Note on the integer geometry of bitwise XOR

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1 European Journal of Combnatorcs 26 (2005) Note on the nteger geometry of btwse XOR Antóno Guedes de Olvera a,,dogo Olvera e Slva b a Departamento de Matemátca Pura, Centro de Matemátca da Unversdade do Porto, Unversdade do Porto, Portugal b Departamento de Matemátca Pura, Unversdade do Porto, Portugal Receved 20 January 2004; accepted 23 March 2004 Avalable onlne 25 May 2004 Abstract We consder the set N of non-negatve ntegers together wth a dstance d defned as follows: gven two ntegers x, y N, d(x, y) s, n bnary notaton, the result ofperformng, dgt by dgt, the XOR operaton on (the bnary notatons of) x and y. Dawson, n Combnatoral Mathematcs VIII, Geelong, 1980, Lecture Notes n Mathematcs, 884 (1981) 136, consders ths geometry and suggests the followng constructon: gven k dfferent ntegers x 1,...,x k N, letv be the set of ntegers closer to x than to any x j wth j, for, j = 1,...,k. LetV = (V 1,...,V k ) and X = (x 1,...,x k ). V s a partton of {0, 1,...,2 n 1} whch, n general, does not determne X. In ths paper, we characterze the convex sets of ths geometry: they are exactly the lne segments. Gven X and the partton V determned by X, we also characterze n easy terms the ordered sets Y = (y 1,...,y k ) that determne the same partton V. Ths,npartcular, extends one of the man results of Combnatoral Mathematcs VIII, Geelong, 1980, Lecture Notes n Mathematcs, 884 (1981) Elsever Ltd. All rghts reserved. 1. Introducton Let us take two non-negatve ntegers n bnary form and consder the result of performng wth them the typcal computer btwse XOR operator. Dawson, n [1], regards ths functon, (, j) ^j usng the C language notaton or (x, y) x y n the notaton heren, geometrcally,asadstance between the two ntegers. Correspondng address: Departamento de Matemátca Pura, Faculdade de Cêncas da U. Porto, Rua do Campo Alegre 687, P Porto, Portugal. E-mal address: agolv@fc.up.pt (A.G. de Olvera) /$ - see front matter 2004 Elsever Ltd. All rghts reserved. do: /j.ejc

2 756 A.G. de Olvera, D.O. e Slva / European Journal of Combnatorcs 26 (2005) He consders also, gven an ordered set X = (x 1, x 2,...,x k ) of such ntegers, the Voronoy cells determned by them, that s, the sets V of elements closer to x than to any x j wth j, forevery, j = 1, 2,...,k. Inpartcular, he proves that there exst sets A B such that an nteger x belongs to V f and only f the set (x) of the postons of the dgts 1 (or 1-bts)ofx satsfes A (x) B. (1.1) Set X := (x ) and let m and M be such that (m ) = A and (M ) = B.Condton (1.1), n ts turn, s true f and only f x contans 1-bts n all the postons where m contans 1-bts, and 0-bts n all the postons where M contans 0-bts. Hence, Dawson s statement can be rephrased, n computer slang, n a sentence lke: x V x matches, as astrng, In [1], a certan dualty s also consdered: let Y := X A B,where by we denote the symmetrc dfference of two sets, and let y be such that Y = (y );then, n partcular, A = X Y and B = X Y.Letuscall X = (x 1, x 2,...,x k ) the ntal k-tuple and Y = (y 1, y 2,...,y k ) the fnal k-tuple. [1, Lemma 1.3] asserts that these rôles are nterchangeable: f we use nstead Y as the ntal k-tuple,weendup wth X as the fnal one, and the Voronoy cells are exactly the same. In ths paper, we proceed further nto the study of ths partcular geometry: Frst, we characterze the lne segments,.e., the sets of form [x y]:={z N: d(x, z) + d(z, y) = d(x, y)}; they are the ntervals,aswecall the sets of the solutons of a condton lke (1.1) above. We also prove that,gvenx, y N,theset S(x, y) := {z N: z x < z y} s convex,nthesense that f t contans both ponts a and b then t contans all the segment [a b].andweprove that any convex set s n fact a lne segment (and vce versa). As the man result, we characterze, gven X = (x 1,...,x k ), the ordered sets Z = (z 1,...,z k ) wth the same partton as X. Moreprecsely (cf. Corollary 2.16), let V(X) = (V 1,...,V k ) (V(X) s then the Voronoy dagram determned by X); we prove that V(Z) = V(X) f and only f: = 1, 2,...,k, j = 1, 2,...,k, j { z x j > z x z j x > z x. Dawson s dualty, referred to above, can be obtaned from here. Fnally, we also prove that takng (m 1, m 2,...,m k ) or (M 1, M 2,...,M k ) as the ntal k-tuple leads to the same Voronoy dagrams, whence makng t easy to reverse Dawson s constructon.

