26 International Conference on Computational Science and Computational Intelligence Computation in the Enumeration of Integer Matrices Shanzhen Gao, Keh-Hsun Chen Department of Computer Science, College of Computing and Informatics University of North Carolina at Charlotte, Charlotte, NC 28223, USA Email: sgao3@uncc.edu, chen@uncc.edu Abstract The enumeration of integer-matrices has been the subject of considerable study and it is unlikely that a simple formula exists. The number in question can be related in various ways to the representation theory of the symmetric group or of the complex general linear group, but this does not make their computation any easier. We will discuss the following three problems: () the number of m n matrices over {, } with each row summing to s and each column summing to t; (2) the number of nonnegative integer matrices of size m n with each row sum equal to s and each column sum equal to t; (3) the number of (, ) - matrices of size n n such that each row has exactly s s and each column has exactly s s and with the restriction that no stands on the main diagonal. We will present many conjectures based on our computation. Keywords: integer matrix, formula, integer sequence, row sum, algorithm I. INTRODUCTION Our motivation of enumeration of integer matrices came from [] and E. Rodney Canfield and Brendan D. McKay s seminal work on the asymptotic enumeration of integer matrices such as [2], [3], [4]. From R. Stanley s book []: Let f(n) be the number of n n matrices M of zeros and ones such that every row and column of M has exactly three ones, f() =, f() = f(2) =, f(3) =. Themost explicit formula known at present for f(n) is f(n) =6 n ( ) β n! 2 (β +3γ)!2 α 3 β α!β!γ! 2 6 γ where the sum is over all (n +2)(n +)/2 solutions to α + β + γ = n in nonnegative integers. This formula gives very little insight into the behavior of f(n), but it does allow one to compute f(n) faster than if only the combinatorial definition of f(n) were used. Hence with some reluctance we accept (ii) as a determination of f(n). Of course if someone were later to prove f(n) =(n )(n 2)/2 (rather unlikely), then our enthusiasm for the above formula of f(n) would be considerably diminished. In the following, we will present the enumeration of certain integer matrices. II. (,)- MATRICES WITH FIXED ROW SUM AND COLUMN SUM Let m, n, s, t be positive integers such that sm = tn. Let f(m, n, s, t) be the number of m n matrices over {, } with each row summing to s and each column summing to t. Equivalently, f(m, n, s, t) is the number of semiregular bipartite graphs with m vertices of degree s and n vertices of degree t. This problem has been the subject of considerable study, and it is unlikely that a simple formula exists. The asymptotic value of f(m, n, s, t) has been much studied but the results are incomplete. Historically, the first significant result was that of Read, who obtained the asymptotic behavior for s = t =3[5]. McKay and Wang (23) solved the sparse case λ( λ) =o((mn) /2) using combinatorial methods [6].Canfield and McKay used analytic methods to solve the problem for two additional ranges. In one range the matrix is relatively square and the density is not too close to or. In the other range, the matrix is far from square and the density is arbitrary. Interestingly, the asymptotic value of f(m, n, s, t) can be expressed by the same formula in all cases where it is known. Based on computation of the exact values for all m; n<3, they got the conjecture that the same formula holds whenever m + n regardless of the density (they defined the density λ = s/m = t/m). We are concerned in this paper with the closed formulas of f(m, n, s, t). The number in question can be related in various ways to the representation theory of the symmetric group or of the complex general linear group, but this does not make their computation any easier. The case s = t =2is solved by Anand, Dumir, and Gupta [7]. A formula for the case s = t =3appears in L. Comtet s Advanced Combinatorics [8], without proof. It is easy to get: f(m, n, s, t) =f(n, m, t, s), (sm = tn) f(m, n, s, t) =f(n, m, n s, m t), (sm = tn) f(m, n,,t)= m! (t!) n (m = tn) f(m, n, s, ) = n! (s!) m (sm = n) We list some conjectures and numbers. We encourage our readers to prove or disprove the following conjectures. f(n, n, 2, 2) = n! n n ( ) n r (2r)! 2 n r = r 2 r r! = 4 (2n)! + n n k= ( 2)k n 2k!(2(n k k))! =4 n n ( 2) i (n!) 2 (2n 2i)! i= i!((n i)!) 2 Example: f(2, 2, 2, 2) =, f(3, 3, 2, 2) = 6 978--59-55-4/6 $3. 26 IEEE DOI.9/CSCI.26.245 3 35
