Graphical Basis Partitions. North Carolina State University. Abstract

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1 Graphical Basis Partitions Jennifer M Nolan jmnolan@eosncsuedu Carla D Savage y cds@cayleycscncsuedu Vijay Sivaraman vsivara@eosncsuedu Pranav KTiwari pktiwari@eosncsuedu Department of Computer Science North Carolina State University Raleigh, NC Abstract Apartitionofaninteger n is graphical if it is the degree sequence of a simple, undirected graph It is an open question whether the fraction of partitions of n which are graphical approaches 0asnapproaches innity A partition is basic if the number of dots in its Ferrers graph is minimum among all partitions with the same rank vector as In this paper, we investigate graphical partitions via basis partitions We showhow to eciently count and generate graphical basis partitions and how to use them to count graphical partitions We give empirical evidence which leads us to conjecture that, as n approaches innity, the fraction of basis partitions of n which are graphical approaches the same limit as the fraction of all partitions of n which are graphical 1 Introduction A partition of a non-negative integer n is a sequence of positive integers =( 1 2 ::: s ) satisfying 1 2 ::: s and :::+ s = n Let P (n) be the set of partitions of n, where P (0) contains only the empty partition, and let p(n) =jp (n)j The partition 2 P (n) is said to be graphical if there eists a simple undirected graph with degree sequence Since the sum of the degrees of the vertices of a graph equals twice the number of edges, a necessary condition for 2 P (n) to be graphical is that n is even (For convenience, we consider to be graphical) Let G(n) denote the set of graphical partitions of an even integer n, and let g(n) = jg(n)j It is an open question, originally y Supported by the National Science Foundation through the CRA Distributed Mentor Project, 1995 Supported in part by National Science Foundation Grant No DMS

2 posed by H Wilf, whether lim n!1 g(n)=p(n) = 0 The best upper bound known is that lim n!1 g(n)=p(n) 0:25, due to Rousseau and Ali 11] Recent methods for eciently counting and generating G(n) 2, 3] allow g(n)=p(n) to be tabulated, but so far these methods have given no insight into the limiting behavior of the ratio Several necessary and sucient conditions to determine whether an integer sequence ( 1 ::: s ) is graphical have been proposed in the literature Seven such criteria have been listed and shown to be equivalent in13] Among these, the most well known is the Erd}os-Gallai condition stated below: kx i k(k ; 1) + nx j=k+1 min(k j ) k =1 ::: s: (1) For a proof, see 8], pp A lesser-known condition is the Nash-Williams condition, which works with the rank vector of the partition For a partition =( 1 ::: s ), the associated Ferrers diagram is an arrayofs rows of dots, where row i has i dots and rows are left justied The conjugate partition of is denoted by 0 =( 0 1 ::: 0 t), where t = 1 and 0 i is the number of dots in the i-th column of the Ferrers diagram of The Durfee square of is the largest square sub-array of dots in the Ferrers diagram of Let d denote the length of a side of the Durfee square The rank vector of, dened in 7], is the vector r =r 1 r 2 ::: r d ] whose entries r i = i ; 0 i are the successive ranks of Atkin 1] The Nash-Williams condition, necessary and sucient for to be graphical, is: kx (r i +1) 0 k =1 ::: d: (2) This condition is shown in 11] to be equivalent to the Erd}os-Gallai condition Since the Nash-Williams condition uses only the rank vector of a partition to determine whether the partition is graphical, it becomes natural to consider families of partitions dened by their rank vectors As an eample, let R(n) be the set of partitions of n for which all rank vector entries are negative It was noted by Erd}os and Richmond 6] that all partitions in R(n) are graphical and hence g(n) r(n) =jr(n)j: They observe that r(n) =p(n) ; p(n ; 1) from a result of Bressoud 5] and that (p(n) ; p(n ; 1))=p(n) p = 6n, from Roth and Szekeres 10], to 2

