Enumeration of certain binary vectors

Enumeration of certain binary vectors Spyros S. Magliveras  Department of Computer Science and Engineering University of Nebraska-Lincoln Lincoln, Nebraska 68588-0115 and Wandi Weiy Department of Mathematics Sichuan University Chengdu, 610064 China December 14, 1995 Abstract

In 1986, W. S. Grith [2] proposed a reliability model and described a Markov chain approach for computing the reliability in the system. F. K. Hwang and S. Papastavridis [4] gave a closed-form formula for this reliability in 1991, by counting the number of n?dimensional binary vectors containing exactly k non-overlapping m-tuples of consecutive ones. In 1988, T. M. Apostol [1] established recursive formulas and a generating function for a related problem, the enumeration of n?dimensional binary vectors containing exactly k isolated m-tuples of consecutive ones. In this paper we study two interrelated elementary enumerator functions in terms of which we are able to address the two enumeration problems mentioned above. This enables us to provide a uni ed approach to deriving closed-form formulas for both problems, and provides solutions to ner enumeration problems. We extend our results to the case of cyclic binary vectors, and brie y discuss the number of cyclic ordered partitions of a positive integer w.

This work was supported in part by NSA grant 904{93{H{3049, and by CCIS { Univ. of Nebraska Lincoln y This work was supported in part by CCIS { Univ. of Nebraska - Lincoln 

1

1 Introduction In 1986, W. S. Grith [2] proposed a reliability model in which a system consists of n independent and identically distributed components arranged in a line. Each component has probability p of being operational and probability q = 1 ? p of having failed. The system fails if and only if at least k nonoverlapping sets of m consecutive components all fail. In his paper Grithmdescribes a probabilistic approach, based on Markov chains, for computing the reliability Rk (n) of such a system. In a binary vector, an m?tuple of consecutive ones is said to be isolated if no other ones m are adjacent to the m?tuple. In 1988, T.M. Apostol [1] studied the number Ak (n) of n?dimensional binary vectors containing exactly k isolated m?tuples of consecutivemones. In his paper Apostol establishes recursive formulas and a generating function for Ak (n), moreover, he tabulates these numbers for n  25, 0  k  5, and m  5. Clearly, a closedform formula for Akm (n) is desirable. In 1991, F.K. Hwang and S. Papastavridis [4] studied a similar enumeration problem to Apostol's. Their problem arises directly from Apostol's by replacing the term isolated by non-overlapping, where two m?tuples of consecutive ones in a binary vector are said to be non-overlapping if they are disjoint. The authors give a closed-form formula for the number Bkm (n) of n?dimensional binary vectors containing exactly k non-overlapping m?tuples of consecutive ones, and also study the cyclic case. Their result directly provides a closed-form m formula to Grith's reliability Rk (n). In Section 2 of this paper we study twom related enumerator functions f and f , de ned below, which are more elementary than Ak (n) and Bkm (n). We obtain closed-form formulas for f and f , and since Akm (n) and Bkm (n) can be expressed in terms of f and f , we obtain closed-form formulas for Akm (n) and Bkm (n). Thus, we are able to obtain a closedform formula for Apostol's problem, and although Hwang and Papastavridis have provided an elegant closed-form formula for Bkm (n), our method provides a uni ed approach to obtaining such closed-form formulas, and solutions to additional ner enumeration questions. We begin with some terminology and establish our notation. For a given n?dimensional binary vector v = (a ; a ;


; an) a 1-component (resp. 0-component) is an element ai which has the value 1 (resp. 0). The number of 1-components of v is called its weight, denoted by w(v), hence (

)

(

)

(

(

(

)

)

)

(

)

(

1

(

(

)

)

(

(

)

)

(

(

1

)

1

)

)

1

2

w(v) =

n X i=1

ai:

