On simple arrangements of lines and pseudo-lines in P^2 and R^2 with the maximum number of triangles
aa r X i v : . [ m a t h . C O ] J un ON SIMPLE ARRANGEMENTS OF LINES AND PSEUDO-LINESIN P AND R WITH THE MAXIMUM NUMBER OFTRIANGLES
NICOLAS BARTHOLDI, J´ER´EMY BLANC, AND S´EBASTIEN LOISEL Introduction
A curve Γ ⊂ P = P ( R ) (respectively ⊂ R ) is a projective (respectively an affine ) pseudo-line if there is a homeomorphism φ : P → P (respectively φ : R → R ) such that φ (Γ) is a line.A projective (respectively an affine ) arrangement of n pseudo-lines is a set of n pseudo-lines in P (respectively in R ), such that any pair of pseudo-lines intersectsin exactly one point. A projective (respectively an affine ) arrangement of lines issuch an arrangement where each pseudo-line is a line. Note that there usually isno homeomorphism φ of the plane turning a pseudo-line arrangement A into a linearrangement φ ( A ) ([Gr¨u], page 42, Theorem 3.2). Our sole interest is with simplearrangements , i.e. arrangements without multiple intersections.A simple projective arrangement A of n lines decomposes the projective plane P into n ( n − / p ( A ) the number of triangles obtained. We can do the same for pseudo-lines, as a triangle is a region delimitedby exactly three pseudo-lines of the arrangement. Similarly, in the Euclidean plane R , we denote by a ( A ′ ) the number of (bounded) triangles delimited by an affinearrangement A ′ .It was originally proposed by Gr¨unbaum [Gr¨u] to look for arrangements withmany triangles, and there is already a substantial literature on this question.Denote by p s ( n ) (respectively a s ( n )) the maximal number of triangles that can beobtained with a simple arrangement of n lines in the projective plane (respectively inthe Euclidean plane). We denote by p s ( n ) and a s ( n ) the same notions for pseudo-lines. A projective arrangement A of n pseudo-lines such that p ( A ) = p ( n )is classically called p -maximal . Here we will only say that the arrangement is maximal , and will use the same terminology for affine arrangements.Then, an easy observation on the number of segments shows that if n ≥
4, thefollowing relations occur:(1) p s ( n ) ≤ p s ( n ) ≤ n ( n − / ≥ ≥ ≥ a s ( n ) ≤ a s ( n ) ≤ n ( n − / . We will say that a segment is used if it is a part of one triangle of the arrangement,and say that it is unused otherwise. An arrangement satisfying the equality withthe bound above is an arrangement whose segments are all used in one triangle– we will say in this case that it is a perfect arrangement . Note that a perfectarrangement is maximal, but the converse is false in general.
Mathematics Subject Classification.
There is currently no known n such that p s ( n ) (cid:9) p s ( n ) or a s ( n ) (cid:9) a s ( n ).Infinitely many examples of integers n ≡ , p s ( n ) = n ( n − / n = 12 (see[Ro2]). A construction in straight lines has been given in [FoRa] to prove that p s ( n ) = n ( n − / n = 2 · t + 2, for any integer t ≥ p s ( n ) = n ( n − / a s ( n −
1) = ( n − n − /
3, and the same if true forarrangements of straight lines (i.e. for p s and a s ). There exist thus infinitely manyexamples of integers n ≡ , a s ( n ) = n ( n − /
3, and we havealso a s ( n ) = n ( n − / n = 2 · t + 1.The projective odd case is worse than the even case: it was observed by J.Granham ([Gr¨u], page 26, Theorem 2.21) that(2) p s ( n ) ≤ n ( n − / if n > is odd . Conversely, the affine even case is worse than the odd case. We give a new boundin this case:
Theorem 1.1. If n is an even integer, then a s ( n ) ≤ ⌊ n ( n − / / ⌋ . The bound of Theorem 1.1 is reached for 4 , , ,
16 pseudo-lines but not for8 , ,
14 pseudo-lines (see Theorem 1.4). Note that adding one line to an affineperfect arrangement of n − ≡ , n ≡ , a s ( n ) ≥ n ( n − / /
3, which isclose to the polynomial of Theorem 1.1. It would be interesting to find the bestpolynomial upper bound, which is between the two above. Note that there existsno even integer n where a s ( n ) > n ( n − / / n ≡ ⌊ n ( n − / ⌋ triangles (Theorem 1.3).However, in the projective even case, the bound may be improved, to give thefollowing result, which seems to be already known (see [Ro3] Table I), but we werenot able to find a proof in the literature. Proposition 1.2. If n ≡ , then p s ( n ) ≤ ⌊ n ( n − / ⌋ − . Note that this bound could be improved, as there is still no known examplewhere the equality occurs.We give then in Proposition 3.1 a way to obtain a good affine arrangementof 2 n − n lines. The construction ishomeomorphic to those of [Ha1] and [FuPa] but has two advantages: it is explicitand may be used starting from non-perfect arrangements. This gives in particularthe following new sequences: Theorem 1.3.
