Hoel, David

Sobel, Milton

Uppuluri, V. R. R.

University of Minnesota Digital Conservancy

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CLUSTER PROBLEMS IN ONE DIMENSION 
by 
D. G. Hoel, Nat. Inst. Environ. Health Sciences 
* v. R.R. Uppuluri, Oak Ridge Nat. Laboratory 
M. Sobel~ University of Minnesota 
Technical Report No. 216 
August, 1973 
Operated by Union Carbide C<1rporation under contract with the U.S. 
Atomic Energy Commission. 
** Supported by NSF Contract No. GP-28922X at the University 
of Minnesota, Minneapolis, Minnesota. 
- __ _!. 
1. Introduction 
Let us consider a one-dimensional lattice with n sites, which are 
indexed by 1, 2, ... , n. The contiguous set [i, i + 1, ... , i + s - 1], is 
called a cluster of sizes. Let Bi= Bi(s) denote the cluster [i, i + 1, ... , 
i + s - 1], of size s. For a given one-dimensional lattice with n sites, 
we have a total of M = n - s + 1 clusters. 
We keep n and s fixed in the following discussion. We may select a 
collection (c > 0) of clusters, of common size, either with or without 
replacement. Let X (Y) denote the number of points of the lattice which belong 
C C 
to the set-theoretic union (intersection) of c clusters. In each of these 
cases, we obtain expressions for the expected coverages E(X) (E(Y )) using 
C C 
a ~ethod given in Robbins [2], and we also obtain expressions for the probability 
of a complete coverage with a collection of c clusters. 
These problems were motivated in understanding the phenomenon of 
degradation of the DNA strand which is exposed to radiation. It is known 
that the DNA strand consists of nucleotide bases which are assumed by some 
to be arranged in a linear strand and by others to be in a circular strand. 
After irradiation a phenomenon described as degradation of the strand takes 
place and it is of interest to be able to predict the e~pected amount of 
degradation and the probability of total degradation. We can study these 
problems in the context of a linear model if the lattice points are arranged 
on a line or in the case of a circular model if the points are arranged on 
a circle. For further results on the distribution of X (Y ), the reader may 
C C 
refer to the report by Roth and Sobel [3]. 
- 1 -
--
-
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2. Expected Coverage 
In this section, in order to apply the known results from the theory 
of continuous variables, we will redefine the cluster as a half-open 
interval. We note that in doing this, we are still considering the n lattice 
sites, and the size of the cluster is the same as was defined before. 
Specifically, we consider the interval [1, n + 1] which contains the n sites 
i=l,2, ... ,n. A cluster B. = B.(s) of size s will be represented by an 
]. ]. 
interval [i, i + s] of length s. We will randomly choose c sites (cluster 
left endpoints) out of n - s + 1 (with or without replacement) and obtain their 
corresponding cluster intervals. Let p(x), x e (1, n + 1), be the probability 
that the point x belongs to at least one of the c cluster intervals. By 
Robbins [2] we then have the expected number of sites in the set theoretic union 
X of 
C 
(;~ .1) 
where 
C 
pi 
clusters given by 
E(X) = Jn + 1 p(x)dx = 
C 1 
is the probability that site 
n 
.6 p. 
i=l l. 
i belongs to at least one of the 
C cluster intervals. For the intersection coverage Y, we have as a 
C 
corollary that 
(2.2) 
where is the probability that the site i belongs to every one of the 
c cluster intervals. 
2.1. Sampling with Replacement 
We shall consider two geometric configurations of the n sites i = 1,2, ... n. 
The first configuration is the linear model which has been previously described 
and the second is the circular model. In this latter situation the sites lie 
on the circumference of a circle and a cluster of length s is againa set of 
- 2 -
--
-
-
.. 
.. 
s contiguous sites along the circle. We begin by giving the values of p. 
l. 
with is (n + 1)/2 for the linear model and when i ~ (n + 1)/2 we use the 
relation p. = p ·+l" 1 n-1 
( c:1) Linear Mode 1: 
The values of the p. 's are easily seen to be as follows: 
l. 
