aa r X i v : . [ m a t h . C O ] O c t The size of maximal systems of brick islands
Tom Eccles ∗ December 11, 2018
Abstract
For integers m , ..., m d > M = [0 , m ] × ... × [0 , m d ] ⊂ R d , a brick of M is a closed cuboid whose vertices have integer coordinates.A set H of bricks in M is a system of brick islands if for each pair of bricksin H one contains the other or they are disjoint. Such a system is maximalif it cannot be extended to a larger system of brick islands. Extending thework of Lengv´arszky, we show that the minimum size of a maximal systemof brick islands in M is P di =1 m i − ( d − C = [0 , m ] d wedefine the corresponding notion of a system of cubic islands, and provebounds on the sizes of maximal systems of cubic islands. The concept of a system of rectangular islands was introduced by Cz´edli in [1].In [4], Pluh´ar generalised this concept to that of a system of brick islands inhigher dimensions, a direction mentioned both in [1] and by Lengv´arszky in [2].To introduce these concepts, let M = [0 , m ] × ... × [0 , m d ] ⊂ R d be a closedcuboid. Then a brick of M is a set of the form [ a , b ] × ... [ a d , b d ], where foreach 1 ≤ i ≤ d we have 0 ≤ a i < b i ≤ m i , and a i , b i ∈ Z . A system of brickislands in M is a set H of bricks in M such that whenever M , M ∈ H , either M ⊆ M , M ⊆ M or M ∩ M = ∅ . We denote the set of systems of brickislands in M by I M , and the maximal elements of I M with respect to inclusionby M ax ( I M ). When M is 2-dimensional, a system of brick islands can also becalled a system of rectangular islands.A related concept is that of a system of square islands , introduced byLengv´arszky in [3]. For m a positive integer, let S = [0 , m ] × [0 , m ] be a closedsquare in the plane. A system of square islands in S is a system of rectangularislands H with every rectangle in H being a square. We denote the set of thesesystems by I S , and the maximal elements of I S with respect to inclusion by M ax ( I S ).We define the higher dimensional analogue of systems of square islands, assuggested in [3]. Let C = [0 , m ] d be a closed cube in d -dimensional space. Thenwe define a system of cubic islands in C to be a system H of brick islands in C such that each brick in H is a cube. We denote the set of these systems by I C ,and the maximal elements of I C with respect to inclusion by M ax ( I C ). ∗ Department of Pure Mathematics and Mathematical Statistics, Centre for MathematicalSciences, Wilberforce Road, Cambridge CB3 0WB, England.
1e shall be concerned with the possible cardinalities of maximal systems ofbrick and cubic islands. For a cuboid M = [0 , m ] × ... × [0 , m d ], we define f d ( m , ..., m d ) = max {| H | : H ∈ M ax ( I M ) } and g d ( m , ..., m d ) = min {| H | : H ∈ M ax ( I M ) } . Similarly, for a cube C = [0 , m ] d , we define f ′ d ( m ) = max {| H | : H ∈ M ax ( I C ) } and g ′ d ( m ) = min {| H | : H ∈ M ax ( I C ) } . We summarise the main results of Cz´edli [1], Lengv´arszky [2], [3] and Pluh´ar [4]that relate to our work. All these results concern the possible cardinalities ofmaximal systems of rectangular, square or brick islands. In [1], Cz´edli provedthat for systems of rectangular islands in M = [0 , m ] × [0 , m ], f ( m , m ) = (cid:22) m m + m + m − (cid:23) . In [2] and [3], Lengv´arszky proved g ( m , m ) = m + m − , and for systems of square islands in S = [0 , m ] × [0 , m ], g ′ ( m ) = m , and f ′ ( m ) ≤ m ( m + 2)3with equality in the last being achieved for k a positive integer and m = 2 k −
1. In[4], Pluh´ar proved that for systems of brick islands in M = [0 , m ] × ... × [0 , m d ], m m ...m d + P m j ...m j d − d − − ≤ f d ( m , ..., m d ) ≤ ( m + 1) ... ( m d + 1)2 d − − d − { } . Using methods similar to those employed in [2] and [3], we shall prove thefollowing theorems about the possible cardinalities of maximal systems of brickand cubic islands. 2 heorem 1.
