REPRESENTATION OF ALGEBRAIC STRUCTURES BY BOOLEAN FUNCTIONS. Logic and Applications 2015 (LAP 2015) September 21-25, 2015, Dubrovnik, Croatia

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1 REPRESENTATION OF ALGEBRAIC STRUCTURES BY BOOLEAN FUNCTIONS SMILE MARKOVSKI Faculty of Computer Science and Engineering, S Ciryl and Methodiu Univerity in Skopje, MACEDONIA mile.markovki@gmail.com Logic and Application 2015 (LAP 2015) September 21-25, 2015, Dubrovnik, Croatia

2 Preface Why Boolean function? Uage of two imple logical function conjunction and excluive dijunction over a two-element et B = {0, 1} Fat computation by uing uitable computer tool (Mathematica, Maple, Matlab, Magma,...) Many application in cryptography, cryptanalyi, coding theory, and o on

3 1 ANF of Boolean function 2 Repreentation of algebra by Boolean function 3 Tranformation defined by groupoid

4 ANF of Boolean function The logical function negation, conjunction, dijunction, excluive dijunction,... are defined on the logical value True and Fale. We conider them on the et B = {0, 1}. An n-ary logical function or a Boolean function i a mapping f : B n B. A et {f 1, f 2,..., f k } of logical function i aid to be complete if any other logical function can be repreented by their uperpoition: {, }, {, }, {,, }, {, +}. The complete ytem {+, } i of pecial interet ince any logical function can be repreented a polynomial in the o called Algebraic Normal Form (ANF).

5 ANF of Boolean function Theorem Every Boolean function f (x 1, x 2,..., x n ) can be repreented in it ANF f (x 1, x 2,..., x n ) = a 0 +a 1 x 1 + +a n x n +a 12 x 1 x 2 + +a n 1n x n 1 x n +a 123 x 1 x 2 x a 12...n x 1 x 2... x n, where a λ B. In a horter way, ANF (f (x 1,..., x n )) = a I x I, I {1,2,...,n} where a I {0, 1}, x {k 1,k 2,...,k p} = x k1 x k2... x kp.

6 ANF Algorithm Step 0. Repreent each element (b 1, b 2,..., b n ) B n a integer k = b 1 + b b b n 2 n 1.( k = (b 1, b 2,..., b n ) B n ) Step 1. Set g(x 1, x 2,..., x n ) = f (0, 0,..., 0). Step 2. For k = 1 to 2 n 1, do Step 3. take the binary repreentation of the integer k, k = b 1 + b b b n 2 n 1 ; Step 4. if g(b 1, b 2,..., b n ) f (b 1, b 2,..., b n ) then et g(x 1, x 2,..., x n ) = g(x 1, x 2,..., x n ) + n i=1 x b i i. Step 5. ANF(f )= g(x 1, x 2,..., x n ). (Here, x 0 denote the empty ymbol and x 1 x.)

7 ANF Algorithm Example 1 The function f and it ANF g are given in Table 1. k x 1 x 2 x 3 f (x 1, x 2, x 3 ) g(x 1, x 2, x 3 ) x x 2 + x 1 x x 2 + x 3 + x 1 x x 2 + x 3 + x 1 x x 2 + x 3 + x 1 x x 2 + x 3 + x 1 x 2 Thu, ANF(f ) = x 2 + x 3 + x 1 x 2. Table: Computation of ANF

8 Partial vectorial Boolean function A partial (n-ary) Boolean function f : D B, D B n. For computing the ANF of a partial function f we can take ome fix value c (0 or 1) and we can enlarge f to a function f 1 by defining f 1 (x 1, x 2,..., x n ) = c for each (x 1, x 2,..., x n ) B n \ D, and we can apply the ANF algorithm to f 1. A mapping f : B n B m i aid to be a vector valued (or vectorial) Boolean function. If f (x 1, x 2,..., x n ) = (y 1, y 2,..., y m ), y i (1 i m) can be conidered a a Boolean function y i = h i (x 1, x 2,..., x n ). So, f : B n B m can be repreented with an m-tuple of Boolean function (h 1, h 2,..., h m ). When D B n, a partial vectorial Boolean function f : D B m i defined.

