# A Study on Characteristic Roots of Lattice Matrices

Authors: Kapiarumalayil Varkey Thomas; Geena Joy
Journal: Journal of Mathematics
 (2016)

---

## Abstract

This paper deals with the characteristic roots of different types of lattice matrices and proves that a matrix and its transpose have the same characteristic roots. Also this paper introduces the concept of similar lattice matrices and proves that similar lattice matrices have the same characteristic roots.


---

## Body

## 1. Introduction

The notion of lattice matrices appeared firstly in the work “Lattice Matrices” [1] by Giveón in 1964. A matrix is called a lattice matrix if its entries belong to a distributive lattice. All Boolean matrices and fuzzy matrices are lattice matrices. Lattice matrices in various special cases become useful tools in various domains like the theory of switching nets, automata theory, and the theory of finite graphs [1].In classical theory of matrices over a field, the characteristic equation of the matrix concerned can be defined determinantally. In the case of lattice matrices, this is not possible. In 1994, Zhang [2] proved the generalization of the Cayley-Hamilton theorem for lattice matrices. Corresponding lattice equation is taken as the characteristic equation of lattice matrix. For matrices over a field, the eigenvector-eigenvalue problems [3] are closely associated with the characteristic equation of the matrix concerned. However, this relationship is vague for lattice matrices.The roots of characteristic equations of matrices over lattices were studied firstly in the work of Tan [4]. In [4] Tan proved the existence of the least element and the greatest element for the set of all roots of the characteristic equation of a matrix over complete and completely distributive lattices with 0 and 1.In the present work, we continue to discuss roots of characteristic equations of matrices (based on [4] by Tan) over a class of complete and completely distributive lattices with 0 and 1. Such lattices are abundant; for example, every finite distributive lattice, a complete chain, and, especially, the real interval [0,1] are all such kind of lattices. In Section 3, we study characteristic roots of different types of lattice matrices. We also introduce the concept of similar lattice matrices and prove that similar lattice matrices have the same characteristic roots.


