From the Euclidean Algorithm for Solving a Key Equation for Dual Reed Solomon Codes to the Berlekamp-Massey Algorithm
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1 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
2 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
3 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 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 ««r 1 f 1 g 1 r 2 f 2 g 2 Return r 1
5 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
6 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
7 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 LC( R 1 Q R 1 ) R 1 F 1 G 1 Return R 1 (aconstant multple of r 1 )
8 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 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 Q (0) 0 1 «(1) 1 Q 0 1 x «««(l) 1 Q x l / 32
9 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 LC( R 1 Q R 1 ) R F G 0 LC( R 1 Q R 1 )! 1 Q (l) x l 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 Q (0) 0 1 «(1) 1 Q 0 1 x «««(l) 1 Q x l / 32
10 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 LC( R 1 Q R 1 ) R F G 0 LC( R 1 Q R 1 )! 1 Q (l) x l 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
11 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 R +1 F+1 G ««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
12 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
13 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 ««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
14 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 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) C A 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)
15 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 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 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 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 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
16 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 )
17 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)
18 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 )
19 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
20 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
21 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
22 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
23 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
24 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
25 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
26 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
27 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)
28 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
29 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
30 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
31 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
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 «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 ««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
33 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 «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 ««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
34 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 «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 ««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
35 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 «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 ««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
36 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 «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 ««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
37 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 «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 ««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
38 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 )
39 From Eucldean to Berlekamp-Massey ALGORITHM: Intalze: d 1 deg( S) d 1 n «F 1 G F 1 G «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 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
40 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
41 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 e(1)x n 1 e (α n k )x k + e (α n k 1 )x k e (α)x n 1 u (α n k )x k + u (α n k 1 )x k u (α)x n 1 Once we have the error postons, we can compute the error values as e Ω(α ) α Λ (α )
42 25 / 32
43 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 )
44 Cash pont 27 / 32 Example The amounts of money one can obtan from a cash pont (dvded by 10)
45 Cash pont 28 / 32 amount amount/ mpossble! mpossble!
46 Countng 29 / 32 Let n g denote the number of numercal semgroups of genus g
47 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
48 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}
49 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,}
50 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
51 Conjecture n g /n g 1 φ n g ng ng 1 + g 1+n g 2 ng ng 2 ng n g / 32
52 Conjecture n g /n g 1 φ Behavor of n g 1+n g 2 n g ng 1+ng 2 ng g Behavor of n g n g 1 ng φ ng g 32 / 32
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