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Computational Geometry Seminar Lecture #10 3-quasi planar and 4-quasi planar graphs David Malachi.

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topological graph G is a graph drawn in the plane. V(G) is a set of distinct points. E(G) is a set of Jordan arcs. (E(G) are straight lines geometric graph) A topological graph is k-quasi-planar if no k of its edges are pairwise crossing. (We shall refer to 3-quasi-planar simply as quasi-planar) It was conjectured that for every fixed k, the maximum number of edges in a k-quasi-planar graph on n vertices is O(n). k = 2 trivial (Planar graphs, for n>2 exists E(G) ≤ 3n – 6).

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Agarwal : “ in 3-quasi-planar graphs E(G) = O(n)”. Pach : “ in 3-quasi-planar graphs E(G) ≤ 65n”. (Eyal Ackerman, Gabor Tardos) Theorem 1 The maximum number of edges in a quasi-planar graph on n ≥ 3 vertices is at most 8n − 20. Theorem 2 For every positive integer n, there is a quasi-planar graph on n vertices with 7n − O(1) edges.

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Proof`s main concept.. (Theorem 1) discharging method Assign “charges” to elements of input. Compute total charge. discharging phase : redistribute charges between elements of input. Compute total charge again.

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Proof (Theorem 1) Let G be a quasi-planar topological graph with n vertices. (WLOG with minimum intersections drawn as possible) Assume G is connected. G doesn’t contain parallel edges, self loops. Let G’ = new graph : V(G’) = V(G) U X(G) X(G) = intersection points of edges of G ( “new” vertices) u X(G), d(u) = 4 E(G’) = subdivided edges of G with respect to X(G) F(G’) = faces of G’ For a face f F(G’) : |f| = number of edges comprising face f. v(f) = # ”original” vertices comprising f.

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We assign “charges” to every face f in F(G’). For a face f, ch(f) is the “charge” of f. ch(f) = |f| + v(f) − 4. G’ does not contain faces of size one or two. No 0-triangles (3 pairwise crossing). Every face in G‘ has a non-negative charge.

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Sum up all the charges, we get : Next, we redistribute the charges without affecting the total charge found in (1). Our target Modify ch(f) to a new charge ch‘ (f) of a face f, which satisfies ch‘ (f) ≥ v(f)/5. Note, The only faces that do not satisfy this bound with the original charge ch are 1-triangles. (ch = 3+1-4 = 0, but v(f)/5 = 1/5 > 0)

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Redistribution begins – charge the 1-triangles… Let f be a 1-triangle. u is the original vertex of f. e1‘ and e2‘ be the two sides of f incident to u. e1 and e2 are edges of segments e1‘ and e2‘. We look for the first face along e1 (e2 symmetric) – which is not a 0-quadrilateral, denote fi. for sure exists, as the last face contains e1 endpoint. fi is not a triangle. (no 0-triangles) We Shift 1/5 units of charge to from fi to f. “ fi lost charge through the edge shared with fi-1 ”

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Current status of charges For every 1-triangle f, ch’(f) = 1/5 = v(f)/5. For every 0-quadrilateral f, ch’(f) = 0 = v(f)/5. Let f be a face of G’. Assume f is not 0-quadrilateral, nor 1-triangle. ch(f) =(|f|+ v(f) – 4) ≥ 1 ch’(f) = ch(f) − xf (xf is the total charge f lost in the redistribution) f lost at most 1/5 unit of charge through any one of its edges only if both endpoints of the edge are new. we have xf ≤ (|f| − v(f))/5 For that f, ch‘ (f) ≥ (2/5)v(f) + (4/5)(ch(f) − 1) ≥ 2v(f)/5 ch‘ (f) ≥ v(f)/5, for all faces f of G‘.

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Second round of redistribution.. Now we collect all the extra charge at faces incident to an original vertex, and place this charge on the vertex. For every face f with v(f) > 0, we compute the extra charge ch‘ (f) − v(f)/5, and distribute it evenly among the original vertices v(f) of f. Each original vertex of f receives (ch‘ (f) − v(f)/5) / v(f) units of charge from f. ch‘‘(f) is the remaining charge at a face f after this step, for any f. ch‘‘(u) is the charge accumulated at an original vertex u.

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By the construction of ch‘‘ we have ch‘‘ (f) ≥ v(f)/5 for every face f. From equation (1) we have Remains to prove a lower bound on ch‘‘ (u) for an original vertex u. Claim For an original vertex u, ch‘‘ (u) ≥ 4/5.

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Suppose Claim is true, Combining our bound for ch’’(f) for any face f F(G’), and ch’’(u) for any u V(G), we get: |E(G)| ≤ 8n – 20. (Theorem) Now on to the proof the Claim.

