1 Intro

We’ve seen previously that we can construct networks based on the regu-lar triangular, rectangular and hexagonal lattices where the total networklength 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 thatR(N) ≤ r and length(N) ∼ O(n). (Alternatively, if we look at the lengthof network per city , we are interested in networks where the normalizedlength is constant.)

Note that for r = 1 this is not possible, since the only network thatachieves this R-statistic is the complete graph on n points. If n

2 points arestationed arbitrarily close to one corner of the square (n+1

2 for n odd) andn2 in the opposite corner, we have for any pair of points (x, y) where x andy are in opposite corners: dist(x, y) ∼

√2n. There are (n2 )(n2 ) = n2

4 suchpoints, so:

length(N) ∼√

2n n2

4∼ O(n

52 )

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 givea such a construction and study its length.

2 Construction

The basic idea is to construct a network that works for any two pointswithin a certain angle of each other (defined below), then superimpose afinite number of rotated copies of this network, so that each pair of pointsis captured by at least one copy of the original construction. To this endwe will construct a network that, for any “good” pair of points, contains apath which only navigates around a certain (fixed) wide angle. Making thisangle wide enough, we ensure that the ratio of shortest path to Euclideandistance is sufficiently small.

Any fixed r > 0 determines such a “wide” angle θr, which we call the“major” angle of our network. Intuitively, we would like θr to be just wideenough so that paths which navigate only around this angle have path ratiosless than or equal to r. How wide does θr have to be? The worst path ratiofor a path navigating around a single angle is achieved when the startingand ending points of the path are at equal distance from the vertex of theangle (see figure 1). Letting a be the distance between each point and the

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vertex, b be the distance between points, and noting that the path ratiopr = 2a

b = ab2

, we have

sin(θr2

) =b2

a=

1pr

or simply

θr = 2 sin−1(1pr

)

Figure 1.

We let ψr = π−θr2 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. Fromeach of these points we draw two lines, forming ψr and −ψr degree angleswith the horizontal, respectively. Note that the horizontal lines (dashed infigure below) are not themselves part of the lattice.

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Figure 2. Lattice roads.

The lines of our lattice intersect to form diamond-shaped cells, wherethe obtuse angle is θr (this was the point of constructing the lattice in thisway). Each of our n points (or cities) inhabits exactly one of these cells andwe now draw two access roads parallel to the edges of the cell in the waypictured below.

The distance y is constrained by two requirements: First, we would likethe construction to form a network –that is for the lines of the constructionto intersect. If we let y be too large the lines will not intersect for smallenough n (this problem applies only to small n, and is dealt with quiteeasily). Second (and this is the more important point) we require that y beconstant. The reason, in short, is that when y grows with, say,

√n the total

length of access road grows with n3/2 and the normalized network lengthis no longer constant (more on this below, where we analyze total networklength).

For any points in the square, x1 and x2, draw a line between them andlet φx,y be the angle at which this line intersects the horizontal. We wouldlike to show that the network we have constructed provides a sufficiently

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short path for all points x1 and x2, such that 0 ≤ φx,y ≤ ψr. Consider inparticular:

Provisional Claim 1: For any points x1 and x2, such that 0 ≤ φx,y ≤ψr, the ratio of shortest path distance to Euclidean distance along lines ofthe network is no greater than r.

Let the region between two upward-pointing parallel lines be called abar (highlighted in pink in figure 2). Looking at figure 3, it turns out theclaim 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 oneempty cell, and are each placed arbitrarily close to the midpoint of oppositeedges (again, as a and b). To correct this, we need to add interior roadsthrough the center of each cell and parallel to the NE-SW pair of edges (seefigure 3). When we do this the worst situation in our new network is whenthe two points are a quarter of the way along opposite edges. Here the path

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ratio isz4+z+ z

4z = 3

2 , where z is the length of the edge of a single cell. Considernext the network formed by adding two lines within each cell parallel to theNE-SW edges and equally spaced apart. Here the worst situation is wherethe two points are 1

6 of the way along opposite edges of the cell and the new

path ratio isz6+z+ z

6z = 4

3 The idea is that if we do this some finite numberof times, we will eventually produce a short enough path between all suchpoints. Indeed, if we partition each cell with n equally spaced lines our newworst situation path ratio is

zn+1

+z+ zn+1

z = n+2n+1 In order to ensure that our

path ratio is smaller than r we pick n large enough so that n+2n+1 ≤ r or simply

n ≥ 2−rr−1 .

Figure 4. Partitioning each cell 0, 1, 2 and 3 times.

Claim 1: In the network obtained by adding n = d2−rr−1e interior roads toeach cell, the path ratio for all points x1, x2 such that 0 ≤ φx,y ≤ θr is atmost r.

Proof. We divide the proof into cases (i) and (ii) listed above (case (iii) isproved in the discussion of interior roads).

In case (i) the two points, x1, x2 belong to the same cell. The latticeroads of such points intersect in one of two ways, exhibited in Figure 3by 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 pairsof points x1, x2 with 0 ≤ φx,y ≤ θr have access roads that intersect like thoseof e and f. (WLOG) Let x1 be the leftmost point. Consider the NE accessroad through x1 and the NW access road through x2. Clearly these roadsintersect (at point x3, say). Moreover, x2 must lie below the NE access roadfor x1, since 0 ≤ φx,y ≤ θr. Take the path from x1 to x3 to x2. This pathnavigates around a single θr-degree angle. So the path ratio is at most r.

In case (ii) the two points x1, x2 belong to different bars. Again, let x1 be

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the leftmost point. Draw a straight line between x1 and x2 and label points(starting from x1 and traveling to x2) where this line intersects successivebars as y1, y2, ..., yn. We show there is a “short” path between yi and yi+1

for any i. Concatenating paths, we get a “short” path from x1 to x2.Fix i, and note that yi and yi+1 are points on the boundary of the same

“bar,” but that yi lies on the upper boundary (call this line b1) and thatyi+1 lies on the lower boundary (call this line b2). (We know that the first ison the upper, the second on the lower and not vice versa, precisely because0 ≤ φx,y ≤ θr.) In general yi+1 is between two “cuts,” and each of thesecuts clearly intersects both b1 and b2 . Thus we can travel from yi alongthe line b1 to the point where it meets the first (leftmost) cut, then followthe cut down to where it intersects line ←−−x1x2. This is the first segment ofour path. The second continues along the cut until it intersects with b2 andthen travels along b2 to the point yi+1. Both segments negotiate a singleψr-degree angle, so the concatenated path from y1 to yi+1 has path rationo worse than r. (We are implicitly appealing to the lemma stated at thebeginning of the previous section, that when we concatenate paths betweencolinear points, the total path ratio is no worse than the path ratio for thesubsidiary paths. Here the yi’s are all colinear, since they lie on the linefrom x1 to x2.)

Figure 5. Path segment from yi to yi+1.

We have constructed (by concatenating all segments yi to yi+1 for 1 ≤i ≤ n − 1) a path from y1 to yn. We need to now show that there is a“short” path connecting x1 to y1 and another “short” path connecting x2 toyi+1. But this is easy, because we have an access road starting from x1 thatintersects b1. Traveling down the access road and then in the direction of x2

on b1 (i.e. rightward), we negotiate a single θr-degree angle, so the path ratioof this segment is no worse than r. The same applies to the segment from x2

to yi+1, completing the proof. (Note that in the path we’ve constructed wealways travel down and rightward, or alternatively, up and leftward. Thisensures that we are using only the “major” angle of the network and not

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the “minor one,” which would give a poor path ratio.) �

To complete the construction, we need to ensure that pairs of pointsx1, x2 for which 0 ≤ φx,y ≤ ψr does not hold are captured by a “copy” of ouroriginal construction. Since ψr is constant, we can simply rotate the lattice(not including access roads) some finite number of times, m = d πψre, throughthe angle ψr and for each of these lattice copies draw the correspondingaccess roads. The kth copy captures all pairs of points (x,y) such that(k − 1)ψr ≤ φx,y ≤ kψr. Moreover, since we have m copies of the originalconstruction, the total network length is simply m times the original networklength, which is clearly still linear in n.

3 Network Length

We show that the normalized length of the network we’ve constructed, whenoptimizing over y, is:

2 d πψre

√tan(ψr)(d2−rr−1e+ 2)

sin(ψr)

In analyzing the total network length it is useful to think separatelyabout the lattice, access roads and the number of rotations of the entireconstruction. Letting x be the length of a “full” line in our lattice (one thatextends from one edge of the square to the opposite edge), we have cos(ψr) =√nx and therefore, x =

√n

cos(ψr) . The number of lines in our construction is

(not exact) (pr + 2)√ny . Multiplying the two gives total lattice length

prior to taking copies of the construction (not exact),

(pr + 2) ny cos(ψr)

We saw above that pr = d2−rr−1e, which gives

(d2−rr−1e+ 2) ny cos(ψr)

(1)

Letting z be the length of one edge of a single cell we have, sin(ψr) =y2z

or simply, z = y2sin(ψr) . The length of access road per point is 2z which gives

total access road length:yn

sin(ψr)(2)

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(We now see why we required that y be constant. If y grows with nor 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)

(d2−rr−1e+ 2) ny cos(ψr)

+yn

sin(ψr)

Optimizing for y we set:

(d2−rr−1e+ 2) ny cos(ψr)

+yn

sin(ψr)= 0

which gives

y =

√tan(ψr)(d

2− rr − 1

e+ 2)

Therefore total optimized network length for a single copy is

(d2−rr−1e+ 2) n√tan(ψr)(d2−rr−1e+ 2) cos(ψr)

+

√tan(ψr)(d2−rr−1e+ 2) n

sin(ψr)

and normalized network length for a single copy is

(d2−rr−1e+ 2)√tan(ψr)(d2−rr−1e+ 2) cos(ψr)

+

√tan(ψr)(d2−rr−1e+ 2)

sin(ψr)

which is just2

√tan(ψr)(d2−rr−1e+ 2)

sin(ψr)(3)

since the two terms in the previous formula are equal.Finally, to “cover” all pairs of points we need m = d πψr

e copies of theoriginal network. Thus the final (normalized) network length is:

2 d πψre

√tan(ψr)(d2−rr−1e+ 2)

sin(ψr)(4)

Noting that,

ψr =π − 2 sin−1(1

r )2

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we plot the normalized network length against r:

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Figure 6. Graph of normalized network length for 1.1 ≤ r ≤ 3 (top) and1 < r ≤ 3 (bottom).

In particular, for values r = 1.5, 2 we get roughly 18.36 and 12.89 respec-tively.

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