Nets in groups, minimum length $g$-adic representations, and minimal additive complements

NETS IN GR OUPS, MINIMUM LENGTH g -ADIC REPRESENT A TI ONS, AND MINIMAL ADDITIVE COMPLEMENTS MEL VYN B. NA THA NSON Abstract. The n umber theoretic analogue of a net in metric geometry sug- gests new problems and results i n com binatorial and additiv e num ber theory . F or example, for a fixed integer g ≥ 2, the study of h -nets in the additive group of integers with respect to the generating set A g = { 0 } ∪ {± g i : i = 0 , 1 , 2 , . . . } requires a kno wledge of the wo rd lengths of integ ers with resp e ct to A g . A g -adic r ep resent ation of an in teger is describ ed that algorithmically produces a represen tation of shortest length. Additive complemen ts and additiv e asymp- totic complements are also discussed, together with their associated minim al it y problems. 1. Nets in metric sp aces Let ( X, d ) b e a metric space. F or z ∈ X and r ≥ 0, the spher e with center z a nd radius r is the set S z ( r ) = { x ∈ X : d ( x, z ) = r } . The op en b al l B z ( r ) of radius r and center z and the close d b a l l B z ( r ) of radius r and ce nter z a re, resp ectiv ely , B z ( r ) = { x ∈ X : d ( x, z ) ≤ r } = [ r ′ 0 ? The metric spa ces ( X, d X ) and ( Y , d Y ) are ca lled bi-Lipschitz e quivalent if there exists a function f : X → Y such that, for p o sitiv e constants K 1 and K 2 , we hav e K 1 d X ( x, x ′ ) ≤ d Y ( f ( x ) , f ( x ′ )) ≤ K 2 d X ( x, x ′ ) for all x, x ′ ∈ X . The metric s paces ( X , d X ) and ( Y , d Y ) ar e called qu asi -isometric if there exis t nets C X in X a nd C Y in Y that are bi-Lips c hitz equiv alent . Thes e are fundamental concepts in metr ic geometry . 2. N ets in groups Let G b e a multiplicative group or semigroup with identit y e . F or subsets A and B of G , w e define the pr o duct set AB = { ab : a ∈ A a nd b ∈ B } . If A = ∅ or B = ∅ , then AB = ∅ . F or b ∈ G , we write Ab = A { b } and bA = { b } A . The set Ab is called the right tr anslation of A by b , a nd the set b A is c a lled the left tr anslatio n of A by b . F or every nonnegative integer h , w e define the pro duct sets A h inductively: A 0 = { e } , A 1 = A , and A h = A h − 1 A for h ≥ 2. Thus, A h = { a 1 a 2 · · · a h : a i ∈ A for i = 1 , 2 , . . . , h } . If e ∈ A , then A i − 1 ⊆ A i for all i ≥ 1, and A h = h [ i =0 A i . Let A b e a set of generator s for a gro up G . Without los s o f gener alit y we can assume tha t A is symmetric , that is, a ∈ A if and only if a − 1 ∈ A . W e define the wor d length fun ctio n ℓ A : G → N 0 as follows: F or x ∈ G and x 6 = e , let ℓ A ( x ) = r if r is the smalles t po sitiv e integer suc h that there e xist a 1 , a 2 , . . . , a r ∈ A w ith x = a 1 a 2 · · · a r . Let ℓ A ( e ) = 0 . The integer ℓ A ( x ) is ca lled the wor d length of x with r esp e ct to A , or, simply , the length of x . Let A b e a s ymmetric generating se t for G . The following pro perties follow immediately fr o m the definition of the word length function: (i) ℓ A ( x ) = 0 if and only if x = e , (ii) ℓ A ( x − 1 ) = ℓ A ( x ) for all x ∈ G , (iii) ℓ A ( xy ) ≤ ℓ A ( x ) + ℓ A ( y ) for all x, y ∈ G , (iv) ℓ A ( x ) = 1 if and only if x ∈ A \ { e } , (v) if x = a 1 · · · a s with a i ∈ A for i = 1 , . . . , s , then ℓ A ( x ) ≤ s , (vi) If A ′ = A ∪ { e } , then ℓ A ′ ( x ) = ℓ A ( x ) for all x ∈ G . Lemma 1. Le t A b e a symmetric gener ating set for a gr oup G . Supp ose that ℓ A ( x ) = r and that the elements a 1 , a 2 , . . . , a r ∈ A satisfy x = a 1 a 2 · · · a r . F or 1 ≤ i ≤ j ≤ r we have ℓ A ( a i a i +1 · · · a j ) = j − i + 1 . NETS IN GR OUPS AND g -ADIC REPRESENT A TIONS 3 Pr o of. By word length prop erties (iii) a nd (v) we have r = ℓ A ( x ) = ℓ A ( a 1 · · · a i − 1 a i · · · a j a j +1 a r ) ≤ ℓ A ( a 1 · · · a i − 1 ) + ℓ A ( a i · · · a j ) + ℓ A ( a j +1 · · · a r ) ≤ ( i − 1) + ℓ A ( a i · · · a j ) + ( r − j ) and s o j − i + 1 ≤ ℓ A ( a i · · · a j ) ≤ j − i + 1 . This co mpletes the pro of.  