## 2. Preliminaries

We recall some basic definitions and results on lattice theory and lattice matrices which will be used in the sequel. For details see [1, 3–7].A partially ordered set(L,≤) is alattice if for all a,b∈L, the least upper bound of {a,b} and the greatest lower bound of {a,b} exist in L. For any a,b∈L, the least upper bound and the greatest lower bound are denoted by a∨b and a∧b (or ab), respectively.A nonvoid subsetK of a lattice L is asublattice of L, if for any a,b∈K, a∨b, a∧b∈K. A lattice L is called acomplete lattice if for any H⊆L, both the least upper bound ∨{y∣y∈H} and the greatest lower bound ∧{y∣y∈H} of H exist in L. A lattice (L,≤,∨,∧) is adistributive lattice if the operations ∨ and ∧ are distributive with respect to each other.An elementa∈L is calledgreatest element of L if x≤a, ∀x∈L. An element b∈L is calledleast element of L if b≤x, ∀x∈L. We use 1 and 0 to denote the greatest element and the least element of L, respectively.For anya,b∈L, the least element x∈L satisfying the inequality b∨x≥a is called therelative lower pseudocomplement of b in a and is denoted by a-b. If for any a,b∈L, a-b exists, then L is said to be adually Brouwerian lattice.A latticeL is said to becompletely distributive if for any x∈L and any {yi∣i∈I}⊆L, where I is an index set,(a)
x∧(⋁i∈Iyi)=⋁i∈I(x∧yi),(b)
x∨(⋀i∈Iyi)=⋀i∈I(x∨yi) holds. It is known [5] that a complete lattice L is dually Brouwerian if and only if (b) is satisfied in L. Therefore, a complete and completely distributive lattice L is dually Brouwerian.In this paper,L denotes a complete and completely distributive lattice with the greatest element 1 and the least element 0, respectively.Lemma 1 (see [7]).
LetL be a complete and completely distributive lattice with 1 and 0. Then for any a,b,c∈L, one has (a)
a-b≤a,(b)
a≤b⇔a-b=0,(c)
(a-b)∨(b-c)=(a∨b)-(bc); in particular, (a-b)∨(b-a)=(a∨b)-(ab).LetMn(L) be the set of all n×n matrices over L (lattice matrices). We shall denote by Aij or aij the element of L which stands in the (i,j)th entry of A∈Mn(L).ForA=(aij),B=(bij),C=(cij)∈Mn(L), defineA≤B⇔aij≤biji,j=1,2,…,n,A∨B=C⇔cij=aij∨biji,j=1,2,…,n,AB=C⇔cij=⋁k=1naikbkji,j=1,2,…,n,A∧B=C⇔cij=aijbiji,j=1,2,…,n,AT=C⇔cij=ajii,j=1,2,…,n,aA=C⇔cij=aaijfora∈L, i,j=1,2,…,n,(1)Iij=1,ifi=j0,ifi≠j,0ij=0,Eij=1,i,j=1,2,…,n,A0=I, Ak+1=AkA for k≥0, k is an integer. (Ak)ij=(aij(k))(i,j=1,2,…,n).Lemma 2 (see [1]).
For anyA,B,C∈Mn(L), one has the following.(a)
The multiplication inMn(L) has the following properties:
(i)
A(BC)=(AB)C,(ii)
AI=IA=A,(iii)
A0=0A=0.(b)
The transposition inMn(L) has the following properties:
(i)
(AB)T=BTAT,(ii)
(AT)T=A.LetA∈Mn(L). If A2=A, then A is calledidempotent; if there exists some integer k≥1 such that Ak=0, then A is callednilpotent; if aij=0, i>j, i,j=1,2,…n, then A is calledupper triangular; if aij=0, i<j, i,j=1,2,…n, then A is calledlower triangular; if A is both upper triangular and lower triangular, then A is calleddiagonal: that is, A=diag(a11,a22,…,ann); if A is diagonal and aii=k∈L, i=1,2,…n, then A is called ascalar matrix.Theorem 3 (see [1]).
LetA∈Mn(L). Then A is nilpotent if and only if An=0.LetA∈Mn(L). Then A is said to beinvertible if there is a B∈Mn(L) such that AB=BA=I. Here B is called theinverse of A and is denoted by A-1. Also, A is calledorthogonal if and only if AAT=ATA=I.Theorem 4 (see [1]).
LetA∈Mn(L). Then A is invertible if and only if A is orthogonal. Hence, A-1=AT.A setS={a1,a2,…,am} of elements of L is adecomposition of 1 in L if and only if ⋁1≤i≤mai=1. The set S is calledorthogonal if and only if aiaj=0, for all i,j=1,2,…,m, i≠j. Hence, S is called anorthogonal decomposition of 1 in L if and only if it is orthogonal and a decomposition of 1 in L.Theorem 5 (see [1]).
LetA∈Mn(L). Then A is invertible if and only if each row and each column of A is an orthogonal decomposition of 1 in L.ForA∈Mn(L), thepermanentA of A is defined as (2)A=⋁σ∈Sna1σ1a2σ2⋯anσn,where Sn denotes the symmetric group of all permutations of the indices {1,2,…,n}.Example 6.
ForI∈Mn(L), (3)I=⋁σ∈SnI1σ1I2σ2⋯Inσn=I11I22⋯Inn=1.Theorem 7 (see [2]).
LetA∈Mn(L). Then AT=A.Theorem 8 (see [8]).
LetA,P,Q∈Mn(L) such that P and Q are invertible. Then PAQ=PAQ.LetVn(L) denote the set of all column vectors (n-vectors) over L. It can be considered as an n×1 lattice matrix over L. Denote e=(1,1,…,1)T∈Vn(L).LetA∈Mn(L). Aneigenvector of A is a vector ξ in Vn(L) such that Aξ=λξ, for some scalar λ in L. The element λ is called the associatedeigenvalue.