For any integer t ≥ we have RIANGLES IN ARRANGEMENTS OF LINES AND PSEUDO-LINES 3 if n = 14 · t + 1 , a s ( n ) = n ( n − / if n = 6 · t + 1 , a s ( n ) = ⌊ n ( n − / ⌋ = ( n ( n − − / if n = 18 · t + 1 , a s ( n ) = ⌊ n ( n − / ⌋ = ( n ( n − − / . In particular, this shows that the best polynomial upper bound for a s ( n ), a s ( n ), p s ( n ) and p s ( n ), n ≡ n ( n − − / ⌊ n ( n − / ⌋ .Finally, we are able to describe the explicit values of a s ( n ), a s ( n ), p s ( n ), p s ( n )for small values of n . A computer program – described in Section 4 – allows usto find explicitly some values of a s ( n ). Using the bounds and sequences describedabove, and the relation between a s ( n ) and p s ( n + 1) – adding the line at infinity –we obtain the following result: Theorem 1.4.
The values of a s ( n ) , a s ( n ) , p s ( n ) , p s ( n ) are given in the table n p s ( n ) −
47 58 −
65 80 a s ( n ) n
17 18 19 20 21 22 23 p s ( n )
85 102
107 124 −
133 154 161 a s ( n )
85 93 −
107 116 −
133 143 − n
24 25 26 27 28 29 30 p s ( n ) 184
191 214 −
225 252
261 290 a s ( n ) 172 −
191 203 −
225 238 −
261 275 −
276 .An entry of the form x − y means that we have an arrangement with x pseudo-lines but that the best upper bound is y . Bold entries are known to be stretchable.Underlined quantities are strictly smaller than the bounds given above. Greyentries were previously known (in particular in [Ro3] for pseudo-lines); we includethem for completeness. Some new bounds – Proof of Theorem 1.1 and Proposition 1.2
We prove Theorem 1.1, i.e. that for any even integer n , the inequality a s ( n ) ≤⌊ n ( n − / / ⌋ holds. Proof of Theorem 1.1.
Let A be an affine simple arrangement of n pseudo-lines,with n ≥ L ∈ A contains n − L touchesexactly n − t , ..., t n − ,such that t i and t i +1 have a common vertex for i = 1 , ..., n −
3, denote by M and N the two pseudo-lines intersecting L in the extremities (such that M touches t and N touches t n − ) and denote by ∆ the region delimited by the three lines L , M , N (which is not a ”triangle” of our arrangement as other pseudo-lines intersectit). The n − L are alternatively inside ∆ and outside it. So,either t or t n − is not contained in ∆. Without loss of generality, we assume that t n − is not contained in ∆ and illustrate the situation in Figure 1. Note that thesegment of the line N which starts from L and which is not contained in t n − isnot used. (On the figure, the segment with an arrow).Then, to every pseudo-line that contains n − L , but it has one of its NICOLAS BARTHOLDI, J´ER´EMY BLANC, AND S´EBASTIEN LOISEL
M NLt ∆ t t t n − t n − ↓ Figure 1.
The situation of the pseudo-line L extremities on it. As the arrangement is simple, a segment cannot be associated tomore than two pseudo-lines.Denote by m the number of pseudo-lines that contain exactly n − m/ ab absurdo , that thereare more than n ( n − / / n ( n − /
3) segments are used,so n ( n − /
3) + m/ ≤ n ( n − m ≤ n . But then, the numberof used segments is at most m · ( n −
2) + ( n − m ) · ( n −
3) = n · ( n −
3) + m ≤ n · ( n − / , which is a contradiction. (cid:3) We prove now Proposition 1.2, i.e. that p s ( n ) < ⌊ n ( n − ⌋ for any positive integer n ≡ Proof of Proposition 1.2.
