1 Introducton Nonlnearty crtera of Boolean functons Cryptograpc transformatons sould be nonlnear to be secure aganst varous attacks. For example, te s

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1 KUIS{94{000 Nonlnearty crtera of Boolean functons HIROSE Souc IKEDA Katsuo Tel Fax E-mal frose, July 14, 1994

2 1 Introducton Nonlnearty crtera of Boolean functons Cryptograpc transformatons sould be nonlnear to be secure aganst varous attacks. For example, te securty of block cpers, suc as DES, wc consst of teratve substtutons and permutatons, strongly depends on te nonlnearty of te substtutons. Several nonlnearty crtera for Boolean functons ave been proposed and nvestgated. Ts paper dscusses te propertes of nonlnearty crtera and te relatonsps among tem. It focuses on te propagaton crteron, te strct avalance crteron, and te nonlnearty. Frstly, a necessary and sucent condton s presented for a Boolean functon wt n varables to satsfy te PC wt respect to all but one elements n f0; 1g n 0f(0;;0)g. From ts condton, t follows tat, for every even n >, Boolean functons wt n varables tat satsfy te PC of degree n 0 1 are perfectly nonlnear, tat s, satsfy te PC of degree n. We also sow tat Boolean functons wt n varables tat satsfy te PC wt respect to all but lnearly ndependent elements are perfectly nonlnear f n > s even and tat tey satsfy te PC wt respect to all but one elements n f0; 1g n 0f(0;;0)g f > 3 s odd. Secondly, we dscuss te constructon of Boolean functons wt n varables tat satsfy te PC wt respect to all but one or tree elements n f0; 1g n 0f(0;;0)g. Seberry, Zang and Zeng[SZZ93] presented metods for te constructon of balanced Boolean functons satsfyng te PC of g degrees. For odd n > 3, tey proposed a metod for constructng balanced Boolean functons wt n varables satsfyng te PC wt respect to all but one elements n f0; 1g n 0f(0;;0)g and constructed balanced Boolean functons satsfyng te PC of degree n01. For even n > 4, tey proposed a metod for constructng balanced Boolean functons wt n varables satsfyng te PC wt respect to all but tree elements n f0; 1g n 0f(0;;0)g and constructed balanced Boolean functons satsfyng te PC of degree about n=3. Ts result s optmal n te sense tat, for even n > 4, Boolean functons wt n varables satsfyng te PC wt respect to all but less tan tree elements n f0; 1g n 0f(0;;0)g are perfectly nonlnear and tat perfectly nonlnear Boolean functons are not balanced. Ts report sows tat, for every odd n > 3, Boolean functons wt n varables tat satsfy te PC wt respect to all but one elements n f0; 1g n 0f(0;;0)g are constructed from all perfectly nonlnear Boolean functons wt n 0 1 varables. It also presents, for every even n > a necessary and sucent condton for Boolean functons to satsfy te PC wt respect all but tree lnearly dependent elements n f0; 1g n 0f(0;;0)g. It sows tat, for every even n > 4, Boolean functons wt n varables tat satsfy te PC wt respect to all but tree lnearly dependent elements n f0; 1g n 0f(0;;0)g are constructed from all perfectly nonlnear Boolean functons wt n 0 varables. Trdly, ts report dscusses Boolean functons wt n varables satsfyng te PC of degree n 0. It sows tat, for every even n > 4, Boolean functons wt n varables satsfyng te PC of degree n 0 are perfectly nonlnear, and tat, for every odd n > 3, tey satsfy te PC wt respect to all but one elements n f0; 1g n 0f(0;;0)g. Lastly, some relatonsps between te PC and te SAC are presented. It s apparent from te denton tat te set of Boolean functons tat satsfy te PC of degree 1 concdes wt tat of Boolean functons tat satsfy te SAC of order 0. It as been sown tat te Boolean functons tat satsfy te SAC of order n 0 are perfectly nonlnear[at90]. Ts report sows, for every odd n > 3, tat te Boolean functons wt n varables tat satsfy te PC of degree n 0 1 satsfy te SAC of order 1, wle tose satsfyng te PC of degree n 0 necessarly not and tat tere exst Boolean functons wt n varables satsfyng te SAC of order and not satsfyng te PC of degree n 0 For every even n >, t sows tat perfectly nonlnear Boolean functons wt n varables do not necessarly satsfy te SAC of order 1. It also sows tat Boolean functons wt n varables tat satsfy te SAC of order n 0 3 do not necessarly satsfy te PC of degree for every n > 3. 1

3 Secton contans te dentons of nonlnearty crtera. Secton 3 s devoted to te dscusson of Boolean functons wt n varables satsfyng te PC wt respect to all but one elements n f0; 1g n 0f(0;;0)g, and tose satsfyng te PC wt respect to all but lnearly ndependent elements n f0; 1g n 0f(0;;0)g. Secton 4 dscusses te constructon of Boolean functons satsfyng te PC wt respect to all but one or all but tree elements n f0; 1g n 0f(0;;0)g. Secton 5 dscusses te Boolean functons wt n varables satsfyng te PC of degree n0. Secton 6 sows te relatonsps between te PC and te SAC. Prelmnares.1 Wals Transform and Boolean Functons Let R denote te set of reals. Denton 1 Te Wals transform of a real-valued functon f f0; 1g n! R s (W(f))(!) = X xf0;1g n f(x)(01)!1x ; were x =(x 1 ;;x n ),! =(! 1 ;;! n ) f0; 1g n and!1x denotes te dot product! 1 x ! n x n. For smplcty, (W(f))(!) s often denoted by F(!). Te nverse Wals transform s X f(x) =(W 01 (F ))(x) = 1 n F(!)(01)!1x!f0;1g n Te Wals transform can be represented as a matrx form[rue91]. For f f0; 1g n! R, letf() denote f(x 1 ;;x n )wenx 1 + x x n n01 =. Let [f] =[f(0);f(1);;f( n 0 1)] and [F] =[F(0);F(1);;F( n 0 1)]. Te Wals transform s represented as [F ]=[f]h n ; were H n denotes te Hadamard matrx of order n. H n s dened recursvely by H 0 = [1]; H n = " # Hn01 H n01 H n01 0H n01 H n s a n n symmetrc non-sngular matrx, and ts nverse s 0n H n. Te nverse Wals transform s represented as [f] = 0n [F ]H n A Boolean functon s a functon of te form f f0; 1g n!f0; 1g m. f f0; 1g n!f0; 1g m s called Boolea functon wt n nputs and m outputs. Let B n;m = ff j f f0; 1g n!f0; 1g m g.for smplcty, we denote B n;1 as B n and call an Boolean functon wt n nputs and 1 output Boolean functon wt n nputs. Boolean functons wt n nputs are also called Boolean functons wt n varables. A form of representaton s dened for Boolean functons wt n varables. Denton Te algebrac normal form of a Boolean functon f B n satype of representaton of f suc tat M f 1 ;; k g}(n) a f1 ;; k gx 1 111x k ; were }(N) s te powersetofn=f1;;ng, and a f1 ;; k g f0; 1g for every f 1 ;; k g}(n).

4 Every Boolean functon can be unquely represented as an algebrac normal form, and any two derent Boolean functons cannot be represented as a same algebrac normal form. Te Wals transform can be appled to Boolean functons n B n wen tey are consdered to be real-valued functons. For te analyss of Boolean functons, t s often convenent to work wt ^f f0; 1g n!f01; 1g, were ^f(x), (01) f(x). Te Wals transform of ^f s ^F (!) = X xf0;1g n ^f(x)(01)!1x = X xf0;1g n (01) f(x)8!1x Denton 3 Te autocorrelaton functon of a Boolean functon f f0; 1g n! f0; 1g s C f f0; 1g n! N suc tat C f (z) = X xf0;1g n ^f(x) ^f(x 8 z); were N s te set of ntegers and x 8 z denotes (x 1 8 z 1 ;;x n 8 z n ). Proposton 1 sows a relatonsp between te autocorrelaton functon of f and te Wals transform of ^f. It states tat te nverse Wals transform of ^F s C f. Proposton 1 For any Boolean functon f, C f = W 01 ( ^F ). f. Proposton sows tat te sum of ^F (!)'s s constant for every Boolean functon wt n varables Proposton For any f B n, X!f0;1g n ^F (!) = n.. Nonlnearty Crtera for Boolean Functons For a set S, let jsj denote te number of elements n S. Denton 4 A Boolean functon f B n s balanced f and only f jfxjf(x) =0gj = jfxjf(x) =1gj = n01. An ane Boolean functon B n s a Boolean functon of te form of (x 1 ;;x n )= x n x n ; were f0; 1g for n. Te set of ane Boolean functons wt n nputs s denoted as A n. Te number of ane Boolean functons wt n nputs s n+1. Te dstance between two Boolean functons, f and g, wt te same number of varables, s d(f; g) =jfx j f(x) 6= g(x)gj. Te nonlnearty of f B n s te mnmum dstance between f and A n. Denton 5 Te nonlnearty of f B n s mn A n d(f;). Te nonlnearty of f B n can be represented wt ^F. Proposton 3 Te nonlnearty off B n s n max!f0;1g n ^F (!) Webster and Tavares [WT86] dened te strct avalance crteron for te desgn crteron of substtuton boxes of DES. For any a f0; 1g n, let W (a) denote te Hammng wegt ofa, tat s, te number of 1's n a. 3