9 Repreentation of algebra by Boolean function We conider finite algebra: groupoid, group, quaigroup, ring. A groupoid (Q, ) i aid to be a quaigroup iff the equation a x = b and a y = b have unique olution x, y Q for each a, b Q. When Q i finite, the main body of the multiplication table of a quaigroup i a Latin quare. We can repreent the function f : A n A, A = {a 1, a 2,..., a k }, 2 p 1 k < 2 p, p 1, by a partial vectorial Boolean function. Repreent a i by the p-tuple ī. Repreent f by the partial vectorial Boolean function f : D B p where D B pn and if f (a i1, a i2,... a in ) = a j then f (ī 1, ī 2,..., ī n ) = j. Repreent f by partial Boolean function (h 1, h 2,..., h p ) correponding to f.

10 Repreentation of algebra by Boolean function Example 2 f : {a, b, c} 2 {a, b, c} i defined by f (a, b) = b, f (b, c) = a, f (c, a) = b, f (c, c) = c and f (x, y) = a for other (x, y) {a, b, c} 2. With the correpondence a 00 = 0, b 10 = 1, c 10 = 2, f i defined by f (0, 0, 1, 0) = (1, 0), f (1, 0, 0, 1) = (0, 0), f (0, 1, 0, 0) = (1, 0), f (0, 1, 0, 1) = (0, 0) and f (x 1, x 2, x 3, x 4 ) = (0, 0) for other (x 1, x 2, x 3, x 4 ) B 4. f i repreented by the pair of Boolean function h 1 (x 1, x 2, x 3, x 4 ) = x 2 + x 3 + x 1 x 2 + x 1 x 3 + x 1 x 4 + x 2 x 4, h 2 (x 1, x 2, x 3, x 4 ) = x 2 x 4. f a b c a a b a By f i given the groupoid. b a a a c b a c

11 Repreentation of algebra by Boolean function Example 3 The ring (Z 4, +, ) ha vectorial Boolean repreentation with the function +, a uual: , It four Boolean function are: h 1 + = x 1 + x 3, h 2 + = x 2 + x 4 + x 1 x 3, h 1 = x 1x 3, h 2 = x 1x 4 + x 2 x 3. The function h 1 + and h+ 2 are Boolean repreentation of the cyclic group C 4 too.

12 Repreentation of algebra by Boolean function Example The quaigroup (Q, ) ha Boolean repreentation with the Boolean function h 1 (x 1, x 2, x 3, x 4 ) = x 1 + x 2 + x 3, h 2 (x 1, x 2, x 3, x 4 ) = 1 + x 1 + x 3 + x 4 + x 1 x 3 + x 2 x 3.

13 Application: olving equation in groupoid Conider the ytem of four equation with four indeterminate in the quaigroup (Q, ) of Example 4: (y 4 (y 2 ((y 2 ((y 2 (((y 3 (y 2 y 4 )) y 4 ) y 4 )) y 1 )) y 1 ))) = 3 (y 1 (y 3 (y 2 (y 1 (y 3 (y 2 ((y 1 (y 2 y 2 )) y 2 ))))))) = 0 ((y 4 (y 3 (y 1 (y 4 (((y 1 (y 2 y 4 )) y 3 ) y 3 ))))) y 1 ) = 2 (y 2 (y 1 ((y 2 ((y 4 (y 4 (y 2 (y 3 y 2 )))) y 3 )) y 2 ))) = 3, where y 1, y 2, y 3, y 4 {0, 1, 2, 3}. The only way to olve thi ytem in (Q, ) i by exhautive earch (neither aociative nor commutative law hold). We have to make 4 4 = 256 check. By Boolean repreentation of the quaigroup and of the variable y 1 = (x 1, x 2 ), y 2 = (x 3, x 4 ), y 3 = (x 5, x 6 ), y 4 = (x 7, x 8 ), we obtain a ytem of Boolean equation.