6 9 24 67 95 3 94 87 53 84 4 398 7 2 37 785 398 2 58 85 387 962 2 959 547 4 77 2 3574 34 599 4 475 2 676 58 33 623 35 84 47 32 988 74 542 99 484 Conjecture. f(m, n, 2, 3) = 2 m n ( ) i m!n!(2m 2i)! i= i!(m i)!(n i)!6 n i 86 9296 382452632 46764764387244 22974728499669349384 339828364899435838656 9345355457243262529337978773248 Conjecture 2. f(m, n, 4, 2) = n! m m r 2 n r = r = m! ( ) 2(m r ) r (4r +2r )! r!r!(m r r )! (n 2m+2r +r )! 24 r 2 (m r ) Conjecture 3. 9 4473 565866 54798875 87998767676 73738375847 982954759537464 2532232525373854963235 f(m, n, 5, 2) = n! m m r 2 n r = r = m! ( ) r +2(m r r ) (4r +2r +m)! r!r!(m r r )! (n+r 2m+2r )!2 r 6 r 2 (m r r ) 756756 2598926974 9647422924949829674 2493577268489633274873267 f(m, n, 6, 2) = n! m m r m r r 2 n r = r = r 2= m! ( ) 3m 3r 2r r 2 r!r!r 2!(m r r r 2)! (n+2r +r 2 3m+3r )! 68 3249656 264964362 6665694498664 4524896548476972584 III. (,)-MATRICES WITH RESTRICTION The number ( f s (n) )of(, ) - matrices of size n n such that each row has exactly s s and each column has exactly s s and with the restriction that no stands on the main diagonal. Conjecture 4. f (n) = n n! k= ( )k k!. 2 9 44 265 854 4 833 33 496 334 96 4 684 57 76 24 84 229 792 932 32 7 49 48 66 55 734 This is also the number of desarrangements of length n. Conjecture 5. f 2 (n) = n k n k k= s= j= ( ) k+j s n!(n k)!(2n k 2j s)!. s!(k s)!((n k j)!) 2 j!2 2n 2k j 9 26 757 357 435 22 4 36 72 632 24 66 26 966 956 9 459 238 879 565 274 82 5 92 545 445 22 898 72 992 496 84 798 39 68 743 38 44 8 567 39 7 438 379 656 47 This also the number of labeled 2-regular digraphs with n nodes. 32 36
Conjecture 6. Conjecture 8. f 3 (n) = n k t n k k= t= s= p= n k p q= k t k t r= w= ( ) k+t+q+r s n!(n k)! 3 2n 2k p 2q r+w 2 2n 2t 2p q r p!q! (k t)!(q + r)! (n k p q)!w!(k t w)! (3n k 3p 2q t 2r s)! r!(k t r)!s!(t s)!(q + r w)!(n k p q r + w)! 44 757 97556 74964945 3993575464 2892363493 2774356645652736 34523964977338385 5459923655258622946 755952384569555562244 This also the number of labeled 3-regular digraphs with n nodes. The number (h(m, n, s, t) ) of nonnegative matrices of size m n with row sum s and column t and the entries on the line from a to a kk ( k =min{m, n} ) are all (sm = nt). Conjecture 7. h(m, n, s, 2) = m k= r + +rp =k r 2 + +r2 q =m k ( ) k m!(n k)! r!r! r p!r2!r2! r2 p! p i= ((s 2i)!i!2i ) r i (ms k 2 p i= ir i 2 q j= jr2 j )! q j= ((s 2j)!j!2j ) r2 j (n k p i= ir i + q j= jr2 j )! 2 (n k p i= ir i +q j= jr2 j ) (2n = sm,p = s 2 and q = s 2 ). IV. NONNEGATIVE INTEGER MATRICES The enumeration of nonnegative integer matrices has been the subject of considerable study, The determination of t(m, n, s, t) is an unsolved problem and it is unlikely that a simple formula exists except for very small s, t. Equivalently, t(m, n, s, t) counts 2 way contingency tables of order m n such that the row marginal sums are all s and the column marginal sums are all t. Another equivalent description is that t(m, n, s, t) is the number of semiregular labelled bipartite multigraphs with m vertices of degree s and n vertices of degree t. The matrices counted by t(m, n, s, t) arise frequently in many areas of sciences, for example enumeration of permutations with respect to descents and statistics. t(n, n, 2, 2) = 4 n n 2 i (n!) 2 (2n 2i)! i= i!((n i)!). 2 3 2 282 62 22 4 935 63 545 7 96 4 54 583 32 393 73 8 2 452 785 322 266 2 62 347 376 347 779 6 2 899 54 33 589 2 98 37 363 842 76 26 4 44 57 594 539 54 672 566 Conjecture 9. t(n, m, 3, 2) = 2 m n i= m!n!