# On the number of unit distances among ‘n’ points Crossing Lemma: For n points and e edges with condition e ≤ 4*n, the crossing number of G on the plane.

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On the number of unit distances among ‘n’ points Crossing Lemma: For n points and e edges with condition e ≤ 4*n, the crossing number of G on the plane is at least e 3 /(100 n 2 ) Crossing Number: ? Ref.[1] Crossing Lemma: For n points and e edges with condition e ≤ 4*n, the crossing number of G on the plane is at least e 3 /(100 n 2 ) Crossing Number: ? Ref.[1]

Keywords: Crossing Number Incidences(with respect to lines and points) Unit Graph Turan Graph T(n,r)

f(n) the maximum number of unit distances possible among n points g(n) the minimum number of different distances possible. Erdós himself said that these are my most striking contribution to Geometry. He also offered \$500 for the magnitude of any. f(n)* g(n) ≥ n C 2 (consider graph which is for g(n) in this graph join all other possible edges their length must be repetition of length’s of g(n) one of the must repeat at least n C 2 /g(n) times)

Erdós showed the lower bound of maximum number (f(n))of unit edges in euclidian planar graph to be n 1+const/log(log(n)) and conjectured that this was about the true magnitude. He also proved the upper bound to be n 3/2 [2] J. Beck and J. spencer prove this to be n 1.444... [4] Finally, Spencer, Szemerèdi and Trotter achieved the best known bound, const*n 4/3 [5] For dimension(d) ≤ 4 the problem is solved by Lenz construction u d = (p-1/2p)n 2 + O(n 2 ) where p = floor(d/2)

For l lines and n points, the maximum number of incidence that can happen Assume all points lie on at least one line(if not then it will not be maximum). If k points lie on a line then it will make k-1 edges. Consider the graph(G) made by these edges and the points. If #i denotes the number of incidences then edges of G is bounded by #i - l so by using Crossing Lemma Cr(G) ≥ (#i - l) 3 /(100*n 2 ) ⇒ #i ≤ const *[(nl) 2/3 +n+l] Note: Cr(G) is at-most l 2

For n points the number of lines containing at least k of them is at most const*n 2 /k 3 (where 2≤k≤√n) [6] Consider those lines which passes through at least k of the points. Consider the graph G formed by the points and the intermediate edges(#edges > l(k-1)) again applying crossing lemma we get l 2 > const* e 3 /n 2 > const[l(k-l)] 3 /n 2 or l(k-1) < 4n Since G has at least l(k-1) edges

The number of unit distances among n points in the plane is at most const*n 4/3 Draw a unit circle around each point now construct a graph G in which we will only include those arcs(from point to point) of circle which has more than 2 points on it’s circumference also if an arc is common between two circle then take only one of them. Claim: Number of edges in G is at most O(n) less than the number of unit distances. Proof: There are n circles, so by not counting those arcs, from circles having less than or equal to 2 points, we have decreases the edge count by at most 2*n from unit distances. Also there are at most n arcs shared by all circles so by counting only once we are again diminishing the edge count by at most 2*n.(proof incremental)

No more than two circle can share same arc since radius of all circle is fixed(unity). If two circle shares an arc then there can’t be any other circle possible which can share arcs with both of them. So it is like clique of size at max 2.So at most 2*n unit distances are not considered if we take only one of the common arcs. The number of crossing in G is at most 2*n 2 because two circle can intersect in at most two points. So by applying Crossing Lemma we get 2*n 2 ≥ [f(n)] 3 /100*n 2 ⇒ f(n) ≤ const * n 4/3

Discussion Can we do better than this using Crossing Lemma. Let us first see the proof of crossing lemma(it is simple) Euler formula (n - e + f = 2 - 2g) gives us cr(G) ≥ e - 3* n + 6 Now we consider subgraph of G and apply Euler formula to them and then combine the estimates using probabilistic method.Consider subgraph formed by choosing all vertices independently with probability p Let us call this graph G p by Euler formula cr(G p ) - e p + 3* n p ≥ 0 Expectation is clearly E(cr(G p ) - e p + 3* n p ) ≥ 0

Since E(n p ) = pn, E(e) = p 2 m and E(crossing) = p 4 cr(G) Which gives cr(G) ≥ p -2 e - 3p -3 n. Now by setting p = 4n/e( ≤ 1 since e ≥ 4n) the inequality becomes cr(G) ≥ 1/64(e 3 /n 2 ). Crossing Lemma gives tight bound to many graphs so we can’t do better than this by just using Crossing Lemma need to consider some more constraints which unit graph has.

Discussion Consider Unit Graph G in a plane, Since two circles can’t share more than two points so the clique of G will be 3.G is also 3 partite graph. So it is T(n,3) graph. G can’t have K 2,3 as it’s subgraph. K 2,3 denotes a complete bipartite graph which has 2 vertices on one side and 3 on other Turan Theorem: Turan number m, is the largest number of edges, a Graph can have which is k partite and has n vertices. m = (k-1)n 2 /2k

Bound by Erdós Assuming that x i points are at unit distance from P i clearly our interest is in calculating Assume that x 1 ≥ x 2 ≥.... ≥ x n then because and Where S i denotes set of points unit distance from P i also which gives but is Now x 1 + x 2 +.... + x a ≤ n + 2 a(a-1)/2 since x i ‘s are in decreasing order so x a ≤ [n + 2 a(a- 1)/2]/a

So, < n + 2 a(a-1)/2 + [n + 2 a(a-1)/2](n-a) by taking a = √n we get bound < n 3/2

[1] Crossing Lemma--Leighton, Ajtai, Chvatal, Newborn and Szemeredi, Crossing free subgraphs, Annals of Discrete Mathematics [2] Erdós paper On sets of distance of n points [3]S. Józsa and E. Szemerèdi, The number of unit distances on the plane in: Infinite and Finite Sets [4] J.Beck and J.Spencer, Unit distances

[5] J. Spencer, E. Szemerèdi and Trotter [6] E. Szemerèdi, W. T. Trotter Extremal Problems in discrete geometry

Thank You tT.

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