14 Application: olving equation in groupoid 1 + x 2 + x 1 x 3 + x 4 + x 3 x 4 + x 1 x 3 x 4 + x 5 + x 1 x 3 x 5 + x 1 x 4 x 5 + x 1 x 6 + x 7 + x 1 x 7 + x 3 x 7 + x 1 x 3 x 7 +x 4 x 7 + x 3 x 4 x 7 + x 1 x 3 x 4 x 7 + x 1 x 3 x 5 x 7 + x 1 x 4 x 5 x 7 + x 6 x 7 + x 3 x 6 x 7 + x 4 x 6 x 7 + x 8 + x 7 x 8 +x 1 x 7 x 8 + x 1 x 3 x 7 x 8 + x 1 x 4 x 7 x 8 = 1, 1 + x 2 + x 1 x 2 + x 3 + x 1 x 3 + x 2 x 3 + x 2 x 4 + x 1 x 3 x 4 + x 1 x 5 + x 6 + x 1 x 6 + x 1 x 3 x 6 + x 1 x 4 x 6 + x 7 +x 2 x 7 + x 3 x 7 + x 4 x 7 + x 3 x 4 x 7 + x 1 x 3 x 4 x 7 + x 5 x 7 + x 3 x 5 x 7 + x 4 x 5 x 7 + x 6 x 7 + x 3 x 6 x 7 +x 1 x 3 x 6 x 7 + x 4 x 6 x 7 + x 1 x 4 x 6 x 7 + x 8 + x 2 x 8 + x 1 x 3 x 8 + x 4 x 8 + x 3 x 4 x 8 + x 1 x 3 x 4 x 8 + x 5 x 8 +x 1 x 3 x 5 x 8 + x 1 x 4 x 5 x 8 + x 1 x 6 x 8 + x 7 x 8 + x 1 x 7 x 8 + x 3 x 7 x 8 + x 4 x 7 x 8 + x 1 x 4 x 7 x 8 +x 3 x 4 x 7 x 8 + x 1 x 3 x 4 x 7 x 8 + x 1 x 3 x 5 x 7 x 8 + x 1 x 4 x 5 x 7 x 8 + x 6 x 7 x 8 + x 3 x 6 x 7 x 8 + x 4 x 6 x 7 x 8 = 1, x 2 + x 4 + x 1 x 4 + x 2 x 4 + x 3 x 4 = 0, 1 + x 1 + x 2 + x 2 x 3 + x 1 x 3 x 4 + x 2 x 3 x 4 = 0, x 2 + x 1 x 2 + x 3 + x 1 x 3 + x 2 x 3 + x 4 + x 5 + x 1 x 2 x 5 + x 3 x 5 + x 1 x 3 x 5 + x 2 x 3 x 5 + x 1 x 6 + x 1 x 3 x 6 +x 2 x 3 x 6 + x 4 x 6 + x 1 x 4 x 6 + x 2 x 4 x 6 + +x 5 x 6 + x 2 x 5 x 6 + x 3 x 5 x 6 + x 1 x 3 x 5 x 6 + x 2 x 3 x 5 x 6 +x 1 x 4 x 5 x 6 + x 2 x 4 x 5 x 6 + x 7 + x 1 x 7 + x 2 x 7 + x 1 x 3 x 7 + x 2 x 3 x 7 + x 1 x 4 x 7 + +x 2 x 4 x 7 + x 5 x 7 +x 1 x 5 x 7 + x 2 x 5 x 7 + x 3 x 5 x 7 + x 1 x 3 x 5 x 7 + x 2 x 3 x 5 x 7 + x 4 x 5 x 7 + x 1 x 4 x 5 x 7 + x 2 x 4 x 5 x 7 +x 1 x 6 x 7 + x 2 x 6 x 7 + +x 3 x 6 x 7 + x 4 x 6 x 7 + x 5 x 6 x 7 + x 1 x 5 x 6 x 7 + x 2 x 5 x 6 x 7 + x 3 x 5 x 6 x 7 +x 4 x 5 x 6 x 7 + x 1 x 8 + x 2 x 8 + x 5 x 8 + x 1 x 5 x 8 + x 2 x 5 x 8 + x 6 x 8 + x 5 x 6 x 8 = 1, x 1 x 3 + x 2 x 3 + x 1 x 2 x 3 + x 4 + x 2 x 4 + x 1 x 5 + x 2 x 5 + x 1 x 2 x 5 + x 3 x 5 + x 1 x 3 x 5 + x 2 x 3 x 5 + x 1 x 2 x 3 x 5 +x 1 x 4 x 5 + x 2 x 4 x 5 + x 1 x 6 + x 1 x 3 x 6 + x 1 x 2 x 3 x 6 + x 1 x 2 x 4 x 6 + x 1 x 5 x 6 + x 1 x 2 x 5 x 6 + x 1 x 2 x 3 x 5 x 6 +x 1 x 4 x 5 x 6 + x 1 x 2 x 4 x 5 x 6 + x 1 x 7 + x 2 x 7 + x 1 x 2 x 7 + x 3 x 7 + x 1 x 3 x 7 + x 1 x 2 x 3 x 7 + x 4 x 7 + x 1 x 4 x 7 +x 1 x 2 x 4 x 7 + x 5 x 7 + x 1 x 5 x 7 + x 2 x 5 x 7 + x 1 x 2 x 5 x 7 + x 3 x 5 x 7 + x 1 x 2 x 3 x 5 x 7 + x 4 x 5 x 7 + x 1 x 2 x 4 x 5 x 7 +x 1 x 6 x 7 + x 1 x 2 x 6 x 7 + x 1 x 3 x 6 x 7 + x 1 x 4 x 6 x 7 + x 1 x 2 x 5 x 6 x 7 + x 1 x 3 x 5 x 6 x 7 +x 1 x 4 x 5 x 6 x 7 + x 8 + x 1 x 8 + x 1 x 2 x 8 + x 5 x 8 + x 1 x 2 x 5 x 8 + x 1 x 6 x 8 + x 1 x 5 x 6 x 8 = 0, x 1 + x 2 + x 4 + x 5 + x 4 x 6 + x 3 x 4 x 6 + x 4 x 5 x 6 = 1, 1 + x 2 + x 1 x 4 + x 2 x 4 + x 1 x 5 + x 2 x 5 + x 1 x 4 x 6 + x 2 x 4 x 6 + x 3 x 4 x 6 + x 1 x 3 x 4 x 6 + x 2 x 3 x 4 x 6 +x 4 x 5 x 6 + x 1 x 4 x 5 x 6 + x 2 x 4 x 5 x 6 = 1.