Suppose that there exists some projective arrangement A of n pseudo-lines with exactly ⌊ n ( n − ⌋ = n ( n − − triangles. Since the numberof segments is not divisible by 3, there exists at least one of them which is nottouching any triangle. We choose then one pseudo-line that touches at most n − n − n ( n − − − ( n −
2) triangles.But this number is strictly bigger than ( n − n − , which is not possible. (cid:3) A way to construct maximal arrangements
Proposition 3.1.
Let n ≥ be an even number and let A = { Y , L , ..., L n } be asimple affine arrangement of n + 1 lines, given by the equations Y := { ( x, y ) ∈ R | y = 0 } ,L i := { ( x, y ) ∈ R | y = m i ( x − a i ) } , where { a , ..., a n − } = (cid:8) tan( α ) (cid:12)(cid:12) ± α ∈ { πn , πn , ..., π − πn } (cid:9) − n < a n − < < a n < n ; and such that the line Y touches exactly n − triangles (which means that everyone of its segments is used in one triangle) of the affine arrangement A .Then, there exist n lines M , M , ..., M n given by the equations M i := { ( x, y ) ∈ R | y = µ i ( x − b i ) } ,where b i = tan( β i ) and { β , ..., β n } = (cid:8) − π + · πn , − π + · πn , ..., − · πn , · πn , ..., π − · πn (cid:9) RIANGLES IN ARRANGEMENTS OF LINES AND PSEUDO-LINES 5 and such that the affine arrangement B = { Y , L , . . . , L n , M , ..., M n } of n + 1 lines is simple and has exactly n triangles more than A ; the line Y touches exactly n − triangles of the arrangement B .Explicitly, we can take µ i := σ · m min n · (cid:0) sin (cid:0) β i (cid:1) + n · b i (cid:1) , where σ := 1 if L n and L n − intersect in the upper half-plane and σ := − other-wise, and where m min := min {| m i | | i = 1 , .., n } .Remark . If | a n − | and | a n | are smaller than 1 / n , then the new arrangement B also satisfies the conditions of the Proposition; this allows us to iterate the processif | a n − | and | a n | are arbitrary small. Proof.
Write ǫ := n , ǫ := n . Multiplying all the slopes by − L n and L n − intersect in the upper half plane.The explicit values of µ i given in the Proposition become thus µ i = ǫ · m min · (cid:0) sin (cid:0) β i (cid:1) + b i · ǫ (cid:1) . We will use the fact that { /b i | i = 1 , ..., n } = { b i | i = 1 , ..., n } .We calculate some simple assertions. For 1 ≤ i ≤ n , we have π/ n < tan( π/ n ) ≤| b i | ≤ tan( π/ − π/ n ) = 1 / tan( π/ n ) < n/π and π/ n < sin( π/n ) ≤ | sin(2 β i ) | ≤
1. This gives – using the equality ǫ = n − – the following relations(3) π/ n < | b i | < n/π ; π/ n < | sin(2 β i ) | ≤ /n < | sin(2 β i ) + b i · ǫ | < /n · m min < | µ i | < m min /n ;and we see that µ i , sin(2 β i ) and b i have the same sign.For 1 ≤ i ≤ n −
2, we obtain similarly the relation(4) π/n < | a i | < n/π. We calculate now some coordinates of intersections of the lines of B .1. The y -coordinate of the intersection of L i and L j (for i = j ) is equal to( a i − a j ) · ( m i − m j ) − . Assuming that { i, j } 6 = { n, n − } , we have | a i − a j | ≥ tan( π/n ) − /n > π/n − /n > /n . Since | m i − m j | ≤ /m min , we obtain thefollowing assertion:(5) The y -coordinate of L i ∩ L j , for { i, j } 6 = { n, n − } is (in absolute value) bigger than m min /n. Note that the lines L n − and L n may intersect at a very small y -coordinate.2. The y -coordinate of the intersection of L i and M j is equal to ( a i − b j ) · ( m i − µ j ) − . We calculate first – using (3) and (4) – that | a i − b j | ≤ max | a k | + max | b k |
