The Frobenius problem: A Geometric Approach

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1 The Frobenius problem: A Geometric Approch Lli Brrière nd Alíci Mirlles Deprtment de Mtemàtic Aplicd IV Universitt Politècnic de Ctluny {lli,lmirll}@m4.upc.edu Abstrct For the well known Frobenius problem, we present new geometric pproch, bsed on the use of the n-dimensionl lttice Z n, where n is the number of genertors. Within this pproch we re ble to study the cses of two nd three genertors. The min feture of our geometric representtion is tht we cn nicely visulize the set of gps, i.e., the non-representble positive integers. In the cse of two genertors, we give description of the set of gps. Moreover, for ny positive integer, m, we derive simple expression for the denumernt d(m;, b). We show tht we cn use the -dimensionl lttice ssocited to the set of genertors {, b} to study the Frobenius problem with genertors {, b, c}. In prticulr, we give, s for two genertors, grphicl representtion of the set of gps. For lrge set of possible vlues of c, this representtion llows us to simplify the computtion of the Frobenius number nd compute the number of gps. MSC000: 11D85, 0M14, 0M30. Keywords: Frobenius problem, integer lttice, semigroup. 1 Introduction Given finite set of positive integers, A = { 1,..., n }, the Frobenius problem for A sks for the mximum nturl number tht cn not be represented s positive combintion of the elements of A. The solution of the Frobenius problem, clled the Frobenius number of A nd denoted by g( 1,..., n ), does exists, provided tht gcd( 1,..., n ) = 1. Severl interesting problems re relted with the Frobenius problem. Nmely, the computtion of the number of non-representble integers, clled gps in the context of numericl semigroups, nd their description, nd the computtion of the denumernt of representble positive integer m, i.e. the number of non-negtive representtions of m on A. This problem ws introduced by Frobenius in its lectures. In 1884 Sylvester set tht the Frobenius number for n = is g( 1, ) = 1 1 nd the number of non-representble positive integers is N( 1, ) = ( 1 1)( 1) [9]. Since then, the Frobenius problem hs been widely Reserch supported by the Ministry of Eduction nd Science (Spin) nd the Europen Regionl Development Fund (ERR) under projects MTM C0-01 nd TEC Corresponding uthor. Postl ddress: Cmpus Nord UPC; c. Jordi Giron, 3-5; Brcelon (Spin) 1

2 studied, from very different points of view, rnging from rthmetics nd lgebr to lgorithmics nd complexity. The Frobenius problem turns out to be difficult problem. For fixed vlues of n, it is known to be polynomil, whilst it is proved to be NP-complete if n is prt of the input [5]. It is shown in [] tht closed formuls doesn t exist for the Frobenius number with fixed number of genertors n 3. We found in the litterture vriety of works giving formuls for the Frobenius number in prticulr cses. One of the clssicl ones is [1], where the uthors lso study the cse of three genertors. Other methods for computing G( 1,, 3 ) re given in [3, 4, 8]. From more lgebric point of view, the set of representble integers is numericl semigroup. The fundmentl gps of semigroup, concept tht inspired prt of our results, re explicitly described in [7]. A review of the existing work up to 005 is the book [6]. We refer the reder to this book nd the references therein for complete overview of the stte of the rt. Nottion Throughout the pper we use the following nottion. We sy tht set of nturl numbers A = { 1,..., n } stisfying gcd( 1,..., n ) = 1 is set of genertors of dimension n. Given such set: R(A) = R( 1,..., n ) = {m = x x + + x n n x i N} denotes the set of integers which re representble s non-negtive liner combintion of elements of A. The expression m = x x + + x n n is representtion of m on A. R(A) = R( 1,..., n ) = N \R(A) denotes the set of non-representble integers, which we will lso cll gps, s R(A) is numericl semigroup. g( 1,..., n ) = mx R(A) denotes the Frobenius number of A. N(A) = R(A) denotes the number of gps. If m is non-negtive integer, the denumert of m nd A is the number of non-negtive representtions of m on A, denoted by d(m;a). Hence, m is gp if nd only if d(m;a) = 0. Our results For the Frobenius problem we present new geometric pproch. Within this pproch, most of the known results for n = cn esily be derived nd visulized. Moreover, we re ble to study the cse of three genertors. Our pproch is bsed in the following ide. Given A = { 1,..., n }, we ssocite to the point (x 1,...,x n ) Z n the integer m = x x n n. This gives lbeling of the integer n-dimensionl lttice. For n =, A = {, b}, we identify prticulr tringle in the plne contining exctly the set of non-representble positive integers. This llows us to give complete description of this set. Moreover, we derive simple expression for the denumernt, d(n;,b). Let A = {,b,c} be set of positive integers, with gcd(,b) = 1 nd < b. We use the -dimensionl lttice nd the lbeling ssocited to {, b} to study the Frobenius problem for A = {,b,c}. In prticulr, if c is non-representble on {,b} nd c is, then we cn esily compute g(,b,c) nd N(,b,c), nd lso chrcterize the set of integers which re non-representble by {,c,b}. The sme results re proved for prticulr configurtion of the multiples of c in the set of gps for {,b}.

