Two Conjectures About Recency Rank Encoding

Transcription

1 Internatonal Journal of Matheatcs and Coputer Scence, 0(205, no. 2, M CS Two Conjectures About Recency Rank Encodng Chrs Buhse, Peter Johnson, Wlla Lnz 2, Matthew Spson 3 Departent of Matheatcs and Statstcs Auburn Unversty Auburn, AL 36849, USA 2 Departent of Matheatcs Texas A&M Unversty College Staton, TX , USA 3 Departent of Matheatcs North Carolna State Unversty Ralegh, North Carolna 27695, USA eal: johnspd@auburn.edu, wlldoath@tau.edu, csps2@ncsu.edu (Receved July 22, 205, Accepted August 5, 205 Abstract In recency rank encodng, a source letter s s replaced by a code word u j, where j s the nuber of dfferent source letters that have appeared snce the last occurrence of s n the source text. Let be the nuber of source letters and suppose that the source s perfectly zeroth order, eanng that the source letters enter the text ndependently wth fxed probabltes f,...,f.fork {0,..., }, let g k = g k (f,...,f be the probablty that, f a source letter s chosen ndependently fro the source text, exactly k dfferent letters other than that letter have appeared snce the last appearance of that letter. The conjectures of the ttle are: ( g 0... g wth equalty at any pont (g j = g j+ fand only f f =...= f = ; Key words and phrases: Huffan s algorth, replaceent codng, source codng, source frequences, zeroth order source. AMS (MOS Subject Classfcatons: 68P30, 94A29, 26D5, 60E5, 60C05, 05A20, 05E05 ISSN , 205, Ths research was partally supported by NSF (DMS grant nuber

2 76 C. Buhse, P. Johnson, W. Lnz, M. Spson (2 f the code words u 0,...,u for recency rank encodng are obtaned by applyng Huffan s algorth to the probabltes g 0,...,g, and the code words w,...,w are obtaned by applyng Huffan s algorth to the probabltes f,...,f then lgth(w j g j lgth(u j. j=0 Recency rank vs. sple replaceent encodng Let S = {s,...,s } and A = {a,...,a n },, n 2, be, respectvely, a source alphabet and a code alphabet. In the odern proble of encodng, we wsh to represent words over S by words over A n such a way that every source word s recoverable fro ts code representatve. Subordnate to ths requreent of unque decodablty are varous optzaton goals: effcency of encodng, effcency of decodng, copresson, error detecton and correcton. See [2] and [3]. The goal of copresson s to nze the average nuber of code letters per source letter n the encodng; we ght call ths average the copresson ndex. In sple replaceent encodng, each source letter s j s assgned a code word w j, and the encodng proceeds by replacng each occurrence of s j n the source text by w j. Thus, the code representatve of source text s s 2...s N wll be the concatenaton w w 2...w N. For unque and effcent decodablty, the lst w,...,w of code words s prefx-free (see [2]. For copresson, coon sense whspers that the w j should be as short as possble, gven the requreent that the lst of w j be prefx-free. If nothng s known about the source text, we ay as well ake the w j as nearly equal n length and as short as possble, so that lgth(w j { log n, log n } for each j =,...,. If the relatve frequences n the source text of s,...,s are known let s call the f,...,f ( s the probablty that a letter plucked at rando fro the source text wll be s j then one can apply Huffan s algorth [2] to the relatve frequences to obtan a prefx-free lst w,...,w for sple replaceent encodng whch nzes the copresson ndex lgth(w j.