Let M = [0 , m ] × ... × [0 , m d ] ⊂ R d be a cuboid. Then the minimalsize of a maximal system of cuboid islands in M , is given by g d ( m , ..., m d ) = d X i =1 m i − ( d − . Theorem 2.
Let C = [0 , m ] d be a cube in d -dimensional space. Then theminimal size of a maximal system of cubic islands in C is given by g ′ d ( m ) = m. Theorem 3.
Let C = [0 , m ] d be a cube in d -dimensional space. Then themaximal size of a system of cubic islands in C , is bounded by f ′ d ( m ) ≤ ( m + 1) d − d − . Moreover, equality can be achieved when m = 2 k − for some positive integer k . The rest of this paper will be organised as follows. In section 4, we shall provethe upper bound for Theorem 1. In section 5, we shall make some preliminaryobservations which will help us in the proof of the lower bound. Then we shallprove the lower bound, in section 6. In section 7, we shall classify the minimalmembers of
M ax ( I M ) for a cuboid M . In sections 8 and 9, we shall proveTheorems 2 and 3 respectively. To establish one direction of Theorem 1, we show that g d ( m , ..., m d ) ≤ d X i =1 m i − ( d − M = [0 , m ] × ... × [0 , m d ], define a set of bricks H by H = { [0 , m ] × ... × [0 , m i − ] × [0 , n i ] × [0 , × ... × [0 ,
1] : 1 ≤ i ≤ d, ≤ n i ≤ m i } . This defines a system of P di =1 m i − ( d −
1) nested bricks. Since each of thesebricks extends the last by 1 in one dimension, H is a maximal system of brickislands in M , which establishes our upper bound on g d . Working towards the lower bound for g d , we shall start with some observationsabout maximal systems of brick islands. For M = [0 , m ] × ... × [0 , m ], we definean elementary cube in M to be a cube of the form [ a , a + 1] × ... × [ a d , a d + 1],where for each 1 ≤ i ≤ d , a i ∈ Z and 0 ≤ a i ≤ m i −
1. We call this cube theelementary cube based at ( a , a , ..., a d ).3 bservation 1. Suppose I ⊂ [ n ] and A is an elementary cube based at ( a , ..., a d ) such that a i = 0 whenever i ∈ I . Suppose A ′ is another elementary cube, basedat ( a ′ , ..., a ′ d ) . If a ′ i = a i for i / ∈ I , and a ′ i ∈ { , } for i ∈ I , then everyelementary cube that intersects A also intersects A ′ . For a system of brick islands H in M , let M ax ( H ) be the set of maximalelements of H \{ M } with respect to inclusion. Corollary 1.
Let M = [0 , m ] × ... × [0 , m d ] and let H ∈ M ax ( I M ) . Supposethat | M ax ( H ) | > and R = [ r , , r , ] × ... × [ r d, , r d, ] ∈ M ax ( H ) . Then no r i, is , and no r i, is m i − ,Proof. Indeed, suppose r , = 1. Let R ′ be [0 , r , ] × ... × [ r d, , r d, ]. From theobservation above, any elementary cube intersecting R ′ intersects R , and henceno element of M ax ( H ) \{ R } intersects R ′ . The brick R ′ cannot be in H alreadyas then we would have R ′ = M and | M ax ( H ) | = 1. This shows that H is notmaximal, since we can add R ′ to it, which is a contradiction. Corollary 2.
Let M = [0 , m ] × ... × [0 , m d ] and let H ∈ M ax ( I M ) . If | M ax ( H ) | > , every vertex of M is occupied by a member of M ax ( H ) .Proof. Given a vertex v of the cuboid M , let C v be the elementary cube whichcontains v . As H is maximal, M ax ( H ) contains some brick R which intersects C v . By the previous result, R must contain C v . Corollary 3.