3 Figure 1: The Frobenius lttice for = 5 nd b = 13. The geometric pproch For given set of genertors, A = { 1,..., n }, we cn consider the liner mp l : Z n Z (x 1,...,x n ) x x n n This mp cn be viewed s n integer lbeling of Z n. If m = x x n n, we lbel the point of coordintes (x 1,...,x n ) by the integer m. Notice tht the function l is liner nd surjective, nd its kernel is the set ker l = {(x 1,...,x n ) Z n x x n n = 0} tht is, the set of integer points lying on the hyperplne of R n of eqution x x n n = 0. In this pper we study the first two cses, n = nd n = 3. (See Figure 1 for simple exmple.).1 The cse of two genertors Let us first concentrte on the cse of two genertors. We define the Frobenius lttice nd study its geometric properties. Definition.1 (Frobenius lttice) Given n ordered set of genertors A = {, b}, with < b, the Frobenius lttice ssocited to A is the pir (Z,l), where l is the mp from the infinite lttice Z to Z l : Z Z (x,y) x + yb We cll l(x,y) the lbel of (x,y). 3

4 F 1 F 0 F 1 F U 1 ( b, ) (0, ) U 0 T 1 U 1 (0, 0) (b, 0) T 0 L 1 (b, ) (b, ) T 1 L 0 L 1 (b, ) Figure : The Frobenius lttice nd its bsic fetures. We next describe the geometric objects ssocited to the Frobenius lttice. A grphicl representtion is given in Figure. The set of points lbeled 0 re the integer points which lie on the line of eqution x + yb = 0 ker l = {(λb, λ) λ Z} As consequence, for every m Z, the set of points lbeled m re the integer points which lie on the line of eqution x + yb = m. If m = u + vb = l(u,v) then for every λ Z l((u,v) + (λb, λ)) = m The lbeling l is covering of Z by Z nd, thus, induces prtition of the plne. We sy tht the plne is decomposed into strips, which re defined s follows. For every m Z nd λ Z, there is exctly one point (x,y) such tht m = l(x,y) nd λb x < (λ + 1)b. This llows us to define the strip ssocited to λ. Definition. (λ-strip ssocited to {,b}) For every λ Z, we define the strip F λ or the λ-strip ssocited to the set {,b} s the subset of the Frobenius lttice F λ = {(x,y) Z λb x < (λ + 1)b} = [0,b) Z The collection of sets {F λ } λ Z constitutes prtition of Z. Indeed, they re pirwise disjoints nd they cover Z. Moreover, for every integer λ, the restriction l Fλ : F λ Z is bijection. 4