## 3. Characteristic Roots of Matrices overL

In this section, we discuss characteristic roots of different types of matrices over a complete and completely distributive lattice with 0 and 1.Theorem 9 (see [2]).
LetA∈Mn(L). Then An∨q1(A)An-1∨p2(A)An-2∨q3(A)An-3 ∨ p4(A)An-4∨⋯∨bI=p1(A)An-1∨q2(A)An-2∨p3(A)An-3∨⋯∨cI, where (4)pkA=⋁1≤r1<r2<⋯<rk≤n⋁σ∈Sr1,r2,…,rk,σisevenar1σr1ar2σr2⋯arkσrk,qkA=⋁1≤r1<r2<⋯<rk≤n⋁σ∈Sr1,r2,…,rk,σisoddar1σr1ar2σr2⋯arkσrk.S(r1,r2,…,rk) is the symmetric group of all permutations of the set {r1,r2,…,rk}, k=1,2,…,n,b=pn(A), and c=qn(A), when n is even; b=qn(A) and c=pn(A), when n is odd.In 1994, Zhang [2] proved Theorem 9, the generalization of the Cayley-Hamilton theorem for matrices over distributive lattice with 0 and 1. In other words, Zhang [2] proved that the lattice equation(5)λn∨q1Aλn-1∨p2Aλn-2∨q3Aλn-3∨p4Aλn-4∨⋯∨b=p1Aλn-1∨q2Aλn-2∨p3Aλn-3∨⋯∨cis satisfied by the matrix A∈Mn(L).Since, in the lattice case, a characteristic equation cannot be defined determinantally it is natural to choose (5) as the characteristic equation of A∈Mn(L).Definition 10 (see [4]).
LetA∈Mn(L). Then the characteristic equation of A is defined as (6)λn∨q1Aλn-1∨p2Aλn-2∨q3Aλn-3∨p4Aλn-4∨⋯∨b=p1Aλn-1∨q2Aλn-2∨p3Aλn-3∨⋯∨c,where(7)pkA=⋁1≤r1<r2<⋯<rk≤n⋁σ∈Sr1,r2,…,rk,σisevenar1σr1ar2σr2⋯arkσrk,qkA=⋁1≤r1<r2<⋯<rk≤n⋁σ∈Sr1,r2,…,rk,σisoddar1σr1ar2σr2⋯arkσrk.S(r1,r2,…,rk) is the symmetric group of all permutations of the set {r1,r2,…,rk}, k=1,2,…,n, b=pn(A) and c=qn(A), when n is even; b=qn(A) and c=pn(A), when n is odd.The following notations are used:(8)Sn+=σ∈Sn∣σiseven,Sn-=σ∈Sn∣σisodd.Fork=n, we have (9)pnA=⋁1≤r1<r2<⋯<rn≤n⋁σ∈Sr1,r2,…,rn,σisevenar1σr1ar2σr2⋯arnσrn=⋁σ∈Sn,σisevena1σ1a2σ2⋯anσn=⋁σ∈Sn+a1σ1a2σ2⋯anσn,qnA=⋁1≤r1<r2<⋯<rn≤n⋁σ∈Sr1,r2,…,rn,σisoddar1σr1ar2σr2⋯arnσrn=⋁σ∈Sn,σisodda1σ1a2σ2⋯anσn=⋁σ∈Sn-a1σ1a2σ2⋯anσn.In [4], Tan proceeds to solve (5), by assuming that the parameter λ in (5) is an element in L. From the idempotency of λ and the absorption law, (5) takes the form (10)λ∨b=λd∨c, where d=p1(A)∨q2(A)∨p3(A)∨q4(A)∨⋯,b=pn(A) and c=qn(A), when n is even; b=qn(A) and c=pn(A), when n is odd.Lemma 11 (see [4]).
b ≤ d.Definition 12.
LetA∈Mn(L) and λ∈L. If λ satisfies (10), λ is called a characteristic root or characteristic value of A.Theorem 13.
LetA∈Mn(L) and S be the set of all characteristic roots of A. Then S is a sublattice of L.Proof.
Letλ,μ∈S. Then λ∨b=λd∨c and μ∨b=μd∨c.
Now(11)λ∨μ∨b=λ∨b∨μ∨b=λd∨c∨μd∨c=λd∨μd∨c=λ∨μd∨c,λμ∨b=λ∨bμ∨b=λd∨cμd∨c=λdμd∨c=λμd∨c.Therefore, λ∨μ, λμ∈S. Hence, the set of all characteristic roots of A is a sublattice of L.Corollary 14.