Let M = [0 , m ] × ... × [0 , m d ] and let H ∈ M ax ( I M ) . Suppose | M ax ( H ) | > and R , R are bricks in M ax ( H ) which intersect an edge E ofthe cube. There is some section of E between the intersections of R and R with E - we shall call this the gap between R and R on E . Suppose that noother member of M ax ( I M ) intersects this gap. Then the length of the gap is atmost . Further, if the length of the gap is exactly , neither of R , R is anelementary cube.Proof. We may assume that E = { ( x, , ...,
0) : 0 ≤ x ≤ m } . Suppose there isa gap of at least 3 between R , R on E - so no member of M ax ( H ) intersects { ( x, , , ...,
0) : a < x < a + 3 } , for some integer 1 ≤ a ≤ n −
4. Then, byapplying Corollary 1 three times, we see that the elementary cube based at( a + 1 , , , ...
0) intersects no member of
M ax ( H ) - otherwise this member of M ax ( H ) would also intersect the gap between R and R on E . This gives riseto a contradiction - H is not maximal, as we can add this elementary cube toit. Now, suppose we have a gap of length 2 between R and R on E - so that( a, , ... ∈ R , ( a + 2 , , ..., ∈ R , and no member of M ax ( H ) intersects { ( x, , , ...,
0) : a < x < a + 2 } , for some integer 1 ≤ a ≤ n −
3. If R was an elementary cube, then we extend it to R ′ by adding in the elementarycube based at ( a, , , ... a + 1 , , ... R ′ intersects no elements of M ax ( H ) other than R . This shows that we can add R ′ into H , contradictingits maximality. Observation 2.
Let M = [0 , m ] × ... × [0 , m d ] , H ∈ M ax ( I M ) and R ∈ H .Then those bricks in H which are contained in R form a set in M ax ( I R ) . Also, itself must be in H . In particular, if R , ..., R k are members of M ax ( H ) ,where R i has side length r ij in dimension j , then | H | ≥ k X i =0 g d ( r i , ..., r id ) Now we are ready to prove the main theorem. Given H ∈ M ax ( I M ), our taskis to show that | H | ≥ P di =1 m i − ( d − d ,and within this by induction on P di =1 m i . First, we establish a slightly strongerresult for d = 1. Lemma 1.
Let M = [0 , m ] ⊂ R be a line segment. Then every maximal systemof cuboid islands in M has size m . In particular, g ( m ) = m .Proof. We prove this by induction on m . For m = 1, the result is trivial. For m ≥ H ∈ M ax ( I M ). There are two different forms that M ax ( H )can take - either it is a single interval [0 , m − M ax ( H ) = { [0 , a ] , [ a + 1 , m ] } for some integer 1 ≤ a ≤ m −
1. In either case, we use Observation 2 andapply the induction hypothesis to the members of
M ax ( H ). This shows that H consists of m − M ax ( H ),together with M itself, and so | H | = m .In d dimensions, our base case is when any side length m i of M is 1. In thiscase, the problem reduces immediately to the ( d − m i ≥ i , and that the theorem holds whenever P di =1 m i is reduced. We shall now proceed in three different ways, dependingon the configuration of M ax ( H ) inside M . The first two cases deal with specialconfigurations which can arise when | M ax ( H ) | is small. | M ax ( H ) = 1 | Without loss of generality,
M ax ( H ) = { [0 , m − × [0 , m ] ... × [0 , m d ] } Applying the induction hypothesis to the sole member of
M ax ( H ), and usingObservation 2, we find that | H | ≥ g d ( m − , m , ..., m d ) = d X i =1 m i + ( d − | M ax ( H ) | > , and M ax ( H ) has an elementwhich divides M into or more regions Let R be a member of M ax ( H ) which divides M , with R = [ r , r ] × [0 , m ] × ... × [0 , m d ] . r = 1 and r = m −
1. If r = 0, then we use observation2 and apply the induction hypothesis in R and in R = [ r +1 , m ] × [ m ] × ... [ m d ],which must be the sole other member of M ax ( H ).This gives | H | ≥ d X i =1 m i − m + r − ( d − ! + d X i =1 m i − m + ( m − r − − ( d − ! = d X i =1 m i − ( d − ! + d X i =1 m i − m − ( d − ! . As every m i is at least 2, this shows that | H | larger than we claim for Theorem1 - and so equality cannot hold in this case.If, on the other hand, 1 < r < r < n −
1, then we must have that
