2012 IEEE International Symposium on Information Theory Proceedings

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1 On the Uncertanty of Informaton Retreval n Assocatve Memores Etan Yaaob an Jehoshua Bruc Electrcal Engneerng Department, Calforna Insttute of Technology, Pasaena, CA 9115, U.S.A Electrcal an Computer Engneerng Department, Unversty of Calforna San Dego, La Jolla, CA 9093, U.S.A. {yaaob, bruc}@caltech.eu Abstract We (people are memory machnes. Our ecson processes, emotons an nteractons wth the worl aroun us are base on an rven by assocatons to our memores. Ths natural assocaton paragm wll become crtcal n future memory systems, namely, the ey queston wll not be How o I store more nformaton? but rather, Do I have the relevant nformaton? How o I retreve t? The focus of ths paper s to mae a frst step n ths recton. We efne an solve a very basc problem n assocatve retreval. Gven a wor W, the wors n the memory that are t-assocate wth W are the wors n the ball of raus t aroun W. In general, gven a set of wors, say W, X an Y, the wors that are t-assocate wth {W, X, Y} are those n the memory that are wthn stance t from all the three wors. Our man goal s to stuy the maxmum sze of the t-assocate set as a functon of the number of nput wors an the mnmum stance of the wors n memory - we call ths value the uncertanty of an assocatve memory. We erve the uncertanty of the assocatve memory that conssts of all the bnary vectors wth an arbtrary number of nput wors. In aton, we stuy the retreval problem, namely, how o we get the t-assocate set gven the nputs? We note that ths paragm s a generalzaton of the sequences reconstructon problem that was propose by Levenshten (001. In ths moel, a wor s transmtte over multple channels. A ecoer receves all the channel outputs an ecoes the transmtte wor. Levenshten compute the mnmum number of channels that guarantee a successful ecoer - ths value happens to be the uncertanty of an assocatve memory wth two nput wors. I. INTRODUCTION One of the nterpretatons of the term assocaton, especally n the context of psychology, s the connecton between two or more concepts. Throughout our lfe, we remember an store an enormous amount of nformaton. However, whle we are not aware of the metho ths nformaton s store, amazngly, t can be accesse an retreve wth relatve ease. The way we thn an process nformaton s manly performe by assocatons. Our memory content retreval process s one by stmulatng t wth an external or nternal nputs. That s, the nowlege of some nformaton gves rse to relate nformaton that was store earler. Mathematcally speang, assume the memory s a set of wors M = {m 1,..., m N }. Then, gven an arbtrary wor x as an nput to the memory M, ts output s another wor or a set of wors from the memory M that are relate or close to the nput wor x. Here, the term close can be nterprete as usng any stance metrc between wors, for example the Hammng stance. A wor s assocate wth another wor or wors whch they agan can be assocate wth more wors an so on, resultng n a sequence of assocatons. From the nformaton theory perspectve, we say that the wors assocate wth an nput wor x are those n stance at most t (a prescrbe value from x. Ths set comprses a 01 IEEE Internatonal Symposum on Informaton Theory Proceengs (a (c Fg. 1. Three example of assocate wors: (a assocate wors wth a sngle wor x, (b assocate wors wth two wors x an y, (c assocate wors wth the sx wors x 1,..., x 6. ball of raus t, see Fg. 1(a. We generalze ths paragm an conser a set of wors that are presente as an nput to the memory. For example, for two nput wors x, y, ther set of assocate wors are the ones that ther stance from both x an y s at most t, see Fg. 