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Multiple-Valued Logic

Lucien Haddad ISMVL 2005

We will discuss :

- Motivation

- Functions on n variables of the k-valued logic & the projections.

- Composition of functions

-What is a Clone ?

- Why “Clone” ?

-Direct Interest for MVLers :

Completeness- Sheffer functions.

- Clones and Relations.

-Completeness Criterion, in particular for the 2 & 3–valued logic.

MotivationLet A be finite set A, F be a binary function and G be a ternary function on A. Thus F assigns to every ordered pair (a1, a2) a unique F(a1, a2) in A. Similarly

G:= (a1,a2,a3) G(a1,a2,a3). Consider the circuit

FG

a1

a2

a3

G(F(a1,a2),a2,a3)

Example.

Let A := {0,1,2};

F(a1,a2) := a1 + 2a2 (mod 3);

G(a1,a2,a3) := a1 + a1a2 + a3 (mod 3)

Then the ternary function

H(a1,a2,a3) := G(F(a1,a2),a2,a3) satisfies

H(a1,a2,a3) = a1 + 2a2 +(a1 + 2a2)a2 + a3

(mod 3)

= a1 + a1 a2 + 2a2 + 2a2 a2 + a3

(mod 3).

In the sequel A := {0,1,…, k-1}, where k

2 .

Functions of n variables.For n 1, an n-ary function f on A assigns

to every ordered n-tuple (a1,…,an) of A an

output f(a1,…,an) in A.

Examples.1- Let n = 4. The function f defined by

f(a1,a2,a3,a4) := a1 + 2a2a4 is a 4-ary

function on A.

e2(3)

e1(4)

a1

a2

a3

a4

a5

a2

a2

2- For 1 i n, define the n-ary function ei

(n) i-th projection by ei

(n) (a1,…,an) := ai for all a1,…,an in A.

Here are two gates realizing projections:

Composition of functions.

Let G be an n-ary function and F1, F2,…, Fn be n m-ary functions on A. Denote by

H := G[F1, F2,…, Fn]

the composition of G with F1, F2,…, Fn .

H is the m-ary function on A defined by H(a1,a2,…,am) := G[F1(a1,a2,…,am) , …,

Fn(a1,a2,…,am) ]

Examples. Let F be a binary function on A. 1 – The unary function F defined by F(a) := F(a,a) can be obtained by composing F with the projections. Indeed F(a) = F(a,a)

= F[e1(1)(a), e1

(1)(a)] = F[e1(1),e1

(1)](a).

2 - We can also obtain the binary function F defined by F(a1,a2) := F(a2,a1) :

F(a1,a2) = F(a2,a1) = F[e2(2)(a1,a2), e1

(2)

(a1,a2)] = F[e2(2),e1

(2)](a1,a2).

2- Define the ternary function K := F[e1

(3), e2(3)]. Then

K(a1,a2,a3) = F[e1(3) (a1,a2,a3) , e2

(3)

(a1,a2,a3)]

= F(a1,a2).

Thus we defined a unary function ( F )

and a ternary function [ K ] from the

binary function F and some projections.

Thus

Composing a function with projections

allows us to create functions of different

arities on A.

3- Let G be a ternary function on A and put

H := G[K, e2(3), e3

(3)].

Then H(a1,a2,a3) =

= G[K(a1,a2,a3),e2(3)(a1,a2,a3),e3

(3)

(a1,a2,a3)] = G(F(a1,a2),a2,a3).

Therefore the ternary function H (defined

earlier) can be obtained by composing the

functions F, G and some projections.

4 - Consider the set of all functions f on

A such that f(0,…,0) = 0.

It is clear that ei(n) (0,…,0) := 0 for all i

and all n, i.e., all projections belong to

that set. Moreover it is not hard to show

that composing functions that assign (0,

…,0) 0 leads to a function with the

same property.

Having said all that, what is a clone ??

Definition.

Let A be a finite set.

A set of functions on A

1 - closed under compositions

2- containing all projection

functions

is called a clone over A.

Why the word “Clone”??

The famous British Group Theorist Philip Hall (1904-1982) used this word for the first time while working on closed sets of functions obtained from functions on a group. He saw an analogybetween cell growth and theway one can build functions from the multiplication andthe inverse functions onthe group.