5 For every pir of integers λ 1, λ, the trnsltion of vector ((λ λ 1 )b, (λ λ 1 )) is bijection between F λ1 nd F λ which pplies every point of F λ1 onto the point inf λ with the sme lbel. For ny given λ, we now give the definition of three subsets in the λ-strip whose respective sets of lbels re the negtive integers, the non-representble positve integers, R(, b), nd the representble integers R(,b) (see Figure ). The points below the line x + yb = 0 re lbeled by the negtive integers. We define L λ = {(x,y) F λ x + yb < 0} The positive integers re the lbels of the points stisfying x + yb 0. Among these points, we distinguish two sets. T λ = {(x,y) F λ x + yb > 0, y < λ} nd U λ = {(x,y) F λ x + yb 0, y λ} When using this nottion, we hve to tke into ccount tht representble integer hs lso non-vlid representtions (m = l(u,v) is vlid representtion of m if nd only if u,v 0). In F 0, the set of representble numbers, R(,b), is the set of lbels of the points stisfying y 0 l(u 0 ) = R(,b) nd the set of gps, R(,b), is the set of lbels of the tringle T 0 l(t 0 ) = R(,b) Thnks to this representtion we cn esily deduce most of the known results on the Frobenius problem with two genertors. In prticulr, both the Frobenius number nd the number of gps follow nicely from the Frobenius lttice nd the definition of T 0. The Frobenius number of A is the lbeling of the up-right corner of T 0, g(,b) = mx{l(x,y) (x,y) T 0 } = l(b 1, 1) = (b 1) b = b b becuse the lbeling function is incresing in both coordintes nd, in T 0, the point of mximum first coordinte is the point of mximum second coordinte. The number of gps generted by A is the number of integer points in T 0, N(A) = R(A) = becuse no points in T 0 lie on the line x + yb = 0. ( 1)(b 1) 5

6 x + yb = 0 (0, 0) ( b, 1) (b 1, 1) y = 1 y = j ( bj, j ) S j (b 1, j) ( bj, j ) ( bj, j ) (b 1, ( 1)) (b, ) Figure 3: The set of gps..1.1 Description of the set of gps Let us see how the elements of R(,b) re distributed in T 0. For grphicl representtion, see Figure 3. Notice tht, for every u 1,u Z, l(u 1,v) l(u,v) bv (mod ). This implies tht R(A) = l(t 0 ) cn be prtitioned into congruence clsses modulo, which correspond to horizontl segments in the geometric representtion. We denote by S j the segment T 0 {(x,y) y = j}. For every j, 1 j 1, the number of points in S j is n j = b bj = By symmetry, n j = bj nd n j + n j = b 1. The points (i, j) S j re ( bj ), j, nd their respective lbels re ( bj ) l + k, j ( bj In S j, the minimum nd the mximum lbels re b ( j) ) + 1, j,...,(b 1, j) bj = + k bj, 0 k n j 1 ) minl(s j ) = l( bj, j = bj bj ) mxl(s j ) = l + n j, j = l(b 1, j) = (b 1) bj ( bj If r = b (mod ), then the smllest non-representble integers re {1,,3,, 1} = {r (mod ),r (mod ),... ( 1)r (mod )} 6

7 m Positive representtions ( b, ) m ( b, ) (0, ) m (0, 0) (b, 0) m (b, ) (b, ) Figure 4: The denumernt of m with respect to {,b}, d(m;,b). We represent by the points which give positive representtion of its lbel. Ech of these points is the minimum of one of the segments S j. Let us concentrte on r = b b. Notice tht r l(s 1) nd thus ( b( 1) ) r = l, ( 1) Moreover, rj S j nd ( b( j) ) rj = l, ( j) In prticulr, the coordintes of 1 in T 0 cn be obtined by solving the eqution Then, l(b kb, ( k)) = The denumernt kr (mod ) = 1 We cn use the Frobenius lttice to give simple expression of the denumernt of given integer. This expression reltes the denumernt of m with the position in the plne of ny of the points lbeled m. Figure 4 illustrtes our result, which is given in the following Proposition. Proposition.3 Let m N nd m = l(u,v). Then, u v d(m;,b) = b Proof. If m = l(u,v) = l(u,v ) then u = u + λb nd v = v λ, for some integer λ. It is cler tht u u v v = + λ nd = λ b b This implies tht u b + v + 1 is the sme for every (u,v) l 1 (m). Hence, we cn ssume w.l.o.g. tht (u,v) F 0, i.e. 0 u < b. In this cse, the set of non-negtive representtions of m is {(u,v) + λ(b, ) 0 λ v }. The number of elements in this set is d(m;,b) = v + 1. Since in this strip u b = 0, we cn write d(m;,b) = u b + v

