OPTIMAL PAIRS OF INCOMPARABLE CLOUDS ... - Semantic Scholar

OPTIMAL PAIRS OF INCOMPARABLE CLOUDS IN MULTISETS Rudolf Ahlswede and Levon H. Khachatrian

Universitat Bielefeld Fakultat fur Mathematik Postfach 100131 33501 Bielefeld Germany

Institute of Problems of Information and Automation Armenian Academy of Sciences Erevan{44 P. Sevak str. 1 Armenian Visiting SFB 343

1

1. Introduction, basic results and problems

Let us be given the partially ordered set Pn ([k]n; ) , where [k] = f0; 1; 2; : : :; k ? 1g and a = (a1; a2; : : :; an )  b = (b1; b2; : : :; bn ) i at  bt for t = 1; 2; : : :; n . In the terminology of our earlier work ([1], [8], [9]) we call a pair (A; B ) with A; B  [k]n a cloud{antichain of length 2, if

a  b; a  b for all a 2 A; b 2 B:

(1.1)

A short expression for (1.1) is: A j B . We denote the set of these pairs by CAC (n) . The objects of our investigation are the functions fn : f0; 1; : : :; kn g ! f0; 1; : : :; kn g , n 2 N , de ned by 

fn () = max jB j : 9(A; B ) 2 CAC (n) with jAj = 



(1.2)

and a characterization of pairs (A; B ) which are optimal, that is, assume this bound. We denote by O(n) the set of all those optimal pairs. In cases where we emphasize the dependence on parameter k we also write CACk (n) , Ok (n) , fn;k () , etc. instead of CAC (n) , O(n) , fn () , etc. Previous work is discussed in [9], where the best results prior to those in this paper can be found. They are all for the binary alphabet, i.e. k = 2 . Familiarity with this paper may be helpful but is not necessary for an understanding of the present results and proofs. We extend here rst the key result of that paper to the case of general k .

Theorem 1. For every ; 0   kn?2 , a pair (A; B ) 2 O(n) with jAj = exists,

such that for some component all a in A have a 0 and for some other component all a in A have a k ? 1 .

From here we derive by an approach similar to (but not identical with) that of [9] the main recursion.

Theorem 2. For every ; 0   kn?2 , (i) fn ( ) = (k ? 1)kn?1 + kfn?2 ( ) ? (k ? 1) : (ii) fn+2s ( ) = (ks ? 1)(kn+s ? ) + ksfn ( ) for s  0: ?




(1.3)

The explicit characterization of all pairs ; fn( ) given in [9] does not seem to allow a reasonably simple extension to general k . Therefore results concerning aspects of this characterization problem are already of interest. Theorem 8 of [1] states that in the case k = 2 for (A; B ) 2 CAC (n) (a) jAjjB j  22n?4 (b) minfjAj; jB jg  2n?2 and that these bounds are best possible. 2

The key observation was that for (A; B ) 2 CAC (n) we have the disjointness properties (A ^ B ) \ (A _ B ) = ?; (A ^ B ) \ (A [ B ) = ? , (A _ B ) \ (A [ B ) = ? , and A\B =? . Therefore jAj + jB j + jA _ B j + jA ^ B j  2n and since by the arithmetic{geometric means inequality
jAj jB j jA _ B j jA ^ B j 4  jAj + jB j + jA4_ B j + jA ^ B j

1

?

we get



n jAj jB j jA _ B j jA ^ B j  24

4

:

(1.4)

jAj jB j  jA _ B j jA ^ B j (see [6])

(1.5)

Now we use the AD{inequality and get

jAj jB j  22n?4 : (b) is an immediate consequence. Inspection shows that the same derivation is valid for all k and thus for (A; B ) 2 CACk (n)  4

jAj jB j  k2 k2n?4 :

(1.6)

This is tight only for even k . In [9] the arithmetic{geometric means inequality was applied to two terms and so do we now for general k . Hence,

kn  jAj + jB j + jA ^ B j + jA _ B j p  jAj + jB j + 2 jA ^ B j jA _ B j 2 p p p  jAj + jB j + 2 jAj jB j = jAj + jB j

(1.7)

jAj 21 + jB j 21  k n2 :

(1.8)

and therefore 3

Theorem 3 of [9] states two consequences in case k = 2 : For all `; 0  `  2n (c) f2n (`2) = (2n ? `)2 (d) f2n+1 (2`2) = 2(2n ? `)2 .

Problem I: For every k describe all (A; B ) 2 CACk (n) with equality in (1.8). In the terminology of [10] this is an equality characterization problem.

Problem II: How does fn;k () behave asymptotically in k; n , and  ?

Next we try to generalize statements of type (a) and (b) to general k .

Problem III: Determine G(n) = (A;B)max min(jAj; jB j): 2CACk (n)

Problem IV: Determine Q(n) = (A;B)max jAj jB j: 2CACk (n) Finally we solve 

Problem V: Let
be any integer, ?kn 
 kn , nd a
(n) = max jAj : (A; B ) 2 CACk (n) , jB j = jAj +
. We completely solve problems III and IV and provide partial results for the other problems.

4

2. Auxiliary Results

We use for the proofs of our Theorems 1, 2 results of Daykin, Kleitman and West [5]. They are described in the abstract of [5]. Except for a reference to these Theorems in brackets, we literally repeat the main part of the abstract: \Let L be a lattice of divisors of an integer (isomorphically, a direct product of chains). We prove jAjjB j  jLjjA ^ B j for any A; B  L where j
j denotes cardinality and A ^ B = fa ^ b : a 2 A; b 2 B g . jA ^ B j attains its minimum for xed jAj; jB j when A and B are ideals (Theorem 2). j
j can be replaced by certain other weight functions (Theorem 3). When the n chains are of equal size k , the elements may be viewed as n{digit k{ary numbers. Then for xed jAj; jB j; jA ^ B j is minimized when A and B are jAj and jB j smallest n{digit k{ary numbers written backwards and forwards, respectively (Theorem 4).1 jA ^ B j for these sets is determined and bounded (Theorem 5)." We don't need Theorem 3. Whereas Theorems 2, 4 are selfexplanatory, we have to give the details of Theorem 5.

Theorem 5. [5] Suppose L is a product of n chains of size k , 0    kn , 0   kn . Let k (n; ; ) = min jA ^ B j : jAj =  , jB j = and "k (n; ; ) = n?1 <   (p + 1)kn?1 and  r mod k , then k (n; ; ) ?  kn . If pk (i) (ii)





k (n; ; ) = k n ? 1;  ? pkn?1 ; ?k p 

"k (n; ; ) = "k n ? 1;  ? pkn?1 ;



?p k





(

+

0

Pp?1 l ?j m

j =0

?

k

; p=0 ; p>0




r 1 ? kn ; 0  r  p + (k ? r) kn ; p < r < k: 

Furthermore,

(iii)

"k (n; kn ? ; kn ? ) = "k (n; ; )

(iv)

k (n; kn ? ; kn ? ) = k (n; ; ) + kn ?  ?

and, nally,

(v) (vi)

0  "k (n; ; )  k4n :

(Immediate from de nitions and added for the ease of reference)

k (n; 0; ) = 0; k (n; kn; ) = ; "k (n; 0; ) = "k (n; kn; ) = 0: any c = (c1 ; : : : ; cn ) 2 [k]n we de ne a forward and a backward value by !v (c) = ni=1 kn?i ci and v (c) = ni=1 ki?1 ci , resp.. !v and v evaluate elements as k{ary numbers when the components are written in natural order or in reversed order. 5

1 For

3. A connection between optimal cloud{antichains and pairs extremal in the meet

In the Introduction we have de ned the set O(n) of optimal cloud{antichains of length 2 in [k]n . Now we de ne the set E (n) of pairs (C; D) with C; D  [k]n which are extremal in the meet C ^ D , that is, 



jC ^ Dj = min jC 0 ^ D0 j : C 0 ; D0  [k]n ; jC 0j = jC j; jD0j = jDj : (3.1) Within O(n) there is the set M(n) of those pairs (A; B ) which are maximal in the sense that for all (C; D) 2 CAC (n) with jC j > jAj necessarily jDj < jB j . Finally, we call (A; B ) 2 CAC (n) A{saturated (resp. B {saturated ), if there is no (A; B 0) 2 CAC (n) with B  B 0 , B = 6 B 0 (resp. (A0; B ) 2 CAC (n) with A  A0 , A= 6 A0 ) and we call (A; B ) saturated (or bisaturated), if it is A{saturated and B {saturated . We denote the set of all these saturated pairs by S (n) . Actually, fewer de nitions would suce for the proofs of our Theorems. However, clarity is gained by analysing in which generality properties used hold. We establish therefore rst simple properties of the classes de ned. It is immediately clear that

M(n) = O(n) \ S (n):

(3.2)

Our rst result is of general nature.

Lemma 1.

?




(i) For A; B  [k]n we have A r (A ^ B ); B r (A ^ B ) 2 CAC (n) . (ii) For (A; B ) 2 CAC (n) we have A \ (A ^ B ) = B \ (A ^ B ) = A \ (A _ B ) = B \ (A _ B ) = ? .

Proof: (i) Suppose that for a 2 A r (A ^ B ) and b 2 B r (A ^ B ) we have a  b , then b = a ^ b 2 A ^ B , contradicting that b 2= A ^ B (the case a  b is symmetrically the same). (ii) If for instance a 2 A \ (A _ B ) , then a b 2 B exists with b  a . This contradicts the assumption (A; B ) 2 CAC (n) . For any C  [k]n we de ne 



'(C ) = c 2 [k]n : c 2= C; 9c0 2 C with c < c0 :

(3.3)

Clearly, C [ '(C ) is an ideal (\downset"). With any pair (A; B ) , A; B  [k]n , we associate a pair ?




(A; B ) = A [ '(A); B [ '(B ) : 6

(3.4)

Lemma 2. For (A; B ) 2 S (n) we have A ^ B  = A \ B  = '(A) = '(B ):

(3.5)

Proof: If for instance a 2 '(A) and a 2= '(B ) , then either a  b for some b 2 B and for an a0 2 A with a0  a the inequality a0  b contradicts (A; B ) 2 CAC (n) , or a is incomparable with all members of B , which contradicts the saturation of (A; B ) . In any case we have '(A) = '(B ) and A ^ B  = A \ B  , because A and B  are ideals. Finally, since A \ B = A \ '(A) = B \ '(B ) = ? , we have

A \ B  = '(A) = '(B ): ?




?




Actually we need a related result. For any ; fn () there is a pair (A; B ) 2 O(n) , ?




jAj; jB j = ; fn() ;

with minimal total weight all such pairs by O (n) .

X

a2A

P W (a) , where W (a) = nt=1 at . We denote the set of

Lemma 3. For (A; B ) 2 O (n) (3.5) holds. Proof: Since (A; B ) is A{saturated the previous argument gives '(B )  '(A) . For a 2 '(A) , a 2= '(B ) the case ? in which
a is incomparable with all members of B is to be discussed. Here A0 = A [ fag r fa0g , a0  a , has smaller total weight than A and since (A0; B ) 2 CAC (n) we get a contradiction. The remaining identities in (3.5) are established as previously.