3 A.G. de Olvera, D.O. e Slva / European Journal of Combnatorcs 26 (2005) Btwse XOR geometry 2.1. Notaton, examples and techncal results Defnton 2.1. Let N be the set of non-negatve ntegers, and fx an nteger n > 0. Let be the bjectve functon defned, for A {1, 2,...,n}, by: (A) := 2 1 A and let = 1. Denote by x y the nteger for whch the bnary representaton has the th dgt (from rght to left) equal to 1 f the th dgts of x and y are dfferent, and equal to 0 f they are equal, for = 1, 2,..., n. Thssthe resultof the btwse XOR operator, whch s used, for example, for fndng a wnnng strategy of the celebrated game of Nm [2, p.44]. More precsely: Defnton 2.2. Let a = n 1 =0 α 2 and b = n 1 =0 β 2 be such that α,β {0, 1} for all = 0,...,n 1. Then, n 1 a b := (α β )2, =0 where 0 0 = 1 1 = 0 and 0 1 = 1 0 = 1. In other words, a b = ( (a) (b)). Followng Dawson, we fx a set {x 1,...,x k } N of k > 0dstnct ntegers smaller than 2 n,anddefne α: {0, 1,...,2 n 1} {x 1,...,x k } so that z α(z), for each z,sas small as possble. Then,forevery {1,..., k}, V := α 1 (x ) ={z N : j = 1,...,k ( j ), z x j > z x }. Note that V := {V 1,...,V k } s a partton of {0,...,2 n 1},.e., the latter set s the unon of the elements of V, that are non-empty and parwse dsjont (snce a x = a y x = y). We call t the partton determned by X = (x 1,...,x k );foremphaszng ts orgn, we also denote t by V(X) and V by V(X, ). Btwse AND and OR are defned smlarly to Defnton 2.2,andwll also be denoted smply by and. Defnton 2.3. Gven non-negatve ntegers a, b N, wesaya s strongly less than b, wrtten a b,fa b = a and a b = b. a, b :={c N: a c b} s also called an nterval. Note that a b (a) (b)( a < b) and that x satsfes condton (1.1)fandonly f x (A ), (B ).

