1 / 32 From the Eucldean Algorthm for Solvng a Key Equaton for Dual Reed Solomon Codes to the Berlekamp-Massey Algorthm Mara Bras-Amorós, Mchael E O Sullvan The Claude Shannon Insttute Workshop on Codng and Cryptography Cork,May18, 2009
Contents 2 / 32 1 Extended Eucldean algorthm (revsted) 2 Key equatons for Reed-Solomon codes (revsted) 3 Extended Eucldean algorthm for the new key equaton 4 From the Eucldean to the Berlekamp-Massey algorthm 5 Other research drectons
Bézout s Theorem 3 / 32 Bézout s Theorem Gven a, b F q[x] there exst f, g F q[x] such that f a + gb gcd(a, b) Extended Eucldean Algorthm Let r 2 a, r 1 b and, for 0 let the Eucldean dvson of r 2 by r 1 be r 2 q r 1 + r Defne f 2 1, g 2 0, f 1 0, g 1 1, and for 0 f f 2 q f 1 g g 2 q g 1 then for all 0 f a + g b r
4 / 32 Extended Eucldean algorthm The extended Eucldean algorthm can be expressed n matrx form as ALGORITHM: Intalze: «r 1 f 1 g 1 b 0 1 r 2 f 2 g 2 a 1 0 «whle deg(r ) 0: end whle q Quotent(r 2, r 1 ) «r f g q 1 r 1 f 1 g 1 1 0 ««r 1 f 1 g 1 r 2 f 2 g 2 Return r 1
Extended Eucldean algorthm 5 / 32 Remark 1 deg(f ) > deg(f 1 ) whle deg(r ) < deg(r 1 ) deg(g ) > deg(g 1 ) (except maybe n the ntal steps); ««r f 2 det ( 1) +1 r 1 f 1 det ( 1) +1 b, so r 1 f 1 r 2 f 2 f r 1 ( 1) b + f 1 r, and, snce deg r < deg r 1 and deg f > deg f 1, then LT(f ) ( 1) LT(b)/LT(r 1 ) deg f deg b deg r 1 ««f g 3 det ( 1) +1 f 1 g det 1 ( 1), so f 1 g 1 f 2 g 2 f g 1 f 1 g ( 1) and the ntermedate Bézout coeffcents are coprme at each step
Extended Eucldean alg wth monc remanders 6 / 32 Defnton For all 1 defne the matrces «R F G 1/LC(r ) 0 R F G 0 ( 1) LC(r ) «r f g r 1 f 1 g 1 «Lemma For all 0, ««««R F G 1/LC( R 1 Q R 1) 0 Q 1 R 1 F 1 G 1, R F G 0 LC( R 1 Q R 1) 1 0 R 1 F 1 G 1 where Q s the quotent of R 1 by R 1
Extended Eucldean alg wth monc remanders 7 / 32 The extended Eucldean algorthm for the new matrces s ALGORITHM: Intalze: R 1 F 1 G 1 R 1 F 1 G 1 «1 0 LC(b) 0 LC(b) b 0 1 a 1 0 «whle deg(r ) 0: end whle Q Quotent( R 1, R 1 ) 0 1 «1 R F G 0 «@ LC( R 1 Q R 1 ) A Q 1 R 1 F 1 G 1 R F G 1 0 0 LC( R 1 Q R 1 ) R 1 F 1 G 1 Return R 1 (aconstant multple of r 1 )
Extended Eucldean alg wth monc remanders The extended Eucldean algorthm for the new matrces s ALGORITHM: Intalze: R 1 F 1 G 1 R 1 F 1 G 1 «1 0 LC(b) 0 LC(b) b 0 1 a 1 0 «whle deg(r ) 0: end whle Q Quotent( R 1, R 1 ) 0 1 «1 R F G 0 «@ LC( R 1 Q R 1 ) A Q 1 R 1 F 1 G 1 R F G 1 0 0 LC( R 1 Q R 1 ) R 1 F 1 G 1 Return R 1 (aconstant multple of r 1 ) If Q Q (0) + Q (1) x + + Q (l) x l then Q 1 1 0 «1 Q (0) 0 1 «(1) 1 Q 0 1 x «««(l) 1 Q x l 0 1 0 1 1 0 7 / 32