(2m 2i)! i!(m i)!(n i)!6 n i. 7 6 26 747 558 25 7 339 672 369 82 64 83 88 24 796 85 853 884 657 44 99 8 23 5 842 995 969 7 7 32 647 865 6 72 289 86 73 359 375 552 628 986 978 4 24 Conjecture. t(n, n, 3, 3) = 6 n n n α α=β= 2 α 3 β (n!) 2 (β+3(n α β))! α!β!(n α β)! 2 6 (n α β). 4 55 28 53 4 2 933 84 4662 857 36 579 6 246 4 772 2 774 683 52 523 853 88 779 443 2 477 36 556 85 6 93 2 569 6 9 292 72 349 4 8 868 7 73 52 923 49 92 728 663 43 727 673 392 444 887 284 377 6 3937 477 62 39 47 28 93 97 36 384 Let h(m, n, s, t) be the number of nonnegative matrices of size m n such that each row sum equal to s and each column sumequaltot, and the entries on the line from a to a 22 are all (sm = nt). The determination of h(m, n, s, t) is an unsolved problem. 33 37
Conjecture. h(n, n, 2, 2) = n n k k= i= ( )k n!(n k)!(k+2i)! k!(n k i)!(i!) 2 2 (n k+i) 3 42 87 27 99 254 33 74 534 4 5663 28 68 535 32 999 6 588 7 52 2 8473 9 537 782 8 373 423 877 85 424 4 259 39 49 657 932 3 2 56 242 83 79 86 289 594 V. ALGORITHM DESCRIPTION FOR t(m, n, s, t) The algorithm used to verify the equations presented counts all possible matrices, but does not construct them. It is a bit involved, so it is best described with an example. Suppose we wanted to compute the number of 4 6 matrices over nonnegative integers with row sum 2 and column 8. Wefirst create a list of all nonincreasing partitions of 2: 2,, 2,, 9 3, etc., and store this in memory. We make sure that each partition stored is not of length greater than the number of columns of the matrix. We then create a state vector of length 6 filled with 8s: #(888888) This state vector symbolizes the sum of integers we must place in each column, and each time the state changes, it is sorted in nondecreasing order. An additional vector, called the cap vector, is created when we deal with a new state. It records the length of the contiguous blocks of numbers found in the state. Here, it is #(6). Next, we iterate over each of the (valid) partitions of 2 that we could possibly use for the choice of the first row of the matrix. Here, our first partition is 8 4. We then create a partition block (pb) vector, which is exactly a cap vector of the partition instead of the state. Here, it is #( ). Finally, we create all the assignment vectors that are valid for this partition and this cap vector. An assignment vector dictates where the indicated element of the partition will be placed in the row. Assignment vectors always have the same length as the partition we are planning to use. The entries of the assignment vector refer to the (zero-based) indices of the cap vector. Since the cap vector in this case only has one index (namely, ) and both 8 and 4 can be elements in the matrix row, we assign 8 and 4 to the th index: #( ) In other words, both the 8 and the 4 will appear in block of the state. Now, there are 6 5 ways of placing the 8 and 4, so we note that when we drop the state vector. We pretend that the firstrowofthematrixwillbe(84), and so, dropping the state vector, the remaining three rows must sum to #(48888) and we record that the number of ways of obtaining a matrix of state #(8 88888)is3times the number of ways we can obtain a matrix of state #( 48888). Of course, we must add to our count the other ways to assign the 8 and 4. Since there are no other ways, no more assignment vectors can be constructed. We then add to our count the ways in which we can use the partition 83(with all applicable assignment vectors), and then 8 2 2 (with all applicable assignment vectors), and so forth. To get a better feel for how the assignment vectors are created, let s say that, in the middle of our counting, we achieve the state #(4666) with two rows left to fill. Our cap vector is then #(23) and suppose we are considering the partition 4 4 3. Its pb is #(2 ). Since the cap vector has length 3, the indices for it are,, and 2, so the entries of each assignment vector can be comprised only of,, and/or 2. To create the first assignment vector, we note that the first element of the partition, 4, cannot be placed in block of the state (the block of two s), since 4 >. A single 4 can be placed in block of the state (the block consisting of the single 4), so the first 4 in the partition can be assigned to block : #(???) But block is only length (as noted by the cap vector s entry of at index ), so no more 4s can go in that block. The second 4 in the partition can also be placed in block 2 of the state (the block of three 6s), since <. Thus, our assignment vector changes to #(2??). 34 38