Case 1: n ~ 2s-l 
\ 1 (n-s+l-i f for 1 s i s s - n-s+l pi = ~ C 1 ( n-2s+l) for s s . s n+l - n-s+l l. - 2 
Case 2: n~ 2s - 1 
l 1 n-s+l-i for ls is n-s+l n-s+l pi = . s n+l 1 for n-s+l s= l. --2 
It then follows from (2.1) that 
(2.3) 
(2.4) 
Case 1: n c? 2s-1 
f s-1 (n-s-i+l)c + (n-2s+2)(n-2s+l)cl E(X) = n - (n-s+l)-c 2 iJ 
C i=l 
Case 2: n s; 2s-1 
n-s 
E(X) = n - 2(n-s+l)-c 2J (n-s-i+l)c 
C i=l 
To obtain E(Y) it is easy to see that for is; (n+l)/2 
C 
Case 1: n ~ 2s 
- 1 
l { i/(n-s+l)} c for is s qi = ( s/(n-s+l)) c for i :i: s 
then by (2.2) and the relation qi = q . 1(i ~ (n+l)/2) n-1.+ 
- 3 -
--
... 
-
(2.5) E(Y) = {n-s+l)-c 
C 
Similarly, for 
Case 2: n ~ 2s - 1 we have 
~ 
1
( i/(n-s+l) f 
qi = t 
and thus 
(n - 2s + 2)sc l . 
i :s: n-s+l 
i :2: n-s+l 
n-s+l 
(2.6) 
(b) 
E(Y) = 2(n-s+l)-c 
C 
z:; 
i=l 
ic + 2(s-1)-n. 
Circular Model 
The circular model is simpler since the adjustments for the internal 
endpoints are not needed. For 
C 
Case 1: n ~ s, we have p. = 1 - (~) and thus 
l. n 
(2,7) E(X ) =nll-(1-!t~ . C 
For E(Y) we use q. = {s/n)c so that 
C l. 
c-1 
(2.8) E(Y ) = s(~) 
C n 
For 
Case 2: n s s, we have pi= 1 and qi= 0 so that 
(2.9) E(X) = n and E(Y) = O. 
C C 
2.2. Sampling without Replacement 
The solution to this problem is analogous to sampling with replacement 
except that the probabilities p. and q. change. Here cs n-s+l and we define 
l. l. 
( j) to be zero if O s j < c. 
C 
{a) Linear Model: 
We again give the p., q. values for is (n+l)/2 and use the relations 
l. l. 
- 4 -
... 
-
.__ 
-
-
-
._ 
.. 
-
-
-
-
-
-
-
-
-
... 
-
Pi= pn-i+l and qi= qn-i+l 
Case 1: n ~ 2s-1 
p. = 
1 
so that 
(2.10) 
\ 1 _ (n-s-i+l)/(n-s+l) 
) C C 
i 1 _ (n-2s+l)/(n-s+l) C C 
E(X J = n _ (n-s+l)-l 
C C 
1 s i s s 
sS iS.E±!_ 
2 
2 ~ (n-s-i+l) + (n- 2s+2 )(n-2s+l)} { 
s-1 ) 
i=l C C J 
Case 2 : n :::; 2 s-1 
p. = 
1 
and thus 
(2.11) 
l 1 -(n-s-i+l)/(n-s+l) C C I 1 s: i s: n-s-c+l 1 . n+l n-s-c+ s is - 2 
n-s 
E(Xc) = n _ 2.0 (!)/(n-s+l) . 
j=c C 
To obtain E(Y) we again consider the two cases. 