Extended Eucldean alg wth monc remanders The extended Eucldean algorthm for the new matrces s ALGORITHM: Intalze: R 1 F 1 G 1 R 1 F 1 G 1 «1 0 LC(b) 0 LC(b) b 0 1 a 1 0 «whle deg(r ) 0: end whle Q Quotent( R 1, R 1 ) 0 1 «1 R F G 0 @ LC( R 1 Q R 1 ) R F G 0 LC( R 1 Q R 1 )! 1 Q (l) x l 0 1 0 1 1 0 A 1 Q(0) 0 1 ««R 1 F 1 G 1 R 1 F 1 G 1! 1 Q (1) x 0 1! Return R 1 (aconstant multple of r 1 ) If Q Q (0) + Q (1) x + + Q (l) x l then Q 1 1 0 «1 Q (0) 0 1 «(1) 1 Q 0 1 x «««(l) 1 Q x l 0 1 0 1 1 0 7 / 32
Extended Eucldean alg wth monc remanders The extended Eucldean algorthm for the new matrces s ALGORITHM: Intalze: R 1 F 1 G 1 R 1 F 1 G 1 «1 0 LC(b) 0 LC(b) b 0 1 a 1 0 «whle deg(r ) 0: end whle Q Quotent( R 1, R 1 ) 0 1 «1 R F G 0 @ LC( R 1 Q R 1 ) R F G 0 LC( R 1 Q R 1 )! 1 Q (l) x l 0 1 0 1 1 0 A 1 Q(0) 0 1 ««R 1 F 1 G 1 R 1 F 1 G 1! 1 Q (1) x 0 1! Return R 1 (aconstant multple of r 1 ) LC(b) LC( R 1 Q R 1 ) Q (j) 9 they all are the LC of the left-most, top-most ; element n the prevous matrx 7 / 32
Extended Eucldean alg wth monc remanders 8 / 32 Splttng the matrx multplcatons we get ALGORITHM: Intalze: «R 1 F 1 G 1 b/lc(b) 0 1/LC(b) R 1 F 1 G 1 LC(b)a LC(b) 0 «whle deg(r ) 0: «R+1 F +1 G +1 0 1 R +1 F+1 G+1 1 0 ««R F G R F G whle deg(r ) deg( R ): R+1 F +1 G +1 R +1 F+1 G+1 «1 LC(R )x (deg(r ) deg( R )) 0 1! «R F G R F G end whle «R+1 F +1 G +1 1/LC(R ) 0 R +1 F+1 G+1 0 LC(R ) ««R F G R F G end whle
Extended Eucldean alg wth monc remanders 9 / 32 We ntroduced a bunch of ntermedate matrces Not all of them satsfy «R F G R F G «R F G 1/LC(r ) 0 R F G 0 ( 1) LC(r ) «r f g r 1 f 1 g 1 «But all of them satsfy F a + G b R, F a + G b R
Extended Eucldean alg wth monc remanders The same algorthm can be expressed as ALGORITHM: Intalze: «R 1 F 1 G 1 LC(b)a LC(b) 0 R 1 F 1 G 1 b/lc(b) 0 1/LC(b) «whle deg(r ) 0: µ LC(R ) p deg(r ) deg( R ) f p 0 or µ 0 then «R+1 F +1 G +1 1 µx p R +1 F+1 G+1 0 1 ««R F G R F G else end f «R+1 F +1 G +1 x p µ R +1 F+1 G+1 1/µ 0 ««R F G R F G end whle µ LC(R ) «R+1 F +1 G +1 1/µ 0 R +1 F+1 G+1 0 µ ««R F G R F G 10 / 32
Reed-Solomon Codes 11 / 32 Prmal Reed-Solomon Code Consder F a fnte feld of sze q p m, α a prmtve element n F, n q 1 the dentfcaton u (u 0, u 1,, u n 1 ) u(x) u 0 + u 1 x + + u n 1 x n 1 (denote u(α) u 0 + u 1 α + + u n 1 α n 1 ) The Reed-Solomon code of dmenson k C (k) s the cyclc code wth generator polynomal (x α)(x α 2 ) (x α n k ) It has generator and party check matrces 0 1 0 1 1 1 1 1 1 α α 2 α n 1 1 α α 2 α n 1 1 α 2 α 4 α 2(n 1) G (k) 1 α 2 α 4 α 2(n 1), H (k) 1 α 3 α 6 α 3(n 1) B @ C A B @ C A 1 α k 1 α (k 1)2 α (k 1)(n 1) 1 α n k α (n k)2 α (n k)(n 1)
Prmal and dual Reed-Solomon Codes Prmal Reed-Solomon Code The Reed-Solomon code of dmenson k C (k) s the cyclc code wth generator polynomal (x α)(x α 2 ) (x α n k ) It has generator and party check matrces G (k) 0 B B B B B @ 1 1 1 1 1 α α 2 α n 1 1 α 2 α 4 α 2(n 1) 1 α k 1 α (k 1)2 α (k 1)(n 1) 1 C C C C C A, H (k) 0 B B B B B B @ 1 α α 2 α n 1 1 α 2 α 4 α 2(n 1) 1 α 3 α 6 α 3(n 1) 1 α n k α (n k)2 α (n k)(n 1) 1 C C C C C C A Dual Reed-Solomon Code The dual Reed-Solomon code of dmenson k C(k) s the cyclc code wth generator polynomal (x α (k+1) ) (x α (n 1) )(x 1) It has generator and party check matrces G(k) 0 B B B B B B @ 1 α α 2 α n 1 1 α 2 α 4 α 2(n 1) 1 α 3 α 6 α 3(n 1) 1 α k α 2k α k(n 1) 1 C C C C C C A, H(k) 0 B B B B B @ 1 1 1 1 1 α α 2 α n 1 1 α 2 α 4 α 2(n 1) 1 α n k 1 α (n k 1)2 α (n k 1)(n 1) 1 C C C C C A 12 / 32