4 758 A.G. de Olvera, D.O. e Slva / European Journal of Combnatorcs 26 (2005) Example 2.4. Take m = ({2, 3, 5, 9}), M = ({1, 2, 3, 4, 5, 8, 9, 11, 12}) (n = 12) and V = m, M. m = 278 = (2) and M = 3487 = (2). The elements of V are exactly the ntegers whose bnary representatons match the pattern: We note that ({0, 1}, ) may be naturally dentfed wth Z/2Z. Inpartcular, for example, (a b) c = a (b c), a b = b a, 0 a = a and a a = 0forall a, b, c N. Notealsothat(a, b) a bdefnes ndeed a dstance n N. Infact, n 1 α 2 n 1 + β 2 = =0 =0 (α,β ) (1,1) 0 n 1 (α + β )2 + 0 n 1 2 2, and α + β = α β and α β = 0except when α = β = 1. Hence, a + b = a b + 2(a b) (2.2) and so a b = (a c) (c b) a c + c b. Lemma 2.5. Let a, b, c N. Then, the followng condtons are equvalent: def c [ab] a b = a c + c b (2.3) a b c a b. (2.4) Proof. Set a = n 1 =0 α 2, b = n 1 =0 β 2 and c = n 1 =0 γ 2 (α,β,γ {0, 1} for all = 0, 1,...,n 1). c [ab] (a c) (c b) = (a c) + (c b) (2.2) (a c) (c b) = 0 = 0, 1,...,n 1,α γ = 0orγ β = 0 = 0, 1,...,n 1,γ = α or γ = β = 0, 1,...,n 1,α β γ α β a b c a b. (2.5) Remark 2.6. By defnton of lne segment (and snce e.g., (c a) (c b) = a b): {x c: c [ab]} = [x ax b]. (2.6) 2.2. The convex sets Proposton 2.7. Let x, y N, x y, and consder S(x, y) := {z N : z x < z y}.

5 A.G. de Olvera, D.O. e Slva / European Journal of Combnatorcs 26 (2005) If a, b S(x, y), then[a b] S(x, y),.e., S(x, y) s convex. V s convex too for every = 1, 2,...,k. Proof. Let m be the bggest element of the set (x y) = (x) (y). Sncem s, by defnton, the leftmost poston of all bts where x and y dffer, x < y holds f and only f m / (x) (or equvalently, f and only f m (y)). Now, snce (z x) (z y) = (x) (y), z S(x, y) f and only f m / (z x). (2.7) Hence, f a, b S(x, y) then m / (a x), (b x). Inordertoprove that also m / (c x) for every c [ab],tssuffcent to show that: (c x) (a x) (b x) c x (a x) (b x). (2.8) But ths condton holds, by (2.6). Fnally, V s also convex because V = j S(x, x j ). V s n fact a lne segment (cf. [1, Lemma 1.3]). More precsely, we have: Proposton 2.8. Let, for = 1, 2,...,k, y N be the element of V at greatest dstance from x,.e.,suchthat, for all 0 z < 2 n,fx z > x y then z / V.Then V =[x y ]. Proof. [x y ] V because the latter s convex. Assume, by contradcton, that there exsts z V \[x y ].ByLemma 2.5 (condton (2.5)fals), for some j wth 1 j k the jth bt of z s dfferent from the jth bt of both x and y (that are equal, consequently). For clearness sake, set x = x, y = y and y = y 2 j 1.Theny s sth bt s equal to y s sth bt for all s j, andsequal to z s sth bt (and hence dfferent from the sth bt of both x and y) fors = j. Agan by condton (2.5) oflemma 2.5, y [zy]. Butthen y V,byconvexty, whch, snce x y > x y, sncontradcton wth the defnton of y. Let K be any convex set, x K and y be the element of K farthest from x, asbefore. Wth the same proof, weobtan that K s a lne segment, exactly [x y]. Theconverse s also true, by Lemma 2.5.Hence, we have: Theorem 2.9. Let K be a subset of {0, 1,...,2 n 1}.Then: K s convex K s a lne segment Explct calculaton of Voronoy dagrams Let us proceed a lttle further n the drecton of the last result. As usual, by a for a real number a we mean the bggest nteger not bgger than a. Defnton Set, for X = (x 1, x 2,...,x k ) and, j = 1, 2,...,k, j, mj X := log 2 (x x j ) +1 (=max( (x ) (x j )); Sm X := {mj X : j = 1, 2,...,k, j }