M ax ( H ) = { R, [ r + 1 , m ] × [ m ] × ... [ m d ] , [0 , r − × [ m ] × ... [ m d ] } Using Observation 2 and applying the induction hypothesis to each of thesethree bricks, we find that | H | ≥ d X i =1 m i − ( d − ! + d X i =1 m i − m − d − − ! . Again, using the fact that each m i is at least 2, this gives the bound we requirefor | H | with strict inequality. | M ax ( H ) | > , and no element of M ax ( H ) di-vides M into regions We define a path P around some edges of the cuboid by P = { (0 , , ..., , x i , m i +1 , ..., m d ) : 1 ≤ i ≤ d, ≤ x i ≤ m i }∪{ ( m , m , ..., m i − , x i , , ...,
0) : 1 ≤ i ≤ d, ≤ x i ≤ m i } We note that P has two edges in each direction, and that these edges are dia-metrically opposite each other in M . Hence no brick in M ax ( H ) intersects bothof these edges, or else it would divide M . The length of P is 2 P di =1 m i , and P has 2 d corners with 2 edges incident at each.Now, consider all the members of M ax ( H ) which intersect P . Suppose thereare k of them, A , ..., A k , with the j th dimension edge length of A i being denoted a ij . By Corollary 3, the gaps between consecutive bricks on P are at most 2.Writing n as the number of gaps of length 2, Corollary 3 tells us that at least n of the A i are not elementary cubes (eg. the ones after the gaps of length 2).Now, the edges of the bricks which lie on P have total length2 d X i =1 m i − k − n k + 2 d such edges (as there are 2 d corners in P ). Hence the A i have between them k ( d − − d edges which are not on P - and so we havethat k X i =1 d X j =1 a ij ≥ d X i =1 m i − k − n + k ( d − − d Now, using Observation 2 and applying the inductive hypothesis in each A i , weobtain | H | ≥ k X i =1 g d ( a i , ..., a id ) ≥ k X i =1 d X j =1 a ij − k ( d −
1) + 1 ≥ d X i =1 m i − k − n − d + 1= d X i =1 m i − d + 1 ! + d X i =1 m i − k − n − d ! . Since the first bracket is the bound we wish to establish for H , this is establishesthe theorem unless k + n > d X i =1 m i − d. In this case, we observe that H contains each of the k bricks A i , at least onefurther brick contained in each A i which is not an elementary cube, and M itself. Since there are at least n bricks A i which are not elementary cubes, weget that | H | ≥ k + n + 1 ≥ d X i =1 m i − d + 2 . This shows that in this final case Theorem 1 holds with strict inequality.
When we showed the upper bound for g d ( m , ..., m d ), we gave one example of asmallest possible maximal system. In this section we classify all such systems. Lemma 2.
Let M = [0 , m ] × ... × [0 , m d ] and let H ∈ M ax ( I M ) . If | H | isminimal among members of M ax ( I M ) , d ≥ and m i ≥ for at least choicesof i , then | M ax ( H ) | = 1 .Proof. We first note that if m d = 1, maximal systems of brick islands in M areprecisely those in M ′ = [0 , m ] × ... × [0 , m d − ]. Using this, we can work insteadin the cuboid given by projecting in all dimensions where the side length of M is 1. So we shall assume that m i ≥ ≤ i ≤ d . When d = 2, this wasproved by Lengv´arszky [2]. Examining the proof of Theorem 1 we note that forequality to hold we must have the following constraints on H :7 M ax ( H ) = { A i : 1 ≤ i ≤ k }• Every A i is an elementary cube or a brick with all sides of length 1 exceptfor one side of length 2. • If some side length a ij of A i is greater than 1, then some side of A i thatlies along P must be in direction j .From these last two constraints we can deduce that every elementary cube con-tained in some A i lies on an edge of P . If d ≥ m i ≥ ≤ i ≤ d ,let v be some vertex of M which is not on P , and C v be the elementary cubewhich contains it. Then no brick A i intersects C v . However, by the first con-straint there are no other members of M ax ( H ); hence we can add C v to H ,contradicting the maximality of H . This contradiction establishes the lemmawhenever d ≥ m i ≥ ≤ i ≤ d .So we may assume that d ≥
3, and that m d = 2. In this case we define setsof bricks H , H , H in d − H = { R × [0 ,
1] : R ∈ H } ∪ { R × [1 ,
2] : R ∈ H } ∪ { R × [0 ,