13 / 32 Prmal and dual Reed-Solomon Codes Propertes Both codes have mnmum dstance d n k + 1 C(k) C (n k) If s a vector of dmenson k and c (c 0, c 1,, c n 1 ) G(k) C(k), c (c 0,c 1,, c n 1) G (k) C (k), then c (c0, αc1, α 2 c2, α n 1 cn 1), c(α ) c0+ αc1α + α 2 c2 α 2 + +α n 1 cn 1α (n 1) c0+ c1α +1 + c2α 2(+1) + +cn 1α (n 1)(+1) c(α ) c (α +1 )
Correcton of RS codes: key polynomals 14 / 32 Suppose c C (k) s the transmtted word, e s the error added to c wth t e d 1 2, and u c + e s the receved word Correcton of RS codes Error locator polynomal Λ Q e 0 (1 α x) Error evaluator polynomal Ω P e 0 e α Q e j 0,j (1 α x)
Correcton of RS codes: key polynomals 14 / 32 Suppose c C (k) s the transmtted word, e s the error added to c wth t e d 1 2, and u c + e s the receved word Correcton of RS codes Error locator polynomal Λ Q e 0 (1 α x) Error evaluator polynomal Ω P e 0 e α Q e j 0,j (1 α x) Error locaton Λ (α ) 0 Error evaluaton (Forney) e Ω (α ) Λ (α )
Correcton of RS codes: key polynomals 14 / 32 Suppose c C (k) s the transmtted word, e s the error added to c wth t e d 1 2, and u c + e s the receved word Correcton of RS codes Error locator polynomal Λ Q e 0 (1 α x) Error evaluator polynomal Ω P e 0 e α Q e j 0,j (1 α x) Error locaton Λ (α ) 0 Error evaluaton (Forney) e Ω (α ) Λ (α ) Syndrome polynomal S e (α) + e (α 2 )x + + e (α n )x n 1 Truncated syndrome polynomal S e (α) + + e (α n k )x n k 1
Correcton of RS codes: key equatons 15 / 32 Key equaton Λ S (1 x n )Ω Truncated key equaton (Berlekamp) Λ S Ω mod x n k
Correcton of RS codes: key equatons 15 / 32 Key equaton Λ S (1 x n )Ω Truncated key equaton (Berlekamp) Λ S Ω mod x n k Bézout-lke presentaton {z} Λ n k S {z} + m(x) x {z } {z } f a (known) g b (known) Ω {z} r
Correcton of RS codes: key equatons 15 / 32 Key equaton Λ S (1 x n )Ω Truncated key equaton (Berlekamp) Λ S Ω mod x n k Bézout-lke presentaton {z} Λ n k S {z} + m(x) x {z } {z } f a (known) g b (known) Ω {z} r Sugyama et al s algorthm solves ths by means of the ext Eucldean algorthm The bound on the degree of Ω states the end of the algorthm Coprmalty of Λ and Ω guarantees uncty
Correcton of RS codes: key polynomals 16 / 32 Suppose c C(k) s the transmtted word, e s the error added to c wth t e d 1, and u c + e s the receved word 2 Correcton of RS codes Correcton of DUAL RS codes Error locator polynomal Λ Q e 0 (1 α x) Error evaluator polynomal Ω P e 0 e α Q e j 0,j (1 α x) Error locaton Λ (α ) 0 Error evaluaton (Forney) e Ω (α ) Λ (α ) Syndrome polynomal S e (α) + e (α 2 )x + + e (α n )x n 1 Truncated syndrome polynomal S e (α) + + e (α n k )x n k 1