1(b. Clearly, ths set of assocatons s strctly smaller then the ball of raus t. For more than two nput wors, the set of assocate wors s gettng even smaller, an n general, the larger the set of nput wors s, the smaller the set of assocate wors s, see Fg. 1(c. For example, assume the nput wor s tall, then many wors can be assocate wth t, such as tree, mountan, tower, laer, etc. But f the nput wors are tall an frut, then out of these four assocate wors, only the wor tree wll be assocate wth tall an frut. Assume that the memory s the set of all bnary vectors, M = {0, 1} n. For any nput wor x, t s mmeate to see that f ts set of assocate wors are the ones of stance at most t, then there are =0 t (n such wors. However, for two nput wors, x, y, the problem of fnng the set of assocate wors of stance at most t from both x an y becomes more complex. In fact, ths problem was propose an solve by Levenshten n [10], [11]. The motvaton came from a completely fferent scenaro n the context of the sequences reconstructon problem. In ths moel, a coewor x s transmtte through multple channels. Then, a ecoer receves all channel outputs an generates an estmaton on the transmtte wor, whle t s guarantee that all channel outputs are fferent from each other, see Fg.. If x belongs to a coe C wth mnmum stance an n every channel there can be at most t > 1 errors, then Levenshten stue the mnmum number of channels N that guarantees the exstence of (b /1/$ IEEE 106

2 Fg.. Channel moel of the sequences reconstructon problem. a successful ecoer. Ths number has to be greater than N = max B t (x 1 B t (x, (1 x 1,x C,x 1 =x where B t (x s the ball of raus t surrounng x. To see that, notce that f the ntersecton of the raus-t balls of x 1 an x contans N wors an the channel outputs are these N wors, then a ecoer cannot etermne what the transmtte wor s. However, f the number of channel outputs s greater than the maxmum sze of the ntersecton of two balls, then there s only one coewor of stance at most t from all receve channel outputs. The motvaton to the moel stue by Levenshten came from fels such as chemstry an bology, where the reunancy n the coewors s not suffcent to construct a successful ecoer. Thus, the only way to combat errors s by repeately transmttng the same coewor. Recently, ths moel was shown to be also relevant n storage technologes [3], [16]. Due to the hgh capacty an ensty of toay s an future s storage meum, t s no longer possble to rea nvual memory elements, but, rather, only a multple of them at once. Hence, every memory element s rea multple tmes, whch s translate nto multple estmatons of the same nformaton store n the memory. Fnng the maxmum ntersecton problem n (1 was stue n [11] wth respect to the Hammng stance an other metrc stances. In [7] [9], t was analyze over permutatons, an n [13], [14] for error graphs. In [1], the equvalent problem for nsertons an eletons was stue an reconstructon algorthms for ths moel were gven n [], [5], [15]. The case of eletons only was stue n the context of trace reconstructon n [4]. Returnng to our orgnal problem, the set of assocate wors wth x an y s B t (x B t (y an the maxmum ntersecton s the value N n (1. The generalze problem of fnng the maxmum sze of assocate wors of m nput wors wth mutual stance s expresse as N t (m, = max { =1 m B t(x }. ( x 1,...,x m, H (x,x j The man goal n ths paper t to analyze the value of N t (m, wth respect to the Hammng stance for fferent values of t, m,. In partcular, we show that f A(D s the sze of a maxmal antcoe of ameter D, that s, the largest set of wors wth maxmum stance D, then N t (A(D, 1 = A(t D. The rest of the paper s organze as follows. In Secton II, we efne the concept of assocatve memores an escrbe the connecton to the sequences reconstructon problem. In Secton III, we solve the problem state n ( for the case = 1. Extensons for arbtrary are gven n Secton IV. In Secton V, we gve effcent ecoers to the reconstructon problem stue by Levenshten. Fnally, Secton VI conclues the paper. II. DEFINITIONS AND BASIC PROPERTIES In ths wor, the wors are bnary vectors of length n. The Hammng stance between two wors x an y s enote by H (x, y an the Hammng weght of a wor x s enote by w H (x. For a wor x {0, 1} n, Bt n (x s ts surrounng ball of raus t, Bt n(x = {y {0, 1}n : H (x, y t}. The sze of Bt n(x, comprsng of length-n wors, s b t,n = =0 t (n. If the length of the wors s clear from the context, we use the notaton B t (x. For 1 n, e s the unt vector where only ts -th bt s one, an 0 s the all-zero vector. Two wors x, y {0, 1} n are calle t-assocate f H (x, y t. Defnton. The t-assocate set of the wors x 1,..., x m s enote by S t ({x 1,..., x m } an s efne to be the set of all wors y that are t-assocate wth x 1,..., x m, ( S t {x1,..., x m } m ={y : H (y, x t, 1 m}= B t (x. Note that for a sngle wor x, we have S t ( {x} = Bt (x. Gven an assocatve memory M, we efne the maxmum sze of a t-assocate set of any m wors from the memory. Defnton. Let M be an assocatve memory an m, t be two postve ntegers. The uncertanty of the assocatve memory M for m an t, enote by N t (m, M, s the maxmum sze of a t-assocate set of m fferent nput wors from M. That s, { ( N t (m, M = max St {x1,..., x m } }. (3 x 1,...,x m M,x =x j In case the assocatve memory M s a coe wth mnmum stance, we wll use the notaton N t (m, nstea of N t (m, M. Then, the value n Equaton (3 becomes N t (m, = max x 1,...,x m, H (x,x j =1 { S t ( {x1,..., x m } }. (4 For example, N t (m, 1 refers to N t (m, {0, 1} n. We now gve the efntons that establsh the connecton wth the channel moel by Levenshten [11]. Assume a coewor x s transmtte over N channels. The channel outputs, enote by y 1, are all fferent from each other (Fg.. A lst ecoer D L receves the N channel outputs an returns a lst of at most L wors x 1,..., x l, where l L. We call t an L-ecoer D L. The L-ecoer D L s sa to be successful f an only f the transmtte wor x belongs to the ecoe output lst,.e., x D L (y 1 = { x 1,..., x l }. In case L = 1 then the ecoer output s a sngle wor an ths s the moel stue by Levenshten [11]. The next Lemma shows the connecton between the value of N t (m, an the econg success of an L-ecoer. Lemma 1. Assume the transmtte wor x belongs to a coe C of mnmum stance. Then, there exsts a successful L- ecoer wth N channels f an only f N N t (L + 1, + 1. Proof: Assume to the contrary that the number of channels s N t (L + 1, an let x 1,..., x L+1 be L + 1 wors 107

3 such that the set S t ({x 1,..., x L+1 } contans N t (L + 1, wors. If one of these L + 1 wors s the transmtte one an the receve channel outputs are the N t (L + 1, wors n S t ({x 1,..., x L+1 }, then any of the L + 1 wors x 1,..., x L+1 coul be the transmtte one an thus can belong to the ecoer s output lst. Hence, the transmtte wor may not belong to the output lst. On the other han, f there are N t (L + 1, + 1 channels, then for any transmtte wor x, there are at most L wors n C, all of stance at least from each other, such that the N t (L + 1, + 1 channel outputs are locate n the ntersecton of ther raus-t balls. For L = 1, the value N t (, was stue by Levenshten [11] an was shown to be N t (, = t =0 ( n t = t+ (. (5 Let us frst remn how ths value was calculate. Assume H (x, y = an the goal s to fn the carnalty of the set S t ({x, y} = {z {0, 1} n : H (z, x, H (z, y t}. For any wor z S t ({x, y}, let S 0,0, S 0,1, S 1,0, S 1,1 be