The typescript of these notes is dated 18 June 1977.
As the last sentence shows, this was in the days before computers became available to all.
A proof is given of the “strip formula” for estimating the number of chance alignments
in a random set of sites. The sites are supposed to be discs, all the same size, and a set of sites is
considered to be aligned if a straight line can be drawn that intersects all the discs.
Equivalently, we can consider the sites to be geometrical points, and a set of points is aligned if
they can be covered by a movable strip of the same width as the discs.
In the 1985 notes
the formula is generalized to allow discs of different sizes
(in which case the term “strip formula” is no longer appropriate).
MB, May 2013
This problem arises from the work of Alfred Watkins,
who asserted that ancient sites such as churches, beacons, mounds, markstones, etc.
could be shown on the map to fall into straight alignments. Watkins believed that
these lines, which he called “leys”, marked the course of ancient trackways that
were constructed in prehistoric times and were still discernible in places.
He tried to show by experiment that the alignments were more numerous than chance
would allow, but his method is questionable (see below).
The problem. Let n points P_{1},P_{2},…,P_{n} be placed
independently and at random (uniform p.d.f. with respect to area) inside a bounded convex
plane region R. A “critical distance” c is decided on, and then a set of points such as
P_{1},P_{2},…,P_{r} is said to form an rpoint alignment if and only if
there exists a straight line passing within distance c of each P_{i}. It is assumed
that c is small compared with the dimensions of R. Given R, c, n, and r,
what is the expected number of rpoint alignments?
Three equivalent statements of the alignment criterion are:
 (a) About each P_{i} as centre draw a circle of radius C; then P_{1}, …, P_{r}
are aligned iff a straight line can be drawn so as to cut all the circles.
 (b) P_{1}, …, P_{r} are aligned iff a movable infinite strip of width 2c
can be placed so as to cover all the P_{i}.
 (c) P_{1}, …, P_{r} are aligned iff the minimum width of their convex hull
is less than 2c.
Although (b) is used below, note that (a) gives a method for dealing with sites of nonzero dimensions,
e.g. churches. Each site can be replaced by a circle of radius c that wholly contains it.
An objective method of fixing the exact position of the circle is to make it concentric with
the smallest circle that wholly contains the site.
If e.g. P_{1}P_{2}P_{3}P_{4}P_{5} is a 5point alignment there will be five 4point subalignments
such as P_{1}P_{2}P_{3}P_{4} and ten subalignments such as P_{1}P_{2}P_{3}. A maximal alignment is
one that is not part of some alignment of higher order. Given R and c, let e(n,r)
be the expected number of rpoint alignments among n points, and let e′(n,r)
be the expected number of maximal rpoint alignments. In these notes, only e(n,r)
is calculated. I do not know how to find e′(n,r), though it is easy to see that
e(n,r)−(r+1)e(n,r+1)<e′(n,r)<e(n,r). 

Calculation of e(n,r). Recall (i) that R is convex, (ii) that c is small
compared with the dimensions of R (e.g. in practice R might be 40 km square
with c = 20 m). Given R and c, let n points P_{1}, …, P_{n} be placed
randomly inside R. Let p(r) be the probability that P_{1}, …, P_{r} form an alignment
with endpoints P_{r}_{−1} and P_{r}. Then
e(n,r)= 
( 
n 
) 
( 
r 
) 
p(r). 
r 
2 

(1) 
Denote the distance P_{r}_{−1}P_{r} by q and let ƒ(q) be the p.d.f. of q. A necessary
condition for P_{1},…,P_{r} to form an alignment with endpoints P_{r}_{−1} and P_{r}
is that P_{1},…,P_{r}_{−2} fall inside the q×4c rectangle shown in Fig. 1.
In view of assumptions (i) and (ii) we may assume
(iii) that q≫c
(iv) that the rectangle in Fig. 1 lies wholly inside r.
That is, we ignore the small proportion of cases in which (iii) or (iv) is false.
Let A be the area of R; then the probability that P_{1},…,P_{r}_{−2} all fall inside the rectangle is
∫ 
^{∞} 
(4cq/A)^{r}^{−2}ƒ(q)dq=(4c/A)^{r}^{−2}M_{r}_{−2} 
_{0} 

where M_{r}_{−2} is the (r−2)nd moment of f. Now assume that P_{1},…,P_{r}_{−2} are placed
independently and at random inside the rectangle, and let the centre line of a movable strip
meet the ends of the rectangle at (−½q,x) and (½q,y). Since q≫c, P_{r}_{−1} and P_{r}
will lie inside the strip iff
If a typical P_{i} (1≤i≤r−2) has coordinates (u_{i},v_{i}) then P_{i} will lie
inside the strip iff
v_{i}−((½q−u_{i})x+(½q+u_{i})y)/q<c. 

(3) 
Setting
and
λ_{i}=v_{i}/c, μ_{i}=2u_{i}/q 

(5) 
we can rewrite (3) as
Each position of the strip corresponds to some point in the (ξ,η) plane according to (4).
Inequalities (2) hold iff (ξ,η) lies inside the infinite strip whose boundaries have slope
μ_{i} and cut the ηaxis at (0,λ±1). Hence the movable strip in the (u,v) plane
covers P_{r}_{−1}, P_{r} and P_{i} iff the corresponding (ξ,η) lies inside the shaded region S_{i}.
The points P_{1},…,P_{r} form an alignment iff
∩ 
^{r}^{−2} 
S_{i}≠∅. 
_{i}_{=1} 

(7) 
We find the probability that (7) holds. Since the (u_{i},v_{i}) have independent uniform p.d.f.s
inside the rectangle, (5) implies that the λ_{i} and μ_{i} have independent uniform p.d.f.s in
λ_{i}<2, μ_{i}<1. Hence the procedure for getting S_{i} can be restated as follows:
The boundary line cutting the square has the equation
where μ_{i}, ν_{i} have independent uniform p.d.f.s in (−1,1); and then S_{i} can lie either
above or below the boundary line, each with probability 1½.
Lemma 1.Given i and j (1≤i<j≤r−2)
the probability that the boundary lines
of S_{i} and S_{j} meet inside the square is 1/3.
Proof.Let these lines meet at (ξ_{0},η_{0}). This point lies inside the square
iff η_{0}−ξ_{0}<1 and η_{0}+ξ_{0}<1. In the (μ,ν)plane the points
I=(μ_{i},ν_{i}) and J=(μ_{j},ν_{j}) have independent uniform p.d.f.s in the square
μ<1, ν<1. It is easy to verify that IJ produced meets the left and right sides
of the square at (−1,η_{0}−ξ_{0}) and (+1,η_{0}+ξ_{0}). Hence we want the probability that IJ
produced meets both sides internally. By symmetry we may assume that I lies in one of the regions
0<ν_{i}<μ_{i}<1 (Fig. 3a) 
0<μ_{i}<ν_{i}<1 (Fig. 3b) 

Then J is required to lie in the shaded region, and the probability that this occurs is
1 
∫ 
^{1} 
∫ 
^{μ} 
(1+μ^{2})dμdν 
+ 
1 
∫ 
^{1} 
∫ 
^{ν} 
(1−ν)(1+μ^{2})dμdν 
2 
_{μ=0} 
_{ν=−μ} 
2(1+μ) 
2 
_{ν=0} 
_{μ=−ν} 
2(1−μ^{2}) 

= 
( 
11 
− 
ln 2 
) 
+ 
( 
ln 2 
− 
7 
) 
= 
1 
. 
12 
12 
3 

Lemma 2. Let C be a plane convex region.
Given a set of s straight lines, all cutting C,
let t be the number of intersection points inside C (no multiple intersections). Then the s
lines divide C into s+t+1 nonempty regions.
Proof. For s=1 or 2 the result is clear.
Suppose the result holds for a given set of s lines.
Add a new line L, and let L meet Δt of the s lines inside C. Then C∩L is divided into
Δt+1 segments, each of which divides one of the s+t+1 regions of C into two parts.
Hence the number of regions is now
(s+t+1)+(Δt+1)=(s+1)+(t+Δt)+1. 

The result follows by induction.
Now take C to be the square of Fig. 2, and fix a set of r−2 boundary lines for the S_{i}.
Let t intersection of the lines be inside C, with no multiple intersections. Then by Lemma 2
C is divided into r−1+t nonempty regions. Since each S_{i} can be either above or below
its boundary line, the set of S_{i} can be defined in 2^{r}^{−2} ways. These are all equally
probable, so that
pr(  ∩  S_{i}≠∅)=(r−1+t)/2^{r}^{−2}. 

If the μ_{i} and ν_{i} are now allowed to vary, Lemma 1 implies that E(t)=(r−2)(r−3)/6.
Multiple intersections have zero probability and can be ignored. Hence
pr(  ∩  S_{i}≠∅)=(r−1+E(t))/2^{r}^{−2}=r(r+1)/(3⋅2^{r}^{−1}). 