Proposition 1. For (A; B ) 2 O (n) we have (A; B ) 2 E (n) . Proof: Set  = jA j and  = jB j . Now clearly

jA ^ B  j  k (n; ; ) (de ned in Section 2) and suppose that here strict inequality holds. By Theorem 2 [5] there are ideals A1 ; B1 with cardinalities jA1j =  , jB1j =  and jA1 ^ B1j = k (n; ; ) . Therefore

jA1 r (A1 ^ B1)j =  ? (n; ;  ) >  ? jA ^ B  j = jAj ? j'(A)j by Lemma 3: By de nition of A jAj ? j'(A)j = jAj and thus jA1 r (A1 ^ B1)j > jAj . Symmetrically jB1 r (A1 ^ B1 )j > jB j . ?




Since A1 r (A1 ^ B1); B1 r (A1 ^ B1 ) 2 CAC (n) (by (i) in Lemma 1) we obtained a contradiction to the optimality of (A; B ) . 7

4. A formula for fn via

E (n)

Let us de ne for 0   kn 



M = (; ) : 0  ;  kn ;  ? k (n; ; ) = :

(4.1)

Here all numbers are non{negative integers. Note that M 6= ? , because ( ; 0) 2 M .

Lemma 4. For 0   kn ?

fn ( ) = (; max ? k (n; ; ) )2M




(4.2)

and the maximum can be achieved for a pair of ideals.

Proof: Let
(?A; B ) ?2 O (n) satisfy jAj = ? , jB j =

fn ( ) . By Proposition ? 
? 
1       jA j; jB j = +k n; jA j; jB j , fn ( )+k n; jA j; jB j and hence jA j; jB j 2

M . ?
?
?  ( n; ; ) . Therefore, fn ( ) = jB j ? k n; jAj; jB j  (; max k )2M

To establish the reverse inequality, suppose the maximum is assumed at (0 ; 0 ) 2 M and let (A0; B 0) satisfy jA0j = 0 , jB 0j = 0 and A0 ^ B 0 = k (n; 0; 0 ) . By Theorem 2 [5] A0 and B 0 can be assumed to be ideals. Then A = A0 r (A0 ^ B 0 ) and B = B 0 r (A0 ^ B 0 ) satisfy (A; B ) 2 CAC (n) (by Lemma 1), jAj = 0 ? k (n; 0; 0) =

, and jB j = 0 ? k (n; 0; 0 )  fn ( ) . ?
Therefore fn ( ) = jB j ? k (n; jAj; jB j) = (; max ?  ( n; ; ) . k )2M

Since A and B  are ideals, the maximum in (4.2) at (0; 0 ) is assumed for a pair of ideals. Actually, with the aid of Theorem 4 [5] a stronger conclusion can be drawn

Lemma 5. If the maximum in (4.2) is assumed at ( ;  ) , then it can be attained by letting A be the  smallest and written backwards, B  be the  smallest n{digit k{ary numbers.

Next we establish a relation between and a pair ( ; ) at which the maximum occurs.

Lemma 6. If the maximum in (4.2) is assumed at ( ;  ) and if  kn?2 , then   kn?1 and   (k ? 1)kn?1 . Proof: This is the rst argument, which not just generalizes that of [9]. However, it starts as previously. Let us consider the pair (k ; (k ? 1)kn?1 ) and derive with (i) in Theorem 5 [5] (for p = k ? 2 ) and k (n; ; ) = k (n; ; ) that (k ; (k ? 1)kn?1) 2 M , because k ? k (n; k ; (k ? 1)kn?1) = k ? k (n ? 1; ; kn?1) ? (k ? 2) = k ? ? (k ? 2) = . 8

Therefore ?
?  ( n; ; )  (k ? 1)kn?1 ? (k ? 1) max k (; )2M

?




= (k ? 1)kn?1 ? (k ? 1) 0 ? k (n; 0; 0) for all (0; 0 ) 2 M . In particular we get at a maximum (;  ) ?
 ? k (n; ;  )  (k ? 1)kn?1 ? (k ? 1)  ? k (n; ;  ) or equivalently  + (k ? 1) ? kk (n; ;  )  (k ? 1)kn?1:

(4.3)

Since for any C; D  [k]n jC jjDj  jC ^ DjjC _ Dj  jC ^ Djkn , we also have   k (n; ;  )  kn : (4.4) Combination of (4.3) and (4.4) yields   (4.5)  + (k ? 1) ? kn? 1  (k ? 1)kn?1: Write now  = (k ? 1)kn?1 +  ,  = kn?1 + " and verify that n?1 n?1    +(k ? 1) ? kn? 1 = (k ? 1)kn?1 +  +(k ? 1)kn?1 +(k ? 1)" ? (k +")((knk??11)k +) = (k ? 1)kn?1 ? kn"?1 . We conclude from (4.5) that "  0 . In view of (4.2)    , but if k ? 1 > 1 this does not imply (as in the case k = 2 ) that   0 and "  0 .

Suppose, for contradiction, that either   0 and " > 0 , or  < 0 and " = 0 . By de nition of (;  ) , fn ( ) =  ? k (n; ;  ) and (;  ) 2 M , so  ? k (n; ;  ) = . Thus fn ( ) ? =  ?  . From the preceding veri cation it follows that  ?  < (k ? 1)kn?1 ? kn?1 = (k ? 2)kn?1 . Thus fn ( ) ? < (k ? 2)kn?1 : (4.6) Next observe that the pair (A; B ) = f0g  fk ? 1g 

n Y t=3

[k]; f1; : : :; k ? 1g  f0; : : :; k ? 2g 

n Y t=3

!

[k]

is a cloud antichain with jAj = kn?2 and jB j = (k ? 1)2 kn?2 , so fn (kn?2 )  (k ? 1)2kn?2 :

(4.7)

It follows from (4.6) and (4.7) that fn (kn?2 ) ? kn?2  (k ? 1)2 kn?2 ? kn?2 = (k ? 2)kn?1 > fn ( ) ? : But fn (
) is a decreasing function and  kn?2 by hypothesis so we have the contradiction fn (kn?2 ) ? kn?2 > fn ( ) ?  fn (kn?2 ) ? kn?2 : 9

5. Proof of Theorem 1

Starting with (A; B ) 2 O (n) , jAj = ; jB j = fn ( ) we know from Proposition 1 that (A; B ) 2 E (n) and that  = jAj = + k (n; ;  ) ,  = jB j = fn ( ) + k (n; ; ) . The same parameter values can be obtained with A; B  being ideals and having the numberings of Lemma 5. Then the sequences in A = A r (A ^ B  ) are all zero in the last component, because jAj =   kn?1 by Lemma 6, and they are all k ? 1 in the rst component, because jB j  (k ? 1)kn?1 again by Lemma 6 and thus all elements of A beginning with an element of f0; 1; : : :; k ? 2g are also in A ^ B  and thus not in A . This means that only elements of A beginning with a k ? 1 can be members of A . 6. Proof of Theorem 2

We make use of the notation 



'(D) = d 2 [k]n : d 2= D; 9 d0 2 D with d < d0 ;  (D) = d 2 [k]n : d 2= D; 9 d0 2 D with d > d0 ;

(6.1) (6.2)

and 



(D) = c 2 [k]n : c 6 d; c 6 d for all d 2 D :

(6.3)

Let now (A; B ) 2 O (n) have an A structured as in Theorem 1: a = (a1 ; a2; : : :; an) 2 A implies a1 = k ? 1; an = 0 . Necessarily B = (A) and it can be described as a union of four disjoint sets: Imagine b 2 [k]n to be written in the form (b1; : : :; bn) and set 



B1 = b 2 [k]n : b1 2 f0; : : :; k ? 2g; bn 2 f1; : : :; k ? 1g; (b2; : : :; bn?1 ) 2 [k]n?2 ;  B2 = b 2 [k]n : b1 2 f0; : : :; k ? 2g; bn = 0; (b2; : : :; bn?1) 2 (A0) [ (A0) ;  where A0 = (a2; : : : ; an?1) 2 [k]n?2 : (k ? 1; a2; : : :; an?1; 0) 2 A ;  B3 = b 2 [k]n : b1 = k ? 1; bn 2 f1; : : :; k ? 1g; (b2; : : : ; bn?1) 2 '(A0 ) [ (A0) ;  B4 = b 2 [k]n : b1 = k ? 1; bn = 0; (b2; : : :; bn?1) 2 (A0) : Then ?
jB j = jB1 [ B2 [ B3 [ B4 j = (k ? 1)2 kn?2 +(k ? 1) j (A0 )j + j'(A0)j +(2k ? 1)j(A0)j , and since j (A0)j + j'(A0)j + j(A0)j = kn?2 ? jA0j = kn?2 ? ; 10

we conclude that fn ( ) = jB j = (k ? 1)2kn?2 + (k ? 1)(kn?2 ? ) + kj(A0)j  (k ? 1)2 kn?2 + (k ? 1)(kn?2 ? ) + k fn?2 ( ) = (k ? 1)kn?1 + k fn?2 ( ) ? (k ? 1) : ?




Conversely, if we choose A0  [k]n?2 , jA0j = such that A0 ; (A0) 2 O (n ? 2) , then j(A0)j = fn?2 ( ) and constructing A and B as before, we obtain jB j = (k ? 1)kn?1 + k fn?2 ( ) ? (k ? 1) : Writing (i) in the form fn+2 ( ) = (k ? 1)(kn+1 ? ) + k fn ( ) (ii) follows by induction on s . 7. A Corollary

Since there exists (A; B ) 2 CAC (n) with jAj = and jB j = fn ( ) , it is immediate that ?
fn fn ( )  : (7.1) Since the function fn is decreasing we have therefore ?
fn fn (fn( ))  fn ( ): (7.2) ?




On the other hand there exists (A0 ; B 0) 2 CAC (n) with jA0j = fn fn ( ) and jB 0j = fn ( ) . Hence ?
fn fn (fn( ))  fn ( ); (7.3) and from (7.2), (7.3) we conclude ?
fn fn (fn( )) = fn ( ) for all  kn : (7.4) For + 1  kn?2 the recursion (i) in Theorem 2 applies and, since fn?2 ( + 1)  fn?2 ( ) , we have fn ( + 1) = (k ? 1)kn?1 + k fn?2 ( + 1) ? (k ? 1)( + 1)  (k ? 1)kn?1 + k fn?2 ( ) ? (k ? 1) ? (k ? 1) = fn ( ) ? (k ? 1) , and hence fn ( + 1) < fn ( ) for all 0  < kn?2 : (7.5) Suppose for some , 0  < kn?2 , strict inequality in (7.1) holds, i.e.
? fn fn ( ) = 1  + 1; then, applying (7.4), we have ?
fn fn (fn( )) = fn ( 1) = fn ( ); which contradicts (7.5). We summarize these ndings

Corollary. (i) (ii) (iii)

?




fn fn ( )  for all  kn ?
fn fn ( ) = for 0  < kn?2 . ?
fn fn (fn ( )) = fn ( ) for all  kn . 11

8. On Problem I: an equality characterization

What can we say about (A; B ) 2 CACk (n) for which

p

p

p

kn = jAj + jB j

(8.1)

holds? First we derive an auxiliary result.