3 conclude that lim n!1 p ng(n) p(n) p 6 : Although every partition has a unique rank vector, the converse it not true For eample, the partitions ( ) and ( ) both have rankvector 1 0 ;1] Gupta, in 7], gives a one-to-one correspondence between rank vectors and a subclass of partitions which he calls basis partitions Letr() be the rank vector of =( 1 ::: s ) and let the weight of be jj = 1 + :::+ s A partition 2 P (n) isbasic i jj = minf j 0 j j r( 0 )=r()g: Informally, is basic if and only if the weight of is minimum over all partitions with the same rank vector as The set B(n) of basis partitions is the set of all partitions of n which are basic For a partition with rank vector r() =r 1 ::: r d ], dene the co-rank vector of, c() =c 1 ::: c d ]by c i = ;r i for 1 i d Then the Nash-Williams condition can be restated as kx (c i ; 1) 0 k =1 ::: d: (3) In Section 2, we survey some results on basis partitions In Section 3, we develop a recurrence for counting graphical basis partitions and compare the fraction of basis partitions of n which are graphical to the fraction of all partitions of n which are graphical In Section 4, we present an algorithm to generate graphical basis partitions The algorithm requires only constant amortized time per partition In fact, this algorithm, without the test for graphical, is the rst constant amortized time algorithm we know of for generating basis partitions Suggestions for further research follow in Section 5 2 Results on Basis Partitions We include here only results on basis partitions which will be required in this paper For further information on basis partitions, including proofs omitted in this section, see 7, 9] We focus rst on the eistence and uniqueness of basis partitions 3

4 Theorem 1 Gupta] Among all partitions with the same rank vector r =r 1 ::: r k ], there is just one with minimum weight The following simple test will determine whether a partition is basic Lemma 1 Apartition with Durfee square ofsized is basic if and only if both d = d or 0 d = d and (4) for 1 i<d : i = i+1 or 0 i = 0 i+1 : (5) Gupta 7] notes the following bijection, where p(n k) denotes the total number of partitions of n into parts of size at most k Theorem 2 Gupta] Let r =r 1 ::: r d ] and let be thebasis partition of r The number of partitions of n with rank vector r is p(m d) where m =(n ;jj)=2 Let B(n d) be the set of basic partitions of n which have a rank vector of length d and let b(n d)=jb(n d)j A partition can be classied according to the length of its rank vector r and the weight n 0 of the basis partition associated with r Combining this with Theorem 2 gives the following Corollary 1 We have p(n) = nx nx d=0 n 0 =0 b(n 0 d)p((n ; n 0 )=2 d) where p(n d)=0if n is not an integer A recurrence for counting B(n d) is given in 9]: Theorem 3 The number b(n d) of basis partitions of n with Durfee square ofsized is: 1, if n = d =0 otherwise, 0, if n 0 or d 0 and otherwise, b(n d)=b(n ; d d)+b(n ; 2d +1 d; 1) + b(n ; 3d +1 d; 1): 4

5 r = (2,1,2) r = (0,-1) Figure 1: Deletion of dots in boes of original partition results in a smaller basis partition It has been shown in 7] thatforagiven rank vector, the corresponding basis partition is easy to construct However, in Section 4, given a rank vector, we will need an eplicit formula for the weight of its basis partition in terms of the rank vector elements We now state and prove such a result: Theorem 4 For a rank vector r =r 1 r 2 ::: r d ], the corresponding basis partition has weight jj = n(r) where X d;1 n(r) =d 2 + djr d j + ijr i ; r i+1 j: (6) Proof Let be the basis partition corresponding to the given rank vector r Assume that r d 0 (the other case, r d < 0 can be handled similarly) Then by (4) above, d = d + r d and 0 d = d Further, by (5) above, either d;1 = d + r d or 0 d;1 = d Nowwe delete the dots in row d and column d of the Durfee square (a total of 2d ; 1 dots deleted), as also r d dots from each i for i d (another dr d deleted) In the new Ferrers diagram (see Figure (1)), either row d ; 1 or column d ; 1 has eactly d ; 1 dots Thus, the new Ferrers diagram satises both properties (4) and (5) above withd ; 1 instead of d, and is therefore a basis for the rank vector r 0 =r 1 ; r d r 2 ; r d ::: r d;1 ; r d ] This gives us a recursive formula for the number of dots in the Ferrers diagram: n(r) =2d ; 1+djr d j + n(r 0 ) (7) which holds similarly in the case that r d < 0 It is easy to verify that the epression in (6) satises this recurrence The recurrence holds for d 1 For d = 0, n(r) is dened to be zero 2 5