Let A be a s ymmetric gene r ating s et for a group G . The length function ℓ A induces a metric d A on G a s follows: d A ( x, y ) = ℓ A ( xy − 1 ) . The distance betw een distinct ele ments of G is a lw a ys a p ositiv e int eger, and so the metric s pace ( G, d A ) is 1-separ ated. Mor eo v er, d A ( x, e ) = ℓ A ( x ) for all x ∈ G , and so, for e very no nneg ativ e in teger h , we hav e S e ( h ) = { x ∈ G : ℓ A ( x ) = h } . Thu s, the set o f a ll group element s of leng th h is precisely the sphere with center e and r adius h in the metr ic space ( G, d A ). If r ≥ 0 and h = [ r ] is the integer pa rt of r , then for every z ∈ G we hav e B z ( r ) = { x ∈ X : d A ( x, z ) ≤ r } = { x ∈ X : d A ( x, z ) ≤ h } = B h ( z ) and so the geo metry of the g roup G is determined b y closed balls with integer radii. If e ∈ A , then A h = ∪ h i =0 A i and B h ( z ) = { x ∈ X : d A ( x, z ) ≤ h } = { x ∈ X : ℓ A ( xz − 1 ) ≤ h } = ( x ∈ X : xz − 1 ∈ h [ i =0 A i ) = { x ∈ X : xz − 1 ∈ A h } = A h z . Theorem 1. L et G b e a gr oup and let A b e a symmetric gener ating set for G with e ∈ A . F or every nonne gative int e ger h , the set C is an h -net in the metric sp ac e ( G, d A ) if and only if G = A h C . The set C is a net if and only if G = A h C for some nonn e gative int e ger h . Pr o of. The set C is an h - net in ( G, d A ) if and only if, for each x ∈ X , there exists z ∈ C with d A ( x, z ) = ℓ A ( xz − 1 ) ≤ h , that is, x ∈ B z ( h ). Equiv alen tly , C is an h -net if and o nly if G = [ z ∈ C B z ( h ) = [ z ∈ C A h z = A h C. Thu s, C is a net if and only if G = A h C for some nonnegative in teger h .  Here a re tw o constructio ns o f nets. 4 MEL VYN B. NA THANSON Theorem 2. L et G b e a gr oup and let A b e a symmetric gener ating set for G with e ∈ A . F or every nonne gative inte g er h , the set C = ∞ [ q =0 S e (( h + 1) q ) is an h -net in the metric sp ac e ( G, d A ) . Note that C = G if h = 0 . Pr o of. By Theor em 1, it suffices to prove that G = A h C . Let x ∈ G with n = ℓ A ( x ). By the divis ion algor ith m, there exist integers q ≥ 0 a nd r suc h that n = r + ( h + 1) q and 0 ≤ r ≤ h. There exist elements a 1 , . . . , a n ∈ A such that x = a 1 · · · a r a r +1 · · · a r +( h +1) q . Since this is a shortest representation of x as a pr o duct o f elements of A , it follows from L e mm a 1 that ℓ A ( a 1 · · · a r ) = r and ℓ A ( a r +1 · · · a r +( h +1) q ) = ( h + 1) q . Therefore, a 1 · · · a r ∈ S e ( r ) ⊆ A r ⊆ A h and a r +1 · · · a r +( h +1) q ∈ S e (( h + 1) q ) ⊆ C hence x ∈ A r C . This c ompletes the pro of.  Theorem 3. L et G b e a gr oup and let A b e a symmetric gener ating set for G with e ∈ A . Supp ose that for every x ∈ G t h er e exist s a ∈ A with (1) ℓ A ( ax ) = 1 + ℓ A ( x ) . F or every nonne gative inte ger h , t h e s et C = ∞ [ q =0 S e ((2 h + 1) q ) is an h -net in the metric sp ac e ( G, d A ) . Pr o of. Let x ∈ G with n = ℓ A ( x ). By the division algorithm, there exist in tegers q ≥ 0 and r s uch that n = r + (2 h + 1 ) q and | r | ≤ h. If r ≥ 0, then the ar gumen t in the pro of of Theorem 2 s ho ws that x ∈ A h C . Suppo se that r < 0. Then n = (2 h + 1) q − | r | and there exist ele ments a | r | +1 , . . . , a (2 h +1) q ∈ A such that x = a | r | +1 · · · a (2 h +1) q . Condition (1 ) implies that there exist element s a 1 , . . . , a | r | ∈ A such that ℓ A ( a | r |− i +1 · · · a | r | x ) = ℓ A ( a | r |− i +1 · · · a | r | a | r | +1 · · · a (2 h +1) q ) = (2 h + 1 ) q − | r | + i NETS IN GR OUPS AND g -ADIC REPRESENT A TIONS 5 for i = 1 , 2 , . . . , | r | . In particular, ℓ A ( a 1 · · · a | r | x ) = (2 h + 1) q and s o a 1 · · · a | r | x ∈ S e ((2 h + 1) q ) ⊆ C. Since a − 1 | r | · · · a − 1 2 a − 1 1 ∈ A | r | ⊆ A h , it follows that x =  a − 1 | r | · · · a − 1 2 a − 1 1   a 1 · · · a | r | x  ∈ A h C. This co mpletes the pro of.  If C is an h -net in G and C ⊆ C ′ , then G = A h C ⊆ A h C ′ ⊆ G and s o C ′ is an h -net in G . Similarly , if C is an h -net in G and y ∈ G , then G = Gy = ( A h C ) y = A h ( C y ) and C y is an h -net in G . Thu s, the set o f h -nets in the metric space ( G, d A ) is closed with resp ect to s upersets and right transla tions. W e mo dify the definitions appropria tely when G is an a dditiv e ab elian gro up with iden tit y element 0. F or subsets A a nd B o f G , we define the sumset A + B = { a + b : a ∈ A a nd b ∈ B } . F or h ≥ 1, the h -fold sumset of A is hA = { a 1 + a 2 + · · · + a h : a i ∈ A for i = 1 , 2 , . . . , h } . W e define 0 A = { 0 } . F or every b ∈ G , ther e is the transla tion A + b = A + { b } . Let A b e a symmetric gener a ting set for G with 0 ∈ A . By Theo rem 1, the set C is a net in G if and only if there is a nonnegative integer h such that G = hA + C. Problem 2. L et A b e a symmetric gener ating set for t he gr oup G with e ∈ A . Describ e and classi fy al l nets in G . Problem 3. The net C