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Proof ( Claim - ch’’(u)≥4/5 ) Let G’ the graph after the second round of distribution, u V(G). G u obtained from G by removing the vertex u and all its incident edges. Let G′ u be the corresponding plane drawing. (With respect to G u ) - f u be the face of G′ u containing u. - Let w be a vertex of f u. There is at least one face f of G′ that touches both u and w. If |f|≥ 5, then f alone contributes at least 4/5 unit of charge to u. Same holds for 4-quadrilateral, 3-quadrilateral, 2-quadrilateral (with the two original vertices not being neighbors).

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As f cannot be a 1-triangle, the following cases remained : - 2-triangle contributing 3/10. - 3-triangle contributing 7/15. - 1-quadrilateral contributing 2/5. - 2-quadrilateral (with neighboring original vertices) contributing at least 7/10.

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Note : -Except for the 1-quadrilateral (and 1-triangle) case we find an edge of G incident to u that is not involved in any crossing, and therefore, the faces on both sides of this edge contribute charge to u. -There is at least one other vertex w′ of f u besides w. ( The minimal value of ch′′(u) = 4/5 is only possible if fu is subdivided into two 1-quadrilaterals and a few 1-triangles in G′. (claim).

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Usage of theorem 1.. Let m0 be the largest value such that the complete graph Km0 can be drawn as a quasi-planar graph. By Theorem 1,m0 ≤ 14. Claim: m0 ≥ 9 for a quasi-planar geometric graph. Proof: One can inspect and see for itself… * By Aichholzer and Krasser K10 cannot be drawn as a quasi-planar geometric graph.

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A Reminder … Theorem 2 For every positive integer n, there is a quasi-planar graph on n vertices with 7n − O(1). Proof A constructive one… - Let be a hexagonal grid such that for each hexagon we draw all the diagonals, but one, as straight line edges. - Add a curved edge for the missing diagonal.

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Finally we add longer edges, two from every vertex. One can verify by inspection the (3-)quasi-planarity of the obtained graph :

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- degree of vertices that are far enough from the boundary is 14. - degree of O(√n) vertices which are close to the boundary is smaller than 14 This quasi-planar graph already has 7n − O(√n) edges. To improve the “error” term: - wrap the graph around a cylinder so that we have three hexagons around the cylinder. - draw five more edges on each of the top and bottom faces. now only O(1) vertices don’t have degree 14. (Theorem 2)

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Part 2 4-quasi planar graphs By Eyal Ackerman

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Theorem : (4-quasi planar) For any integer n > 2, every topological graph on n vertices with no four pairwise crossing edges has at most 36(n - 2) edges. Proof : ( lets start…) Number of a 4-quasi planar graph on n=3 vertices, Is 3 < 36(3 - 2) = 36. Assume n ≥ 4.

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A few definitions e| p,q - the segment of e between p and q. - e E(G) - p and q are points on e. lens – e1,e2 E(G) which intersect at least twice. Region bounded by segments of e1 and e2 that connect two consecutive intersection points is called a lens. A wedge is a triplet w = (v, l, r) such that - v V (G). - l and r are edges emerging from v. - l immediately follows r in a clockwise order. (of the edges touching v).

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Assumptions and markings - G drawn with least possible number of crossings. (with respect to quasi planarity) - No three edges crossing at the same point. - For every v V(G), d(v)≥2. -G does not contain self loops/no parallel edges. -G does not contain Empty Lenses. (Lens which does not contain a vertex of G). G’ as before… V(G’) = V(G) U X(G) (X(G) “new” vertices) E(G’) = subdivided edges of G with respect to X(G). (We will refer E(G’) edges as p-edges) F(G’) = Faces of G’. |f| = number of p-edges f. v(f) = number of “original” vertices.

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Proof`s main idea is… Discharging method Assign “charges” to faces of G’. Redistribute the charges, with certain logic. No faces with negative charge. All wedges have least of 1/18 unit of charge. Total wedges is charge (2|E(G)| / 18 ) ≤ 4n – 8.

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Starting… As before, we assign charges to F(G’): ch(f) = |f|+v(f) – 4 Total sum of charges is : Equality is due to Euler`s formula (f=e-v+2) + the next equality :

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Minimum face size is 3. Face of size one yields a self-intersecting edge. Face of size two yields an empty lens or two parallel edges Only faces with negative charge are 0-triangles. (3 + 0 – 4 = -1) Solution : Charge the 0-triangles !

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Let t be a 0-triangle - e1 be one of its p-edges. - f1 be the other face incident to e1 |f1| > 3 If v(f1) > 0 or |f1| > 4 ( f1`s charge is ≥ 1) move 1/3 units of charge from f1 to t. ( f1 contributed 1/3 units of charge to t through p-edge e1 ) Otherwise, f1 must be a 0-quadrilateral.

