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High School Generalized Continued Denominators

Understand and apply the HGCD definition and procedure, illustrated with an example and steps for calculating GCD. Explore the concept through equations and explanations.

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High School Generalized Continued Denominators

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  1. An example of HGCD Defintion If j > I, then Lemma • for i < j < k • for 0 ≦ j < k

  2. An example of HGCD Example a0=57 a1=33 57 * (-4) + 33 * 7 = 3

  3. An example of HGCD

  4. An example of HGCD Procedure GCD(a1,a0) if a1 divides a0 then return a1 else R ← HGCD(a1,a0) ; ← R * ; if b1 divides b0 then return b1 else c ← b0 mod b1 ; return GCD (b1,c)

  5. An example of HGCD • HGCD(a0,a1) a0=139 a1=53 deg(a0) = a0=b02m+c0 , m = ,deg(c0) < m a1=b12m+c1 , deg(c1) < m Note: deg(a1)≦ deg(a0), return

  6. An example of HGCD 139 = 17*23+3 , m= 53 = 6*23+5

  7. An example of HGCD f = d mod e = 53 mod 33 = 20 e=g02 +h0 →33 = 17*21 + 1 f=g12 +h1 → 20 = 10*21 + 0 Note: e, f 是連續的餘數項目且deg最大 為 ,也就是3/4的deg(a0) Note: g0與g1的deg最大為m+1

  8. An example of HGCD

  9. An example of HGCD Procedure GCD(a1,a0) if a1 divides a0 then return a1 else R ← HGCD(a1,a0) ; ← R * ; if b1 divides b0 then return b1 else c ← b0 mod b1 ; return GCD (b1,c)

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