Partitions with parts in a finite set and the non-intersecting circles problem

0.13 MB·English

by  Madjid Mirzavaziri

PARTITIONS WITH PARTS IN A FINITE SET AND THE
NON-INTERSECTING CIRCLES PROBLEM
5
1 MADJID MIRZAVAZIRI AND DANIEL YAQUBI
0
2
Abstract. Let n be a non-negative integer and A = a1,...,ak be a multi-set with
n { }
a k not necessarily distinct members, where a1 6 ... 6 ak. We denote by ∆(n,A) the
J
number of ways to partition n as the form a1x1 +...+akxk, where xi’s are distinct
8
positive integers and xi < xi+1 whenever ai = ai+1. We give a recursive formula for

]
O ∆(n,A) and some explicit formulas for some special cases. Using this notion we solve
C the non-intersecting circles problem which asks to evaluate the number of ways to draw
. n non-intersecting circles in a plane regardlessto their sizes. The latter also enumerates
h
t the number of unlabelled rooted tree with n+1 vertices.
a
m
[

1
v 1. Introduction
7
9
Let n be a positive integer. A partition of n is a representation of n as a sum of
9
1 positive integers. The partition function, denoted by Π, enumerates the partition of the
0
. positive integer n. In the case of partitioning n into a specified number of parts, say k,
1
0 the terminology partitions of n into k parts is used. Since the order of parts in a partition
5
1 doesnotcount, theyareregisteredinadecreasing order ofmagnitude. Thus, thepartition
:
v
of n into k parts is the number of solutions in positive integers of the linear equation
i
X
r n = x +x +...+x , x x ... x 1
a 1 2 k 1 2 k
≥ ≥ ≥ ≥
which is denoted by Π(n,k). For example, 7 = 5+1+1 = 4+2+1 = 3+3+1 = 3+2+2,
so that Π(7,3) = 4. We see that Π(n,k) equals the number of solutions of
n k = y +y +...+y , y y ... y 0.
1 2 k 1 2 k
− ≥ ≥ ≥ ≥
If exactly s of the integers y are positive, there are Π(n k,s) solutions (y ,y ,...,y ).
i 1 2 k
−
Therefore we have the recursive formula Π(n,k) = k Π(n k,s). Since we have
s=0 −
the trivial initial conditions Π(n,n) = Π(0,0) = 1 anPd Π(n,k) = 0 for n < k, we can
2010 Mathematics Subject Classification. 05A18.
Key words and phrases. Set partition; the non-intersecting circles problem; rooted tree.
1
2 MADJID MIRZAVAZIRIAND DANIELYAQUBI
recursively calculate the numbers Π(n,k). Clearly Π(n,1) = 1 and Π(n,2) = n . It can
⌊2⌋
be easily shown that Π(n) is the number of solutions of
n = x +x +...+x , x x ... x 0.
1 2 n 1 2 k
≥ ≥ ≥ ≥
This can also be formulated as the number of solutions of n = y +2y +...+ny , where
1 2 n
y 0, for 1 i n.
i
≥ ≤ ≤
A multi-set A is a pair (S,m), where S is a set and m is a function from S to N. The
set S is called the set of underlying elements of A and is denoted by S(A). For each a
a S the multiplicity of a is given by m(a). A multi-set A is called an ℓ-multi-set if
∈
m(a) = ℓ for some ℓ N. A formal definition of a multi-set can be found in [1].
a A ∈
∈
NPow, let A = a ,a ,...,a be a multi-set of k positive integers. A partition of n with
1 2 k
{ }
respect to the multi-set A is a representation of n as a sum of elements of A. More on
integer partitions can be found in [2, 3, 4, 5, 6, 9, 8] and [11].
2. The Results
Definition 2.1. Let n be a non-negative integer and A = a ,...,a be a multi-set
1 k
{ }
with k not necessarily distinct members. We denote by D(n,A) the number of ways to
partition n as the form a x +...+a x , where x ’s are positive integers and x 6 x
1 1 k k i i i+1
whenever a = a . The number of ways to partition n as the form a x + ... + a x ,
i i+1 1 1 k k
where x ’s are non-negative integers and x 6 x whenever a = a , is also denoted
i i i+1 i i+1
by D (n,A). The numbers D(n,A) and D (n,A) are called the natural partition number
0 0
and the arithmetic partition number of n with respect to A, respectively.
Proposition 2.2. Let n be a non-negative integer and A be a multi-set with the multi-
plicity mapping m. Then for each a A
∈
D(n,A) = D(n am(a),A aI ), (2.1)
ℓ
− \
06ℓX6m(a)
am(a)6t
where D(0, ) = 1. In particular, D(n,I ) = Π(n,k). Furthermore,
k
∅
D (n,A) = D(n+σ(A),A).
0
Proof. Let A = a ,...,a . We have m(a) occurrence of a in the equation n = a x +
1 k 1 1
{ }
...+a x . Let x ,...,x have coefficients a in the equation, x = ... = x = 1
k k i+1 i+m(a) i+1 i+ℓ
and x > 1, where ℓ = 0,1,...,m(a). If we subtract am(a) from the both sides
i+ℓ+1
of the equation n = a x + ... + a x then we reach into the equation n am(a) =
1 1 k k
−
PARTITIONS WITH PARTS IN A FINITE SET 3
a x +...+a x +a x +...+a x . The natural partition number of the latter
1 1 i i i+ℓ+1 i+ℓ+1 k k
equation is D(n am(a),A aI ). The other parts of the above assertion is clear. (cid:3)
ℓ
− \
Corollary 2.3. Let n be a positive integer. Then D(n, 1,2 ) = n 1 , D(n, 1,2,2 ) =
−
{ } 2 { }
n−1 ( n+1 n+3 ) and D(n, 1,1,2 ) = 3 n−1 + 1 n−1 (cid:4) 1 3(cid:5) n+2 + 1+(−1)n .
4 2 − 4 { } 2⌊ 3 ⌋ 2 2 − 2 2⌊ 3 ⌋ 2
(cid:4) (cid:5) (cid:4) (cid:5) (cid:4) (cid:5) (cid:4) (cid:5)(cid:0)(cid:4) (cid:5) (cid:4) (cid:5) (cid:1)
Proof. Let n = 2k +r, where r = 1,2. Using Proposition 2.2 we can write
D(n, 1,2 ) = D(n 2, 1,2 )+D(n 2, 1 ) = D(n 2, 1,2 )+1
{ } − { } − { } − { }
= D(n 4, 1,2 )+D(n 2, 1 )+1 = D(n 4, 1,2 )+2
− { } − { } − { }
= D(n 6, 1,2 )+3 = ... = D(n 2k, 1,2 )+k
− { } − { }
n 1
= 0+k = − .
⌊ 2 ⌋
For the second assertion, let n = 4k +r, where r = 1,2,3,4. then
D(n, 1,2,2 ) = D(n 4, 1,2,2 )+D(n 4, 1,2 )+D(n 4, 1 )
{ } − { } − { } − { }
n 5
= D(n 4, 1,2,2 )+ − +1
− { } ⌊ 2 ⌋
n 7 n 3
= D(n 8, 1,2,2 )+ − + −
− { } ⌊ 2 ⌋ ⌊ 2 ⌋
= ...
k
n (4i 1)
= D(n 4k, 1,2,2 )+ − −
− { } ⌊ 2 ⌋
Xi=1
k
n (4i 1)
= D(r, 1,2,2 )+ − −
{ } ⌊ 2 ⌋
Xi=1
k
n (4i 1)
= 0+ − −
⌊ 2 ⌋
Xi=1
n+1
= k k(k +1)
⌊ 2 ⌋−
n 1 n+1 n+3
= − .
⌊ 4 ⌋ ⌊ 2 ⌋−⌊ 4 ⌋
(cid:0) (cid:1)
4 MADJID MIRZAVAZIRIAND DANIELYAQUBI
Now, let n = 3k +r where r = 1,2,3, then
D(n, 1,1,2 ) = D(n 2, 1,1,2 )+D(n 2, 1,2 )+D(n 2, 2 )
{ } − { } − { } − { }
n 3 1+( 1)n
= D(n 2, 1,1,2 )+ − + −
− { } ⌊ 2 ⌋ 2
n 5 n 3 1+( 1)n
= D(n 4, 1,1,2 )+ − + − +2( − )
− { } ⌊ 2 ⌋ ⌊ 2 ⌋ 2
= ...
3k +1
= D(n 4( ), 1,1,2 )
− ⌊ 2 ⌋ { }
3k+1
⌊ 2 ⌋ (n 2i) 1 3k +1 1+( 1)n
+ − − + ( − )
⌊ 2 ⌋ ⌊ 2 ⌋ 2
Xi=1
3k+1
⌊ 2 ⌋ (n 2i) 1 1+( 1)n
= 0+ − − +k( − )
⌊ 2 ⌋ 2
Xi=1
3k +1 n 1 3k+1 ( 3k+1 +1) 3k +1 1+( 1)n
= − ⌊ 2 ⌋ ⌊ 2 ⌋ + ( − ).
⌊ 2 ⌋⌊ 2 ⌋− 2 ⌊ 2 ⌋ 2
It is enough to note that k = n 1 . (cid:3)
−
⌊ 3 ⌋
Example 2.4. We evaluate D(17, 1,2,2,3 ) and D (17, 1,2,2,3 ). Using Proposi-
0
{ } { }
tion 2.2 and Corollary 2.3 we can write
D(17, 1,2,2,3 )
{ }
= D(14, 1,2,2,3 )+D(14, 1,2,2 )
{ } { }
= D(11, 1,2,2,3 )+D(11, 1,2,2 )+9
{ } { }
= D(8, 1,2,2,3 )+D(8, 1,2,2 )+6+9
{ } { }
= 1+2+6+9
= 18.
PARTITIONS WITH PARTS IN A FINITE SET 5
Moreover,
D (17, 1,2,2,3 )
0
{ }
= D(17+8, 1,2,2,3 )
{ }
= D(22, 1,2,2,3 )+D(22, 1,2,2 )
{ } { }
= D(19, 1,2,2,3 )+D(19, 1,2,2 )+25
{ } { }
= D(16, 1,2,2,3 )+D(16, 1,2,2 )+20+25
{ } { }
= D(13, 1,2,2,3 )+D(13, 1,2,2 )+12+20+25
{ } { }
= D(10, 1,2,2,3 )+D(10, 1,2,2 )+9+12+20+25
{ } { }
= D(7, 1,2,2,3 )+D(7, 1,2,2 )+4+9+12+20+25
{ } { }
= 0+2+4+9+12+20+25 = 72.
Proposition 2.5. Let n be a positive integer. Then
n(n+3)
D( , 1,2,...,n ) = D(2n, 1,1,...,1 ).
2 { } { }
Lemma 2.6. Letnbe a positiveintegerandA = a ,a . Ifa = a then D(n, a ,a ) =
1 2 1 2 1 2
{ } { }
n , and if a = a then D(n, a ,a ) = n 1 .
⌊2a1⌋ 1 6 2 { 1 2} ⌊a1−a2⌋
Corollary 2.7. Let n be a positive integer. Then
n 4 2 n−3 n+2 2⌊n−24⌋ 2⌊n−24⌋−3
D(n, 1,2,3 ) = − ⌊ 2 ⌋−⌊ 6 ⌋ 3 3
{ } 6 2 − (cid:4) (cid:5)(cid:4)2 (cid:5)
(cid:4) (cid:5)(cid:0) (cid:1)
2 n−4 +3 2 n−2 2⌊n−24⌋+9
+ ⌊ 2 ⌋ ⌊ 2 ⌋−⌊ 6 ⌋ .
6 2
(cid:4) (cid:5)(cid:0) (cid:1)
Proof. By Proposition 2.5, Lemma 2.6 and the last part of Proposition 2.2, we have
D (n 6, 1,2,3 ) = D (n 6, 1,1,1 ). Thus D(n, 1,2,3 ) = D(n 3, 1,1,1 ). Put
0 0
− { } − { } { } − { }
6 MADJID MIRZAVAZIRIAND DANIELYAQUBI
n = n 3 and let n = 2k +r, where r = 1,2. Hence
1 1
−
D(n , 1,1,1 ) = D(n 3, 1,1,1 )+D(n 3, 1,1 )+D(n 3. 1 )
1 1 1 1
{ } − { } − { } − { }
n 3
= D(n 3, 1,1,1 )+ 1 − +1
1
− { } ⌊ 2 ⌋
n 6 n 3
= D(n 6, 1,1,1 )+ 1 − + 1 − +1+1
1
− { } ⌊ 2 ⌋ ⌊ 2 ⌋
= ...
2k
2k ⌊ 3 ⌋ n 3i 2k
= D(n 3( , 1,1,1 )+ 1 − +
1
− ⌊ 3 ⌋ { } ⌊ 2 ⌋ ⌊ 3 ⌋
Xi=1
2k
⌊ 3 ⌋ n 3i 2k
= D(r, 1,1,1 )+ 1 − +
{ } ⌊ 2 ⌋ ⌊ 3 ⌋
Xi=1
2k
⌊ 3 ⌋ n i 2k 2k 3
= 0+ 1 − ⌊ 3 ⌋⌊ 3− ⌋
⌊ 2 ⌋− 2
Xi=1
2k 2k+3
⌊ 6 ⌋ n ⌊ 6 ⌋ n +1 2k 2k 3
= 1 i + 1 i ⌊ 3 ⌋⌊ 3− ⌋
⌊ 2 ⌋− ⌊ 2 ⌋− − 2
Xi=1 (cid:0) (cid:1) Xi=1 (cid:0) (cid:1)
k 2 n1 k+3 2k +3 2 n1+1 2k+9 2k 2k 3
−
= ⌊ 2 ⌋−⌊ 3 ⌋ + ⌊ 2 ⌋−⌊ 6 ⌋ ⌊ 3 ⌋⌊ 3 ⌋.
3 2 6 2 − 2
(cid:4) (cid:5)(cid:0) (cid:1) (cid:4) (cid:5)(cid:0) (cid:1)
Now note that k = n1 1 and n = n 3. (cid:3)
2− 1 −
(cid:4) (cid:5)
Definition 2.8. Let n be a non-negative integer and A = a ,...,a be a multi-set
1 k
{ }
with k not necessarily distinct members, where a 6 ... 6 a . We denote by ∆(n,A)
1 k
the number of ways to partition n as the form a x +...+a x , where x ’s are distinct
1 1 k k i
positive integers and x < x whenever a = a . The number of ways to partition n
i i+1 i i+1
as the form a x +...+a x , where x ’s are distinct non-negative integers and x < x
1 1 k k i i i+1
whenever a = a , is also denoted by ∆ (n,A). The numbers ∆(n,A) and ∆ (n,A) are
i i+1 0 0
called the natural distinct partition number and the arithmetic distinct partition number
of n with respect to A.
P. A. Macmahan [7] was the first mathematician interested in set partitions. Also, E.
Deutsch studied the number of partitions of n into exactly two odd size of parts (see
A117955 of [10]) and the number of partitions of n into exactly two sizes parts, one odd
and one even (see A117756 of [10]).
PARTITIONS WITH PARTS IN A FINITE SET 7
Proposition2.9. Letn bea non-negativeintegerand Abe amulti-set withthe background
set S(A) = b ,...,b . Then
1 ℓ
{ }
ℓ
∆(n,A) = ∆(n σ(A),A)+ ∆(n σ(A),A b ). (2.2)
i
− − \{ }
Xi=1
Moreover, if A = a ,...,a with a 6 ... 6 a then ∆(n,A) = 0 when n < k (k +
{ 1 k} 1 k i=1
1 i)a . Furthermore, P
i
−
∆ (n,A) = ∆(n+σ(A),A).
0
Proof. At most one of x ’s can be 1. If there is no x with x = 1 then we can write
i i i
n σ(A) = a (x 1)+...+a (x 1) and there are ∆(n σ(A),A) solutions for this
1 1 k k
− − − −
equation under the required conditions. Moreover, if x = 1 for some j, then other x ’s
j i
are greater that 1 and thus we can write n σ(A) = a (x 1)+...+a (x 1)+
1 1 j 1 j 1
− − − − −
a (x 1)+...+a (x 1). There are ∆(n σ(A),A b ) solutions for the latter
j+1 j+1 k k i
− − − \{ }
equation, where b = a . The other parts are obvious. (cid:3)
i j
Corollary 2.10. Letnbe apositiveinteger. Then∆(n, 1,1 ) = n 1 and ∆(n, 1,2 ) =
−
{ } ⌊ 2 ⌋ { }
n 1 + n 1 .
− −
⌊ 3 ⌋ ⌊ 6 ⌋
Proof. Let n = 2k +r, where r = 1,2. Using Proposition 2.9 we can write
∆(n, 1,1 ) = ∆(n 2, 1,1 )+∆(n 2, 1 )
{ } − { } − { }
= ∆(n 2, 1 )+1
− { }
= ∆(n 4, 1,1 )+∆(n 4, 1 )+1
− { } − { }
= ...
= ∆(n 2k, 1,1 )+k
− { }
= 0+k
n 1
= − .
⌊ 2 ⌋
8 MADJID MIRZAVAZIRIAND DANIELYAQUBI
Now let n = 3k +r, where r = 1,2,3. Thus
∆(n, 1,2 ) = ∆(n 3 1,2 )+∆(n 3, 1 )+∆(n 3, 2 )
{ } − { } − { } − { }
= ∆(n 3, 1,2 )+∆(n 3, 2 )+1
− { } − { }
= ∆(n 6, 1,2 )+∆(n 6, 1 )+∆(n 6, 2 )+∆(n 3, 2 )+1
− { } − { } − { } − { }
= ∆(n 6, 1,2 )+∆(n 6, 2 )+∆(n 3, 2 )+2
− { } − { } − { }
= ...
n−1
⌊ 3 ⌋
= ∆(n 3k, 1,2 )+ ∆(n 3i, 2 )+k
− { } − { }
Xi=1
n−1
⌊ 3 ⌋ n 1
= 0+ ∆(n 3i, 2 )+ − .
− { } ⌊ 3 ⌋
Xi=1
If n = 3k then k i is even and so
−
n−1
⌊ 3 ⌋ n 1 n 1 n 1
∆(n 3i, 2 )+ − = − + − .
− { } ⌊ 3 ⌋ ⌊ 6 ⌋ ⌊ 3 ⌋
Xi=1
Similarly, we have the result for the cases n = 3k +1 and n = 3k +2. (cid:3)
Example 2.11. We evaluate ∆(18, 1,2,2,3 ). Regarding to Proposition 2.9 and corol-
{ }
lary 2.10 we have
∆(18, 1,2,2,3 )
{ }
= ∆(10, 1,2,2,3 )+∆(10, 2,2,3 )+∆(10, 1,2,3 )+∆(10, 1,2,2 )
{ } { } { } { }
= 0+0+∆(4, 1,2,3 )+∆(4, 2,3 )+∆(4, 1,3 )+∆(4, 1,2 )
{ } { } { } { }
+∆(5, 1,2,2 )+∆(5, 2,2 )+∆(5, 1,2 )
{ } { } { }
= 0+0+0+0+0+1+0+0+2
= 3.
The 3 solutions are
18 = 1 3+2 2+2 4+3 1
× × × ×
= 1 5+2 2+2 3+3 1
× × × ×
= 1 4+2 1+2 3+3 2.
× × × ×
PARTITIONS WITH PARTS IN A FINITE SET 9
3. Application
To solve the non-intersecting circles problem, let us assume the following notations.
Let n be a positive integer. We denote the set of all multi-sets A = a ,...,a
1 k
{ }
such that there are distinct positive integers x ,...,x with n = a x + ... + a x ,
1 k 1 1 k k
where x < x whenever a = a , by . Recall that for an A there are
i i+1 i i+1 n,k n,k
A ∈ A
∆(n,A) solutions (x ,...,x ) satisfying the above condition. We denote the set of these
1 k
(x ,...,x ) by .
1 k A
X
Remark 3.1. We also recall that the number of (n ,...,n ) with 1 6 n 6 ... 6 n 6 s
1 r 1 r
is equal to k r 1 s = k r 1 s = r+s 1 .
i=1 i−1 i i=1 r−i i r−
P (cid:0) − (cid:1)(cid:0) (cid:1) P (cid:0) − (cid:1)(cid:0) (cid:1) (cid:0) (cid:1)
The non-intersecting circles problem asks to evaluate the number of ways to draw n
non-intersecting circles in a plane regardless to their sizes. For example, if we use the
symbol ( ) for a circle then there are 4 formats for 3 circles
( )( )( ),(( )( )),(( ))( ),((( )))
and 9 formats for 4 circles
( )( )( )( ),(( )( )( )),(( )( ))( ),((( )( ))),
(( ))( )( ),((( ))( )),((( )))( ),(((( )))),
(( ))(( )).
If we denote the answer by C then we can see that C = C = 1,C = 2,C = 4,C = 9
n 0 1 2 3 4
and C = 20.
5
Theorem 3.2. Let C be the number of ways to draw n non-intersecting circles in a plane
n
regardless to their sizes. Then
√2n k
⌊ ⌋ C +a 1
Cn = xi−1 i − .
(cid:18) a (cid:19)
Xk=1 A= a1,.X..,ak n,k(x1,..X.,xk) AYi=1 i
{ }∈A ∈X
Proof. Given n, let we draw our circles in ℓ parts with y circles in i-th part. We can
i
assume that y 6 ... 6 y . Thus n = y + ... + y . We can rewrite it as the form
1 ℓ 1 ℓ
n = a x +...+a x such that x < x whenever a = a . This shows that we have
1 1 k k i i+1 i i+1
a parts with x circles of the form (x 1) where ( ) denotes a circle containing x 1
i i i i
− −
circles. We can form the ai parts of the form (xi − 1) in Cxi−1a+iai−1 ways. The latter
(cid:0) (cid:1)
10 MADJID MIRZAVAZIRIAND DANIELYAQUBI
is true since we may put r = a and s = C in Remark 3.1. Note that a single form
i xi 1
−
(x 1) can be drawn in C ways.
i − xi−1
Now notice the fact that the maximum of k occurs when a = ... = a = 1. Since we
1 k
have 1 6 x < ... < x in this case, we can therefore deduce that k(k+1) 6 n. Thus the
1 k 2
maximum value of k is √2n . (cid:3)
⌊ ⌋
Example 3.3. For n = 6 we have
= 1 , 2 , 3 , 6
6,1
A {{ } { } { } { }}
= 1,1 , 1,2 , 1,3 , 1,4 , 2,2
6,2
A {{ } { } { } { } { }}
= 1,1,1
6,3
A {{ }}
We can therefore write
6 = 1 6 = 2 3 = 3 2 = 6 1
× × × ×
= 1 1+1 5 = 1 2+1 4
× × × ×
= 1 4+2 1
× ×
= 1 3+3 2
× ×
= 1 2+4 1
× ×
= 2 1+2 2
× ×
= 1 1+1 2+1 3.
× × ×
Thus
C C +1 C +2 C +5
C = 5 + 2 + 1 + 0
6
(cid:18) 1 (cid:19) (cid:18) 2 (cid:19) (cid:18) 3 (cid:19) (cid:18) 6 (cid:19)
C C C C
0 4 1 3
+ +
(cid:18) 1 (cid:19)(cid:18) 1 (cid:19) (cid:18) 1 (cid:19)(cid:18) 1 (cid:19)
C C +1 C C +2 C C +3 C +1 C +1
3 0 2 1 1 0 0 1
+ + + +
(cid:18) 1 (cid:19)(cid:18) 2 (cid:19) (cid:18) 1 (cid:19)(cid:18) 3 (cid:19) (cid:18) 1 (cid:19)(cid:18) 4 (cid:19) (cid:18) 2 (cid:19)(cid:18) 2 (cid:19)
C C C
0 1 2
+
(cid:18) 1 (cid:19)(cid:18) 1 (cid:19)(cid:18) 1 (cid:19)
= 20+3+1+1+9+4+4+2+1+1+2
= 48.
Corollary 3.4. Let n be a positive integer. Then C is the number of unlabelled rooted
n
tree with n+1 vertices.