A 1-block (resp. 0-block) of v is a segment of consecutive 1-components (resp 0-components) of maximal length. Thus, a 1-block, or a 0-block can not be extended in v, into a longer segment of consecutive 1's (resp. 0's). Let f (n; m; k; w; b) (resp. f (n; m; k; w; b)) be the number of n?dimensional binary vectors of weight w, with b 1-blocks, and k isolated (resp. non-overlapping) m?tuples of consecutive 1-components, Further, let g(n; m; k; w) (resp. g (n; m; k; w)) be the number of n?dimensional binary vectors of weight w with 1

1

2

k isolated (resp. non-overlapping) m?tuples of consecutive 1-components, and h(n; m; k; b) (resp. h (n; m; k; b)) be the number of n?dimensional binary vectors with b 1-blocks and k isolated (resp. non-overlapping) m?tuples of consecutive 1-components. Clearly, 1

n X m) Ak (n) = g(n; m; k; w) w=0

(1.1)

(

d n2 e X m ) Ak (n) = h(n; m; k; b) b=0 d n2 e X

(1.2)

(

g(n; m; k; w) = h(n; m; k; b) = Bkm (n) = (

)

Bkm (n) = (

)

w=0 n X

w=0 d n2 e X

1

h (n; m; k; w) = 1

f (n; m; k; w; b)

(1.3)

f (n; m; k; w; b)

(1.4)

g (n; m; k; w)

(1.5)

h (n; m; k; b)

(1.6)

1

b=0 ne dX 2

g (n; m; k; w) = and,

b=0 n X

b=0 n X w=0

1

f (n; m; k; w; b)

(1.7)

f (n; m; k; w; b):

(1.8)

1

1

Therefore, formulas for f (n; m; k; w; b), and f (n; m; k; w; b), will provide formulas for all g(n; m; k; w), g (n; m; k; w), h(n; m; k; b), h (n; m; k; b)), Akm (n) and Bkm (n), and consequently, nding formulas for f (n; m; k; w; b) and f (n; m; k; w; b) is a more fundamental problem than the problems solved respectively by Apostol and by F.K. Hwang and S. Papastavridis. In Section 3 we discuss the cyclic version of these problems. In Section 4, we apply a lemma established in Section 3 to develop the cyclic counterpart of the well-known formula for the number of integer solutions of the equation 1

1

(

1

)

(

)

1

x + x + : : : + xb = w 1

(1.9)

2

xi  1 (1  i  b): (1.10)   Throughout this paper the combinatorial meaning of uv is de ned only for integers u and    v, and uv = 0 if u < 0 or v < 0 or u < v, with the exception that we take ?? = 1. 1 1

3

2 Formulas for f (m; n; k; w; b) and f1(n; m; k; w; b) We rst prove the following result:

Theorem 2.1 For any given positive integers n; m; k; w; b, we have ! ! ! ! X w ? rm ? 1 b r n ? w + 1 r ? k (2.1) f (n; m; k; w; b) = (?1) k r b ? r ? 1 : b rk Proof: In general, an n?dimensional binary vector counted in the number f (n; m; k; w; b) has the pattern 0y1 1x1 0y2 1x2


0y 1x 0y +1 , that is: (0| ; :{z: : ; 0}; 1| ; :{z: : ; 1}; 0| ; :{z: :; 0}; 1| ; :{z: : ; 1};


,0| ; :{z: : ; 0}; 1| ; :{z: : ; 1}; 0| ; :{z: : ; 0}) b

y

x

1

y

1

with

b

b

x

2

yb

2

xb

yb

+1

y + y +


+ yb + yb = n ? w y  0; yi  1 (2  i  b); yb  0 1

2

and

(2.3) (2.4)

+1

1

(2.2)

+1

x + x +


+ xb = w (2.5) xi  1; (1  i  b) (2.6) where exactly k of the xi are equal to m: (2.7) Clearly, f (n; m; k; w; b) is nothing but the number of patterns (2.2) subject to (2.3)-(2.7) . Let y0 = y + 1, yi0 = yi (2  i  b) , and yb0 = yb + 1. It follows that the number of integer solutions to (2.3) subject to (2.4) is exactly the same as the number of integer solutions of (2.8) y0 + y0 +