- the ternary function

H(a1,a2,a3) := G(F(a1,a2),a2,a3)

Remark. If a clone, C, contains a binary function F and a ternary function G, then as C is closed under composition and contains all projections, it contains:- the unary function F, - the binary function F and

- the ternary function H := G[K, e2(3), e3

(3)]

Examples of clones.

Denote by OA(n)

the set of all n-ary functions on A and let

OA := OA(n)

Thus OA is the set of all functions on A.

1 - OA is clone over A. It is the “ largest ” clone on A, i.e., every clone on A is contained in it.

2- The set of all projections is a clone over A. It is the “ smallest ”   clone on A, i.e., it is contained in every clone on A.

3 – Let A = {0,1}. The set

{f OA | f(0,…,0) = 0} is a clone called the clone of all 0-preserving functions over {0,1}.

Generalization. Let A = {0,1,…,k-1} and let B be a subset of A. Then the set

Φ:= {f OA| f(a1,a2,…,an) B

whenever a1, a2,…, an B}

is a clone over A. Notice that Φ = OA for B = { } or B = A.

Why Φ is a clone?

Let a1, a2,…, am B. As ei(m) (a1,…,am) :=

ai B, we have that ei(m) belongs to Φ,

i.e., Φ contains all projections.

Now let g, f1, f2,…, fm Φ.

As f1, f2,…, fm are in Φ, f1(a1,a2,…,am)  B, …, fm(a1,a2,…,am) B. Now since g is in

Φ, g[f1(a1,a2,…,am) , …, fm(a1,a2,…,am) ] B. Thus g[f1, f2,…,

fm] belong to Φ and so Φ is closed under

compositions, i.e., Φ is a clone.

4 – Let A = {0,1}. The set

{f OA| f(a1,a2,…,an) f(b1,b2,…,bn) whenever a1 b1 and a2 b2 and … an bn

}

is a clone over {0,1} (the clone of all monotonic functions).

Generalization. Let A = {0,1,…,k-1} and let R be a binary relation on A. The set

Φ := {f OA| if (a1, b1) R,…,

(an, bn) R, then (f(a1,a2,…,an), f(b1,b2,

…,bn)) R } is a clone over A.

5 – Let A = {0,1}. The set {f OA| if a1 = ~ b1 ,…, an = ~ bn

then f(a1,a2,…,an) = ~ f(b1,b2,…,bn) } is a clone over {0,1} called the clone of all self-dual functions.

Generalization. Let A = {0,1,…,k-1}. The set Φ := {f OA| if a1 b1 and … and an

bn , then f(a1,a2,…,an) f(b1,b2,…,bn) }

is a clone over A. For example the ternary function defined byg(a1,a2):= 1 + a1 (mod k) belongs to Φ.

Indeed let (a1,a2) and (b1,b2) be such that

a1 b1 and a2 b2. Then

g(a1,a2) := 1 + a1 and g(b1,b2):= 1 + b1

(mod k). Now as a1 b1 and as a1 , b1

{0,1,…,k-1}, we have that 1 + a1 1 + b1

(mod k), i.e., g(a1,a2) g(b1,b2). Thus the binary function g

belongs to the set

Φ := {f OA| if a1 b1 and … an bn ,

then f(a1,a2,…,an) f(b1,b2,…,bn) } .

Problem.

Let X be a set of functions on A.

What is the set Clo(X) of all

functions from OA which can be

obtained as finitary compositions of functions from X (and the projections)?

Stated differently,

What is the set Clo(X) of all functions on A which can be realized by circuits constructed exclusively from gates realizing functions from X (assuming that we have a potentially unlimited supply of gates of each type f X) ?

Definitions. A set of functions X is (functionally) complete in OA if Clo(X) = OA , i.e., if every function on A can be obtained as compositions of functions from X (and the projections).

A function f is Sheffer in OA if the set {f}

is complete in OA , i.e., if any function on

A can be realized by circuits constructed

from the gate realizing f.

The completeness problem was first studied for A = {0,1}. The first general completeness criterion was given by Emile Post in 1921,and was

rediscovered

many times. It is

expressed in terms

of the five maximal

clones on A =

{0,1}. This will be

discussed later on.

Examples. 1 - Let A = {0,1}. {~,} is complete in OA . Moreover the function

f(a1,a2) := 0 if (a1,a2) (0,0) and f(0,0) := 1(“NOR” function); is Sheffer in OA .

Generalization. Let A = {0,1,…,k-1}.