Correcton of RS codes: key polynomals 16 / 32 Suppose c C(k) s the transmtted word, e s the error added to c wth t e d 1, and u c + e s the receved word 2 Correcton of RS codes Correcton of DUAL RS codes Error locator polynomal Error locator polynomal Λ Q e 0 (1 α x) Λ Q e 0 (x α ) Error evaluator polynomal Error evaluator polynomal Ω P e 0 e α Q e j 0,j (1 α x) Ω P e 0 e Q e j 0,j (x α ) Error locaton Λ (α ) 0 Error evaluaton (Forney) e Ω (α ) Λ (α ) Syndrome polynomal S e (α) + e (α 2 )x + + e (α n )x n 1 Truncated syndrome polynomal S e (α) + + e (α n k )x n k 1
Correcton of RS codes: key polynomals 16 / 32 Suppose c C(k) s the transmtted word, e s the error added to c wth t e d 1, and u c + e s the receved word 2 Correcton of RS codes Correcton of DUAL RS codes Error locator polynomal Error locator polynomal Λ Q e 0 (1 α x) Λ Q e 0 (x α ) Error evaluator polynomal Error evaluator polynomal Ω P e 0 e α Q e j 0,j (1 α x) Ω P e 0 e Q e j 0,j (x α ) Error locaton Error locaton Λ (α ) 0 Λ(α ) 0 Error evaluaton (Forney) Error evaluaton (Forney) e Ω (α ) Λ (α ) Ω(α ) Λ (α ) Syndrome polynomal S e (α) + e (α 2 )x + + e (α n )x n 1 Truncated syndrome polynomal S e (α) + + e (α n k )x n k 1
Correcton of RS codes: key polynomals 16 / 32 Suppose c C(k) s the transmtted word, e s the error added to c wth t e d 1, and u c + e s the receved word 2 Correcton of RS codes Correcton of DUAL RS codes Error locator polynomal Error locator polynomal Λ Q e 0 (1 α x) Λ Q e 0 (x α ) Error evaluator polynomal Error evaluator polynomal Ω P e 0 e α Q e j 0,j (1 α x) Ω P e 0 e Q e j 0,j (x α ) Error locaton Error locaton Λ (α ) 0 Λ(α ) 0 Error evaluaton (Forney) Error evaluaton (Forney) e Ω (α ) Λ (α ) Ω(α ) Λ (α ) Syndrome polynomal S e (α) + e (α 2 )x + + e (α n )x n 1 Truncated syndrome polynomal Syndrome polynomal S e(α n 1 ) + e(α n 2 )x + + e(1)x n 1 Truncated syndrome polynomal S e (α) + + e (α n k )x n k 1 S e(α n k 1 )x k + + e(1)x n 1
Correcton of RS codes: key polynomals 17 / 32 If e and e are such that e(α ) e (α +1 ) then Λ x t Λ (1/x) Ω x t 1 Ω (1/x) S x n 1 S (1/x) S x n 1 S (1/x)
Correcton of RS codes: key equatons 18 / 32 Key equaton Key equaton Λ S (1 x n )Ω ΛS (x n 1)Ω Truncated key equaton (Berlekamp) Truncated key equaton Λ S Ω mod x n k deg(λ S (x n 1)Ω)< n d/2 Bézout-lke presentaton {z} Λ n k S {z} + m(x) x {z } {z } f a (known) g b (known) Ω {z} r Sugyama et al s algorthm solves ths by means of the ext Eucldean algorthm The bound on the degree of Ω states the end of the algorthm Coprmalty of Λ and Ω guarantees uncty