Lemma 7. For an (A; B ) 2 CACk (n) satisfying (8.1) necessarily we have (i) (ii) (iii) (iv)

A and B are convex. (A; B ) 2 O(n) . (A; B ) is bisaturated. '(A) = '(B ) = A ^ B and (symmetrically) (A) = (B ) = A _ B .

Proof: Statements p (i), (ii) and (iii) are trivially true, because otherwise we could p increase jAj + jB j .

Lemma 2 implies

'(A) = '(B ) = A ^ B   A ^ B:

(8.2)

Also we have by the disjointness properties and the inclusion (A)  A _ B

kn  jAj + jB j + j'(A)j + j (A)j  jAj + jB j + jA ^ B j + jA _ B j p p
 ? jAj + jB j 2 (by (1.7)) and (8.1) yields

j'(A)j + j (A)j = jA ^ B j + jA _ B j:

Thus (8.2) gives nally (iv).

Lemma 8. If (8.1) holds, then or

(i)  + = kn and

p

p

p

 ? k (n; ; ) + ? k (n; ; ) = kn

(ii)  + = kn and k (n; ; ) =  kn .

Proof: By Lemma 4 we know that for (A; B ) 2 O(n) we can assume that

jAj =  ? k (n; ; ); jB j = ? k (n; ; ) and there are ideals A ; B  with jAj =  , jB j = , jA ^ B j = k (n; ; ) (0  ;  kn ) . Since (8.1) holds, by (iii) in Lemma 7 (A; B ) 2 S (n) and hence by 12

Lemma 2 k (n; ; ) = jA ^ B j = j'(A)j . Now by (iv) in Lemma 7 we conclude that k (n; ; ) = j'(A)j = jA ^ B j: (8.3) p

Also jA _ B j = k (n; ; ) , because jA ^ B j + jA _ B j = 2 jA ^ B jjA _ B j , i jA ^ B j = jA _ B j . Recall also that under (8.1) kn = jAj + jB j + jA ^ B j + jA _ B j and by (8.3) we have jAj + jA ^ B j = ; jB j + jA _ B j = :

Theorem 3. p p p jAj + jB j = kn holds for (A; B ) 2 CACk (n) if and only if, for some integer

`, 0  `  km (i) n = 2m , jAj = `2 , jB j = (km ? `)2 or

(ii) n = 2m + 1 , jAj = k`2 , jB j = k(km ? `)2 .

Proof: Suciency of (i) or (ii) follows by a construction, which generalizes that of

Section 8 in [9] from k = 2 to general k . In case (i) it goes as follows.

Let Am (s) (resp. Am (s) ) be the s smallest (resp. largest) weight elements of [k]m . Set  A = a 2 [k]2m : a = (a1; a2); a1 2 Am (`); a2 2 Am(`) ,  B = b 2 [k]2m : b = (b1; b2); b1 2 Am (km ? `); b2 2 Am (km ? `) , and verify that (A; B ) 2 CAC (n) , jAj = `2 , and jB j = (km ? `)2 . ?
In case (ii) choose (A0 ; B 0) = A  [k]; B  [k] . To see the necessity we use the following result, which is a special case of our forthcomming Theorem 2 in [10], which corrects Theorem 6 of [5]: i i+1 n?i "k (n; ; ) = k (n; ; ) ?  kn = 0 i (i) k j , k -  ) k j or (ii)  or equals kn or 0 . n

In our case k (n; ; ) = k (n; ; kn ? ) = (kkn?) =  ? kn2 is possible only when kn j2 and we can use the relations ki j; ki+1 -  ) ki j ; = kn ? : However, we have also ki j , ki+1 -  ) kn?i jkn ?  and ki+1 - kn ?  and therefore necessarily i  n ? i . Hence (ii) in Lemma 8 holds i ki j where 1.) i  m , if n = 2m 2.) i  m + 1 , if n = 2m + 1 . 13

Now we have jAj =  ? k (n; ; ) = k22 and

1.) n = 2m ,  = km
` ( ` integral), jAj = `2 , jB j = (km ? `)2 2.) n = 2m + 1 ,  = km+1 ` , jAj = k`2 , jB j = k(km ? `)2 . 9. An asymptotic result

The recursive formula of Theorem 2 in conjunction with Theorem 3 allow to estimate the growth of fn .

Theorem 4.

p p (i) For any 2 N (the natural numbers) such that p + fn ( ) < kn

p

n+2s ? p 2 ? fn+2s ( ) = 1: k lim s!1


?

(ii) For any 2 N nlim !1

p

kn ? p 2 fn ( ) = 1:

?




?p
Proof: (i) We choose n such that  kn and let "n ( ) = kn ? p 2 ? fn ( ) > 0 .

The recursion (ii) in Theorem 2 can be written in the form

fn+2s ( ) = which implies ?

p

?

kn+2s ? p 2 ? ks


p

??

p

kn ? p 2 ? fn ( ) ;


kn+2s ? p 2 ? fn+2s ( ) = ks "n ( )


and ks"n ( ) ! 1(s ! 1) . (ii) For any and n so large that < kn we write

p p ( kn+2s ? p )2 ( kn+2s ? p )2 fn+2s( ) = (ks ? 1)(kn+s ? ) + ksfn ( )

and notice that

p

( kn+2s ? p )2 lim = 1: s!1 (ks ? 1)(kn+s ? ) + ks fn ( )

14




10. A divisibility property of optimal and bisaturated pairs

We present here a result, which we later use and which is interesting in itself.

Theorem 5. Let (A; B ) be optimal and bisaturated, that is, (A; B ) 2 O(n) \S (n) = M(n) . Then a maximum pair ( ;  ) with  ?k (n; ;  ) = jAj ,  ?k (n; ;  ) = jB j has the properties kj and kj . In particular k j (jAj ? jB j):

Proof: Since by de nition of optimality for a jAj = and jB j = fn ( ) we know from Lemma 4 that

?




fn ( ) = (; max ? k (n; ; ) ; )2M

(10.1)

where





M = (; ) : 0  ;  kn ;  ? k (n; ; ) = :

We also know from Lemma 5 that at a maximum assuming pair ( ; ) there is a ?
      realisation (A; B ) = A r (A ^ B ); B r (A ^ B ) , where A are the  smallest and written backwards, B  are the  smallest n{digit k{ary numbers. First we show that integer  ? 1 represented lexicographically backwards does not appear in the list of B  . Namely otherwise we have

k (n;  ? 1; ) = k (n; ;  ) ? 1 ?




and thus  ? 1 ? k (n;  ? 1; ) =  ? 1 ? k (n; ;  ) ? 1 = . Hence ( ? 1;  ) 2 M . But we then have the contradiction ?




 ? k (n;  ? 1;  ) = fn ( ) + 1 > fn ( ) = (; max ? k (n; ; ) : )2M

Also wise

 ? 1 written lexicographically forward does not appear in A , because otherk (n; ;  ? 1) = k (n; ; ) ? 1 and thus  ? 1 ? k (n; ;  ? 1) = fn ( );  ? k (n; ;  ? 1) = + 1

in contradiction to bisaturation. We interpret now these properties in terms of the structures of A and B  under the assumption k -  and derive a contradiction. By our supposition k -  for the last digit dn (  ? 1) (i.e. the n{th component of  ? 1 ) of integer  ? 1 (the biggest number in B  ) we have

dn (  ? 1) 6= k ? 1: 15

(10.2)

Let integer  ? 1 be the `{th lexicographic element, when it is read backwards (as the numbers in A ). We just proved that it does not appear in A and so `   . Since dn (  ? 1) 6= k ? 1 , we get dn (  ) = dn (  ? 1) + 1 and so integer  is the (` + kn?1 ){th element when read lexicographically backwards. Therefore integer  is also not in the list A . This means that k (n; ;  +1) = k (n; ;  ) and thus

 ? k (n; ;  + 1) = ;  + 1 ? k (n; ;  + 1) = fn ( ) + 1 , which contradicts optimality and completes the proof.

Remark: The relation k j (jAj?jB j) need not hold, if we require only (A; B ) 2 O(n) or only (A; B ) 2 S (n) . Example 1: n = 4 , k = 2 , = 8 , f4(8) = 1 because f4(8) < 2 by (1.7) and f4(8)  f4(9) = 1 by Theorem 3 (i), but 2 - 8 ? 1 . Example 2: n = 6 , k = 2 , (A; B ) 2 S (6) (by inspection), jAj = 3 , jB j = 20 , 2 - (20 ? 3) . : A = f111 100; 110 011; 001 111g , B = B1 [ B2 , B1 = f101 010; 101 001; 100 110; 100 101; 011 010; 011 001; 010 110; 010 101; 111 010; 111 001g , B2 = f110 110; 11 0101; 10 1110; 10 1101; 01 1110; 01 1101; 10 1011; 10 0111; 01 1011; 01 0111g . Note also that here '(A) = '(B ) = 6 A ^ B , because (000 000) 2= A ^ B . 11. Solution of Problem III

It suces to study the set ?








Z G(n) = (A; B ) 2 CACk (n) : jAj = G(n)  jB j ;

(11.1)

where G(n) is de ned in the Introduction. ?




Lemma 9. The set Z G? (n)
consists only of pairs (A; B ) for which jAj = jB j = G(n) or equivalently fn G(n) = G(n) .

Proof: Clearly there are (A; B ) 2 Z (G) which are optimal and bisaturated. If suces to show that for such a pair jAj = jB j . We know from Lemma 5 that there exist integers  ;  , 0   ;   kn such that  ? k (n; ; ) = jAj = G;  ? k (n; ; ) = jB j = fn (jAj); 16

(11.2) (11.3)

!

where A and B are the not intersected parts of ideals A and B  with jAj =  , ! ! jB j =  . Here A are the  smallest and written backwards, B  are the  smallest n{digit k{ary numbers. Then jB j > jAj would mean  >  and from Theorem 5    + k . At rst we claim that

k (n; ; ) = k (n; ; ):

(11.4)

Otherwise, since k (n; ; ) is an increasing function of , we'd have k (n; ; ) < ! k (n; ; ) . Then if A and A denote respectively the rst written!backward !and the rst written forward  elements of [k]n , and A0 = A n A ^ A , B 0 = A n ! A ^ A , then (A0 ; B 0) 2 CAC (n) and min(jA0j; jB 0j) = jA0 j =  ? k (n; ; ) >  ? k (n; ;  ) = G , contradicting the maximality of G . We also claim that

k (n; ;  ) ?  = k (n; ;  ) ?  :

(11.5)

Otherwise, since k (n; ; )? is a decreasing function of  , we'd have k (n; ; )? ! ! !  > k (n;  ; ! ) ?  . Then with A0 = B  n B  ^ B  and B 0 = B  n B  ^ B  , where B  and B  are respectively the rst written backward and the rst written forward  elements of [k]n , we'd have min(jA0j; jB 0j) = jA0j =  ? k (n; ;  ) >  ? k (n; ; ) = G again contradicting the maximality of G . From (11.4) and (11.5) it follows that for every i , 0  i <  ?  ,

k (n;  + i;  + i + 1) = k (n;  + i;  + i) and k (n;  + i + 1;  + i + 1) = k (n;  + i;  + i) + 1: Therefore, for every i , 0  i <  ?  , the ( + i + 1){th forward lexicographic element is simultaneously the ( + i + 1){th backward lexicographic element, i.e. !

?!

?

?!

?

 =  ;  + 1 =  + 1;  ? 1 =  ? 1: It! is easy to verify, that no? two consecutive integers  ;  + 1 ;  < kn ; satisfy ?!  =  and  + 1 =  + 1 . Also, we note that  + 1   ? 1 , since  ?   k  2 . This is a contradiction and hence  =  . 17

Theorem 6. (i) For n  2 , 2 j k

G(n) = k4n and fn k4n = k4n (ii) For n = 2m , m  1 , 2 - k
? G(n) = kn4?1 and fn kn4?1 = kn4?1 (iii) For n = 2m + 1 , m  1 , 2 - k  2m  2m 2m G(n) = k(k 4 ?1) and f2m+1 k(k 4 ?1) = k(k 4 ?1) . ?




Proof: (i) Recall (1.6), which was obtained by the Ahlswede/Zhang method: n

G(n)  k4 ; n  2:

(11.6)

This bound is achievable for all even k and all n  2 by the following construction:    A = (a1; a2; : : :; an ) : a1 2 0; 1; : : :; k2 ? 1 ; a2 2 k2 ; : : :; k ? 1 ,    B = (b1 ; b2; : : :; bn ) : b1 2 k2 ; : : :; k ? 1 ; b2 2 0; 1; : : :; k2 ? 1 .

2m (ii) It follows from (11.6) that G  k 4?1 and this bound is achievable by this construction: ?

?  A = (a1; : : :; am; am+1; : : :; a2m) : (a1; : : :; am) 2 R km2?1 ; (am+1; : : :; a2m) 2 R km2+1 ,  ?

? B = (b1 ; : : :; bm; bm+1; : : :; b2m) : (b1; : : : ; bm) 2 R km2+1 ; (bm+1; : : :; b2m) 2 R km2?1 ,

where R(s) (resp. R(s) ) is the set of s lexicographic smallest (resp. largest) elements of [k]m . ?




(iii) From Lemma 9 it follows that G(n) = max  ? k (n; ; ) and from Theorem 5  it follows that if G(n) =  ? k (n; ; ) , then k j  . Let us denote g() =  ? k (n; ; ) .

Lemma 10. If  < km2?1 km+1 or  > km2+1 km+1 , then 







m m 2m g() < g k 2? 1 km+1 = g k 2+ 1 km+1 = k(k 4 ? 1) :

Proof: We use from Theorem 5 in [5] k (n; ; )  k
n , from where we get 2

g() =  ? k (2m + 1; ; )   ? k2m+1 = P (); say: 18

Also from our Theorem 2 in [10] it follows that g() = P () if and only if km+1 j  . Hence, 









2m m m m g k 2? 1 km+1 = P k 2? 1 km+1 = k(k 4 ? 1) = g k 2+ 1 km+1   m k + 1 m +1 : =P 2 k



h

2m+1 Let us note that the function P () is monotonically increasing in the interval 0; k 2 i h and monotonically decreasing in k2m2+1 ; k2m+1 . Therefore, if  < km2?1 km+1 (or  > km +1 km+1 ), then

2









m m g()  P () < P k 2? 1 km+1 = g k 2? 1 km+1 :

So Lemma 10 shows, that max g() = G is achieved, when

km ? 1 km+1    km + 1 km+1 : 2 2

(11.7)

Proof of (iii): Let us proceed by induction on m .

m = 1 : We need to consider only  divisible by k (since (A; B ) is bisaturated) 2 and satisfying k?2 1 k2    k+1 2 k . 2 If  = k?2 1 k2 or  = k+1 2 k , then we have 







2 g k ?2 1 k2 = g k +2 1 k2 = k(k 4? 1) : 2 If k?2 1 k2 <  < k+1 2 k , then we apply Theorem 5 of [5] and get 







 ? k (3; ; ) =  ? k 2;  ? k ?2 1 k2; k + k ?2 1
k       k ? 1  k ? 1 k ? 1  k ? 1 k ? 1  =  ? k 1; k ? 2 k; k ? 2 k + 2
k + 2 k ? 2 k 2 = k(k 4? 1) 2

and G(3) = g() = k(k 4?1) for all  = k?2 1 k2 + k
s , s = 0; 1; : : :; k . 19

i

Suppose it is true for m ? 1 and for  = km?21 ?1 km + km?1
s , s = 0; 1; : : :; k ; ?
2m?2 max  ? (2m ? 1; ; ) = k(k 4 ?1) and let us prove it for m . ?




Ifm max 0 ? (2m +m 1; 0; 0 ) =  ? (2m + 1; ; ) , it follows from Lemma 10 that k ?1 km+1    k +1 km+1 and therefore k?1 k2m <  < k+1 k2m . 2 2 2 2 Now 







 ? k (2m + 1; ; ) =  ? k 2m;  ? k ?2 1 k2m ; k + k ?2 1
k    k ? 1  k ? 1  2 m ? 1 2 m ? 1 + k ?2 1
k ;k? 2 k =  ? k 2m ? 1; k ? 2 k    k ? 1 k ? 1 2 m ? 1 + 2 k? 2 k   k ? 1  k ? 1  k ? 1  2 m ? 1 2 m ? 1 2 m ? 1 + k ?2 1 k2m?1 ? k 2m ? 1; k ? 2 k ; k ? 2 k =k? 2 k 2 2 + (k ?4 1) k2m?1 = 1 ? k (2m ? 1; 1; 1) + k ?2 1 k2m?1 + (k ?4 1) k2m?1; where 1 = k ? k?2 1 k2m?1 , and, since km2?1 km+1    km2+1 km+1 we get km?1 km  1  km?1 +1 km . It follows that  ? k (2m +1; ; ) is maximum i 1 ? 2 2 ? k (2m ? 1; 1; 1) is maximum. According to the induction hypothesis max1 1 ?
2m?2 k (2m ? 1; 1; 1) = k(k 4 ?1) and it is achieved for 1 = km?21 ?1 km + km?1
s ; ?
2m s = 0; 1; : : :; k . Hence, max  ? k (2m + 1; ; ) = k(k 4 ?1) and it is achieved for   ?
 = 1 + k?2 1 k2m?1 k = km?21 ?1 km + km?1
s + k?2 1 k2m?1 k = km2?1 km+1 + km
s ; s = 0; 1; : : :; k . Finally, we give a construction of (A; B ) achieving this bound: 





m A = (a1; : : :; am; am+1; : : : ; a2m; a2m+1) : (a1; : : :; am) 2 R k 2? 1 ;    m k + 1 ; a2m+1 2 [k] (am+1; : : :; a2m) 2 R 2 





m B = (b1; : : :; bm; bm+1; : : :; b2m; b2m+1) : (b1 ; : : :; bm) 2 R k 2+ 1 ;    m k ? 1 ; b2m+1 2 [k] : (bm+1; : : : ; b2m) 2 R 2

20

12. Solution of Problem IV

Theorem 7. The function Q(n) = (A;B)max jAjjB j satis es 2CAC (n) k

(i) For n  2 , 2 j k Q(n) = k162n n

2

(ii) For n = 2m , m  1 , 2 - k Q(n) = (k 16?1) and this value is assumed only for (A1; B1) and (A2; B2) ; jA1j  jB1j , jA2j  jB2j with the cardinalities jA1j = (km4?1)2 , jB1j = (km4+1)2 , jA2 j = k2m4?1 , jB2j = k2m4?1 . 2m

2m+2

(iii) For n = 2m + 1 , m  1 , 2 - k Q(n) = (k ?1)(16k ?1) and this value is assumed only for (A3; B3) and (A4; B4) jA3j  jB3j , jA4j  jB4j with the cardinalities m m+1 m m+1 jA3j = (k ? 1)(4k + 1) ; jB3j = (k + 1)(4k ? 1) m m+1 m m+1 jA4j = (k ? 1)(4k ? 1) ; jB4j = (k + 1)(4k + 1) :

Proof of (i): By the Ahlswede/Zhang method, jAjjB j 

k2n , so (i) follows from 16

Theorem 6 (i). The remaining cases are more complicated. We need further auxiliary results.

Lemma 11. Let (A; B ) 2 CACk (n) ; jAj  jB j , n = 2m , 2 - k and let ` be a positive integer. 2m (a) If jAj < `2 < k 4?1 , then

jAj
jf2m(jAj) < `2
f2m(`2) = `2
(km ? `)2: 2m (b) If `2 < jAj  k 4?1 , then

jAj + jf2mj(A)j < `2 + f2m(`2) = `2 + (km ? `)2: Proof: (a) We use inequality (1.8), that is,

p p jAj + f2m(jAj)  k2m :

p

Hence

p p p p jAj + f2m(jAj)  k2m = `2 + (km ? `)2:

p

From Theorem 3 we have

f2m(`2) = (km ? `)2: 21

(12.1)

Therefore, from the condition jAj < `2 < f2m(`2)  f2m(jAj) and (12.1) one has (a). (b) In this case we have `2 < jAj  f2m(jAj)  f2m(`2):

(12.2)

Inequality (12.1) is equivalent to p

p

(12.3) jAj + jf2m(jAj) + 2 jAj
f2m (jAj)  `2 + f2m(`2) + 2 `2f2m(`2): Now, if jAj + f2m(jAj)  `2 + f2m(`2) , then from (12.2) and by arithmetic{geometric 2
f2m (`2 ) . Hence jAj+jB j+pjAj
f2m (jAj) > mean inequalities we have j A j
f ( j A j ) > ` 2 m p

`2 + f2m(`2) + `2f2m (`2) which contradicts (12.3). Proof of (ii): Suppose Q(n) is achieved for2some pair (A; B ) with jB j = f2m(jAj) ; m ?1 k jAj  f2m(jAj) . From (1.8) we have jAj  4 and we have even  m  k ? 1 2  jAj  k2m ? 1 ; 2 4 because `2 = km2?1 2 > jAj would contradict (a) in Lemma 11.
?
? If jAj = km2?1 2 , then by Theorem 3 f2m(jAj) = km2+1 2 and this gives the cardinalities jA1j; jB1j of the Theorem. ?
Now, if jAj > km2?1 2 , then from Lemma 11 (b) we have


?

2 2  m 2  m  m m ? 1 2 k2m + 1 k ? 1 k ? 1 k + 1 k + f = + = jAj+f2m(jAj) < 2 m 2 2 2 2 2 or equivalently jAj + f2m(jAj)  k2m2+1 ? 1 = k2m2?1 , from where 

 2m 2 k ?1

with possible equality only for jAj = f2m(jAj) = k2m4?1 . 4 However, f2m k2m4?1 = k2m4?1 follows from Theorem 6 (ii) and this gives the cardinalities of (A2; B2) .

jAj
jf2m(jAj)j

Clearly an (A; B ) 2 CACk (n) with jAjjB j = Q(n) is optimal and bisaturated. Therefore for some ;

 ? k (n; ; ) = jAj; ? k (n; ; ) = jB j

(12.4)

k j ; k j :

(12.5)

and by Theorem 5

22

Lemma 12. Suppose that n = 2m + 1 , jAj
jB j = Q(2m + 1) , jAj  jB j and jAj =  ? k (2m + 1; ; ) , jB j = ? k (2m + 1; ; ) . Then k ? 1 k2m     k + 1 k2m: 2

2

Proof: Let us introduce ?


