On Time-Bounded Incompressibility of Compressible Strings and Sequences

Time-Bounde d Incompress ibilit y of Compress ible Strin g s and Sequence s Edgar G. Da ylight a , 1 W outer M. Ko o len b P aul M.B. Vit´ an yi b , c a University of A mster dam, Institute of L o gic, L anguage, and Computation ,Amster da m, Th e Netherland s b Centrum vo or Wiskunde en Informatic a, Scienc e Park 123, 1098 X G Amster da m, The N etherlands c University of Amster dam, De p artment of Computer Scienc e, Amster da m, The Netherlands Abstract F or ev ery total recursive t ime b ound t , a constan t fraction of all compressible (lo w Kolmogoro v complexit y) strings is t -boun ded incompressible (high t ime-b ounded Kolmogoro v complexit y); there are uncoun tably man y infin ite s equ ences of wh ic h ev ery initial s egment of length n is compressible to log n y et t -b ounded incom- pressible b elo w 1 4 n − log n ; and th er e are a countably infinite num b er of recursive infinite sequences of which ev ery initial segment is similarly t -b ound ed incompress- ible. These results and their pro ofs are related to, b u t d ifferen t from, Barzdins’s lemma. Key wor ds: Kolmogoro v complexit y, compressibilit y, time-b ounded incompressibilit y, Barzdins’s lemma, finite strings and infinite sequences, computational complexit y 1 In tro duction Informally , the Kolmogo r o v complexit y of a finite binary string is the length of t he shortest string fr om whic h t he origina l can b e losslessly reconstructed b y an effectiv e general-purp ose computer suc h as a particular univ ersal T uring Email addr esses: e gdaylight @yahoo.c om (Edgar G. Da yligh t), W.M.Kool en-Wijkst ra@cwi.nl (W outer M. Ko olen), P aul.Vitan yi@cwi.nl (P aul M.B. Vit´ an yi). 1 a.k.a. Karel v an Oudh eu sden Preprint submitted to Elsevier 29 O ctob er 2018 mac hine U . Hence it constitutes a low er b ound on how far a lo ssless compres- sion pr o gram can compress. F ormally , the c onditional Kolmo gor ov c omplexity C ( x | y ) is the length of the shortest input z suc h that the un iv ersal T uring mac hine U on input z with auxiliary information y outputs x . The unc ondi - tional Kolmo gor ov c om p lexity C ( x ) is defined b y C ( x | ǫ ) where ǫ is the empty string (of length 0). Let t b e a total recursiv e function. Then, the time-b ounde d c onditional Kolmo go r ov c omplexity C t ( x | y ) is the length of the shortest input z such that the univ ersal T uring mac hine U on input z with auxiliary in- formation y outputs x within t ( n ) steps where n is the length in bits of x . The time -b ounde d unc onditional Kolmo gor ov c omplexity C t ( x ) is defined b y C t ( x | ǫ ). F or an in tro duction to the definitions and notions of Kolmog oro v complexit y (algorithmic information theory) see [3]. 1.1 R elate d Work Already in 1968 J. Barzdins [2] obtained a result known as B arzdins’s lemm a , probably the first result in resource-b ounded Kolmogorov complexit y , of whic h the lemma b elo w quotes the ite ms that a re relev an t here. Let χ denote the c haracteristic sequence of an a r bitrary recursiv ely en umerable (r.e.) subset A of the natural num b ers. That is, χ is an infinite sequence χ 1 χ 2 . . . where bit χ i equals 1 if a nd only if i ∈ A . Let χ 1: n denote the first n bits of χ , and let C ( χ 1: n | n ) denote the conditional Kolmogorov complexit y o f χ 1: n , giv en the n um b er n . Lemma 1 (i) F or every cha r acteristic se quenc e χ of a r.e. set A ther e exists a c onstant c such that fo r al l n we h ave C ( χ 1: n | n ) ≤ log n + c . (ii) Ther e exists a r.e. set A with c h ar acteristic se quenc e χ such that for eve ry total r e cursive function t ther e is a c onstant c t with 0 < c t < 1 such that for al l n we have C t ( χ 1: n | n ) ≥ c t n . Barzdins actually prov ed this statemen t in terms of D.W. Lo v eland’s v ersion of Kolmog o ro v complexit y [4], whic h is a sligh tly different setting. He also pro v ed that there is