15 Application: olving equation in groupoid There are everal tool for olving Boolean equation, for example we can ue Gröbner bae to olve the above ytem immediately. Thi example i a mall one, but for a little more complex groupoid with 8 element underlying et and a ytem of equation with 10 indeterminate we need to make exhautive earch for finding the olution, while Gröbner bae will olved the ytem in econd.

16 Application: claification of et of algebra of ame type The Boolean repreentation of algebra allow the et of all algebra of the ame type to be claified according to their Boolean function repreentation. It i done by the degree of the ANF polynomial of the Boolean function. An algebra i of degree d when the highet degree of ome of it Boolean function i d. Every of the algebra in Example 1 4 ha degree 2.

17 Application: claification of et of algebra of ame type Thi quaigroup of order 8 ha alo preentation by quadratic Boolean function: x 1 + x 3 + x 5 + x 1 x 4 + x 2 x 4 + x 3 x 4 + x 1 x 5 + x 2 x 5 + x 3 x 5 + x 1 x 6 + x 2 x 6 + x 3 x 6, 1 + x 2 + x 3 + x 4 + x 1 x 4 + x 2 x 4 + x 3 x 4 + x 1 x 5 + x 2 x 5 + x 3 x 5 + x 1 x 6 + x 2 x 6 + x 3 x x 2 + x 5 + x 6 + x 3 x 4 + x 3 x 5 + x 1 x 6 + x 2 x 6 + x 3 x 6.

