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A SYMMETRIC FUNCTIONS APPROACH TO STOCKHAUSEN’S PROBLEM Lily Yen Department of Combinatorics and Optimization, University of Waterloo, Waterloo, Ontario, N2L 3G1, Canada lyen@jeeves.uwaterloo.ca Submitted July 9, 1995. Accepted February 5, 1996. We consider problems in sequence enumeration suggested by Stockhausen’s prob- lem, and derive a generating series for the number of sequences of length k on n available symbols such that adjacent symbols are distinct, the terminal symbol occurs exactly r times, and all other symbols occur at most r − 1 times. The analysis makes extensive use of tech- niques from the theory of symmetric functions. Each algebraic step is examined to obtain information for formulating a direct combinatorial construction for such sequences. 1. Introduction The score of the piano work nr. 7 Klavierst¨uck XI by Karlheinz Stockhausen (1957) [S] consists of 19 fragments of music. The performer is instructed to choose at random one of these fragments, and play it; then to choose another, different, fragment and play that, and so on. If a fragment is chosen that has already been played twice, the performance ends. We can state a more general problem as follows: Denote each fragment of music by asymbol.Thenanr-Stockhausen sequence on n symbolsisasequencesuchthat (1) adjacent symbols are distinct, (2) the terminal symbol occurs exactly r times, (3) no symbol occurs more than r times, (4) exactly one symbol occurs r times, (5) the symbol alphabet is N n = {1, 2, ,n}. The original problem posed by Stockhausen is then the case when r = 3 and n = 19. We shall refer to 3-Stockhausen sequences simply as Stockhausen sequences.Weletc r (n, k) be the number of r-Stockhausen sequences of length k on n symbols, and s r (n):= (r−1)n+1  k=2r−1 c r (n, k), The research for this paper was supported by the Natural Sciences and Engineering Research Council of Canada under a postdoctoral fellowship. Typeset by A M S-T E X 1 2 for the minimum (resp. maximum) length of an r-Stockhausen sequence is 2r − 1(resp. (r − 1)n + 1). The expected length of a performance under the assumption that each fragment is equally likely is given in [RY]. The method of generating series is a major technique in enumeration. In this case, the determination of the exact number of Stockhausen sequences on n symbols is an enumera- tive question that can be approached using the theory of symmetric functions for coefficient extraction in the generating series approach. A generalization of the problem leads to a combinatorial construction for the Stockhausen sequences. Although only known tech- niques [G] are used, these are sophisticated, and the problem serves as a useful study of these techniques, and the algebraic analysis gives partial information about the formulation of a bijective proof. The paper is organized as follows. In Section 2, we define a few classical symmetric functions, and state some properties they satisfy. Section 3 contains the derivation of the generating series for the number of Stockhausen sequences. In Section 4 we generalize the method given in Section 3 and give an analogous derivation for the generating series forthenumberofr-Stockhausen sequences. By examining the generating series for r- Stockhausen sequences, we give in Section 5 a series of combinatorial constructions that leads to r-Stockhausen sequences. 