Next in the partition, we have a 3, which is also greater than, so it too cannot go into block. Block has already been taken by the 4. Hence the only remaining place for it is in block 2: #(22?) Finally, the last element of the partition is a, which can go anywhere in the state. We begin by assigning it to block, giving the resulting assignment vector as #( 2 2 ). How many ways could these assignments be carried out? The first 4 has only one way. The second 4 and the 3 are both in block 2, but they are different numbers, so they can be inserted in 3 2 ways. Finally, the has 2 ways to be inserted into block. Hence we multiply to get 2 ways for this assignment vector, and dropping the state, we get #( 2 3 6). Sorting it, it becomes #( 2 3 6), which we will process after we deal with the remaining assignment vectorspossiblefor443. To get the next assignment vector, we note that we can keep everything the same, but the in the partition can be put in block 2. This gives #(222) and to compute the number of ways, we have 3 2 =6. To get the next assignment vector, we note we ve exhausted all possibilities for #( 2??), so we then find the next way to assign the two 4s in the partition. The only remaining option is to put them both in block 2, so we start with #(22??). Now, the 3 can go in block and the can go in block, giving #(22) and total number of ways 3 2 =6. Now, we think of a block of the assignment vector as the entries that correspond to an equal number in the partition; here, the first two entries correspond to the partition entry 4, so they form a block. The pb tells us the length of each block of the assignment vector. For example, recall that here, pb is #(2 ), so each assignment vector corresponding to this partition has three blocks, the first of which has length two, and the remaining two have length one. We construct assignment vectors that are nondecreasing in each block, though we can have a decrease when we move to a new block from an old one. The remaining three assignment vectors and the number of ways to make the assignment are then #(222)withways 3 2 =3 #(222)withways 3 2 2 =6 #(222)withways 3 2 =3. Let s consider a larger example. Suppose the state was #(2223333455) with row sum 8. This state will produce a cap vector of #(4 3 4 2) (since zeroes in the state are ignored). Let s suppose we were considering the partition 333222, which gives a pb of #(3 3 3). There are 433 total assignment vectors for this partition. The first one we could construct is #(222)withways 4 3 3 3 4 3 =6, an intermediate one we could construct is #(23422) with ways 4 2 (for placing the three 3s) 3 3 2 (for placing the three 2s) 4 2 (for placing the three s) totalling 576, and the last one we could construct is #(3442222) with ways 2 2 (for placing the three 3s) 4 3 (for placing the three 2s) 3 3 2 (for placing the three s) totalling 2. Notice that each block of each assignment vector has its entries in nondecreasing order, but often there is a decrease when we move from block to block. Since the state vectors are nondecreasing, this is to be expected. In general, for each state vector that is achieved, this algorithm will iterate over all assignment vectors for each valid partition, multiplying cofactors and adding the results. When fitting the last row, though, the calculation is surprisingly easy: continuing the example we had above, if we examine the state #(236),weseethatthereisonlyonepossiblepartition of 2 that fits it (namely 632)andthereisonlyoneway to fit it in. Hence, there is only one way to achieve this state. The situation is the same for every state with one row left to be filled. For further speedup, a fast storage object must be used, so that if a given state is seen again, we can recall from memory how many partially-filled matrices can produce it. This speedup is necessary, for without it, the algorithm will take too long. Other approaches and improvements are certainly possible, such as storing all possible assignment vectors for each partition and later recalling the applicable ones. 35 39