Combining the above results we find that
Values of M_{r}_{−2} in two particular cases
(1) Circle of radius a:
M_{r}_{−2}= 
2Γ(r)a^{r}^{−2} 
Γ(½r+1)Γ(½r+2) 

(2) Rectangle of sides a and b:
Let d=√(a^{2}+b^{2}), α=a/d, β=b/d,
H_{1}= 
1 
ln 
( 
1+β 
) 
, 
K_{1}= 
1 
ln 
( 
1+α 
) 
, 
β 
α 
α 
β 


H_{r}= 
1+(r−2)α^{2}H_{r}_{−2} 
, 
K_{r}= 
1+(r−2)β^{2}K_{r}_{−2} 
(r=3,4,…). 
(r−1) 
(r−1) 

Then
M_{r}_{−2}= 
4d^{r}^{−2} 
( 
H_{r}+K_{r}− 
1−α^{r}^{+2}−β^{r}^{+2} 
) 
. 
r(r+1) 
(r+2)α^{2}β^{2} 

Practical calculation
In practice it is convenient to denote the area of R by k^{2}.
The expected number of rpoint alignments among n points, with each of the r points allowed
to be offline by a distance not exceeding c, is given by
e(n,r)=n(n−1) … (n−r+1)⋅(c/k)^{r}^{−2}⋅G(r) 

where the factor G(r) depends only on r and the shape of the region R.
For any particular shape G(r) can be tabulated once for all, thus:

Circle  Square*  9:8 rect.†  3:2 rect.  2:1 rect. 
G(3) 
1.0217  1.0428  1.0456  1.0754  1.1381 
G(4) 
1.0610  1.1111  1.1188  1.2037  1.3889 
G(5) 
0.7433  0.8024  0.8126  0.9269  1.1888 
G(6) 
0.3940  0.4407  0.4494  0.5488  0.7907 
G(7) 
0.1683  0.1962  0.2016  0.2651  0.4312 
G(8) 
0.06020  0.07365  0.07630  0.1084  0.1992 
G(9) 
0.01855  0.02398  0.02505  0.03851  0.07982 
G(10) 
0.005019  0.006910  0.007280  0.01209  0.02822 
G(11) 
0.001211  0.001790  0.001901  0.003402  0.008921 
G(12) 
0.0002637  0.0004217  0.0004513  0.0008680  0.002551 
*O.S. 1:50 000 maps cover 40×40 km.
†O.S. 1inch (1:63 360) maps cover 45×40 km.
Example. On 1:50000 Sheet 154 there are 165 old churches, excluding those in Cambridge city.
If the critical distance c is 20 m, find the expected number of 5point alignments.
According to the formula, the expected number is
165×164×163×162×161×(20/40000)^{3}×0.8024=11.5 

Watkins’s experiments. To find to what extent alignments of sites might occur by chance,
Watkins carried out a few rough practical tests.
One of these is described in The Old Straight Track as follows:
“This is in the Andover map (Sheet 283) which contains 51 churches, and there are on it no less than
8 separate instances of 4 churches falling in a straight line, in addition to the example of 5 churches
in alignment already mentioned. To test the argument that this might be accidental I took a similar sized
sheet of paper and marked 51 crosses in haphazard distribution over it. Only one instance could be found on this of
4 points aligning, and none of 5.”
The map Watkins was using measures 18×12 inches, but he doesn’t say how accurate an alignment had to be
before he included it. If we take his result at face value, we can get a rough idea by working out what
value of c would make the number of 4point alignments equal to 1. This turns out to be surprisingly small,
namely c=0.0055 in. = 0.14 mm.
In Archaic Tracks Round Cambridge Watkins describes a similar experiment and states that
“only one or two”
4point alignments were found on “a large sheet with a hundred points”.
This is vague, but we can get an idea
of the tolerance allowed, by assuming a sheet 18×12 inches as before and the expected number of 4pointers
equal to 1.5. Then c=0.0017 in. = 0.043 mm, a distance scarcely perceptible
to the naked eye.
But Watkins also says here that on average a 3pointer comes when only nine marks are made on a sheet.
This implies c=0.027 in = 0.69 mm, which is not only much larger than before
but also seems more likely to indicate Watkins’s true standard of accuracy. Why then the discrepancy?
The suspicion must be that with 50 or 100 crosses on his sheet Watkins thought he had counted all the chance
4pointers when in fact many more remained to be discovered. To find all the alignments among a large number of points
is a very long job. (I once decided to find all 3point alignments of mounds on the 1:50 000 Cambridge map.
Chance alone would yield about 60, but it took so long to find even 5 or 6 that I abandoned the task.
A computer is really needed.)