?




G2m+1 (; ) =  ? k (2m + 1; ; )
? k (2m + 1; ; ) : Since k (2m + 1; ; )  k2 m+1 (by (v) Theorem 5 [5]), stated in Section 2, we have 

G2m+1 (; )   ? k2 m+1 = P (; ); say:







  
1? ? k2 =  1 ? m+1 k2m+1 k2m+1

It is easy to see that P (1; ) < P (2; ) , if 1 < 2 < k2m2+1 or 1 > 2 > k2m2+1 and P (; 1) < P (; 2) , if 1 < 2 < k2m2+1 or 1 > 2 > k2m2 +1 . 2m Suppose that  < k?2 1 k2m (or  > k+1 2 k ), then 





Q(2m + 1) = G2m+1(; )  P (; ) < P k ?2 1 k2m; : ?




?




On the other hand since k j by (12.5), then P k?2 1 k2m ; = G? 2m+1 k?2 1
k2m; , (we apply Theorem 2 of [10]) and hence Q(2m + 1) < G2m+1 k?2 1 k2m; , which contradicts maximality.

Proof of (iii): Let us proceed by induction on m . m = 1 : From Lemma 12 one has k ? 1 k2     k + 1 k2 : 2 2 2 Case:  = k?2 1 k2 (or = k+1 2 k ). ?




Since k2 j  and k j , we have k 3; k?2 1 k2; = k?2 1
k and 













2 ? k ?2 1 k = (k 4?k 1) k2 ? k : G3 (12.6)   Let us consider the function T ( ) = k2 ? k . We verify that T ( 1 ) < T ( 2 ) if 1 < 2 < k23 or 1 > 2 > k23 .

k ? 1 k2 ; = k ? 1 k2 ? k ? 1 2 2 2 k

23

2 k2 +1) and Hence, taking account of 2 - k , k j one has maxkj ;2 k T ( ) = k(k ?1)( 4 the maximum is achieved for 1 = k32?k , 2 = k32+k only. -

?
2 k4 ?1) and So, max G3 k?2 1 k2 ; = (k ?1)( 16 



2 jA3j = k ?2 1 k2 ? k 3; k ?2 1 k2; 1 = (k ? 1)(4k + 1)   > (k + 1)(k2 ? 1) ; k ? 1 > 2 : jB3 j = 1 ? k 3; k ; = 1 2 4 or   8 2 ? 1) k ? 1 k ? 1 ( k ? 1)( k 2 2 > > < jA4 j = 2 k ? k 3; 2 k ; 2 = 4   2 > k ? 1 k2 ; = (k + 1)(k + 1) : > : jB4 j = 2 ? k 3; 2 2 4 8 > >
x2 > k2 . With regards to 2 - k , k j  , k j one has   maxkjx L(x) = k22?k 1 ? k22k?2k = (k?1)(4 k+1) and it is achieved only for x1 = k22?k , x2 = k22+k . 24

Therefore from (12.9) 2 2 2 2 1)(k + 1) = (k2 ? 1)(k4 ? 1) : G3 (; ) = k (k 16? 1) + (k ? 1)(k4 ?
4 16

Furthermore, substitution of ?  = x1 (or ?  = x2 ) in (12.8) give us 2 2 k (3; ; ) =  ? (k ? 1)(4k + 1) or  ? (k ? 1)(4k ? 1)

and hence the same parameters for (A3 ; B3) and (A4; B4) occur. Now suppose that 2m?2 ?1)(k2m ?1) ( k the statement is true for n = 2m ? 1 that is S2m?1 = and it 16 assumed only for 04 j = (km?1 ?1)(km ?1) 03 j = (km?1 ?1)(km +1) j A j A 4 4 (A03 ; B30 ); (A04; B40 ) : m?1 +1)(km ?1) ; m?1 +1)(km +1) : (12.10) ( k ( k 0 0 jB3j = jB4j = 4 4 (

(

According to Lemma 12 one has

k ? 1 k2m     k + 1 k2m: 2 2 2m Case:  = k?2 1 k2m (or = k+1 2 k )

Since k2m j  , k j then k k?2 1 k2m; = (k?k1) and     ?
 k+1)
k2m ? . G2m+1 k?2 1 k2m; = (k?2 1) k2m ? (k?2 1) k ? (k?2 1) k = (k?1)( 4k k ?




?k or 2 = We verify that max G2m+1 k?2 1 k2m; is assumed for 1 = k2m+1 2 k2m+1 +k and 2 ?

? 2m 2m+2 2m 2m 2 G2m+1 k?2 1 k2m; 1 = G2m+1 k?2 1 k2m; 2 = (k ?1)(k 16 +1)(k ?1) < (k ?1)(16k ?1) , if m > 1 and so  > k?2 1 k2m . ?




2m Case: k?2 1 k2m <   < k+1 2 k

We apply again Theorem 5 of [5]: 



k (2m + 1; ; ) = k 2m;  ? k ?2 1 k2m ; k + k ?2 1 k   2 k ? 1 k ? 1 k ? 1 k ? 1 ( k ? 1)   2 m ? 1 2 m ? 1 + 2 k + 2 k ? 4 k2m?1 ;k ? 2 k = k 2m ? 1; k ? 2 k and so 25

 ? k (2m + 1; ; ) = k +2 1 1 ? k (2m ? 1; 1; 1) ? (k ?2 1) 1 ? k (2m ? 1; 1; 1 ) 2 (12.11) + (k 4? 1) k2m?1 and ?




?




? k (2m + 1; ; ) = (k +2 1) 1 ? k (2m ? 1; 1; 1) ? (k ?2 1) 1 ? k (2m ? 1; 1; 1 ) 2 + (k 4? 1) k2m?1; ?




?

where 1 = k ? (k?2 1) k2m?1 , 1 = k ? k?2 1 k2m?1 . Substituting (12.11) in G2m+1 (; ) , after simpli cation one has 2 ? 1)2 k4m?2 G2m+1 (; ) = G2m?1 (1; 1)
k2 + (k 16 2 ?
?
+ (k 4? 1) 1 + 1 ? 2k (2m ? 1; 1; 1 )
k2m?1 ? (1 + 1 ? 2k (2m ? 1; 1; 1)) : Therefore, 2 ? 1)2 2 max G2m+1 (; )  (k 16 k4m?2 + max G2m?1(1; 1 )
k2 + k 4? 1
max R(y); y

where y = 1 + 1 ? 2k (2m ? 1; 1; 1 )2m?and R2(my) = y(k2m?1 ? y) . By the induc2 tion hypothesis max G2m?1 (1; 1) = (k ?161)(k ?1) and is achieved only for (see (12.10)) 

 00 01 ? k (2m ? 1; 01; 10 ) = jA03j 1 ? k (2m ? 1; 001 ; 100 ) = jA04j and 00 : (12.12) 10 ? k (2m ? 1; 01; 10 ) = jB30 j 1 ? k (2m ? 1; 001 ; 100 ) = jB40 j

2m?1 2m?1 Also we verify that maxy R(y) = (k ?1)(4 k +1) and the maximum is assumed for y1 = k2m?2 1 ?1 , y2 = k2m?2 1 +1 and they are compatible with (12.12).

Hence 2 2m?2 ? 1)(k2m ? 1) (k2 ? 1)2 k4m?2 G2m+1 (; )  k (k + 16 16 2 2m?1 2m?1 + 1) (k2m ? 1)(k2m+2 ? 1) = : + (k ? 1)(k 16? 1)(k 16

26







?

?




On the other hand we take  = 1 + k?2 1 k2m?1 k , = 1 + k?2 1 k2m?1 k and 2m 2m+2 verify G2m+1(; ) = (k ?1)(16k ?1) . The optimal constructions are: 











m A3 = (a1; : : : ; am; am+1; : : :; a2m+1) : (a1; : : :; am) 2 Rm k 2? 1 ;   m+1 k + 1 (am+1; : : :; a2m+1) 2 Rm+1 2    m k + 1 ; B3 = (b1; : : : ; bm; bm+1; : : :; b2m+1) : (b1; : : :; bm) 2 Rm 2   m+1 k ? 1 (bm+1; : : :; b2m+1) 2 Rm+1 2

and m?1 k ; A4 = (a1; : : : ; am; am+1; : : :; a2m+1) : (a1; : : :; am) 2 Rm 2  m+1  k ? 1 (am+1; : : :; a2m+1) 2 Rm+1 2    m k + 1 ; B4 = (b1; : : : ; bm; bm+1; : : :; b2m+1) : (b1; : : :; bm) 2 Rm 2  m+1  k + 1 : (bm+1; : : :; b2m+1) 2 Rm+1 2

13. Solution of Problem V

From Lemma 4 in Section 4 we conclude that ?




a
(n) = 0max  ? k (n; ;  +
) : kn

(13.1)

It is convenient to introduce the function

Fn (; ) =  ? k (n; ; )

(13.2)

and write for any
, ?kn 
 kn

a
(n) = max F (;  +
):  n 27

(13.3)

Theorem 8.

(i) n = 2m : For any
and t with ?k2m  k2m ? 2tkm ? km 
 k2m ? 2tkm + km  k2m a
(n) = F2m (tkm; tkm +
):

(ii) n = 2m + 1 : For any
and t with ?k2m+1  k2m+1 ? 2tkm+1 ? km+1 
 k2m+1 ? 2tkm+1 + km+1  k2m+1

a
(n) = F2m+1(tkm+1 ; tkm+1 +
): The proof uses three auxiliary results (Lemmas 13, 14, and 15 below) concerning the function Fn , which are obtained by a thorough analysis. Essential use is made of properties of the function k , which were obtained by Daykin, Kleitman, and West. We rely upon their Theorem 5 in [5], which is restated in Section 2 and from now on will be refered to as Theorem DKW. We present and prove now our auxiliary results.