8 F 0 F 0 + k X k = [u, b 1] [ v, 1] k = l(u, v) Figure 5: Adding gp k to the set of genertors {,b}. Remrk tht d(m;,b) = 0 if nd only if m is gp. 3 Some results on the cse of three genertors In this section we del with the Frobenius poblem for {,b,c} with gcd(,b) = 1, < b nd c R(,b). For this purpose, we will use the -dimensionl lttice ssocited to {,b} insted of the 3-dimensionl lttice ssocited to {,b,c}. First, we give the plne construction tht llows us to represent g(,b,c) in Z. Our concern is to determine the distribution of the multiples of c in the tringle T 0, for which we know l(t 0 ) = R(,b). In fct, the Frobenius problem with three genertors {,b,c} is esy to solve if c is gp but c is not. This cse covers hlf of the possible vlues of c. We show lso tht, in two prticulr cses, the set of gps R(,b,c) hs nice plne represention s subset of T 0, nd both the Frobenius number nd the number of gps re esy to compute. 3.1 The representtion of R(, b, c) in Z For three genertors, {,b,c}, the lbeling of (x,y,z) Z 3 is defined s l(x,y,z) = x + yb + zc. This gives lbeling of the points of the three dimensionl lttice. Given c = u vb R(,b), with (u, v) T 0, the trnsltion of vector (u, v) pplied to F 0 gives new strip, denoted by F 0 + c. We denote by U 0 + c the set obtined from U 0 by the sme trnsltion. Since (u, v) T 0, (U 0 + c) T 0 is rectngle contining the gps of R(,b) but hve non-negtive representtion of the form x + yb + c. Notice tht x + yb + c cn be represented in Z 3 by (x,y,1). In this wy, the set of gps for {,b,c} cn be obtined from T 0, by removing the points in U 0 + c, U 0 + c, U 0 + 3c, nd so on. In Figure 5 we show the representtion of the 3-dimensionl lttice in the plne. More formlly, we hve the following. Definition 3.1 (k-representble integer) Let {, b} be set of two positive integers with < b, 8

9 gcd(,b) = 1. For given non-negtive integer k we define the set of k-representble integers by R k (,b) = {m Z m = λ 1 + λ b + k,λ 1 0,λ 0} = l(u 0 + k) In prticulr, if k = 0, R 0 (,b) = R(,b). Definition 3. (k-rectngle) Let {,b} be set of two positive integers with < b, gcd(,b) = 1. For given non-negtive integer k we define the k-rectngle by the set of points X k = (U 0 + k) T 0 Notice tht, with this definition l(x k ) = R k (,b) \ R(,b). Tht is, if k is gp on {,b}, then the k-rectngle contins the points in T 0 which re gps on {,b} but hve representtion on {,b,k} of the form x + yb + k. The k-rectngle is empty if k is representble on {,b}. If k is gp, esy computtions give the following property (see Figure 5). Property 1 Let {,b} be set of two positive integers with < b, gcd(,b) = 1. If k = u vb is the representtion of k in T 0 then X k = [u,b 1] [ 1, v] (1) As the two following properties show, in order to describe R(,b,c), we re interested on the ic-representble integers, {R ic (,b)} i 0 nd the ic-rectngles, {X ic } i 1. Property The set of integers which hve non-negtive representtion by {,b,c} is R(,b,c) = R ic (,b) Property 3 The set of points in T 0 with lbels in R(,b,c) is In wht follows we show tht the set of gps on {,b,c} depends only on reltively smll number of multiples of c. i=0 X ic Definition 3.3 Let {,b} be set of two positive integers with < b, gcd(,b) = 1. For given c R(,b), we define the first representble multiple s the positive number, denoted by s > 1, such tht c, c,...,(s 1)c R(,b) nd sc R(,b) Proposition 3.4 If c R(, b) nd s is the first representble multiple of c on {, b}, then R(,b,c) = s 1 i=0 R ic (,b) 9