2] : R ∈ H } We note that no element of H intersects an element of H . Now, H ∪ H ∪ H is a maximal system of brick islands in the ( d − M ′ =[0 , m ] × ... × [0 , m d − ], and so by Theorem 1 | H ∪ H ∪ H | ≥ d − X i =1 m i − d + 2 = d X i =1 m i − d. Thus if we have equality in Theorem 1 for H , then there is at most one inter-section between any of H , H and H . We observe that any minimal memberof H must be in H ∪ H , and any maximal member of H ∪ H must be in H . So for equality to hold, H has a unique minimal element R , which is alsothe unique maximal element of H ∪ H . We also know that H has a uniquemaximal element M ′ corresponding to M ∈ H , and so the bricks in H must benested. If | H | ≥
2, then the second largest element of H corresponds in H to the unique element of M ax ( H ); if | H | = 1, then R in H ∪ H correspondsin H to the unique element of M ax ( H ).Using this lemma, we can classify the minimal elements of M ax ( I M ). Asystem of brick islands is a minimal element of M ax ( I M ) if and only if it canbe obtained by the following procedure: • Take any brick R in M such that every side length of R is 1 except forone dimension, in which it is r . • Take any system of r brick islands in R (the largest of which is R itself). • All other bricks are nested, with R being the smallest and M being thelargest, such that each brick extends the last by 1 in one direction.We count the bricks in such a system. There are r bricks within R , and P di =1 m i − ( r + d −
1) to extend each dimension to m i , giving a system of therequired size. We prove that these are all the minimal elements of M ax ( I M ) byinduction on P di =1 m i . The base case is when M has at most one dimension ofsize at least 2, in which case we can take R = M . If m i > m i , then M ax ( H ) has a unique element H max by lemma 2. Applyingthe induction hypothesis in H max , we obtain the result for M .8 Proof of Theorem 2
Before we prove our results about systems of cubic islands, we observe theobvious analogue of Observation 2 for cubic islands.
Observation 3.
Let C = [0 , m ] d , H ∈ M ax ( I C ) and R ∈ H . Then those bricksin H which are contained in R form a set in M ax ( I R ) . Also, C itself must bein H . In particular, if R , ..., R k are the members of M ax ( H ) , where R i hasside length r i k X i =0 g ′ d ( r i ) ≤ | H | ≤ k X i =0 f ′ d ( r i )Now we prove Theorem 2, on the minimal size of maximal systems of cubicislands. We wish to show that g ′ d ( m ) = m . We first note that g ′ d ( m ) ≥ m , asa sequence of m nested cubes is maximal in C = [0 , m ] d . To prove the upperbound for g ′ d ( m ) we will proceed by induction on m . For m ≤ d = 2, the theorem was proved in [3]. So we shall assume d ≥ m ≥
3, and that the theorem holds ∀ m ′ ≤ m . Given C = [0 , m ] d , and H ∈ M ax ( I C ), our task is to show that | H | ≥ m . We proceed in three differentways, depending on the size of the largest element of H . H contains an element of size m − In this case, the result follows immediately from the inductive hypothesis, to-gether with Observation 3. H is of size m − Denote the element of H of size m − R . Consider the bottom left corner of R , with coordinates ( r , ..., r d ), with each of the r i in { , , } . Note that theyare not all 1 - if they were we could extend the system H by adding in a cubeof size [0 , m − d . Now, for 1 ≤ i ≤ d , set a i = (cid:26) r i = 2 m i − r i = 0 or 1Then we note that the elementary cube based at ( a , ..., a m ) does not intersect R . This shows that, by the maximality of H , there is at least one cube R ′ otherthan R in M ax ( H ). Thus H contains C , R ′ and ( m −
2) cubes contained in R (by Observation 3, and the inductive hypothesis applied to R ). Consequently | H | ≥ m , as required. H are of size at most m − Consider the path P as in the proof of Theorem 1; P = { (0 , , ..., , x i , m, ..., m ) : 1 ≤ i ≤ d, ≤ x i ≤ m }∪{ ( m, m, ..., m, x i , , ...,