Lemma 13. Assume that 0 < s < k;   s
kn?1 , and  (k ? s)kn?1 . Then (i) Fn (; )  Fn ( ? rk; ? rk) for all r ; 0  r  min(k; ) . (ii) If k j  , then Fn (; )  Fn ( ? ; ? ) for all  ; 0    min(; ) . Proof: (i) From Theorem 4 in [5] (see Section 2) it follows that k (n; ; ) can be assumed for a pair of ideals A; B with jAj =  , jB j = , where A are the  smallest and written backwards, B are the smallest n{digit k{ary numbers. ! Further, we continue to denote by x (resp. x ) the x{th lexicographic vector written backwards (resp. the x{th lexicographic vector). Now let A = A1 [ A2 with jA1j =  ? rk; jA2j = rk and let B = B1 [ B2 with jB1j = ? rk and jB2j = rk; where A1 (resp. B1 ) is the set of the ( ? rk) smallest lexicographic vectors written backwards (resp. ( ? rk) smallest lexicographic vectors). Then Fn (; ) = jA1 [ A2 j ? j(A1 [ A2 ) \ (B1 [ B2)j = jA1j + jA2j ? jA1 \ B1j ? jA1 \ B2j ? jA2 \ (B1 [ B2 )j and Fn ( ? rk; ? rk) = jA1 j ? jA1 \ B1j . We notice in view of the bound on  that the last component of a vector in the ideal A is never one of the (k ? s) integers s; s + 1; : : :; k ? 1 while each of the digits 0; 1; : : :; k ? 1 must occur exactly jBk2 j = r times in the last component of a vector in B2 because the vectors in B2 are consecutive. It follows that jA1 \ B2j  jB2j?r(k ?s) . Similarly, in view of the bound on , it follows that the rst component of a vector in the ideal B is never one of the s digits k ? s; k ? s +1; : : : ; k ? 1 while each of the digits 0; 1; : : :; k ? 1 must occur exatly jAk2j = r times as a rst component of a vector in A2 since the vectors in A2 are consecutive (in the backward order). It 28

follows that jA2 \ (B1 [ B2 )j  jA2j? rs and therefore Fn (; )  jA1j?jA1 \ B1j = Fn ( ? rk; ? rk) . (ii) Suppose that A = A1 [ A2 with jA1 j =  ?  , jA2 j =  and B = B1 [ B2 with jB1j = ?  and jB2j =  , where A1 (resp. B1 ) is the set of ( ? ) smallest lexicographic vectors written backwards (resp. ( ? ) smallest lexicographic vectors). From (i) it follows that it is sucient to consider  ; 0 <  < k . We consider the rst components of vectors A2 . As k j  the rst component of the {th vector is equal to k ? 1 . Hence, the rst components of vectors A2 are k ? ; k ?  + 1; : : :; k ? 1 . We consider separately cases a)   s and b)  > s . a)   s : Then A2 \ (B1 [ B2) = ? , since  (k ? s)kn?1 implies that the rst components of vectors B = B1 [ B2 are smaller than or equal to k ? s ? 1 < k ? s  k ?  . Also, clearly jA1 \ B2 j  jB2j = jA2 j =  . Therefore, Fn (; ) = jA1 [ A2 j?j(A1 [ A2 ) \ (B1 [ B2)j = jA1 j?jA1 \ B1j + jA2 j?jA1 \ B2j? jA2 \ (B1 [ B2)j  jA1 j ? jA1 \ B1j = Fn ( ? ; ? ) .

b)  > s : Then jA2 \ (B1 [ B2)j   ? s since the rst components of the  vectors in A2 are k ? ; k ?  + 1; : : :; k ? 1 while the rst components of vectors in B are  k ? s ? 1 in view of  (k ? s)kn?1 . Now suppose that the last components of the  vectors in B2 are i; i +1; : : :; i +  ? 1( mod k) . These integers are distinct because  < k . It follows that jA1 \ B2 j  s since otherwise some vector in A would have last coordinate  s , which would imply jA1 j  (s + 1)kn+1 , contradicting that jA1 j =  ?  <   skn?1 . Therefore, Fn (; ) = jA1 j?jA1 \ B1j + jA2 j?jA1 \ B2 j?jA2 \ (B1 [ B2 )j  jA1 j?jA1 \ B1j +  ? ( ? s) ? s = jA1 j ? jA1 \ B1j = Fn ( ? ; ? ) .

Corollary. Let 0 < s < k ;   skn?1 and  (k ? s)kn?1 , then (i) Fn (; )  Fn ( + rk; + rk) (ii) k j  implies Fn (; )  Fn ( + ; + ) .

Proof: We use the identity (iv) of Theorem DKW, that is k (n; kn ? ; kn ? ) = k (n; ; ) + kn ?  ? . One has Fn (; ) =  ? k (n; ; ) =  ? k (n; kn ? ; kn ? ) + kn ?  ? = kn ? ? k (n; kn ? ; kn ? ) = Fn (kn ? ; kn ? ) and Fn ( + rk; + rk) = Fn (kn ? ? rk; kn ?  ? rk) . We introduce  = kn ? and  = kn ?  and observe that   s
kn?1 ,   (k ? s)kn?1 . According to (i) in Lemma 13 one has Fn (; ) = Fn (kn ? ; kn ? ) = Fn (;  )  Fn ( ? rk;  ? rk) = Fn (kn ? ? rk; kn ?  ? rk) = Fn ( + rk; + rk) . Annalogously one can prove (ii). 29

Lemma 14. Assume that k -  and   1 mod k , 0 < 1 < k . Then  Fn (; )  max Fn ( ? 1; ? 1); Fn ( + k ? 1 ; + k ? 1) . Proof: As in the proof of Lemma 13 let A , jAj =  , be the ideal of the  smallest lexicographic vectors written backwards, and B , jB j = , be the smallest n{digit k{ary vectors. The rst component of  ? 1 , being biggest vector in A , is equal to 1 ? 1 , since  = 1 ( mod k) . We distinguish two cases:  ? 1 2 B and  ? 1 2= B . ?

1)  ? 1 2 B : Then  ? 1 = ! for some < . It is easy to see, that  ? i = ?! ?

? ikn?1 for all i = 1; : : :; 1 . Hence  ? i 2 B for all i = 1; : : :; 1 . Therefore, if we remove the last 1 vectors from A , we decrease A \ B by 1 . This is equivalent to k (n; ; ) = k (n;  ? 1; ) + 1  k (n;  ? 1; ? 1) + 1 , from where one has Fn ( ? 1; ? 1 ) =  ? 1 ? k (n;  ? 1 ; ? 1 )   ? k (n; ; ) = Fn (; ) . ?

2)  ? 1 2= B : Then clearly  + i 2= B , i = 0; : : :; k ? 1 ? 1 . We add the next (k ? 1) lexicographic vectors to the ideals A and B : A = A [ B 0 , B  = B [ B 0 ; ? ?!  ! A0 = ; : : :;  + k ? 1 ? 1 , B 0 = ; : : :; + k ? 1 ? 1 .

? Let us consider A \B  = (A \B )[(A \B 0 ) . As  + i 2= B for i = 0; : : :; k ?1 ?1 , then A \ B = A \ B:

Also, obviously jA \ B 0j  jB 0j = k ? 1 . Therefore, k (n;  + k ? 1; + k ? 1) = jA \ B  j = jA \ B j + jA \ B 0 j = jA \ B j + jA \ B 0j = k (n; ; ) + jA \ B  j  k (n; ; ) + k ? 1 or equivalently Fn ( + k ? 1 ; + k ? 1 ) =  + k ? 1 ? k (n;  + k ? 1 ; + k ? 1 )   ? k (n; ; ) = Fn (; ) . Now we try to estimate max Fn (;  +
) for arbitrary xed
. We concentrate on the case n = 2m . Obviously, for any
, there exists a unique t , such that

k2m ? 2km t ? km <
 k2m ? 2kmt + km :

(13.4)

Lemma 15. (i) If skm?1 < t < (s + 1)km?1 , then max f2m(;  +
) can be

attained when

k j ; sk2m?1    (s + 1)k2m?1 and (k ? s ? 1)k2m?1   +
 (k ? s)k2m+1: (ii) If t = skm?1 , then in case (a) k2m ? 2kmt <
 k2m ? 2kmt + km 30

max f2m(;  +
) can be attained, when

k j ; sk2m?1 ? km    sk2m?1 and (k ? s)k2m?1   +
 (k ? s)k2m?1 + km : and in case (b) k2m ? 2km t ? km <
 k2m ? km t max Fn (;  +
) can be obtained when

k j ; sk2m?1    sk2m?1 + km and (k ? s)k2m?1 ? km   +
 (k ? s)k2m?1:

Proof: (i) k j  follows from Lemma 14. As skm?1 < t < (s + 1)km?1 , then sk2m?1 + km  tkm  (s + 1)k2m?1 ? km and using (13.4) one has

k2m ? 2(s + 1)k2m?1 + km <
 k2m ? 2sk2m?1 ? km : Now, if  < sk2m?1 , then

 +
< k2m ? sk2m?1 ? km = k2m?1 (k ? s) ? km and we apply Lemma 13. Thus

F2m(sk2m?1; sk2m?1 +
)  F2m(;  +
): If  > (s +1)k2m?1 , then  +
> (k ? s ? 1)k2m?1 + km and we apply the Corollary above. Thus ?




F2m (s + 1)k2m?1; (s + 1)k2m?1 +
 F2m(;  +
): If sk2m?1 <  < (s + 1)k2m?1 , but (k ? s ? 2)k2m?1 <  +
< (k ? s ? 1)k2m?1 or (k ? s)k2m?1 <  +
< (k ? s + 1)k2m?1 , then we apply again Lemma 13 or the Corollary, respectively. The proof of (ii) is analogous.

Proof of Theorem 8: Suppose that sk2m?1  tkm < (s + 1)k2m?1 . We proceed by induction on m : m=1: k2 ? 2kt ? k 
 k2 ? 2kt + k: 31

(13.5)

From Lemma 15 (ii) it follows that max F2(;  +
) can be attained, when  2  (t ? 1)k; tk; (t + 1)k . Using (ii) and (vi) of Theorem DKW we verify ?

F2 (t ? 1)k; (t ? 1)k +





F2(tk; tk +
) ?

F2 (t + 1)k; (t + 1)k +





t?2 
? j  X 2 = (t ? 1)k ? (t ? 1) ? ; k j =0 tX ?1 
? j  = tk ? t2 ? ; and k j =0 t

X = (t + 1)k ? (t + 1)2 ? j =0






?j : k

From here, using (13.5), one has 



F2 (tk; tk +
) ? F2 (t ? 1)k; (t ? 1)k +
= k ? 2t + 1 ?
? (kt ? 1)  0 ?

and








F2 (tk; tk +
) ? F2 (t + 1)k; (t + 1)k +
= 2t + 1 ? k +
k? t  0: ?




So, max F2(;  +
) = F2 (kt; kt +
) for all
satisfying (13.5).

m ? 1 ! m : Suppose (i) is true for m ? 1 , with respect to any
1 ; t1 satisfying k2m?2 ? 2t1 km?1 ? km?1 
1  k2m?2 ? 2t1 km?1 + km?1 : That is

max F (;  +
1 ) = F2m?2(t1km?1 ; t1km?1 +
1 ):  2m?2

1) Case tkm = sk2m?1; m  2 :

(To prove this case we do not need to use induction hypothesis.) a) k2m ? 2sk2m?1 
 k2m ? sk2m?1 + kn : According to Lemma 15 (case (a) in (ii)) it follows that max F2m (;  +
) can be attained when

k j ; sk2m?1 ? km    sk2m?1; (k ? s)k2m?1   +
 (k ? s)k2m?1 + km : If  +
= (k ? s)k2m?1 , then we verify (using formula (iv) of Theorem DKW) that ?