2. Symmetric Functions We review some classical symmetric functions, and recall some properties of the ring of symmetric functions. The reader is directed to [M] for a fuller account. Afunctionf(x 1 ,x 2 , ) in infinitely many indeterminates is a symmetric function if f is invariant under any permutation of any finite number of the variables. We shall consider such symmetric functions over the rationals. The following symmetric functions are used in the derivation of the generating series for c r (n, k). The complete symmetric function h n is defined by h n (x 1 ,x 2 , )=  i 1 ≤i 2 ≤···≤i n x i 1 x i 2 ···x i n . Moreover, h 0 =1. Ifλ =(λ 1 ,λ 2 , ,λ k ) is a partition, i. e. a non-increasing sequence of positive integers, we let h λ denote h λ 1 h λ 2 ···h λ k . The weight of λ is |λ| :=  i λ i . The monomial symmetric function m λ is the sum of all distinct monomials of the form x λ 1 i 1 ···x λ k i k ,wherei 1 , ,i k are distinct. The power sum symmetric function p n is defined by p n (x 1 ,x 2 , )=  i x n i . Similar to h λ , p λ is defined to be p λ 1 p λ 2 ···p λ k . It is known that [M] each of the sets {h λ }, {m λ },and{p λ },whereλ ranges over all partitions of n, is a basis of the vector space over of symmetric functions homogeneous of degree n. 3 Let m j (λ) denote the number of j’s in the partition λ. The complete symmetric function h n can be expressed in terms of the power sum symmetric functions as follows: h n =  λn z −1 (λ)p λ , where z(λ)=   j≥1 j m j (λ) m j (λ)!  and λ  n indicates that λ is a partition of n. For instance, h 1 = p 1 , h 2 =(p 2 1 + p 2 )/2, and h 3 =(p 3 1 +3p 2 p 1 +2p 3 )/6. There is [M] an inner product ·, · defined on the vector space of symmetric functions such that m λ ,h µ  = δ λ,µ ,sop λ ,p µ  = z(λ)δ λ,µ for any two partitions λ and µ,where δ λ,µ is 1 if λ = µ and zero otherwise. It follows that if f and g are symmetric functions, then p n f,g = f,n ∂g ∂p n , so the action adjoint to multiplication by p n is n ∂ ∂p n . For any partition λ,letl(λ) denote the number of parts of λ. Then the number of monomials on n (≥ l(λ)) symbols of the form x λ 1 i 1 ···x λ l(λ) i l(λ) ,wherei 1 , ,i l(λ) are distinct, is (2.1) n! (n − l(λ))! m 1 (λ)! m 2 (λ)! ···m λ 1 (λ)! = m λ (1, ,1    n 1’s , 0, )= (n) l(λ) m 1 (λ)! m 2 (λ)! ···m λ 1 (λ)! , where (n) j = n(n − 1) ···(n − j + 1) denotes the jth falling factorial. The following proposition is needed and can be regarded as the adjoint form of Taylor’s Theorem on the ring of symmetric functions. Proposition 2.1. Let f and g be symmetric functions and u arealnumber,andsuppose that f = f(p 1 ,p 2 , ).Then f(p 1 ,p 2 , ),e up j g = f(p 1 ,p 2 , ,p j−1 ,p j + ju,p j+1 ,p j+2 , ),g. Proof. We use the adjoint action to multiplication by p j , f(p 1 ,p 2 , ),e up j g =  k≥0 u k k!  j k ∂ k ∂p k j f(p 1 ,p 2 , ),g  =   k≥0 (uj) k k! ∂ k ∂p k j f(p 1 ,p 2 , ),g  =  exp  ju ∂ ∂p j  f(p 1 ,p 2 , ),g  = f(p 1 ,p 2 , ,p j + ju, ),g using the formal version of Taylor’s Theorem. 4 3. A solution for the original problem We derive a generating series for the number c 3 (n, k) of Stockhausen sequences on n symbols of length k. To reach this goal, we begin with the generating series for sequences satisfying condition (1) for Stockhausen sequences (Lemma 3.1), then we derive the gener- ating series of sequences satisfying conditions (1) and (2) for r = 3 (Proposition 3.2); and finally we address the generating series for Stockhausen sequences by imposing appropriate restrictions in order to satisfy conditions (3) and (4). Lemma 3.1. The generating series for all strings on n symbols such that adjacent symbols are different is D(z,x 1 , ,x n )= 1 1 −  n i=1 zx i 1+zx i , where z is an ordinary marker for the length of the string and x i marks the occurrence of symbol i. See [RY,§2] for a proof. Proposition 3.2. Let D n (z, x 1 , ,x n ) be the generating series of sequences on