compute the remainder of the Euclidean division of x by each ni. Thus, to find the solution, it suffices to check successively the integers from 0 to N until finding the solution.
Although very simple this method is very inefficient: for the simple example considered here, 40 integers (including 0) have to be checked for finding the solution, which is 39. This is an exponential time algorithm, as the size of the input is, up to a constant factor, the number of digits of N, and the average number of operations is of the order of N.
Therefore, this method is rarely used, neither for hand-written computation nor on computers.
The search of the solution may be made dramatically faster by sieving. For this method, we suppose, without loss of generality, that (if it were not the case, it would suffice to replace each by the remainder of its division by ). This implies that the solution belongs to the arithmetic progression
By testing the values of these numbers modulo one eventually finds a solution of the two first congruences. Then the solution belongs to the arithmetic progression
Testing the values of these numbers modulo , and continuing until every modulus has been tested gives eventually the solution.
This method is faster if the moduli have been ordered by decreasing value, that is if For the example, this gives the following computation. We consider first the numbers that are congruent to 4 modulo 5 (the largest modulus), which are 4, , , ... For each of them, compute the remainder by 4 (the second largest modulus) until getting a number congruent to 3 modulo 4. Then one can proceed by adding at each step, and computing only the remainders by 3. This gives
This method works well for hand-written computation with a product of moduli that is not too big. However, it is much slower than other methods, for very large products of moduli. Although dramatically faster than the systematic search, this method also has an exponential time complexity and is therefore not used on computers.
The constructive existence proof shows that, in the case of two moduli, the solution may be obtained by the computation of the Bézout coefficients of the moduli, followed by a few multiplications, additions and reductions modulo (for getting a result in the interval ). As the Bézout's coefficients may be computed with the extended Euclidean algorithm, the whole computation, at most, has a quadratic time complexity of where denotes the number of digits of
For more than two moduli, the method for two moduli allows the replacement of any two congruences by a single congruence modulo the product of the moduli. Iterating this process provides eventually the solution with a complexity, which is quadratic in the number of digits of the product of all moduli. This quadratic time complexity does not depend on the order in which the moduli are regrouped. One may regroup the two first moduli, then regroup the resulting modulus with the next one, and so on. This strategy is the easiest to implement, but it also requires more computation involving large numbers.
Another strategy consists in partitioning the moduli in pairs whose product have comparable sizes (as much as possible), applying, in parallel, the method of two moduli to each pair, and iterating with a number of moduli approximatively divided by two. This method allows an easy parallelization of the algorithm. Also, if fast algorithms (that is algorithms working in quasilinear time) are used for the basic operations, this method provides an algorithm for the whole computation that works in quasilinear time.
On the current example (which has only three moduli), both strategies are identical and work as follows.
Bézout's identity for 3 and 4 is
Putting this in the formula given for proving the existence gives
for a solution of the two first congruences, the other solutions being obtained by adding to -9 any multiple of . One may continue with any of these solutions, but the solution is smaller (in absolute value) and thus leads probably to an easier computation
Bézout identity for 5 and 3×4 = 12 is
Applying the same formula again, we get a solution of the problem:
The other solutions are obtained by adding any multiple of , and the smallest positive solution is .
The system of congruences solved by the Chinese remainder theorem may be rewritten as a system of simultaneous linear Diophantine equations:
where the unknown integers are and the Therefore, every general method for solving such systems may be used for finding the solution of Chinese remainder theorem, such as the reduction of the matrix of the system to Smith normal form or Hermite normal form. However, as usual when using a general algorithm for a more specific problem, this approach is less efficient than the method of the preceding section, based on a direct use of Bézout's identity.
In § Theorem statement, the Chinese remainder theorem has been stated in three different ways: in terms of remainders, of congruences, and of a ring isomorphism. The statement in terms of remainders does not apply, in general, to principal ideal domains, as remainders are not defined in such rings. However, the two other versions make sense over a principal ideal domain R: it suffices to replace "integer" by "element of the domain" and by R. These two versions of the theorem are true in this context, because the proofs (except for the first existence proof), are based on Euclid's lemma and Bézout's identity, which are true over every principal domain.
However, in general, the theorem is only an existence theorem and does not provide any way for computing the solution, unless one has an algorithm for computing the coefficients of Bézout's identity.
The statement in terms of remainders given in § Theorem statement cannot be generalized to any principal ideal domain, but its generalization to Euclidean domains is straightforward. The univariate polynomials over a field is the typical example of a Euclidean domain, which is not the integers. Therefore, we state the theorem for the case of a ring of univariate domain over a field For getting the theorem for a general Euclidean domain, it suffices to replace the degree by the Euclidean function of the Euclidean domain.
The Chinese remainder theorem for polynomials is thus: Let (the moduli) be, for i=1, ..., k, pairwise coprime polynomials in . Let be the degree of , and be the sum of the If are polynomials such that or for every i, then, there is one and only one polynomial , such that and the remainder of the Euclidean division of by is for every i.
The construction of the solution may be done as in § Existence (constructive proof) or § Existence (direct proof). However, the latter construction may be simplified by using, as follows, partial fraction decomposition instead of extended Euclidean algorithm.