LetA∈Mn(L). Then the set of all characteristic roots of A is a subset of the set of all eigenvalues of A.In 1998, Tan [4] proved the existence of the least element and the greatest element for the set of all characteristic roots of A by the following theorem.Theorem 15 (see [4]).
LetA∈Mn(L). Then the set of all characteristic roots of A is [(b-c)∨(c-b),d∨c]={λ∈L∣(b-c)∨(c-b)≤λ≤d∨c}, where d∨c=eTAne.For simplicity, the greatest and the least characteristic roots ofA are denoted by Ru(A) and rl(A), respectively. Thus, Ru(A)=d∨c=eTAne and rl(A)=(b-c)∨(c-b)=(pn(A)-qn(A))∨(qn(A)-pn(A)).Theorem 16.
LetA∈Mn(L). Then the permanent of A is a characteristic root of A.Proof.
We have(12)A=⋁σ∈Sna1σ1a2σ2⋯anσn=⋁σ∈Sn+a1σ1a2σ2⋯anσn∨⋁σ∈Sn-a1σ1a2σ2⋯anσn=pnA∨qnA=b∨c,ifnisevenc∨b,ifnisodd=b∨c.Now (13)rlA=b-c∨c-b=b∨c-bcbyLemma1(c)≤b∨cbyLemma1(a)=A.Also by Lemma 11, b≤d⇒A=b∨c≤d∨c=Ru(A).
Therefore,rl(A)≤A≤Ru(A). Hence, by Theorem 15, permanent of A is a characteristic root of A.Theorem 17.
LetA∈Mn(L). Then A is nilpotent if and only if the only characteristic root of A is 0.Proof.
First assume thatA is nilpotent. Then by Theorem 3, An=0. Therefore, by Theorem 15, Ru(A)=eTAne=eT0e=0. Hence, the only characteristic root of A is 0. Conversely assume that the only characteristic root of A is 0. Then by Theorem 15, Ru(A)=eTAne=⋁1≤i,j≤naij(n)=0. Therefore, aij(n)=0, for i,j=1,2,…,n. That is, An=0. Hence, by Theorem 3, A is nilpotent.Proposition 18.
LetA∈Mn(L) be an idempotent matrix. Then the greatest characteristic root of A is ⋁1≤i,j≤naij.Proof.
We haveRu(A)=eTAne=eTAe=⋁1≤i,j≤naij. Hence, the proof is complete.Remark 19.
LetA∈Mn(L) be idempotent. Then 0 need not be a characteristic root of A.Example 20.
Consider the latticeL={0,a,b,c,d,1}, whose diagrammatical representation is shown in Figure 1.
It is easy to see thatL is a distributive lattice.
LetA=caac∈M2(L). Then A2=caac=A. Hence, A is idempotent. Now the characteristic roots of A are those values of λ satisfying λ∨c=(λ∧c)∨a⇒λ=b,c.Figure 1Remark 21.
LetA,B∈Mn(L). Then the characteristic roots of AB and BA need not be the same.Example 22.
In Example20, let A=abcd, B=bcd1∈M2(L). Then AB=0c11 and BA=cccd. Now the characteristic roots of AB are those values of λ satisfying λ∨0=(λ∧1)∨c⇒λ=λ∨c⇒λ=c,1 and the characteristic roots of BA are those values of λ satisfying λ∨a=(λ∧1)∨c⇒λ∨a=λ∨c⇒λ=b,c,1.Lemma 23 (see [9]).
LetA∈Mn(L). Then pn(AT)=pn(A) and qn(AT)=qn(A).Theorem 24.
LetA∈Mn(L). Then the characteristic roots of A and AT are the same.Proof.
By Lemma2(b), we have (14)RuAT=eTATne=eTAnTe=eTAnTeT=eTAne=RuA,and by Lemma 23, (15)rlAT=pnAT-qnAT∨qnAT-pnAT=pnA-qnA∨qnA-pnA=rlA.
Therefore, by Theorem15, the characteristic roots of A and AT are the same.Theorem 25.
LetA∈Mn(L) be a triangular (upper or lower) matrix. Then the characteristic roots of A are those values of λ∈L satisfying(16)⋀1≤i≤naii≤λ≤⋁1≤i≤naii.Proof.