the sym- bols {1, 2, ,n} (marked by x 1 , ,x n ) of length k (marked by z) such that the terminal symbol occurs exactly three times (non-terminal symbols occur arbitrarily many times) and adjacent symbols are distinct. Then D n (z,x 1 , ,x n )=z 3      n  j=1 x 3 j   n i=1 i=j zx i 1+zx i  2  1 −  n i=1 i=j zx i 1+zx i  3      . Proof. Consider such a sequence which terminates with the symbol j.Thesequencede- composes into AjBjB  j,whereA, B,andB  are strings on the symbols {1, 2, ,n}\{j} with distinct adjacent elements, A possibly empty but B and B  non-empty. By Lemma 3.1, the generating series for all such sequences is z 3 x 3 j L w 1 1 − w  n i=1 i=j zx i 1+zx i , where L w is the linear transform defined by L w f(w)= 1 2! ∂ 2 f ∂w 2     w=1 . Although we introduce L w here to simplify the expression for the generating series, it will be seen later that it has combinatorial significance. To get the sequences described in the statement, we take the union over j, and so, summing the above generating series over j we obtain D n = z 3 n  j=1 x 3 j L w 1 1 − w  n i=1 i=j zx i 1+zx i . 5 We remark that a very easy direct proof is possible, but L w is introduced in order to illustrate how its counterpart, L (r) w , will be used in Section 4, where for r>3, this linear transform helps in carrying out the calculations with symmetric functions. To restrict the frequency of occurrence of symbols, we let P 3 be the set of all partitions with exactly one 3 as the largest part. For α in P 3 ,letx α denote x α 1 1 x α 2 2 ··· and let [x α ]f denote the coefficient of x α in f;then [x α ]D n =[x α ]z 3 p 3 L w 1 1 − w(zp 1 − z 2 p 2 ) forthepowersumsp 1 ,p 2 ,p 3 in x 1 ,x 2 , ,x n . Let (3.1) G(z,x 1 , ,x n )=z 3 p 3 L w 1 1 − w(zp 1 − z 2 p 2 ) , where p 1 , p 2 ,andp 3 are power sum symmetric functions in infinitely many indeterminates. Then [x α ]D n =[x α ]G(z,x 1 ,x 2 , ,x n , 0, 0, ). Therefore the generating series for the set of all Stockhausen sequences on n symbols is F n (z,x 1 , ,x n )=  α∈P 3  β [x β ]G(z, x 1 ,x 2 , ,x n , 0, 0, ), where the inner sum is over all distinct permutations β of α. The sequences defined in the statement of Proposition 3.2 satisfy conditions (1) and (2). Conditions (3) and (4) are imposed by the restriction that all partitions α are in P 3 . The last condition is imposed by evaluating G with x n+1 = x n+2 = ··· = 0 to exclude n +1,n+2, from the alphabet, and by the sum over β so that each element of the alphabet N n is permitted to occur. Since G is a symmetric function in the x i ’s, we may expand it in terms of the monomial symmetric functions: G =  θ∈P 3 m θ (x 1 ,x 2 , )a θ z |θ| , where the a θ ’s are scalars. Then another way of expressing F n is F n =  α∈P 3 m α (1, 1, ,1    n , 0, 0, )[m α ]G =  α∈P 3 (n) l(α) m 1 (α)! m 2 (α)! a α z |α| , because from (2.1) m α (1, 1, ,1    n , 0, 0, ) is the number of terms in the monomial symmet- ric function m α on x 1 ,x 2 , ,x n , which is equal to the number of sequences over {0, 1, 2, 3} of length n with n − l(α) occurrences of 0, m 1 (α) occurrences of 1, m 2 (α) occurrences of 2 and one occurrence of 3 (as specified by P 3 ). 