18 Application: claification of et of algebra of ame type Quaigroup of order a b c d e f 0 a e 1 2 c d f 3 8 b b c 8 4 e f a 1 9 d 6 2 c 5 2 d f 8 a e b d 3 e 2 1 b c f 6 a a b d 3 e 6 c 5 f 5 4 a 8 b d 2 c 6 e f e d c b a f 1 7 b d a 2 3 f 1 e c 8 d c a f 0 b e 9 2 f c 5 b d a e 4 3 a 6 c b 7 a f e 0 d b 8 1 f a e c 5 2 d 0 b c f e 6 2 d c b a 5 d e b d f 6 0 a c e 3 0 e c a f 1 b d f 9 7 a f d 4 1 b e 6 c 8

19 Application: claification of et of algebra of ame type The previou quaigroup i of degree 6. One of it Boolean function look like follow: h 1 = 1 + x 2 + x 6 + x 7 + x 8 + x 1 x 2 + x 1 x 3 + x 2 x 4 + x 1 x 5 + x 2 x 5 + x 4 x x 1 x 2 x 5 + x 2 x 3 x 5 + x 2 x 5 x 8 + x 1 x 2 x 5 x 6 + x 2 x 3 x 4 x x 2 x 3 x 6 x 7 + x 3 x 5 x 7 x x 1 x 2 x 3 x 5 x 8 + x 2 x 4 x 5 x 7 x x 1 x 2 x 4 x 6 x 7 x 8 + x 2 x 3 x 4 x 6 x 7 x 8. It conit of 121 member.

20 Application: claification of et of algebra of ame type The claification ha ene to be made when the et all algebra of the ame type i enough large, like in the cae of quaigroup. There are 576 quaigroup of order 4, of order 5, of order 6, and o on. The et of all quaigroup of order 4 i claified into three clae. The cla of linear quaigroup, h 1 and h 2 are both linear (of degree 1), and it contain of 144 quaigroup. The cla of emi-linear quaigrou contain 288 quaigroup, one of the function h 1 or h 2 i linear, the other i quadratic (of degree 2). The cla of quadratic quaigroup contain 144 quaigroup, both of the function h 1 and h 2 are quadratic. Why thi claification i important? Application in coding theory, cryptography, cryptanalyi, etc.

21 Application: matrix preentation of quaigroup The repreentation of quaigroup by Boolean function allow repreentation of ome quaigroup in a matrix form. A illutration, a matrix form of a linear quaigroup (Q, ) of order 4 with Boolean function h 1 = a 0 + a 1 x 1 + a 2 x 2 + a 3 x 3 + a 4 x 4, h 2 = b 0 + b 1 x 1 + b 2 x 2 + b 3 x 3 + b 4 x 4, where a i, b i B, i given by x y = [ ] a0 b 0 [ ] [ ] [ ] [ ] a1 a + 2 x1 a3 a + 4 y1 b 1 b 2 x 2 b 3 b 4 y 2 where x = (x 1, x 2 ), y = (y 1, y 2 ) are binary preentation of the variable x and y. Shortly, it can be written a x y = m + Ax T + By T.

22 Application: matrix preentation of quaigroup Thi matrix repreentation i poible iff the binary matrice A and B are noningular. We can characterize all linear quaigroup (Q, ) of order 2 n in matrix form x y = m + Ax T + By T, where m i n 1 binary matrix, A, B are noningular n n binary matrice and x = (x 1, x 2,..., x n ), y = (y 1, y 2,..., y n ) are the binary preentation of the variable x and y.

23 Application: matrix preentation of quaigroup An algebra (Q, f ) with a k-ary operation f i aid to be a k-ary quaigroup if given k of the element a i uch that f (a 1, a 2,..., a k ) = a k+1, the k + 1t one i uniquely determined. The matrix repreentation of linear k-ary quaigroup of order 2 n i f (x 1, x 2,..., x n ) = m + A 1 x T 1 + A 2 x T A n x T n, m i n 1 binary matrix, A i are noningular n n binary matrice, x i = (x i1,..., x in ) are the binary preentation of the variable x i.

24 Application: matrix preentation of quaigroup We have the following reult for quaigroup of order 4. Theorem Each quaigroup (Q, ) of order 4 ha a matrix repreentation of form x y m + Ax T + By T + (CAx T ) (CBy T ), (1) where m i ome contant from B 2, A and B are noningular [ 2-dimenional ] [ ] binary [ matrice, ] [ C] i one of the matrice ,,,, and denote the component-wie multiplication of vector.