6 3 Counting Graphical Basis Partitions Let H(n) be the set of basis partitions of n which are graphical and let H(n d) be the set of graphical basis partitions of n with Durfee square of size d Denote the size of these sets by h(n) andh(n d), respectively By the Nash-Williams condition (2), is a graphical partition of n 0 with rank vector r if and only if any partition of n with rank vector r is graphical By Theorem 2, the number of such partitions of n is p((n ; n 0 )=2 d) where d is the length of r Thus classifying graphical partitions of n according to the length d of their rank vector and the weight of the corresponding basis partition gives g(n)= d=n X d=0 g(n d)= nx nx d=0 n 0 =0 h(n 0 d)p((n ; n 0 )=2 d): Since p(n k) is easy to compute, a fast algorithm for computing h(n d) can be used for ecient computation of g(n) Let H(n d t s) be the set of basic partitions of n with Durfee square of size d whose co-rank vector c =c 1 ::: c d ] satises: and kx (c i ; 1 ; t) 0 k =1 ::: d (8) dx Lemma 2 For even n, H(n d 0 0)=H(n d) (c i ; 1 ; t) s: (9) Proof When s = 0, condition (9) is implied by condition (8) and when t = 0, the condition (8) is the Nash-Williams criterion (3) for graphical partitions 2 Let h(n d t s) denote the size of H(n d t s) The recurrence below allows h(n d t s) to be computed within time polynomial in n Theorem 5 For integers n d t s with s n d 0, h(n d t s) can be denedrecursively as follows If n<d 2 : h(n d t s)=0: 6

7 - - - d d d d-1 d-1 d-1 i i a b c B B B B B B A A A A A A Figure 2: The mapping for cases (a), (b), and (c) in the proof of Theorem 5 7

8 If n = d 2 : If n>d=1: h(n d t s)= h(n d t s)= ( 1 if t ;1 and s ;d(1 + t) 0 otherwise 8 >< >: 2 if n + t + s 0 1 if ;n + t + s otherwise If n>d 2 > 1: h(n d t s)= b(n;d X 2 )=dc j=;b(n;d 2 )=dc h(n ; d(2 + jjj)+1 d; 1 t+ j maf0 s+ t + j +1g): Proof There are no partitions of n with Durfee square of size d if n<d 2 If n = d 2, the unique partition of n with Durfee square of size d has co-rank vector c =0 ::: 0], so conditions (8) and (9) become 0 and kx (c i ; 1 ; t) =;k(1 + t) k =1 ::: d s dx (c i ; 1 ; t) =;d(1 + t): These are satised i t ;1ands;d(1 + t): The only basis partitions of n with Durfee square of size 1 are =(n) and its conjugate 0 =(1 1 ::: 1) For, the co-rank vector is c() =1; n] so conditions (8) and (9) become 0 c 1 ; 1 ; t = 1; n ; 1 ; t and s c 1 ; 1 ; t = 1; n ; 1 ; t which are both satised i n + t + s 0, since s 0 For 0, the co-rank vector is c( 0 )=n; 1] so conditions (8) and (9) become n ; 1 ; 1 ; t 0andn ; 1 ; 1 ; t s, which are both satised i ; n + t + s +2 0 (10) 8