Correcton of RS codes: key equatons 18 / 32 Key equaton Key equaton Λ S (1 x n )Ω ΛS (x n 1)Ω Truncated key equaton (Berlekamp) Truncated key equaton Λ S Ω mod x n k deg(λ S (x n 1)Ω)< n d/2 Bézout-lke presentaton {z} Λ n k S {z} + m(x) x {z } {z } f a g b (known) (known) Ω {z} r n Bézout-lke presentaton {z} Λ {z} S {z} Ω (x {z 1) } f a g (known) b (known) m(x) {z } r Sugyama et al s algorthm solves ths by means of the ext Eucldean algorthm The bound on the degree of Ω states the end of the algorthm Coprmalty of Λ and Ω guarantees uncty
Correcton of RS codes: key equatons 18 / 32 Key equaton Key equaton Λ S (1 x n )Ω ΛS (x n 1)Ω Truncated key equaton (Berlekamp) Truncated key equaton Λ S Ω mod x n k deg(λ S (x n 1)Ω)< n d/2 Bézout-lke presentaton {z} Λ n k S {z} + m(x) x {z } {z } f a g b (known) (known) Ω {z} r n Bézout-lke presentaton {z} Λ {z} S {z} Ω (x {z 1) } f a g (known) b (known) m(x) {z } r Sugyama et al s algorthm solves ths by means of the ext Eucldean algorthm The bound on the degree of Ω states the end of the algorthm Coprmalty of Λ and Ω guarantees uncty Goal: solve ths by means of the ext Eucldean algorthm The key equaton tself states the end of the algorthm Coprmalty of Λ and Ω guarantees uncty
Correcton of RS codes: key equatons 19 / 32 Lemma Suppose that at most d 1 2 errors occurred Then Λ and Ω are the unque polynomals λ, ω satsfyng the followng propertes 1 deg(λ S ω(x n 1)) < n d/2 2 deg(λ) d/2 3 λ, ω are coprme 4 λ s monc
20 / 32 Extended Eucldean alg for the dual key equaton ALGORITHM: Intalze: «R 1 F 1 G 1 S 1 0 R 1 F 1 G 1 x n 1 0 1 «whle deg(r ) n d/2: µ LC(R ) p deg(r ) deg( R ) f p 0 or µ 0 then «R+1 F +1 G +1 1 µx p R +1 F+1 G+1 0 1 ««R F G R F G end whle else end f «R+1 F +1 G +1 x p µ R +1 F+1 G+1 1/µ 0 ««R F G R F G Return F, G
20 / 32 Extended Eucldean alg for the dual key equaton ALGORITHM: Intalze: «R 1 F 1 G 1 S 1 0 R 1 F 1 G 1 x n 1 0 1 «whle deg(r ) n d/2: µ LC(R ) p deg(r ) deg( R ) f p 0 or µ 0 then «R+1 F +1 G +1 1 µx p R +1 F+1 G+1 0 1 ««R F G R F G end whle else end f «R+1 F +1 G +1 x p µ R +1 F+1 G+1 1/µ 0 ««R F G R F G Return F, G 1 deg(f S G (x n 1)) deg(r ) < n d/2
20 / 32 Extended Eucldean alg for the dual key equaton ALGORITHM: Intalze: «R 1 F 1 G 1 S 1 0 R 1 F 1 G 1 x n 1 0 1 «whle deg(r ) n d/2: µ LC(R ) p deg(r ) deg( R ) f p 0 or µ 0 then «R+1 F +1 G +1 1 µx p R +1 F+1 G+1 0 1 ««R F G R F G end whle else end f «R+1 F +1 G +1 x p µ R +1 F+1 G+1 1/µ 0 ««R F G R F G Return F, G 2 deg(f ) n deg(r 1 ) n (n d/2) d/2
Extended Eucldean alg for the dual key equaton ALGORITHM: Intalze: «R 1 F 1 G 1 S 1 0 R 1 F 1 G 1 x n 1 0 1 «whle deg(r ) n d/2: µ LC(R ) p deg(r ) deg( R ) f p 0 or µ 0 then «R+1 F +1 G +1 1 µx p R +1 F+1 G+1 0 1 ««R F G R F G end whle else end f «R+1 F +1 G +1 x p µ R +1 F+1 G+1 1/µ 0 ««R F G R F G Return F, G «F G 3 F, G coprme Indeed, det 1 F ( G ) + G ( F ) 1 F G 20 / 32