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We look at the face incident to the opposite edge to e1,and continue on until we find a face which is not a 0-quadrilateral, Denote - f i. f i will contribute 1/3 units of charge to t through e i-1. In a similar way t obtains 2/3 units of charge from its other p-edges.

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After re-distributing charges this way, the charge of every 0-triangle is 0. A face can contribute through each of its p-edges at most once every face f such that |f|+v(f) ≥ 6 still has a non-negative charge. Remains to verify 1-quadrilaterals and 0-pentagons, which had only one unit of charge to contribute, also have a non-negative charge. Observe 1-quadrilateral contributes to at most two 0-triangles (endpoints of a p-edge through which it contributes must be vertices from X(G) ) 0-pentagon, on the other hand, can contribute to at most three 0- triangles by Observation 2.1

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Observation 2.1 0-pentagon contributes charge to at most three 0-triangles. Moreover, if it contributes to three 0-triangles, then the contribution must be done through consecutive p-edges. Proof : (in waving hands) One can easily inspect that a contribution to three 0-triangles, through non-consecutive p-edges implies four pairwise crossing edges. At this stage, faces with positive charge are called good faces, and faces with 0 charge are called bad faces. (except for 0-triangles and 0-quadrilateral). 0-pentagons that contributed to three 0-triangles, and 0-hexagons that contributed to six 0-triangle == BAD

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At this point, ALL faces have non-negative charge. Next stage is to put charge on the wedges.. First, a few definitions … A-crossing - Let w = (v,l,r) be a wedge. An A-crossing of w is a triplet cr = (e,p,q) such that : - e is an edge crossing l at p, and r at q - e|p,q does not intersect l or r. - the endpoints of e are not in w|e,p,q cr is an uncut A-crossing of w, if e|p,q is a p-edge. cr is farther than cr`, for 2 A-crossings cr and cr` of w, if p` l|v,p and q` r|v,q. cr is an empty A-crossing of w, if there are no original vertices in w|cr (w|cr = w|e,p,q )

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X-crossing Let w be a wedge, cr1 = (e1,p1,q1) and cr2 = (e2,p2,q2) be two A-crossings of w. x = (cr1,cr2) is an X-crossing of w if e1|p1,q1 and e2|p2,q2 intersect exactly once. We say that x is empty if both cr1 and cr2 are empty A-crossings. w|x represents the area w|cr1 U w|cr2. y is the intersection point of e1|p1,q1 and e2|p2,q2. Vis(x) is the curve (e1|p1,y) U (e2|y,q2). Vis(x)l is e1|p1,y. Vis(x)r is e2|y,q2.

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Observation 2.4 x = ( cr1 = (e1, p1, q1), cr2 =(e2,p2, q2)) is an empty X-crossing of a wedge w = (v, l, r),Vis(x)l part of e1. an edge e’ that crosses Vis(x)l (resp., Vis(x)r) must cross l (resp., r) and must not cross e2 (resp., e1). Let z be the crossing point of e’ and e1|p1,y. Since x is an empty X-crossing of w, e’ must cross the boundary of w|x at least one more time. (no original vertex inside) If it crosses e1|p1,q1 at another point different from z, then we have an empty lens. if e’ crosses r|v,q2 then it must also cross e2|p2,y four pairwise crossing edges: e0, e1,e2, r. e’ must cross l. e’ must not cross e2. (four pairwise crossing).

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Charge the wedges Let x and x’ be two X-crossings of a wedge w. x is a farther X-crossing of w than x’, if one A-crossing of x is farther than one A-crossing of x’ and the other A-crossing of x’ is not farther than the other A-crossing of x. An uncut A-crossing is farther (resp., closer) than an X-crossing of the same wedge, if it is farther (resp., closer) than both A-crossings of the X-crossing. Given a wedge w = (v, l, r), we look for the farthest empty uncut A-crossing or X-crossing of w.

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Start charging wedges If there is no such uncut A-crossing or X-crossing, then the face incident to w is not a 1-triangle. Its charge is at least 1/3 units, from which we use 1/18 units to charge w. Assume we find such A-crossing, cr = (e, p, q). Let f be the face incident to e|p,q outside w|cr. -f is not 0-triangle as this would yield an empty lens or parallel edges. -f is not 0-quadrilateral since this would imply an empty uncut A- crossing farther than cr. -If f is a bad pentagon, then it follows from Observation 2.1 that there is an empty X-crossing, farther than cr. f must be a good face which will contribute 1/18 units of charge to w through e|p,q.