Define the unary function f(a):= 1 + a (mod k) and the binary function g(a1,a2):= max {a1,a2}.

The set {f,g} is complete in OA .

2 - Let A = {0,1}. Define the three binary functions: f1(a1,a2) := min{a1,a2}; f2(a1,a2) := 1 if a1 a2 and f2(a,a) := 0; ( XOR ) f3(a1,a2) := a1a2 , i.e.,

f3(a1,a2) := 1 if (a1,a2) (1,0) and f3(1,0) := 0. The set {f1,f2,f3} is complete

in OA

Generalization. Sierpinski 1945 Let A = {0,1,…,k-1}. The set of

all binary functions on A is complete in OA .

What is the use of clones in all this ?

Definition. A clone is a maximal clone if it is properly contained in no other clone but OA .

Using results from Universal Algebra, it can be shown that

A set of functions X is complete in OA if and only if X is contained in NO maximal clone over A.

In particular a function f is Sheffer in OA iff it belongs to no maximal clone over A.

The set X is complete if and only if to every maximal clone M in OA, X contains a function f not belonging to M.

So the most general completeness criterion reduces to determine all maximal clones over A.

A general completeness criterion was found by

- E. Post (1921) for A= {0,1};

- S. V. Jablonskii (1954) for A = {0,1,2};

- I. G. Rosenberg (1970) for A = {0,1,…, k-1}.

One can describe Clones using relations. Definition.

An h-ary relation on A is a subset of Ah.Examples. Let A = {0,1,2} and Ω := {(0,0),(1,1),(2,2)}.

- Ω {(0,1),(1,0)} is an equivalence

relation

- Ω {(0,1),(1,2),(0,2)} is a partial order

relation on A.- R:= {(a,b,c,d) A4 | a + b = c + d (mod 3)}is a 4-ary relation on A. Here (1,0,2,2) belongs to R while (1,0,2,0) does not.

We have seen that a function on {0,1} is called monotonic if f(a1,a2,…,an) f(b1,b2,…,bn)

whenever a1 b1, a2 b2 … and an bn .

This can be expressed as:

a1 a2 … an f(a1, a2,…,an)

If b1 b2 … bn Then f(b1, b2,…,bn)

Definition. We say that the n-ary function f preserves the h-ary relation R if for every matrix (aij) whose columns are all in R the row values of f form an h-tuple belonging to R, i.e.,

a11 a12 … a1n f(a11 a12 … a1n)

If : : : Then :

ah1 ah2 … bhn f(ah1 ah2 … ahn)

.

R R R R

Notation.

Pol(R) := {f OA | f preserves R}.

Fact. Let A be a finite set and R be any relation on A. Then Pol(R) is a clone over A.Example. Let A = {0,1} and

ρ:={(a,b,c,d) A4 | a+b+c+d = 0 (mod

2)}.Then Pol(ρ) is the clone of all linear

functions over A. Thus f Pol(ρ) iff

f(a1, a2,…,an) := β1a1 + β2a2 +… + βnan

+ c

for some β1 ,…,βn , cA and + is mod 2 (E.

Post)

Theorem. Let A = {0,1}. There are

exactly five maximal clones over A. These are

- Pol({0}), the 0-preserving functions;

- Pol({1}), the 1-preserving functions;

- Pol(), the monotonic functions;

- Pol({(0,1),(1,0)}), the self-dual functions;

- Pol(ρ), the linear functions.

The completeness criterion due to E. Post:

Theorem. A set of function X is complete in {0,1} if and only if to each of the 5 relations above, X contains a function not preserving it.

Application 1. The set {~, } is complete.

As ~ 0 = 1 and ~ 1 = 0, ~ preserves neither {0} nor {1} . 0 1 but ~ 1 ~ 0,

thus ~ Pol(). It is not hard to show that ~ is not linear either. As 0 1 = 1 0 = 1, is not self-dual. By the above result {~, } is complete.

Application 2. The NOR function is Sheffer in {0,1}.

Indeed let us denote the NOR function by f.

From f(0,0) = 1 and f(1,1) = 0, NOR preserves neither {0} nor {1}.

As f(1,0) = f(0,1) = 0, NOR is not self-dual. Since f(0,0) = 1 and f(1,0) = 0, NOR is not monotonic.

It is not hard to show that NOR is not linear.