the followng four sets: S 0,0 = { : y = z = x }, S 0,1 = { : y = x, z = x }, S 1,0 = { : y = x, z = x }, S 1,1 = { : y = z = x }. Note that S 0,0 + S 0,1 = n an S 1,0 + S 1,1 =. Snce H (z, x t an H (z, y t we get that S 0,1 + S 1,1 t, S 0,1 + S 1,0 t, or S 0,1 + S 1,1 t, S 0,1 + S 1,1 t. Denote S 0,1 = an S 1,1 = so we get + t, + t, or 0 t /, + t t. Therefore, the number of wors n the ntersecton of these two spheres s gven by t / ( n t ( S t ({x, y} = N t (, =. =0 =+ t where ( a b = 0 f b < 0 or b > a. If we substtute the orer of, n the last term, we get 0 mn{, t}, 0 t max{, }, an mn{,t} ( t max{, } ( n S t ({x, y} =N t (, =. (6 =0 =0 Ths last representaton of N t (, wll be helpful n showng the followng property. Due to the lac of space, we omt the proof of the next lemma as well as the followng one, whch extens the soluton of N t (, for m = 3. Lemma. Let t, be two postve ntegers such that s even, then N t (, = N t (, 1. Lemma 3. Let t, be such that t > 1, an n large enough. The value of N t (3, s gven by ( n 3 ( N t (3, = ( (, 1,, 3, where 1,, 3, 4 satsfy the followng constrants: t, 1 + t 4 t 1, 3 t t ( 1 + 4, 4 max{ t, t} t ( III. THE CASE = 1 In ths secton, we analyze the value of N t (m, for = 1. A frst observaton on the value of N t (m, 1 s state n the next lemma. Lemma 4. For m, t 1, f N t (m, 1 l an N t (m + 1, 1 < l, then N t (l, 1 = m. Proof: Snce N t (m, 1 l, there exst m fferent wors x 1,..., x m such that S t ({x 1,..., x m } = B t (x 1 B t (x m l an assume y 1,..., y l are l wors whch belong to ths ntersecton. Therefore, H (x t for all 1 m an 1 j l, an thus {x 1,..., x m } S t ({y 1,..., y l } = B t (y 1 B t (y l, an hence N t (l, 1 m. Assume to the contrary that N t (l, 1 m + 1 an let z 1,..., z l be l wors such that S t ({z 1,..., z l } = B t (z 1 B t (z l m + 1. As n the frst part, we get that N t (m + 1, 1 l, whch s a contracton. Hence, N t (l, 1 = m. In general, for a gven set of wors x 1,..., x m, the closer the wors are, the larger the sze of the set S t ({x 1,..., x m } s. In case = 1, we loo for a set of wors that are all close to each other, or equvalently - the maxmum stance between all pars of wors s mnmze. An antcoe of ameter D s a set A {0, 1} n of wors such that the maxmum stance between every two wors n A s at most D. That s, for all x, y A, H (x, y D. For D 1, A(D s the sze of the largest antcoe of ameter D. It was shown n [6] that the value of A(D s gven by { b D A(D =,n f D s even, b D 1,n 1 f D s o. Our next goal s to show that for all D 1, N t (A(D, 1 = A(t D. That s, the t-assocate set of a maxmum antcoe of ameter D s a maxmum antcoe of ameter t D. Lemma 5. For all 0 D t n, N t (A(D, 1 A(t D. Proof: Assume that D s even. We tae the A(D wors n B D (0 an conser the set S t (B D (0 = B t (x. x B D (0 Then, B t D (0 S t (B D (0 an hence N t (A(D, 1 A(t D for even D. In case that D s o, let = (D 1/. Let us start wth a maxmal antcoe of ameter + 1. Let X be the set 108

4 X=B (0 B (e 1 ={aw : a {0, 1}, w B n 1 (0}, an let Y=B t 1 (0 B t 1 (e 1 ={bu : b {0, 1}, u B n 1 t 1 (0}. Then, for every x X, y Y, H (x, y t. Therefore, N t (A(D, 1 A(t D for o D as well. The equvalent upper boun s prove n the next two lemmas. Lemma 6. For all 0 D t an n (t D (D+1 + 1, where D s even, N t (A(D + 1, 1 < A(t D. Proof: Let X = {x 1,..., x A(D+1 } be a set of A(D + 1 wors. Snce the largest antcoe wth ameter D has sze A(D, there exst two wors, say x 1, x, where H (x 1, x D + 1. Hence, the sze of S t (X s no greater than the sze of S t ({x 1, x }, whch, accorng to (5, s at most M = Note that t ( D +1 t ( D +1 ( n D 1 M < j t ( D +1 t j =D+1 t+ j ( n j D+1. ( D + 1 For 0 j t ( D + 1 an (t D (D+1 + 1, we have ( n j D+1 ( n j+1 