On F-squares and their critical sets

We define the notion of critical set of an F-square, following the definition of critical set in latin squares, and then give critical sets for certain classes of F-squares. We also generalise certain results obtained for critical sets of latin squares, and look at minimal such sets. We show that critical sets of F-squares need to be studied as well as critical sets for latin squares as the techniques used differ considerably. We obtain theorems for the sizes of critical sets of types F(n; 1,n — 1), F(n;1, 1, n — 2), and F(n; 2, 2, ... , 2). Publication Details This article was originally published as Fitina, LF, Seberry, J and Sarvate, D, On F-Squares and their Critical Sets, Australasian Journal of Combinatorics, 19, 1999, 209-230. This journal article is available at Research Online: http://ro.uow.edu.au/infopapers/339 On F-Squares and their Criti al Sets L F Fitina, Jennifer Seberry Centre for Computer Se urity Resear h Department of Computer S ien e University of Wollongong NSW 2522 Australia Dinesh Sarvate Department of Mathemati s College of Charleston Charleston SC 29424 USA May 24, 1999 Abstra t We de ne the notion of riti al set of an F-square, following the de nition of riti al set in latin squares, and then give riti al sets for ertain lasses of F-squares. We also generalise ertain results obtained for riti al sets of latin squares, and look at minimal su h sets. We show that riti al sets of F-squares need to be studied as well as riti al sets for latin squares as the te hniques used di er onsiderably. We obtain theorems for the sizes of riti al sets of types F (n; 1; n 1); F (n; 1; 1; n 2); and F (n; 2; 2; : : : ; 2). 1 Introdu tion Let n = 0 + 1 + : : :+ v 1, where i is a natural number for ea h i. A frequen y square or F-square of type F = F (n; 0; 1; : : : ; v 1) and of order n is an n n array with entries hosen from the set N = f0; 1; : : : ; v 1g, su h that ea h element i o urs i times in ea h row and in ea h olumn. An F-square F an also be thought of as the set of ordered triples F = f(i; j; k)g where element k o urs in position (i; j). A subset of F will also be alled a subsquare or partial F-square. A subsquare of an F-square is also said to be embeded in the F-square. Two subsquares of an F-square are alled isotopi , if one an be transformed onto the other by rearranging rows, rearranging olumns, and renaming elements. A non-empty subset S of F = F (n; 0; 1; : : : ; v 1) is a riti al set (of F ) if, 1. F is the only F-square of order n whi h has element k in position (i; j) for ea h (i; j; k) 2 S. (We then say that F is uniquely ompletable from S, and that S is uniquely ompletable to F ), 2. (a) every proper subset of S is ontained in at least two F-squares of type F (n; 0; 1; : : : ; v 1) or (b) for every (i; j; k) 2 S; l 2 N; l 6= k =) there does not exist any F-square of type F (n; 0; 1; : : : ; v 1) whi h ontains (S=f(i; j; k)g) [ f(i; j; l)g We note that a latin square is an F-square of type F = F (n; 1; 1; : : : ; 1)). A latin square L = f(i; j; k)g of order n is alled ba k ir ulant if (i+ j) mod n = k for every triple (i; j; k) 2 L. An F-square will usually have more than one riti al set. We say that a riti al set is minimal if it has minimum ardinality. The notation r or r(F ) will denote the size of a riti al set. We denote by s r(F (n; 0; 1; : : : ; v 1)) the size of the smallest riti al set of 1 (F (n; 0; 1; : : : ; v 1)). The size of the smallest riti al set in any F-square of order n is denoted by fs s(n), and that of the largest riti al set in any F-square of order n is denoted by fl s(n). In the ase of latin squares only, these sizes are denoted by s s(n) and l s(n) respe tively. Curran and van Rees [5℄ evaluated s s(n) for n = 1; : : : ; 5, and re ently Howse[9℄ extended this result to n = 6. The results they obtained are summarised as follows: n 1 2 3 4 5 6 s s(n) 0 1 2 4 6 9 We onje ture that the size of the smallest riti al set of type (F(n; 1; : : : ; 1; n i)) where there are i zeroes satis es s r(F (n; 1; : : : ; n i)) s r(F (n; 1; : : : ; 1; n i+ 1)) s r(F (n; 1; 1; : : : ; 1)) = s s(n): Criti al sets of latin squares were rst studied by Smetaniuk [12℄, Curran and van Rees [5℄, and Cooper, Donovan and Seberry[1℄. If n is even, let Ce be the set Ce = f(i; j; i+ j) : i = 0; : : : ; n2 1; and j = 0; : : : ; n2 1 ig [ f(i; j; i+ j) : i = n2 + 1; : : : ; n 1 and j = n2 i; : : : ; n 1g where addition is taken modulo n. If n is odd, let Co be given by Co = f(i; j; i+ j) : i = 0; : : : ; (n 3)=2 and j = 0; : : : ; (n 3)=2 ig [ f(i; j; i+ j) : i = (n 1)=2 + 1; : : : ; n 1 and j = (n 1)=2 i; : : : ; n 1g where addition is redu ed modulo n. Curran and van Rees showed that sets Ce and Co both satisfy ondition 1 of the de nition of riti al set, and that Ce was in fa t riti al. Cooper, Donovan and Seberry [1℄ later veri ed that both Ce and Co are riti al sets, and that in fa t Ce is minimal. In summary they showed that: Theorem 1 Ce is a minimal riti al set for a ba k ir ulant square of even order n, and jCej = n2 4 . Co is a riti al set for a ba k ir ulant square of odd order n, and jCoj = n2 1 4 . Smetaniuk[12℄, using a di erent method, showed that the size of a minimal riti al set of a latin square of order 2n is at most n2. It is not yet established whether for n odd jCoj = n2 1 4 is minimal, but the eviden e so far, is that this may be the ase, see Howse[9℄. Donovan and Cooper [7℄ later established another family of riti al sets in ba k ir ulant latin squares, thus settling a onje ture by Nelder [11℄ that in the latin square representing the addition table of the integers modulo n, the upper triangle of entries bounded by, but not in luding the main right-to-left diagonal, is a riti al set. The riti al set in question is A = f(i; j; i+ j) : i = 0; : : : ; n 2 and j = 0; : : : ; n 2 ig: They further proved that: Lemma 1 (Donovan and Cooper) If L is any ba k ir ulant latin square of order n, and r is some integer su h that (n 3)=2 r n 2, then the set B = f(i; j; i+ j) : i = 0; : : : ; r and j = 0; : : : ; r ig [ f(i; j; i+ j) : i = r + 2; : : : ; n 1 and j = r + 1 i; : : : ; n 1g is a riti al set in L. 2 If P is a partial latin square then an element p 2 P is said to be 2-essential if there exists a 2 2 latin subsquare S, su h that S \ P = p. P is alled 2riti al if it is uniquely ompletable to a latin square, and every element of P is 2-essential. Stinson and van Rees give a produ t onstru tion on latin squares, and show that given a 2riti al set, a riti al set of higher order an be onstru ted. Their onstru tion ould not be applied to ertain types of riti al sets in ba k ir ulant latin squares of odd order (among others). Cooper, Donovan and Gower[3℄ determined a family of riti al sets for latin squares that are the produ t of a latin square of order 2 with a ba k ir ulant latin square of odd order. In [16℄ Peddada, Seberry and Chaudhry showed that the uniquely ompletable sets of Cooper, Donovan and Gower are not minimal, and also gave a number of omputer generated riti al sets. Let P = f(i; j; k) : i; j; k 2 Ng be a partial F-square of order n. jP j is said to be the size of the partial square and the set f(i; j) : (i; j; k) 2 P for some k 2 Ng is alled the shape of P . We now look at general onstru tions of riti al sets for some types of F-squares. 