F2m(sk2m?1; sk2m?1 +
) = F2m (k ? s)k2m?1 ?
; (k ? s)k2m?1 : Let  +
> (k ? s)k2m?1 and hence sk2m?1 ? km <  . Then

k j ; sk2m?1 ? km <   sk2m?1 ; (k ? s)k2m?1 <  +
 (k ? s)k2m?1 + km (13.6) 32

We compare values F2m (sk2m?1; sk2m?1 +
) and F2m(;  +
) , where  satis es (13.6). Using formulas (ii) and (vi) of Theorem DKW one has

F2m(sk2m?1; sk2m?1 +
) = sk2m?1 ? k (2m; sk2m?1; sk2m?1 +
)    s?2 
? j  X
? s + 1 2 m ? 2 2 m ? 1 2 m ? 1 2 m ? 2 ? (s ? 1)sk ? = sk ?  2m ? 1; k ; sk + k k j =0 s?1 
? j  X 2 m ? 2 = s(k ? s)k ? k j =0

(13.7)

and

  2 m ? 1 F2m(;  +
) =  ? k (2m; ;  +
) =  ? k 2m ? 1; k ;  +
? (k ? s)k     
? s ? 1  2 m ? 1 2 m ? 2 + ? (k ? s) k =  ? k 2m ? 2; k ? (s ? 1)k ; k ? (k ? s)k k 

s?2
? j X   2 m ? 2 ? ? (k ? s) k ? (s ? 1) k + (s ? 1)(k ? s)k k j =0 



s?2
? j X = k + (s ? 1)(k ? s)k2m?2 ? k j =0     
? s + 1 2 m ? 1 2 m ? 2 ? k 2m ? 2; k ? (s ? 1)k ; k ? (k ? s)k : + k 



From (13.6) if follows that k2m?2 ? km?1 < k ? (s ? 1)k2m?1  k2m?2 and 0 <  ? (k ? s)k2m?2 +
?s+1  km?1 . k k Let us show that for all ;  such that k2m?2 ? km?1 <  k2m?2 , 0 <   km?1 it is true that k (2m ? 2; ; ) =  . It is known that k (2m ? 2; ; )  k2 m
?2 implies 2m?2 m?1 k (2m ? 2; ; )  k2 m
?2 > (k k2m??k 2 ) =  ? km?1   ? 1 . So,   k (2m ? 2; ; ) >  ? 1 implies k (2m ? 2; ; ) =  . Therefore, ?  ? (s ? 1)k2m?1 ;  ? (k ? s)k2m?2 + 
?s+1 
=  ? (k ? s)k2m?2 +  2 m ? 2 ; k k k k k 
?s+1  and hence for all  staisfying (13.6) one has k s?1 
? j  X 2 m ? 2 = F2m (sk2m?1; sk2m?1 +
): F2m (;  +
) = s(k ? s)k ? k j =0

b) k2m ? 2sk2m?1 ? km 
 k2m ? 2sk2m?1 : This case can be treated analogously. So, if tkm = sk2m?1 , then max F (;  +
) = F2m(sk2m?1; sk2m?1 +
):  2m

2) Case sk2m?1 < tkm < (s + 1)k2m?1; m  2 : 33

Using again formula (ii) of Theorem DKW one has

F2m(tkm ; tkm +
) = tkm ? k (2m; tkm; tkm +
) ?
= tkm ? (k ? s ? 1)tkm?1 ? k 2m ? 1; tkm?1; tkm +
? (k ? s ? 1)k2m?1 s?1 
? j  X m m ? 1 m ? 1 2 m ? 2 = tk ? (k ? s ? 1)tk ? stk + s(k ? s ? 1)k ? k j =0   
? s m ? 1 2 m ? 2 m ? 1 2 m ? 2 ? k 2m ? 2; tk ? sk ; tk ? (k ? s ? 1)k + k s?1 
? j  X m ? 1 2 m ? 2 = k t + sk (k ? s ? 1) ? k j =0   
? s m ? 1 2 m ? 2 m ? 1 2 m ? 2 ? k 2m ? 2; tk ? sk ; tk ? (k ? s ? 1)k + k : (13.8) Lemma 15 says that in this case max F2m(;  +
) can be attained, when

k j ; sk2m?1    (s + 1)k2m?1 and (k ? s ? 1)k2m?1   +
 (k ? s)k2m?1:

At rst let us consider the extremal cases, that is (*)  = sk2m?1 and (**)  +
= (k ? s ? 1)k2m?1 . (*)  = sk2m?1 : Using (13.7) and (13.8) one has

F2m(tkm ; tkm +
) ? F2m(sk2m?1; sk2m?1 +
) = ?  
km?1 t ? sk2m?2 ? k 2m ? 2; km?1t ? sk2m?2; km?1t ? (k ? s ? 1)k2m?2 +
k?s  0 , since k (m; ; )  min( ; ) for all ;  . (**)  +
= (k ? s ? 1)k2m?1 : As k j  , we have k j
and hence F2m(tkm ; tkm +
) = km?1 + sk2m?2(k ? s ? 1) ? sk
 
m ? 1 2 m ? 2 m ? 1 2 m ? 2 ? k 2m ? 2; k t ? sk ; k t ? (k ? s ? 1)k +k

(13.9)

and we verify that

F2m (k ? s ? 1)k2m?1 ?
; (k ? s ? 1)k2m?1 = (k ? s ? 1)k2m?2
(s + 1) ?
(sk+ 1) : ?




Therefore

?




F2m(tkm ; tkm +
) ? F2m (k ? s ? 1)k2m?1 ?
; (k ? s ? 1)k2m?1 = km?1 t ? k2m?2(k ? s ? 1) +
k  
m ? 1 2 m ? 2 m ? 1 2 m ? 2 ? k 2m ? 2; k t ? sk ; k t ? k (k ? s ? 1) + k  0: 34

Now let

k j ; sk2m?1 <   (s + 1)k2m?1 and (k ? s ? 1)k2m?1 <  +
 (k ? s)k2m?1: Then (we use (ii) of Theorem DKW) F2m(;  +
) = k + sk2m?2(k ? s ? 1)  s?1     X
?j 
? s  2 m ? 2 2 m ? 2 ? : + k ? k 2m ? 2; k ? sk ; k ? (k ? s ? 1)k k j =0 Now we compare the values F2m(tkm; tkm +
) and F2m(;  +
) . One has F2m (tkm; tkm +
) ? F2m (;  +
) = tkm?1 ? k ? (1) + (2) ; (13.10) where 



(1) = k 2m ? 2; km?1t ? sk2m?2; km?1t ? (k ? s ? 1)k2m?2 +
k? s 



(2) = k 2m ? 2; k ? sk2m?2; k ? (k ? s ? 1)k2m?2 +
k? s





:

We use abbreviatione t1 = t ? skm?1 and 1 = k ? sk2m?2 . Then (1) = k (2m ? 2; km?1t1 ; km?1t1 +
1 ) and (2) = k (2m ? 2; 1; 1 +
1 ) ,   where
1 = (2s + 1 ? k)k2m?2 +
k?s . We verify that

k2m?2 ? 2t1 km?1 ? km?1 
1  k2m?2 ? 2t1 km?1 + km?1 : (13.10) can be written in the form:

F2m(tkm ; tkm+
)?F2m (; +
) = F2m?2 (t1km?1 ; t1km?1+
1 )?F2m?2(1; 1+
1 )  0; according to the induction hypothesis. We remark that (ii) can be proved in the same way. Finally we introduce now some simpli cation for the determination of the value F2m(tkm ; tkm +
) . If
satis es (i), then

tkm +
= k2m ? km t +  for some ? km    km : 35

Lemma 16. a) If ?km    0 , then

F2m (tkm; tkm +
) = t2 + t ? k (m; t; km + )

b) If 0    km , then

F2m(tkm ; tkm +
) = t2 ? k (m; t; ):

Proof: a) ?km    0 :

Let tkm = s1k2m?1 + s2 k2m?2 +


+ smkm . Then tkm +
= k2m ? km t +  = (k ? s1 ? 1)k2m?1 +


+ (k ? sm ? 1)km + km + : We use again formula (ii) of Theorem DKW. One has k (2m; tkm; (k ? s ? 1)k2m?1 +


+ (k ? sm ? 1)km + km + ) ?
= k 2m ? 1; km?1t; (k ? s2 ? 1)k2m?2 +


+ km +  + : : :

m X m ? 1 m + (k ? s1 ? 1)k t =


= k (m; t; k + ) + (k ? si ? 1)km?it i=1 m X = k (m; t; km + ) + kmt ? t ? t si km?i = k (m; t; km + ) + kmt ? t ? t2; i=1 m X si km?i = t . Therefore we have because i=1 F2m(tkm; tkm +
) = tkm ? k (tkm ; tkm +
) = t2 + t ? k (m; t; km + ):

(b) is proved analogously.

14. On optimal and bisaturated pairs

Recall that (A; B ) 2 CACk (n) is called optimal exactly if jB j = fn (jAj) . An optimal pair (A; B ) is always A{saturated , but not necessarily B {saturated . We try to describe pairs of cardinalities (jAj; jB j) for all pairs (A; B ) , which are optimal and bisaturated. Earlier we denoted this set by M(n) = O(n) \S (n) . Clearly for (A; B ) 2 CACk (n) ?
(A; B ) 2 M(n) i fn fn (jAj) = jAj: (14.1) We also know (see Corollary (iii) in Section 7), that for all  ? ?

fn fn fn (a) = fn (a) (14.2) and hence for all a ?
(C; D) 2 O(n); jC j = fn (a); jDj = fn fn (a) implies (C; D) 2 M(n): 36

(14.3)

Theorem 9. Let (A; B ) 2 O(n); jAj = a; jB j = fn (a) . Then (i) (A; B ) 2 M(n) implies k j (jAj ? jB j) . (ii) For all a  kn?2 (A; B ) 2 M(n) . (iii) For all a , kn?2  a  kn?2 (k ? 1)2 , (A; B ) 2 M(n) i k j (jAj ? jB j) . (iv) For any integer
; ?(kn ? skn?1 ) 
 kn ? 2kn?1 ; with k j
there is a unique a ; kn?2  a  (k ? 1)2kn?2 for which (A; B ) 2 M(n) , jAj = a , jB j = fn (a) and jB j ? jAj =
. (v) Let M (n) equal the number of dierent pairs of integers (jAj; jB j) , where jAj  jB j and (A; B ) 2 M(n) . Then M (n) = kn?1 ? kn?2 + 1: At rst let us introduce an auxiliary result.

Lemma 17. Let kn?2  a  (k ? 1)2kn?1 and let  ? k (n; ; ) = a; ? k (n; ; ) = fn (a):

(14.4)

Then kn?1  ;  (k ? 1)kn?1 .

