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I.
Graph compositions - Basic enumeration
proofs follow [1]
Notation:
G = E(G), V(G) denotes a labelled graph.
A composition of G is a partition of the
vertex set V(G) into vertex subsets of connected induced subgraphs of G i.e. partition of
G provides a set of connected subgraphs of G  G1, G2, ...,Gm , Gi = E(Gi), V(Gi) , i
= 1,...,m .
Note:
Because the same vertex subset may be spanned by different edge subsets – therefore to the
same composition of G which is a partition of vertex set V(G) there may correspond
different families G1, G2, ...,Gm   G~1, G~2, ..., G~m of connected subgraphs.
1
2
5
6
3
4
8
7
1
2
5
6
3
4
8
7
Biparite complete graph
K2,3 compositions` example:
the same compositions of K2,3
Kn,m complete biparite graph has nm edges linking the first row n dots in
such a way that each dot of this row is linked to every one of the second row
of m dots (dots = vertices).
Notation:
C(G) = the number of distinct compositions of a graph G.
Why the name: composition of G ?

Exercise 1.
Let Pn be the path with n vertices. Prove that C(Pn) = 2n-1 , n> 0, C(Po)  1.
Exercise 2.
Let Kn be the complete graph on n vertices. Prove that
C(Kn) = Bn, n> 0, C(K0)  1.
Bn = Bell numbers = exponential numbers.
Note: the same composition of G which is a partition of vertex set V(G) - may give rise to
different families G1, G2, ...,Gm   G~1, G~2, ..., G~m of connected subgraphs.
Comment
These above are two extreme cases:
no connected graph G with n vertices can have fewer than C(Pn) and no connected
graph G with n vertices can have more than
2n-1  C(Fn)  Bn, where Fn
C(Kn) compositions. i.e.
is any connected graph with n vertices.
Exercise 3.
Let G = G1  G2 and there are no edges from vertices of
G1 to vertices of G2.
Then C(G) = C(G1) C(G2) . The same holds for G1 and G2 having exactly one vertex in
common. Prove or rather see this.
Note: one obtains compositions of G by pairing compositions of G1 and G2 in all
possible ways.
Exercise 4.
Let G = G1  G2 and there is exactly one edge from a vertex of
G1 to a vertex of G2
whose removal disconnects G . Then C(G) = 2 C(G1) C(G2) . Prove or rather see this.
Note: Let e be the distinguished edge between vertices vj and vj . Any composition of G
can be obtained in exactly two ways.: either e is included to supply the component vi in
G1 and the component vj in G2 or not. Thus the count from Exercise 3. is now doubled.

Exercise 5.
Let Tn be any tree with n vertices. Prove that C(Tn) = 2n-1 , n> 0.
Proof: Use induction. Consider Tn+1. Remove one edge. This disconnects Tn+1 into two
parts for which the formula holds. By the result of Exercise 4. the proof is accomplished.
Exercise 6.
Let K-n be the complete graph on n vertices with one edge removed.. Prove that C(K-n)
= Bn - Bn-2
, n > 1.
Proof: Let
e
be the deleted
edge between vertices
vj and vj. Its deletion affects a
composition counted by C(Kn) = Bn only when the component containing vertices vj and vj
consists of exactly these two vertices vj and vj. Otherwise there is a by-pass in Kn
connecting these two vertices vj and vj. Therefore from the number of composition counted
by C(Kn) = Bn one must substract those compositions for which one of the partition
component is {vj , vj}. This restriction rules out exactly C(Kn-2) = Bn-2 compositions of Kn.
Exercise 7.
Let Cn be the cycle graph with n vertices. Prove that C(Cn) = 2n – n , n> 0.
Proof: Delete any edge. The resulting graph becomes Pn with C(Pn) = 2n-1. Any
composition of Pn is also a composition of Cn . The deleted edge may be reinserted ,
providing a new composition of Cn - previously not counted unless the composition of Pn
had been obtained by deleting from Pn either no edge or exactly one edge.
In these cases , reinserting the original deleted edge gives the same composition of Cn:
namely the composition consisting of all n vertices. Therefore the total number of
of distinct compositions of Cn is equal to C(Cn) = 2 2n-1 – n
II. Compositions
with parts constrained
= 2n – n.
by the
leading
summand
we follow [2]; here now composition  ordered partition of a natural number
ordered partition of a natural number  constrained composition of a natural number
constrained? yes ! constrained by...
Of course:
Compositions constrained by the requirement i1  i 2 ... i k  a are partitions
Note that parts of compositions might be zero if a = 0 .
According to:
Clark Kimberling. Enumeration of paths, compositions of integers, and
Fibonacci numbers. The Fibonacci Quarterly, 39(5):430-435, 2001