2 F-squares of type F = F (n; t; n t). Theorem 2 Let t = 1, so F is of the form F (n; 1; n 1). For every natural number n 2, there exists a minimal riti al set of size fs s(n) = n 1. Proof. In su h an F-square 0 o urs pre isely on e in every row and on e in every olumn, while 1 o urs n 1 times in every row and n 1 times in every olumn. Consider the short diagonal D0 = f(1; n 1; 0); (2; n 2; 0); : : : ; (n 1; 1; 0)g. We laim that it is a riti al set. Denote by D the shape of D0. Sin e 0 o urs on e in rows 2 to n 1, therefore 1 an ll every other ell in these rows. Also, 0 o urs on e in olumns 2 to n 1, so 1 an ll every other ell (that is, those olumn ells that were not lled above). Thus the only ell not yet lled is (0; 0), and this an be uniquely lled with the element 0. That is, D0 is uniquely ompletable. Consider the subset D=f(1; n 1; 0) of D. We an uniquely ll ea h ell in the rows 2; : : : ; n 1 and in the olumns 2; : : : ; n 2, be ause the element 0 o urs on e ea h in ea h of these rows and olumns. Only ells (0; 0); (0; n 1); (1; 0); (1; n 1) remain un lled. Consider ell (1; n 1). There are exa tly two possible entries that an ll this ell, 0 or 1. If 0 then we obtain the previously lled F-square. If the entry is 1 then element 0 must be for ed into ells (0; n 1) and (1; 0), in whi h ase 1 must again ll ell (0; 0). But this then gives a omplete F-square of the form F (n; 1; n 1). Thus D=f(1; n 1; 0) has at least two ompletions. A similar argument an be made for ea h of the other entries in D. Thus D0 is a riti al set. Let E be any subset of F of size n 2, and with every ell lled with element 0. Then ea h row i su h that there exists (i; j; 0) 2 E an be lled uniquely, as an every olumn in E than ontains a 0. There will be two rows that annot be uniquely lled, and two olumns, giving rise to exa tly 4 ells that haven't been lled (the interse tions of these rows and olumns). Ea h of these un lled rows/ olumns ontains only 10s. Thus lling any one of these ells will for e the other ells to be lled uniquely. That is, ea h su h E has at least two ompletions. Thus there annot be any riti al set of size n 2 (or less), and D is of minimal size. Clearly fs s(n) = n 1. 2 Remark. This is obviously the smallest riti al set size taken over all riti al sets of order n. That is, fs s(n) = n 1. Further work, whi h will appear elsewhere, indi ates that the largest riti al set size for all riti al sets of order n, is fl s(n) 7(n4 )2 2 when n is even, and is fl s(n) 7(n 1 4 )2 + 7(n 1 4 ) 1 when n is odd. 3 Remark. In the ase of latin squares a uniquely ompletable partial latin square is not a riti al set if and only if any subset of the partial latin square has at least two ompletions. In the ase of F-squares, as an be seen in the proof of the theorems above and below, the situation di ers in general. A uniquely ompletable partial F-square F may not be a riti al set be ause every non-trivial subset of F does not lead to any legitimate F-square. Thus the idea of latin inter hanges as dis ussed and used by several authors (see for example [1℄) is not parti ularly useful for F-squares in general. Theorem 3 Every F-square of type F (n; 2; n 2); n 4 has a riti al set of size 2n 3. For this type the element 0 o urs twi e in ea h row and twi e in ea h olumn, whereas element 1 o urs n 2 times in ea h row and in ea h olumn. Proof. Let D0 = f(1; n 1; 0); (2; n 2; 0); : : : ; (n 1; 1; 0)g[f(2; n 1; 0); (3; n 2; 0); : : : ; (n 1; 2; 0)g be the union of the two short diagonals indi ated. Denote by D the shape of D0. Note that in D, rows 2; : : : ; n 1 and olumns 2; : : : ; n 1 ontain the element 0 twi e ea h. Ea h ell in these rows/ olumns an be lled uniquely, with the element 1. Fill these. So far, only ells (0; 0); (0; 1); (1; 0) and (1; 1) haven't yet been lled. But row 0 has the element 1 o uring in ea h of its ells, ex ept for ells (0; 0) and (0; 1). Thus these two ells an be lled uniquely with element 0 ea h. Column 0 now ontains 1 in ea h of the ells (2; 0); : : : ; (n 1; 0), and element 0 in ell (0; 0). Thus 0 an uniquely ll ell (1; 0). Element 1 is now for ed uniquely into ell (1; 1). Thus D0 is uniquely ompletable. Consider the set E = D=f(2; n 1)g. Element 0 o urs twi e in ea h of the rows/ olumns 3; : : : ; n 1. Thus the rest of the ells in these rows/ olumns an be uniquely lled with the element 1. Fill these. Consider olumn 0. The ells ommon to this olumn and the union of the rows 3; : : : ; n 1 are (3; 0); : : : ; (n 1; 0), and these must ne essarily now have been lled with the element 1. But there are n 2 o uran es of 1 in this olumn, so the only two remaining ells, (0; 0) and (1; 0), must be lled with element 0. So far, only ells (0; 1); (0; n 1); (1; 1) and (1; n 1) are un lled. Ea h of these ells may be lled with either 0 or 1. If for example 0 goes into (1; n 1) then element 1 goes into ell (0; n 1), for ing 0 into (0; 1) and 1 into (1; 1). We obtain an F-square of type F (n; 2; n 2). On the other hand if 1 goes into ell (1; n 1) then 0 goes into ell (0; n 1), 1 into ell (0; 1) and 0 into ell (1; 1), giving another F-square of type F (n; 2; n 2). That is, E has two ompletions. Similar arguments an be made for the partial F-square D=(n 1; 1). That is, D=(n 1; 1) has at least two ompletions. Consider the set D=f(2; n 2). Ea h of rows 3; : : : ; n 1 ontain element 0 twi e. Ea h of olumns 3; : : : ; n 3 and n 1 ontain the element 0 twi e. Thus the remaining ells in these rows/ olumns an be lled uniquely, with the element 1. Fill these. The as yet un lled ells will be the interse tions of the un lled rows and olumns: These are the ells (0; 0); (0; 1); (0; n 2); (1; 0); (1; 1); (1; n 2); (2; 0); (2; 1); (2; n 2), nine in all. If ell (2; n 2) is lled with 0 then we an uniquely omplete to the above F-square. We will show that there is at least one other ompletion. Fill ell (2; 0) with element 1, and ell (2; n 2) with element 1. Now olumn 1 ontains 1 in all its ells, ex ept ells (0; 0) and (1; 0). These an now be lled with element 0. Row 1 now ontains 0 in ells (1; 0) and (1; n 1). Thus any remaining ell in this row an be lled, with element 1. Fill these. Only ell (0; n 2) in olumn n 2 hasn't yet been lled. Sin e 1 already o urs n 2 times in this olumn, therefore element 0 an uniquely ll this ell. This for es element 0 into ell (0; 1), and element 1 into ell (2; 1). The result is an F-square of type F (n; 2; n 2). That is, D=f(2; n 2; 0)g has at least two ompletions. Similar arguments an be made for ea h of the sets D=f(3; n 3)g; : : : ;D=f(n 2; 2)g, as well as the deletion of any of the entries in the lower small diagonal. Thus D0 is a riti al set. The size of the riti al set is r = n 1 + n 2 = 2n 3. 2 4 The above theorem is for any integer n. For n even, we have: Theorem 4 If n 4 is even, then ea h F-square of type F = F (n; 2; n 2) has a riti al set D0 of size r = 2n 4. Furthermore D0 is minimal. Proof. Let D0 be the set f(0; 0; 0); (0; 1; 0); (1; 0; 0); (1; 1; 0); (2; 2; 0); (2; 3; 0); (3; 2; 0); (3; 3; 0); : : : ; (n 4; n 4; 0); (n 4; n 3; 0); (n 3; n 4; 0); (n 3; n 3; 0)g. Let D be the shape of D0. Then 0 o urs twi e ea h in rows 0; : : : ; n 3, and twi e ea h in olumns 0; : : : ; n 3. Thus ea h of the remaining ells in these rows and olumns an be lled uniquely with element 1. Having lled these, only four ells remained un lled, namely: (n 2; n 2); (n 2; n 1); (n 1; n 2); (n 1; n 1). But these an be lled uniquely with element 0. Thus D0 is uniquely ompletable. Suppose (0; 0; 0) is deleted from D0. Then rows 1; : : : ; n 3 ontains 0 twi e, and olumns 0; 2; : : : ; n 3. Every empty ell in these