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2 nd case Assume we find such X-crossing x = (cr1 = (e1,p1,q1), cr2 = (e2,p2,q2)). Vis(x)l = e1|p1,y. y is the intersection point of e1|p1,q1 and e2|p2,q2. Let f1 be the face that is incident to y and outside w|x. Several cases for f1 : - Suppose f1 is a 0-triangle, e3 be the third edge incident to it. By Observation 2.4 that e3 must cross l, thus l, e1, e2, e3 are pairwise crossing.

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Suppose f1 is a bad pentagon, then we consider the possible cases: none, one, or both of e1|p1,y and e2|y,q2 are p-edges : -In case both of them are p-edges,then there is an empty uncut A-crossing of w that is farther than x. -In case one of them, WLOG e2|y,q2, is a p-edge, then there is an empty X-crossing farther than x. -In case none of them is a p-edge,then there must be four pairwise crossing edges.

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In a similar way, if f1 is a bad hexagon then there must be four pairwise crossing edges, or an X-crossing of w farther than x. So, if f1 is a bad face, it must be a 0-quadrilateral. Let f2 be the face outside of w|x that shares a p-edge with f1 and is incident to Vis(x)l. If there is no such face, or there is no face outside w|x that shares a p- edge with f1 and is incident to Vis(x)r, then there is an empty X-crossing farther than x.

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f2 cannot be a 0-triangle. (four pairwise crossing edges). if f2 is a bad pentagon (or a bad hexagon), then again (as before) there must be four pairwise crossing edges or an empty X-crossing farther than x.

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So, lets assume f2 be a 0-quadrilateral, Lets examine the next face, that is, the face f3 != f1 such that f3 is outside w| x, shares a p-edge with f2, and is incident to Vis(x)l. If there is no such a face, then again, we have an empty X-crossing farther than x. Else – Apply the arguments for f2 and so on.. Until we find such a good face, which we must encounter, otherwise we have an empty X-crossing farther than x. Denote : f i -the first good face we encounter along Vis(x)l, e i -be the p-edge of fi that is contained in Vis(x)l. f i contributes 1/18 units of charge to w through e i.

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Left is to prove that after charging every wedge of an original vertex, there are no faces with a negative charge. We show that by proofing a face cannot contribute to “too many" wedges. Definition For a (good) face f and one of its p-edges m, we say that f is a possible X-contributor to a wedge w through m, if there is an empty X-crossing of w, x, such that f is outside w|x and m Vis(x)l. Observation 2.5. Let f be a face and let m be one of its p-edges. Then f is a possible X-contributor through m to at most one wedge.

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Proof (Observation 2.5) Let f be a face. Suppose f is a possible X-contributor through one of its p-edges m, to two wedges, w1 = (v1, l1, r1) and w2 = (v2, l2, r2). Let e be the edge containing m. There are four points p1, q1, p2, q2, such that cr1 = (e, p1, q1) is an A-crossing of w1. cr2 = (e, p2, q2) is an A-crossing of w2. x1 =((e, p1, q1),(e1‘, p1‘, q1‘)) - the empty X-crossing of w1, such that m Vis(x1)l. x2 =((e, p2, q2),(e2‘, p2‘, q2‘)) - the empty X-crossing of w2,such that m Vis(x2)l.

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Lets we sort p1, q1, p2, q2 by the order in which they appear when traversing e from one of its endpoints to the other. (such that when traversing m the face f is to our right.) pi must precede qi, for i = 1,2 since f is outside of w|xi. Assume, WLOG, that p1 precedes p2. (Other case – symmetric) Notice, By Observation 2.4 l2 must cross l1. Since m ( e| p1,q1 U e| p2,q2 ), the order of the four points is p1, p2, q1, q2 or p1, p2, q2, q1

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Suppose q1 precedes q2 : w1|x1 and w2|x2 do not contain any original vertex. We have 2 options: - l2 must cross l1 and r1. - r2 must cross l1 and r1. The first case yields an empty lens. In the second case, note that e1’ must cross either l2 or r2, yielding four pairwise crossing edges.

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Suppose q2 precedes q1 : Then the edge e2’ must cross e twice and creating an empty lens, or cross l1 and yield four pairwise crossing edges.

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We conclude that f cannot be a possible X-contributor through m to more than one wedge. (Observation 2.5) face f, such that |f| ≥ 7, ends up with a charge of at least (when v(f)=0) |f| - 4 - |f|( 1/3 +1/18 ) > 0. k-quadrilaterals, for k > 0, and good pentagons and hexagons, end up with a non-negative charge, as their charge after charging the 0-triangles was at least 1/3. Summing up the charge over all the wedges we have 2|E(G)|/18 ≤ 4n – 8 |E(G)| ≤ 36n – 72

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The End…. Thank you !

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