0) : 1 ≤ i ≤ d, ≤ x i ≤ m } Given two points p and p on P which are seperated on P by at least twovertices of C , and elementary cubes C and C containing p and p respectively,9e note that p and p differ by m in (at least) 1 dimension. Hence no cube ofside at most m − C and C .Let A , ...A k be those cubes in M ax ( H ) which contain a point of the form p + ( c , ..., c d ), with p ∈ P and | c i | ≤ ≤ i ≤ d . We project A i on to thosepoints p ∈ P for which A i has a point of this form. Then each A i is projectedonto at most 2 edges of P (which occurs precisely when A i is at most 1 awayfrom a corner of P in every direction). The gaps between adjacent projectionsare at most 2, very similarly to in the cuboid case - if there is a gap of 3, wecan put an elementary cube on P in the middle of it to extend H . We may alsoget extra gaps at the 2 d corners of P , as the cubes closest to the corners neednot project into them. These gaps have size at most 2. Writing a i for the sidelength of the cube A i , this gives2 k X i =1 a i + 2 k + 4 d ≥ md Thus at least one of the following must hold: k ≥ m − k X i =1 a i ≥ m − d − ≥ m ( d − m ≥ d ≥
3. If 1 holds, then we note that | H | ≥ k + 1, as each of the A i and C itselfare in H . If 2 holds, we use Observation 3 and apply the inductive hypothesisto each A i to obtain | H | ≥ P ki =1 a i + 1. In either case, we get that | H | ≥ m . Finally, we prove Theorem 3, on the maximal size of a system of cubic islands.As in the setup of the theorem, let C = [ m ] d and H ∈ M ax ( I C ). Then our taskis to show that | H | ≤ ( m +1) d − d − , and to demonstrate an H for which equalityholds when m = 2 k −
1. We do the latter first. For C k = [2 k − d we definesystems of cubic islands H k recursively. C has H = { C } . To form H k , divide C k into 2 d subcubes of side 2 k − − d hyperplanes passing through themiddle of the cube. Place a copy of H k +1 in each of these subcubes, and add C k to obtain H k . This gives | H k | = 2 d | H k − | + 1= 2 d ( k − d − d − k d − d − | H | ≤ ( m +1) d − d − , we use induction on m . The result is trivial for m = 1. Given m ≥
2, we split into 2 cases, depending on the size
M ax ( H ).10 .1 Case 1: | M ax ( H ) | = 1 Here
M ax ( H ) has a unique member R of side length m −
1. Applying theinduction hypothesis in R , | H | ≤ m d − d − ≤ ( m + 1) d − d − | M ax ( H ) | ≥ In this case,
M ax ( H ) has no elements of side length greater than m −
2. Now,order the vertices of C and consider each in turn. If the elementary cube C v which contains the vertex v intersects no element of M ax ( H ), then add C v into H . If C v intersects some element R of M ax ( H ) but is not contained in it, thenmove R into the corner, together with every cube in H that it contains. Note R cannot have contained any other vertex of C , as it has side length at most m −
2. After applying this process to every vertex, we have a family H ′ with | H ′ | ≥ | H | , such that every vertex of C is occupied by a different element of M ax ( H ′ ). Hence | M ax ( H ′ ) | ≥ d . Now we use the same argument as appliedin [1] and [3].Suppose A , ..., A k are the elements of M ax ( H ′ ). Then extend each A i by in every direction. The interiors of the extended cubes are disjoint, and arecontained within an extended version of C of side m +1. Thus their total volume P ki =1 ( a i +1) d is at most ( m +1) d . We write n ( A i ) for the number of cubes of H ′ which are contained in A i , and a i for the side length of A i . Using Observation3, and applying the inductive hypothesis in each A i , we get that | H | ≤ | H ′ | ≤ k X i =1 n ( A i ) ≤ k X i =1 ( a i + 1) d − d − ≤ m + 1) d d − − d d − n + 1) d − d − | H | , and our proof is complete.
10 Further work
As mentioned in [3], we could consider the problem of cubic islands in a cuboid;the members of our system H would be cubes, while M remains a cuboid witharbitrary sides. While we have got a best polynomial upper bound on f ′ d ( m ), wehave not found a reasonable lower bound, and this is another possible extension. References [1] G. Cz´edli, The number of rectangular islands by means of distributive lat-tices,
European Journal of Combinatorics , (2009), 208-215.[2] Z. Lengv´arszky, The minimum cardinality of maximal systems of rectangularislands, European Journal of Combinatorics , (2009), 216-219.113] Z. Lengv´arszky, The size of maximal systems of square islands, EuropeanJournal of Combinatorics , (2009), 889-892.[4] G. Pluh´ar, The number of brick islands by means of distributive lattices, Acta Universitatis Szegediensis ,75