5 Denton 6 A Boolean functon f B n s sad to satsfy te strct avalance crteron(sac) f and only f f(x) 8 f(x 8 a) s balanced for any a f0; 1g n suc tat W (a) =1. For a Boolean functon satsfyng te SAC, any 1-bt cange of nputs causes te cange of te output wt probablty 1=. Let f(x 1 ;;x n ) B n.for any 1 ;; m suc tat < < 111< m 6 n and b 1 ;;b m f0; 1g, letf j x1 =b 1 ;;x m =b m B n0m denote te subfuncton of f obtaned by substtutng b 1 ;;b m for x 1 ;;x m, respectvely. Forre [For90] extended te noton and dened te SAC of ger orders. Te orgnal denton by Forre was smpled by Lloyd. Denton 7 [Llo91] A Boolean functon f B n s sad to satsfy te strct avalance crteron of order m f and only f, for any 1 ;; m suc tat < < 111< m 6 n and b 1 ;;b m f0; 1g, f j x1 =b 1 ;;x m =bm B n0m satses te SAC. It s obvous from te denton tat te orgnal SAC of Denton 6 s equvalent to te SAC of order 0. Te value of a functon satsfyng te SAC depends on all of ts varables. Lloyd[Llo91] proved tat te functons satsfyng te SAC of order m also satsfy te SAC of order k(< m). Let SAC n (m) denote te set of f B n satsfyng te SAC of order m. It s apparent from Denton 6 tat every f B 0 [ B 1 does not satsfy te SAC. SAC n (n 0 1) = SAC n (n) =forevery n. Denton 8 [MS90] A Boolean functon f B n s perfectly nonlnear f and only f f(x) 8 f(x 8 a) s balanced for any a f0; 1g n suc tat 1 6 W (a) 6 n. For a perfectly nonlnear Boolean functon, any cange of nputs causes te cange of te output wt probablty 1=. Te followng proposton drectly follows from te denton of te autocorrelaton functon and te perfect nonlnearty. Proposton 4 Let f B n. f s perfectly nonlnear f and only f C f (z) = 0 for every z f0; 1g n 0 f(0;;0)g. Meer and Staelbac[MS90] proved tat te set of perfectly nonlnear Boolean functons concdes wt te set of Boolean bent functons dened by Rotaus[Rot76]. ^F (!) Denton 9 f B n s dened to be a Boolean bent functon f and only f = n= for every! f0; 1g n. Proposton 5 f B n s perfectly nonlnear f and only f ^F (!) = n= for every! f0; 1g n. Preneel, et al.[pllgv91] extended te noton of te perfect nonlnearty and dened te propagaton crteron. Denton 10 A Boolean functon f B n s sad to satsfy te propagaton crteron(pc) of degree k f and only f f(x) 8 f(x 8 a) s balanced for any a f0; 1g n suc tat 1 6 W (a) 6 k. Let PC n (k) denote te set of Boolean functons wt n varables satsfyng te propagaton crteron of degree k. PC n (n) s te set of perfectly nonlnear Boolean functons wt n varables. Denton 11 A Boolean functon f B n s sad to satsfy te propagaton crteron(pc) wt respect to A V n f and only f f(x) 8 f(x 8 a) s balanced for every a A. Proposton 6 Let f B n and A V n. f satses te PC wt respect to A f and only f C f (z) =0 for every z V n 0 A. 4

6 3 Propagaton crtera of Boolean functons 3.1 A necessary and sucent condton for te propagaton crteron of degree n 0 1 In ts secton, we nvestgate Boolean functons tat satsfy te PC of degree n 0 1. We begn by presentng a bt general teorem tat gves a necessary and sucent condton for f B n to satsfy te PC wt respect to all but one elements n V n. Before presentng te teorem, we prove two smple lemmas. For a =(a 1 ;;a n ) f0; 1g n,letdec(a) =a 1 +a n01 a n. Lemma 1 Let m > 0beannteger. Te ntegers x; y > 0 satsfyng te equaton s, x + y = m for even m, x = m= and y =0,orx = 0 and y = m=, for odd m, x = y = (m01)=. (Proof) If one of x and y s 0, ten m s even and te oter s m=. If we assume tat x 6= 0andy 6= 0, ten, we can represent x and y as x = ex q x, y = ey q y ; respectvely, weree x > 0, e y > 0, and q x > 1, q y > 1 are odd. Wtout loss of generalty, t can be assumed tat e y > e x > 0. Tus, ex q x + ey q y q x + (e y0e x ) q y = m = m0e x Snce q x + (e y0e x ) q y >, m 0 e x > 1, wc mples tat q x + (e y0e x ) q y s even. e y 0 e x = 0 snce q x and q y are odd. For Tus, q x + q y = m0e x ; snce q x + q y s a multple of but not of 4, m 0 e x =1. Hence, e x = e y =(m01)= and q x = q y = 1. Ts mples m s odd and x = y = (m01)=. Te lemma as been proved. Let V n denote f0; 1g n 0f(0;;0)g. Lemma For every f B n, ^F (0);; ^F ( n 0 1) =[C f (0);;C f ( n 0 1)] H n (Proof) Ts lemma drectly follows from Proposton 1. Te followng teorem presents a necessary and sucent condton for a Boolean functon to satsfy te PC wt respect to all but one nonzero vectors. For every b =(b 1 ;;b n ) f0; 1g n,letv b denote te dec(b)+1-tcolumnvector of H n, and let l b (x 1 ;;x n )=b 1 x b n x n. Teorem 1 Let b V n. f B n satses te PC wt respect to V n 0fbg f and only f, for even n >, ^F (!) = n= for every! f0; 1g n, 5

7 for odd n > 3, ^F (!) = ( (n+1)= f b 1! =0 0 f b 1! =1, or ^F (!) = ( (n+1)= f b 1! =1 0 f b 1! =0. (Proof) f B n satses te PC wt respect to V n 0fbg f and only f C f (a) = 0 for every a V n 0fbg. Tus, from Lemma, ^F can be represented as ^F = C f (0)v 0 + C f (b)v b Let u 0 =(v 0 T + v b T )=; u 1 =(v 0 T 0 v b T )=; were v 0 T and v dec(b) T are te transposes of v 0 and v dec(b), respectvely. Ten, represented as ^F = c 0 u 0 + c 1 u 1 ; ^F s able to be were c 0 = C f (0) + C f (b) and c 1 = C f (0) 0 C f (b). Snce u 0 = [1 8 l b (0);;1 8 l b ( n 0 1)]; u 1 = [l b (0);;l b ( n 0 1)]; Let Snce ^F (!) =( c0 f b 1! =0, c 1 f b 1! =1. ^F (!) X = ^F 0 for every! suc tat b 1! =0,and!f0;1g n ^F (!) = n, ^F 0 + ^F 1 = n+1 Hence, from Lemma 1, Wen n s even, ^F 0 = ^F 1 = n=. Wen n s odd, ^F0 =0, ^F 1 = (n+1)=,or ^F 0 = (n+1)=, ^F 1 =0. ^F (!) = ^F 1 for every! suc tat b 1! =1. Te teorem as been proved. Boolean functons n B n satsfyng te PC wt respect to V n 0f(1;;1)g are te ones satsfyng te PC of degree n 0 1. Tus, te followng two corollares are mmedately derved from Teorem 1. Corollary 1 For even n >, PC n (n 0 1) = PC n (n). 6

8 Corollary For odd n > 3, f PC n (n 0 1) f and only f, ( ^F (!) (n+1)= f W (!) s even = 0 f W (!) sodd, or ^F (!) ( (n+1)= f W (!) sodd = 0 f W (!) seven. Corollary 3 Let n > 3 be odd and b V n.iff B n satses te PC wt respect to V n 0fbg, ten f(x) 8 f(x 8 b) 0or1 (Proof) For odd n > 3, f f B n satses te PC wt respect to V n 0fbg, ten, from te proof of Teorem 1, or C f (0) + C f (b) = n+1 C f (0) 0 C f (b) = 0; C f (0) + C f (b) = 0 C f (0) 0 C f (b) = n+1 For te former case, C f (b) = n, and for te latter case C f (b) =0 n. C f (b) = n and C f (b) =0 n mples tat f(x) 8 f(x 8 b) 0andf(x) 8 f(x 8 b) 1, respectvely. From Teorem 1 and Proposton 3, te followng corollary can be derved mmedately. Corollary 4 Let n > 3 be odd. If f B n satses te PC wt respect to all but one elements n V n, ten te nonlneartes of f s n01 0 (n01)=. Te above corollary states tat, for every odd n > 3, te nonlneartes of f B n wc satses te PC wt respect to all but one elements n V n are g and unquely determned. Te partcular case of Corollary 4 s as follows. Corollary 5 Let n > 3 be odd. If f PC n (n 0 1), ten te nonlneartes of f s n01 0 (n01)=. 3. A necessary and sucent condton for te propagaton crteron wt respect to all or all but one nonzero elements Ts secton s devoted to a necessary and sucent condton for Boolean functons n B n to satsfy te PC wt respect to all nonzero vectors for even n and wt respect to all but one nonzero vectors for odd n. Lemma 3 Let k be any nteger suc tat 1 6 k 6 n and b 1 ;;b k f0; 1g n be lnearly ndependent. let r 1 ;;r k f0; 1g. Te number of elements n f0; 1g n satsfyng te followng equatons are n0k. 8 >< > l b1 (x 1 ;;x n ) = r 1. l bk (x 1 ;;x n ) = r k Teorem Let n and k be any ntegers suc tat n > and 1 6 k 6 n. Let b 1 ;;b k f0; 1g n be lnearly ndependent. If f B n satses te PC wt respect to V n 0fb 1 ;;b k g, ten, 7