C 
Case 1: n ~ 2s-l 
q. = 
l 
and thus 
(2.12) 
( 
~ (!)/(n-:+1) lS:iS:s 
( <:>,cn-~+1, s s i s: ( n+ 1 ) / 2 
E(Y ) = (n- s+l) 2 ~ ( 1 ) + ( n-2s+2 )( s) . -1 Is-1 . l
C C • C C 
l.=C 
- 5 -
--
-
-
Case 2: n s 2s - 1 
1 s: i s; n-s+l 
= 
n-s+l s: i s: (n+l )/2 
and hence 
(2.13) 
n-s ~ (i)+2s-n. 
i=c C 
(b) Circular Model: 
For this case we have by similar reasoning that for 
Case 1: n ~ s, it follows that pi= 1 - (n: 5 )/(:) and hence 
(2.14) E(Xc) = n }1 - (n: 8 )/(:) f. 
For E(Yc) we use qi=(;)/(;) and thus 
(2.15) E(Y) = c(s)/(n-l) 
C C c-1 
Case 2: n ::a s 
We have that pi= 1 and q1 c O which yieldi 
(2. 16) E(X) = n and E(Y) = O. 
C C 
3. Probability of Complete Coverage 
To develop a formula for the probability of a complete coverage 
in the linear case with replacement, we start with the case of odd s and 
point out later that the same result holds for even s. For odd s, we can 
identify each cluster with its center point. For n=s, we always get complete 
coverage. To obtain complete coverage for n > s, we need first of all to. 
- 6 -
--
-
-
-
-
to select the two end clusters, whose center points are distinct; we denote 
them by c1 and c2 . Our analysis focuses attention on the n-s spaces between 
c1 and c2 . We obtain complete coverage if and only if the remaining c-2 
center points to be selected partition these n-s spaces into c-1 parts 
so that each part is of size at most s. We separate these partitions 
first according to the number of zeros in the partition and then according 
to the sizes of. the subsets of contiguous zeros in the partition. Thus 
A~ 2 1(n-s,c-1,s) denotes the number of ordered partitions with a total 
.) ' ' 
of 6 zeros of which 3 are contiguous in one set, 2 are contiguous in another 
and one zero is separated from both sets. This term has to be multiplied 
by c!/(4!3!2!) since we are selecting c clusters in all and each set of j 
contiguous zeros corresponds to j selections of the same cluster. Thus the 
probability P{CC} of complete coverage is given by 
(3 .1) P( cc} C ! = ( 
A1 (n-s ,c-1,s) { A2(n-s ,c-1,s) A (n-s,c-1,s) + 2 , + 3 , Mc o • • 
+ 
A1 1(n-s,c-1,s) } { + A3(n-s,c-l,s) + A2 1(n-s,c-1,s) 
(2!)2 4! 3!2! 
+ 
A1 , 1 , 1 ( n- s, c-1, s) l ) + ... 
(2! )3 
where M = (n-s+l)c is the total possible number of ordered partitions. The 
terms indicated by braces in (3.1) terminate as soon as the total number of 
zeros J is such that (c-1-J)s < n-s or J > C n - ;, i.e., we sum on J up 
[c n the of n Let j = (j 1, j2, jr) denote to - ;L integer part C - - ... , r s N 
subsets of zeros, the ith one containing j. contiguous zeros and let 
l. 
r 
J = 2; j .. 
i=l l. 
The problem of finding ordered partitions A.(t,p,m) with J zeros 
~ 
- 7 -
... 
is equivalent to finding partitions of t into p-J parts, with each part 
at most m and no zeros, except for the positions of the sets of zeros. 
Hence we obtain for p ~ J 
(3.2) A.{t,p,m) = {p-J+l)// r )i\A (t,p-J,m) 
,l r \\Q' 1 ,Q' 2 ' · · · ,Q' a ~ 0 
where Q'l is the frequency of the number 1 in the vector ,.i,· .. ,Q'a is the 
frequency of the number a, in ,L, so that Q'1-fa2+ ... -+aa = r, and double parentheses 
indicate the usual multinomial coefficient. Thus for J = 3 we have 
(3.3) A2 , 1{t,p,m) 
Substituting these in (3.l) gives the result 
[c-.!!] 