6 760 A.G. de Olvera, D.O. e Slva / European Journal of Combnatorcs 26 (2005) a X := (Sm X ); b X := ({1,...,k}\Sm X ); y X := x b X ; m X := x a X ; M X := x b X. (We drop the symbol X whenever t s not necessary.) Remark Denote by z the btwse complement of z N, z := (2 n 1) z. Then b X = a X,anda X = (x x ) a X = (x a X ) (x a X ) = m X M X. We have the followng theorem: Theorem 2.12 (cf. [1, Lemmas 1.3 and 1.4] ). For every X = (x 1, x 2,...,x k ) and every = 1, 2,...,k, V = V(X, ) =[x y X ]= m X, M X. (2.9) Proof. We have seen before (condton (2.7)) that the condton z S(x, x j ) s equvalent to m j / (x z),or,nother words, to 2 mj 1 (x z) = 0. Hence, z V = S(x, x j ) a (x z) = 0. j By Proposton 2.8, however,v =[x y ] where y s the element z V for whch the value of x z s maxmum. But the maxmum value of w for whch a w = 0sclearly b = a (2 n 1), thecomplement of a.thus,the maxmum s attaned for z such that x z = b z = x b.hence, ths s the value of y.itsnoweasy to see, btwse, that m = x (x b ) = x a and that M = x (x b ) = x b (e.g., x (x b ) s 1 exactly when x = 1andb = 0). Theorem2.13. Let X = (x 1, x 2,...,x k ) for a subset {x 1, x 2,...,x k } of {0, 1,...,2 n 1} wth k (dstnct) elements and X = (x 1, x 2,...,x k ) for another subset {x 1, x 2,...,x k } of the same set. Then V(X ) = V(X) f and only f, for every = 1, 2,...,k, x V(X, ); (2.10) Sm X = Sm X. (2.11) Proof. Suppose frst V(X ) = V(X). Then x V(X, ) = V(X, ) and, by Theorem 2.12, m X = m X and M X = M X.Moreover, by Remark 2.11, a X = m X M X = a X.Ths mples condton (2.11). Conversely, suppose that x a X = m X x M X = x b X and a X = a X.Then x a X x a X (x b X ) a X.But(x b X ) a X x a X,andsom X = m X.Theproof that M X = (x a X ) (b X a X ) = = M X proceeds n a smlar way. Corollary Let X be as n Theorem Then the Voronoy dagram determned by X, V(X), equals the Voronoy dagram determned by any of the collectons Y, A or B defned below: Y = (y1 X, y 2 X,...,y k X ); A = (m1 X, m 2 X,...,m k X ); B = (M1 X, M 2 X,...,M k X ).

7 A.G. de Olvera, D.O. e Slva / European Journal of Combnatorcs 26 (2005) Proof. We prove that n all three cases s := mj X equals m X j for all, j = 1, 2,...,k such that j. Frst,note that, for every s > s, s Sm X f and only f s Sm X j snce the s th bts of x and x j are equal, and thus the s th bts of a X and a X j are also equal. Denote, for x N and s such that 1 s k, thesth bt of x by s x,andnote that s y = s y j exactly when s x = s x j,sncey = x b X and y j = x j b X j.thsproves that m X j s for X = (y1 X, y 2 X,...,y k X).Thesamehappens for the other defntons of X,bythesame reasons. Now, n order to show that also m X j s, tssuffcent to prove that s y and s y j (respectvely, sm and s m j,and s M and s M j )are dfferent. But by defnton of s = m j, s x and s x j are ndeed dfferent, and s Sm Sm j.itfollows that s a = 1 = s a j and so s b = 0 = s b j.fnally, s y = s x 0 = s x, sm = s x 1 = s x, sm = s x 0 = s x,andasmlar stuaton occurs when we replace by j. Remark By the defnton of y X,weobtan n the frst case [1,Corollary 1.6]: when X = (x 1,...,x k ) s replaced n Dawson s constructon by Y = (y 1,...,y k ),asdefned above, we also fnd Y replaced by X. Ths s so because b X = b Y,byTheorem Corollary Let X = (x 1, x 2,...,x k ) for a subset {x 1, x 2,...,x k } of {0, 1,..., 2 n 1} wth k (dstnct) elements and X = (x 1, x 2,...,x k ) for another subset {x 1, x 2,...,x k } of the same set. Then V(X ) = V(X) f and only f, for every = 1, 2,...,k, and x V(X, ) x V(X, ) or, equvalently, f and only f (2.10) (2.12), j = 1, 2,...,k, j { x x j > x x x j x > x x. (2.13) Proof. By symmetry, all we have to prove s that condton (2.12)(together wth condton (2.10)) mples condton (2.11). Let us fx {1, 2,...,n} and set more smply x := x, x := x, a := a X, b := b X, a := a X and b := b X.Snce Condton (2.10) reads x a x x b, f,for s = 1, 2,...,k, wedenote agan by s a the sth bt of a and suppose s a = 1(andhence s b = 0), then s x = s x 1 s x s x 0 = s x,andso s x = s x.inthe same way, by condton (2.12), s a = 1alsomples s x = s x.comng back to our former notaton, what we have shown s that x and x concde n all the 1-bts of a X and n all the 1-bts of a X,whchare the elements of Sm X and Sm X,respectvely. Now suppose, for a contradcton, that condton (2.11)fals. Wthout loss of generalty we may then suppose that there exst, j ( j) such that r := mj X < s := m X j.then s Sm X Sm X j,ands / (x x j ),buts (x x j ),whchmeans that x and x j have equal sth bts but the sth bts of x and x j are dfferent. But ths s mpossble snce by our prevous argument the sth bts of x and x are equal, and the same happens wth the sth bts of x j and x j.

