ХНУРЭ,кафедра ПО ЭВМ, Тел , N.V. Bilous Факультет компьютерных наук Кафедра ПО ЭВМ, ХНУРЭ Discrete mathematics. Jegalkin Algebra.

2 Jegalkin algebra. Jegalkin algebra is an algebra, that uses the conjunction (x  y = x  y), the eXclusive OR (x  y) and the constant of unity 1 as an initial (or elementary) functions.

3 Identities of Jegalkin algebra. Properties of conjunction:  Associative law – х  (y  z) = (х  y)  z;  Commutative law – х  y=y  х;  Idempotence law – х  х=х;  Actions with the constants – x  0=0, x  1=x.

4 Identities of Jegalkin algebra. XOR operation properties ( addition by module 2):  The commutative law: x  y=y  x; xy xyxy Proof the commutative law : yxyx

5 Identities of Jegalkin algebra.  The associative law: х  (y  z)= (х  y)  z xyz yzyzx  (y  z)xyxy(x  y)  z Proof of the associative law :

6 Identities of Jegalkin algebra. Rules of the summand elimination:  x  x=0  x  0=x x xxxxx0x Proof of identity:

7 Identities of Jegalkin algebra.  The distributive law (  accordingly  ): x(y  z)=xy  xz xyz yzyzx(y  z) xyxz xy  xz Proof of distributiveness  accordingly 

8 Transition formulae from logic algebra to Jegalkin algebra.  Introducing of the negation in Jegalkin algebra: Prove this identity by the truth table:

9 Transition formulae from logic algebra to Jegalkin algebra.  I ntroducing of disjunction in Jegalkin algebra: Prove given formula analytically:

10 Jegalkin polynomial. Jegalkin polynomial is final addition by module 2 of two different elementary conjunctions above set of variables {x 1, x 2,…, x n }. The quantity of variables including into elementary conjunction is called the range of elementary conjunction. The quantity of two different elementary conjunctions in polynomial is called the length of polynomial.

11 Jegalkin polynomial and the rule of its building. For the building of Jegalkin polynomial for any function given by the formula of Jegalkin algebra it is needed to open all brackets in giving formula using the distributive law and do all possible simplifications with the help of the law for the constants, idempotence law and the rules of the summand elimination.

12 Jegalkin polynomial and the rule of its building. Example. Build Jegalkin polynomials for implication (  ) and equivalence (~). Solution.

13 Linear Boolean functions. A Boolean function is called linear if its Jegalkin polynomial does not contain conjunctions of variables. Example. Verify the function as for the linearity

14 Linear Boolean functions. Solution. Build Jegalkin polynomial of given function using the following identities:, х  у=xу  х  у, Transform the given result using the building rule of Jegalkin polynomial.

15 Linear Boolean functions. Continuation of example. Function is nonlinear.

16 Jegalkin polynomial and the rule of its building. Example. Build Jegalkin polynomial for the implication function using the method of the indefinite coefficients.

17 Jegalkin polynomial and the rule of its building. Solution. Write down polynomial for the given function in the form of sum by module 2 for all possible elementary conjunctions for x, y without the negation: f 13 (x,y) = x  y = a 1 xy  a 2 x  a 3 y  a 4,

18 Jegalkin polynomial and the rule of its building. Continuation of example. f 13 (0,0) = 0  0 = 1 1 = a 1  0  0  a 2  0  a 3  0  a 4 = a 4 f 13 (0,1) = 0  1 = 1 1 = a 1  0  1  a 2  0  a 3  1  1 = a 3  1, from here follows that a 3 = 0 f 13 (1,0) = 1  0 = 0 0 = a 1  1  0  a 2  1  a 3  0  1 = a 2  1, from here follows that a 2 = 1

19 Jegalkin polynomial and the rule of its building. Continuation of example. f 13 (1,1) = 1  1 = 1 1 = a 1  1  1  1  1  0  1  1 = a 1  1  1= a 1 Substitute given values of the coefficients in the polynomial: x  y=a 1 xy  a 2 x  a 3 y  a 4 =1  xy  1  x  0  y  1 =xy  x  1