9 since s 0 Note that if n + t + s 0, then since n 1, ;n + t + s +2=(n + t + s) ; 2(n ; 1) ;2(n ; 1) 0 so that if 2 H(n 1 s t) then 0 2 H(n 1 s t) and therefore h(n 1 s t) = 2 in this case Otherwise, H(n 1 s t) contains only 0 if (10) holds and is empty if (10) does not hold If n>d 2 > 1, assume inductively that the theorem holds for any (n 0 d 0 t 0 s 0 ) satisfying the hypotheses of the theorem, with n 0 <n Let =( 1 ::: s ) 2 H(n d t s) Since is basic, by (4) either d = d or 0 d = d So falls into one of the three mutually eclusive cases (a) d = 0 d = d, (b) d >d, or (c) 0 d >d Dene () =, where is the partition obtained from the Ferrers diagram of as follows: (a) If d = 0 d = d: delete row d and column d of (See Figure 2(a)) (b) If d >d:deleterowdand columns d through d of (See Figure 2(b)) (c) If 0 d >d: delete rows d through 0 d and column d of (See Figure 2(c)) Note that in all cases, has Durfee square of size d ; 1 Further, in case (a), since is basic, either d;1 = d or 0 d;1 = d, so that the resulting is basic Similarly, in (b) and (c), either d;1 = d or 0 = d;1 0 d, so that d;1 = d ; 1or 0 = d ; 1, so will be basic d;1 Let c =c 1 ::: c d ] be the co-rank vector of Weshownowhowtocount the partitions which fallinto each of the three cases For case (a), () = 2 B(n ; 2d +1 d; 1) and c() =c 1 ::: c d;1 0] and c() = c 1 ::: c d;1 ] Note that 2 H(n d t s) if and only if satises conditions (8) and (9) for s and t and this occurs if and only if satises and d;1 X kx (c i ; 1 ; t) 0 k =1 ::: d; 1 (c i ; 1 ; t) s ; c d +1+t s +1+t that is, if and only if 2 H(n ; 2d +1 d; 1 t maf0 s+ t +1g) For case (b), let j = d ; d Then () = 2 B(n ; d(2 + j)+1 d; 1) and c() = c 1 c 2 ::: c d;1 ;j]andc() =c 1 + j c 2 + j ::: c d;1 + j] As in case (a), it can be checked that 2 H(n d t s) if and only if 2 B(n ; d(2 + j)+1 d; 1 t+ j maf0 s+1+t + j)g 9

10 For case (c), let j = 0 d ; d Then () = 2 B(n ; d(2 + j)+1 d; 1) and c() = c 1 c 2 ::: c d;1 j] and c() =c 1 ; j c 2 ; j : : : c d;1 ; j] As in case (a), it can be checked that 2 H(n d t s) if and only if 2 B(n ; d(2 + j)+1 d; 1 t; j maf0 s+1+t ; jg) Combining the contributions from cases (a), (b), and (c), respectively, we get h(n d t s) = h(n ; 2d +1 d; 1 t maf0 s+ t +1g) (11) + X j1 h(n ; d(2 + j)+1 d; 1 t+ j maf0 s+1+t + jg) (12) + X j1 h(n ; d(2 + j)+1 d; 1 t; j maf0 s+1+t ; jg): (13) To get an upper bound on j, note that n d 2 + dj, so that j b(n ; d 2 )=dc Then the right-hand-side terms, (11), (12), and (13) can be combined as 2 h(n d t s)= j=b(n;d X 2 )=dc j=;b(n;d 2 )=dc h(n ; d(2 + jjj)+1 d; 1 t+ j maf0 s+1+t + jg): With the recurrence of Theorem 5, the number of graphical basis partitions of n, h(n) = bx p nc d=0 h(n d) = bx p nc d=0 h(n d 0 0) can be computed using a 4-dimensional table for the values h(n d t s) Each entry can be computed in O(n) time, thereby giving an algorithm which is polynomial in n, even though h(n) appears to grow eponentially Although the amount of storage can be reduced to (n 3 )bykeeping entries only for the two most recent values of d, the storage became prohibitive for us at just over n = 200 The values obtained by implementing the recursion of Theorem 5, for even n 200, are shown in Tables 1 and 3 at the end of this paper, along with the ratios h(n)=b(n), showing the fraction of basis partitions of n which are graphical The surprising observation is that this ratio appears to be approaching the same limit as g(n)=p(n), the fraction of all partitions which are graphical These values are given for comparison in Tables 2 and 4 Both ratios appear to be non-increasing for n suciently large and we conjecture that the limits eist and: h(n) lim n!1 b(n) = lim g(n) n!1 p(n) : 10