in the metric sp ac e ( G, d A ) is c al le d minimal if n o pr op er subset of C is a net. Determine if the metric s p ac e ( G, d A ) c ontains minimal nets, and, if so, c o nstruct examples of minimal nets. Is it p ossible to classify the minimal nets in a m etric sp a c e of t he form ( G, d A ) ? Problem 4 . Supp ose that minimal nets ex ist in the metric sp ac e ( G, d A ) . Do es every net c ont a in a minimal net? Problem 5. F or every inte ger g ≥ 2 , c onsider t he additive gr oup Z of inte g ers with gener a ting set A g = { 0 } ∪ {± g i : i = 0 , 1 , 2 , . . . } . L et ℓ g and d g denote, r esp e ctively, the wor d length function and the met ri c induc e d on Z . Classify the nets in the metric sp ac e ( Z , d g ) . Do es this sp ac e c ontain minimal nets? The metrics d 2 and d 3 ar e p articularly inter esting. 6 MEL VYN B. NA THANSON 3. An algorithm to compute g -adic length Fix an in teger g ≥ 2, and consider the additive gr oup Z with generating set A g = { 0 } ∪ { ± g i : i = 0 , 1 , 2 , . . . } . W e deno te b y ℓ g ( n ) the w ord length of an in teger n with resp ect to A g . A partition of an integer n as a sum of not neces sarily distinct elements o f A g will b e called a g -adic r epr esentation of n . In o rder to understand the metric geometry of the group Z with gener ating set A g , it is useful to hav e an algorithm to compute the g -adic length ℓ g ( n ) of a n integer n in ( Z , d g ). In this section w e construct a spec ial g -adic repre s en tation that has shortest length with resp ect to the generating set A g . Note that the shortest length representation of an integer with res pect to the gener ating set A g is no t unique. F or example, fo r even g w e hav e n = −  g 2  g i +  1 − g 2  g i +1 + g i +2 =  g 2  g i +  g 2  g i +1 . These a re g -adic re presen tations of n of s ho rtest length g . Similarly , for o dd g ,  g + 1 2  g i = −  g − 1 2  g i + g i +1 are g -adic repr esen tations of s hortest length ( g + 1) / 2. W e co nsider separately the re pr esen tations of integers as sums a nd differences of powers of g for g even and for g o dd. Theorem 4. L et g b e an even p ositive inte ger. Every inte ger n has a unique r ep r esentation in the form n = ∞ X i =0 ε i g i such that (i) ε i ∈ { 0 , ± 1 , ± 2 , . . . , ± ( g / 2) } for al l nonn e gative int e gers i , (ii) ε i 6 = 0 for only finitely many nonne g ative inte gers i , (iii) if | ε i | = g / 2 , then | ε i +1 | < g / 2 and ε i ε i +1 ≥ 0 . Mor e over, n has lengt h ℓ g ( n ) = ∞ X i =0 | ε i | in the metric sp ac e ( Z , d g ) asso ciate d with the gener ating set A g = { 0 } ∪ {± g i : i = 0 , 1 , 2 , . . . } . A representation o f the in teger n that satisfies conditions (i), (ii), and (iii) will be called the minimum length g -adic r epr esentation of n . Pr o of. W e begin by des c ribing a “standardizing and shortening” algor ithm that, for every no nzero integer n , pro duces a g -adic repres e ntation that satisfies condi- tions (i), (ii), and (iii) and that ha s length ℓ g ( n ). There are five op erations that we ca n p erform on an a rbitrary r e presen tation of an integer as a sum of elements of the genera ting set A g . Ea ch of these oper a tions pro duces a new representation with a str ictly smaller num ber of summands. (a) If 0 o ccurs as a summand in the representation of a no nzero integer n , then delete it. (b) If g i and − g i bo th app ear as summands, then delete them. NETS IN GR OUPS AND g -ADIC REPRESENT A TIONS 7 (c) If g i (resp. − g i ) o ccurs m ≥ g times for some i , then apply the division algorithm to write m = q g + s with 0 ≤ s ≤ g − 1, and r eplace q g o ccurrences of g i (resp. − g i ) with q summands g i +1 (resp. − g i +1 ). This oper ation reduces the nu mber of summands in the r epresen tation by q ( g − 1). (d) If g i o ccurs m times for so me i , wher e g / 2 < m < g , then r eplace mg i with ( g − m )( − g i ) + g i +1 . Similarly , if − g i o ccurs m times for some i , where g / 2 < m < g , then replace m ( − g i ) with ( g − m ) g i + ( − g i +1 ). These substitutions r educe the num ber of summands in the representation of n by m − ( g − m + 1) = 2 m − g − 1 ≥ 1. W e can iterate o perations (a)–(d) o nly finitely many times, since the nu mber of summands strictly decreases with each iteration. At the end of the proc e ss, we ha v e a representation n = P ∞ i =0 ε i g i with co efficien ts ε i ∈ { 0 , ± 1 , ± 2 , . . . , ± g / 2 } for all i and