s N+1 G = r 1 [ 1] N+1 and t N+1 G = r 0 [ 1] N.

Transcription

1 Euclidean algorithm in Lightning-Bolt form Jonathan L.F. King University of Florida, Gainesville FL , USA Webpage 17 July, 2018 (at 11:44) The Euclidean algorithm, EU, is often presented by a series of equations. I have found the following table-form convenient, both because it organises the computation, and gives a name to each number in the table. Because the update-rule follows the shape of a lightning-bolt, I call it the LBolt algorithm. Henceforth, all variables are integers unless explicitly stated otherwise. Given integers r 0 and r 1 (for the time being, assume each is positive) we will compute G := Gcd(r 0, r 1 ) as well as a pair S,T of Bézout multipliers satisfying 1: G = S r 0 + T r 1. [There is a one-parameter family of Bézout-pairs; the algorithm will compute a particular pair. I ll explain via an example. Suppose we want the Gcd of r 0 := 114 and r 1 := 33. Then initialize the table as: In order to compute a BP (Bézout-Pair), we ll need 1 : r n = s n t n 33 to hold, for every n. Notice that it already holds, trivially, for n=0 and n=1. At stage n, divide r n into r n 1 to get a quotient, q n, and a remainder, r n+1. That is, 2: r n 1 = [q n r n + r n+1. Now use this value of q n to update three columns: 3: Doing this for n=1 gives r n+1 = r n 1 q n r n ; s n+1 := s n 1 q n s n ; t n+1 := t n 1 q n t n Continue until you get a 0 in the r-column; I ll compute the resulting quotient and write in the q-column, obtaining The GCD-row [shown here red and italicized is the row above the -row The numbers we sought lie in the GCD-row. In this instance, G = r 3, S = s 3 and T = t 3. And indeed, 3 = Why the extra row? You wonder Why bother to compute s 4 and t 4? It isn t necessary, but they provide verification-data. Consider finding (G, S, T ) when r 0 := 98 and r 1 := 51. Initialize: Now compute This r 5, which is 1, is indeed Gcd(98, 51). And 1 = [ Now examine the -row; here, the 6 th row. Note that s 6 equals r 1 upto ±. And t 6 equals r 0 upto ±. In general, letting G := Gcd(r 0, r 1 ), this extra row satisfies 1 that 4: s N+1 G = r 1 [ 1 N+1 and t N+1 G = r 0 [ 1 N. 1 This is stated formally, and proven, in (9c), further below. Webpage Page 1 of 5

2 Prof. JLF King Proving the Euclidean Alg. works Page 2 of 5 If you made a computational error earlier in the table, a glance at this [N+1 th -row will usually shout Error!. Convention. Depending on context, agree to use GCD-row to mean both its index, and its contents. E.g, for the preceding LBolt table, expression Let N := GCD-row makes N = 5. I might also say In the GCD-row, the t-value is 25. Related pamphlets. Our Teaching page has link practice sheet for the LBolt alg with premade tables. There, too, is link Algorithms in Number Theory which uses LBolt iteratively to compute the G := Gcd(M 1, M 2,..., M L ) of a list of integers, computing also Bézout multipliers S 1, S 2,..., S L st. L 5: S l=1 lm l = G. We call S := (S 1,..., S L ) a Bézout tuple for the given tuple M := (M 1,..., M L ). Exer: Fix an L-tuple M which is not the all-zero tuple. Prove that the set of Bézout tuples for M is [L 1-dimensional. The 2 nd page of Algorithms in NT describes an algorithm for solving linear congruences such as 33 x , and has a worked-example. Proving the Euclidean Alg. works I ll leave this as an Exer: The Euclidean-Alg always halts. Define the divisor and common-divisor sets, D(K) := { } d Z d K C(K, N) := D(K) D(N). [Below, LC stands for Linear Combination. and 6: LC Lemma. Consider integers α, β, γ, M such that 6a: Then α + [M β = γ. : C(α, β) = C(β, γ). Proof. Each d C(α, β) necessarily divides α + [Mβ, since M Z. Thus C(α, β) D(γ). By its definition, C(α, β) D(β). Consequently 6b: C(α, β) C(β, γ). OTOHand, we can rewrite (6a) as γ + [ M β = α. The above reasoning hands us 6c: C(α, β) C(β, γ). This, together with (6b), yields ( ). 6d: Corollary. Consider an LBolt seeded with integers r 0 and r 1. Then C(r k, r k+1 ) = C(r 0, r 1 ), for each index k. Consequently, 6e: Gcd(r k, r k+1 ) = Gcd(r 0, r 1 ). Letting N be the GCD-row index, then, 6f: r N = Gcd(r 0, r 1 ), since r N+1 is zero.