9 1. wen n s even, f PC n (n);. wen n s odd, for some X suc tat k, f satses te PC wt respect to V n 0fb g. (Proof) Snce C f (z) = ^f(x) ^f(x 8 z), f Bn satses te PC wt respect to V n 0fb 1 ;;b k g xf0;1g n f and only f C f (a) = 0 for every a V n 0fb 1 ;;b k g. Tus from Lemma, ^F can be represented as ^F = C f (0)v 0 T + C f (b 1 )v b1 T C f (b k )v bk T ; Let u 0 = v T 0 ; u = (v T 0 + v T b )= for k, ^F as ten we can rewrte ^F = c 0 u 0 + c 1 u c k u k ; were Snce c 0 = C f (0) 0 c = C f (b ) u 0 = [1;;1]; kx =1 C f (b ) u = [1 8 l bk (0);;1 8 l bk ( n 0 1)] for k; and l b s balanced for every b V n, X!f0;1g n ^F (!) = n c 0 + n01 (c c k )= n Tus, c 0 + c c k = n+1 From Lemma 3, tere exst some! f0; 1g n suc tat ^F (!) =c 0 Tere also exst some! f0; 1g n suc tat, for any j suc tat 1 6 j 6 k and 1 ;; j suc tat < 111< j 6 k, ^F (!) =c 0 + c c j For te case were n s even. Snce c 0 + c c k = n+1, c 0 +(c 0 + c c k ) = n+1 (c 0 + c 1 )+(c 0 + c + + c k ) = n (c 0 + c k )+(c 0 + c c k01 ) = n+1 ; from Lemma 1, c 0 = c 0 + c 1 = 111= c 0 + c k = n ; 8

10 Tus, c 0 = n ;c 1 = 111= c k =0 Hence, for every! f0; 1g n, j ^F(!)j = n= For te case were n s odd. From Lemma 1, c 0 =0or n+1 ; and, for any j suc tat 1 6 j 6 k and 1 ;; j suc tat < 111 < j 6 k, c 0 + c c j =0or n+1 () If we assume c 0 = 0, ten c 0 + c c k = n+1 Snce c 0 + c = 0 or n+1 for every suc tat k, c = 0 or n+1 Tus, only any one of c 1 ;;c k s n+1 and te oters are all 0. Hence, for some b, ( j ^F (n+1)= f b (!)j = 1! =0 0 f b 1! =1. () If we assume c 0 = n+1,ten c 0 + c c k =0 Snce c 0 + c = 0 or n+1 for every suc tat k, c =0or 0 n+1 Tus, only any one of c 1 ;;c k s 0 n+1 and te oters are all 0. Hence, for some b, ( j ^F (n+1)= f b (!)j = 1! =1 0 f b 1! =0. Hence, te teorem as been proved. 4 Boolean functons satsfyng te PC wt respect to all but one or tree nonzero elements Ts secton gves an exact caracterzaton of Boolean functons wt te odd number of nputs tat satsfy te PC wt respect to all but one nonzero vectors. Te motvaton of ts researc s a metod n [SZZ93] to construct balanced Boolean functons wt te odd number of nputs tat satsfy te PC wt respect to all but one nonzero vectors and tat to construct balanced Boolean functons wt te even number of nputs tat satsfy te PC wt respect to all but tree nonzero vectors. 9

11 4.1 Boolean functons wt te odd number of varables Ts secton presents, for odd n > 3, a spectral property of Boolean functons n B n tat satsfy te PC wt respect to all but one elements n V n. Ts s an exact caracterzarton of suc Boolean functons. Seberry, et al.[szz93] presented a smple metod tat, for any odd n > 3, generates balanced Boolean functons n B n satsfyng te PC wt respect to all but one elements n V n from Boolean functons n PC n01 (n 0 1). In ts secton, t s sown tat, for every odd n > 3, one can construct all Boolean functons tat satsfy te PC wt respect to all but one elements n V n from all Boolean functons n PC n01 (n 01). It also gves a constructon metod tat s slgtly derent from te metod of Seberry, et al. and tat reects spectral propertes. Some results are presented for te number of Boolean functons satsfyng te PC wt respect to all but one nonzero vectors. A lemma s rstly proved wc s a bass of te followng dscusson. It states tat, for any a V n, for eac column v of te matrx constructed from -t rows of H n suc tat te dot product of a and te bnary representaton of s equal to 0 or 1, tere exsts a column n H n01 tat s equal to v or 0v. We dene some notatons. For a matrx M, let col(m;)bete-t column of M. For a = (a 1 ;;a n ) and n, leta denotes (a 1 ;;a ). Lemma 4 For every a V n,letk n (a; 0) and K n (a; 1) be n01 n matrces tat are constructed by removng all (dec(!) + 1)-t rows of H n, were a 1! = 1 and a 1! = 0, respectvely. Ten, for eac column v of H n01, K n (a; 0) as two columns tat s equal to v, andk n (a; 1) as v and 0v, for every suc tat n, col(k n (a; 0);) = col(k n (a; 1);)or col(k n (a; 0);) = 0col(K n (a; 1);) (Proof) We prove te teorem by nducton. Wen n = 1, snce H 0 = [1] and " # 1 1 H 1 = ; 1 01 K 1 (1; 0) = [1; 1] and K 1 (1; 1) = [1; 01]. Te teorem s proved for n =1. For n >, we consder te followng two cases One s te case were a n = 0 and te oter s te one were a n =1. For te Case were a n =0. Snce a 1 (! 1 ;;! n01 ; 0) = a 1 (! 1 ;;! n01 ; 1); and H n = " # Hn01 H n01 H n01 0H n01 for c =0; 1, " Kn01 (a K n (a; c) = n01 ;c) K n01 (a n01 ;c) K n01 (a n01 ;c) 0K n01 (a n01 ;c) # Wen c = 0, from te nductve assumpton, for every column of H n0, K n01 (a n01 ; 0) as exactly two columns wc are equal to t. Tus, by permutng te columns of K n (a; 0), " # Hn0 H n0 H n0 H n0 H n0 H n0 0H n0 0H n0 10

12 s obtaned. Ts mples tat, for eac column of H n01, K n (a; 0) as exactly two columns wc are equal to t. Wen c = 1, for every column v 0 of H n0, K n01 (a n01 ; 0) as v 0 and 0v 0. Tus, for eac column v of H n01, K n (a; 0) as v and 0v. It s also easly derved from te nductve assumpton tat, for every suc tat n, col(k n (a; 0);) = col(k n (a; 1);)or col(k n (a; 0);) = 0col(K n (a; 1);) For te Case were a n =1. K n (a; 0) = K n (a; 1) = H n01 H n01 If a =(0;;0; 1), ten H n01 ; 0H n01 It s apparent tat te teorem olds for ts case. If a 6= (0;;0; 1), snce a 1 (! 1 ;;! n01 ; 0) = a 1 (! 1 ;;! n01 ; 1) 8 1; K n (a; c) = " K n01 (a n01 ;c) K n01 (a n01 ;c) K n01 (a n01 ; 1 8 c) 0K n01 (a n01 ; 1 8 c) # for c =0; 1. Snce, for every j suc tat 1 6 j 6 n01, col(k n01 (a; 0);j) = col(k n01 (a; 1);j)or col(k n01 (a; 0);j) = 0col(K n01 (a; 1);j); for every a, tere exsts some n n nondegenerate matrx 5 suc tat " # Kn01 (a K n (a; c) 5= n01 ;c) K n01 (a n01 ;c) K n01 (a n01 ;c) 0K n01 (a n01 ;c) 5 s a matrx tat excanges l-t and (l + n01 )-t columns of K n (a; c) for every l suc tat 1 6 l 6 n01 and col(k n01 (a; 1 8 c);l)=0col(k n01 (a; c);l) Ts s te same case as te one were a n = 0. Hence, te teorem as been proved. An example of Lemma 4 s gven. Example 1 Let n =4anda =(0; 1; 0; 1). Let H 4 =[v 4 1;;v 4 16] and H 3 =[v 1 ;;v 8 ]. Ten, K 4 (a; 0) = = K 4 (a; 1) = = v 4 1 v 4 3 v 4 6 v 4 8 v 4 10 v 4 1 v 4 13 v 4 15 T v 1 v v 5 v 6 v 3 v 4 v 7 v 8 v 5 v 6 v 1 v v 7 v 8 v 3 v 4 ; v 4 v 4 4 v 4 5 v 4 7 v 4 9 v 4 11 v 4 14 v 4 16 T v 1 v 0v 5 0v 6 v 3 v 4 0v 7 0v 8 v 5 v 6 0v 1 0v v 7 v 8 0v 3 0v 4 For eac column v of H 3, K 4 (a; 0) as two columns tat are equal to v,andk 4 (a; 1) as a column tat s equal to v and a column tat s equal to 0v. ( col(k4 (a; 1);) for =1; ; 5; 6; 9; 10; 13; 14; col(k 4 (a; 0);)= 0col(K 4 (a; 1);) for =3; 4; 7; 8; 11; 1; 15; 16 11