The composition of the natural number
(D*)
n
is the vector
i1 + i 2 +...+ i k =
n,
 i1 , i 2 ,..., i k  solution of
a  i1 , i 2 ,..., i k  b.
C(n,k,a,b) = the number of the vector solutions of the Diophantic (*) equation
(D*)
for example for C (n,k,0,), C (n,k,1,) one has
 n  k  1
,
C ( n, k ,0, )  
k

1


 n  1

C ( n, k ,1, )  
k

1


Exercise 8.
Consider the number fn (k) of compositions of a natural number n into k parts with the
strictly largest part in the first position i.e.
(>)
i1 + i2 +...+ i k =
1  i1 , i 2 ,..., i k ,
n,
i1 > i k
for k>1
.
Observe-show that the following formula gives the ordinary generating function for these
numbers :
 f n (k ) 
n0
Proof:
k2
(1  z ) z k
.
1  2z  z k
 f n (k )  z k   (nk 1) z n 
n0
n1
k 2
zk
, where higher order
1  z  z 2  ...  z k 1
Fibonacci sequences are defined  (nkk1)1    (nki1) ... (HEY-ou! what are the initial values?)
i 0
Exercise 9.
Consider the number f*n (k) of compositions of a natural number n into k parts with the
largest part in the first position i.e.
()
i1 + i2 +...+ i k =
1  i1 , i 2 ,..., i k ,
n,
i1  i k
for k>1
.
Observe-show that the following formula gives the ordinary generating function for these
numbers :
 f n* (k ) 
n0
Proof:
k2
(1  z ) z k
.
1  2 z  z k 1
 f n* (k )  z k   (nk ) z n 
n0
n1
zk
, where higher order
1  z  z 2  ...  z k
k 1
Fibonacci sequences are defined  (nk)k    (nk)i ... (HEY-ou! what are the initial values?)
i 0
Exercise 10.
Let Ck(n) be the number of compositions of
in which at least one k occurs. Prove
z
z  z k  z k 1
, n  k  1.
C k ( n) z 


1  2 z 1  2 z  z k  z k 1
n0
n
then that
Proof:
n
Consider C*k(n) to be the number of compositions of
equals to k.
Consider
n
in which no part
C*k(n,m) to be the number of compositions of
with no part equal to k.
Note: by the product law od generating functions
C
n0
*
k
(n, m) z n  ( z  z 2  ...  z k 1  z k 1  ...) m
n
sum the above over m
via

m 1
n
inito m parts
Observe that
z  z k  z k 1
q
.

k
k 1
1 q
1  2z  z  z
That is all ... as of course
 C ( n) z
n
n0
 ... 
q=?
z
.
1  2z
\bibitem{1}
A. Knopfmacher, M.E.Mays {\it Graph compositions I: Basic enumeration } Integers
The Electronic Journal of Combinatorial Number Theory {\bf 1} (2001), #A04. 
\bibitem{2}
A. Knopfmacher and N Robbins {\it Compositions with parts constrained by the leading
summand } to appear in Ars Combinatoria 
\bibitem{3}
A. Knopfmacher, M.E.Mays {\it Compositions with m distinct parts} Ars Combinatoria
{\bf 53} (1999) 111-128. 
and here now come
Foot-Notes:
(*)
Diophantic equations are equations with integer solutions. A well-known example is
the problem of integer solutions for the equation of Pythagoras a2=b2+c2.
A notorious example is the last theorem of Fermat (recently solved), an=bn+cn for n>2.
see:
GREEK MATHEMATICS
Maria Fragoulopoulou
University of Athens
http://www.indianshm.com/articles/pdf/Greek%20Mathematics.pdf
Antal E. Fekete
<--- see it!
\bibitem {4}
Antal E. Fekete {\it Apropos Bell and Stirling Numbers } Crux Mathematicorum with
Mathematical Mayhem {\bf 25} No. 5 (1999), 274-281
http://www.journals.cms.math.ca/cgi-bin/vault/crux.pl?Volume=25&Issue=5&Year=1999
q =?
acha! ad Exercise 10...
q=
produced
z  z k  z k 1
.
1 z
 by
a.krzysztof kwaśniewski

the member of the intercontinental
Editorial Committee Member
Editorial Board Member of
in
Institute of Combinatorics and its Applications
Advanced Studies in Contemporary Mathematics
JOURNAL
OF ANALYSIS AND APPLICATIONS (INDIA)

a.k.k.