Türkçe Bilgi: Çinlilerin kalan teoremi

's are not pairwise coprime. A solution x exists if and only if: : $a_i \equiv a_j \pmod \qquad \mboxi\mboxj . \,\!$ All solutions x are then congruent modulo the least common multiple of the n_i. Versions of the Chinese remainder theorem were also known to Brahmagupta (7th century), and appear in Fibonacci's Liber Abaci (1202). Çözümü Bulmak için Yapıcı bir algoritma Aşağıdaki algoritma sadece aralarında ikili asal olan $n_i$ sayıları için geçerlidir. (Eşzamanlı kongrüanslar için modül asal ikili değilse, peşpeşe ekleme yöntemi kullanarak çözüme ulaşılabilir.) Aşağıdaki gibi verilmiş olan kongrüanslar sistemi için bir çözümün gerekli olduğunu varsayalım: : $x \equiv a_i \pmod \quad\mathrm\; i = 1, \ldots, k.$ İlk adım olarak $N=n_1n_2\ldots n_k$ çarpımı tanımlanır. Sonra aşağıdaki yöntem izlenerek bir x çözümü bulunabilir. Bütün ideğerleri için $n_i$ ve $N/n_i$ sayılarının asal olduğunu biliyoruz. Genişletilmiş Öklid algoritması yardımıyla $r_in_i + s_iN/n_i = 1$ eşitliğini sağlayan $r_i$ ve $s_i$ sayılarını bulabiliriz. Ardından $e_i=s_iN/n_i$ değerine ulaşılır ve yukarıda verilen eşitlik şu hale getirilir: : $r_i n_i + e_i = 1 \,\!$ $e_i$ sayısı göz önünde bulundurulursa, yukarıdaki eşitliğin $n_i$ ile bölünmesi halinde kalanın 1 olması gerektiğini görürüz. Diğer taraftan, $s_iN/n_i$ şeklinde ifade edildiğinden, elimizdeki $N$ değeri $j\ne i$ olduğu sürece bu eşitliğin $n_j$ ile eşit olarak bölüneceğini garanti eder. : $e_i \equiv 1 \pmod \quad \mathrm \quad e_i \equiv 0 \pmod \quad \mathrm ~ i \ne j$ Bu nedenle, kongrüanslar için kullanılan çarpım kuralları da göz önünde bulundurularak, bu kongrüans sistemine bulunabilecek çözümlerden bir tanesi: : $x = \sum_^k a_i e_i.\!$ Örneğin, aşağıda verilen kongrüans sistemi için bir x tamsayısı bulmaya çalışalım: : $x \equiv 2 \pmod, \,\!$ : $x \equiv 3 \pmod, \,\!$ : $x \equiv 1 \pmod. \,\!$ Genişletilmi Öklid algoritması yardımıyla x için mod 3 ve 20 [1] kullanarak (−13) × 3 + 2 × 20 = 1, yani e₁ = 40 değeri bulunur. Aynı yöntemle x için mod 4 ve 15 [2] kullanarak (−11) × 4 + 3 × 15 = 1, yani e₂ = 45 değeri bulunur. Son olarak, x için mod 5 ve 12 [3] kullanarak 5 × 5 + (−2) × 12 = 1, yani e₃ = −24 değerini elde ederiz. Bulduklarımızı birleştirecek olursak muhtemeel x sayılarından bir tanesi 2 × 40 + 3 × 45 + 1 × (−24) = 191 olur. Diğer bütün çözümler 191'in mod 60 × 4 × 5 = 60 değerine eşleniktir, bir diğer ifadeyle olası x değerlerinin hepsi 11'in mod 60 değeriyle eşleniktir. NOT: Genişletilmi Öklid algoritması çeşitli yaklaşımlarla uygulanabilir. Bu soru için farklı uygulamalar kullanılarak $e_1=-20$ , $e_2=-15$ , ve $e_3=-24$ değerleri elde edilebilir. Fakat, bunlar da bizi aynı sonuca ulaştıracaktır, yani (-20)2+(-15)3+(-24)1=-109=11 mod 60. Temel ideal bölgeler için Açıklama For a principal ideal domain R the Chinese remainder theorem takes the following form: If u₁, ..., u_k are elements of R which are pairwise coprime, and u denotes the product u₁...u_k, then the quotient ring R/uR and the product ring R/u₁R× ... × R/u_kR are isomorphic via the isomorphism : $f : R/uR \rightarrow R/u_1R \times \cdots \times R/u_k R$ such that : $f(x +uR) = (x + u_1R, \ldots , x +u_kR) \quad\mbox x\in R.$ This isomorphism is unique; the inverse isomorphism can be constructed as follows. For each i, the elements u_i and u/u_i are coprime, and therefore there exist elements r and s in R with : $r u_i + s u/u_i = 1. \,\!