+ yb0 + yb0 = n ? w + 2 subject to yi0  1 (1  i  b + 1): (2.9) In turn, the number of integer solutions to (2.8) subject to (2.9) is well known to be ! n?w+1 (2.10) b (for example, see [3], page 31.) When r < b, for any given r?set fp ; p ; : : : ; pr g of indices such that 1  p < p < : : : < pr  b, the number of solutions of (2.5) subject to (2.6) and subject to xp1 = xp2 = : : : = xp = m is the number of solutions of 1

1

2

1

+1

+1

1

+1

2

1

1

2

2

r

xq1 + xq2 + : : : + xq ? = w ? rm with xq  1; (1  j  b ? r) b r

j

4

(2.11)

where,

fq ; q ; : : : ; qb?r g [ fp ; p ; : : :; pr g = f1; 2; : : : ; bg: 1

2

1

(2.12)

2

Hence, by the same principle as for (2.10) this number of solutions is !

w ? rm ? 1 (2.13) b?r ?1 which is dependent on r but independent of the choice of p ; p ; : : : ; pr . When r = b and w = rm, there is a single solution and the above formula yields ! ! w ? rm ? 1 = ?1 = 1: ?1 b?r ?1 When r = b and w > rm then there are no solutions to (2.11) and formula (2.13) is still correct. By the inclusion-exclusion principle, the number of solutions of (2.5) subject to (2.6) and (2.7) is ! ! ! X b w ? rm ? 1 r r ? k (?1) k r b ? r ? 1 : 1

2

rk

Therefore, the number of vectors of (2.2) subject to (2.3) { (2.7) is ! ! ! ! n ? w + 1 X(?1)r?k r b w ? rm ? 1 k r b?r?1 b rk which is f (n; m; k; w; b). 2 We will need the following: Lemma 2.1 Let a; b; c; p; q 2 Z, such that a; b; c; p + q > 0, Then, the number of integral solutions (u ; : : : ; up; v ; : : : ; vq) to equation 1

1

u +


+ up + v +


+ vq = c 1

(2.14)

1

subject to the conditions

1  ui  a; (1  i  p) ; is

1  vj  b; (1  j  q) !

!

(2.15)

!

p q c ? sa ? tb ? 1 : (2.16) N (a; b; c; p; q) = s t p+q?1 s;t Proof: For a non-negative integral vector x = (u ; : : :; up; v ; : : :; vq ), we say that x satis es property Pi if ui > a, 1  i  p. Similarly x is said to satisfy property Qj if vj > b. To evaluate the number of solutions to equation (2.14), subject to (2.15) we need to determine the number of solutions to (2.14), satisfying none of the properties P ; : : :; Pp; Q ; : : :; Qq. Let N (Pi1 ; : : :; Pi ; Qj1 ; : : :; Qj ) be the number of solutions x to (2.14) satisfying properties Pi1 ; : : :; Pi ; Qj1 ; : : : ; Qj , and let X

0

(?1)s+t

1

1

1

s

s

t

t

5

1

X

Ns;t =

N (Pi ; : : :; Pis ; Qj1 ; : : : ; Qjt ):  i1  :: :  is  p1 1  j1  : ::  jt  q

(2.17)

1

Then, it is clear that

!

!

!

Ns;t = ps qt c ?psa+ ?q ?tb1? 1 : Hence, by the inclusion-exclusion principle the lemma follows.

(2.18)

2

Theorem 2.2 For arbitrary positive integers n, m, k, w and b, ! ! ! X b k ? 1 n ? w + 1 N (m ? 1; m; w ? km + b ? r; r; b ? r): f (n; m; k; w; b) = b r r b ? r ? 1 1

0

(2.19)

Proof: An n?dimensional binary vector counted in f (n; m; k; w; b) has pattern (2.2) subject to (2.3), (2.4) and ti + zi  1 (1  i  b) (2.20) t + t + : : : + : : :tb = k (2.21) where, xi c: xi = ti m + z i ; t i = b m (2.22) This is because there are exactly b mx c non-overlapping m?tuples of consecutive ones in a 1-block of length x. Suppose there are exactly r of the ti's that are zero, say tp1 = tp2 = : : : = tp = 0. Then, (2.21) becomes 1