Extended Eucldean alg for the dual key equaton ALGORITHM: Intalze: «R 1 F 1 G 1 S 1 0 R 1 F 1 G 1 x n 1 0 1 «whle deg(r ) n d/2: µ LC(R ) p deg(r ) deg( R ) f p 0 or µ 0 then «R+1 F +1 G +1 1 µx p R +1 F+1 G+1 0 1 ««R F G R F G end whle else end f «R+1 F +1 G +1 x p µ R +1 F+1 G+1 1/µ 0 ««R F G R F G Return F, G «R F 4 F s monc Indeed, det 1 F ( R ) + R ( F ) 1 R F 20 / 32
20 / 32 Extended Eucldean alg for the dual key equaton ALGORITHM: Intalze: «R 1 F 1 G 1 S 1 0 R 1 F 1 G 1 x n 1 0 1 «whle deg(r ) n d/2: µ LC(R ) p deg(r ) deg( R ) f p 0 or µ 0 then «R+1 F +1 G +1 1 µx p R +1 F+1 G+1 0 1 ««R F G R F G end whle else end f «R+1 F +1 G +1 x p µ R +1 F+1 G+1 1/µ 0 ««R F G R F G Return F, G Theorem: If t d 1 2 then the algorthm outputs Λ and Ω
From Eucldean to Berlekamp-Massey 21 / 32 The only reason to keep the polynomals R (and R ) s that we need to compute ther leadng coeffcents (the µ s) Lemma LC(R ) LC(F S) Proof On one hand, the remander R F S G (x n 1) F S x n G + G has degree at most n 1 for all 0 Ths means that all terms of x n G cancel wth terms of F S and that the leadng term of R must be ether a term of F S or a term of G or a sum of a term of F S and a term of G On the other hand, the algorthm only computes LC(R ) whle deg(r ) n d/2 We want to see that n ths case the leadng term of R has degree strctly larger than that of G Indeed, one can check that for 0, deg(g ) < deg(f ) and that all F s n the algorthm have degree at most d/2 So deg(g ) < deg(f ) d/2 n d/2 deg(r )
From Eucldean to Berlekamp-Massey ALGORITHM: Intalze: d 1 deg( S) d 1 n «F 1 G 1 1 0 F 1 G 1 0 1 «whle d n d/2: µ Coeffcent(F S, d ) p d d f p 0 or µ 0 then «F+1 G +1 1 µx p F +1 G+1 0 1 d +1 d 1 d +1 d ««F G F G else end f «F+1 G +1 x p µ F +1 G+1 1/µ 0 d +1 d 1 d +1 d ««F G F G end whle 22 / 32
From Eucldean to Berlekamp-Massey 23 / 32 Ths last algorthm s the Berlekamp-Massey algorthm that solves the lnear recurrence tx Λ j e(α +j 1 ) 0 for all > 0 j0 Ths recurrence s derved from Λ S beng a polynomal and thus havng no x n 1 terms of negatve order n ts expresson as a Laurent seres n 1/x, and from the equalty S x n 1 1 e(1) + e(α) «+ e(α2 ) + x x x 2
Movng back to prmal Reed-Solomon codes 24 / 32 Suppose c C (k) s the transmtted word, e s the error added to c and u c + e s the receved word Then c (c 0,αc 1, α 2 c 2,,α n 1 c n 1) C(k) and e (e 0, αe 1, α 2 e 2,, α n 1 e n 1) has the same non-zero postons as e Let u : c + e (u0, αu1, α 2 u2,, α n 1 un 1) The error values e can be computed from the error values e added to u by u e e /α added to Now we can use the prevous algorthm wth S e(α n k 1 )x k + e(α n k 2 )x k+1 + + e(1)x n 1 e (α n k )x k + e (α n k 1 )x k+1 + + e (α)x n 1 u (α n k )x k + u (α n k 1 )x k+1 + + u (α)x n 1 Once we have the error postons, we can compute the error values as e Ω(α ) α Λ (α )
25 / 32
Other research drectons: numercal semgroups 26 / 32 Defnton A numercal semgroup s a subset Λ of N 0 satsfyng 0 Λ Λ + Λ Λ N 0 \ Λ s fnte (genus:g: N 0 \ Λ )