$ Set e_i = s u/u_i. Then the inverse of f is the map : $g : R/u_1R \times \cdots \times R/u_kR \rightarrow R/uR$ such that : $g(a_1+u_1R,\ldots ,a_k+u_kR)= \left( \sum_^k a_i e_i \right) + uR \quad\mboxa_1,\ldots,a_k\in R.$ Note that this statement is a straightforward generalization of the above theorem about integer congruences: the ring 'Z' of integers is a principal ideal domain, the surjectivity of the map f shows that every system of congruences of the form : $x \equiv a_i \pmod \quad\mathrm\; i = 1, \ldots, k$ can be solved for x, and the injectivity of the map f shows that all the solutions x are congruent modulo u. Genel halkaları için Açıklama The general form of the Chinese remainder theorem, which implies all the statements given above, can be formulated for commutative rings and ideals. If R is a commutative ring and I₁, ..., I_k are ideals of R which are pairwise coprime (meaning that I_i + I_j = R whenever i ≠ j), then the product I of these ideals is equal to their intersection, and the quotient ring R/I is isomorphic to the product ring R/I₁ x R/I₂ x ... x R/Ik via the isomorphism : $f : R/I \rightarrow R/I_1 \times \cdots \times R/I_k$ such that : $f(x + I) = (x +I_1, \ldots , x+I_k) \quad\mbox x\in R.$ Uygulamalar * In the RSA algorithm calculations are made modulo $n$ , where $n$ is a product of two large prime numbers $p$ and $q$ . 1024-, 2048- or 4096-bit integers $n$ are commonly used, making calculations in $\Bbb/n\Bbb$ very time-consuming. By the Chinese Remainder Theorem, however, these calculations can be done in the isomorphic ring $\Bbb/p\Bbb \oplus \Bbb/q\Bbb$ instead. Since $p$ and $q$ are normally of about the same size, that is about $\sqrt$ , calculations in the latter representation are much faster. Note that RSA algorithm implementations using this isomorphism are more susceptible to fault injection attacks. * The Chinese Remainder Theorem may also be used to construct an elegant Gödel numbering for sequences, which is needed to prove Gödel's incompleteness theorems. * The following example shows a connection with the classic polynomial interpolation theory. Let r complex points ("interpolation nodes") $\textstyle \lambda_1,\cdots,\lambda_r$ be given, together with the complex data $a_\$ , for all $\textstyle 1\leq j\leq r$ and $\textstyle 0\leq k<\nu_j$ . The general Hermite interpolation problem asks for a polynomial $\textstyle P(x)\in \C [4]$ taking the prescribed derivatives in each node $\lambda_j\$ : : $P^(\lambda_j)=a_\quad\forall 1\leq j\leq r\ , 0\leq k<\nu_j$ . :Introducing the polynomials $\textstyle A_j(x):=\sum_^\frac}(x-\lambda_j)^k$ , the problem may be equivalently reformulated as a system of $r\$ simultaneous congruences: : $P(x)\equiv A_j(x)\pmod},\quad\forall 1\leq j\leq r\$ . :By the Chinese remainder theorem in the principal ideal domain $\textstyle\C [5]$ , there is a unique such polynomial $\textstyle P(x)$ with degree $\textstyle\deg(P) . A direct construction, in analogy with the above proof for the integer number case, can be performed as follows. Define the polynomials \textstyle Q:=\prod_^(x-\lambda_i)^and \textstyle Q_j:=\frac} . The partial fraction decomposition of \textstyle \frac gives$ r polynomials $\textstyle S_j$ with degrees $\textstyle\deg(S_j)<\nu_j$ such that : $\frac=\sum_^\frac}$ , :so that $\textstyle 1=\sum_^S_i Q_i$ . Then a solution of the simultaneous congruence system is given by the polynomial : $\sum_^ A_i S_i Q_i = A_j + \sum_^(A_i-A_j) S_i Q_i\equiv A_j\pmod}\quad\forall 1\leq j\leq r\$ ; :and the minimal degree solution is this one reduced modulo $\textstyle Q$ , that is the unique with degree less than n. * 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. Non-commutative case: a counter-example The Chinese remainder theorem does not hold in the non-commutative case. Consider the ring $R$ of non-commutative real polynomials in $x$ and $y$ . Let $I$ be the principal two-sided ideal generated by $x$ and $J$ the principal two-sided ideal generated by $xy+1.$ Then $I+J=R$ but $I\cap J \neq IJ.$ 'ISPAT:' Observe that $I$ is formed by all polynomials with an $x$ in every term and that every polynomial in $J$ vanishes under the substitution $y=-1/x$ . Consider the polynomial $p=(xy+1)x$ . Clearly $p\in I\cap J$ . Define a term in $R$ as an element of the multiplicative monoid of $R$ generated by $x$ and $y$ . Define the degree of a term as the usual degree of the term after the substitution $y=x$ . On the other hand, suppose $q\in J$ . Observe that a term in $q$ of maximum degree depends on $y$ otherwise $q$ under the substitution $y=-1/x$ can not vanish. The same happens then for an element $q\in IJ$ . Observe that the last $y$ , from left to right, in a term of maximum degree in an element of $IJ$ is preceded by more than one $x$ . (We are counting here all the preceding $x$ s. e.g. in $x^2yxyx^5$ the last $y$ is preceded by $3$ $x$ s.) This proves that $(xy+1)x\notin IJ$ since that last $y$ in a term of maximum degree ( $xyx$ ) is preceded by only one $x$ . Hence $I\cap J\neq IJ$ . On the other hand, it is true in general that $I+J = R$ implies $I \cap J = IJ + JI$ . To see this, note that $I \cap J = (I \cap J) (I+J) \subset IJ + JI$ , while the opposite inclusion is obvious. Also, we have in general that, provided $I_1, \ldots, I_m$ are pairwise coprime two-sided ideals in $R$ , the natural map : $R / (I_1 \cap I_2 \cap \ldots \cap I_m) \rightarrow R/I_1 \oplus R/I_2 \oplus \cdots \oplus R/I_m$ is an isomorphism. Note that $I_1 \cap I_2 \cap \ldots \cap I_m$ can be replaced by a sum over all orderings of $I_1, \ldots, I_m$ of their product (or just a sum over enough orderings, using inductively that $I \cap J = IJ + JI$ for coprime ideals $I, J$ ). Bunlara da bakınız * Covering system * Hasse principle * Residue number system * Secret Sharing using the Chinese Remainder Theorem Referanslar * Donald Knuth. The Art of Computer Programming, Volume 2: Seminumerical Algorithms, Third Edition. Addison-Wesley, 1997. ISBN 0-201-89684-2. Section 4.3.2 (pp.286–291), exercise 4.6.2–3 (page 456). * Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. Introduction to Algorithms, Second Edition. MIT Press and McGraw-Hill, 2001. ISBN 0-262-03293-7. Section 31.5: The Chinese remainder theorem, pp.873–876. * Dış bağlantılar * Chinese remainder theorem at cut-the-knot * "Chinese Remainder Theorem" by Ed Pegg, Jr., Wolfram Demonstrations Project, 2007. * * C# program and discussion at codeproject * University of Hawaii System CRT by Lee Lady
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