( 3 .4) P( cc} = 7 J~o s ~((011 ,a2'. ~~: a ,c-J-r)) A (n-s c-1-J s) 0 ' ' r 
11' (j. + 1)! 
i:t:l l. 
where the inside sum in (3.4) is over all the unordered partitions of J into 
positive parts. In more detail this can be written as 
- 8 -
-... 
-
(3.5) ,;! ( ((c-1)} P{cc} - Mc A0 (n-s,c-1,s) + A0 (n~s,c-2,s) \ ~! 
{ 
(c-2) (c-2)} { (-3) 
+ A (n-s,c-3,s) ~' + 2 2 + A (n-s,c-4,s) [, o • (2!) O • 
n this shows the terms for J s 4 and J continues up to [c - -]. 
s 
The quantity A (t,p,m) used throughout (3.5) is the number of ordered 
0 
partitions oft into p positive parts each at most m; it is well known 
(see for example [1], pp. 227-228) that for p ~ 1 
(3.6) 
min( [ !::.E.], p} 
( ) !I\ ( )i (p)(t-1-mi) A t ,p ,m = t.:,, -1 . l · 
0 i=O l p-
Fort~ p and m ~ 1 the formula (3.6) also gives the number of ordered partitions 
oft-pinto p nonnegative parts, each part being of size at most m-1. 
For even values of the integer s we select the spaces (associating the 
· dd 1 · h h 1 ) d · · th · f h (-s2 + 1 ) 
8 t m1 e space wit eac c uster an partition e n-s points rom t e 
to the (n 
-
~ - 1)9t . W · 1 2 d · · h · 2 e again se ect c- spaces an partition t e n-s points 
into c-1 parts, and for complete coverage each part must be at mosts. Hence, 
the derivation is exactly the same for both s odd and s even. 
To illustrate the computation, suppose we want the P(CC} for n = 11, s = 3 
and c = 5. The [c - .!!] = 1 and needs only 2 terms. 
s 
Considering the 3 partitions 
(3,3,1,1),(3,2,2,l) and (2,2,2,2) with their multiplicities {or by using (3.6)), 
we obtain A (8,4,3) = 19 and from the partition (3,3,2) with its multiplicity 
0 
we obtain A {8,3,3) = 3. Hence, for this case 
0 
( 3. 7) 3000 = o8 59049 · 05 ... 
- 9 -
-3.1 Circular With Replacement 
In the case of the circular chain with replacement (with n points, c 
clusters, each of sizes), we again considers odd and later show that the same 
result holds for seven. The analysis is reduced to that of a linear chain 
with replacement by breaking the circle at the center point of the first 
cluster selected and linearizing it. We arbitrarily assign this midpoint 
to the right end of the linear chain and, since we are partitioning then 
spaces by selecting points, the partition cannot start with a zero. A set of 
Z contiguous zeros in general indicates that a particular cluster was selected 
Z+l times, but a set of Z terminal zeros indicates that the initial cluster 
was selected a total of Z+l times or Z times after the first choice. 
As before, A(n,c,s) denotes the number of ordered partitions of n into 
c nonnegative parts with each part of size at mosts. We need the number of 
ordered partitions of then spaces (there is no space after the last point) 
by c-1 points into c nonnegative parts, each part at mosts, and with no 
zero at the outset. This is given by 
(3.8) B(n,c,s) = A(n,c,s) - A(n,c-1,s) , 
the total number of ordered partitions of n into c parts, each at mosts, and 
with the first part positive. To separate these according to the structure 
of the non-terminal zeros, we first note that the number of ordered partitions 
with no zeros in B(n,c,s) is 
[.!!.:£] 
s 
(3.9) B (n,c,s) = A (n,c,s} = 2J (-l)i(~}(n-!:~1) 
o o i=o 
By substracting 1 from each part of any partition we note that A (n,c,s) = 
0 
A(n-c,c,s-1). In order to include with the non-zero cases in (3.9) those cases 
- 10 -
--
with only 1 terminal zero, we define 
(3.10) a* (n,c,s} = B (n,c,s) + B (n,c~l,s) = A (n,c,s) + A (n,c-1,s). 
0 0 0 0 0 
* To obtain B1(n,c,s), the number of ordered partitions with exactly 1 
non-extreme zero and at most 1 terminal zero_£!: with 2 terminal zeros, we 
note that we can insert a new zero in exactly c-1 places in the partitions 
associated with the two B functions in (3.10). This is because each extreme 
0 
is either ruled out·or lea4s to no new partition. Hence, the number of unordered 
partitions corresponding to B~(n,c,s) is (c-l)!B;(n,c,s)/2!. Similarly, 
* let B2(n,c,s) denote the number of partitions either with two zeros which 
are contiguous and both non-extreme .2!: with 3 terminal zeros. This is 
multiplied by (c-l)!/3! to get the unordered partitions. Hence, the 
probability P(CC} of a complete coverage is given by 
(3.11) P(cc} 
* * (c-l)! ( * B1(n,c,s) { B2(n,c,s) 
= _1 B (n,c,s) + 2 , + 3, C O • • n 
* + B1 , 1 ( n , c , s ) } 
{2! )2 
* * * 
{
B3(n,c,s) B2 1(n,c,s) B1 1 1{n,c,s)} +------+ 2 + 22 + 
4! 3!2! (2! )3 . ·) 
* For each B (n,c,s) in (3.11) we count the number of ways we can put in the 
zeros and note that 
* c-2) *( ) Bl {n ,c, s) = ( 1 B0 n,c-1,s 
* c-3) *c ) B2(n,c,s) = ( 1 B0 n,c-2,s 
(3.12) 
* (c;3 )B:(n,c-2,s) B1 , 1(n,c,s) = 
* {ci:4 )B: (n ,c-3, s) B3'n,c ,a) = 
- 11 -
-* B2 , 1 ( n, c , s ) 
* B1 , 1 ,1(n,c,s) 
= 2(c;4 )B:(n,c-3,s) 
= (c;4 )B:(n,c-3,s), etc. 
In general, we can write 
(3.13) * B.(n,c,s) 
;L 
(( c-J-1 a )~ *( ) = B n,c-J, s a 1 ,a2 , ... ,ex ,c-J-1- LJ ex. o r i=l 1 
r 
where,J, = (j 1 ,j 2 , ... ,jr)' J =LJj. and ex. is the frequency of i i=l l l. 
contiguous zeros (non-extreme) or i+l terminal zeros (i = 1,2, ... ,a) as 
a 
in ( 3. 9) and LJ ex i = r. Hence. the probability of complete coverage is 
i=l 
[ c-.!!J * 
(3.14) P(ccJ = (c-1)! Lis LJl(i c-J-1 )) Bo(n,c-J,s) 
n c-1 J =o ~ ex 1 ,ex 2 , ... ,ex a , c-J -1- r r . 1 TT (J.+1). 
i=l ]. 
where the inside sum is over all the unordered partitions of the J zeros 
into positive parts. In more detail this can be written as 
(3.15) 
(c-1)! ( * f (ci2 )} * P{CC} = nc-l B0 (n,c,s) + { 2 ! B0 (n,c-l,s) 
+ l + _g__ B*( 2 ) { 
(-3) (c-3)} 
3, 2 n,c- ,s 
. (2!) 0 
(c-4) } 
3 * 
3 B (n,c-3,s) (2!) 0 
+ ... ) . 
For example, for n=lO, c=4, and s=3, we note that 
- 12 -
A (10,4,3) 
0 
= fJ (-1/ (~) (9- 3i) = 84-80+6 = 
i=o i 3 
10, 
(3.16) * B ( 10 ,4, 3) = o for x < 4; 
0 
* B (10,4,3) = 10, 
0 
P{cc} = ll. (10) = .06 . 
103 
The ten partitions in this example are (3,3,3,1) with multiplicity 4 and 
L~,.~,.L~~) with multiplicity 6. 
For evens we select spaces and partition the points. Except for this, 
the duality is complete and the resulting formula is the same for all s. 