13 Te followng teorem mples an njectve mappng from te set of Boolean functons n B n tat satsfy te PC wt respect to all but one nonzero vectors to PC n01 (n 0 1) for odd n > 3. Teorem 3 Let n > 3beodd. Letf B n and b f0; 1g n. Suppose f satses te PC wt respect to V n 0fbg. For 1 ;; n01 f0; 1g n suc tat 0 6 dec( 1 ) < 111 <dec( n01) 6 n 0 1and ^F( ) 6= 0for166 n01,letf W B n01 be dened as ^fw (0);; ^f W ( n01 0 1) = 1 Ten, f W s perfectly nonlnear. n+1 ^F(1 );; ^F ( n01) (Proof) From te denton of te nverse Wals transform, 1 ^F (0);; ^F( n n 0 1) H n = ^f(0);; ^f( n 0 1) Snce f satses te PC wt respect to V n 0fbg, ( ( ^F (!) (n+1)= f b 1! =0 (n+1)= f b 1! =1 = or 0 f b 1! =1 0 f b 1! =0. Tus, 1 ^F (1 n );; ^F( n01) K n (b; c) = ^f(0);; ^f( n 0 1) ^fw (0);; ^f W ( n01 0 1) K n (b; c) = n01 ^f(0);; ^f( n 0 1) were c = 0 and c = 1, respectvely. From Lemma 4, for K n (b; c), tere exsts a nondegenerate n n -matrx 5 suc tat K n (b; c) 5= (01) c H n01 H n01 5 excanges columns of matrces wen operated from te rgt of tem. Hence, ^fw (0);; ^f W ( n01 0 1) n01 H n01 (01) c H n01 = ^f(0);; ^f( n 0 1) 5 Ts equaton sows tat, for every! f0; 1g n01, W( ^f W ) (!) = n01 Ts completes te proof. Te followng teorem states tat te mappng n Teorem 3 s surjectve. Teorem 4 Let n > 3 be odd and g B n01. Let 1 ;; n01 f0; 1g n, b V n and c f0; 1g suc tat 0 6 dec( 1 ) < 111 < dec( n01) 6 n 0 1 and b 1 = c for n01. Let ^F f0; 1g n! N be dened as ^F (!) =( (n+1)=^g( 0 1) f! =, 0 oterwse, and ^f =(W 01 ( ^F )). If g s perfectly nonlnear, ten ^f f0; 1g n!f01; 1g and f satses te PC wt respect to V n 0fbg. 1

14 (Proof) Snce ^F (!) = 0 wen! 6= and b 1 = c for n01, ^f = 1 n ^F (0);; ^F( n 0 1) H n = 1 ^F (1 n );; ^F( n01) K n (b; c) 1 = [^g(0);;^g( n01 0 1) K n (b; c) n01 From Lemma 4, for K n (b; c), tere exsts a nondegenerate n n -matrx 5 suc tat K n (b; c) 5= (01) c H n01 H n01 5 excanges columns of matrces wen operated from te rgt of tem. Hence, ^f 1 5 = ^g(0);;^g( n01 0 1) H n01 n01 (01) c H n01 1 = ^G (01) c ^G ^G(!) n01 Snce = n01 for every! f0; 1g n01, ^f f0; 1g n!f01; 1g and, from Teorem 1, f satses te PC wt respect to V n 0fbg. From Teorem 3 and 4, t s obvous tat te algortm below generates all te Boolean functons n B n tat satsfy te PC wt respect to all but one nonzero vectors from all te Boolean functons n PC n01 (n 0 1) for odd n > 3. Algortm 1 nput p PC n01 (n 0 1), b V n for odd n > 3. output f 0 ;f 1 B n tat satsfy te te PC wt respect to V n 0fbg. procedure 1. Let c f0; 1g and c 1 ;;c n01 f0; 1g n suc tat 0 6 dec( c 1) dec( c n01) 6 n 0 1; and, for n01, b 1 c = c. Let ^F c (!) =( (n+1)= ^p( 0 1) f! = c 0 oterwse, were ^Fc = W( ^fc ). 3. Let ^fc = 1 n ^Fc H n 13

15 For Algortm 1, snce ^F (0) = X xf0;1g n (01) f(x) =0; f s balanced, and g s not balanced snce ^G(0) 6= 0. For every p PC n01 (n 0 1) and b V n,let Alg n (p; b) denotes te set of te Boolean functons obtaned by te above algortm, wc satsfy te PC wt respect to V n 0fbg. SnceH n s nondegenerate, for any derent pars (p; b) and (p 0 ;b 0 ), Alg n (p; b) \ Alg n (p 0 ;b 0 ) =. Tus te followng corollary can be obtaned. Corollary 6 For every odd n > 3, te number of Boolean functons n B n wc satsfy te PC wt respect to all but one elements n V n s ( n 0 1)jPC n01 (n 0 1)j, and te alf of tem are balanced. In partcular, for te Boolean functons wt n varables satsfyng te PC of degree n 0 1, te followng corollary s derved. Corollary 7 For every odd n > 3, jpc n (n 0 1)j =jpc n01 (n 0 1)j, te number of balanced functons n PC n (n 0 1) s jpc n01 (n 0 1)j. 4. Boolean functons wt te even number of varables Teorem says tat, for even n >, Boolean functons wc satsfy te PC wt respect to all but one or two nonzero vectors are perfectly nonlnear, because less tan tree derent nonzero vectors are always lnearly ndependent. Perfectly nonlnear Boolean functons are not balanced. Seberry, et al.[szz93] presented a metod for constructng balanced Boolean functons satsfyng te PC wt respect to all but tree elements n V n for every even n > 4. Ter result s optmal n te sense tat tere exst no balanced Boolean functons wc satsfy te PC wt respect to all but less tan tree nonzero vectors. Proposton 7 [SZZ93] Let n > 4 be even. For any par of b 1 ;b V n suc tat b 1 6= b, tere exst balanced Boolean functons n B n satsfyng te PC wt respect to V n 0fb 1 ;b ;b 1 8 b g. In ts secton, for even n > 4, an exact caracterzaton s presented of Boolean functons n B n satsfyng te PC wt respect to all but tree lnearly dependent elements n V n. A metod of constructon of suc Boolean functons are also presented, and some relatonsps between te number of tem and tat of perfectly nonlnear Boolean functons are gven. Frstly, we present two smple lemmas. Lemma 5 Tere exst no postve ntegers x, y, z and m suc tat x + y + z = m. (Proof) Suppose tat x, y, z are postve ntegers. Ten, x, y, z can be represented as x = e 1 q 1, y = e q, z = e 3 q 3 ; were e 1 ;e ;e 3 > 0, and q 1 ;q ;q 3 are odd ntegers. Wtout loss of generalty, we may assume tat 0 6 e 1 6 e 6 e 3.Ifx + y + z = m, ten e 1 q 1 + e q + e 3 q 3 q 1 + (e 0e 1 ) q + (e 30e 1 ) q 3 = m = m0e 1 Snce te left-and sde of te above equaton s greater tan 3, m 0 e 1 >, wc mples tat te left-and sde s even. Tus, e 0 e 1 = 0 and e 3 0 e 1 > 1. Ten, q 1 + q = m0e 1 0 (e 30e 1 ) q 3 Snce bot of q 1 and q are odd, q 1 + q s a multple of but not of 4. Ts contradcts tat m 0 e 1 > and (e 3 0 e 1 ) >. Hence, te lemma as been proved. 14