3 Two Conjectures About Recency Rank Encodng 77 In recency rank encodng, ntroduced by Elas [] n 987 (although he credts others for ndependently havng the dea, we use a prefx-free lst u 0,...,u of code words; an occurrence of s S s replaced by u k when exactly k eleents of S \{s} have appeared n the source text snce the last occurrence of s. (In theory, the source text has no begnnng. In practce, every actual chunk of source text has a begnnng, so there wll have to be soe conventon for gettng started, n actual applcaton of recency rank encodng. Decodng s unque and not terrbly hard, although t s a lttle ore te-consung than sply recognzng code words, as n sple replaceent. Recency rank encodng was touted as an effectve on-lne ethod, requrng no knowledge of the statstcs of the source text, but only the nuber of source letters. However, t does requre user agreeent on the code words u 0,...,u and, f nothng s known about the source text, t sees sensble to ake the u j all of length around log n. What, then, s the advantage of recency rank encodng over sple replaceent of the source letters s j wth code words w j, of the lengths also around log n, especally n vew of the fact that decodng of sple replaceent code s notceably easer than recency rank decodng? Perhaps because of such consderatons, recency rank codng has not survved as a practcal ethod n dgtal councaton. However, as a source of nterestng atheatcal questons, recency rank codng s a rch, htherto untapped (so far as we know resource. Our a here s to pose a couple of atheatcal questons about recency rank codng, n the for of conjectures, and to confr part of one of these conjectures. Let the source be perfectly zeroth order; ths eans that letters enter the source text ndependently, as though they were drawn wth replaceent fro an urn, n whch s,...,s occur n proportons f,...,f, respectvely. (But note that the are not requred to be ratonal. For text fro such a source, f a block of k consecutve letters s chosen fro the source text at rando, then for any,..., k {,...,}, the probablty that the block chosen wll be s...s k s the product k f j. Text fro such a source s the sae, statstcally, whether read for-

4 78 C. Buhse, P. Johnson, W. Lnz, M. Spson ward or backwards. Therefore, we can defne the probabltes of nterest n analyzng recency rank codng as follows. For k {0,,..., }, let g k = g k (f,...,f be the probablty that, f a letter s s selected at rando fro the source text, exactly k dfferent letters of S \{s} wll appear n the source text (readng forward before the next occurrence of s. Thus, g 0 = f 2 g = j f f t j f = t= f 2 j and the expressons for g 2,...,g as forulas n f,...,f are a bt ore coplcated. For nstance: g 2 = = f j <j 2 j j 2 j <j 2 j j 2 2 f [ ( + 2 k k=2 f k j k=2 ] f k j 2 f k=2 [ ( f 2 j f ] 2 j In general, for 2 k, the forula for g k s g k = f 2 Q {,...,}\{} Q =k ( k j Q j Q + k t= ( k t S Q S =t ( k j S j S

5 Two Conjectures About Recency Rank Encodng 79 Conjecture.. If 2, f,...,f > 0, and f =, then for all k {,..., }, g k (f,...,f g k (f,...,f, wth equalty f and only f f =...= f = Conjecture.2. Suppose f,...,f are as n Conjecture and that g k = g k (f,...,f, k =0,...,. Suppose that the applcaton of Huffan s algorth to f,...,f resultsncodewordsw,...,w, and an applcaton of Huffan s algorth to g 0,...,g results n code words u 0,...,u. Then f lgth(w g k lgth(u k k=0 In Conjecture.2, lgth stands for length, eanng the nuber of code letters n the word. The conjecture s that for a zeroth order source, the best possble copresson ndex achevable by recency rank encodng, by a shrewd choce of code words, s no better (saller than the best possble copresson ndex achevable by sple replaceent. We do not have a good guess about condtons for equalty n Conjecture.2, although, f Conjecture. holds, then f =... = f = ples g 0 =...= g = (because the g ust su to, whch ples equalty n Conjecture.2. It s straghtforward to see that g 0 (,..., =. Therefore, f the nequaltes g 0... g hold for all f,...,f,thenf =... = f = ples that g 0 =...= g =. We do not have strong reasons for these conjectures. They have wthstood testng for sall values of. In secton 3, we wll prove that g 0 g wth equalty f and only f f =... = f =. In secton 2, we wll develop the analyss to be used n the proof n secton 3; soe ay fnd ths analyss of nterest n tself. 2 Useful nequaltes Lea 2.. If α, β, γ, δ R, α β and γ δ, thenαγ + βδ αδ + βγ wth equalty f and only f ether α = β or γ = δ.

6 80 C. Buhse, P. Johnson, W. Lnz, M. Spson Proof. The proposed nequalty s equvalent to (α β(γ δ 0, whch obvously follows fro the hypotheses. The suffcency and necessty of the condtons for equalty also follow fro ths equvalence. Theore 2.2. Suppose that 2, a... ( a > 0, andb... ( b > 0. Then the functon h defned by h(x = a x j b j s strctly ncreasng on [0,, unlessethera =... = a or b =... = b. Clearly, f ether a =...= a or b =...= b,thenh s constant. Proof. Suppose that y>x 0. We a to show that h(y h(x andthat equalty ples that ether a =...= a or b =...= b. Each of the followng after the frst s clearly equvalent to the nequalty precedng: a x ( h(y h(x; ( ( (2 a y j b j a x ( a x j b j ; a y (3 (4 (5 <j <j a x+y b +,j j ( a x a y j b j + a x j a y b (a a j x [ a y x j a x a y j b j ] b j +a y x b <j a x+y b +,j j (a y axj b j + a yj ax b ; <j (a a j x [ a y x a y ax j b j ; b j +a y x j b ]. Now, y x>0anda a j for <j ples that a y x Therefore, by Lea 2., wth α = a y x, β = a y x each par, j such that <j, a y x a y x j. j, γ = b, δ = b j, for b + a y x j b j a y x b j + a y x j b,