10 Proof. By Property, we need only to prove tht R(,b,c) s 1 i=0 R ic (,b) Since sc R(,b), there re two integers m 1 0 nd m 0 such tht sc = m 1 + m b. Let m = x + yb + zc R(,b,c), with x 0, y 0 nd z 0. If z < s, then m s 1 i=0 R ic(,b). Assume tht z s, nd let z = ps + q with 0 q s 1 the integer division of z by s. We cn write m = x + yb + zc = x + yb + (ps + q)c = x + yb + p(m 1 + m b) + qc = (x + pm 1 ) + (y + pm )b + qc, where 0 q s 1. Thus, m s 1 i=0 R ic(,b). Corollry 3.5 c R(,b), if c,c,,(s 1)c R(,b) nd sc R(,b), s 1 R(,b,c) = R(,b)\ l(x ic ). Proof. It follows strightforwrd from Proposition 3.4 nd the definition of X ic. 3. Solving the Frobenius problem Although the Frobenius number for three genertors is known, we hve defined nice geometric description of the set of gps which will be used to give more intuitive computtion of g(,b,c), toghether with the computtion of the number of gps N(,b,c). In fct, given c R(,b), if the first representble multiple of c is smll, we cn give simple expressions for both the Frobenius number g(,b,c) nd the number of gps N(,b,c). We strt by completely chrcterize the cse s =, tht is, c is gp on {,b} nd c is not. An exmple of this cse is shown in Figure 6. Proposition 3.6 Let c R(,b) nd c = u vb the representtion of c in T 0. Then, the first representble multiple of c is s = if nd only if = b u v. Under these conditions, the number of gps nd the Frobenius number corresponding to the set of genertors {,b,c} re, respectively, N(,b,c) = N(,b) (b u)v nd g(,b,c) = mx{l(u 1, 1),l(b 1, v 1)} Proof. If c = u vb is the representtion of c in T 0, then 1 u < b nd 1 v <. The first representble multiple of c is if nd only if (u, v) U 1. This is equivlent to u < b u nd v 1 < b u nd v = b u v 10

11 (u 1, 1) (b 1, v 1) (0, 0) X c c = l(u, v) c = l(u, v) Figure 6: For c = l(u, v), with = b u v, the Frobenius number of {,b,c} is g(,b,c) = mx{l(u 1, 1),l(b 1, v 1)} nd the number of gps is N(,b,c) = N(,b) (b u)v. By Corollry 3.5, the set of gps R(,b,c) re represented in the plne by the set T 0 \ X c. Consequently, N(,b,c) = T 0 \ X c = N(,b) X c nd Property 1 implies X c = (b u)v nd, thus g(,b,c) = mx l(t 0 \ X c ) N(,b,c) = N(,b) (b u)v Let us compute the mximum lbel of the points in T 0 \X c. By the monotonicity of the lbeling function l, we cn distinguish two locl mxim (see Figure 6). In the set {(x,y) T 0 \ X c y [ v, 1]}, the mximum lbel is ttined in the point C 1 = (u 1, 1), since C 1 + (0,1) U 0 nd C 1 + (1,0) X c. In the set {(x,y) T 0 \ X c y [, v 1]}, the mximum lbel is ttined in the point C = (b 1, v 1), since C + (0,1) X c nd C + (1,0) U 1. This gives which completes the proof. g(,b,c) = mx{l(c 1 ),l(c )} = mx{l(u 1, 1),l(b 1, v 1)} We hve shown tht, if c is gp for {,b}, but c is not, then we cn use simple geometric resoning to compute the number of gps for {,b,c} s well s the Frobenius number. If s similr resoning cn be pplied in some prticulr cses. Roughly speking, if c is represented in T 0 by (u, v), Lemm 3.7 gives necessry nd sufficient condition to ensure tht (u, v) is in T 0 (remrk tht it could be in T 1 ) nd the points (iu, iv) re lso in T 0, for i =,...,s 1. The 11