16 Lemma 6 Let w, x, y, z and m be postve ntegers. w + x + y + z = m f and only f m s even and w = x = y = z = (m0)=. (Proof) Suppose tat w, x, y, z are postve ntegers. Ten, tey are able to be represented as w = e 1 q 1, x = e q, y = e 3 q 3, z = e 4 q 4 ; were e 1 ;e ;e 3 ;e 4 > 0, and q 1 ;q ;q 3 ;q 4 are odd ntegers. Wtout loss of generalty, wemay assume tat 0 6 e 1 6 e 6 e 3 6 e 4.Sncew + x + y + z = m, e 1 q 1 + e q + e 3 q 3 + e 4 q 4 q 1 + (e 0e 1 ) q + (e 30e 1 ) q 3 + (e 40e 1 ) q 4 = m = m0e 1 Snce te left-and sde of te above equaton s greater tan 4, m 0 e 1 >. Snce te left-and sde s even, e 0 e 1 =0. Tus, q 1 + q + (e 30e 1 ) q 3 + (e 40e 1 ) q 4 = m0e 1 Snce q 1 + q s a multple of but not of 4 and m0e 1 s a multple of 4, e 3 0 e 1 = e 4 0 e 1 =0and q 1 + q + q 3 + q 4 = m0e 1 For =1; ; 3; 4, q can be represented as q =r + 1, were r > 0sannteger. Hence, 4 Because 4X =1 4X =1 r (r +1)+1! = m0e 1 r (r +1)+1s odd, m 0 e 1 = and r 1 = r = r 3 = r 4 = 0. Hence, m s even and w = x = y = z = (m0)=. Te followng teorem presents a necessary and sucent condton for f B n to satsfy te PC wt respect to all but tree lnearly dependent elements n V n for even n > 4. Teorem 5 Let n > 4 be even and f B n.letb 1, b, b 3 be derent elements n V n and be lnearly dependent. f 6 PC n (n) satses te PC wt respect to V n 0fb 1 ;b ;b 3 g f and only f or j ^F(!)j = ( n=+1 f b 1 1! = b 1! = b 3 1! =0 0 oterwse, j ^F(!)j = ( n=+1 f b 1! =0,b j 1! = b k 1! = 1 for derent, j, k 0 oterwse. (Proof) f B n satses te PC wt respect to V n 0fb 1 ;;b k g f and only f C f (a) = 0 for every a V n 0fb 1 ;b ;b 3 g.from Lemma, ^F can be represented as ^F = C f (0)v T 0 + C f (b 1 )v T b1 + C f (b )v T b + C f (b 3 )v T b3 ; Let u 0 = v T 0 ; u = (v 0 T + v b T )= for =1; ; 3, ten we can rewrte ^F as ^F = c 0 u 0 + c 1 u 1 + c u + c 3 u 3 ; 15

17 were Snce c 0 = C f (0) 0 (C f (b 1 )+C f (b )+C f (b 3 )) c = C f (b ) u 0 = [1;;1]; u = [1 8 l b (0);;1 8 l b ( n 0 1)] for =1; ; 3; and l b s balanced for every b V n, X!f0;1g n ^F (!) = n c 0 + n01 (c 1 + c + c 3 )= n Tus, Snce (c 0 + c 1 + c + c 3 )+(c 0 + c 1 )+(c 0 + c )+(c 0 + c 3 )= n+ ^F (!) = 8 >< > c 0 + c f b =0andb j 1! =1 for =1; ; 3 and j f1; ; 3g0fg c o + c 1 + c + c 3 f b 1! = 0 for =1; ; 3; from Lemma 1, 5, 6, tere are followng two cases Case 1 c 0 + c 1 + c + c 3 = c 0 + c 1 = c 0 + c = c 0 + c 3 = n, Case only one of c 0 + c 1 + c + c 3, c 0 + c 1, c 0 + c and c 0 + c 3 s n+ and te oters are 0. For Case 1, f PC n (n). For Case, f c 0 + c 1 + c + c 3 = n+, ten ^F (!) = n+ wen b 1 1! = b 1! = b 3 1! =0. If c 0 + c = n+,ten ^F (!) = n+ wen b 1! = 0 and b j 1! = b k 1! = 1 for derent, j, k. Hence, te teorem as been proved. Te next corollary can be proved n te same way as Corollary 3. Corollary 8 Let n > 4beeven. Let b 1, b, b 3 be derent elements n V n and be lnearly dependent. If f B n s balanced and satses te PC wt respect to V n 0fb 1 ;b ;b 3 g, ten, for eacof f1; ; 3g, f(x) 8 f(x 8 b ) 0or1 (Proof) For even n > 4, f f B n 0 PC n (n) satses te PC wt respect to V n 0fb 1 ;b ;b 3 g, ten, from te proof of Teorem 5, C f (0) + C f (b 1 )+C f (b )+C f (b 3 ) = n+ C f (0) + C f (b 1 ) 0 C f (b ) 0 C f (b 3 ) = 0 C f (0) 0 C f (b 1 )+C f (b ) 0 C f (b 3 ) = 0 C f (0) 0 C f (b 1 ) 0 C f (b )+C f (b 3 ) = 0 for derent ; j; k f1; ; 3g; or, for derent ; j; k f1; ; 3g, C f (0) + C f (b 1 )+C f (b )+C f (b 3 ) = 0 C f (0) + C f (b ) 0 C f (b j ) 0 C f (b k ) = n+ C f (0) 0 C f (b )+C f (b j ) 0 C f (b k ) = 0 C f (0) 0 C f (b ) 0 C f (b j )+C f (b k ) = 0 For te former case, C f (b 1 )=C f (b )=C f (b 3 )= n, and for te latter case, C f (b )= n and C f (b j ) = C f (b k ) = 0 n. C f (b) = n and C f (b) = 0 n mples tat f(x) 8 f(x 8 b) 0 and f(x) 8 f(x 8 b) 1, respectvely. 16

18 Te followng corollary can be easly derved from Teorem 5. Ts presents a spectral property of te balanced Boolean functons satsfyng te PC wt respect to all but tree elements n V n for every even n > 4. Corollary 9 Let n > 4 be even. Let b 1, b, b 3 V n be derent and lnearly dependent. f B n s balanced and satses te PC wt respect to V n 0fb 1 ;b ;b 3 g f and only f ( ^F (!) n=+1 f b = 1! =0,b j 1! = b k 1! = 1 for derent, j, k 0 oterwse. Corollary 10 Let n > 4beeven. Te nonlnearty of any balanced Boolean functon n B n satsfyng te PC wt respect to all but tree elements n V n s n01 0 n=. (Proof) Ts corollary drectly follows from Proposton 3 and Teorem 5. Seberry, et al.[szz93] proved tat te nonlneartes of balanced Boolean functons satsfyng te PC wt respect to all but tree nonzero vectors are at least n01 0 n=. Corollary 10 determnes te nonlnearty of balanced Boolean functons satsfyng te PC wt respect to all but tree nonzero vectors unquely, and sows tat te lower bound of nonlnearty of Seberry, et al. s optmal. In te followng, t s sown tat, for every even n > 4, one can construct all Boolean functons tat satsfy te PC wt respect to all but tree lnearly dependent vectors n V n from all Boolean functons n PC n0 (n 0 ). A lemma s rstly proved for te bass of te followng dscusson. Lemma 7 Let n >, a =(a 1 ;;a n );b =(b 1 ;;b n ) V n suc tat a 6= b and c; d f0; 1g. Let K n (a; c; b; d) bea n0 n matrx tat s constructed by removng all (dec(!) + 1)-t rows of H n, were a 1! 6= c or b 1! 6= d. Ten, for eac column v of H n0, K n (a; c; b; d) as four columns tat s equal to v f c = d = 0, and as two columns tat s equal to v and two columns tat s equal to 0v f c 6= 0ord 6= 0, for every suc tat n, col(k n (a; c; b; d);) = col(k n (a; c 0 ; b; d 0 );)or col(k n (a; c; b; d);) = 0col(K n (a; c 0 ; b; d 0 );); and, for f(c 1 ;d 1 ); (c ;d ); (c 3 ;d 3 ); (c 4 ;d 4 )g = f0; 1g, col(k n (a; c 1 ; b; d 1 );) = col(k n (a; c ; b; d );) m col(k n (a; c 3 ; b; d 3 );) = col(k n (a; c 4 ; b; d 4 );) (Proof) Ts lemma can be proved by nducton. For every (a; c) and (b; d), K n (a; c; b; d) =K n (b; d; a; c). Wen n =, snce H = K n ((1; 0); 0; (0; 1); 0) = K n ((1; 0); 0; (0; 1); 1) = K n ((1; 0); 1; (0; 1); 0) = K n ((1; 0); 1; (0; 1); 1) = ; ; 17