7 Two Conjectures About Recency Rank Encodng 8 wth equalty f and only f ether a y x = a y x j ( a = a j orb = b j. Therefore, nequalty (5 holds because each ter on the left s greater than or equal to the correspondng ter (ndexed by (, j on the rght. Therefore, equalty holds f and only f t holds for each ter; consequently, f equalty holds, then for each par (, j, <j, ethera = a j or b = b j.ths ples that, f equalty holds n (, and thus n (5, then ether a =...= a or b =... = b. For (assung equalty holds f the a are not all equal, then a >a, whch ples b = b, whence b =...= b. Although we wll ake no use of t here, t would be churlsh of us not to pont out the followng corollary. Corollary 2.3. Suppose that k, 2 and that A =[a j ] s a k atrx of postve real nubers such that each row s non-constant and non-ncreasng. Then the functon H : { (x,...,x k R k x > 0, =,...,k } (0, defned by H(x,...,x k = [ k ( a x j ] [ ( k s strctly ncreasng n each varable x, =,...,n. Proof. If k of the varables wthout loss of generalty, say x 2,...,x k are fxed, then the resultng functon of the reanng varable, x,s where c = p(x =c a x j ( k a x j and b j = =2 ( a x j b j, k =2 a x j ] a x j, j =,...,. Snce a... a and b... b, and nether fnte sequence s constant by assuptons about A, wehavep(x =c h(x, wth h beng of the for gven n Theore 2.2, strctly ncreasng, and c>0.

8 82 C. Buhse, P. Johnson, W. Lnz, M. Spson Both Theore 2.2 and Corollary 2.3 have generalzatons to arbtrary fnte, postve easure spaces. We wll gve, wthout proof, the generalzaton of Theore 2.2. Theore 2.4. Suppose that (M,μ s a postve easure space, 0 <μ(m < and a, b : M (0, are easurable functons such that for all s, t M, a(s a(t f and only f b(s b(t. Thenh :[0, (0,, defnedby ( h(x = M ( a(t x dμ(t a(t x b(tdμ(t M s strctly ncreasng on [0, unless one of a, b s essentally constant. Theore 2.2 s the specal case of Theore 2.4 n whch M = {,...,} and μ s the countng easure. 3 g 0 g, and a necessary and suffcent condton for equalty Theore 3.. Suppose that 2, f,...,f > 0, and =. Then g 0 (f,...,f g (f,...,f, wth equalty f and only f f =... = f =. Proof. We ay as well suppose that >f... f > 0, whch ples that f... f > 0. Fro Secton, g 0 (f,...,f = f 2 and

9 Two Conjectures About Recency Rank Encodng 83 g (f,...,f = f 2 j [( ] f = f 2 j f f j f f = f 2 j f 2 Therefore, the nequalty g 0 g s equvalent to ( f 2 f 2 f or f f. f 2 f 2 j = f,snce f =. Settng f = a and f = b and referrng to Theore 2.2, the left-hand sde of the nequalty above s h(2, and the rght-hand sde s h(. By Theore 2.2, therefore, we have g 0 g, wth equalty only f (and also f ether a =...= a or b =...= b ; each s equvalent to f =...= f =.

10 84 C. Buhse, P. Johnson, W. Lnz, M. Spson References [] Peter Elas, Interval and recency rank source codng, two on-lne adaptve varable-length schees, IEEE Transactons n Inforaton Theory, 33, no., 987, 3-0. [2] Darrel Hankerson, Greg Harrs, Peter Johnson, Introducton to Inforaton Theory and Data Copresson, 2nd edton, Chapan & Hall/CRC, Boca Raton, [3] D. C. Hankerson, D. G. Hoffan, D. A. Leonard, Charles C. Lndner, K.T. Phelps, C.A. Rodger, J. R. Wall, Codng Theory and Cryptography: The Essentals, 2nd edton, Taylor & Francs, 2000.