Proof: We know that fn (kn?2 ) = (k ? 1)2kn?2 (follows from Theorem 3) and that (A; B ) 2 M(n) , where jAj = kn?2 , jB j = (k ? 1)2kn?2 . Therefore, for any a , kn?2  a  (k ? 1)2 kn?2 ,

jfn(a) ? aj  kn ? 2kn?1 = fn (kn?2 ) ? kn?2 : Suppose, for some a , kn?2  a  fn (a)  (k ? 1)2kn?2 ,  < kn?1 . We denote
a = ?  and together with (14.4) let us consider kn?1 ? k (n; kn?1; kn?1 +
a ); kn?1 +
a ? k (n; kn?1; kn?1 +
a ): (14.5) As  < kn?1 ,
a = fn (a) ? a  kn ? 2kn?1 , then kn?1 +
a  (k ? 1)kn?1 . Therefore, from Lemma 13 it follows that

kn?1 ? k (n; kn?1; kn?1 +
a )   ? k (n; ;  +
a ) =  ? k (n; ; ) = a
a + kn?1 ? k (n; kn?1; kn?1 +
a )  ? k (n; ;  +
a ) = fn (a) Now, if kn?1 ? k (n; kn?1; kn?1 +
a ) = a1 > a , then

fn (a1)  kn?1 +
a ? k (n; kn?1; kn?1 +
a ) > fn (a): This is a contradiction, since fn (
) is a monotonically decreasing function. 37

Hence

kn?1 ? k (n; kn?1 ; kn?1 +
a ) = a; kn?1 +
a ? k (n; kn?1 ; kn?1 +
a ) = fn (a):

(14.6)

Suppose that
a = k
r +
1 , 0 
1 < k . If 1) 0 <
1 < k , we verify (using (ii) of Theorem DKW) that

k (n; kn?1; kn?1 +
a ) =



kn?1 +
a =  (n; kn?1; kn?1 +
+ k ?
) k a 1 k 

and hence

kn?1 ? k (n; kn?1; kn?1 +
a + k ?
1 ) = a; kn?1 +
a + k ?
1 ? k (n; kn?1; kn?1 +
a + k ?
1 ) = fn (a) + k ?
1 > fn (a); a contradiction. (+
a ) If 2)
a = 0 , then, as k (n; ; )   kn ,0 one has ?00k (n; ; +
a0 )  00? knn?1 = Tn(;
a) , say, and we verify that T ( ;
a ) < T ( ;
a ) , if  <   k  k ?
a . Therefore 2  ? k (n; ;  +
a)  T (;
a ) < T (kn?1 ;
a ) =n?k1n?n1??1 k (n; kn?1; kn?1 +
a ) and since k j
a we have k (n; kn?1; kn?1 +
) = k (kkn +
) . Hence a =  ? k (n; ;  +
a ) < kn?1 ? k (n; kn?1; kn?1 +
a ) and this is also a contradiction, because we assumed (14.6). ;  (k ? 1)kn?1 is proved analogously.

Proof of Theorem 9: (i) This restates Theorem 5. (ii) Let us show that for all a < kn?2 fn (a) > fn (a + 1) , from where the claim follows. We use our main recursive formula: for a  kn?2

fn (a) = (k ? 1)kn?1 + kfn?2 (a) ? (k ? 1)a: Hence, as fn?2 (a)  fn?2 (a + 1)

fn (a)  (k ? 1)kn?1 + kfn?2 (a + 1) ? (k ? 1)a > (k ? 1)kn?1 + kfn?2 (a + 1) ? (k ? 1)(a + 1) = fn (a + 1): (iii) follows from (i) and (iv). (iv) Let M1(n)  M(n) be de ned by 

M1(n) = (Ai ; Bi)Ni=1 : (Ai; Bi) 2 M(n); kn?2 = jA1j < jA2j <


< jAN j = (k ? 1)2kn?2 ; (k ? 1)kn?2 = jB1j > jB2j >


> jBN j = kn?2 : 38

Obviously, jAij = jBN ?i j . We show that for all 1  i < N jAi+1j?jAi j < k (or equivalently for all 1  j < N jBj j ? jBj+1j < k) , from where, using (i), (iv) follows. Assume it does not and there exists i , 1  i < N for which jAi+1j ? jAij  k . Suppose that for some ;

 ? k (n; ; ) = jAi+1j and ? k (n; ; ) = jBi+1j = fn (jAi+1 j): According to Lemma 17 kn?1  ;  (k ? 1)kn?1 . Also, as (Ai+1; Bi+1) 2 M(n) , then k j  and k j . Now, we consider

 ? k ? k (n;  ? k; ) = a and ? k (n;  ? k; ) = b: Let us show that k (n;  ? k; ) < k (n; ; ) for  kn?1 , k j  , k j . If = kn?1 , then using (ii) of Theorem DKW one has
? k (n; ?k; kn?1) = k n ? 1; k ? 1; kn?1 = k ?1 = k (n; ; kn?1)?1 < k (n; ; kn?1) . Let skn?1 <  (s + 1)kn?1 , s > 0 . Then ?
?
?
k (n; ?k; ) = k n ? 1; k ? 1; ? skn?1 +s k ? 1  k n ? 1; k ; ? skn?1 + s k ? s = k (n; ; ) ? s < k (n; ; ) . Therefore,

a =  ? k ? k (n;  ? k; ) >  ? k ? k (n; ; ) = jAi+1j ? k  jAi j and

f (a)  b = ? k (n;  ? k; ) > ? k (n; ; ) = f (Ai+1): ?




We consider (A ; B ) 2 M(n) , where jAj ?= fn f
n (a ) , jB j = fn (a) . Now jAj < jAi+1j is impossible, since jA j = fn fn (a)  a > jAi j and hence the next pair of (Ai; Bi) is not (Ai+1 ; Bi+1) . Also jAj  jAi+1j is impossible, since jB j = fn (a) > f (Ai+1) and hence (Ai+1; Bi+1) 2= M(n) . Finally, (v) follows from (ii) and (iv), Now for all
; ?(kn ? 2kn?1 ) 
 kn ? 2kn?1 ; and
j k let us nd (A; B ) 2 M1(n) for which jB j = jAj +
. From the de nition of Fn (;  +
) it follows that for all 0  a  kn max F (;  + fn (a) ? a) = a:  n We formulate a consequence of Theorems 8, 9 and Lemma 17. 39

Theorem 10. For n = 2m (the case n = 2m + 1 is similar) 8 > > > > > > >
(A; B ) : > > > > > > :

(

jAj = t2 ? 

k (m; t;  )

jB j = (km ? t)2 ? k (m; t; ) + 

8 > >
> > > > m 2 > ; > jB j = (k ? t) + t? > = > :

k (m; t; km + 1 ) + 1 where for t; ; 1 : km?1  t  (k ? 1)km?1; k j ; 0    km; k j 1 ; ?km  1  0; (A; B ) 2 CACk (2m):

15. Explicit values of fn (a)

We return to our original question: what is fn (a) ?

Theorem 11.

(i) f2m(t2 ) = (km ? t)2 , for all 0  t  km . (ii) f2m(t2 ? t) = (km ? t)(km ? t + 1) , for all 1 < t .
? (iii) f2m (skm?1)2 ? sk = (km ? skm?1)2 + (kk?s) for all 0 < s < k ; k j  ; ?km    km .

Moreover, all pairs (A; B ) with above mentioned parameters are optimal and bisaturated.

Proof: (i) equals (i) of Theorem 3. It can also be proved inductively on m as the case

(ii) to follow. (ii) If a) km?1  t  (k ? 1)km?1 , then we apply Theorem 10. We put  = km and get

jAj = t2 ?k (m; t; km) = t2 ?t; jB j = (km ?t)2 ?k (m; t; km)+km = (km ?t)(km ?t+1) and as (A; B ) 2 M(2m) , then f2m(t2 ? t) = (km ? t)(km ? t + 1) . If b) (k ? 1)km?1 < t < km , then we abbreviate t1 = km ? t + 1 ; 1 < t1  km?1 , and consider f2m(t21 ? t1) . Now, if f2m(t21 ? t1 ) = (km ? t1 )(km ? t1 + 1) = t2 ? t , then f2m(t2 ? t) = t21 ? t1 = (km ? t)(km ? t + 1) according to (ii) of Theorem 9.

Therefore it is sucient to consider c) 1 < t < km?1 : We proceed by induction on m . m = 1 : As 1 < t < k (condition of (ii), then t is from the interval a) which is settled. 40

> > > > > > > ;

m?1!m : f2m?2(t2 ? t) = (km?1 ? t)(km?1 ? t + 1) for all 1 < t < km?1:

(15.1)

Now we use the induction hypothesis (15.1) and our recursive formula for 1 < t < km?1 : f2m(t2 ? t) = (k ? 1)k2m?1 + kf2m?2(t2 ? t) ? (k ? 1)(t2 ? t) = (k ? 1)k2m?1 + k(km?1 ? t)(km?1 ? t + 1) ? (k ? 1)(t2 ? t) = (k2 ? t)(km ? t + 1) . (iii) follows from Theorem 10. We put t = skm?1 and verify that for 0    km , kj jAj = (skm?1)2 ? k (m; skm?1; ) = (skm?1) ? sk , jB j = (km ? skm?1)2 + (kk?s) and for ?km  1  0 , m j 1 jAj = (skm?1)2 + skm?1 ? s(kmk+1 ) = (skm?1)2 ? sk1 , jB j = (km ? skm?1 )2 + 1 (kk?s) and so (iii) holds.

Remark: For n odd there are similar results. 16. An algorithm for computing fn (a)

We distinguish three cases

Case a  kn?2 :

Here we apply our recursive formula.

Case kn?2  a  (k ? 1)2 kn?2 :

We do the case n = 2m (the case n odd is simular). We nd the unique t for which t2 ? t < a  t2 + t ; km?1  t  (k ? 1)km?1 .

Subcase 1) t2 ? t < a  t2 : We nd the unique  , 0   < km , k j  for which t2 ? k (m; t;  + k) < a  t2 ? k (m; t; ) . Then f2m(a) = (km ? t)2 ? k (m; t; ) +  . Subcase 2) t2 < a  t2 + t : We nd the unique 1 , k j 1 ; ?km  1 < 0 for which t2 +t?k (m; t; km +1 +k) < a  t2 + t ? k (m; t; km + 1 ) . Then f2m(a) = (km ? t)2 + t ? k (m; t; km + 1 ) + 1 . Let us note that, as (m; t; ) is a strictly monotonically increasing function in  for any xed t ; km?1  t  (k ? 1)km?1 ; and also note that k j  . Therefore one needs to nd  by at most log2km?1 = (m ? 1) log2 k trials. 41

Case a > (k ? 1)2k2m?2 : We nd the unique t ; t  (k ? 1)km?1 for which t2 ? t < a  t2 + t . Subcase 1) t2 < a  t2 + t : According to (ii) of Theorem 11 one has

(km ? t)(km ? t ? 1)  f2m(a) < (km ? t)2: We nd the unique b ; (km ? t)(km ? t ? 1)  b < (km ? t)2 , for which

f2m(b + 1) < a  f2m(b): Then f2m(a) = b .

Subcase 2) t2 ? t < a  t2 : Here (km ? t)2  f2m(a) < (km ? t)(km ? t + 1) and we nd b1 , (km ? t)2  b1 < (km ? t)(km ? t + 1) , for which f2m(b1 + 1) < a  f2m(b1) . Then f2m(a) = b1 :

Acknowledgement

The authors thank George F. Clements for his admirable eorts in helping to improve the presentation.

42

References

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