points are stationed arbitrarily close to one corner of the square ( n+1

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1 1 Intro We ve seen previously that we can construct networks based on the regular triangular, rectangular and hexagonal lattices where the total network length is linear in n and the R-statistic for the network is, respectively,... The question naturally arises how well we can do for any fixed r > 1 in particular whether, fixing r, we can always find a network N such that R(N) r and length(n) O(n). (Alternatively, if we look at the length of network per city, we are interested in networks where the normalized length is constant.) Note that for r = 1 this is not possible, since the only network that achieves this R-statistic is the complete graph on n points. If n points are stationed arbitrarily close to one corner of the square ( n+1 for n odd) and in the opposite corner, we have for any pair of points (x, y) where x and n y are in opposite corners: dist(x, y) n. There are ( n )( n ) = n 4 such points, so: n n length(n) O(n 5 ) 4 It thus makes sense to ask, given that we cannot find a construction for r=1, whether we can find one for r arbitrarily close to 1. In what follows we give a such a construction and study its length. Construction The basic idea is to construct a network that works for any two points within a certain angle of each other (defined below), then superimpose a finite number of rotated copies of this network, so that each pair of points is captured by at least one copy of the original construction. To this end we will construct a network that, for any good pair of points, contains a path which only navigates around a certain (fixed) wide angle. Making this angle wide enough, we ensure that the ratio of shortest path to Euclidean distance is sufficiently small. Any fixed r > 0 determines such a wide angle θ r, which we call the major angle of our network. Intuitively, we would like θ r to be just wide enough so that paths which navigate only around this angle have path ratios less than or equal to r. How wide does θ r have to be? The worst path ratio for a path navigating around a single angle is achieved when the starting and ending points of the path are at equal distance from the vertex of the angle (see figure 1). Letting a be the distance between each point and the 1

2 vertex, b be the distance between points, and noting that the path ratio p r = a b, we have or simply = a b sin( θ b r ) = a = 1 p r θ r = sin 1 ( 1 p r ) Figure 1. We let ψ r = π θr be the minor angle of our construction, and build a lattice as follows. Moving along one of the vertical edges of our area-n square, we mark off points at distance y (as yet undefined) from one another. From each of these points we draw two lines, forming ψ r and ψ r degree angles with the horizontal, respectively. Note that the horizontal lines (dashed in figure below) are not themselves part of the lattice.

3 Figure. Lattice roads. The lines of our lattice intersect to form diamond-shaped cells, where the obtuse angle is θ r (this was the point of constructing the lattice in this way). Each of our n points (or cities) inhabits exactly one of these cells and we now draw two access roads parallel to the edges of the cell in the way pictured below. The distance y is constrained by two requirements: First, we would like the construction to form a network that is for the lines of the construction to intersect. If we let y be too large the lines will not intersect for small enough n (this problem applies only to small n, and is dealt with quite easily). Second (and this is the more important point) we require that y be constant. The reason, in short, is that when y grows with, say, n the total length of access road grows with n 3/ and the normalized network length is no longer constant (more on this below, where we analyze total network length). For any points in the square, x 1 and x, draw a line between them and let φ x,y be the angle at which this line intersects the horizontal. We would like to show that the network we have constructed provides a sufficiently 3

4 short path for all points x 1 and x, such that 0 φ x,y ψ r. Consider in particular: Provisional Claim 1: For any points x 1 and x, such that 0 φ x,y ψ r, the ratio of shortest path distance to Euclidean distance along lines of the network is no greater than r. Let the region between two upward-pointing parallel lines be called a bar (highlighted in pink in figure ). Looking at figure 3, it turns out the claim is true in the following two cases: (i) the two points belong to the same cell (as d and e) (ii) the two points are in different bars (as b and c) But is not true in the case, (iii) the two points belong to the same bar, but are separated by some number k > 1 of cells (as a and b) Figure 3. Points in the network. The worst scenario in case (iii) is when the points are separated by one empty cell, and are each placed arbitrarily close to the midpoint of opposite edges (again, as a and b). To correct this, we need to add interior roads through the center of each cell and parallel to the NE-SW pair of edges (see figure 3). When we do this the worst situation in our new network is when the two points are a quarter of the way along opposite edges. Here the path 4

5 ratio is z 4 +z+ z 4 z = 3, where z is the length of the edge of a single cell. Consider next the network formed by adding two lines within each cell parallel to the NE-SW edges and equally spaced apart. Here the worst situation is where the two points are 1 6 of the way along opposite edges of the cell and the new path ratio is z 6 +z+ z 6 z = 4 3 The idea is that if we do this some finite number of times, we will eventually produce a short enough path between all such points. Indeed, if we partition each cell with n equally spaced lines our new z n+1 +z+ z n+1 z = n+ n+1 worst situation path ratio is path ratio is smaller than r we pick n large enough so that n+ n+1 n r r 1. In order to ensure that our r or simply Figure 4. Partitioning each cell 0, 1, and 3 times. Claim 1: In the network obtained by adding n = r r 1 interior roads to each cell, the path ratio for all points x 1, x such that 0 φ x,y θ r is at most r. Proof. We divide the proof into cases (i) and (ii) listed above (case (iii) is proved in the discussion of interior roads). In case (i) the two points, x 1, x belong to the same cell. The lattice roads of such points intersect in one of two ways, exhibited in Figure 3 by the pairs {e, f}and{e, f }. The path along access roads from e to f is short while the path from e to f is long. We need to show that all pairs of points x 1, x with 0 φ x,y θ r have access roads that intersect like those of e and f. (WLOG) Let x 1 be the leftmost point. Consider the NE access road through x 1 and the NW access road through x. Clearly these roads intersect (at point x 3, say). Moreover, x must lie below the NE access road for x 1, since 0 φ x,y θ r. Take the path from x 1 to x 3 to x. This path navigates around a single θ r -degree angle. So the path ratio is at most r. In case (ii) the two points x 1, x belong to different bars. Again, let x 1 be 5