ε i = 0 for a ll s ufficien tly large i . (e) Supp ose that ε i = − g/ 2 and ε i +1 ≥ 1 for some i . W e replace − ( g / 2) g i + ε i +1 g i +1 with ( g / 2) g i + ( ε i +1 − 1) g i +1 . Similar ly , if ε i = g / 2 and ε i +1 ≤ − 1 for some i , then we re pla ce ( g / 2) g i + ε i +1 g i +1 with − ( g / 2) g i + ( ε i +1 + 1) g i +1 . Each of these op e r ations reduces the n um be r o f summands b y 1. W e re peat this o p eration a s often as p ossible. Again, at the end o f the pr ocess, w e hav e a representation n = P ∞ i =0 ε i g i , where ε i ∈ { 0 , ± 1 , ± 2 , . . . , ± g / 2 } for all i a nd ε i = 0 for a ll sufficiently lar ge i . Mor e o v er, if ε i = | g / 2 | , then ε i ε i +1 ≥ 0. The construction of a minim um length g -adic representation is a lmost complete. W e must still eliminate co nsecutiv e co efficien ts of g / 2 or − g / 2. Suppose that ε i = ε i +1 = g / 2 for s ome nonnegative integer i . Cho ose the smalle s t such integer i a nd, for this i , the lar gest in teger k ≥ 2 such that ε i = ε i +1 = · · · = ε i + k − 1 = g 2 . W e a pply the iden tit y ε i − 1 g i − 1 + i + k − 1 X j = i  g 2  g j + ε i + k g i + k = ε i − 1 g i − 1 +  − g 2  g i − i + k − 1 X j = i +1  g 2 − 1  g j + ( ε i + k + 1 ) g i + k to eliminate the k success iv e digits of g / 2. This reduces the num ber of summands by g k 2 −  g 2 + ( k − 1)  g 2 − 1  + 1  = k − 2 ≥ 0 . Observe that ε i − 1 6 = ± g / 2, and that ε i + k ≤ g / 2. Similarly , the identit y i + k − 1 X j = i  − g 2  g j =  g 2  g i + i + k − 1 X j = i +1  g 2 − 1  g j − g i + k allows us to eliminate k succes siv e digits of − g / 2 and reduce the num ber of sum- mands by k − 2 ≥ 0. It may still happ en that the representation o f n con tains consecutive digits of g / 2 or − g / 2. How ev er, if ℓ is the least in teger such that ε ℓ = ε ℓ +1 = ± g / 2, then ℓ ≥ i + k . It follows that the pro cess of replacing c on- secutive digits of g / 2 or − g / 2 must termina te, and we obtain a minimum length 8 MEL VYN B. NA THANSON g -adic r epresen tation of n . Moreov er, if w e initiate the standa rdizing and shor ten- ing algorithm with any g -a dic r epresen tation of n of length ℓ g ( n ), then we obtain a minimum leng th g -adic re pr esen tation with exactly the same length. W e shall prove that the minim um length g -adic repr esen tation is unique. Let n = P ∞ i =0 ε i g i be a minimum length g -adic r epresen tation, and let r = max { i ∈ N 0 : ε i 6 = 0 } . W e ca ll ε r g r the le ading term of the representation. If ε i ∈ { 0 , ± 1 , ± 2 , . . . , ± ( g / 2) } for i = 0 , 1 , . . . , r − 1, then      r − 1 X i =0 ε i g i      ≤ g ( g r − 1 ) 2( g − 1) < g r . It follows that n is po sitiv e if the leading term of n is p ositiv e, and n is nega tiv e if the leading term of n is negative. Th us, 0 = P ∞ i =0 0 · g i is the unique minimum length r epresen tation of 0. W e observe that if n = P ∞ i =0 ε i g i is a minim um length g -adic repre s en tation of n with leading term ε r g r , then − n = P ∞ i =0 ( − ε i ) g i is a minimum length g -adic representation of − n with leading term ( − ε r ) g r . Therefore, it suffices to pr o v e the uniqueness of the minimum length g -adic repre sen tation for p ositive integers. Let n ≥ 1 have lea ding term ε r g r . If 1 ≤ ε r ≤ ( g / 2) − 1, then conditio n (iii) gives the upp er b ound n = ε r g r + r − 1 X i =0 ε i g i ≤ ε r g r + g 2 [( r − 1) / 2] X i =0 g r − 2 i − 1 +  g 2 − 1  [ r / 2] X i =1 g r − 2 i = ε r g r + g 2 r − 1 X i =0 g i − [ r / 2] X i =1 g r − 2 i . If ε r = g / 2, then co ndit ion (iii) gives the upper bound n =  g 2  g r + r − 1 X i =0 ε i g i ≤  g 2  g r +  g 2 − 1  [( r − 1) / 2] X i =0 g r − 2 i − 1 + g 2 [ r / 2] X i =1 g r − 2 i = g 2 r X i =0 g i − [( r − 1) / 2] X i =1 g r − 2 i − 1 . NETS IN GR OUPS AND g -ADIC REPRESENT A TIONS 9 Condition (iii) also gives a low er b ound for n . Since ε r ≥ 1, we ha ve ε r − 1 6 = − g / 2, and s o n = ε r g r + r − 1 X i =0 ε i g i ≥ ε r g r −  g 2 − 1  [( r − 1) / 2] X i =0 g r − 2 i − 1 − g 2 [ r / 2] X i =1 g r − 2 i = ε r g r − g 2 r − 1 X i =0 g i + [( r − 1) / 2] X i =0 g r − 2 i − 1 . Therefore, if 1 ≤ ε r ≤ ( g / 2) − 1 and if n ′ and n a re pos itive integers whose minim um length g -adic representations hav e leading terms ( ε r + 1 ) g r and ε r g r , resp ectiv ely , then n ′ − n ≥   ( ε r + 1 ) g r − g 2 r − 1 X i =0 g i + [( r − 1) / 2] X i =0 g r − 2 i − 1   −   ε r g r + g 2 r − 1 X i =0 g i − [ r / 