1

2

r

tq1 + tq2 + : : : + tq ? = k tq  1 (1  i  b ? r)

(2.23)

b r

(2.24)

i

and,

zp1 + : : : + zp + zq1 + : : : + zq ? = w ? (tq1 + : : : + tq ? )m = w ? km (2.25) m ? 1  zp  1; (1  i  r); m ? 1  zq  0; (2.26) where p ; : : :; pr , q ; : : : ; qb?r satisfy (2.12). For a given choice of indices 1  p < p < : : : < pr  b, the number of solutions of (2.23) subject to (2.24) is ! k?1 : (2.27) b?r?1 By de ning new variables zq0 = 1+ zq , for 1  j  b ? r, we obtain a system of equations equivalent to (2.25), (2.26) as follows: zp1 + : : : + zp + zq0 1 + : : : + zq0 ? = w ? km + b ? r (2.28) r

b r

b r

i

1

j

1

1

j

j

r

b r

6

2

m ? 1  zp  1; (1  i  r); m  zq0  1; (1  j  b ? r): (2.29) Hence, the number of solutions of (2.25) subject to (2.26) is equal to: N (m ? 1; m; w ? km + b ? r; r; b ? r): (2.30) It follows that for any particular choice of p ; p ; : : : ; pr , the number of solutions of (2.20)(2.22) is ! k ? 1 N (m ? 1; m; w ? km + b ? r; r; b ? r): (2.31) b?r ?1 i

j

1

Since there are

 b r

2

ways of choosing p ; p ; : : :; pr , we obtain (2.19) by (2.10) and (2.31). 2 1

2

We illustrate the statement of Theorem (2.2) by some examples for small values of k and b.

Example 2.1 When k = b = 1, n  w  1, m  1 the expression for f (n; m; 1; w; 1) in Theorem (2.2) reduces to

1

!

!

!

!

!

X (?1)s t 0s 1t w ? m0 ? tm (n ? w + 1) 10 00 s;t ! ! ! ! X w ? m ? m w ? m w ? m ? tm 1 t ): = (n ? w + 1)( 0 ? = (n ? w + 1) (?1) t 0 0 t This evaluates to 0 if w < m or w  2m, and n ? w + 1 if m  w < 2m. This is the correct count as can be easily veri ed from the gure below. (0| ; :{z: :; 0}; 1| ; :{z: : ; 1}; 0| ; :{z: : ; 0}) y w y +

0

0

1

2

Example 2.2 When k = 3; b = 2, n  w  1, m  1 the expression for f1(n; m; 1; w; 1) in Theorem (2.2) reduces to ! ! ! ! ! X 1 w ? m ? tm 1 0 s +t 0 (?1) s t (n ? w + 1) 0 0 0 s;t0 ! ! ! ! X 1 w ? m ? tm w ? m w ? m ? m t = (n ? w + 1)( 0 ? ): = (n ? w + 1) (?1) t 0 0 t0 As analyzed in Section 1, all functions displayed in (1.1) - (1.8) now have closed-form formulas. We just exhibit the one for (1.1) since it was asked for in the literature. Theorem 2.3 If n, m, k, w, and b are arbitrary positive integers, then: d n2 e n X X m ) Ak = w=0 b=0 (

n?w+1 b

!

X

rk

(?1)r?k

7

!

!

!