a r.e. set suc h that its c ha racteristic sequence χ = χ 1 χ 2 . . . satisfies C ( χ 1: n ) ≥ log n for ev ery n . Kummer [5 ], The orem 3.1, solving the op en problem in Exercise 2.59 of the first edition of [3] prov ed that there exists a r.e. set suc h that its c ha r a cteristic sequence ζ = ζ 1 , ζ 2 , . . . satisfies C ( ζ 1: n ) ≥ 2 log n − c f or some constan t c and infinitely man y n . The conv erse of item (i) do es not hold. T o see this, consider a sequenc e χ = χ 1 χ 2 . . . and a constant c ′ ≥ 2, suc h that for ev ery n w e ha v e C ( χ 1: n | n ) ≥ n − c ′ log n By item (i), χ can not b e the c hara cteristic sequence of a r.e. set. T ra nsform χ in to a new sequence ζ = χ 1 α 1 χ 2 α 2 . . . with α i = 0 2 i , a string of 0s of length 2 i . While ob viously ζ can not b e the c haracteristic sequence of a 2 r.e. set, there is a constant c suc h that for eve ry n we hav e that C ( ζ 1: n | n ) ≤ log n + c . Item (i) is easy to prov e a nd item (ii) is hard to prov e. Putting it ems (i) and (ii) together, there is a c haracteristic sequence χ of a r .e. set A whose initial segmen ts are b o t h logarithmic compressible and time-b ounded lin- early incompress ible, for ev ery to tal recursiv e time b ound. Belo w, w e iden- tify the natural n um b ers with finite binary strings a ccording to the pairing ( ǫ, 0) , (0 , 1) , (1 , 2) , ( 00 , 3) , (01 , 4) , . . . , where ǫ again denotes the empty string. 1.2 Pr esent R esults Theorem 1 L et k 0 , k 1 b e p o sitive in te ger c onstants and t a total r e c ursive function. (i) A c onstant fr action o f al l s trings x of len g th n with C ( x | n ) ≤ k 0 log n satisfies C t ( x | n ) ≥ n − k 1 . ( L emma 2 ) . (ii) L et t ( n ) ≥ cn for c > 1 sufficiently lar ge. A c onstant fr action of al l strings x of length n with C ( x | n ) ≤ k 0 log n satisfies C t ( x | n ) ≤ k 0 log n ( L emma 3 ) . (iii) Ther e exist unc ountably many ( a c tual l y 2 ℵ 0 ) in finite b i n ary se quenc es ω such that C ( ω 1: n | n ) ≤ log n and C t ( ω 1: n | n ) ≥ 1 4 n − log n for every n ; mor e over, ther e exist a c o untably in fi nite numb er of ( that is ℵ 0 ) r e cursive infinite binary se quenc es ω ( henc e C ( ω 1: n | n ) = O ( 1 )) such that C t ( ω 1: n | n ) ≥ 1 4 n − log n for every n ( L emma 5 ) . Note that the order of quan tification in Barzdins’s lemma is “there exists a r.e. set such that f or every total recursiv e f unction t there exists a constan t c t .” In contrast, in it em (iii) w e prov e “ there is a p ositiv e constan t suc h that for ev ery total r ecursiv e function t t here is a sequence ω .” While Ba rzdins’s lemma prov es the existence of a single c haracteristic sequence of a r.e. set that is time-limited linearly incompress ible, in item (iii) w e pro v e the exis- tence of uncoun tably man y sequenc es that are logarithmically compre ssible o v er the initial segmen ts, and the existence of a coun tably infinite num b er of recursiv e seque nces, suc h that all t ho se sequen ces ar e time-limited linearly incompressible. W e generalize item (i) in Corollaries 1 and 2. Section 2 presen ts preliminaries. Section 3 g iv es the results on finite strings. Section 4 give s the results on infinite sequence s. Finally , conclusions are presen ted in Section 5. The pro of s for t he results are differen t fro m Barzdins’s pro o f s. 3 2 Preliminaries A (binary) program is a concatenation of instructions, and an instruction is merely a string. Hence, w e ma y view a program as a string. A program and a T uring machin e (or mac hine for short) are used synon ymously . The length in bits of a string x is denoted b y | x | . If m is a natural n um ber, then | m | is the length in bits of the m th binary string in length-increasing lexicographic order, starting with the empt y string ǫ . W e also use the notation | S | to denote the car dina lity o f a set S . Consider a standard en umeration o f all T uring machines T 1 , T 2 , . . . . Let U denote a univ ersal