3. The x -coordinate of the intersection of M i and M j is equal to x ij = µ i b i − µ j b j µ i − µ j = sin(2 β i ) b i − sin(2 β j ) b j sin(2 β i ) − sin(2 β j ) + ǫ · ( b i − − b j − ) , where b i = tan( β i ), b j = tan( β j ). We study now three cases:3a) If β i + β j = 0, the x -coordinate x ij is equal to 0, and the y -coordinate isnegative.3b) Assume that β i + β j = ± π/
2, which implies that sin(2 β i ) = sin(2 β j ) and b i b j = 1, whence 1 /b i − /b j = b j − b i . We find x ij = sin(2 β i ) · ( b i − b j ) / ( ǫ · ( b j − b i )) = − sin(2 β i ) /ǫ , which implies – with (3) – that | x ij | > ( π/ n ) /n − > n .3c) Assume that β i + β j / ∈ { , ± π/ } . The trigonometric identities leads tosin(2 β i ) tan( β i ) − sin(2 β j ) tan( β j ) = tan( β i + β j )(sin(2 β i ) − sin(2 β j )), which impliesthat x ij − tan( β i + β j ) = − tan( β i + β j ) · ǫ · (1 /b i − /b j )sin(2 β i ) − sin(2 β j ) + ǫ · (1 /b i − /b j ) . We bound the values of this expression: π/n < π/ n ) ≤ | /b i − /b j | ≤ / tan( π/ n ) < n/π , and 8 /n = 2 /π · (2 π/n ) < − cos(2 π/n ) = sin( π/ − sin( π/ − π/n ) ≤ | sin(2 β i ) − sin(2 β j ) | ≤
2, and π/n < tan( π/n ) ≤ | tan( β i + β j ) | ≤ tan( π/ − π/n ) = 1 / tan( π/n ) < n/π . We obtain – since ǫ = n − – the followingbounds(7) π/n < | /b i − /b j | < n/π ;8 /n < | sin(2 β i ) − sin(2 β j ) | ≤ /n < | sin(2 β i ) − sin(2 β j ) + ǫ (1 /b i − /b j ) | < π/n < | tan( β i + β j ) | < n/π, and see that the expressions sin(2 β i ) − sin(2 β j ) + ǫ (1 /b i − /b j ) and sin(2 β i ) − sin(2 β j ) have the same sign.The bounds (7) yield a minimal bound for | x ij − tan( β i + β j ) | , which is ( π/n ) · ǫ · ( π/n ) / π / n · ǫ > n − . Similarly, the maximal bound is ( n/π ) · ǫ · (4 n/π ) / (7 /n ) = 4 / π · n · ǫ < n − . We obtain the following relation(8) 3 /n < | x ij − tan( β i + β j ) | < /n . We study now the sign of x ij − tan( β i + β j ), which is the same as those of − tan( β i + β j ) · (1 /b i − /b j ) · (cid:0) sin(2 β i ) − sin(2 β j ) (cid:1) = − (cid:0) sin(2 β i ) tan( β i ) − sin(2 β j ) tan( β j ) (cid:1) · (cid:0) / tan( β i ) − / tan( β j ) (cid:1) .Note that the function x sin(2 x ) tan( x ) = 2 sin( x ) on ] − π/ π/
2[ actslike x x (it is an even function, growing on [0; π/ x ) tan( x ) by x in the above expression without changing the sign. Similarly,we may replace 1 / tan( x ) by 1 /x . The sign of x ij − tan( β i + β j ) is thus the sameas the sign of − ( β i − β j ) / (1 /β i − /β j ) = ( β i + β j ) · β i · β j .(9) The sign of x ij − tan( β i + β j ) is the same as the sign of ( β i + β j ) · β i · β j .
4. We describe now the order of the x -coordinates of the intersections of someline M i with the other lines of B . Assume that b i = tan( β i ) > D ∈ B , D = M i , denote by x D the x -coordinate of the intersection of M i and D . Recall– see (6) – that x L j ∈ ] a j − /n ; a j + 3 /n [ for j = 1 , ..., n . Furthermore, since RIANGLES IN ARRANGEMENTS OF LINES AND PSEUDO-LINES 7 L n − and L n intersect on the upper half-plane and a n − < < a n , we see that m n − > > m n . Since b i > µ i >
0, and since | µ i | < | m n | , | m n − | – see (3) –the intersection of M i with L n − (respectively with L n ) has negative (respectivelypositive) x -coordinate. Thus, x L n ∈ ]0; 1 /n + 3 /n [ and x L n − ∈ ] − /n − /n ; 0[.The positions of the X L j , for j = 1 , ..., n are given in Figure 2. x L n − x L n tan( − π/ π/n ) tan( − π/n ) tan( π/n ) tan(2 π/n ) tan( π/ − π/n )0 · · ·· · · Figure 2.
