[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-0-en-105":3,"doc-seo-326872-105":59,"doc-detail-326872-en":130},{"code":4,"msg":5,"data":6},0,"success",[7,13,18,23,28,33,38,43,48,51,55],{"id":8,"doc_module":4,"doc_module_name":9,"category_name":10,"show_sort_weight":11,"slug":12},1,"Document","Story & Novel",90,"story-novel",{"id":14,"doc_module":4,"doc_module_name":9,"category_name":15,"show_sort_weight":16,"slug":17},2,"Literature",80,"literature",{"id":19,"doc_module":4,"doc_module_name":9,"category_name":20,"show_sort_weight":21,"slug":22},4,"Exam",70,"exam",{"id":24,"doc_module":4,"doc_module_name":9,"category_name":25,"show_sort_weight":26,"slug":27},5,"Comic",60,"comic",{"id":29,"doc_module":4,"doc_module_name":9,"category_name":30,"show_sort_weight":31,"slug":32},6,"Technology",50,"technology",{"id":34,"doc_module":4,"doc_module_name":9,"category_name":35,"show_sort_weight":36,"slug":37},7,"Healthcare",40,"healthcare",{"id":39,"doc_module":4,"doc_module_name":9,"category_name":40,"show_sort_weight":41,"slug":42},8,"Research & Report",30,"research-report",{"id":44,"doc_module":4,"doc_module_name":9,"category_name":45,"show_sort_weight":46,"slug":47},9,"Religion & Spirituality",20,"religion-spirituality",{"id":46,"doc_module":4,"doc_module_name":9,"category_name":49,"show_sort_weight":46,"slug":50},"World Cup","world-cup",{"id":52,"doc_module":4,"doc_module_name":9,"category_name":53,"show_sort_weight":52,"slug":54},10,"Lifestyle","lifestyle",{"id":56,"doc_module":4,"doc_module_name":9,"category_name":57,"show_sort_weight":24,"slug":58},19,"General","general",{"code":4,"msg":60,"data":61},"ok",{"site_id":62,"language":63,"slug":64,"title":65,"keywords":66,"description":67,"schema_data":68,"social_meta":123,"head_meta":125,"extra_data":127,"updated_unix":129},105,"en","lecture-notes-on-computer-algebra","Lecture Notes on Computer Algebra","","Seven-lecture course notes on computer algebra from the fall of 2004 focus on modular arithmetic and the integer Chinese remainder algorithm. The material covers canonical homomorphisms, modular inversion and modular exponentiation complexity, evaluation of polynomials via Horner’s form, and theoretical results guaranteeing uniqueness of solutions under pairwise coprime moduli. It also presents constructive algorithms (CRA and CRA2) with cost analysis and worked examples, plus exercises on extensions to non-coprime moduli and interpolation.",{"@graph":69,"@context":122},[70,84,105],{"@type":71,"itemListElement":72},"BreadcrumbList",[73,77,79,82],{"item":74,"name":75,"@type":76,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":78,"name":9,"@type":76,"position":14},"https://docshare.wps.com/document/",{"item":80,"name":40,"@type":76,"position":81},"https://docshare.wps.com/document/research-report/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/lecture-notes-on-computer-algebra/326872/",{"url":83,"name":65,"@type":85,"image":86,"author":91,"headline":65,"publisher":94,"fileFormat":97,"inLanguage":63,"description":67,"dateModified":98,"datePublished":99,"encodingFormat":97,"isAccessibleForFree":100,"interactionStatistic":101},"DigitalDocument",{"url":87,"@type":88,"width":89,"height":90},"https://docshare.wps.com/thumbnails/lecture-notes-on-computer-algebra/326872.png","ImageObject",300,407,{"name":92,"@type":93},"Rowan","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-10-24","2026-09-21",true,{"@type":102,"interactionType":103,"userInteractionCount":24},"InteractionCounter",{"@type":104},"ViewAction",{"@type":106,"mainEntity":107},"FAQPage",[108,114,118],{"name":109,"@type":110,"acceptedAnswer":111},"What