T uring mach ine suc h that f o r ev ery y ∈ { 0 , 1 } ∗ and i ≥ 1 w e hav e U ( i, y ) = T i ( y ). That is, fo r all finite binary strings y and ev ery mac hine index i ≥ 1, w e ha ve that U ’s execution on inputs i and y results in the same output as that obta ined by executing T i on input y . Let t b e a total recursiv e function. Fix U and define that C ( x | y ) e quals min { | p | : p ∈ { 0 , 1 } ∗ and U ( p, y ) = x } . F o r t he same fixed U , define that C t ( x | y ) eq uals min {| p | : p ∈ { 0 , 1 } ∗ and U ( p, y ) = x in t ( | x | ) steps } . (By definition the sets o v er whic h is minimized are coun ta ble and not empt y). 3 Finite Strings Lemma 2 L et k 0 , k 1 b e p ositive inte ger c on stants and t b e a total r e cursive function. Ther e is a p ositive c o n stant c t such that for sufficiently lar ge n the strings x of length n satisfying C t ( x | n ) ≥ n − k 1 form a c t -fr action o f the strings y of length n satisfying C ( y | n ) ≤ k 0 log n . Pro of. The pro o f is b y diagonalization. W e use the following algorithm with inputs t, n, k 1 and a natural num b er m . Algorithm A ( t, n, k 1 , m ) Step 1. Using the univ ersal reference T uring ma chine U , recursiv ely enume r- ate a finite list of a ll binary programs p of length | p | < n − k 1 . There are at most 2 n / 2 k 1 − 1 suc h programs. Execute eac h of these prog r ams o n input n . Consider the set of all programs that halt within t ( n ) steps and whic h o utput precisely n bits. Call t he set of these o utputs B . Note that | B | ≤ 2 n / 2 k 1 − 1 and it can b e computed in time O (2 n t ( n ) / 2 k 1 ). Step 2. Output the ( m + 1)th string of length n , say x , in the lexicographic order of all strings in { 0 , 1 } n \ B and halt. If there is no suc h string then halt with o utput ⊥ . End of Algorithm 4 Because of the selection pro cess in Step 1 , |{ 0 , 1 } n \ B | ≥ 2 n − 2 n / 2 k 1 + 1 a nd ev ery x ∈ { 0 , 1 } n \ B has time-b ounded complexit y C t ( x | n ) ≥ n − k 1 . (1) F or | m | ≤ k 0 log n − c , where the constan t c is defined b elow , and provide d { 0 , 1 } n \ B is sufficien tly large, that is, n k 0 / 2 c ≤ 2 n  1 − 1 2 k 1  + 1 , (2) there are at least n k 0 / 2 c strings x of length n that will b e output b y the algorithm. Call this set D . Eac h string x ∈ D satisfies C ( x | t, n, k 1 , A , p ) ≤ | m | ≤ k 0 log n − c. (3) Since w e can describ e the fixed t, k 0 , k 1 , A , a program p to reconstruct x from these data, and the means to tell them apart, in a n additional constant n um- b er of bits, sa y c bits (in t his w ay the quan tity c can be deduced from the conditional), it follow s that C ( x | n ) ≤ k 0 log n . F or giv en k 0 , k 1 , and c , inequal- it y (2) holds for ev ery s ufficien tly large n . F or suc h suffic ien tly large n , the cardinalit y of the set of strings of length n satisfying b oth C ( x | n ) ≤ k 0 log n and C t ( x | n ) ≥ n − k 1 is at least | D | = n k 0 / 2 c . Since the n um ber of strings x of length n satisfying C ( x | n ) ≤ k 0 log n is at most P k 0 log n i =0 2 i < 2 n k 0 , the lemma follo ws with c t = 1 / 2 c +1 . ✷ Cor ollar y 1 Let k 0 b e a p o sitiv e in t eger constant and t b e a total recursiv e function. F or ev ery sufficien tly large natural num b er n , the set of strings x of length n such that C t ( x | n ) 6≤ k 0 log n is a p ositiv e constan t fraction of the strings y of length n satisfying C ( y | n ) ≤ k 0 log n . W e can generalize Lemma 2. Let t b e a total recursiv e function, and f , g b e total recursiv e functions such that (4) b elo w is satisfied. Cor ollar y 2 F o r ev ery sufficien tly larg e natural n umber n , the set of strings x of length n t ha t satisfy b o th C ( x | n ) ≤ f ( n ) and C t ( x | n ) ≥ g ( n ) is a p ositive constan t fra ction of the strings y of length n satisfying C ( y | n ) ≤ f ( n ). Pro of. Use a similar algorithm A ( t, n, g , m ) with | p | < g ( n ) in Step 1, and | m | ≤ f ( n ) − c in the analysis. Require 2 