2] X i =1 g r − 2 i   = g r − g r − 1 X i =0 g i + r − 1 X i =0 g i =1 . If n ′ and n are p ositive integers whose minim um length g -adic repr esen tations have leading terms g r +1 and ( g / 2) g r , resp ectiv ely , then n ′ − n ≥   g r +1 − g 2 r X i =0 g i + [ r / 2] X i =0 g r − 2 i   −   g 2 r X i =0 g i − [( r − 1) / 2] X i =0 g r − 2 i − 1   = g r +1 − g r X i =0 g i + r X i =0 g i =1 . Therefore, if n = P r i =0 ε i g i and n = P r ′ i =0 ε ′ i g i are t w o minimum length g -adic representations of the p ositiv e in teger n with leading terms ε r g r and ε ′ r ′ g r ′ , resp ec- tively , then these representations hav e the same leading terms, that is , r = r ′ and ε r = ε ′ r ′ Since n − ε r g r = r − 1 X i =0 ε i g i and n − ε r g r = r − 1 X i =0 ε ′ i g i are als o minim um length g -adic r epresen tations, their leading terms m ust b e equa l. Contin uing inductively , we see that every in teger has a t most o ne minimum length 10 MEL VYN B. NA THANSON g -adic representation, and so every in teger ha s exactly o ne minimum leng th g -adic representation. This completes the pr o of.  Theorem 5. Every inte g er n has a unique r epr esentation in the form n = ∞ X i =0 ε i 2 i such that (i) ε i ∈ { 0 , ± 1 } for al l nonne ga tive inte gers i , (ii) ε i 6 = 0 for only finitely many nonne g ative inte gers i , (iii) if ε i = ± 1 , then ε i +1 = 0 . F or every inte ger n , ℓ 2 ( n ) = ∞ X i =0 | ε i | in the metric sp ac e ( Z , d 2 ) asso ciate d with the gener ating set A 2 = { 0 } ∪ { ± 2 i : i = 0 , 1 , 2 , . . . } . Pr o of. This is the case g = 2 of Theo rem 4.  Theorem 6. L et g b e an even p ositive inte ger. Consider the metric sp ac e ( Z , d g ) asso ciate d with the gener ating s et A g = { 0 } ∪ {± g i : i = 0 , 1 , 2 , . . . } . F or every nonne gative inte ger h , t h e set C = ∞ [ q =0 S e ((2 h + 1) q ) is an h -net in the metric sp ac e ( Z , d g ) . Pr o of. By Theorem 3, it suffices to prove that for every integer n there exists g k ∈ A g such that ℓ g ( n + g k ) = ℓ g ( n ) + 1 or ℓ g ( n − g k ) = ℓ g ( n ) + 1. Le t ε r g r be the leading term in the minimum length g -a dic repres en tation of n . Let k ≥ r + 2. If n ≥ 0, then the minimum length g -adic r e presen tation of n + g k satisfies ℓ g ( n + g k ) = ℓ g ( n ) + 1. Similarly , if n < 0, then the minimum length g -adic r epresen tation of n − g k satisfies ℓ g ( n − g k ) = ℓ g ( n ) + 1. This completes the pro of.  Theorem 7 . L et g b e an o dd inte ger, g ≥ 3 . Every nonzer o inte ger n has a unique r ep r esentation in the form n = ∞ X i =0 ε i g i wher e (i) ε i ∈ { 0 , ± 1 , ± 2 , . . . , ± ( g − 1) / 2 } for al l n o nne gative inte gers i , (ii) ε i 6 = 0 for only finitely many nonne g ative inte gers i . Mor e over, n has lengt h ℓ g ( n ) = ∞ X i =0 | ε i | in the metric sp ac e ( Z , d g ) asso ciate d with the gener ating set A g = { 0 } ∪ {± g i : i = 0 , 1 , 2 , . . . } . A representation of n that satisfies conditions (i) and (ii) will b e called the minimum length g -adic r epr esent atio n of n . NETS IN GR OUPS AND g -ADIC REPRESENT A TIONS 11 Pr o of. Let n = P ∞ i =0 ε i g i be a representation that satisfies conditions (i) and (ii). Since − n = P ∞ i =0 ( − ε i ) g i is a lso a representation of − n that satisfies conditions (i) and (ii), we conclude that it suffices to prov e that every nonnegative integer has a unique minimal leng th g -adic repr esen tation. If ε i 6 = 0 for some i and r = max { i : ε i 6 = 0 } , then n = ε r g r + n ′ where | n ′ | =      r − 1 X i =0 ε i g i      ≤  g − 1 2  r − 1 X i =0 g i = g r − 1 2 . Therefore, (2)  ε r − 1 2  g r + 1 2 ≤ n ≤  ε r + 1 2  g r − 1 2 . It follows that ε r ≥ 1 if n ≥ 1 and ε r ≤ − 1 if n ≤ − 1. In par ticular, 0 = P ∞ i =0 0 · g i is the unique minimum leng th g -adic repr esen tation of 0. If n ≥ 1, then ε r ∈ { 1 , 2 , . . . , ( g − 1) / 2 } and inequa lit y (2) implies that (3) g r + 1 2 ≤ n ≤ g r +1 − 1 2 . Suppo se that n = ∞ X j =0 ε ′ j g j is ano ther representation of n that s atisfies conditions (i) and (ii), with r ′ = max { i : ε ′ i 6 = 0 } . Inequalities (2) and (3) imply that r = r ′ and ε r = ε ′ r ′ . It follo ws inductively tha t ε i = ε ′ i for all nonneg a tiv e integers i . Thu s, a minimal length g -adic repre s en tation is unique. Next we prove that every positive in teger has a minimal length g -adic repr e- sentation. F or every ε ∈ { 1 , 2 , . . . , ( g − 1) / 2 } , the num ber of integers n that can