8 762 A.G. de Olvera, D.O. e Slva / European Journal of Combnatorcs 26 (2005) An nterestng queston arses as to whether all parttons of the set {0, 1,...,2 n 1} n k ntervals can be constructed n ths way from a set {x 1, x 2,...,x k },whenreorderngs of {1, 2,...,n} are consdered. We fnsh ths secton by showng through three small examples that the answer to ths queston s negatve, and that condtons (2.10)and(2.11), separately, are not suffcent for forcng V(X) = V(Y ): Example Let n = 3and consder the partton of {0 = 000 (2),...,7 = 111 (2) } represented below. Suppose that the elements of form x are those we have underlned and, for a certan order < n of the elements of {1, 2, 3},theydetermne the partton. We fnd a contradcton: 1 < n 2snce 010 s closer to 011 than to 000; 2 < n 1snce 101 s closer to 111 than to 100. The other three possble choces of elements of X = (x 1,...,x 6 ) that could generate ths partton can be dscarded n a smlar way. Example Consder X = (x 1, x 2 ) := (2 = 10 (2), 3 = 11 (2) ) and X = (x 1, x 2 ) := (2 = 10 (2), 1 = 01 (2) ) and the parttons they determne n {0, 1, 2, 3}. Thenx 1 = x 1 and x 2 V(X, 2) but V(X) V(X ). Example Fnally, consder X = (0 = 00 (2), 1 = 01 (2), 2 = 10 (2) ) and X = (2 = 10 (2), 3 = 11 (2), 0 = 00 (2) ) and the parttons they determne n {0, 1, 2, 3}. Although they have the same m j for every j (n fact, as shown below, x x j = x x j ),the parttons are dfferent. Acknowledgements Ths research was partally supported by Fundação Calouste Gulbenkan through the Program Novos Talentos em Matematca. The frst author s research was partally

9 A.G. de Olvera, D.O. e Slva / European Journal of Combnatorcs 26 (2005) supported by FCT and PRAXIS XXI through the Research Unts Plurannual Fundng Program, and by the project POCTI/36563/MAT/2000. References [1] J.E. Dawson, A constructon for a famly of sets and ts applcaton to matrods, n: Combnatoral Mathematcs VIII, Geelong, 1980, Lecture Notes n Mathematcs, vol. 884, Sprnger, Berln, New York, 1981, pp [2] E.R. Berlekamp, J. Conway, R.K. Guy, Wnnng Ways for Your Mathematcal Plays, Games n General, vol. 1, Academc Press, London, 1982.