By the above Theorem NOR is Sheffer in {0,1}. Here is a better proof of this result:

Application 3. An n-ary function f is Sheffer in {0,1} iff

f(0,…,0) = 1, f(1,…,1) = 0 and f(a1, a2,…,an) = f(~a1,

~a2,…, ~an)

for some a1, a2,…,an {0,1}. The first two conditions give that f is not monotonic and is neither 0 nor 1 preserving. The third condition gives that f is neither self-dual nor linear.

This result allows to show that almost 25% of functions are Sheffer on {0,1}.

We turn to the case A = {0,1,…, k-1}.

As mentioned earlier Jablonskii established a completeness criterion for {0,1,2} and I.G. Rosenberg for the very

general case. It has been

reported that the famous

Algebraist A. I. Mal’cev

had a completeness

criterion for A = {0,1,2,3}.

In the Rosenberg classification maximal clones belong to 6 different families; all determined by relations. We will not give their full description here due to lack of space. We rather apply the Rosenberg classification to the case A = {0,1,2}.

Put Ω := {(0,0),(1,1),(2,2)}.

Theorem: There are exactly 18 maximal

clones over {0,1,2}. These are polymorphs

of the following relations:

Central relations:

B1 := {0}; B2 := {1};

B3 := {2}; B4 := {0,1};

B5 := {0,2}; B6 := {1,2}.

Equivalence relations:

E1 := Ω {(0,1),(1,0)};

E2 := Ω {(0,2),(2,0)};

E3 := Ω {(1,2),(2,1)};

E4 := Ω {(0,1),

(1,0),(0,2),(2,0)}; E5 := Ω {(0,1),(1,0),(1,2),(2,1)}; E6 :=

Ω {(0,2),(2,0),(1,2),(2,1)};

Permutation with cycles of prime length: σ := {(0,1),(1,2),(2,1)}.

Bounded order relations:

O1 := Ω {(0,1),(1,2),(0,2)};

O2 := Ω {(0,2),(2,1), (0,1)};

O3 := Ω {(1,0),(0,2),(1,2)};

Prime affine relation:

R:= {(a,b,c,d) A4 | a + b = c + d (mod 3)}.

h-Regular relation:

θ := {(a,b,c) A3 | |{a, b, c}| 2}.

We describe two of these clones.

1 - Let B6 = {1,2}. Then f Pol(B6) iff

f(a1 ,…, an) {1, 2} whenever a1,…,an = 1

or 2.

Thus the constant function c1(a1,…,an ) := 1

and the function h(a1,a2):= (a1 + a2 (mod

2)) +1 both belong to Pol(B6).

However the function g(a1,a2) := a1 + a2

(mod 3) does not belong to Pol(B6) as

g(1,2) = 0 {1,2}.

One can show that the set Pol(B6){g} is

complete in OA.

2 - Let θ := {(a,b,c) A3 | |{a, b, c}| 2}.

Thus (0,2,0) θ while (2,0,1) θ.

Pol(θ) is called the Slupeckiclone on A. It contains all unary and all constant functions on A. A function h belongs to Pol(θ) iff h takes at most two values or it can be

defined from a unary function, i.e., h(a1 ,

…, an) := φ(ai) for some unary function φ,

like for example h(a1 ,a2, a3) := a2 + 1

(mod 3).

Theorem: A set of function X is complete in {0,1,2} if and only if to each of the 18 relations above, X contains a function not preserving it.

Remarks:

1- It is not hard to construct, for each of the the 18 relations above, a binary function that do not preserve it. This shows that the set of all binary functions is complete (Sierpinski’s Theorem for the case A = {0,1,2}).

2- We mentioned earlier that the set {f,g} is complete in OA where

f(a):= 1 + a (mod k) and g(a1,a2):= max

{a1,a2}. This can be verified for A

= {0,1,2} using the 18 relations above. For example

one can check that f preserves none of the Central relations Bi, none of the

Equivalence relations Ei and none of the

Order relations Oi.

On the other hand g does not preserve neither σ (permutation), nor R (the p-group) nor θ.

To see that g does not preserve θ take

0 0 max(0,0) = 0

1 0 max(1,0) = 1

1 2 max(1,2) = 2

The two columns on the left hand side belong to θ while the column on the right hand side does not. Thus g Pol(θ).

In the same direction:

Clones of …

- Partial Functions;

- Uniformly Delayed Functions;

- Hyperfunctions and

Partial Hyperfunctions

Thanks for your attention

Questions ? Comments ?

Working with clones…