12 computtion of the number of gps nd the Frobenius number under the conditions of Lemm 3.7 re given in Theorem 3.9. The geometric representtion of the Frobenius lttice gives n intuitive pproch to our results (see Figures 6 nd 7). The bsic ide is tht computtions re esy if the points in T 0, the lbels of which re the multiples of c, re ligned. In fct, the hypothesis of Theorem 3.9 nd Theorem 3.10 re symmetric, s shown in Figures 7 nd 8. Lemm 3.7 Let c R(,b) nd c = u vb the representtion of c in T 0. Then, s = b u v if nd only if (u, v), (u, v),... ((s 1)u, (s 1)v) T 0 nd (su, sv) U 1 () Proof. Assume tht tht (u, v) T 0. This is equivlent to 0 < u < b nd 0 < v <. Then, By definition of T 0 nd U 1, this is equivlent to And, since (u, v) T 0, this is equivlent to (). s = b u v s 1 < b u s nd 0 < s v (s 1)u < b su nd 0 < sv < ((s 1)u, (s 1)v) T 0 nd (su, sv) U 1 Next we give technicl Lemm tht will llow us to do esy computtions in the proof of Theorem 3.9. Lemm 3.8 Let c R(,b), c = u vb the representtion of c in T 0, nd r > 0 such tht (u, v), (u, v),...(ru, rv) T 0 (3) For every i = 1,...,r, let Y ic = [iu,b 1] [ iv, (i 1)v 1], then r X ic = nd the rectngles {Y ic }...r re pirwise disjoint. r Y ic (4) Proof. Eqution (4) follows from Eqution (1) in Property 1. The pirwise disjointness of the collection of rectngles is cler from its definition. Theorem 3.9 Let c R(,b) nd c = u vb the representtion of c in T 0. If s = b u v, then s is the first representble multiple of c. Moreover, nd N(,b,c) = N(,b) (s 1)bv + s(s 1) uv g(,b,c) = mx{l((s 1)u 1, (s )v 1),l(b 1, (s 1)v 1)} 1

13 (0, 0) c = l(u, v) c = l(u, v) (3u 1, v 1) (b 1, 3v 1) Y c Y c Y 3c 3c = l(3u, 3v) 4c = l(4u, 4v) Figure 7: For c = l(u, v), with 4 = b u v, the Frobenius number of {,b,c} is g(,b,c) = mx{l(3u 1, v 1), l(b 1, 3v 1)} nd the number of gps is N(,b,c) = N(,b) (b u)v (b u)v (b 3u)v. Proof. Lemm 3.7 implies tht s is the first representble multiple of c (on {,b}). To prove the Theorem, we hve to compute the number of gps nd the Frobenius number, for the set of genertors {,b,c}. (See Figure 7 for grphicl explntion.) By Corollry 3.5, s 1 N(,b,c) = N(,b) X ic Now, by Lemm 3.8, with Y ic = [iu,b 1] [ iv, (i 1)v 1] s in the mentioned Lemm, s 1 s 1 s 1 s 1 s(s 1) X ic = Y ic = Y ic = (b iu)v = (s 1)bv uv Thus, N(,b,c) = N(,b) (s 1)bv + s(s 1) uv s 1 The Frobenius number is the mximum lbel of the points in T 0 \ X ic. By Lemm 3.8, g(,b,c) = mx l A resoning nlogous to the proof of Proposition 3.6 gives tht the mximum is ttined in one of the locl mxim C i = (iu 1, (i 1)v 1), for i = 1,...,s 1 ( T 0 \ s 1 Y ic ) nd C s = (b 1, (s 1)v 1) 13