r b w ? rm ? 1 : k r b?r?1

3 The cyclic case In the cyclic case, we are dealing with a binary vector v = (a ; a ; : : : ; an), where an is considered to be adjacent to a . It is clear that two elements which are distinct when viewed as linear vectors, may be equivalent under a cyclic shift. For example the two vectors (0; 0; 0; 1; 1; 1; 1; 0; 0; 0; 1; 1; 1; 1) and (1; 1; 1; 1; 0; 0; 0; 1; 1; 1; 1; 0; 0; 0) are equivalent. Let (n; m; k; w; b) (resp (n; m; k; w; b)) be functions de ned just as f and f but now for the cyclic case. For example, (n; m; k; w; b) counts the number of equivalent classes under cyclic shifts of n?dimensional binary vectors of weight w with b 1-blocks and k isolated m?tuples of 1's. Similarly, let (n; m; k; w) and (n; m; k; w)) be the cyclic analogues of g(n; m; k; w) and g (n; m; k; w) respectively, and (n; m; k; b) and  (n; m; k; b) the cyclic analogues of functions h(n; m; k; b) and h (n; m; k; b). We begin by considering (n; m; k; w; b). Since we are considering the cyclic case, we may restrict our attention to vectors of the Pb x y x y x y 1 1 2 2 form 1 0 1 0 : : : 1 0 where i (xi + yi) = n, xi  1; yi  1. These are equivalent to vectors in (2.2) where yb = 0. It is convenient to denote vector 1x1 0y1 1x2 0y2 : : : 1x 0y simply by (x ; y ; x ; y ;


; xb; yb) = [x; y] (3.1) where x = (x ; x ; : : : ; xb), y = (y ; y ; : : : ; yb) with the xi and yi positive integers. For a given xed w  n let X denote the set of all positive integer vectors x = (x ; x ; : : :; xb) satisfying x + x +


+ xb = w (3.2) and similarly let Y denote the set of all positive integer vectors y = (y ; y ; : : : ; yb) satisfying 1

2

1

1

1

1

1

1

1

b

b

=1

b

+1

1

1

1

2

1

2

b

2

2

1

1

2

2

1

2

y + y +


+ yb = n ? w: (3.3) Then, the collection of all binary vectors of the form 1x1 0y1 : : : 1x 0y , of weight w with b 1-blocks, can be viewed as the cartesian product Z = X  Y . If G is any permutation group on the symbols f1; 2; : : : ; bg de ne a group action GjX of G on X by (3.4) (x ; x ; : : : ; xb)g = (x ; x ; : : :; xb ); g 2 G: Similarly, we de ne a corresponding group action GjY . We proceed to state the following: 1

2

b

1

2

2g

1g

b

g

Lemma 3.1 Let b, w be positive integers, and z a divisor of b. Let  = (1; 2; : : :; b) and

 be any element of order z in the cyclic group generated by . If z divides w, then, the number of vectors x = (x ; x ; : : :; xb ) xed by  such that x + x + : : : + xb = w is equal to: w ? 1! z : (3.5) b z ?1 This number is 0 if z does not divide w. 1

2

1

2

Proof: Let s = b=z. Without loss of generality let  = s = (1; 1 + s; 1 + 2s; : : : ; 1 + (z ? 1)s)(2; 2 + s; 2 + 2s; : : :; 2 + (z ? 1)s) : : : (s; 2s; : : : ; zs). Then, x = (x ; x ; : : : ; xb) 2 X is xed by  if and only if xi = xj whenever i and j are in the same cycle of . If this happens 1

8

2

we say that x 2 X is constant on the cycles of . It follows that  xes (x ; : : :; xb) if and only if 1

x = x s = x s = : : : = x z? x = x s = x s = : : : = x z? . . . ::: . . . . ::: . . . . ::: . xs = x s = x s = : : : = xzs . 1

1+

1+2

1+(

2

2+

2+2

2+(

2

s 1)s

1)

(3.6)

3

That is, if and only if,

zx + zx + : : : + zxs = w (3.7) which occurs if and only if (x ; x ; : : : ; xs) is a solution to x + x + : : : + xs = w=z: (3.8) If z does not divide w, then, trivially, there are no solutions to (3.7), otherwise the number of solutions to (3.7) in X is: w ? 1! z b ?1 : z Since every element of order z in G has exactly b=z cycles of length z and no other cycles, formula (3.5) holds for any element of order z in G. 2 Let G be the cyclic group generated by  = (1; 2; : : : ; b) as above. G acts on Z = X  Y by [x; y]g = [xg; yg ] for x 2 X , y 2 Y , g 2 G. It is not dicult to see that (n; m; k; w; b) is the number of G-orbits on a certain proper subset of Z . We are now able to state the following: Theorem 3.1 If n, m, k, b, and w are arbitrary positive integers, and D = gcd (n; k; b; w), 1

1

2

2

1

then

X (n; m; k; w; b) = 1b (z) zjD

2

! ! b! w n?w ? 1! X ? rm ? 1 k r r ? z z : (?1) z k rz b b ?1 ? r ? 1 z z rk=z z

(3.9)

where the outer sum is taken over all divisors z of D, and (z ) is Euler's totient.