an hence, ( ( n n M < < = A(t D. j + 1 j t D An equvalent property can be shown for D o. We sp ts etals ue to ts long proof an the lac of space. Lemma 7. For all 0 D t, where D s o, an n large enough, N t (A(D + 1, 1 < A(t D. We summarze ths result n the followng corollary. Corollary 8. For all 0 D t an n large enough, N t (A(D, 1 = A(t D. Proof: From Lemma 5 we get that N t (A(D, 1 A(t D an from Lemma 6 an Lemma 7 N t (A(D + 1, 1 < A(t D. The contons of Lemma 4 hol an thus N t (A(t D, 1 = A(D, or N t (A(D, 1 = A(t D. We note that the result shown by Levenshten for = 1 s a specal case of Corollary 8 for D = 1. IV. EXTENSIONS FOR ARBITRARY Our goal n ths secton s to use the results foun n Secton III n orer to erve bouns on N t (m, for arbtrary. Frst, we state a useful Theorem from [1]. Theorem 9. [1] Let C be a coe n the Hammng graph Γ wth stances from D = { 1,..., s } {1,..., n}. Further let L D (B be a maxmal coe n B Γ wth stances from D. Then, one has C Γ L D(B. B For all 1, we enote by ρ to be the maxmal rate of a coe C of mnmum stance{ an} length n, that s, C ρ = max C n. Theorem 9 wll serve us to prove the next lemma.. Lemma 10. Let B be a set an let L(B be a maxmal coe n B wth mnmum stance, then L(B ρ B. Proof: We tae the coe C n Theorem 9 to be a coe C of mnmum stance an maxmal rate ρ. Then, we get L(B C B n = ρ. Now, we can erve a connecton wth N t (m,. Lemma 11. For all m, t, an n large enough, N t ( ρ m, N t (m, 1. Proof: Assume that X s a set of m wors such that S t (X has sze A. Accorng to Lemma 10, let L(X be a coe n X of mnmum stance an sze ρ m. Then, S t (X S t (L(X an thus N t ( ρ m, N t (m, 1. Fnally, n case m, t, are fxe we erve the followng. Lemma 1. For any fxe m, t, an n large enough, such that m 3 an t, N t(m, = Θ(n t. Proof: Snce N t (m, N t (, an N t (, = Θ(n t, then N t (m, s at most O(n t. To show the other recton of ths equalty, we show an example of a set X such that the carnalty of S t (X s O(n t. Let e j be the vector whch ts l-th bt s one f an only f l {,..., j}. For 1 m, let 0 = ( an 1 =, an x = e 1 0. Then, for all = j, H (x, x j =. For any vector y of weght at most t, such that ts frst m bts are zero, y S t (X. Snce there are t such vectors we l=0 ( n m l get that for n large enough, N t (m, s at least O(n t. Together we conclue that N t (m, = Θ(n t. V. SEQUENCES RECONSTRUCTION DECODERS The man goal n [11] was to fn the necessary an suffcent number of channels n orer to have a successful ecoer for a coe wth mnmum stance. Ths number was stue for fferent error moels n [7] [14], however the only ecoer constructons, whch we are aware of, were gven n [], [5], [15] for channels wth nserton an eletons, an n [4] for eletons only. In ths secton, we show how to construct ecoers for substtuton errors, where the ecoer has to output the transmtte wor (an not a lst of wors. The case = 1 was solve n [11] where the majorty algorthm on each bt successfully ecoes the transmtte wor. Accorng to Lemma, ths algorthm wors for = as well snce the number of channels has to be the same. However, f s greater than two, then the majorty algorthm on each bt no longer wors. In general, accorng to Lemma, f s even then the number of channels for a coe wth mnmum stance or 1 s the same. Hence, we only nee to solve here the case of o mnmum stance. For the rest of ths secton, we assume that the transmtte wor belongs to a coe C wth o mnmum stance, there are at most t > 1 errors n every channel, an the number of channels s N = N t (, + 1. The N channel outputs are enote by y 1. Furthermore, the coe C has a ecoer D C, whch can successfully correct 1 errors. 109