14 5c = l( 5u,5v ) ( 4u 1, 1) ( 3u 1,4v 1) 4c = l( 4u,4v ) 3c = l( 3u,3v ) c = l( u,v ) c = l( u, v ) (0, 0) Figure 8: For c = l( u,v ), with 5 = v b u, the Frobenius number of {,b,c} is g(,b,c) = mx{l( 4u 1, 1), l( 3u 1,4v 1)} nd the number of gps is N(,b,c) = N(,b) u v u v 3u v 4u ( 4v ). By noticing tht l(c i ) = l(iu 1, (i 1)v 1) = (iu 1) (i 1)vb b = ic b + vb, which is incresing in i, we cn conclude tht g(,b,c) = mx{l((s 1)u 1, (s )v 1),l(b 1, (s 1)v 1)} Theorem 3.10 Let c = l(u, v) R(,b). If s = v b multiple of c. Moreover, b u, then s is the first representble nd s(s 1) N(,b,c) = N(,b) (b u)(s 1) + (b u)( v) g(,b,c) = mx{l( (s )(b u) 1,(s 1)( v) 1),l( (s 1)(b u) 1, 1)} Proof. Let u = b u nd v = v. Then, ( u,v ) is the representtion of c in the tringle of gps T 1. Anlogously to Lemm 3.7, we hve tht s = v b u if nd only if ( u,v ), ( u,v ),... ( (s 1)u,(s 1)v ) T 1 nd ( su,sv ) U 1 Moreover, for 1 i s 1, we cn replce the rectngles Y ic, introduced in Lemm 3.8, by Z ic = [ iu, (i 1)u 1] [iv, 1] nd compute the number of gps nd the Frobenius number for {,b,c} in T 1, s we did in T 0 when proving Theorem 3.9. (See Figure 8 for grphicl explntion.) This gives s 1 s 1 N(,b,c) = N(,b) Z ic = N(,b) u ( iv ) 14

15 nd thus On the other hnd, which is ttined in one of the points N(,b,c) = N(,b) u (s 1) + u v s(s 1) g(,b,c) = mx l ( T 1 \ s 1 Z ic ) (5) nd C i = ( (i 1)u 1,iv 1), for i = 1,...,s 1 C s = ( (s 1)u 1, 1) By noticing tht l(c i ) = l( (i 1)u,iv 1) = (i 1)u + iv b b = ic b + u, which is incresing in i, we cn conclude tht g(,b,c) = mx{l( (s )u 1,(s 1)v 1),l( (s 1)u 1, 1)} (6) Replcing u by b u nd v by v in Equtions (5) nd (6), we get nd s(s 1) N(,b,c) = N(,b) (b u)(s 1) + (b u)( v) g(,b,c) = mx{l( (s )(b u) 1,(s 1)( v) 1),l( (s 1)(b u) 1, 1)} This concludes the proof. 4 Conclusion For the Frobenius problem with genertors { 1,..., n } we hve proposed geometric pproch. This pproch is bsed on lbeling of the n-dimensionl integer lttice, were n is the number of genertors. Our results in the cses nd 3 follow from n explortion of the set of gps which cn nicely be visulized thnks to our geometric pproch. For three genertors {,b,c}, with < b nd gcd(,b) = 1, we propose technique tht llows to compute in simple fshion both the Frobenius number nd the number of gps. As finl remrk, notice tht our computtions need only O(log ) opertions, s they rely on the solution of the diophntine eqution u + bv = 1. Moreover, our results re proved in geometric flvor which gives deeper understndig of the Frobenius problem. A nturl question tht rises from Theorem 3.9 nd Theorem 3.10 is: Given pir of genertors {,b}, with < b nd gcd(,b) = 1, which vlues of c R(,b) meet the hypothesis of one of the mentioned theorems? In other words, for which vlues of c R(,b) re the Frobenius problem, nd mybe other problems relted with, geometriclly esy to solve? In Figure 10 we show with simple exmple n intuitive nswer to this question. In fct, if c = u vb is the representtion of c in T 0 nd the multiples of c re not ll ligned in T 0, we cn not pply our technique. 15