Assume thatA is an upper triangular matrix. Then aij=0, i>j, i,j=1,2,…,n.
Therefore, we have(17)RuA=eTAne=⋁1≤i1,i2,…,in+1≤nai1i2ai2i3⋯ainin+1=⋁1≤i1≤i2≤⋯≤in+1≤nai1i2ai2i3⋯ainin+1.
For any1≤i1≤i2≤⋯≤in+1≤n, there must exist some k∈{1,2,…,n} such that ik=ik+1. Then Ru(A)≤⋁1≤i≤naii. On the other hand, since aii=aiiaii⋯aii is a term in the expression ⋁1≤i1≤i2≤⋯≤in+1≤nai1i2ai2i3⋯ainin+1, we have Ru(A)=⋁1≤i≤naii.
For anyσ∈Sn with i≤σ(i), we have σ(n)=n and σ(i)≠σ(j) for any i≠j, and so σ(i)=i for any i. In this case, a1σ(1)a2σ(2)⋯anσ(n)=a11a22⋯ann. Then (18)pnA=⋁σ∈Sn+a1σ1a2σ2⋯anσn=⋁σ∈Sn+,i≤σia1σ1a2σ2⋯anσn=a11a22⋯ann,qnA=⋁σ∈Sn-a1σ1a2σ2⋯anσn=⋁σ∈Sn-,i≤σia1σ1a2σ2⋯anσn=0.Hence, (19)rlA=pnA-qnA∨qnA-pnA=⋀1≤i≤naii.
By Theorem15, the characteristic roots of A are those values of λ∈L satisfying (20)⋀1≤i≤naii≤λ≤⋁1≤i≤naii.
IfA is a lower triangular matrix, then AT is an upper triangular matrix. Proceedings as above, we get that the characteristic roots of A are those values of λ∈L satisfying (21)⋀1≤i≤naii≤λ≤⋁1≤i≤naii.Corollary 26.
LetA∈Mn(L) be a diagonal matrix (i.e., A=diag(a11,a22,…,ann)). Then the characteristic roots of A are those values of λ∈L satisfying (22)⋀1≤i≤naii≤λ≤⋁1≤i≤naii.Corollary 27.
LetA∈Mn(L) be a scalar matrix (i.e., A=diag(a,a,…,a), where a∈L). Then “a” is the only characteristic root of A.Proposition 28.
LetA∈Mn(L). Then the set of all characteristic roots of A is L if and only if b=c and d=1.Proof.
First assume that the set of all characteristic roots ofA is L.
Then by Theorem15, we have (23)rlA=b-c∨c-b=0⇒b-c=0,c-b=0⇒b≤c,c≤bbyLemma1(b)⇒b=c,RuA=d∨c=1⟹d∨b=1⟹d=1byLemma11.Hence, b=c and d=1.
Conversely assume thatb=c and d=1. Then the characteristic equation (10) may be written as (24)λ∨b=λd∨c⟹λ∨b=λ∨b,which is true for all λ∈L.
Hence, the set of all characteristic roots ofA is L.Proposition 29.
LetA∈Mn(L). Then 0 is a characteristic root of A if and only if b=c.Proof.
First assume that 0 is a characteristic root ofA. Then (10) becomes (25)0∨b=0d∨c⟹b=c.Conversely assume that b=c. Then (10) can be written as (26)λ∨b=λd∨b=λ∨bd∨b=λ∨bdbyLemma11⟹λ∨b≤d,which is true when λ=0, by Lemma 11. Hence, 0 is a characteristic root of A.Example 30.
ConsiderE∈Mn(L),n≧2. We have (27)pkE=⋁1≤r1<r2<⋯<rk≤n⋁σ∈Sr1,r2,…,rk,σisevenEr1σr1Er2σr2⋯Erkσrk=1,k=1,2,…,n,q1E=0qkE=⋁1≤r1<r2<⋯<rk≤n⋁σ∈Sr1,r2,…,rk,σisoddEr1σr1Er2σr2⋯Erkσrk=1,k=2,3,…,n.Therefore, b=c=1 and d=1.
Hence, by Proposition28, the set of all characteristic roots of E is L.Lemma 31.
IfA∈Mn(L) is invertible, then A=1.Proof.
AssumeA is invertible. Then by Theorem 4, AAT=I. Using Theorems 7 and 8, we have 1=I=AAT=AAT=AA=A. Hence, A=1.Theorem 32.
LetA∈Mn(L) be an invertible matrix. Then the only characteristic root of A is 1.Proof.