be represented in the form n = P ∞ i =0 ε i g i with r = max { i : ε i 6 = 0 } , ε r = ε , and ε i ∈ { 0 , ± 1 , ± 2 , . . . , ± ( g − 1) / 2 } for i = 0 , 1 , . . . , r − 1 is exac tly g r . Each of these in tegers sa tisfies inequality (2). Since the num ber of integers that satisfy this inequality is exactly g r , it follows fr om the pigeonhole principle and fro m the uniqueness of a minimal length g -a dic r epresen tation that every integer satisfying inequality (2) has a minimal length g -adic representation. Ther efore, e very integer satisfying inequality (3) must hav e a minimal length g -a dic repres en tation for every r ≥ 0, a nd so every integer has such a representation. Finally , we pr o v e that the minimal length g -adic representation of n ha s length ℓ g ( n ). Given any repres en tation o f an int eger n as a sum of elements o f the g ener- ating s et A g , we can obtain another representation with an equal or smaller n um ber of summands as follows: (a) Delete all o ccurrences o f 0. (b) If g i and − g i bo th o ccur, delete them. (c) If g i (resp. − g i ) o ccurs g times, r eplace these g summands with the one summand g i +1 (resp. − g i +1 ). 12 MEL VYN B. NA THANSON (d) After applying the first three o perations as o ften as p o ssible, we obtain n = P ∞ i =0 ε i g i with ε i ∈ { 0 , ± 1 , . . . , ± ( g − 1) } for all i . If ( g + 1) / 2 ≤ ε i ≤ g − 1 for some i , then we choose the smallest such i and apply the iden tit y ε i g i = − ( g − ε i ) g i + g i +1 to r eplace these ε i summands with g − ε i + 1 ≤ ε i summands. Similarly , if − ( g − 1) ≤ ε i ≤ − ( g + 1 ) / 2 for s o me i , then we apply the identit y ε i g i = ( g + ε i ) g i − g i +1 to replace | ε i | summands with g + ε i + 1 ≤ | ε i | summands. Iterating this pro cess, we obtain a minimum length g -adic representation of n . If we apply this algo rithm to a representation of n of length ℓ g ( n ), then w e obtain a minim um length g -adic representation of n of length at mos t ℓ g ( n ), hence o f length exactly ℓ g ( n ). This completes the pro of.  Theorem 8. Every inte g er n has a unique r epr esentation in the form n = ∞ X i =0 ε i 3 i such that (i) ε i ∈ { 0 , ± 1 } for al l nonne ga tive inte gers i , (ii) ε i 6 = 0 for only finitely many nonne g ative inte gers i . F or every inte ger n , ℓ 3 ( n ) = ∞ X i =0 | ε i | in the metric sp ac e ( Z , d 3 ) asso ciate d with gener a ting set A 3 = { 0 } ∪ {± 3 i : i = 0 , 1 , 2 , . . . } . Pr o of. This is Theor em 7 in the ca s e g = 3.  Let ( Z , d 2 ) and ( Z , d 3 ) b e the metric spaces o n the additiv e group of in tegers asso ciated with the generating sets A 2 = { 0 } ∪ {± 2 i : i = 0 , 1 , 2 , . . . } and A 3 = { 0 } ∪ {± 3 i : i = 0 , 1 , 2 , . . . } , res p ectively . Ther e is a canonical leng th-preserving function from ( Z , d 2 ) onto ( Z , d 3 ) constructed as follows. Every integer n has leng th ℓ 2 ( n ) = h fo r some h ≥ 0, and s o n ∈ S (2) e ( h ). By Theorem 5 , e v ery n ∈ S (2) e ( h ) has a unique r epresen tation in the form n = h − 1 X i =0 ε k i 2 k i where k 0 , k 1 , . . . , k h − 1 is a sequence of nonnegative integers such that k i − k i − 1 ≥ 2 for i = 1 , 2 , . . . , h − 1 and ε k i = ± 1 for i = 0 , 1 , 2 , . . . , h − 1. F or i = 0 , 1 , 2 , . . . , h − 1, we define ˜ k i = k i − i Then ˜ k 0 = k 0 and ˜ k i = k i − i ≥ k i − 1 + 2 − i = ˜ k i − 1 + 1 for i = 1 , 2 , . . . , h − 1. Therefor e, ˜ k 0 , ˜ k 1 , . . . , ˜ k h − 1 is a str ic tly increa sing sequence of nonneg ativ e int egers. NETS IN GR OUPS AND g -ADIC REPRESENT A TIONS 13 Define f : Z → Z b y f h − 1 X i =0 ε k i 2 k i ! = h − 1 X i =0 ε k i 3 ˜ k i = h − 1 X i =0 ε k i 3 k i − i . Theorems 5 and 8 imply that the function f : Z → Z is one-to- one and on to, and that f is length-preserving , that is , ℓ 2 ( n ) = ℓ 3 ( f ( n )) for all integers n . In pa r ticular, the function f maps the sphere S (2) e ( h ) onto the sphere S (3) e ( h ) for a ll h ≥ 0. F or a n y p ositiv e integer r , define the integers m = r X i =0 2 3 i and n = r − 1 X i =0 2 3( i +1) . Then m − n = 1 and so d 2 ( m, n ) = ℓ 2 ( m − n ) = ℓ 2 (1) = 1 . How ev er, f ( m ) = r X i =0 3 3 i − i = 1 + r X i =1 3 2 i and f ( n ) = r − 1 X i =0 3 3( i +1) − i = r − 1 X i =0 3 2 i +3 . Therefore, f ( m ) − f ( n ) = 1 + r X i =1 3 2 i − r − 1 X i =0 3 2 i +3 = 1 + 2 r +1 X i =2 ( − 1) i 3 i and s o d 3 ( f ( m ) , f ( n )) = ℓ 3 ( f ( m ) − f ( n )) = 2 r + 1 . It follows that d 3 ( f ( m ) , f ( n ) d 2 ( m, n ) = 2 r + 1 and s o lim sup  d 3 ( f ( m ) , f ( n ) d 2 ( m, n )) : m, n ∈ Z and m 6 = n  = ∞ Therefore, the function f is not a bi-Libschitz equiv alence. Problem 6 . Richar d E. Schwartz [3] aske d the fol lowing b e autiful question: Ar e the met ri c sp ac es ( Z , d 2 ) and ( Z , d 3 ) qu asi-isometric? It is n o t even known if t he y ar e bi-Lipschitz e quivalent. This is one r e aso n why it is imp ortant to classify the nets in the metric sp ac es ( Z , d g ) . Problem 7. J oh n H . Conway [1] suggeste d c ombining the gener ating sets A 2 and A 3 . Consider the additive gr oup Z of inte gers with gener ating set A 2 , 3 = { 0 } ∪ {± 2 i : i = 0 , 1 , 2 , . . . } ∪ { ± 3 i : i = 0 , 1 , 2 , . . . } . L et ℓ 2 , 3 and d 2 , 3 denote, r esp e ctively, the c orr esp onding wor d length function and metric induc e d on Z . Conway aske d: Is the diameter of t hi s metric sp ac e infinite? 14 MEL VYN B. NA THANSON If the diameter of the metric space ( Z , A 2 , 3 ) is infinite, then a theorem o f Nathanson [2 , T he o rem 1 ] implies that for every p ositiv e in teger h ther e ar e in- finitely many integers of length exac tly h . Equiv alently , the spher e S e ( h ) is infinite. F or every positive integer h , let λ 2 , 3 ( h ) denote the smallest p ositiv e integer of length h , that is , the smallest positive in teg er that can b e represented a s the sum or differ- ence of exactly h pow ers of 2 a nd p o wers of 3, but that ca nno t b e repres en ted as the sum or difference of fewer than h p o wers of 2 and p ow ers of 3 . W e hav e λ 2 , 3 (1) = 1, λ 2 , 3 (2) = 5, and λ 2 , 3 (3) = 21. A shor t calc ulation s hows that λ 2 , 3 (4) ≥ 150, but the exact v alue of λ 2 , 3 (4) has not yet been deter mined. Problem 8. Find al l solut io n s in p ositive inte gers of the exp onential diophantine e quations 2 a − 3 b = 14 9 and 2 c − 3 d = 15 1 . These e qu a tions have no solutions if and only if λ 2 , 3 (4) = 15 0 . Problem 9. L et P b e a finite or infinite set of prime nu mb ers and c onsider the additive gr oup Z of inte gers with gener ating set A P = { 0 } ∪ { ± p i : p ∈ P and i = 0 , 1 , 2 , . . . } . L et ℓ P and d P denote, r esp e ctively, the c orr esp onding wor d lengt h fun ctio n and metric induc e d on Z . F or every p ositive inte ger h , let λ P ( h ) denote the smal lest p ositive inte ger of length h , t h at is, the sm a l lest p ositive inte ger that c an b e r ep- r esente d as t h e su m or differ enc e of ex actly h elements of A P , but t hat c annot b e r epr esente d as the sum or differ enc e of fewer than h elements of A P . Compute the function λ P ( h ) . Problem 10. L et P b e a finite or infinite set of prime numb ers, and let S P b e the semigr oup of p ositive int e gers gener ate d by P . Consider the additive gr oup Z of inte gers with gener ating set A S ( P ) = { 0 } ∪ {± s : s ∈ S ( P ) } . L et ℓ S ( P ) and d S ( P ) denote, r esp e ctively, the c orr esp onding wor d length function and metric induc e d on Z . F or every p ositive inte ger h , let λ S ( P ) ( h ) denote the smal lest p ositive inte ger of length h , t h at is, t h e smal lest p ositive inte ger that c an b e r epr esente d as the su m or differ enc e of exactly h elements of the set S ( P ) , but that c annot b e r epr esen t e d as the s u m or differ enc e of fewer than h elements of the S ( P ) . Compute the funct i on λ S ( P ) ( h ) . 4. Additive complements In this section we co nsider a natural additive n umber theor etic genera lization of the metric concept of h -nets in g roups. Let W be a nonempty subs e t of a g roup o r semigroup G . The set C in G will b e called a c omplement to W if G = W C . If A is a symmetric gener ating set fo r a g roup G with e ∈ A , then an h -net in the metric space ( G, d A ) is a complement to the pro duct set A h . Let C ( W ) denote the set of all c o mplemen ts to W . Then (i) C ( W ) 6 = ∅ s ince G ∈ C ( W ), (ii) If C ∈ C ( W ) and C ⊆ C ′ , then C ′ ∈ C ( W ), (iii) If C ∈ C ( W ) and x ∈ G , then C x ∈ C ( W ). A complement C to W is minimal if no pro per subs e t o f C is a complemen t to W . If C is a minimal complement, then the rig h t tr anslation C x is also a minimal complement for a ll x ∈ G . NETS IN GR OUPS AND g -ADIC REPRESENT A TIONS 15 Suppo se that W is a subset of a group and that C is a complement to W that do es not contain a minimal co mplemen t to W . If D is an y subset of C suc h