3 Prof. JLF King Proving the Euclidean Alg. works Page 3 of 5 7: Bézout Lemma. Consider an LBolt seeded with integers r 0 and r 1. For each k, then, B(k): r k = [s k r 0 + [t k r 1 holds. I ll refer to assertion [ k N: B(k) as the Bézout row-property or LBolt row-property. With N := GCD-index, consequently, 7a: Gcd(r 0, r 1 ) = [s N r 0 + [t N r 1. Proof. The LBolt-seeding gives B(0) and B(1). Now fix a posint n st. B(n 1) and B(n). Courtesy (3), s n+1 r 0 + t n+1 r 1 = [[s n 1 q n s n r 0 + [[t n 1 q n t n r 1 = [ s n 1 r 0 + t n 1 r 1 qn [s n r 0 + t n r 1, since multiplication distributes-over addition. Assertions B(n 1) and B(n) now give us that s n+1 r 0 + t n+1 r 1 = [ r n 1 qn [r n which, by (3) again, equals r n+1. We ve thus inductively established k 1 : [ B(k 1) & B(k) = B(k+1).

4 Prof. JLF King Alternate initialization Page 4 of 5 Alternate initialization Consider an LBolt seeded with integers r 0 and r 1. Define matrices M n := [ s n t n s n+1 t n+1 and R n := [ r n r n+1. Up till now, our initialization matrix M 0 has the identity matrix I := [ However, our Bézout Lemma proof only used B(0) and B(1), i.e that : M 0 R 0 = R 0. and so other values of M 0 are possible. As an example, the usual LBolt for Gcd(3, 2) is 8a: 8b: Another initial-matrix is M 0 := [ , yielding Row-2 gives us a different Bézout pair. We might conjecture that check-pair ( 6, 9) equals the check-pair ( 2, 3) from the first table, but multiplied by Det(M 0 ). Yet another init-matrix is M 0 := [ , producing 8c: Row-2 gives us a third Bézout pair. And the checkpair ( 8, 12) indeed equals Det( ) times the ( 2, 3) from our first table. Check-row. We study an LBolt seeded with a coprime pair r 0 r 1, and initial-matrix M 0 st. ( ) holds. For k 1, let [ 0 1 Q k := 1 q k and observe Det(Q k ) = 1. Define product matrix P n := Q n Q 2 Q 1 ; hence P 0 is the identity matrix I. Update-rule (3) tells us that R k = Q k R k 1 and M k = Q k M k 1. Consequently, 9a: Moreover, 9b: R n = P n R 0, M n = P n M 0 M n R 0 and Det(M n ) = [ 1 n Det(M 0 ). = P n M 0 R 0 by ( ) ===== P n R 0 = R n. Letting N := GCD-index, we have that recall M N R 0 = R N ==== [ : 1, since r 0 r 1. Have (S, T ) := (s N, t N ) denote the Bézout-pair, and let (α, β) := (s N+1, t N+1 ) be the pair whose values we wish to determine. Finally, set δ := Det(M 0 ). Our ( ) gives the top two lines of Sr 0 + T r 1 = 1, αr 0 + βr 1 = 0. Notice that αt + βs = δ [ 1 N follows by computing Det(M N ). Multiplying the middle eqn by T and the bottom by r 0 gives Adding them yields αt r 0 + βt r 1 = 0 and αt r 0 + βsr 0 = r 0 δ[ 1 N. β note == β [Sr 0 + T r 1 = r 0 δ [ 1 N. Finally, plugging this into the middle eqn gives 0 = αr 0 + r 0 δ[ 1 N r 1. When r 0 0, then 0 = α + r 1 δ[ 1 N. Hence 0

5 Prof. JLF King Alternate initialization Page 5 of 5 α = r 1 δ [ 1 N = r 1 δ [ 1 N+1. We have proven the following theorem. 9c: Check-value Theorem. Consider an LBolt seeded with integers r 0 0 and r 1, together with an initialmatrix M 0 satisfying 9d: M 0 [r 0 r 1 = [ r 0 r 1. Let N := GCD-index and G := Gcd(r 0, r 1 ). Then 9e: s N+1 G = r 1 Det(M 0 ) [ 1 N+1 and t N+1 G = r 0 Det(M 0 ) [ 1 N. (Recall that our standard LBolt has Det(M 0) = 1.) As of: Thursday 06Jan2011. Typeset: 17Jul2018 at 11:44.