Cash pont 27 / 32 Example The amounts of money one can obtan from a cash pont (dvded by 10)
Cash pont 28 / 32 amount amount/10 0 0 10 mpossble! 20 2 30 mpossble! 40 + 4 50 5 60 + + 6 70 + 7 80 + + + 8 90 + + 9 100 + 10 110 + + + 11
Countng 29 / 32 Let n g denote the number of numercal semgroups of genus g
Countng 29 / 32 Let n g denote the number of numercal semgroups of genus g n 0 1, snce the unque numercal semgroup of genus 0 s N 0
Countng 29 / 32 Let n g denote the number of numercal semgroups of genus g n 0 1, snce the unque numercal semgroup of genus 0 s N 0 n 1 1, snce the unque numercal semgroup of genus 1 s N 0 \ {1}
Countng 29 / 32 Let n g denote the number of numercal semgroups of genus g n 0 1, snce the unque numercal semgroup of genus 0 s N 0 n 1 1, snce the unque numercal semgroup of genus 1 s N 0 \ {1} n 2 2 Indeed the unque numercal semgroups of genus 2 are {0, 3, 4, 5,}, {0, 2, 4, 5,}
Conjecture n g /n g 1 φ 30 / 32 Conjecture 1 n g n g 1 + n g 2 2 lm g n g 1 +n g 2 n g 1 n 3 lm g g φ n g 1 At the moment t has not even been proved that n g s ncreasng
Conjecture n g /n g 1 φ n g ng ng 1 + g 1+n g 2 ng ng 2 ng n g 1 0 1 1 1 1 2 2 2 1 2 3 4 3 075 2 4 7 6 0857143 175 5 12 11 0916667 171429 6 23 19 0826087 191667 7 39 35 0897436 169565 8 67 62 0925373 171795 9 118 106 0898305 176119 10 204 185 0906863 172881 11 343 322 0938776 168137 12 592 547 0923986 172595 13 1001 935 0934066 169088 14 1693 1593 0940933 169131 15 2857 2694 0942947 168754 16 4806 4550 0946733 168218 17 8045 7663 0952517 167395 18 13467 12851 0954259 167396 19 22464 21512 0957621 166808 20 37396 35931 0960825 166471 21 62194 59860 0962472 166312 22 103246 99590 0964589 166006 23 170963 165440 0967695 165588 24 282828 274209 0969526 165432 25 467224 453791 0971249 165197 26 770832 750052 0973042 164981 27 1270267 1238056 0974642 164792 28 2091030 2041099 0976121 164613 29 3437839 3361297 0977735 164409 30 5646773 5528869 0979120 164254 31 9266788 9084612 0980341 164108 32 15195070 14913561 0981474 163973 33 24896206 24461858 0982554 163844 34 40761087 40091276 0983567 163724 35 66687201 65657293 0984556 163605 36 109032500 107448288 0985470 163498 37 178158289 175719701 0986312 163399 38 290939807 287190789 0987114 163304 39 474851445 469098096 0987884 163213 40 774614284 765791252 0988610 163128 41 1262992840 1249465729 0989290 163048 42 2058356522 2037607124 0989919 162975 43 3353191846 3321349362 0990504 162906 44 5460401576 5411548368 0991053 162842 45 8888486816 8813593422 0991574 162781 46 14463633648 14348888392 0992067 162723 47 23527845502 23352120464 0992531 162669 48 38260496374 37991479150 0992969 162618 49 62200036752 61788341876 0993381 162570 50 101090300128 100460533126 0993770 162525 31 / 32
Conjecture n g /n g 1 φ Behavor of n g 1+n g 2 n g ng 1+ng 2 ng 1 0 50 g Behavor of n g n g 1 ng φ ng 1 0 50 g 32 / 32