rows and olumns an be uniquely lled with element 1. So far ells (0; 0); (0; n 2); (0; n 1); (n 2; 0); (n 2; n 2); (n 2; n 1); (n 1; 0); (n 1; n 2); (n 1; n 1) are not lled. None of these ells an be lled uniquely. Similar arguments an be made for any other proper subset of D0. Thus D0 is a riti al set. Consider D0 again. Ea h non-empty row and olumn ontains 0 exa tly twi e. Any row and olumn isotope of D0 will again have this property. Consider row 0. Suppose we remove the entry 0 in ell (0; 1) and pla e it elsewhere in the same row. Call the new set of triples F . There are only two ells that we an pla e this 0 element in, without destroying the form F (n; 2; n 2), and these are ells (0; n 2) and (0; n 1). Suppose 0 is pla ed in ell (0; n 2). Then as above we an ll all the rows and olumns ontaining the element 0 twi e. These are rows 0; : : : ; n 3, and olumns 0; 2; : : : ; n 3. Sin e in olumn n 1 ells (0; n 1); : : : ; (n 3; n 1) ea h ontain the element 1 on e ea h, therefore ells (n 2; n 1) and (n 1; n 1) an be lled uniquely with entry 0. Thus olumn n 1 too an be uniquely lled. So far, rows n 2; n 1 and olumns 1 and n 2 haven't been lled ompletely. The interse tions of these rows and olumns are ells (n 2; 1); (n 2; n 2); (n 1; 1) and (n 1; n 2). None of these an be uniquely lled. Similar arguments an be made if entry 0 is removed from ell (0; n 1). Similarly, no isotope of F will be uniquely ompletable. The di eren e between D0 and F is in the set D0, every non-empty row and olumn ontains 0 twi e, whereas in F , olumns 1 and n 1 ontained 0 only on e ea h. It an be easily he ked that if an extra 0 is pla ed either in olumn 1 or n 1 without destroying the form F (n; 2; n 2), then the new set will be uniquely ompletable to an F-square of the form F (n; 2; n 2). That is, no non-isotope of D0 that has the same size as D' or that has a smaller size an uniquely omplete. That is, D0 is minimal. The size of the riti al set is r = 2:(n 2) = 2n 4: 2 By modifying the above proofs we an generalise the above result as follows: Theorem 5 Let tjn; t 6= n. Then there is an F-square of type F (n; t; n t) with riti al set of size r = tn t2. 2 Example 1 Let n = 8. Then for t = 1; 2; 4 we have the following riti al sets: 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 Theorem 6 If t k then F (tk+1; t; t(k 1)+1) has a riti al set of size r = (k 1)t2+ 1 2 t(t+1). Proof. Constru t a set D0 of ells as follows: 1. From rows 0; : : : ; u; : : : ; t 1, pi k ells (u; v), where 0 v t 1. 2. From rows t; : : : ; u; : : : ; 2t 1, pi k ells (u; v), where t v 2t 1. : : : : : : 3. From rows t(k 2); : : : ; u; : : : ; t(k 1) 1, pi k ells (u; v), where t v t(k 1) 1. 4. (a) From row t(k 1), pi k ells (t(k 1); v), where t(k 1) v tk 1. (b) From row t(k 1) + 1, pi k ells (t(k 1); v), where t(k 1) v tk 2. : : : : : : ( ) From row tk 1, pi k ell (tk 2; t(k 1)). We laim that the set ontaining all of the above ells, ea h lled with entry 0, is a riti al set.Ea h of the rows 0; : : : ; t(k 1) ontains the element 0 t times. Thus the other entries in these rows an be lled with entry 1. Similarly for olumns 0; : : : ; t(k 1). Row tk and olumn tk now ea h ontain element 1 t(k 1)+1 times. So the other ells in these row and olumn an be lled with element 0. Row tk+1 and olumn tk+1 now ea h ontains the element 0 t times, and so the empty ells in these rows and olumns an be lled with element 1. Continuing this way, we an uniquely ll the F-square. The proof that the above set is riti al is similar to the proofs of the above theorems. On inspe tion r = (k 1)t2 + 12t(t+ 1). 2 Example 2 Consider n = 13. 13 = 2 6+1 and also 13 = 3 4+1. Then for types F (13; 2; 11) and F (13; 3; 10) we have the following riti al sets respe tively: