The Number of Canalyzing Functions over Any Finite Set

(1)

The Number of Canalyzing Functions over Any Finite Set

*

Yuan Li1#, David Murrugarra2, John O. Adeyeye1, Reinhard Laubenbacher3

1_{Department of Mathematics, Winston-Salem State University, Winston-Salem, USA} 2_{School of Mathematics, Georgia Tech, Atlanta, USA}

3_{Virginia Bioinformatics Institute, Virginia Tech, Blacksburg, USA} Email: #_{[email protected]}

Received March 1, 2013; revised May 20, 2013; accepted June 16, 2013

permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

ABSTRACT

In this paper, we extend the definition of Boolean canalyzing functions to the canalyzing functions of multi-state case. Namely, _{f Q}: n__Q_{, where}



_{. We obtain its cardinality and the cardinalities of its various subsets} (They may not be disjoint). When , we obtain a combinatorial identity by equating our result to the formula in [1]. For a better understanding to the magnitude, we obtain the asymptotes for all the cardinalities as either or

.



1, , ,2 q

Q a a  a

2 q

n 

q 

Keywords: Canalyzing Function; Inclusion and Exclusion Principle

1. Introduction

The idea of canalization was initiated from Waddington, C. H. [2]. When comparing the class of canalyzing functions to other classes of functions with respect to their evolutionary plausibility as emergent control rules in genetic regulatory systems, it is informative to know the number of canalyzing functions with a given number of input variables [1]. However, the Boolean network modeling paradigm is rather restrictive, with its limit to two possible functional levels, ON and OFF, for genes, proteins, etc. Many discrete models of biological networks therefore allow variables to take on multiple states. Common used discrete multi-state model types are so- called logical models [3], Petri nets [4], and agent- based models [5].

In this paper, we generalize the concept of Boolean canalyzing rules to the multi-state case. By generalizing the results in [1], we provide formulas for the cardinalities of various subsets of canalyzing functions. We also obtain the asymptotes of these cardinalities as either or q . We obtain a combinatorial identity by equating our result to the formula in [1].

n   

2. Preliminaries

In this section we introduce the definition of a canalyzing

function.

Let

 

n 



1, 2, , n Q



, 



a a1, , ,2 aq



and : n

f Q Q.

A function is canalyzing if there is a variable xi and an element aQ so that the value of the function is fixed once variable xi is fixed at . More precisely, we have the following definitions.

a

Definition 2.1

1) The function f x x



₁, , ,₂  xn



is i a b: : canalyzing if f x



1, ,xi 1, , i 1, , n



1 1, 1,

a x x

  

 

, n

b, for all , , i i

x  x_ x_  x 2) The function

.



1, , ,2 n



f x x  x is i a: : canalyzing if there exists bQ such that



1, , i 1, , i 1, , n



f x  x_ a x_ x b

1, , 1, 1, 

,

, for all

i i n

.



1, , ,2 n



f x x  x is : :a b canalyzing if there exists i

 

n such that



1, , i 1, , i 1, , n



f x  x_ a x_ x b

1, , i1, i1, , n

 , for all

x  x_ x_  x .



1, ,2 ,xn



is i: : b canalyzing if there exists aQ such that



1, , i 1, , i 1, , n



f x  x_ a x_ x b

1, , 1, 1, 

,

, for all

i i n

.



1, , ,2 n



f x x  x is : :a  canalyzing if there exist i

 

n b, Q such that



1, , i 1, , i 1, , n



f x  x_ a x_ x b

1, , i1, i1, , 

n

, for all x  x_ x_  x .

*_{Supported by an award from the USA DoD # W911NF-11-10166.}

#_{Corresponding author.} 6) The function



, , ,



1 2 n

(2)

canalyzing if there exist a, bQ such that



1, , i1, , i 1, , n



f x  x_ a x_

1, , i1, i1, , n

x 

 b, for all

x  x_ x_  x .



1, , ,2  xn



is  : :b canalyzing if there exist i

 

n ,a



1, , i1, , i 1, , n



Q

such that f x  x_ a x_

1, , i1, i1, , n

x 

 b, for all

x  x_ x_  x .

8) The function is f x x



1, , ,2  xn



is   : : canalyzing if there exist i

 

n , ,a b



1, , i1, , i 1, , n



Q such that

f x  x_ a x_

1, , i1, i1, , n

x 

 b, for all

x  x_ x_  x .

By abuse of notation, we also use i: : b to stand for the set of all the i: : b canalyzing functions,

: :

i a b will stand for the set of all the i a b: : canalyzing functions and etc. We use  to stand for the empty set.

By the definitions, we immediately have the following propositions.

Proposition 2.2 If b1b2, then

1 2

: : : :

i a b i a b  b b

 .

Proposition 2.3 If 1 2 and i1i2, then 1: : 1 2: :b2

i  b  i  .

By the definitions, we have

   

 

: : : : : : : : ,

: : : : : : ,

: : : : ,

: : : : .

b Q a Q i n

i n a Q

b Q i n

a Q b Q

a Q

b Q

i n

b a i

b i b a b

a a b i a

i i a i b

i b i a b

i a i a b

a b i a b

  

 



           

     

 



For any set S , we use S to stand for its cardinality.

We use

 



!



,

! !

n C n k

k n k



 to stand for the binomial

coefficients. As usual, should be explained as

zero once .



, C n k



kn

Obviously, for the above notations, the cardinality are same for different values of i a, and . In other words, b we have i a b1: :1 1  i2:a2:b2 ,

: : : :

i  b  j  c , : :a b  : :c d and etc.

3. Enumeration

Theorem 3.1 Given i

 

n a b, , Q, the number of

: :

i a b canalyzing functions is qqnqn1. In other words, we have _{i a b}: : __qqnqn1

.

Proof: A function in the set i a b: : is uniquely determined by its value on inputs



x1, , xn



with

i

x a . There are



q1



q  q

1

n n n

q q

 _ 1_{such inputs,}

and the function can take different values. Thus

1

: : qn qn i a b q   . □

Because : : 



i a b: : b Q i a



, by Proposition 2.2,

we get

Theorem 3.2 The number of all the i a: : canaly- zing function is



1



1₁

.

n n n n

q q q q

q q   q  

Lemma 3.3 We have k₁ : : qn kqn1 j

j i a b q



 



for

any



a a1, , ,2  ak



Q.

Proof: A function in the set



_j_₁ is uni

quely determined by it values on inputs : : k

i a_j b



x1, , xn



with



1, ,



i k

x a  a . There are such

inputs. □





qk qn1_qn_kqn1

Theorem 3.4 Given i

 

n and bQ, the number

of i: : b canalyzing functions is qq n

other



1



1

n _qn q

q 

  . I

words, we have : : n



n1 1



q

q q

i  b q  q   .

and Exclusion Principle, we ve

Proof: By Inclusion ha

 

 

 

 





1 2

1 1

1

1 2

,

1 1

, , , 1 2

1 1

1

: : : :

: : : : : :

1 : :

,1 , 2

1 , 1

1 ,

,

1

k

n n n n

n n

n

a Q

a a Q

a Q

k

k q

j

a a a Q j

q q q q

k _q _kq q

q

k q kq

k

q k

q q

k

q

i b i a b

i a b i a b i a b

i a b

C q q C q q

C q k q

q C q k q

q

 



 

 

1

 

 

 

  

 







 

 

     

 

     

 

 

 _  _

 









 







1





1



1 _qn q _qn _qn 1q

q  q q 

 _{ } _ _ _

 

  . 

Similar to Lemma 3.3, we have

Lemma 3.5 If



i i_{1 2}, , , ik





 

n , then



k

q1 q

1

: : k n k

j j

i a b q  





.

Based on this lemma, we can get the following result.

Theorem 3.6 We have

 

1

 

 1

1

: : 1k , q kqn k.

k n

a b  C n k q  

 

 





(3)

 

 

_{ }

 

_{ }

 

  1

2

1 2

1 1

1

1 1 1

1 1

1 1 1

1 1

1

: : : :

: : : : : :

1

: : 1 : :

,1 , 2

1 , 1

1 , .

k

n n

k n k

i n

i n i j n

k

n j

i i n j j

q q q q

k q q q

k q q

k n

a b i a b

i a b i a b j a b

i a b j a b

C n q C n q

C n k q q

C n k q

 



 

    



     

 

  

 

 

 

 

  

   

 

     

 









 







1n

n

n 

From the above theorem, we can get the following re

m 3.7 We have

sult.

Theore

 

1

 

 1

1

: : 1k , q kqn k

k n

a q  C n k q  

 

  





Proof: Because : : : :

b Q

a a



  

_

 b , by Theorem

3.6, we just need to show : :a b₁  : :a b₂  if 1 2

b b . Suppose f :a: 1 : : 2 ere

1

i and



b   a b , then th

exist i₂ n



such that

1: : 1 2: : 2

f i a b  i a b since

 

: : : :

i n

a b i a b



 

_

.

If i1i2, we get a contradiction by Proposition 2.2. If

1i2, w

i e get a contradiction by Proposition 2.3 since 1: :b1  i1: : b1 and

i a i2: :a b2  i2: : b2 . □

to find umb

Now, we are going the formula for the n er

of all the canalyzing functions with given canalyzed value b. In other words, the formula of  : :b .

Let Sb



i: :a b i

 

n a, Q



for any bQ. By Inclusion and Exclusion Principle, we have

 

 

1

: : : : nq 1 k .

k k

i n a Q

b i a b  N

  

  

_{ }







where

,

.

b

k

s S s k T s

N T

  





_

In order to evaluate Nk, we write all the members in

b as the following n q matrix. S

1 2

1: : 1: : 1: :

2 : : 2 : : 2 : :

: : : : : :

q

a b a b a b

A

n a b n a b n a b

  

 

 



 

 



   



For any , we will choose ele-

m

k

ents from the above matrix to form s.

Suppose k1 of its elements are fro the first row m

(th

b sS

ere are



q k, 1



ways to do so). Let these k1

elements be C

1

12 1

, 1: : , , 1: k :

b a b  a b .

Suppose 11 ond row

1:a :

2

k of its elements are from the sec (there are



q k, 2



elements be C

2

21

2 :a :b , 2 :a₂₂: , , 2 :b  a₂_k :b .



Su ose of its elements are from the last row (th

pp kn

ere are



q k, n



ways to do so). Let these kn elements be

C

1 b , n a: n2: , ,b  n a: nkn:b .

1 2

k k

: n a_n :

,0 , 1, 2, ,

n i

k k k q i n

        .

Similar to Lemma 3.3, we have

Lemma 3.8 Let s be the subset of Sb as mentioned

above, then _T _qq k1q k2 q kn_.

T s



Hence,

.

We get

3.9 For any









 1   1

1

, , n

n

q k q k

k n

k k k

C q k C q k q  

  





N 

Theorem bQ, we have

 

1 

1 2

1

1 1

: :b  

1 , .

n j j n

q k

nq n

k

j

k k k k k j

C q k q  

     



 _ _ 

 

  _ _

 _ _ 

 







In order to evaluate : :i   , we need two more

lemmas. Their proofs are similar to that of Lemma 3.3 and we omit them.

Lemma 3.10 If



a a1, , ,2  ak



Q and



b1, , ,b2 bk



Q, then

  1

1

: : n

k

q k q

j j

j

i a b q  





.

Lemma 3.11 If rk

ar 1 2

11, , 1k; 21, , 2k ; ; r1, , r

a  a a  a  a  a

e k1  kr distinct elements of Q,



b1,,br



Q. Then,

 

1 2

1 1 2

1 1 2 2

1 1 1

: : : : r : :

n r

k k k

j j

j j j

q k k k q

i a b i a b i a b

q 

  

   

 

rj r

  

     

     

 

 



Now, we are ready to find the cardinality of



: : i   .

 





1

1 ,0 1 2

: :

i  

1 1

! .

! ! ! !

n

q i

k q k q

q

k k k k k q q

q q

q k k k k

  

     

 



 _ _

 _ _



 

Proof: First, we have : : : :

a Q b Q

i i

 

  

_{ }

a b .

(4)

Let Si



i a b a b: : , Q



, we get

 

2

1 1

: :

i   1 N ere

q k

k k







 . Wh

,

.

i

k

s S s k T s

N T

  





_

In order to evaluate , we write all the elements in as the following k matrix.

N q q i

S

1 1 2 1

: : :

i a b a b

 

  1

1 2 2 2 2

1 2

: : :

: : : : : :

q

q q q q

i i a b

i a b i a b i a b

B

i a b i a b i a b

 



 

 



   



For any



i

sS with s k , we will choose

elements fro e matrix to form

k

m the abov s.

Suppose k1 it elem re from the first ro

(There are C q k



, 1



ways to do so). Let these

of ents a w

elements be 1

k

11 1 12 1 1₁ 1

: : , : : , , : k :

i a b i a b  i a b .

Suppose k elements are from the second row,

we must cho2 lu

of its os e i

e these elements from different co mns, otherwise th ntersection will be  by Proposition 2.2 (There are C q



k k1, 2



elements be i a: 21:b2 , :i a22:b2 , :i a2k₂:b2 

Suppose

,

of its elements are from the last row

(The ways to do so).

Le

q

k C q

e

re are



   k1 k2  kq1,kq



t these kq lements be

1 2

: q : q , : q : q , ,

i a b i a b  : : .

q

qk q

i a b

where . We

have 1 2

,0 , 1, 2, ,

q i

k   k  k k  k q i  q



 





1

1 2

,

1 1 2

,0

1 1

, ,

,

i

q i

q



s S s k T s

k k k k q

q q k k k

T

C q k C q k k

C q k k k I

  

    

 

 

  





 



where k N

1 2 q

k k k

I _

1 2

1 1 2 2

1 1 1

: : : : q : :

k

k k

j j qj q

j j j

i a b i a b i a b

  



 

   

 

    _ _





 









By Lemma 3.11, we know

This number is zero if

A straightforward computing shows that



 1 2

1 2 q

q k k k k k

I _ q       

1 1

.

n n

q

k q q k q

q

 _ 



kq.



 







1 1 2 1

!

C k

q

  ₁

1 2

,

.

! ! ! !

q q

q

k k

k k k q k

   



 

Hence, we get

, ,

C q k C qk k q

 

_

_

 





2

q q

1 1

1

1 1

1

1 ,0 1 2

1 1

1 ,0 1 2

: :

1 1

! 1

! ! ! !

1 1

! .

! ! ! !

n

q i

k k

q

k q k q

k k k k k q q

n

k q k q

q

k k k k k q q

i

N N

q

k k k q k

q q

q k k k k



 

 

 

     

  

     

 

   

 

  _ _



 

 







□ Now we begin to evaluate   : : .

 

1

 

1

1 1

: :

  

1 1 .

nq

k k

U V

 

 





 





where

q





 





 

1 1 2

1 2

1 1 2

,0

,0 1 2

!

! ! ! !

! 1

! !

n

q q i

n

q i

k

q k q

t t t k t q

q k q

t t t k t q q

U

q

n q

t t t q k

nq q

q k t t ! t !



     



      



 





 



and

Proof: Let

 

1 

1 ,0 1,0 1

,

n j j n i i

q k n

k j

k k k k k k q j

V q C q k q



        



 

 _ _

 







 



: : , ,



.

S  i a b a bQ i n

We have

 

: : : :

i n a Q b Q i a b

      

_{  }

 

2

1 1

: : nq 1k

k k

N 



   



 , where

and

,

.



k

s S s k T s

N T

  





We write all the elements of as the following matrices.

2

nq S

(5)

1 1 1 2 1

2 1 2 2 2

1

1 2

1 1 1 2 1

2 1 2 2 2

2

1 2

1 1 1 2

1: :a b 1: :a b 1: :a bq

 

 



1: : 1: : 1: :

2 : : 2 : : 2 : :

: : : : :

q

q q q q

q

q q q q

n

a b a b a b

M

a b a b a b

a b a b a b

a b a b a b

M

a b a b a b

n a b n a b n a

M

 

 



 

 

 

 

 



 

 





   

  

   

 



 

1

2 1 2 2 2

1 2

:

: : : : : :

q

q q q q

b

n a b n a b n a b

 

 

 



   



We combine all the above Mi to form a nq q matrix M whose first q rows are M₁, the sec ro

ond q ws are M2, , the last q rows are Mn. In words, we have

other

1 1 1:a b1 2  1 1:bq

2 1 2 2 2

1 2

1 1 1 2 1

2 1 2 2 2

1 2

1 1 1 2 1

2 1 2 2 2

1: : : :

1: : 1: : 1: :

2 : : 2 : : 2 : :

: : : : : :

: :

q

q q q q

q

q q q q

q

a b a

a b a b a b

a b a b a b

a b a b a b

M

a b a b a b

n a b n a b n a b

n a b n a b n a b

n a b





   

  

   



   

 

   

1 n a: q:b2 n a: q:bq

 



 



 

  

We are going to choose elements from

 



 



k M to

form the intersection. In or o get a possi non

em

der t ble

s m pty intersection, we know all these k element ust

come from either the same Mi (for some fixed i) or all of them from the same column of M by Proposition 2.3.

Each Mi is in fact the transpose f B and each column of

o

M is all the elements of A (As sets, they are equal). Hence, a typical intersection is e her the one eorem .9 or the one in Theorem 3.1 t these

two cases ar ot disjoint.

Suppose we choose ki elements from

1 2

, 1, 2, , , ,

i n

it

in Th 3 2. Bu

e n

0 i , 1,2, ,

M i n k k k k

k k i n

     

    .

If there exist such that i kik, then kj  0, j i.

e one in Lemma

This implies the intersection h

3.11 and

looks like t kq.

If 0    ki k 1, i

 

n , th e intersection looks

like the one in Lemma 3.8 and knq. The above two cases are disjoint now. By Le

en th

mma 3.11 and Lemma 3.8, we get

1

,

k

,0 , 1, 1, ,

n i

i i

s S s k

N T

 





_





k k k k k

T s

k k

i k k k k i n

U V

     

    









 



where (Note: there are n matrices M M₁, ₂, , Mn and columns of

q M )





 





 

1 1 2 ,0

!

! ! !

1 1 2

1 1

1 2

,0 1 2

,0 1,0 1

!

! 1

! ! ! !

,

n

q i n

j j n i i

q k q

t q q

q k q

t t t k t q q

q k n

k j

k k k k k k q j

q k

nq q

q k t t t

V q C q k q







 



     



        





 

 

  

 







Hence, k

t t k t

q

U n

t t t q

   





 

 

  



 

2

1 1

1 1 1

: : 1

1 1

nq k

k k

nq q nq

k k

k k k k

k k k

N

U V U V





 

  

    

      



1

1 k , □

In the following, we will reduce the formula : :

  

when q2 and compare it with the one in [1]. We have

 

1

 

1

: : 1k 1 k .

k k

U V

 

  



2  



2 

1 1

n

k k



where





 

1 1 2

1 1

2 2 ,0 2 1 2

2 ,0 1,0 2 1

2! ₂

! ! 2 !

2 2,

n

i

n

2j j

n i i

k k

t t k t

k n

k j

k k k k k k j

U n

t t k

V C



 

   



        







 

 _ _

 







k

A simple calculation shows that

2

 

1 1

2 2

1 4 2 ,1 2 2

n n

U  n  C n 

(6)

2 4 .

U  n

1

V 0 since the condition of the sum is not satisfied.

2

 

1

2

3 2

2 2,

, 2 2 2

n

1

2

2,0 1 1

j j

n

k n

V C k

C n







 

 _ _







n i

j k   k  k j 

Note, is the number of solutions of the

equation When

 

, 2

C n

1 k   k

3 k 2nn, 2,0 ki 1.

 

  



 

1

1 2

k

k 

 



2

2 2

n j j

k n

V





 



,0 1

1 2 2 1

1 2

2,

, 2 2 , , 2 2

n i n k

j

k k k j

k k

k t

C k

C n k C n t C n t k t





   

 

    _{ }



   









Note, 2 is the number of solu

tions of , with

exactly

he en ,

t

  

, ,



C n t C n t k  t the equation ₁ 

t components equal to

-2

n

k k 

2.

0 ki 2

nce, wh q2

 

  



1 1 2

1

1 2 1

3 2 ₁

2

: :

k 

4n 1 , 2 2

1 , , 2 2

n k

k

k n

k k t

k

k n

t

C n k

C n t C n t k t



 

  

   

  _{ } 









 

When , one can obtain (without calculator)

the sequ 120, 3514. These results are con-

sistent with t By [1], the cardinality of

  

  



1, 2,3, 4

n

ence 4, 14, hose in [1]. : :

   should be

 

1

: :

1n k n k

n   

  

  





 

1 2

1

2 1 , 2 2k .

k n

C n k  







So, we obtain the following combinatorial identity(for any positive integer n).

 

  



 





1 2 1

3 2 ₁

1 , , 2 2

k

k n

t

C n t C n t k t

2

2 1

k k t

n

  



  

The left sum should be explained as 0 if

  _{ }  

  



1

n . As

usual, is 0 if

Fro

 

,

C n k

m Theorem

kn.

3.1, we know : : _qn _qn1

q

i a b    ,

since

 

: : : :

i n a

i a b  

   

_{  }



, we obtain b

 ₁ 1

2

: : q qn

nq q  

    .

In order to get an intuitive idea about the magnitude of all the cardinality numbers, We will find their asymptote

as or

We have the followin n

n  q .

g notatio

Definition 3.14 We call

   

x

f x g x if

 

lim 1

x

f x

_{g x}  .

Now, we can list all the cardinalities asymptotically.

t

Theorem 3.15 If n4 and q2, hen

   

 

1 1

1

1 1

1

2 2

: : ; : : ;

: : , : : ;

: : , : :

n n

n

q q q q

q

q n

q q q q

q n

q q q q

q n

q q q

i a b q i a qq

a b nq a b nq

a nqq a nqq

b nqq b nqq

i q q i q q

 



 



 

   

     

       

1 1

: : n q qn , : : q q qn ;

i  b qq   i  b qq  

n

q q q q



   

1

1 1

1

1 1

2 2

;

: : , : : .

n

n n

q

q n

q q q q

nq q nq q



 



 

       

Proof:

The first two rows are Theorem 3.1 and Theorem 3.2. We will give a proof of the last row, the others are similar and easier.

 





 

1 2 1

1 1

1,0 1 2

1 2

! ! !

.

q i n

q nq

k k

t t t t q q

q q

t t t

nq q 

 

     

 

 

When

1

1 1

! 1

1 !

n

q q

nq

U q

q



 





1 1

: 1 k 1 k.

k k

U V

 

  



 





:



2 k q, we have





 

   

_

_

1 1 2

1 1

,0 1 2

2 2

0 , 1,2, ,

! 1

i

t q i q

nq

  







! 1

! ! ! !

! 1 .

n

q i

n n

q k q k

t t t k t q q

q

q q q q

nq

U q

q k t t t

q nq q q



 



     

 

 

 



 

Hence,

 

_

_

   





 

1

1 1

1

2 1

2

1 2 1

2

1 3

1

! 1 1

.

q

n n

q k

q q q q

k q

k

q q q

q

q q

U

qnq q q q q

U _{nq q} _q

q q q

q q



 



 





 

 

 



So,

 

1 2

1

lim 0

q k

k k

n

U



 







.

1 0

V  When

(7)

 



  

1 1

2 2

2

,0 1,0 1 0 , 1,2,

,

1 ! .

n

 

_!  1 2

, ( 1)

j j n i i

n

i

q k n

k k k k k k j

n q q

k q i n

V q C q k q

q q q q





        



  





 

 _ _

 



 







n

n q q

k

q

j

Note, . Hence,

q



q q









2 2

1

1 n

n j

j

q k q q 



  



 

₁



  

 ₁2 2

1

1 1 ! n

nq

k n n q q

k k

V nqq q q q 

 



  



.

We obtain

 

_

_{  }

   



  

 





 

  2 2 1

2 2 2

1

1 1

1 2 1

2

1 1

1

1 !

1 ! 1

.

n

n n n

k

n n q q

k k

q q

n n n qn q n

q q q q q

V

nqq q q q

U _{nq q}

q q q q

q

q q



  



 





 



 



 



Hence, nq

q



 

1 1

1

0. lim

nq k

k k

n

V

U





 





In summary, we obtain

 ₁ 1

2 : :

1.

lim _q _qn

n_{nq q}  

   

In other words,

 ₁ 1

2 : :

   n q qn . nq q  



From the above proof, it is also clear that we have

 ₁ 1

2 : :

1.

lim _q _qn

q_{nq q}  

   

In other words,

 ₁ 1

2

: : q q qn .

nq q  

   

□ When , the first equation of the last row in the

ab been obtained in [1].

4. Conclusion

multi-state case. nd Exclusion Principle, we get for- inality all such functions and the car-

cs Institute at Virginia ank Alan Veliz-Cuba and any useful discussions. The

na, “The Number

Physica D, Vol.

197, No. 3-4, 20 doi:10.1016/j.

In this paper, we generalized the definition of Boolean canalyzing functions to the functions of

Using Inclusion a mulas for the card

dinalities of its various subsets. When q2, we derive an interesting combinatorial identity by equating our formula to the one in [1]. Finally, for a better understanding to the magnitudes, we provide all the asymptotes of these cardinalities as either n  or q 

5. Acknowledgements

This work was initiated when the first and the third authors visited Virginia Bioinformati

.

Tech in June 2010. We th Franzeska Hinkelmann for m

first and the third authors thank Professor Reinhard Laubenbacher for his hospitality and for introducing them to Discrete Dynamical Systems. The third author was supported in part by a Research Initiation Program (RIP) award at Winston-Salem State University.

We greatly appreciate an anonymous reader. Because of his insightful comments, in this paper, the proofs for many lemmas are simplified, the results are more general (On any finite set instead of finite field).

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