6 We use the property m α ,h θ  = δ α,θ of the inner product to extract the coefficient of m α in G.Inordertohavem 1 (α)! and m 2 (α)! in the denominator and one part of size 3 in α, we consider uh 3 e u(h 1 +h 2 ) =  α∈P 3 h α u l(α) m 1 (α)! m 2 (α)! . Clearly,  G, uh 3 e u(h 1 +h 2 )  =  α∈P 3 u l(α) m 1 (α)! m 2 (α)! a α z |α| . With the help of the mapping Ξ n : u j → (n) j ,forj =0, 1, 2, , extended linearly to the power series ring in u, we obtain F n =  k≥0 c 3 (n, k)z k =Ξ n  G, uh 3 e u(h 1 +h 2 )  . We are now in a position to work on Φ 3 (z,u):=  n F n (z)u n /n!, our goal for this section. Lemma 3.3. Let c be independent of u.ThenΞ n e uc u k =(n) k (1 + c) n−k . If, moreover, g(u)=1+g 1 u + g 2 u 2 + ··· is a power series in u,thenΞ n e uc g(u)=  u n n!  e u(1+c) g(u). Proof. The first statement follows from direct computation. Now Ξ n e uc g(u)=Ξ n  k≥0 e uc g k u k =  k≥0 g k · (n) k (1 + c) n−k = n  k=0 g k n! (n−k)! (1 + c) n−k , and the second statement follows. Note that (n) k =0ifn<k. It now follows that the generating series for c 3 (n, k)is Corollary 3.4. Φ 3 (z,u):=  k,n≥0 c 3 (n, k)z k u n n! = e u G, uh 3 e u(h 1 +h 2 ) . Proof. We know that  k≥0 c 3 (n, k)z k = G, Ξ n uh 3 e u(h 1 +h 2 )  = G,  u n n!  e u uh 3 e u(h 1 +h 2 )  by Lemma 3.3. Since G is independent of u, the last expression is  u n n!  G, e u uh 3 e u(h 1 +h 2 )  =  u n n!  e u G, uh 3 e u(h 1 +h 2 ) . 7 It follows that Φ 3 (z,u)=e u G, uh 3 e u(h 1 +h 2 ) . To evaluate the inner product, we expand G and uh 3 e u(h 1 +h 2 ) in terms of the power sums, and use the property p λ ,p µ  = z(λ)δ λ,µ of the inner product, the adjoint action to multiplication by p j , and Taylor’s Theorem on the ring of symmetric functions. We know that upon substitution of G using (3.1) Φ 3 (z, u)=e u  L w z 3 p 3 1 − w(zp 1 − z 2 p 2 ) , u 6 (p 3 1 +3p 2 p 1 +2p 3 )e u(p 1 +(p 2 1 +p 2 )/2)  by linearity of ,  and ignoring terms in the second argument that are independent of p 3 , we get = L w e u  z 3 p 3 1 − w(zp 1 − z 2 p 2 ) , u 3 p 3 e u(p 1 +(p 2 1 +p 2 )/2)  by adjoint action of p 3 ,wehave = L w e u  z 3 1 − w(zp 1 − z 2 p 2 ) ,ue u(p 1 +(p 2 1 +p 2 )/2)  using Proposition 2.1 we obtain = L w e u  z 3 1 − w(z(p 1 + u) − z 2 (p 2 + u)) ,ue up 2 1 /2  again by the property of the inner product we ignore terms in the first argument indepen- dent of p 1 and reach = L w e u  z 3 1 − wzu + wz 2 u − wzp 1 ,ue up 2 1 /2  . Since 1 1 − wzu + wz 2 u − wzp 1 =  j≥0 (wzp 1 ) j (1 − wzu + wz 2 u) j+1 , 8 we have that Φ 3 (z,u)=z 3 ue u L w  j≥0 (wz) j (1 − wzu + wz 2 u) j+1  p j 1 ,e up 2 1 /2  = z 3 ue u L w  j≥0 (wz) 2j (1 − wzu + wz 2 u) 2j+1  p 2j 1 ,e up 2 1 /2  = z 3 ue u L w  j≥0 (wz) 2j (1 − wzu + wz 2 u) 2j+1  p 2j 1 ,u j p 2j 1 2 j j!  = z 3 ue u L w  j≥0 u j (wz) 2j (1 − wzu + wz 2 u) 2j+1 (2j)! 2 j j! . We summarize the result in the following. Proposition 3.5. Let L (r) w (f(w)) = 1 r! ∂ r f ∂w r    w=1 ,then Φ 3 (z,u)=z 3 ue u L w  j≥0 u j (wz) 2j (1 − wzu + wz 2 u) 2j+1 (2j)! 2 j j! . We remark that the original Stockhausen number s 3 (n)is  u n n!  Φ 3 (1,u)=  u n n!  1 2 ue u  j≥0 (2j)(2j − 1) (2j)! 2 j j! u j , so s 3 (n)=n n−1  j=0  n − 1 j  2j 2  (2j)! 2 j . A direct combinatorial proof of this expression is given in [RY]. 4. A generating series for r-Stockhausen sequences We use the method suggested by the previous section to derive a generating series for r-Stockhausen sequences. Theorem 4.1. Let Φ r (z,u):=  n,k≥0 c r (n, k)z k u n n! be the generating series for the number c r (n, k) of sequences of length k on n symbols such that adjacent symbols are distinct, the terminal symbol occurs exactly r times, and all other symbols occur at most r − 1 times. Then Φ r (z,u)=uz r L (r−1) w Θ w exp  u[t r−1 ] e wzt/(1+zt) 1 − t  , 9 where z is the ordinary marker for the length of the sequence, u is the exponential marker for the number of available symbols, L (r) w (f(w)) = 1 r! ∂ r f ∂w r    w=1 and Θ w is the inverse Laplace transform Θ w : w j → j! w j . We may take advantage of the analysis of the original problem (when r =3). Proof. Let F n,r := z r p r   j≥1 (−1) j−1 z j p j  r−1  1 −  j≥1 (−1) j−1 z j p j  r . Then c r (n, k)=[z k ]Ξ n  F n,r ,uh r e u(h 1 +h 2 +···+h r−1 )  . Now F n,r = z r L (r−1) w p r  1 − w  j≥1 (−1) j−1 z j p j  −1 , and  k≥0 c r (n, k)z k =Ξ n z r L (r−1) w  p r  1 − w  j≥1 (−1) j−1 z j p j  −1 ,uh r e u(h 1 +h 2 +···+h r−1 )  =  u n n!  z r L (r−1) w  p r  1 − w  j≥1 (−1) j−1 z j p j  −1 ,uh r e u(1+h 1 +h 2 +···+h r−1 )  =  u n n!  