19 and, for a =(1; 0);b=(1; 1) and a =(0; 1);b=(1; 1), K n (a; 0; b; 0) = K n (a; 1; b; 0) = K n (a; 0; b; 1) = K n (a; 1; b; 1) = Tus te teorem s proved for n = because H 0 =[1]. For a n = b n =0. Snce a 1 (! 1 ;;! n01 ; 0) = a 1 (! 1 ;;! n01 ; 1); b 1 (! 1 ;;! n01 ; 0) = b 1 (! 1 ;;! n01 ; 1); and H n = " # Hn01 H n01 H n01 0H n01 K n (a; c; b; d) = " Kn01 (a n01 ;c; b n01 ;d) K n01 (a n01 ;c; b n01 ;d) K n01 (a n01 ;c; b n01 ;d) 0K n01 (a n01 ;c; b n01 ;d) From te nductve assumpton, for every column v 0 # of H n03, K n01 (a n01 ;c; b n01 ;d) as four columns tat are equal to v 0 f c = d = 0, and as two columns tat are equal to v and two columns tat are equal to 0v 0 f c =1ord =1. Tus, for every column v of H n0, K n (a; c; b; d) as four columns tat are equal to v f c = d = 0, and as two columns tat are equal to v and two columns tat are equal to 0v f c =1ord =1. It s apparent from te nductve assumpton tat, for every suc tat n, col(k n (a; c; b; d);) = col(k n (a; c 0 ; b; d 0 );)or col(k n (a; c; b; d);) = 0col(K n (a; c 0 ; b; d 0 );) It s also apparent tat and, for f(c 1 ;d 1 ); (c ;d ); (c 3 ;d 3 ); (c 4 ;d 4 )g = f0; 1g, col(k n (a; c 1 ; b; d 1 );) = col(k n (a; c ; b; d );) m col(k n (a; c 3 ; b; d 3 );) = col(k n (a; c 4 ; b; d 4 );) For a n =1;b n =0. a 1 (! 1 ;;! n01 ; 0) = a n01 1! n01 a 1 (! 1 ;;! n01 ; 1) = a n01 1! n () Wen a =(0;;0; 1), K n (a; 0; b; 0) = K n (a; 0; b; 1) = K n (a; 1; b; 0) = K n (a; 1; b; 1) = K n01 (b n01 ; 0) K n01 (b n01 ; 0) K n01 (b n01 ; 1) K n01 (b n01 ; 1) K n01 (b n01 ; 0) 0K n01 (b n01 ; 0) ; K n01 (b n01 ; 1) 0K n01 (b n01 ; 1) ; ; 18

20 From Lemma 4, for every column v of H n0, K n01 (b n01 ; 0) as two columns tat are equal to v, and K n01 (b n01 ; 1) as v and 0v. Tus, K n (a; c; b; d) as four columns equal to v f c = d =0, and as two columns equal to v and two columns equal to 0v f c =1ord =1. Snce, for every j suc tat 1 6 j 6 n01, col(k n01 (b n01 ; 0);j) = col(k n01 (b n01 ; 1);j)or col(k n01 (b n01 ; 0);j) = 0col(K n01 (b n01 ; 1);j); for every suc tat n, col(k n (a; c; b; d);) = col(k n (a; c 0 ; b; d 0 );)or col(k n (a; c; b; d);) = 0col(K n (a; c 0 ; b; d 0 );); and also, for f(c 1 ;d 1 ); (c ;d ); (c 3 ;d 3 ); (c 4 ;d 4 )g = f0; 1g, col(k n (a; c 1 ; b; d 1 );) = col(k n (a; c ; b; d );) m col(k n (a; c 3 ; b; d 3 );) = col(k n (a; c 4 ; b; d 4 );) () Wen a 6= (0;;0; 1), K n (a; c; b; d) = " K n01 (a n01 ;c; b n01 ;d) K n01 (a n01 ; 1 8 c; b n01 ;d) K n01 (a n01 ;c; b n01 ;d) 0K n01 (a n01 ; 1 8 c; b n01 ;d) # From te nductve assumpton, for every suc tat n01, col(k n01 (a n01 ;c; b n01 ;d);) = col(k n01 (a n01 ; 1 8 c; b n01 ;d);)or col(k n01 (a n01 ;c; b n01 ;d);) = 0col(K n01 (a n01 ; 1 8 c; b n01 ;d);); and, for every (c 1 ;d 1 ); (c ;d ) f0; 1g suc tat (c 1 ;d 1 ) 6= (c ;d ), col(k n01 (a n01 ;c 1 ; b n01 ;d 1 );) = col(k n01 (a n01 ; 1 8 c 1 ; b n01 ;d 1 );) m col(k n01 (a n01 ;c ; b n01 ;d );) = col(k n01 (a n01 ; 1 8 c ; b n01 ;d );) Tus, tere exsts some n n -matrx 5, wc permutes te columns of matrces, suc tat, for every c and d, K n (a; c; b; d) 5= " Kn01 (a n01 ;c; b n01 ;d) K n01 (a n01 ;c; b n01 ;d) K n01 (a n01 ;c; b n01 ;d) 0K n01 (a n01 ;c; b n01 ;d) # Tus, ts case can be proved n te same way as te case were a n = b n =0. For a n = b n =1. () Wen a =(0;;0; 1), K n (a; 0; b; 0) = K n (a; 0; b; 1) = K n (a; 1; b; 0) = K n (a; 1; b; 1) = K n01 (b n01 ; 0) K n01 (b n01 ; 0) K n01 (b n01 ; 1) K n01 (b n01 ; 1) K n01 (b n01 ; 1) 0K n01 (b n01 ; 1) ; K n01 (b n01 ; 0) 0K n01 (b n01 ; 0) ; ; Ts case can be proved n te same way as te case were a =(0;;0; 1) and b suc tat b n 6=0. 19

21 () Wen a 6= (0;;0; 1) and b 6= (0;;0; 1), K n (a; c; b; d) = " K n01 (a n01 ;c; b n01 ;d) K n01 (a n01 ;c; b n01 ;d) K n01 (a n01 ; 1 8 c; b n01 ; 1 8 d) 0K n01 (a n01 ; 1 8 c; b n01 ; 1 8 d) # In te same way asforteabove case, t can be sown tat tere exsts some n n -matrx 5, wc permutes te columns of matrces, suc tat, for every c and d, K n (a; c; b; d) 5= " Kn01 (a n01 ;c; b n01 ;d) K n01 (a n01 ;c; b n01 ;d) K n01 (a n01 ;c; b n01 ;d) 0K n01 (a n01 ;c; b n01 ;d) # Ts case can be proved n te same way as te case were a n = b n =0. Ts completes te proof. Example Let n = 4 and a = (0; 1; 0; 1), b = (1; 0; 0; 1). Let H 4 = [v1 4;;v4 16] and H = [v 1 ;v ;v 3 ;v 4 ]. Ten, K 4 (a; 0; b; 0) = = v 4 1 v 4 5 v 4 1 v 4 16 T v 1 v 3 v 3 v 1 v v 4 v 4 v v 3 v 1 v 1 v 3 v 4 v v v 4 K 4 (a; 0; b; 1) = = K 4 (a; 1; b; 0) = = K 4 (a; 1; b; 1) = = v 4 v 4 6 v 4 11 v 4 15 T v 1 0v 3 v 3 0v 1 v 0v 4 v 4 0v v 3 0v 1 v 1 0v 3 v 4 0v v 0v 4 v 4 3 v 4 7 v 4 10 v 4 14 T v 1 v 3 0v 3 0v 1 v v 4 0v 4 0v v 3 v 1 0v 1 0v 3 v 4 v 0v 0v 4 v 4 4 v 4 8 v 4 9 v 4 13 T v 1 0v 3 0v 3 v 1 v 0v 4 0v 4 v v 3 0v 1 0v 1 v 3 v 4 0v 0v v 4 For eac column v of H, K 4 (a; 0; b; 0) as four columns tat are equal to v, and eac of K 4 (a; 0; b; 1), K 4 (a; 1; b; 0) and K 4 (a; 1; b; 1) as two columns tat are equal to v and two columns tat are equal to 0v. Te followng teorem mples an njectve mappng from te set of Boolean functons n B n tat satsfy te PC wt respect to all but tree lnearly dependent nonzero vectors to PC n0 (n 0 ) for even n > 4. Teorem 6 Let n > 4 be even. Let f B n and b 1 ;b ;b 3 f0; 1g n suc tat b 1, b, b 3 are derent and lnearly dependent. Suppose f satses te PC wt respect to V n 0fb 1 ;b ;b 3 g and s not 0

