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Erd Ös asks if it is possible to have n points in general position in the plane (no three on a line or four on a circle) such that for every i (1≤i≤n-1 ) there is a distance determined by the points that occur exactly i times. So far some examples have been discovered for 2≤n≤8 [1] [2]. A solution for the 8 point is provided by I. Palasti [3]. Here two other possible solutions for the 8 point case as well as all possible answers to 4 - 7 point cases are provided and finally a brief discussion on the generalization of the problem to higher dimensions is given.

It seems certain that if n is large enough the Erdős problem mentioned above has no solution [

1) Is there an example for n = 9 or no such examples exists? [

2) How many independent patterns are there for

3) Is it possible to have n arbitrarily large for the same problem in

The aim of this article is to analyze the second question and give a brief discussion about the third question. Indeed all possible solutions for

Let _{i} repeats i times and the distance between the points

will be algebraic numbers. It is possible to consider a graph pattern for each solution so that the weight of the edges of the graph corresponds to the number of relevant repeated lengths appeared in the geometric solution of the problem which is a number between 1 to n.

The solution of the 2 point case is single line and in 3 point case it is an isosceles triangle which is not an equilateral triangle. For 4 point case, the following 5 solutions are available that the last two solutions have a unique graph pattern in

In order to generalize the problem, if the sequence of distinct lengths which is the set

The sequences listed above for 4 point case, have one free variable. Other possible cases for the 4 point problem include sequences {1,1,1,3} and {1,1,2,2} that have two degrees of freedom (two free variables) and the sequences {1,5}, {2,4} and {3,3} the degree of freedom of which is zero and in this case we have four fix points:

4 & 5 | 3 | 2 | 1 | |
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Note: For four point cases, we have maximum 4 degrees of freedom, but in this problem, we only need to discuss about zero, one and two degrees of freedom cases. To find the solution to the main problem for more than 4 points a computational algorithm is required which is discussed below.

Method 1: Suppose that the points

1) All intersection points of perpendicular bisectors of the sides related to two entrees of points

2) All intersection points of the circles with radius

3) All intersection points of the circles with radius

Note: The main problem is to review various patterns of length sequence

Our general strategy to solve Erdős problem is to start with a previously known example, i.e. 4 point or 5 point case and add new points using method 1. If this method fails we want to show that adding one new point with free variable coordinates will enable us complete this strategy by using method 1. Suppose we constricted b points in the plane with d distinct distances given by

Because in other cases, this new point can be found using method 1. Note that _{t}’s are all new distinct distances. All of the above cases, have at least b-1 new distinct lengths. So if we add j new points step by step in this manner we have:

It is obvious that at each step of adding new point, if we have:

Number of existing points >

then we can obtain all remain points only by using method 1. The maximum number of new points added by the above process is called

We want to start with b = 4 points. For 4 point pattern with m free variables, we have

Regardless of the dimension, a total of 118 independent weighted graphs are considered for the 5 point case. These 118 cases can be classified based on six general cases to have four equal lengths in this graph. To find all the possible 5 point patterns, it is necessary to analyze the equation system of 118 independent graphs. In 2D case some of these cases may either not lead to a solution or lead to more than one solution. The following

After solving the equations a total of 89 five point patterns are obtained that other than the first example given below, which has one degree of freedom, for the rest of 88 solutions 5 fixed points are obtained. But there is another way to analyze all possible scenarios of five point case which is using method 1 because if there is a five point pattern with a sequence of lengths

Total | |||||||
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118 | 32 | 32 | 32 | 12 | 6 | 4 | No. of graph patterns |

89 | 43 | 30 | 13 | 3 | 0 | 0 | No. of 2D solutions |

1 - 43 | 44 - 73 | 74 - 86 | 87 - 89 | Indexes of the solutions in |

Our first pattern with one degree of freedom for 5 points that can produce six, seven and eight point patterns is called pattern (*) and is as

value other than 0 and −75 degree within the range of −150 < a < 30. The coordinates of

and the coordinates of the other two points in terms of a are:

Also, we have the following angle relations:

All 89 five point 2D solutions are presented in

First we show that all 6 point cases any sequence lengths

The rule is proved by removing the marked vertex in the above figure. Otherwise, the graph is as follows:

So then one of the following cases is true that either two vertices are removed that are presented in the figure:

Or there will be the following case:

The last case requires a little more analysis. If four edges with the weights 4 are connected to vertices a and f, the solution is presented by the same vertices, otherwise there are at least 3 edges with the weights of 5 among the edges

Otherwise, there must be at least two edges with a weight of 4 among the edges

Now, using method 1 and also four and five point patterns it is possible to search for 6 point patterns. 54 solutions are obtained in this way 47 of which are obtained by adding one point to the 5 point patterns and only 7 solutions are obtained by adding two points to the 4 point patterns. But we guess that, there must be 55 solutions in this case based on the probable relation between the number of solutions and Fibonacci numbers which this issue has been more explained in section 2.7.

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Seven patterns obtained by pattern (*) are presented:

1) In the three first patterns the value of angle can be selected as a free variable. Since the first 5 points are the same as pattern (*), only the coordinates of the new point

2) The next four results are related to two specific angles a discussed in the following

The 40 remaining solutions are presented in the following

The six cases that are not directly obtained by 5 point cases but they are obtained by adding two new points to four point patterns are presented in

Pattern 48 in the above table, was presented by I. Palasti in 1989 [

In 7-point case, in addition to the 4 point sequences of one degree of freedom, the 4 point sequences of two degrees of freedom will be required i.e. the sequences {1,1,1,3} and {1,1,2,2}. Obviously it is always possible to remove the vertices connected to the edges with the weight of 1, 2 and 2 that accordingly a maximum of 3 vertices will be removed and there will be 4, 5 and 6 point pattern with a maximum of 4 distinct lengths. Of course, in 5 and 6 point cases it is possible to remove two other vertices so that the maximum remaining lengths are 4 distinct lengths. Therefore it is just needed to analyze 4 point case with a maximum of 4 distinct lengths.

j | n | ||
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1 | 1 | ||

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1 | 3 |

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Then using method 1 and also 4, 5 and 6 point patterns, it is possible to search for 7 point patterns. 13 solutions are obtained in this way 9 of which are obtained by adding a new point to the 6 point patterns and two solutions are obtained by adding two new points to the 5 point pattern (*) and two of them are obtained by adding three points to 4 point patterns. Among all solutions only the first solution has the free variable a and in the rest of solutions all points are fixed and non parametric. The coordinates of the points

And also we have:

The above solution is obtained by the reunion of patterns 1 and 2 in

The remaining 8 solutions extracted from 6 point patterns are presented in the following

Pattern 9 in the above table, was presented by Andy Liu in 1987 [

And at the end of this section two solutions of 4 point pattern obtained by adding three new points are as follows

Pattern 13 in the above table, was presented by I. Palasti in 1989 [

In the eight point case we have to show that it is always possible to obtain a 4 point pattern by removing 4 vertices with maximum 4 distinct lengths that can be one of the length sequences of {1,2,3}, {2,2,2}, {1,1,4}, {1,1,1,3} or {1,1,2,2}. We know that there are 6 edges with the weights of 1 - 3. Thus there is a graph with 6 edges and 8 vertices. Therefore at least one of the vertices is connected to two edges, by removing this edge, we

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3 | 8 | ||

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88 | 46 |

89 | 47 |
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will obtain a graph with 7 vertices and 4 edges. Thus in this case there is at least one vertex that is connected to two edges. By removing this vertex two other edges will be removed. Then two edges are remained that it is possible to remove these two edges by removing two other vertices. So we will get a graph with 4 vertices that had a maximum of four distinct lengths. The problem in 8 point case has two other solutions other than the solution discussed at the beginning both of which are directly obtained by the pattern 1 in 7-point case by adding a new point that all three solutions are presented in