Φ r (z, u), by definition. So (4.1) Φ r (z, u)=uz r L (r−1) w  p r  1 − w  j≥1 (−1) j−1 z j p j  −1 ,h r e u(1+h 1 +h 2 +···+h r−1 )  . In order to extract the coefficients from (4.1), we perform some technical maneuvers using the properties of symmetric functions stated in Section 2. It is convenient for expository purposes to isolate these into a series of steps. Step 1. Apply the adjoint action to multiplication by p j to remove p j from the first argument of the inner product. Since h r =  α∈P z −1 (α)p α ,wherez(α)=1 m 1 (α) m 1 (α)! 2 m 2 (α) m 2 (α)! ···, it follows that ∂h r ∂p r = 1 r .Also,e u(1+h 1 +···+h r−1 ) expanded in terms of the power sums using the formula above is independent of p r ,so Φ r = uz r L (r−1) w    1 − w  j≥1 (−1) j−1 z j p j   −1 ,e u(1+h 1 +···+h r−1 )  = uz r L (r−1) w  1,   1 − w  j≥1 (−1) j−1 z j j ∂ ∂p j   −1 e u(1+h 1 +···+h r−1 )  . 10 Step 2. Introduce the mapping Θ w : w j → j! w j to express   1 − w  j≥1 (−1) j−1 z j j ∂ ∂p j   −1 as an exponential function, and h j ’s in terms of the power sums so that the ring theoretic Taylor’s Theorem can be used. Since  1 − w  j≥1 (−1) j−1 z j p j  −1 =Θ w e w j≥1 (−1) j−1 z j p j , Φ r = uz r L (r−1) w Θ w  1,e w j≥1 (−1) j−1 z j j ∂ ∂p j e u(1+h 1 +···+h r−1 )  The generating series for the complete symmetric functions is  k≥0 h k t k =  ∞ i=1 (1 − tx i ) −1 ,so 1+h 1 + ···+ h r−1 =[t r−1 ] 1 1 − t ∞  i=0 (1 − tx i ) −1 =[t r−1 ] 1 1 − t exp  i≥1 ln(1 − tx i ) −1 =[t r−1 ] 1 1 − t e p 1 t+p 2 t 2 +··· . Thus e u(1+h 1 +···+h r−1 ) =exp  u[t r−1 ] e p 1 t+p 2 t 2 2 +··· 1−t  . Step 3. Apply Taylor’s Theorem to get Φ r = uz r L (r−1) w Θ w  1,e w j≥1 (−1) j−1 z j j ∂ ∂p j exp  u[t r−1 ] e p 1 t+p 2 t 2 2 +··· 1 − t  = uz r L (r−1) w Θ w  1, exp  u[t r−1 ] e (wt+p 1 )+(−2wz 2 +p 2 ) t 2 2 +··· 1 − t  = uz r L (r−1) w Θ w  1, exp  u[t r−1 ] e wzt−wz 2 t 2 +··· 1 − t  since the inner product is with 1, all p i , i ≥ 1, are set to 0 in the second argument of the inner product. We conclude that Φ r = uz r L (r−1) w Θ w exp  u[t r−1 ] e wzt/(1+zt) 1−t  . We remark that the exponential e wzt/(1+zt) is the generating function for Laguerre polynomials. A short proof of the r-Stockhausen problem that is inspired by this analysis isgivenin[RY,§5]. An explicit formula derived from this expression for the number of r-Stockhausen sequences and a direct combinatorial proof are given in [Y]. [...]... the alphabet, and by the sum over β so that each element of the alphabet Nn is permitted to occur Since G is a symmetric function in the xi ’s, we may expand it in terms of the monomial symmetric functions: G= mθ (x1 , x2 , )a z |θ| , θ∈P3 where the a ’s are scalars Then another way of expressing Fn is Fn = mα (1, 1, , 1, 0, 0, )[mα ]G = α∈P3 n α∈P3 (n)l(α) a z |α| , m1 (α)! m2 (α)! because... for the length of the string and xi marks the occurrence of symbol i See [RY,§2] for a proof Proposition 3.2 Let Dn (z, x1 , , xn ) be the generating series of sequences on the symbols {1, 2, , n} (marked by x1 , , xn ) of length k (marked by z) such that the terminal symbol occurs exactly three times (non-terminal symbols occur arbitrarily many times) and adjacent symbols are distinct Then... sequences satisfying conditions (1) and (2) for r = 3 (Proposition 3.2); and finally we address the