11 4 Generating Basis and Graphical Basis Partitions In this section we giveanecient algorithm to generate H(n) and prove that the algorithm works in constant amortized time per graphical basis partition We rstshowhowto eciently generate basis partitions of n, B(n), and then modify the algorithm to generate H(n) Since there is a one-to-one correspondence between basis partitions of n and rank vectors whose corresponding basis partition has weight n, we can represent a basis partition by its rank vector We found it more natural to work with the co-rank vector, which is the negative of the rank vector It follows then from Theorem 4 that for a co-rank vector c =c 1 c 2 ::: c d ], the corresponding basis partition has weight jj = n(c) where n(c) =d 2 + d;1 X Since c() =r( 0 ), we get the following from (14) ijc i ; c i+1 j + djc d j (14) Corollary 2 For a co-rank vector c =c 1 c 2 ::: c d ], n(c 1 c 2 ::: c d ]) ; n(c 2 ::: c d ]) = 2d ; 1+p where p = jc 1 ; c 2 j + jc 2 ; c 3 j + + jc d;1 ; c d j + jc d j (15) The algorithm of this section generates graphical co-rank vectors for a given n, that is, co-rank vectors whose associated basis partitions have weight n and are graphical It works by successively prepending entries to partially constructed co-rank vectors If = ( 1 2 ::: d;1 ) is a co-rank vector, then the weight of the basis corresponding to is given by (14) For any integer, let =( + 1 ) be the co-rank vector obtained by prepending 1 + to By (15), the dierence in weights, n( ) ; n() is given by 2d ; 1+jj + p, where p = j 1 ; 2 j + j 2 ; 3 j + :::+ j d;2 ; d;1 j + j d;1 jletb( n) be the set of co-rank vectors (of any possible length) with weight n() + n, having su In the Ferrers diagram, we refer to the L-shaped gure formed by row 1 and column 1 as the outermost right angle Then the intuition for construction of B( n)isasfollows The 11

12 Ferrers diagram corresponding to the partially constructed co-rank vector has n() dots and n more dots are to be added as successive outermost right angles Then B( 0) = fg and for n>0, B( n) can be written as the disjoint union, B( n) = = jjn;2d+1;p jj=n;2d+1;p jj<n;2d+1;p B( n; 2d +1;jj;p) (16) B( 0) B( n; 2d +1;jj;p): (17) In each step, the scheme increases the size of the co-rank vector by one, ie adds a new outermost right angle to the Ferrers diagram, consuming some or all of the n available dots The rst term considers the case when all the dots are consumed and the second term takes care of the case when has to be further augmented to consume the remaining dots Note that each set in the rst term is non-empty if the left hand side is non-empty From Corollary 2 it follows that prepending + 1 to the co-rank vector consumes 2d;1+jj+p dots Adding a new outermost right angle would require at least 2d+1+jj+p remaining dots Therefore each set in the second term on the right side of (17) will be nonempty if and only if n ; 2d +1;jj;p 2d +1+jj + p () jj n=2 ; 2d ; p: Equation (17) can be rewritten as B( n) = jj=n;2d+1;p jjn=2;2d;p B(( + 1 ) 0) B(( + 1 ) n; 2d +1;jj;p): (18) In (18), each set on the right-hand side is non-empty, and therefore in the recursion tree based on (18), there is a one-to-one correspondence between leaves and basis partitions in B( n) To generate graphical basis partitions, the tree needs to be pruned For a vector v =v 1 v 2 ::: v k ], dene residual(v) = P k (v i ; 1), andneed(v) as the 12