Closely related to the length sequence {1,2,3,4,5,6,7} there are 5 other examples with one degree of freedom. The sequence of distinct lengths of two initial solutions is {1,1,2,2,4,5,6,7} and these two solutions are presented in

Using the discussed method no pattern was obtained for 9 point case, however this does not guarantee that there will be no solution for 9 point case. Here some solutions are presented that are very close to the intended solution and in fact they were the solutions that only needed one additional condition to be a candidate for the main solution. The initially expressed solutions have one free variable. The following solution is the only one that is not obtained by 5 point (*) pattern. This pattern still has the free variable a and its distinct length sequence includes {1,2,2,3,4,5,6,6,7}. Although there is a free variable and only one other equation is required that can be any of the following equations that are related to the equations with the repetitions of 2 and 6:

The following cases that are obtained by five point pattern (*) they still have the free variable a. The sequence of distinct lengths of the first two solutions is {1,2,3,3,3,4,5,7,8} and in case of the other two solutions this sequence includes {1,1,2,2,4,5,6,7,8} (

All solutions that are presented below have fixed points and they have no free variable. First the solutions with independent lengths of {1,1,2,2,4,5,6,7,8} (

And solution of sequence {1,1,2,3,3,5,6,7,8} (

And the last solution close to the original answer to the problem by distinct lengths {1,1,1,3,4,5,6,7,8} is as follows (

Based on the results presented in the previous sections it can be observed that in 2D case the number of n point solutions for the value of n between 1 and 8, it is based on Fibonacci sequence [

On the other hand (

Therefore, the solutions can be presented in the following chart where the horizontal axis represents the number of points, and the vertical axis presents the Fibonacci number index corresponding to the number of points (

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54 |
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1 & 4 | 2 | |||

1 & 5 | 3 |

1 & 6 | 4 | |||
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1 & 7 | 5 | |||

1 | 6 |

4 | 7 | |||
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7 | 8 | |||

3 | 9 |

a | n | ||
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−133.2262 | 11 |

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13 |
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1 | |||

2 |

3 |
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Note: In the above table we have

We do not know why Fibonacci number appears in the above table and an explanation of this fact seems to be very interesting.

When the number of dimensions increases the dimension of the solutions is increased as well but the seemingly independent solutions are united together. For example, in cases where two distinct solutions were obtained from the intersection of two circles on the 2D plane, this solution will be the result of the intersection of two spheres which is a circle in 3D space. Thus, two distinct points that produce two distinct solutions in two dimensional cases give way to the same solution with a higher dimension.

Therefore, by increasing the number of dimensions, it is just enough to consider the graph pattern cases to find the n-point independent solutions, so the limited sequence of number of solutions can be obtained, only by counting the number of graph patterns.

For example the 4 point state that only has 4 independent cases can be easily analyzed. The 4 point case solution was presented for a desired number of dimensions greater than 2 in table 1. By increasing the number of dimensions, the number of solutions remains 4 but the dimension of the solutions increases at each stage. Also as mentioned in section 2-1 the number of independent 5 point cases graph patterns was 118 cases.

To determine the number of independent solutions of 5 point 3D case it should be noted that one can always fix 4 points with arbitrary pattern of a sequence related to distinct lengths. Thus, only the fifth point should be found. If there is a vertex the attached sides of which are

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Number of points | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
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Number of independent solutions | 0 | 1 | 1 | 5 | 89 | 55 | 13 | 3 |

cases for

So, the sequence of the number of solutions in 3D case is:

On the other hand the (1,3)-Fibonacci sequence is [

So, based on all above discussions, we ask is there any relation between the number of solutions of the sequence length of

Amir Jafari,Amin Najafi Amin, (2016) On the ErdÖs Distance Conjecture in Geometry. Open Journal of Discrete Mathematics,06,109-160. doi: 10.4236/ojdm.2016.63012