VI. ENUMERATION OF (-,,)-MATRICES Let r(m, n, s, t) be the number of (,, ) matrices of size m n with each row sum equal to s and each column sum equal to t (sm = nt). r(m, n, s, t) is related to the following two problems: Can you obtain a formula for the number of connected labeled 2 regular pseudodigraphs? ("connected" means there is a path between any pair of vertices disregard the directions of edges, i.e. only one component). Can you obtain a formula for the number of non-isomorphic connected labeled 2 regular pseudodigraphs? This interest stems from some research in theoretical physics on the fractional quantum Hall effect. The graphs essentially arise in a Wick expansion of some physical quantity, and in physical terms the CONNECTED labeled 2-regular pseudodigraphs are Feynman diagrams. The graphs must be 2-regular because each vertex represents a 2-body interaction. For the non-isomorphic graphs, they are really physically different. To compute the number of connected labeled 2-regular pseudodigraphs (multiple arcs and loops allowed) of order n. You start with the numbers of all labeled 2-regular pseudodigraphs and apply the standard recurrence relation to produce the numbers of those which are connected (weakly connected, to be more precise). That recurrence can be found in. Then you can get a sequence:,,2,4,2,474,689,.... The enumeration of (, ) matrices is a special case of enumeration of (-,,)-matrices. Some Easy Questions: Question. The number of (,, ) matrices with matrix sum n? Matrix sum=the sum of all the entries of the matrix. Answer : + Question 3. The number of (,, ) matrices with m entries and matrix sum n? Answer: (m+n)/2 i=n d(m)m! i!(i n)!(m + n 2i)!. Question 3. The number of r r (,, ) matrices with matrix sum n? Answer: (r 2 +n)/2 i=n (r 2 )! i!(i n)!(r 2 + n 2i)!. Let f 2 (n) be the number of (,, ) matrices with each row and each column with exactly one "" and one " ". It is clearly that f 2 (n) is even. f 2 (n) =n(n ) 2 f 2 (n 2) + n(n )f 2 (n ) f 2 () =,f 2 (2) = 2. VII. OPEN PROBLEMS Problem 2. How many n n matrices in F q ( q is a prime) exist up to similarity? Problem 3. Up to similarity, compute the number of n n matrices with entries in F 2 with each column and each row with exactly one "". That s the number of partitions of n with ordering. Problem 4. How many n n trace zero matrices in F q ( q is a prime) exist up to similarity? REFERENCES [] R. P. Stanely, Enumerative Combinatorics, Volumes, Cambridge University Press (first edition 986, second edition 2). [2] E.R. Canfield and B.D. McKay,Asymptotic enumeration of dense - matrices with equal row sums and equal column sums, Electron. J. Combin., 2 (25) R29. [3] E.R. Canfield, C. Greenhill, and B.D. McKay, Asymptotic enumeration of dense - matrices with specified line sums, J. Combin. Theory Ser. A 5 (28), no., 32 66. [4] E.R. Canfield and B.D. McKay,Asymptotic enumeration of contingency tables with constant margins, arxiv:math/736 [5] R.C. Read, Some enumeration problems in graph theory, Doctoral Thesis, University of London, (958). [6] B. D. McKay and X. Wang, Asymptotic enumeration of - matrices with equal row sums and equal column sums, Linear Alg. Appl., 373 (23) 273-288. [7] Anand, Dumir, and Gupta in Duke Math J., 33 (966) 757-769. [8] L. Comtet, Advanced Combinatorics (page 236),Kluwer Academic Publishers,974(page 236). Question 4. The number of a r (,, ) matrices with each row sum n? Answer: (r+n)/2 r! ( i!(i n)!(r + n 2i)! )a i=n 36 32