13 minimum non-negative integer ` satisfying, ` + jx (v i ; 1) 0 j =1 2 ::: k: Note that by the Nash-Williams condition, (3), v is a graphical co-rank vector if and only if need(v) = 0 Denote need() by s, and dene H( n s) to be the set of co-rank vectors in B( n) which are graphical Note that need(( + 1 ) ) = maf0 s+1; ( 1 + )g so that we can directly adapt recurrence (18): H( n s) = jj=n;2d+1;p jjn=2;2d;p for which the base case is given by, H(( + 1 ) 0 maf0 s+1; ( 1 + )g) H(( + 1 ) n; 2d +1;jj;p maf0 s+1; ( 1 + )g) (19) H( 0 s) = ( fg if s =0 fg otherwise (20) Clearly, H(n) =H( n 0) Thus, the recursion of (19) can be used to generate H(n) In the remainder of this section, we show how to implement the recursion eciently, so that the total time spent iso(jh(n)j) = O(h(n)), disregarding the output For the vector =( 1 2 ::: d;1 ), dene p = j 1 ; 2 j + j 2 ; 3 j + :::+ j d;2 ; d;1 j + j d;1 j Then the following hold Lemma 3 For all 2 B( n), residual( ) is maimized when =( 1 + ) where = n ; 2d +1; p Proof By Corollary 2, the outermost right angle in the Ferrers diagram corresponding to has 2(d ; 1) ; 1+p dots To add a new outermost right angle, at least 2d ; 1+p dots are needed The residual will be maimized if after adding these dots, the remaining = n ; 2d +1; p dots are added to the rst column, in which case =( 1 + ) 2 13

14 Lemma 4 H( n s) is non-empty if and only if 1 + ; 1 s, where = n ; 2d +1; p Proof For all 2 B( n), if the maimum residual( ) (obtained by setting = n ; 2d +1; p (from Lemma 3)) is less than s = need(), then H( n s) is empty If however 1 + ; 1 s, then 2 H( n s) where = ( 1 + ) and hence H( n s) is non-empty 2 Lemma 5 On the right side of equation (19), if H((+ 1 ) n;2d+1;jj;p ma(0 s+ 1 ; ( 1 + ))) is empty for = m then it is empty for all <m Proof Let s 0 =ma(0 s+1; ( 1 + )) and n 0 = n ; 2d +1;jj;p Ifs +1; ( 1 + m) < 0 then 1 + m>1 (since s is non-negative) Therefore prepending another element to the co-rank vector with as large a value as possible (by choosing the appropriate value of as indicated in Lemma 4) will yield a graphical vector, contradicting our assumption that H( n s) is empty Hence s 0 = s +1; ( 1 + m) If H((m + 1 ) n 0 m s0 ) is empty, then by 4,( 1 + m) +(n 0 m ; 2(d +1)+1; (p + jmj)) ; 1 < s 0 Substituting values of s 0 and n 0 m gives 2(m ;jmj) + < s, where is independent ofm Since, ;jj m ;jmj for <m,2( ;jj)+<sfor <m, which implies that ( 1 + )+(n 0 ; 2(d +1)+1; (p + jj)) ; 1 <s 0 Therefore by Lemma 4, H(( + 1 ) n; 2d +1;jj;p s 0 )isempty for <m 2 Based on the above results we present an algorithm in Figure 3 which, as we will prove, generates all elements of H(n) in total time O(h(n)) In each step of the recursion, the algorithm prepends an element, + 1 to the partially constructed co-rank vector It considers the values of in descending order, in the range as indicated in recursion (19) If a particular value of = m generates an empty set, subsequent values of need not be checked (by Lemma 5) Claim 1 In the recursion tree of the above algorithm, the total number of leaves is at most twice thenumber of graphical basis partitions of n, and every node with one child has a sibling 14