Thus, we want to find a polynomial , which satisfies the congruences
Consider the polynomials
The partial fraction decomposition of gives k polynomials with degrees such that
Then a solution of the simultaneous congruence system is given by the polynomial
In fact, we have
This solution may have a degree larger that The unique solution of degree less than may be deduced by considering the remainder of the Euclidean division of by This solution is
They are pairwise coprime if the are all different. The remainder of the division by of a polynomial is
Now, let be constants (polynomials of degree 0) in Both Lagrange interpolation and Chinese remainder theorem assert the existence of a unique polynomial of degree less than such that
Lagrange interpolation formula is exactly the result, in this case, of the above construction of the solution. More precisely, let
The partial fraction decomposition of is
In fact, reducing the right-hand side to a common denominator one gets
and the numerator is equal to one, as being a polynomial of degree less than which takes the value one for different values of
Using the above general formula, we get the Lagrange interpolation formula:
Hermite interpolation is an application of the Chinese remainder theorem for univariate polynomials, which may involve moduli of arbitrary degrees (Lagrange interpolation involves only moduli of degree one).
The problem consists of finding a polynomial of the least possible degree, such that the polynomial and its first derivatives take given values at some fixed points.
More precisely, let be elements of the ground field and, for let be the values of the first derivatives of the sought polynomial at (including the 0th derivative, which is the value of the polynomial itself). The problem is to find a polynomial such that its jth derivative takes the value at for and
Consider the polynomial
This is the Taylor polynomial of order at , of the unknown polynomial Therefore, we must have
Conversely, any polynomial that satisfies these congruences, in particular verifies, for any
therefore is its Taylor polynomial of order at , that is, solves the initial Hermite interpolation problem. The Chinese remainder theorem asserts that there exists exactly one polynomial of degree less than the sum of the which satisfies these congruences.
There are several ways for computing the solution One may use the method described at the beginning of § Over univariate polynomial rings and Euclidean domains. One may also use the constructions given in § Existence (constructive proof) or § Existence (direct proof).
The Chinese remainder theorem can be generalized to non-coprime moduli. Let be any integers, let , and consider the system of congruences:
If , then this system of equations has a unique solution modulo . Otherwise, it has no solutions.
If we use Bézout's identity to write , then the solution is
This defines an integer, as g divides both m and n. Otherwise, the proof is very similar to that for coprime moduli.
The Chinese remainder theorem can be generalized to any ring, by using coprime ideals (also called comaximal ideals). Two ideals I and J are coprime if there are elements and such that This relation plays the role of Bézout's identity in the proofs related to this generalization, which, otherwise are very similar. The generalization may be stated as follows.
between the quotient ring and the direct product of the where "" denotes the image of the element in the quotient ring defined by the ideal Moreover, if is commutative, then the ideal intersection of pairwise coprime ideals is equal to their product; that is
if Ii and Ij are coprime for i ? j.
The prime-factor FFT algorithm (also called Good-Thomas algorithm) uses the Chinese remainder theorem for reducing the computation of a fast Fourier transform of size to the computation of two fast Fourier transforms of smaller sizes and (providing that and are coprime).
Most implementations of RSA use the Chinese remainder theorem during signing of HTTPS certificates and during decryption.
The Chinese remainder theorem can also be used in secret sharing, which consists of distributing a set of shares among a group of people who, all together (but no one alone), can recover a certain secret from the given set of shares. Each of the shares is represented in a congruence, and the solution of the system of congruences using the Chinese remainder theorem is the secret to be recovered. Secret sharing using the Chinese remainder theorem uses, along with the Chinese remainder theorem, special sequences of integers that guarantee the impossibility of recovering the secret from a set of shares with less than a certain cardinality.
Dedekind's theorem on the linear independence of characters. Let M be a monoid and k an integral domain, viewed as a monoid by considering the multiplication on k. Then any finite family ( fi )i?I of distinct monoid homomorphisms fi : M -> k is linearly independent. In other words, every family (?i)i?I of elements ?i ? k satisfying
must be equal to the family (0)i?I.
Proof. First assume that k is a field, otherwise, replace the integral domain k by its quotient field, and nothing will change. We can linearly extend the monoid homomorphisms fi : M -> k to k-algebra homomorphisms Fi : k[M] -> k, where k[M] is the monoid ring of M over k. Then, by linearity, the condition
Next, for i, j ? I; i ? j the two k-linear maps Fi : k[M] -> k and Fj : k[M] -> k are not proportional to each other. Otherwise fi and fj would also be proportional, and thus equal since as monoid homomorphisms they satisfy: fi (1) = 1 = fj (1), which contradicts the assumption that they are distinct.
Therefore, the kernels Ker Fi and Ker Fj are distinct. Since k[M]/Ker Fi ? Fi(k[M]) = k is a field, Ker Fi is a maximal ideal of k[M] for every i ? I. Because they are distinct and maximal the ideals Ker Fi and Ker Fj are coprime whenever i ? j. The Chinese Remainder Theorem (for general rings) yields an isomorphism:
Consequently, the map
is surjective. Under the isomorphisms k[M]/Ker Fi -> Fi(k[M]) = k, the map ? corresponds to:
for every vector (ui)i?I in the image of the map ?. Since ? is surjective, this means that
for every vector
Consequently, (?i)i?I = (0)i?I. QED.