generating series for Stockhausen sequences by imposing appropriate restrictions in order to satisfy conditions (3) and (4) Lemma 3.1 The generating series for all strings on n symbols such that adjacent symbols are different is 1 D(z, x1 , , xn ) = n zxi , 1 − i=1 1+zxi where z is an ordinary marker for... set of all Stockhausen sequences on n symbols is [xβ ]G(z, x1 , x2 , , xn , 0, 0, ), Fn (z, x1 , , xn ) = α∈P3 β where the inner sum is over all distinct permutations β of α The sequences defined in the statement of Proposition 3.2 satisfy conditions (1) and (2) Conditions (3) and (4) are imposed by the restriction that all partitions α are in P3 The last condition is imposed by evaluating... functions such that mλ , hµ = δλ,µ , so pλ , pµ = z(λ)δλ,µ for any two partitions λ and µ, where δλ,µ is 1 if λ = µ and zero otherwise It follows that if f and g are symmetric functions, ∂g ∂ then pn f, g = f, n ∂pn , so the action adjoint to multiplication by pn is n ∂pn For any partition λ, let l(λ) denote the number of parts of λ Then the number of λl(λ) monomials on n (≥ l(λ)) symbols of the form xλ11... formal version of Taylor’s Theorem
            4 3 A solution for the original problem We derive a generating series for the number c3 (n, k) of Stockhausen sequences on n symbols of length k To reach this goal, we begin with the generating series for sequences satisfying condition (1) for Stockhausen sequences (Lemma 3.1), then we derive the generating series of. .. remark that a very easy direct proof is possible, but Lw is introduced in order to (r) illustrate how its counterpart, Lw , will be used in Section 4, where for r > 3, this linear transform helps in carrying out the calculations with symmetric functions To restrict the frequency of occurrence of symbols, we let P3 be the set of all partitions with exactly one 3 as the largest part For α in P3 , let... Although we introduce Lw here to simplify the expression for the generating series, it will be seen later that it has combinatorial significance To get the sequences described in the statement, we take the union over j, and so, summing the above generating series over j we obtain n 1 Dn = z 3 x3 Lw j zxi
1 − w n 1+zxi i=1 j=1 i=j             5 We remark that a. .. the number of j’s in the partition λ The complete symmetric function hn can be expressed in terms of the power sum symmetric functions as follows: z −1 (λ)pλ , hn = λ n where j mj (λ) mj (λ)! z(λ) = j≥1 and λ n indicates that λ is a partition of n For instance, h1 = p1 , h2 = (p2 + p2 )/2, and 1 h3 = (p3 + 3p2 p1 + 2p3 )/6 1 There is [M] an inner product ·, · defined on the vector space of symmetric... i=j 1−   3  Proof Consider such a sequence which terminates with the symbol j The sequence decomposes into AjBjB j, where A, B, and B are strings on the symbols {1, 2, , n} \ {j} with distinct adjacent elements, A possibly empty but B and B non-empty By Lemma 3.1, the generating series for all such sequences is z 3 x3 Lw j 1 1−w n zxi i=1 1+zxi i=j , where Lw is the linear transform defined by . are used, these are sophisticated, and the problem serves as a useful study of these techniques, and the algebraic analysis gives partial information about the formulation of a bijective proof. The. to choose another, different, fragment and play that, and so on. If a fragment is chosen that has already been played twice, the performance ends. We can state a more general problem as follows:. A SYMMETRIC FUNCTIONS APPROACH TO STOCKHAUSEN’S PROBLEM Lily Yen Department of Combinatorics and Optimization, University of Waterloo, Waterloo, Ontario, N2L 3G1, Canada lyen@jeeves.uwaterloo.ca Submitted