Proof: Let  = (1; 2; 3; : : : ; b). Then, the cyclic group G generated by  contains (z) elements of order z for each divisor z of b. The cycle decomposition of an element  of order z in G consists of exactly s = b=z cycles of length z, and no other cycles. By an application of lemma 3.1 the number of elements y = (y ; : : : ; yb) 2 Y such that y = (y ; : : : ; yb) is xed by  (3.10) and y + y + : : : + yb = n ? w (3.11) is 0 if z does not divide n ? w, and n?w ? 1! z (3.12) b z ?1 1

1

1

2

9

otherwise. We now count the number of elements of X such that:

x = (x ; x ; : : : ; xb) is xed by  1

(3.13)

2

and

exactly k of the xi are equal to m: (3.14) An element x = (x ; x ; : : : ; xb) is xed by  if and only if x is constant on the cycles of . Thus, if z does not divide k there are no x 2 X , subject to (3.14), xed by . On the other hand, if zjk, let k = tz, and select an integer r 1

2

k=z = t  r  s = b=z:

(3.15)

Suppose x = (x ; : : :; xb) is xed by  and has some r cycles of  lled with m's. Then, on the rz indices i ; i ; : : : ; irz of these cycles 1

1

2

xi1 = xi2 = : : : = xi = m:

(3.16)

rz

Moreover, the complementary indices fq ; : : :; qb?rz g = f1; 2; : : : ; bg ? fi ; : : : ; ir g fall into s ? r cycles of  on which x is constant. Let qj1 ; : : :; qj ? be a system of representatives of these s ? r cycles. Then we have 1

1

z

s r

zxq 1 + : : : + zxq

?

js r

j

= w ? rmz:

(3.17)

If z does not divide w then there are no solutions to (3.17). Otherwise, (3.17) becomes (3.18) xq 1 + : : : + xq ? = wz ? rm: In all, (3.18) can be achieved in js r

j

!

!

b=z w=z ? rm ? 1 (3.19) b=z ? r ? 1 r distinct ways. Now, by applying the inclusion-exclusion principle, the number of elements x 2 X xed by  with exactly k xi equal to m is: ! ! ! X r b=z w=z ? rm ? 1 r ? (?1) (3.20) k r b=z ? r ? 1 : z rk=z k z

Hence, if  is an element of order zjD, it xes in all ! ! ! n?w ? 1! X w=z ? rm ? 1 b=z r r ? z (?1) k b b=z ? r ? 1 r z z ? 1 rk=z k z

(3.21)

vectors v = (x ; y ; x ; y ; : : : ; xb; yb) of weight w, where exactly k of the xi are equal to m. An application of the Cauchy - Frobenius Theorem yields the conclusion of our theorem 2. 1

1

2

2

10

Next, we turn our attention to the non-overlapping cyclic case. Here, we are interested in determining the number of equivalence classes, under cyclic shifts, of binary vectors of dimension n, weight w, with b 1-blocks and k non-overlapping m-tuples of 1-components. As before, these can be viewed as positive integer vectors of the form v = (x ; y ; x ; y ; : : :; xb; yb) = [x; y], where x 2 X and y 2 Y subject to the additional conditions (2.13),(2.14), and (2.15), counted up to equivalence under cyclic shifts. Hence, we are interested in the number of [x; y] 2 Z xed by  2 hi, of order z dividing b, where  = (1; 2; : : : ; b). We have: 1

1

x = x = (x ; : : : ; xb); and y = y = (y ; : : :; yb) 1

2

2

(3.22)