25 Application: matrix preentation of quaigroup The proof of thi theorem ue the following reult. Lemma Let h 1 and h 2 be Boolean repreentation of an n-ary quaigroup (Q, f (x 1, x 2,..., x n )) of order 4, and let x i = (x i1, x i2,..., x in ) be binary preentation of the variable x i. Then in each term x j1 x j2... x jt, t 2, of h 1 or h 2 there are no 2 binary variable x jα, x jβ both belonging to ome of the et {x i1, x i2,..., x in }.

26 Tranformation defined by groupoid Conider an alphabet (i.e., a finite et) G, and denote by G + = {a 1 a a n a i Q} the et of all nonempty word (i.e., finite tring) formed by the element of G. Let be a binary operation on the et G, i.e., conider a groupoid (G, ). For each a G we define a function e a, : G + G + a follow. Let a i G, α = a 1 a 2... a n. Then e a, (α) = b 1 b 2... b n b 1 = a a 1, b 2 = b 1 a 2,..., b n = b n 1 a n.

27 Tranformation defined by groupoid The function e a, i called e-tranformation of Q + baed on the operation with leader a, and it graphical repreentation i on the next figure. a 1 a 2... a n 1 a n a b 1 b 2... b n 1 b n Figure: Graphical repreentation of an e a, function

28 Tranformation defined by groupoid Example 5 a b c a a b a Conider the groupoid ({a, b, c}, ): b a a a c b a c 2, and take a leader c. The tring α = b a c c a b b a c b a c c a b b c a a b i tranformed into the tring e c, (α) = a a a a a b a a a b a a a a b a a a a b. of Example

29 Tranformation defined by groupoid The following theorem i true. Theorem The tranformation e c, defined by a groupoid (G, ) i a permutation of G + if and only if the groupoid i a quaigroup.

30 Application: pattern of quaigroup We preent four conecutive application of an e-tranformation on Table 2. leader = α = e 0, (α) = e 2 0, (α) = e 3 0, (α) = e 4 0, (α) Table: Conecutive e-tranformation We can replace the cipher 0, 1, 2, 3 in Table 2 by pixel in different color, and then ome pattern will appear. The pattern are epecially uable if we take enough long periodical tring and if we apply enough time the e-tranformation.

The left pattern i obtained from the cyclic group C 4, and the right one

31 Application: pattern of quaigroup In Figure 2 two pattern are obtained by 60 tranformation of the periodic tring of length 100. The left pattern i obtained from the cyclic group C 4, and the right one from a quaigroup. Figure: Pattern preentation of the cyclic group C 4 and a quaigroup

32 Application: pattern of quaigroup It can be een that the left pattern how a kind of fractalne, while the right one a kind of randomne (or non-fractalne). In the figure both quaigroup are emi-linear, C 4 ha Boolean repreentation (x 1 + x 3, x 2 + x 4 + x 1 x 3 ) and the quaigroup ha Boolean repreentation (x 2 + x 4 + x 1 x 3, x 1 + x 3 ).All linear quaigroup ha fractal pattern, and all quadratic quaigroup ha non-fractal pattern. There are 192 fractal quaigroup of order 4, and 384 are non-fractal.

33 Application: pattern of quaigroup The pattern characteritic of quaigroup how again why ome of them are uitable for deigning code, and the other are uitable for deigning cryptographic primitive. We emphaize that non-fractalne i not enough when quaigroup are ued for cryptographic deign. There are alo everal other propertie that have to be atified by their Boolean function repreentation: non-linearity, differential property, correlation coefficient, and other.

34 Thank you for your attention!

7.2 INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES 281

7.2 INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES 281 72 INVERSE TRANSFORMS AND TRANSFORMS OF DERIVATIVES 28 and i 2 Show how Euler formula (page 33) can then be ued to deduce the reult a ( a) 2 b 2 {e at co bt} {e at in bt} b ( a) 2 b 2 5 Under what condition