15 1 int generate( n d p s) /* Generates rank vectors of n() + n with su = ( 1 2 ::: d;1 ) and p = j 1 ; 2 j + j 2 ; 3 j + :::+ j d;2 ; d;1 j + j d;1 j */ 2 begin 3 if (n=0) then output() 4 if (( 1 + n ; 2d +1; p) ; 1 <s) return 0 5 tmp1 = n ; 2d +1; p =tmp1 6 proceed = generate(( 1 + ) 0 d+1 p+ jj ma(0 s+1; ( 1 + ))) 7 if (proceed == 0) return 0 8 tmp2 = min(n=2 ; 2d ; p tmp1;1) 9 for ( =tmp2 -tmp2 ;;) 10 proceed = generate(( 1 + ) n; 2d +1;jj;p d +1 p+ jj ma(0 s+1; ( 1 + ))) 11 if (proceed==0) return 1 12 = -tmp1 13 if (( 1 + ; 1 s) and ( 6= 0) 14 proceed = generate(( 1 + ) 0 d+1 p+ jj ma(0 s+1; ( 1 + ))) 15 return 1 16 end Figure 3: Algorithm for generating H(n) Proof If a node is not a leaf, its leftmost child generates a valid object (follows from Lemma 4 and Step 4 of Figure 3) Further, at most one of its children can be a leaf which fails to generate a valid object (follows from Lemma 5 and the use of variable \proceed" in the algorithm) Thus for every \failure" leaf there is a corresponding \good" leaf Therefore the number of leaves 2 number of objects For the second claim, note that a node u, if it is the only child of its parent v, hastobe the child corresponding to the call in step 6 Since u was called with n = 0, it cannot have any children Therefore any node u without a sibling cannot have a child, and so every node with a child has a sibling 2 From the above claim, we conclude that the number of nodes is bounded by a constant times the number of leaves which is O(h(n)) Moreover, each node involves a constant amount of work Therefore we conclude that the algorithm, if we eclude time to output the results, averages constant time per item generated 15

16 5 Directions for Further Research The open question remains as to whether g(n)=p(n) approaches 0, aswell as our new conjecture that the fraction of basis partitions which are graphical approaches the same limit as the fraction of all partitions which are graphical Perhaps generating functions can be found for these quantities which would give insight into their asymptotic behaviour, although they do not seem to be easily derivable from the recurrences here in Theorem 5 or in 2] In order to be able to collect data for larger values of n, we need faster ways to count (ordinary or basic) graphical partitions, for eample, by asymptotically reducing the storage requirements References 1] AOL Atkin, A note on ranks and conjugacy of partitions, Quart J Math 17, No 2 (1966) ] TM Barnes and CD Savage, A recurrence for counting graphical partitions, Electronic Journal of Combinatorics 2, R11 (1995) 3] TM Barnes and CD Savage, Ecient generation of graphical partitions, preprint (1995), submitted for publication 4] JA Bondy and USR Murty, Graph TheorywithApplications, North-Holland, ] D Bressoud Etension of the partition sieve, J Number Th 12 (1980) ] P Erd}os and L Richmond, On graphical partitions, Combinatorica 13, No 1 (1993) ] H Gupta, The Rank-vector of a partition, The Fibonacci Quarterly 16, No 6, (1978) ] F Harary, Graph Theory, Addison-Wesley Publishing Company, ] JM Nolan, CD Savage, and HS Wilf, Basis partitions, preprint (1996), submitted for publication 16

17 10] K F Roth and Szekeres, Some asymptotic formulas in the theory of partitions, Quart J Math Oford Ser (2) 5 (1954) ] C Rousseau and F Ali, On a conjecture concerning graphical partitions, Congressus Numerantium 104 (1994) ] F Ruskey, RCohen,P Eades and A Scott, Alley CATs in search of good homes preprint (1994) 13] G Sierksma and H Hoogeveen, Seven criteria for integer sequences being graphic, Journal of Graph Theory 15, No 2 (1991)

18 n h(n) b(n) h(n)=b(n) Table 1: Fraction of basis partitions of n which are graphical (2 n 100) n g(n) p(n) g(n)=p(n) Table 2: Fraction of all partitions of n which are graphical (2 n 100) 18

19 n h(n) b(n) h(n)=b(n) Table 3: Fraction of basis partitions of n which are graphical (102 n 200) n g(n) p(n) g(n)=p(n) Table 4: Fraction of all partitions of n which are graphical (102 n 200) 19