1

y + y + : : : + yb = n ? w (3.23) x + x + : : : + xb = w (3.24) xi c; 0  z  m ? 1; 1  i  b (3.25) xi = ti m + zi ; ti = b m i t + t + : : : + tb = k (3.26) ti + zi  1; 1  i  b: (3.27) We state the following: Theorem 3.2 If n, m, k, b, and w are arbitrary positive integers, and D = gcd (n; k; b; w), then (n; m; k; w; b) = ! ! ! 1 X (z) (n ? w ? z)=z X zb (k ? z)=z N (m ? 1; m; s ? r +(w ? km)=z; r; s ? r) b zjD (b ? z)=z r r (b ? rz ? z)=z (3.28) where N (a; b; c; p; q) is de ned in (2.16), the outer sum is taken over all divisors z of D, and (z) is Euler's totient. Proof: The number of y 2 Y , subject to (3.23), and xed by  has already been calculated and is given by (3.12). Moreover, if x 2 X satis es (3.24)-(3.27), and x is xed by , then x is constant on the cycles of , so if we revise our notation and denote by xi the common values of x on the ith cycle of , x can be loosely denoted by x = xz xz : : :xzs , with s = b=z and xi not necessarily distinct from xj when i 6= j . Thus, certain z entries of x are equal to x , certain z entries equal to x , etc. We have: 1

2

1

2

1

2

1

0

1

1

2

zx + zx + : : : + zxs = w:

(3.29)

z(t m + z ) + z(t m + z ) + : : : + z(ts + zs) = w

(3.30)

1

That, is

1

which implies: where,

2

1

2

2

m

s X i=1

2

ti + z

s X i=1

s X i=1

zi = w=z

ti = k: 11

(3.31) (3.32)

If z does not divide k then there are no solutions to (3.32). Suppose that zjk, then, t + t + : : : + ts = k=z: (3.33) For a given r, such that k=z  r  s = b=z, suppose that exactly r of the ti are 0, say tp1 = tp2 = : : : = tp = 0. Then, on the complementary set on indices, fq ; : : :; qs?r g = f1; 2; : : : ; sg ? fp ; : : :; pr g we have: tq1 + tq2 + : : : + tq ? = k=z; with tq  1: (3.34) So, (3.31) and (3.34) imply sX ?r (3.35) zp1 + : : : + zp + zq1 + : : : + zq ? = w=z ? m tq = w ?z km j 1

2

1

r

1

s r

r

i

s r

j

=1

where,

m ? 1  zp  1; 1  i  r; and m ? 1  zq  0; 1  j  s ? r: (3.36) Now, (3.34) can be done in k ?1 ! z (3.37) b ?r?1 z ways. Moreover, by de ning a new set of variables zq0 = 1 + zq we obtain: (3.38) zp1 + : : :zp + zq0 1 + : : : + zq0 ? = w ?z km + s ? r where m ? 1  zp  1; 1  i  r; the number of solutions to (3.38) is: N (m ? 1; m; s ? r + (w ? km)=z; r; s ? r): (3.39) Therefore, for a given r and a particular choice of p ; p ; : : :; pr , the number of solutions is: ! k=z ? 1 N (m ? 1; m; s ? r + (w ? km)=z; r; s ? r): (3.40) b=z ? r ? 1 Hence, over all possible choices for p ; : : :; pr the number of solutions is ! ! s k=z ? 1 N (m ? 1; m; s ? r + (w ? km)=z; r; s ? r): (3.41) r b=z ? r ? 1 Consequently, the number of x 2 X xed by , subject to (3.24) - (3.27) is: ! ! X b ( k ? z ) =z z (3.42) (b ? rz ? z)=z N (m ? 1; m; s ? r + (w ? km)=z; r; s ? r): r r i

j

j

j

r

s r

i

1

2

1

0

Hence, putting together (3.12) and (3.42), we obtain the number of [x; y] 2 Z xed by  as: ! ! ! (k ? z)=z N (m ? 1; m; s ? r + (w ? km)=z; r; s ? r): (3.43) (n ? w ? z)=z X zb (b ? z)=z r r (b ? rz ? z)=z 0

12

Now, by an application of the Cauchy-Frobenius Theorem we obtain that (n; m; k; w; b) = 1

1 X (z) (n ? w ? z)=z b zjD (b ? z)=z

!