modular representations are discussed for Z/(m) ?","Question",{"text":112,"@type":113},"The notes describe a nonnegative representation and a symmetric representation for elements of Z/(m). They then define a canonical simplifier Hm mapping from Z onto Zm for computations.","Answer",{"name":115,"@type":110,"acceptedAnswer":116},"How is the modular inverse in Zm computed and what is its time complexity?",{"text":117,"@type":113},"For a with gcd(a,m)=1, the inverse is found using a half-extended Euclidean algorithm for m and a. The notes state the inverse can be computed in time complexity O(L(m)^2).",{"name":119,"@type":110,"acceptedAnswer":120},"What does the Chinese remainder algorithm (CRA) guarantee and how is its cost estimated?",{"text":121,"@type":113},"Under pairwise co-prime nonunit moduli, the congruence system has a unique solution in [0,M) with M=∏mi. The notes give a recursive constructive approach (CRA and CRA2) and conclude that when moduli and residues have length one, the overall cost of CRA is O(k^2).","https://schema.org",{"og:url":83,"og:type":124,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":126,"canonical":83},"index,follow",{"doc_id":128,"site_id":62},326872,1790636493,{"code":4,"msg":5,"data":131},{"doc_id":128,"user_id":132,"nickname":92,"user_avatar":133,"doc_module":4,"category_id":39,"category_name":40,"doc_title":65,"doc_description":67,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":24,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":139,"language":140,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":67,"update_tm":144,"read_time":145,"preview_page_count":4},1099514067415,"https://ap-avatar.wpscdn.com/avatar/100002539d78ffe74a7?x-image-process=image/resize,m_fixed,w_180,h_180&k=1779092875211072502","Lecture Notes on Computer Algebra  \nZiming Li  \nAbstract  \nThese notes record seven lectures given in the computer algebra course in the fall of 2004 . The theory of subresultants is not required for the 􀀌nal exam due to its complicated constructions.  \n1 Chinese remainder algorithm  \n1.1 Modular arithmetic  \nLet m be a positive integer greater than one. We discuss basic operations in Z=(m) . The elements in Z=(m) can be represented in two ways:  \n􀀏 f0; 1; : : : ; m 􀀀 1g (nonnegative representation);  \n􀀏 􀀈a 2 Z j 􀀀 ~~m~~2 \u003C a 􀀔 ~~m~~2 􀀉 (symmetric representation) .  \nLet us 􀀌x a representation Z m e.g. the nonnegative representation. Via division we have a canonical simpli􀀌er Hm from Z onto Zm.  \nProposition 1 For n 2 Z with L (n) 􀀕 L (m), the time for computing Hm(n) is dominated by O (L (m)(L (n) 􀀀 L(m) + 1)), where L (n) denotes the length of the integer n.  \nProof. It follows from the cost estimation of integral division (see Theorem  \n2.1.7 in [8]) .  \nFor two elements a, b in Zm, we compute a + b, a 􀀀 b, and ab as if they were integers, and then apply Hm to the results. Thus, the costs for computing (a+b) and (a􀀀b) are dominated by O (L (m)), and the cost for ab dominated by O (L (m)2 ) . The latter estimation kills any hope to use fast multiplications for modular multiplication.  \nA particular operation in Z m is the inversion. Given a 2 Zm with gcd(a; m) = 1, compute an element b 2 Zm such that ab 􀀑 1 mod m.  \n1  \nThis can be done by half-extended Euclidean algorithm for m and a. Thus, the time complexity for computing the inverse is O (L (m)2 ) (see Theorem 2.1.9 in [8]) .  \nExample 1 Compute the inverse of 4 mod 7 .  \nf7 = 4 + 3; v = 0 􀀀 1 􀀁 1 = 􀀀1; g f4 = 3 + 1 1 􀀀 1 􀀁 (􀀀1) = 2g: It follows that  \n2 􀀁 4 􀀑 1 mod 7 :  \nAt last, we consider modular exponentiation. Given a 2 Zm and n 2 N, compute an mod m. The time to compute an mod m by repeated multiplication will be proportional to nL (m)2 .  \nProposition 2 The time to compute an mod m is L (n)L (m)2 . Proof. By binary exponentiation we get the recursion  \nEn 􀀘 Ebn=2c + 2L (m)2 with E2 = L (m)2 : It follows that En 􀀘 L (n)L (m)2 .  \nExercise. What is the time to compute an , where a 2 N, by binary exponentiation ?  \n1.2 Modular homomorphisms  \nThe canonical simpli􀀌er Hm is a ring homomorphism from Z to Z=(m) . Let R be a ring. For r 2 R, 􀀞 r : R [x] ! R is an evaluation that sends f (x) to f (r) . Write  \nf (x) = (: : : (fnx + fn􀀀1)x + : : :)x + f0 ;  \nwhich is called Horner's form of f. It takes n multiplications and n additions to compute f (r) .  \nExample 2 Two ways to map Z[x] to Z=(m):  \nZ [x] ! Z=(m)[x] ! Z=(m)  \nand  \nZ [x] ! Z ! Z=(m):  \nHorner's rule is good for evaluating a polynomial at a single point. When evaluating a polynomial at many points, e.g. , the set of consecutive integers, there are better ways (see [7]) .  \nExercise. What is the time to evaluate a polynomial f 2 Z[x] modulo m ?  \n2  \n1.3 The integer Chinese remainder algorithm  \nThe Chinese Remainder Problem: Given m 1 ; : : : ; mk 2 Z+ andr 1 ; : : : rk 2 N, 􀀌nd integers that solve the congruence system:  \n>8 x 􀀑 r 1 mod m 1  \n\u003C 􀀁 􀀁 􀀁 (1)  \n:> x 􀀑 rk mod mk  \nCongruence systems often appears in the combining phase in the applications of the divide-conquer-combine principle.  \nTheorem 1 If the nonunit moduli m 1 , . . . , mk are pairwise co-prime, then (1) has a unique integral solution r in [0; M ), where M = Qki=1 mi . Moreover, every integral solution of (1) is in form (nM + r), for all n 2 Z.  \nProof. Let Mi = M=mi for i = 1; : : : ; k. Since m 1 ; : : : ; mk are pairwise coprime, gcd(M1 ; : : : ; Mk) = 1 . It follows that there are a 1 ; : : : ; ak 2 Z such that  \na 1 M1 + 􀀁 􀀁 􀀁 akMk = 1 :  \nConsequently,  \naiMi 􀀑 1 mod mi ; for i = 1; : : : ; k. (2)  \nSet R = r 1 a 1 M1 + 􀀁 􀀁 􀀁 + rkakMk . By (2) we deduce  \nR 􀀑 air iMi 􀀑 ri mod m i for i = 1,   , k.  \nLet r be the remainder of R and M. Then r is a solution of (1) in [0; M) . If r0 is another solution of (1) in [0; M), and suppose that r","cbCaip5sGokqXNHt","https://ap.wps.com/l/cbCaip5sGokqXNHt","pdf",227940,31,"English","# 1 Chinese remainder algorithm\n## 1.1 Modular arithmetic\n## 1.2 Modular homomorphisms\n## 1.3 The integer Chinese remainder algorithm\n## 1.4 Interpolation","[{\"question\":\"What modular representations are discussed for Z/(m) ?\",\"answer\":\"The notes describe a nonnegative representation and a symmetric representation for elements of Z/(m). They then define a canonical simplifier Hm mapping from Z onto Zm for computations.\"},{\"question\":\"How is the modular inverse in Zm computed and what is its time complexity?\",\"answer\":\"For a with gcd(a,m)=1, the inverse is found using a half-extended Euclidean algorithm for m and a. The notes state the inverse can be computed in time complexity O(L(m)^2).\"},{\"question\":\"What does the Chinese remainder algorithm (CRA) guarantee and how is its cost estimated?\",\"answer\":\"Under pairwise co-prime nonunit moduli, the congruence system has a unique solution in [0,M) with M=∏mi. The notes give a recursive constructive approach (CRA and CRA2) and conclude that when moduli and residues have length one, the overall cost of CRA is O(k^2).\"}]","Lecture Notes on Computer Algebra | PDF",1789975607,78]