5 A frst observaton n constructng a ecoer s that we can always etect whether the output wor s the transmtte one. Ths can smply be one by checng f the maxmum stance from all channel outputs s at most t. Lemma 13. For any ĉ C, ĉ = c f an only f max { H(ĉ, y } t. 1 N Proof: If ĉ = c then every channel suffers at most t errors an thus max 1 N { H (ĉ, y } t. In case ĉ = c, let us assume to the contrary that max 1 N { H (ĉ, y } t. Then the set S t ({ĉ, c} contans at least N = N t (, + 1 wors n contracton to the efnton of N t (,. A nave algorthm can choose any of the channel outputs an a all error vectors of weght at most t 1. For one of these error vectors we wll get a wor wth at most 1 errors whch can be ecoe by the ecoer of the coe C. We wll show how to mofy an mprove the complexty of ths algorthm. Assume for example that t = Then, there are two channel outputs, say y 1 an y, that are fferent n at least one bt locaton. If we flp ths bt n both y 1 an y, then n exactly one of them the number of errors reuces by one an thus s at most 1, whch can be ecoe by D C. We show how to generalze ths ea for arbtrary t. We let ρ = t 1. Frst, we prove the followng Lemma. Lemma 14. There exst two channel outputs y such that H (y ρ 1. Proof: Assume to the contrary that there o no exst such wors. Then, the wors y 1, y form an antcoe of ameter ρ. Accorng to [6], the maxmum sze of such an antcoe s b ρ 1,n = ρ 1, whle accorng to (5 the =0 (n value of N satsfes t +1 ( n t N > N t (, = =0 = t+ ρ 1 ( n t ( = > b ρ 1,n. =0 = t+ ( Algorthm 15. The nput to the ecoer are the N wors y 1 an t returns an estmaton ĉ on c. Step 1. Fn two wors y such H (y ρ 1, an let 1,,..., ρ 1 be ρ 1 fferent nces that the two vectors are fferent from each other. Step. For all vectors e of weght ρ on these ρ 1 nces, a D(y 1 + e = ĉ 1, D(y + e = ĉ. b If max 1 N { H (ĉ 1, y } t, ĉ = ĉ 1. c If max 1 N { H (ĉ, y } t, ĉ = ĉ. Theorem 16. The output of Algorthm 15 satsfes ĉ = c. Proof: The success of Step 1 s guarantee accorng to Lemma 14. For every nex j, 1 j ρ 1, exactly one of the channel outputs y 1 or y has an error. Therefore, ether y 1 or y has at least ρ errors on these nces. Wthout loss of generalty assume t s y 1 an let I { 1,..., ρ 1 } be a subset of ts error locatons, where I = ρ. In Step we exhaustvely search over all error vectors e of weght ρ on these ρ 1 nces. For every error vector e let I e = { : e = 1}. Therefore, there exsts an error vector e 1 such that ts the set of nces wth value one s covere by the set I,.e. I e1 I. Hence, H (c, y 1 + e 1 1, so the ecoer n Step.b succees. Hence the algorthm succees an ĉ = c. The complexty of Algorthm 15 s sgnfcantly better than the nave approach. However, the larger the value of ρ s, the larger the algorthm s complexty s. We report on another algorthm wth better complexty, O(nN, for the case = 3. VI. CONCLUSION Ths paper propose a moel of an assocatve memory: Two wors are assocate f ther Hammng stance s no greater than some prescrbe value t. Our man goal was to stuy the maxmum sze of the assocatve memory output as a functon of the number of nput wors an ther mnmum stance. We observe that ths problem s a generalzaton of the sequences reconstructon problem that was propose by Levenshten. Fnally, we presente a econg algorthm for the sequences reconstructon problem. VII. ACKNOWLEDGEMENT The authors than Tuv Etzon for helpful scussons on antcoes. Ths research was supporte n part by the ISEF Founaton, the Lester Deutsch Fellowshp, an the NSF Expetons n Computng Program uner grant CCF REFERENCES [1] R. Ahlswee, H.K. Aynan, an L.H. Khachatran, On perfect coes an relate concepts, Desgns, Coes an Cryptography, vol., pp. 1 37, 001. [] T. Batu, S. Kannan, S. Khanna, an A. 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