! X zb r0 r

!

(k ? z)=z N (m ? 1; m; s ? r +(w ? km)=z; r; s ? r): (b ? rz ? z)=z

2.

Let Akm (n) denote the cyclic analog of Akm (n). Thus, Akm (n) is the number of equivalence classes under cyclic shift of n?dimensionalmbinary vectors containing exactlym k isolated m?tuples of consecutive ones. Similarly, let Bk (n) be the cyclic analog of Bk (n). The following are straight forward: (

)

(

)

(

(

Akm (n) = (

n X

)

w=0

)

)

(

(n; m; k; w)

d n2 e X m) Ak (n) = (n; m; k; b) b=0 n dX 2e (

(n; m; k; w) = (n; m; k; b) =

Bkm (n) = (

)

Bkm (n) = (

)

w=0 n X

w=0 d n2 e X

1

 (n; m; k; w) = 1

(3.44) (3.45)

(n; m; k; w; b)

(3.46)

(n; m; k; w; b)

(3.47)

(n; m; k; w)

(3.48)

 (n; m; k; b)

(3.49)

1

b=0 ne dX 2

(n; m; k; w) = and,

b=0 n X

)

b=0

1

n X w=0

(n; m; k; w; b)

(3.50)

(n; m; k; w; b):

(3.51)

1

1

Equations (3.44) { (3.51) together with Theorems 3.1 and 3.2 allow us to give a cyclic analog of Theorem 2.3. The resulting explicit formulas for (n; m; k; w), (n; m; k; w), Akm ,

(n; m; k; w),  (n; m; k; w), and Bkm are rather cumbersome and we do not present them here. (

1

1

(

)

)

4 Number of cyclic ordered partitions It is well known that the formula for the number of integer solutions of equation (1.9) subject to (1.10) has many applications in combinatorial enumeration. These integer solutions are 13

also called ordered partitions of w with b parts. Two ordered partitions (x ; x ; : : : ; xb) and (y ; y ; : : :; yb) are considered to be cyclically equivalent if 1

1

2

2

(y ; y ; : : : ; yb) = (xi; xi ; : : :; xb; x ; : : : ; xi? ) 1

2

+1

1

1

for some i. Clearly, this relationship is an equivalence relation according to which all the integer solutions are partitioned into equivalence classes. An equivalence class is called a cyclic integer solution of (1.9) subject to (1.10), or a cyclic ordered partition of w. Let c(w; b) be the number of such cyclic solutions. Cyclic ordered partitions of w into b parts also have the following combinatorial interpretation. Let there be w points on a circle, and let b bars be placed normally to the circle so that there is at most one bar between any two adjacent points. Two such con gurations of points and bars are considered to be equivalent if rotating one produces the other. Then, the number of inequivalent such con gurations is clearly also c(w; b). By lemma 3.1 in Section 3 and the Cauchy-Frobenius Theorem we have:

Theorem 4.1 If w and b are arbitrary positive integers, and if D = gcd (w; b), then c(w; b) is given by: w ? 1! X 1 : (4.1) c(w; b) = b (z) zb z ?1 zjD Acknowledgement The authors would like to thank the anonymous referee for valuable

comments and the Journal Editor for carefully reading the manuscript and making helpful and constructive suggestions.

References [1] T. M. Apostol, Binary vectors with prescribed subsets of consecutive ones, JCT (A), 47 (1988), 176 { 190 [2] W. S. Grith, On consecutive-k-out-of-n failure systems and their generalizations, in Reliability and Quality Control, A. P. Basu (ed.), Elsevier, New York, 1986, 157-165. [3] Marshall Hall, Jr., Combinatorial Theory, Willey-Interscience in Discrete Math., John Willey & Sons, 1986 [4] F.K. Hwang and S. Papastavridis, Binary vectors with exactly k non-overlapping m?tuples of consecutive ones, Discrete Math., 30 (1991), 83 { 86.

14