22 perfectly nonlnear. For 1 ;; n0 f0; 1g n suc tat 1 6 dec( 1 ) < 111 < dec( n0) 6 n 0 1 and ^F ( ) 6= 0for16 6 n0,letf W B n01 be dened as ^fw (0);; ^f W ( n0 0 1) = 1 Ten f W s perfectly nonlnear. n +1 ^F(1 );; ^F ( n0) (Proof) Snce f satses te PC wt respect to V n 0fb 1 ;b ;b 3 g and s not perfectly nonlnear, or j ^F(!)j = ( n=+1 f b 1 1! = b 1! = b 3 1! =0 0 oterwse, ( n=+1 f b j ^F(!)j = 1! =0,b j 1! = b k 1! = 1 for derent, j, k 0 oterwse. Wtout loss of generalty, we can x =1,j =,k = 3. Tus, 1 ^F (0);; ^F ( n n 0 1) H n = ^f(0);; ^f( n 0 1) 1 ^F (1 n );; ^F( n0) K n (b 1 ; 0; b ;c) = ^f(0);; ^f( n 0 1) ^fw (0);; ^f W ( n0 0 1) K n (b 1 ; 0; b ;c) = n 01 ^f(0);; ^f( n 0 1) were c = 0 and c = 1, respectvely. From Lemma 7, tere exsts a nondegenerate n n -matrx 5 suc tat K n (b 1 ; 0; b ;c)= H n0 H n0 (01) c H n0 (01) c H n0 5 excanges columns of K n (b 1 ; 0; b ;c). Hence, ^fw (0);; ^fw ( n0 0 1) wc sows tat n 01 ^f(0);; ^f( n 0 1) W( ^fw ) (!) = n 01 H n0 H n0 (01) c H n0 (01) c H n0 = 5; for every! f0; 1g n0. Ts completes te proof. Te followng teorem states tat te mappng n Teorem 6 s surjectve. Teorem 7 Let n > 4 be even and g B n0. Let 1 ;; n0 f0; 1g n, b 1 ;b ;b 3 V n and c; d f0; 1g suc tat06 dec( 1 ) < 111 < dec( n0) 6 n 0 1 and b 1 1 = c, b 1 = d and b 3 1 = c 8 d for n0.let^f f0; 1g n! N be dened as ^F (!) =( n=+1^g( 0 1) f! =, 0 oterwse, and ^f = W 01 ( ^F ). If g s perfectly nonlnear, ten ^f f0; 1g n!f01; 1g and f satses te PC wt respect to V n 0fb 1 ;b ;b 3 g. (Proof) Snce ^F (!) = 0 wen! 6= and b 1 1 = c and b 1 = d for n0, ^f = 1 n ^F (0);; ^F( n 0 1) H n = 1 ^F (1 n );; ^F( n0) K n (b 1 ;c; b ;d) 1 = [^g(0);;^g( n0 0 1) K n (b 1 ;c; b ;d) n01 1 ;

23 From Lemma 7, for K n (b 1 ;c; b ;d), tere exsts a nondegenerate n n -matrx 5 suc tat K n (b 1 ;c; b ;d)5= H n0 H n0 (01) c_d H n0 (01) c_d H n0 5 excanges columns of K n (b 1 ;c; b ;d). Hence, ^f 1 5 = n 01 ^g(0);;^g( n01 0 1) H n0 H n0 (01) c_d H n0 (01) c_d H n0 1 = ^G ^G n (01) 01 c_d ^G (01) c_d ^G Snce ^G(!) = n 01 for every! f0; 1g n0, ^f f0; 1g n!f01; 1g and, from Teorem 5, f satses te PC wt respect to V n 0fb 1 ;b ;b 3 g. From Teorem 6 and 7, t s obvous tat te algortm below generates all te Boolean functons n B n tat satsfy te PC wt respect to all but tree nonzero vectors and tat s not perfectly nonlnear from all te Boolean functons n PC n0 (n 0 ) for even n > 4. Algortm nput p PC n0 (n 0 ), b 1 ;b V n for even n > 4. output f (0;0) ;f (0;1) ;f (1;0) ;f (1;1) B n tat satsfy te PC wt respect to V n 0fb 1 ;b ;b 1 8 b g. procedure 1. Let (c; d) f0; 1g and (c;d) 1 ;; (c;d) n0 f0; 1g n suc tat 0 6 dec( (c;d) 1 ) < 111<dec( (c;d) n0 ) 6 n 0 1; and, for every suc tat n0, b 1 1 (c;d) = c; b 1 (c;d) = d. Let ^F (c;d) (!) = ( n=+1^p( 0 1) f! = (c;d) 0 oterwse; were ^F(c;d) = W( ^f (c;d) ). 3. Let ^f(c;d) = 1 n ^F(c;d) H n For Algortm, ^F(c;d) (0) = 0 only f (c; d) =(0; 0). Tus, f (c;d) s balanced f (c; d) =(0; 0) and not balanced oterwse. Te followng corollary presents te relatonsp between te number of balanced Boolean functons satsfyng te PC wt respect to all but tree elements n V n and tat of perfectly nonlnear Boolean functons n B n0. Corollary 11 Let n > 4 be even. Te number of balanced Boolean functons n B n satsfyng te PC wt respect to all but tree elements n V n s n 0 1 jpc n0 (n 0 )j.!

24 4.3 Examples Ts secton gves examples of Algortm 1 and Algortm. Example 3 Two Boolean functons n B 5 are constructed tat satsfy te PC wt respect to all but one nonzero vectors. Let p PC 4 (4) be p(x 1 ;x ;x 3 ;x 4 )=x 1 x 8 x 3 x 4 Let b =(0; 1; 1; 1; 1). Te elements!'s n f0; 1g 5 tat satsfy b 1! =0are 0; 1; 6; 7; 10; 11; 1; 13; 18; 19; 0; 1; 4; 5; 30; 31; were eac of te numbers represents dec(!). Tus, ^F0 =[0; 0; 8; 8; 8; 08; 0; 0; 8; 8; 0; 0; 0; 0; 8; 08; 8; 8; 0; 0; 0; 0; 8; 08; 0; 0; 08; 08; 08; 8; 0; 0] ^f0 = 1 5 ^F0 H 5 = [1; 1; 1; 01; 1; 01; 1; 1; 1; 1; 01; 1; 1; 01; 01; 01; 1; 1; 01; 1; 1; 01; 01; 01; 01; 01; 01; 1; 01; 1; 01; 01] Te algebrac normal form of f 0 s f 0 (x 1 ;;x 5 )=x 1 x 8 x 1 x 3 8 x x 4 8 x x 5 8 x 4 x 5 f 1 can be generated n te same way as te above. f 1 (x 1 ;;x 5 )=x 8 x 1 x 8 x 1 x 3 8 x x 4 8 x x 5 8 x 4 x 5 Te trut tables of f 1 and f are sown n Fgure 1. f 0 s balanced, wle f 1 s not balanced. f 0 ;f 1 PC 5 (4) snce tey satsfy te PC wt respect to V 5 0f(0; 1; 1; 1; 1)g. For b =(0; 1; 1; 1; 1), f 0 (x) 8 f 0 (x 8 b) 1 f 1 (x) 8 f 1 (x 8 b) 0 Example 4 Four Boolean functons n B 6 are constructed tat satsfy te PC wt respect to all but tree nonzero vectors. Let p PC 4 (4) be p(x 1 ;x ;x 3 ;x 4 )=x 1 x 8 x 3 x 4 Let b 1 =(1; 1; 1; 1; 0; 0), b =(0; 0; 1; 1; 1; 1) and b 3 =(1; 1; 0; 0; 1; 1). Te elements!'s n f0; 1g 6 tat satsfy b 1 1! = b 1! = b 3 1! = 0 are 0; 3; 1; 15; 1; ; 5; 6; 37; 38; 41; 4; 48; 51; 60; 63 were eac of te numbers represents dec(!). Tus, ^F(0;0) = [16; 0; 0; 16; 0; 0; 0; 0; 0; 0; 0; 0; 16; 0; 0; 016; 0; 0; 0; 0; 0; 16; 16; 0; 0; 16; 016; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 16; 16; 0; 0; 16; 016; 0; 0; 0; 0; 0; 016; 0; 0; 016; 0; 0; 0; 0; 0; 0; 0; 0; 016; 0; 0; 16] 3

25 x x x x x f ( x 1,..., x 5 ) 0 x x x x x f ( x 1,..., x 5 ) 1 Fgure 1 Trut tables of f 0 and f 1. 4

26 ^f0 = 1 6 ^F(0;0) H 6 = [1; 01; 1; 01; 01; 01; 1; 1; 1; 1; 01; 01; 01; 1; 01; 1; 1; 1; 1; 1; 1; 01; 01; 1; 1; 01; 01; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 01; 01; 1; 1; 01; 01; 1; 1; 1; 1; 1; 01; 1; 01; 1; 1; 1; 01; 01; 01; 01; 1; 1; 1; 01; 1; 01] Te algebrac normal form of f (0;0) s f (0;0) (x 1 ;;x 6 ) = x 1 8 x 3 8 x 1 x 3 8 x x 3 8 x 1 x 4 8 x x 4 8 x 1 x 5 8 x 3 x 5 8 x 1 x 6 8 x 3 x 6 8 x 5 x 6 f (0;1), f (1;0) and f (1;1) can be generated n te same way as te above. f (0;1) (x 1 ;;x 6 ) = x 1 x 3 8 x x 3 8 x 1 x 4 8 x x 4 8 x 1 x 5 8 x 3 x 5 8 x 1 x 6 8 x 3 x 6 8 x 5 x 6 f (1;0) (x 1 ;;x 6 ) = x 3 8 x 1 x 3 8 x x 3 8 x 1 x 4 8 x x 4 8 x 1 x 5 8 x 3 x 5 8 x 1 x 6 8 x 3 x 6 8 x 5 x 6 f (1;1) (x 1 ;;x 6 ) = x 1 8 x 1 x 3 8 x x 3 8 x 1 x 4 8 x x 4 8 x 1 x 5 8 x 3 x 5 8 x 1 x 6 8 x 3 x 6 8 x 5 x 6 f (0;1) f (1;0) and f (1;1) are balanced, wle f (0;0) s not balanced. Te trut table of f (1;1) s presented n Fgure. f (0;0), f (0;1), f (1;0), f (1;1) PC 6 (3) snce tey satsfy te PC wt respect to V 6 0fb 1 ;b ;b 3 g and te Hammng wegts of b 1 ;b ;b 3 are all 4. Table 1 sows te values of f (c1 ;c )(x)8 f (c1 ;c )(x 8 b ) for (c 1 ;c ) f0; 1g and =1; ; 3. 5 Boolean functons satsfyng te PC of degree n Boolean functons wt even number of varables In ts secton, t s proved tat, for every even n > 4, PC n (n 0 ) = PC n (n). Frstly,we present a smple lemma. For u f0; 1g n and f0; 1g n, let [u] denote te dec()+1- t elementofu. For b f0; 1g n,letv b denote te dec(b)+1-t column vector of te Hadamard matrx H n. Lemma 8 Let n >. Let b =(0;;0; (1;;1). For v b1 ;;v bn+1 and =( 1 ;; n ) f0; 1g n, f W () s even, ten [v bn+1 ] =1; ( 1 f =0 [v b ] = 01 f =1, f W () s odd, ten [v bn+1 ] = 01; ( 1 f =1 [v b ] = 01 f =0, 1; 0;;0) for every suc tat n and b n+1 = Table 1 Te value of f (c1 ;c )(x) 8 f (c1 ;c )(x 8 b ). b 1 b b 3 f (0;0) f (0;1) f (1;0) f (1;1)