16 ... s = 5 s = 4 Theorem 3.9 s = 3 s = s = 3 s = 4 s = 5 Theorem Figure 9: The portion of T 0 which is covered by our technique. Another nturl question, relted with our previous results, is: Given pir of genertors {, b}, with < b nd gcd(,b) = 1, how mny vlues of c R(,b) meet the hypothesis of one of the mentioned theorems? In other words, which portion of the tringle of gps T 0 corresponds to points for which the Frobenius problem, nd mybe other problems relted with, re geometriclly esy to solve? In Figure 9 we represent the nswer to this question. Relted to this question, it s worth to mention tht Proposition 3.6 sys tht in cse s = Theorem 3.9 nd Theorem 3.10 re equivlently fullfilled. Indeed, it is esy to see tht = b u = v v b b u Moreover, point (u, v) T 0 stisfies this condition if nd only if b u b 1 nd 1 v This gives rectngle in T 0 of size b which we know to be exctly the points corresponding to vlues of c stisfying s =. This rectngle contins lmost hlf of the points in T 0, nd is the big white rectngle in Figure 9. In fct, we cn sy more. By simple geometric resoning, it is esy to see tht if s 4 then we re under the hypothesis of either Theorem 3.9 or Theorem Open problems We let some open problems, concerning the use of the technique we propose. We hve studied the cse {,b,c} with < b nd gcd(,b) = 1. To pply the sme technique to the generl cse needs the considertion of (t lest) d = gcd(,b) different -dimensionl lttices. 16

17 (0, 0) 3c c = l(u, v) 4c c = l(u, v) 3c = l(3u, 3v) 4c = l(4u, 4v) 5c = l(5u, 5v) Figure 10: For c = l(u, v) with s = 5, (u, v),(u, v) T 0 nd (3u, 3v),(4u, 4v) T 1, the Frobenius number nd the number of gps for {,b,c} is not so esy to compute. The points in T 0 corresponding to 3c nd 4c re obtined from (3u, 3v) nd (4u, 4v) by the trnsltion of vector ( b,). Also, we think tht some work cn be done using the 3-dimensionl lttice. This pproch hs the drwbck tht visuliztion becomes more difficult. For instnce, the line x+yb = 0 in the plne trnsltes to plne in the spce, the plne x + yb + zc = 0. The tringle T 0 becomes tetrhedron. But, obviously, only smll portion of the tetrhedron correspond to the set of gps. A possible question is: is there nice representtion of the set of gps, s subset of the tetrhedron determined by x + yb + zc = 0? The generliztion of our method to higher dimension could probbly give some new result. References [1] A. Bruer nd J.E. Shockley, On problem of Frobenius, J. Reine Angew. Mth., 11 (196) pp (see Mth. Rev. 47#17) [] F. Curtis, On formuls for the Frobenius number of numericl semigroup, Mth. Scnd., 67 (1990) [3] M. Hujter nd B. Vizvri, The exct solution to the Frobenius problem with three vribles, J. Rmnujn Mth. Soc., 55() (1987), [4] I.D. Kn, B.S. Stechkin, nd I.V. Shrkov, Frobenius problem for three rguments, Mth. Notes, 6(4) (1997), [5] J.L. Rmírez-Alfonsín, Complexity of the Frobenius Problem, Combintoric, 16(1996) [6] J.L. Rmírez-Alfonsín, The Diophntine Frobenius Problem, Oxford University Press (005). 17

18 [7] J.C. Rosles, Fundmentl gps of numericl semigroups generted by two elements, Liner Algebr Appl., 405 (005), [8] E. S. Selmer nd Ö. Beyer, On the liner diophntine problem of Frobenius in three vribles, J. Reine Angew. Mth. 301 (1978), [9] J.J. Sylvester, Problem 738, Eductionl Times, 37 (1884), 6. Repreinted in: Mthemticl questions with their solution, Eductionl Times, 41 (1884), 1. 18