Extended Euclidean Algorithm - SlideServe
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Extended Euclidean Algorithm - SlideServe
Extended Euclidean Algorithm. Presented by Lidia Abrams Anne Cheng. Euclidean Algorithm THEOREM. If m and n are any integers, not both zero, then the Greatest Common Divisor of m and n , denoted gcd(m,n) is the largest of the common divisors of m and n. FORMULA.
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SOLVED:The extended Euclidean algorithm can be used to express gcd(a, b) as a linear combination with integer coefficients of the integers a and b . We set s0=1, s1=0, t0=0, and t1=1 and let sj=sj-2-qj-1 sj-1 and tj=tj-2-qj-1 tj-1 for j=2,3, …, n, where the qj are the quotients in the divisions used when the Euclidean algorithm finds gcd(a, b), as shown in the text. It can be shown (see [Ro 10]) that gcd(a, b)=sn a+tn b . The main advantage of the extended Euclidean algorithm is that it uses one
SOLVED:The extended Euclidean algorithm can be used to express gcd(a, b) as a linear combination with integer coefficients of the integers a and b . We set s0=1, s1=0, t0=0, and t1=1 and let sj=sj-2-qj-1 sj-1 and tj=tj-2-qj-1 tj-1 for j=2,3, …, n, where the qj are the quotients in the divisions used when the Euclidean algorithm finds gcd(a, b), as shown in the text. It can be shown (see [Ro 10]) that gcd(a, b)=sn a+tn b . The main advantage of the extended Euclidean algorithm is that it uses one
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SOLVED:The extended Euclidean algorithm can be used to express gcd(a, b) as a linear combination with integer coefficients of the integers a and b . We set s0=1, s1=0, t0=0, and t1=1 and let sj=sj-2-qj-1 sj-1 and tj=tj-2-qj-1 tj-1 for j=2,3, …, n, where the qj are the quotients in the divisions used when the Euclidean algorithm finds gcd(a, b), as shown in the text. It can be shown (see [Ro 10]) that gcd(a, b)=sn a+tn b . The main advantage of the extended Euclidean algorithm is that it uses one
SOLVED:The extended Euclidean algorithm can be used to express …
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