* As another example we take n=5, c=4, s=3 and obtain B (5,4,3) = 10, 
0 
* B (5,2,3) = 2 and hence 
0 
(3.17) 6 2 110 22 P(cc} = 53 (10 + 8 + 6) = 125 = 25 = .88. 
This can also be (and has been) checked by independent calculations. 
In addition to the formula for A (n,c,s) in (3.9) we also note that 
0 
( ) n-c A n,c,s is the coefficient of x 
0 
in 
C 
(3.18) n::s) = (1 s-1 )c +x+ ... +x 
- 13 -
A similar formula is also obtained by David and Barton (1, p. 227, 228] 
for the probability that the maximum white ball run~ m when r 1 white 
balls and r 2 = r - r 1 red balls are randomly aligned. They obtained 
(3.19) 
where a= min(f2+1, (r-r2 )/(m+l)) and hence we find that 
(3.20) 
given by (3.6). 
where A is 
0 
From the left side of (3.18) we easily obtain (3.9) and from the right side 
we note that 
(3.2r) A (n,c,s) = 
0 
for n < c and for n > cs 
for n = c and any s ~ 1 
for c < n < cs 
and by (3.9) the same result holds for B
0
(n,c,s). It follows from (3.10) 
that for s :t 1 
- 14 -
0 for n < c- 1 and for n > cs 
* ( •I •"'l•"'I) 
. 'I ••• B (n,c,s) = 0 1 for n = c- l, for n = c and for n = cs 
>1 for c < n < cs 
Hence, we have the two speciai results 
l 0 for n > cs. (3.23) P(cc) = ( c-1 c-1)!/n for n = cs 
3.2. Linear Case Without Replacement 
Assuming s is odd again, in order to have complete coverage we must 
first of all select the two clusters with centers (s+l)/2 and n-(s-1)/2. 
Between these two centers there are n-s spaces which we have to partition 
(with c-2 additianal selections) into c-1 parts in such a way that each 
-1 
part is at most s. Each partition (or selection) has probability (n-s+~) 
C 
since there are n-s+l clusters in all and we are selecting c of them. 
Hence, using (3.9) 
P(cc) = A {n-s,c-1,s)/(n-s+l) 
0 C 
(3. 24) [n-s-c+l] -1 s 
= (n-s+l) 6 · (-l)\c-1) 
C i=O i 
where it is understood that (:)=o for o ~a< b. 
(n-s(i+l)-1) 
c-2 
For even s we again select spaces and partition points, and by this 
duality it is easily seen that the same result holds for odd and even 
values of s. 
In particular, we note from (3.21) and (3.24) that 
P(CC) = { 01 
for n > cs 
for c = n-s+l 
- 15 -
3.3 Circular Case Without Replacement 
In this case we use the first cluster to unfold or linearlize the circle. 
Hence, (assuming s odd), we wish to partition the n spaces using the 
c-1 additional selections into c parts in such a manner that each part is 
(n-1 at most s. The total number of possible selections is c-l) and the number 
of partitions is A (n,c,s). Hence, 
0 
P( cc) 
(3 .26) 
( (n-1 = A n,c,s)/ 1 ) 0 c-
-i 
= cn-1) 
c-1 
~] . 
6 (-l/ (c)(n-si-1) i c-1 i=o 
Again, the same result holds for s even. In particular, we note as above 
that 
{: for n > cs ( 3 .2 7) P(cc) = for c ~ n-s+l 
" 
where C can only go up to n if we sample clusters without replacement. 
• 16 -
-• ial 
References 
[1] David, F. N. and Barton, D. E. (1962). Combinatorial Chance. 
Charles Griffin and Company, London. 
(2) Robbins, H. E. (1944). On the measure of a random set I. 
Ann. Math. Statist. 12 70-74. 
(3) Roth, A. and Sobel, M. (1973). Distribution of the Size of 
Unions and Intersections of Clusters in one Dimension. Univ. 
of Minnesota Technical Report No. 217. 
- 17 -


Cluster Problems in One Dimension

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Cluster Problems in One Dimension

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