The disposition of x L j for j = 1 , ..., n .We describe now the values of x M j , for j = i . Writing z := β i + β j , the discussionmade above – in particular 3a), 3b), (8) and (9) – shows the following:1] if z = 0 then x M j = 0 , x L n − < x M j < x L n ;2] if z = π/ then x M j < − n ;3] if z < or if π/ < z < π then tan( z ) + 3 /n < x M j < tan( z ) + 1 /n ;4] if < z < β i then tan( z ) − /n < x M j < tan( z ) − /n ;5] if β i < z < π/ then tan( z ) + 3 /n < x M j < tan( z ) + 1 /n . We obtain the situation of Figure 3 (note that x Y = b i and that there is no x M j near tan(2 β i )). 1]2] 3] 3] 4] 4] 5] x Y − π/ π/n ) tan( − π/n ) tan( π/n ) tan( β i ± π/ n ) tan( π/ − π/n )0 · · ·· · · · · · Figure 3.
The places of x M j , depending on the value of z = β i + β j .In particular, every element of U = { x M j | j = 1 , ..., n, j = i } ∪ { x Y } is betweentwo consecutive x L j ’s. Furthermore, between two x L j ’s there is exactly one elementof U , except for one place (near tan(2 β i )), where there is no element of U .Doing the same for every M i (the situation for β i < M j ’s is symmetric), we see that between two consecuting intervals con-taining one a j (of the form ] a j − /n ; a j +3 /n [ or ]0; 1 /n +3 /n [ or ] − /n − /n ; 0[), exactly n/ M , ..., M n , Y intersect. Furthermore, these inter-sect only one triangle of A , which is the triangle touching Y (see 5 and 6). Weobtain the situation of Figure 4.Each of the n − A that touches Y is replaced in B by n +1 triangles,and we obtain also n triangles at the extremities (formed by the lines M i and M j with β i + β j = ± π/ B has thus exactly n triangles more than A , and every segment of Y ⊂ B is used in one triangle. This achieves to prove theProposition. (cid:3) Corollary 3.3.
Let n > be an odd integer, write m := 2 n − and assume that a s ( n ) > n ( n − / .We have a s ( m ) ≥ a s ( n )+( n − , i.e. m ( m − / − a s ( m ) ≤ n ( n − / − a s ( n ) .In particular, if a s ( n ) = ⌊ n ( n − / ⌋ then a s ( m ) = ⌊ m ( m − / ⌋ . NICOLAS BARTHOLDI, J´ER´EMY BLANC, AND S´EBASTIEN LOISEL Y Y Figure 4.
The configurations of the lines M , M , . . . , M n and Y – with at the left also some of the lines L , ..., L n . Proof.
Let A be an affine arrangement of n pseudo-lines with more than n ( n − / Y ∈ A that touches n − y = 0, arrange the intersections of Y with the other pseudo-lines L , ..., L n − to satisfy the conditions of Proposition 3.1,and stretch every pseudo-line L i ∈ A\ Y so that the segments of L i that touch Y become segments of lines. Then, adding the lines M , ..., M n of Proposition 3.1 toour arrangement gives an arrangement of 2 n − n − trianglesmore than A . (cid:3) We are now able to prove Theorem 1.3, with the help of the following newmaximal arrangement:
Figure 5.
A maximal affine arrangement of 19 pseudo-lines with107 triangles
Proof of Theorem 1.3.
1. There exists a maximal affine arrangement of n = 15lines with n ( n − / ǫ >
0, let A ǫ be the arrangement { L , ..., L } of 15 lines given by L i := { ( x, y ) ∈ R | y = m i ( x − a i ) } ,where RIANGLES IN ARRANGEMENTS OF LINES AND PSEUDO-LINES 9 a = tan( − π/ m = 1 . a = tan(2 π/ m = − . a = tan( − π/ m = 4 . a = tan(3 π/ m = − a = tan( − π/ m = 3 . a = tan(4 π/ m = − . a = tan( − π/ m = 14 . a = tan(5 π/ m = − . a = tan( − π/ m = 13 . a = tan(6 π/ m = − . a = tan( − π/ m = − a = − ǫ m = 50 a = 0 m = 0 a = ǫ m = − a = tan( π/ m = − Figure 6.