We have(28)pnAqnA=⋁σ∈Sn+a1σ1a2σ2⋯anσn⋁σ∈Sn-a1σ1a2σ2⋯anσn.
Each termu of pn(A)qn(A) is of the form a1σ(1)⋯anσ(n)a1μ(1)⋯anμ(n), where σ∈Sn+, μ∈Sn-. Then there must exist some j∈{1,2,…,n} such that σ(j)≠μ(j). Corresponding to that j, ajσ(j)ajμ(j)=0, by Theorem 5. Hence, u=0. Therefore, we get (29)pnAqnA=⋁σ∈Sn+a1σ1a2σ2⋯anσn⋁σ∈Sn-a1σ1a2σ2⋯anσn=0.Similarly we can prove that qn(A)pn(A)=0.
Therefore,(30)bc=pnAqnA,ifnisevenqnApnA,ifnisodd=0.
Hence,(31)rlA=b-c∨c-b=b∨c-bcbyLemma1(c)=b∨c-0=b∨c=A=1byLemma31.
Therefore,rl(A)=1=Ru(A).
Hence, the only characteristic root ofA is 1.Definition 33.
LetA,B∈Mn(L). If there exists an invertible matrix P∈Mn(L) such that B=P-1AP, then B is said to besimilar to A.LetA,B∈Mn(L). If there exists an orthogonal matrix Q∈Mn(L) such that B=Q-1AQ, then B is said to beorthogonally similar to A.Note that, IfB is similar to A, then A is similar to B. From Theorem 4, B being similar to A is the same as B being orthogonally similar to A.Theorem 34.
Similar lattice matrices have the same characteristic roots.Proof.
LetA,B∈Mn(L). Assume that B is similar to A. Then there exists an invertible matrix P∈Mn(L) such that B=P-1AP=PTAP, by Theorem 4.
We have(32)RuB=eTBne=eTPTAPne=eTPTAnPe=PeTAnPe=eTAnebyTheorem5=RuA.
We claim thatpn(B)=pn(A) and qn(B)=qn(A).
To do this, we first provepn(AP)=pn(A)pn(P)∨qn(A)qn(P) for any A,P∈Mn(L) and P is invertible.
We have(33)pnAP=⋁σ∈Sn+AP1σ1AP2σ2⋯APnσn=⋁σ∈Sn+⋁1≤k1,k2,…,kn≤na1k1a2k2⋯anknPk1σ1Pk2σ2⋯Pknσn=⋁σ∈Sn+⋁τ∈Sna1τ1a2τ2⋯anτnPτ1σ1Pτ2σ2⋯Pτnσn=⋁τ∈Sna1τ1a2τ2⋯anτn⋁σ∈Sn+Pτ1σ1Pτ2σ2⋯Pτnσn=⋁τ∈Sn+a1τ1a2τ2⋯anτn⋁σ∈Sn+Pτ1σ1Pτ2σ2⋯Pτnσn∨⋁τ∈Sn-a1τ1a2τ2⋯anτn⋁σ∈Sn+Pτ1σ1Pτ2σ2⋯Pτnσn=⋁τ∈Sn+a1τ1a2τ2⋯anτnpnP∨⋁τ∈Sn-a1τ1a2τ2⋯anτnqnP=pnApnP∨qnAqnP.Similarly, we can prove pn(PA)=pn(P)pn(A)∨qn(P)qn(A) and qn(AP)=qn(PA)=pn(P)qn(A)∨qn(P)pn(A).
Therefore,(34)pnB=pnPTAP=pnPTApnP∨qnPTAqnP=pnPTpnA∨qnPTqnApnP∨pnPTqnA∨qnPTpnAqnP=pnPpnA∨qnPpnAbyLemma23andTheorem32=pnP∨qnPpnA=PpnA=pnAbyLemma31,qnB=qnPTAP=pnPTAqnP∨qnPTApnP=pnPTpnA∨qnPTqnAqnP∨pnPTqnA∨qnPTpnApnP=qnPqnA∨pnPqnAbyLemma23andTheorem32=qnP∨pnPqnA=PqnA=qnAbyLemma31.Hence, (35)rlB=pnB-qnB∨qnB-pnB=pnA-qnA∨qnA-pnA=rlA.Therefore, by Theorem 15, the characteristic roots of A and B are the same.


## 4. Conclusion

In this paper, we have studied the characteristic roots of different types of lattice matrices which have further strengthened the theory of lattice matrices. There is a great scope for further development in this area. We hope the research in this direction would be able to throw light on a few new concepts in lattice matrices, thereby paving the way for applications in areas ranging from switching circuits to theoretical computer science.



---

*Source: 3964351-2016-10-31.xml*