that C \ D is a complement to W , then there exists c ∈ C \ D such that C \ ( D ∪ { c } ) is a complement to W . If G is an additive g r oup and W is a subset of G , then the subset C of G is a complement to W if W + C = G . Theorem 9. L et W b e a nonempty, fin i te set of inte gers. In the additiv e gr oup Z , every c omplement to W c ontains a minimal c omplement to W . Pr o of. Let C b e a complemen t to W . Then C is infinite since W is finite. Let w ′ = min( W ) a nd w ′′ = max( W ). F or every integer n , there exis ts w ∈ W and c ∈ C such that n = w + c . It follows that n − w ′′ ≤ c = n − w ≤ n − w ′ . W rite C = { c i } ∞ i =0 . W e construct a decrea sing sequence o f sets { C i } ∞ i =0 as follows: Let C 0 = C . F or i ≥ 0, define C i +1 = ( C i \ { c i } if C i \ { c i } is a co mplemen t to W C i otherwise. Then { C i } ∞ i =0 is a sequenc e of co mplemen ts to W and C i +1 ⊆ C i for all i ≥ 0. Let C ∗ = ∞ \ i =0 C i . F or every in tege r n and nonnega tiv e integer i , there ex is t in tegers w i,n ∈ W and c i,n ∈ C i such that n = w i,n + c i,n , and n − w ′′ ≤ c i,n ≤ n − w ′ . The pigeo nhole principle implies that there is an in tege r c such that n − w ′′ ≤ c ≤ n − w ′ and c = c i,n for infinitely many i . If c = c i,n , then n − c = n − c i,n = w i,n ∈ W . Therefore, c ∈ C i for all i ≥ 0, that is, c ∈ C ∗ , and n − c ∈ W , hence n ∈ W + C . Therefore, C ∗ is a complement to W . Suppo se that there exists an integer c j ∈ C ∗ such that C ∗ \ { c j } is a complement to W . Since C ∗ ⊆ C j , it w ould follow tha t C j \ { c j } is also a complement to W . In this case , how ever, at s tep j in our inductiv e construction, w e would hav e defined C j +1 = C j \ { c j } , and so c j / ∈ C ∗ , w hich is absurd. Therefore, the remov al of any element fro m C ∗ results in a set that is no longer a co mplemen t to W , and so C ∗ is minimal. This completes the pro of.  Problem 11. L et W b e an infinite set of int e gers. Do es t h er e exist a minimal c omplement to W ? Do es ther e exist a c omplement to W that do es not c ontain a minimal c omplement? Problem 12. L et G b e an infinite gr oup, and let W b e a fi n ite subset of G . Do es ther e exist a minimal c omplement to W ? Do es ther e exist a c omplement to W t ha t do es not c ont a in a minimal c omplement? Problem 1 3. L et G b e an infinite gr oup, and let W b e an infin ite subset of G . Do es t he r e exist a minimal c omplement to W ? Do es ther e exist a c omplement to W that do es not c ont a in a minimal c omplement? 16 MEL VYN B. NA THANSON 5. Asymptotic complements Let W be a nonempt y subset of a group or semigroup G . The set C in G will b e called an asymptotic c omplement t o W if all but at mo st finitely many elements of G b elong to the pr oduct se t W C , tha t is, | G \ W C | < ∞ . Let AC ( W ) denote the set o f all asymptotic complements to W . Then (i) AC ( W ) 6 = ∅ since G ∈ AC ( W ), (ii) If C ∈ AC ( W ) and C ⊆ C ′ , then C ′ ∈ AC ( W ), (iii) If C ∈ AC ( W ) and x ∈ G , then C x ∈ AC ( W ). An a symptotic complement C to W is minimal if no prop er subset of C is an asymptotic complement to W . If C is a minimal a s ymptotic complement, then C x is a lso a minimal asymptotic complement for a ll x ∈ G . Problem 1 4. L et W b e a finite or infinite set of inte gers. Do es ther e exist a minimal asymptotic c omplement to W ? Do es ther e exist a c omplement to W that do es not c ont a in a minimal c omplement? Problem 15. Consider the additive semigr oup N 0 of nonne gative inte gers. L et W b e a fi n ite or infinite subset of N 0 . Do es ther e exist a minimal asymptotic c omplement to W ? Do es ther e exist an asymptotic c omplement to W that do es not c ontain a minimal asymptotic c omplement? Problem 16 . L et G b e an infinite gr oup, and let W b e a finite or infinite s ubset of G . Do es ther e exist a minimal asymptotic c omplement to W ? Do es t he r e ex- ist an asymptotic c omplement t o W that do es not c ontain a minimal asymp totic c omplement? References [1] J. H. Conw ay , p ersonal comm unication, 2008. [2] M. B. Nathanson, Phase tr ansitions in infinitely gener ate d gr oups, and r elate d pr oblems in additive numb er the ory , arXiv: 0811.3990. [3] R. E. Sch wartz, pers on al communication, 2008. CUNY (Lehman College a nd the Graduat e Center) E-mail addr ess : melvyn.n athanson@ lehman.cuny.edu Curr ent addr ess : Pri nc eton University E-mail addr ess : melvyn@p rinceton. edu