The perfect arrangement A ǫ of 15 lines, beginning ofthe induction.We can verify by inspection that the configuration is perfect (see Figure 6),if ǫ is small enough. Furthermore, we can apply Proposition 3.1 to get a perfectarrangement of 29 lines. By iterating, starting from A ǫ , for a small ǫ (which dependson t ), one obtains an arrangement of 14 · t + 1 lines, for any integer t ≥ ǫ >
0, let A ǫ be the arrangement { L , ..., L } of 7 lines given by L i := { ( x, y ) ∈ R | y = m i ( x − a i ) } ,where a = tan( − π/ m = 3 a = tan(2 π/ m = − a = tan( − π/ m = 1 a = − ǫ m = − a = tan 0 m = 0 a = ǫ m = 7 a = tan( π/ m = − Figure 7.
The maximal arrangement A ǫ of 7 lines, beginning ofthe induction.We see that the arrangement has 11 triangles, and use Proposition 3.1 to get anmaximal arrangement of 13 lines. By iterating, one gets – for any integer t ≥ n = 6 · t + 1 lines, with ⌊ n ( n − / ⌋ triangles.3. The arrangement of Figure 5 is a maximal arrangement of 19 pseudo-lines with107 triangles. Iterating Corollary 3.3 we find – for any integer t ≥ n = 18 · t + 1 pseudo-lines, with ⌊ n ( n − / ⌋ triangles. (cid:3) Description of the computer algorithm
In this Section, we discuss a computer algorithm to search for affine pseudo-linearrangements with many triangles. The problem of finding line arrangements withmany triangles is a geometrical one. It is possible to formulate a related combina-torial problem for pseudo-line arrangements. We will work with wiring diagrams (introduced by Goodman [Goo]), see Figures 5 and 8. In this representation the n curves are x -monotone and are restricted to n y -co y -coordinates except for somelocal switches where adjacent lines cross.The information of an affine arrangement of n pseudo-lines is stored into a ( n − × m matrix M , where m is some positive integer. Each column contains some X ’sand describes the crossings at some x -coordinate; an ” X ” at the height i means thatthe pseudo-lines i and i + 1 intersect there. We typically add suggestive horizontallines to these matrices to obtain pseudo-line diagrams as seen in Figure 8. Figure 8.
Pseudo-line Diagrams describing the line configura-tions above
RIANGLES IN ARRANGEMENTS OF LINES AND PSEUDO-LINES 11
The polygons of pseudo-line affine arrangements represented by the matrices areeasy to compute and the notation lends itself to several “pruning” ideas. From nowon, we will write M = ( M , ..., M m ) and refer to this matrix as the pseudo-linediagram. The search algorithm is: Function depth first search( M )Denote M = ( M , ..., M k ) .If M is a pseudo-line affine arrangementCount its triangleselseGenerate the list L of all possible choices of M k +1 .For each M k +1 ∈ L ,depth first search( ( M , ..., M k , M k +1 ) )End forEnd if We add some ”pruning criteria” to reduce the search:(1) In any given column, no two crosses may be adjacent.(2) It is not permitted to put a cross between two pseudo-lines that have alreadycrossed.(3) Without loss of generality, we may impose that all crosses be placed as farto the left as possible.A vector ( M , ..., M k ) with insufficiently many intersections but otherwise satisfyingthe three above properties is called a partial or incomplete affine arrangement.Although an arrangement is incomplete we are able to compute its triangles and tosee that some segments are already unused (i.e. not touching a triangle of the futurecomplete arrangement). Since we are looking for diagrams with many triangles, wemust have few unused edges – this allows us to discard some partial arrangementswithout compromising the search.(1) If on column k we put a cross in row j that closes a triangle, then thepolygons in column k and rows j − j + 1 cannot be triangles.(2) If the budget of unused segments is exhausted, we will have some forceddispositions of the crosses, to ensure that every remaining segment willtouch one triangle. 5. Computer Results
We have looked for maximal affine arrangements of n pseudo-lines. Perfectarrangements are only possible when n ≡ n = 3 , , , , , , , ,
29. Because several of our heuristics exploit the lownumber of unused edges, as the unused edge budget increases, the search quicklybecomes intractable when looking for imperfect arrangements.
Proposition 5.1.
The maximum number of affine triangles in a pseudo-line ar-rangement found by our algorithm are given in the following tables: n a s ( n ) 1 2 5 7 11 14 21 25 32 37 47 n
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