27 x x x x x x Fgure Trut table of f (1;1). 6

28 (Proof) Ts lemma can be proved from te fact tat, [v bn+1 ] =(01) n ; and, for eac suc tat n, [v b ] =(01) n Teorem 8 For every even n > 4, PC n (n 0 ) = PC n (n). (Proof) Suppose tat f PC n (n 0 ). Ten, C f (a) = 0 for any a f0; 1g n suc tat 1 6 W(a) 6 n 0. Tus, ^F s able to be represented as ^F = C f (0)v T 0 + C f (b 1 )v T b C f (b n )v T bn + C f (b n+1 )v T bn+1 Let u 0 = v 0 T ; u = (v 0 T + v b T )=; for every n + 1. Ten, ^F canberepresented as ^F = c 0 u 0 + c 1 u c n+1 u n+1 ; were c 0 = C f (0) 0 (C f (b 1 )+111+ C f (b n+1 )); c = C f (b ) From Lemma 8, for any odd s suc tat 1 6 s 6 n and 1 ;; s suc tat < 111 < s 6 n 0 1, " # sx sx ^F ( k=1 k ) = c 0 + c k +1; k=1 and, for any even t suc tat 1 6 t 6 n and j 1 ;;j t suc tat 0 6 j 1 < 111<j t 6 n 0 1, " # tx tx ^F ( k=1 j k ) = c 0 + c c n+1 0 c jk +1 k=1 Tus, for every par of odd ntegers s and t suc tat 1 6 s; t 6 n and s + t 6 n and 1 ;; s and j 1 ;;j t suc tat f 1 ;; s g\fj 1 ;;j t g = and < 111 < s 6 n 0 1 and 0 6 j 1 < 111 < j t 6 n 0 1, sx tx sx tx ^F (0) + ^F ( k=1 k )+ ^F ( l=1 = 4c 0 +(c c n+1 ) = n+ j l )+ ^F ( k=1 k + Snce n + s even, from Lemma 1, 5, 6, all of ^F (0), ^F ( l=1 j l ) sx k=1 k ), ^F ( tx l=1 j l ), ^F ( sx k=1 k + tx l=1 j l )are equal to n, or only one of tem s equal to n+ and te oters are equal to 0. In te former case, f s perfectly nonlnear. In te latter case, f ^F (0) = n+, ten ^F (!) = 0 for every! 6= 0, wc contradcts tat X!f0;1g n ^F (!) = n. 7

29 If ^F (0) = c 0 + c c n+1 = 0, ten c 0 = n+1.for ts case, ^F (1; 0;;0) + ^F (0; 1; 0;;0) + ^F (1; 1; 0;;0) = (c 0 + c 1 )+(c 0 + c )+(c 0 + c c n+1 ) = c 0 +(c 0 + c c n+1 ) = n+ From Lemma 1, 5, and c 0 = n+1, tere are followng tree cases (Case 1) c 1 = n+1, c = 0 n+1, c 0 + c c n+1 =0, (Case ) c 1 = 0 n+1, c = n+1, c 0 + c c n+1 =0, (Case 3) c 1 = 0 n+1, c = 0 n+1, c 0 + c c n+1 = n+. For Case 1. Snce ^F (0; 0; 1; 0;;0) + ^F (0; 0; 0; 1; 0;;0) + ^F (1; 1; 1; 1; 0;;0) = (c 0 + c 3 )+(c 0 + c 4 )+(c 0 + c c n+1 ) = c 0 +(c 0 + c c n+1 ) = n+ ; te same argument as te above one sows tat (Case 1.1) c 3 = n+1, c 4 = 0 n+1, c 0 + c c n+1 =0. (Case 1.) c 3 = 0 n+1, c 4 = n+1, c 0 + c c n+1 =0. (Case 1.3) c 3 = 0 n+1, c 4 = 0 n+1, c 0 + c c n+1 = n+. For Case 1.1, 1. and 1.3, te followng tree equatons can be derved, respectvely, ^F (1; 0; 1; 0; 0;;0) = c 0 + c c n+1 0 (c 1 + c 3 )=0 n+ ; ^F (1; 0; 0; 1; 0;;0) = c 0 + c c n+1 0 (c 1 + c 4 )=0 n+ ; ^F (0; 1; 1; 1; 0;;0) = c 0 + c + c 3 + c 4 = 0 n+ ; wc are contradctons. For Case. Ts case can be proved n te same way ascase1. For Case 3. Snce c 0 + c c n+1 = n+, ^F (0; 0; 1; 0;;0) + ^F (0; 0; 0; 1; 0;;0) + ^F (1; 1; 1; 1; 0;;0) = (c 0 + c 3 )+(c 0 + c 4 )+(c 0 + c c n+1 ) = c 0 +(c 0 + c c n+1 ) = n+3 From Lemma 1, 5, tere are followng tree cases (Case 3.1) c 3 = n+1, c 4 = 0 n+1, c 0 + c c n+1 = n+, (Case 3.) c 3 = 0 n+1, c 4 = n+1, c 0 + c c n+1 = n+, (Case 3.3) c 3 = n+1, c 4 = n+1, c 0 + c c n+1 =0. 8

30 Table Bounds of te degree of te PC of balanced Boolean functons number of varables upper bound lower bound For Case 3.1, 3. and 3.3, te followng tree equatons can be derved, respectvely, ^F (1; 1; 0; 1; 0;;0) = c 0 + c 1 + c + c 4 = 0 n+ ; ^F (1; 1; 1; 0; 0;;0) = c 0 + c 1 + c + c 3 = 0 n+ ; ^F (0; 0; 1; 1; 0;;0) = c 0 + c c n+1 0 (c 3 + c 4 )=0 n+ ; wc are contradctons. Hence, te teorem as been proved. Snce perfectly nonlnear Boolean functons are not balanced, te followng corollary can be derved. It presents an upper bound of te degree of te PC of a balanced Boolean functons wt te even number of nputs. Corollary 1 For every even n > 4, te degree of te PC of balanced Boolean functons s less tan n 0. As for te lower bound, te followng as been proved. Proposton 8 [SZZ93] Let n > 4 be even. Suppose tat n =3t + c, werec =0; 1; or. Ten tere exst balanced Boolean functons n B n tat satsfy te PC of degree t 0 1 wen c =0; 1ort wen c =. Table sows te bounds of te degree of te PC of balanced Boolean functons. Te bounds are tgt for n =4; Boolean functons wt odd number of varables In ts secton, t s sown tat, for every odd n > 3, every f PC n (n 0 ) satses te PC wt respect to all but one elements n V n. Lemma 9 Let x; y; z > 0 be ntegers and m > 0 be an even nteger. x + y + z =31 m f and only f x = y = z = m=. (Proof) x; y; z can be represented as x = e 1 q 1, y = e q, z = e 3 q 3 ; were e 1 ;e ;e 3 > 0, and eac of q 1 ;q ;q 3 s 0 or odd. We may assume tat x > y > z > 0. () If we assume tat y = z =0,tenx =31 m, wc contradcts tat x s an nteger. () Suppose tat x 6= 0,y 6= 0,z =0. Ten, e 1q 1 + e q =31 m. Snce, wtout loss of generalty, we can assume tat e > e 1 > 0, q 1 + (e 0e 1 ) q =31 m0e 1 If e 1 = e, ten q 1 + q s a multple of but not of 4. Ts mples tat m 0 e 1 =1,wc contradcts tat m s even. If e 1 <e, ten te left-and sde s odd and m 0 e 1 = 0. Tus, q 1 + (e 0e 1 ) q =3,wc mples tat (e 0 e 1 ) = 1. Ts contradcts tat e 1 and e are ntegers. 9