SpecAlg - A-152 A.3 A.3.1 Special Topics The Euclidean...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
A-152 A.3 Special Topics A.3.1 The Euclidean algorithm Let F be a field. In Theorem A.2.16 we gave a nonconstructive proof for the existence of the greatest common divisor of two polynomials a ( x ) and b ( x ) of F [ x ]. The Euclidean algorithm is an algorithm that constructs gcd( a ( x ) ,b ( x )) explicitly. The basic method is simple. If q ( x ) is any polynomial, then gcd( a ( x ) ,b ( x )) = gcd( a ( x ) - q ( x ) b ( x ) ,b ( x )) . In particular, a ( x ) can be replaced in the calculation by its remainder r ( x ) upon division by b ( x ). Assuming that a ( x ) has degree at least as big as that of b ( x ), the remainder r ( x ) will have smaller degree than a ( x ); so the gcd of the original pair of polynomials will be equal to the gcd of a new pair with smaller total degree. We can continue in this fashion decreasing the degree of the remainder at each stage until the process stops with remainder 0, and at this point the gcd becomes clear. In fact the approach we take is a little different. From our proof of Theorem A.2.16 we know that gcd( a ( x ) ,b ( x )) is the monic polynomial of minimal degree within the set G = { s ( x ) a ( x ) + t ( x ) b ( x ) | s ( x ) ,t ( x ) F [ x ] } Thus we examine all equations of the form p ( x ) = s ( x ) a ( x ) + t ( x ) b ( x ) , looking for one in which nonzero p ( x ) has minimal degree. The unique monic scalar multiple of this p ( x ) is then equal to gcd( a ( x ) ,b ( x )). If we have two suitable equations: m ( x ) = e ( x ) a ( x ) + f ( x ) b ( x ) ; (A.1) n ( x ) = g ( x ) a ( x ) + h ( x ) b ( x ) ; (A.2) then we can find a third with lefthand side of smaller degree. Assume that the degree of m ( x ) is at least as big as that of n ( x ). By the Division Algorithm A.2.5 there are q ( x ) and r ( x ) with m ( x ) = q ( x ) n ( x )+ r ( x )and deg( r ( x )) < deg( n ( x )). Subtracting q ( x ) times equation (2) from equation (1) we have the desired r ( x ) = m ( x ) - q ( x ) n ( x ) = (A.3) ± e ( x ) - q ( x ) g ( x ) ² a ( x ) + ± f ( x ) - q ( x ) h ( x ) ² b ( x ) . Next we may divide r ( x ) into n ( x ) and, using equations (2) and (3), further reduce the degree of the lefthand side. Continuing as before we must ultimately arrive at an equation with 0 on the left. The lefthand side of the previous equation will then have the desired minimal degree. A benefit of this method of
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
A.3. SPECIAL TOPICS A-153 calculation is that the appropriate polynomials s ( x ) and t ( x ) are produced at the same time as the gcd. To succeed with this approach we must have two equations to begin with. These are provided by: a ( x ) = 1 · a ( x ) + 0 · b ( x ); (A.4) b ( x ) = 0 · a ( x ) + 1 · b ( x ) . (A.5) (A.3.1) Theorem. ( The Euclidean Algorithm. ) Assume that deg( a ( x )) deg( b ( x )) with a ( x ) 6 = 0 . At Step i we construct the equation E i : r i ( x ) = s i ( x ) a ( x ) + t i ( x ) b ( x ) . Equation E i is constructed from E i - 1 and E i - 2 , the appropriate initialization being provided by (4) and (5): r - 1 ( x ) = a ( x ); s - 1 ( x ) = 1; t - 1 ( x ) = 0; r 0 ( x ) = b ( x ); s 0 ( x ) = 0; t 0 ( x ) = 1 . Step i . Starting with r i - 2 ( x ) and r i - 1 ( x ) ( 6 = 0) use the Division Algorithm A.2.5 to define q i ( x ) and r i ( x ) : r i - 2 ( x ) = q i ( x ) r i - 1 ( x ) + r i ( x ) with deg( r i ( x )) < deg( r i - 1 ( x )) .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/13/2011 for the course CRYPTO 101 taught by Professor Na during the Spring '11 term at Harding.

Page1 / 15

SpecAlg - A-152 A.3 A.3.1 Special Topics The Euclidean...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online