6 the leftmost point. Draw a straight line between x 1 and x and label points (starting from x 1 and traveling to x ) where this line intersects successive bars as y 1, y,..., y n. We show there is a short path between y i and y i+1 for any i. Concatenating paths, we get a short path from x 1 to x. Fix i, and note that y i and y i+1 are points on the boundary of the same bar, but that y i lies on the upper boundary (call this line b 1 ) and that y i+1 lies on the lower boundary (call this line b ). (We know that the first is on the upper, the second on the lower and not vice versa, precisely because 0 φ x,y θ r.) In general y i+1 is between two cuts, and each of these cuts clearly intersects both b 1 and b. Thus we can travel from y i along the line b 1 to the point where it meets the first (leftmost) cut, then follow the cut down to where it intersects line x 1 x. This is the first segment of our path. The second continues along the cut until it intersects with b and then travels along b to the point y i+1. Both segments negotiate a single ψ r -degree angle, so the concatenated path from y 1 to y i+1 has path ratio no worse than r. (We are implicitly appealing to the lemma stated at the beginning of the previous section, that when we concatenate paths between colinear points, the total path ratio is no worse than the path ratio for the subsidiary paths. Here the y i s are all colinear, since they lie on the line from x 1 to x.) Figure 5. Path segment from y i to y i+1. We have constructed (by concatenating all segments y i to y i+1 for 1 i n 1) a path from y 1 to y n. We need to now show that there is a short path connecting x 1 to y 1 and another short path connecting x to y i+1. But this is easy, because we have an access road starting from x 1 that intersects b 1. Traveling down the access road and then in the direction of x on b 1 (i.e. rightward), we negotiate a single θ r -degree angle, so the path ratio of this segment is no worse than r. The same applies to the segment from x to y i+1, completing the proof. (Note that in the path we ve constructed we always travel down and rightward, or alternatively, up and leftward. This ensures that we are using only the major angle of the network and not 6

7 the minor one, which would give a poor path ratio.) To complete the construction, we need to ensure that pairs of points x 1, x for which 0 φ x,y ψ r does not hold are captured by a copy of our original construction. Since ψ r is constant, we can simply rotate the lattice (not including access roads) some finite number of times, m = π ψr, through the angle ψ r and for each of these lattice copies draw the corresponding access roads. The kth copy captures all pairs of points (x,y) such that (k 1)ψ r φ x,y kψ r. Moreover, since we have m copies of the original construction, the total network length is simply m times the original network length, which is clearly still linear in n. 3 Network Length We show that the normalized length of the network we ve constructed, when optimizing over y, is: π ψ r tan(ψ r )( r r 1 + ) sin(ψ r ) In analyzing the total network length it is useful to think separately about the lattice, access roads and the number of rotations of the entire construction. Letting x be the length of a full line in our lattice (one that extends from one edge of the square to the opposite edge), we have cos(ψ r ) = n x and therefore, x = n. The number of lines in our construction is cos(ψ r) n y. Multiplying the two gives total lattice length (not exact) (p r + ) prior to taking copies of the construction (not exact), (p r + ) n y cos(ψ r ) We saw above that p r = r r 1, which gives ( r r 1 + ) n y cos(ψ r ) (1) Letting z be the length of one edge of a single cell we have, sin(ψ r ) = y z y or simply, z = sin(ψ r). The length of access road per point is z which gives total access road length: yn () sin(ψ r ) 7

8 (We now see why we required that y be constant. If y grows with n or even n the total access road length is not linear in n.) Thus the total length of a single copy of the network is (not exact) Optimizing for y we set: which gives ( r r 1 + ) n + yn y cos(ψ r ) sin(ψ r ) ( r r 1 + ) n + yn y cos(ψ r ) sin(ψ r ) = 0 y = tan(ψ r )( r r 1 + ) Therefore total optimized network length for a single copy is ( r r 1 + ) n tan(ψ r )( r r 1 + ) n + tan(ψ r )( r r 1 + ) cos(ψ sin(ψ r) r ) and normalized network length for a single copy is which is just ( r r 1 + ) + tan(ψ r )( r r 1 + ) cos(ψ r) tan(ψ r )( r r 1 + ) sin(ψ r ) tan(ψ r )( r r 1 + ) sin(ψ r ) since the two terms in the previous formula are equal. Finally, to cover all pairs of points we need m = π ψ r copies of the original network. Thus the final (normalized) network length is: π ψ r tan(ψ r )( r r 1 + ) (4) sin(ψ r ) (3) Noting that, ψ r = π sin 1 ( 1 r ) 8

9 we plot the normalized network length against r: 9

10 Figure 6. Graph of normalized network length for 1.1 r 3 (top) and 1 < r 3 (bottom). In particular, for values r = 1.5, we get roughly and 1.89 respectively. 10