f ( n ) − c ≤ 2 n − 2 g ( n ) + 1 . (4) ✷ 5 Lemma 3 L et t b e a total r e cursive function with t ( n ) ≥ cn for some c > 1 and k 0 b e a p ositive in te ger c o n stant. F or every sufficiently lar ge n a tur a l numb er n , ther e is a p os i tive c onstant c t such that the set of strings x of leng th n satisfying C t ( x | n ) ≤ k 0 log n is a c t -fr action of the s et of s tring s y of le ngth n satisfying C ( y | n ) ≤ k 0 log n . Pro of. W e use the follo wing algorithm that take s p ositive in tegers n, m a s inputs and computes a string x of length n satisfying C t ( x | n ) ≤ k 0 log n − c . Algorithm B ( n, m ) Output the string 0 n −| m +1 | ( m + 1) (where | m + 1 | is the length of the string represen tation of m + 1) and halt. E nd of Algorithm Let k 0 b e a p ostiv e in teger and c a positiv e in teger constan t c hosen be- lo w. Consider strings x that are o utput b y algorithm B and t ha t satisfy C t ( x | n, B , p ) ≤ | m | ≤ k 0 log n − c with c the n um b er of bits to con tain de- scriptions of B and k 0 , a program p to reconstruct x fro m these da t a , a nd the means to tell the constituent items apart. Hence, C t ( x | n ) ≤ k 0 log n . The r un- ning t ime of algorithm B is t ( n ) = O ( n ), since the output strings are length n and to output the m th string with m ≤ 2 k 0 log n − c w e simply tak e the binary represen tation of m and pad it with nonsignificant 0s to length n . Ob viously , the strings tha t satisfy C t ( x | n ) ≤ k 0 log n ar e a subset of the strings that sat- isfy C ( x | n ) ≤ k 0 log n . There are at least n k 0 / 2 c strings of the first kind while there a r e at mo st 2 n k 0 strings o f the second kind. Setting c t = 1 / 2 c +1 finishes the pro of. ✷ It is we ll know n t ha t if we flip a f a ir coin n times, that is, giv en n random bits, then w e obtain a string x of length n with Kolmogorov complexit y C ( x | n ) ≥ n − c with pro ba bilit y at least 1 − 2 − c . Such a string x is algorit hmically random. W e can also get by with less random bits to obtain resource-b ounded algorithmic randomness from compressible strings. Lemma 4 L et a, b b e c ons tants a s in the pr o of b elow. Given the se t of strings x of length n satisfying C ( x | n ) ≤ k 0 log n , a total r e cursive function t , the c onstant k 1 as b efor e, and O ( ab log n ) fa ir c oin flips, we obtain a set o f O ( ab ) strings o f le n gth n such that with p r ob ability at le ast 1 − 1 / 2 b one string x in this set satisfies C t ( x | n ) ≥ n − k 1 . Pro of. By L emma 2, a c t th fraction of the set A of strings x of length n that ha v e C ( x | n ) ≤ k 0 log n also ha v e C t ( x | n ) ≥ n − k 1 . Therefore, by c ho osing, uniformly at random, a constan t n umber a of strings from the set A we increase (e.g. b y means of a Chernoff b ound [3 ]) the pro babilit y that (at least) one of those strings cannot b e compressed b elo w n − k 1 in time t ( n ) to at least 1 2 . T o c ho ose an y one string from A requires O (log n ) random bits b y dividing A in 6 t w o equal size parts and r ep eating this with the c hosen half , and so o n. The selected a elemen ts take O ( a log n ) random bits. Applying the previous step b times, the pro babilit y that at least one of the ab c hosen strings cannot b e compressed b elo w n − k 1 bits in time t ( n ) is a t least 1 − 1 / 2 b . ✷ 4 F rom Finite Strings to Infinite Sequences W e pro v e a result reminiscen t o f Barzdins’s lemma, L emma 1. In Barzdins’s v ersion, characteristic sequenc es ω of r.e. sets a re considered whic h b y Lemma 1 ha v e complexit y C ( ω 1: n | n ) ≤ log n + c . Here, we consider a wider class of se- quences of whic h the initial segmen ts are logarithmically compressible (suc h sequence s are not neces sarily c haracteristic sequences of r.e. sets as explained in Section 1.1). Lemma 5 L et t b e a total r e cursive function. (i) Ther e ar e unc ountably ma n y ( actual ly 2 ℵ 0 ) se quenc es ω = ω 1 ω 2 . . . such that b oth C ( ω 1: n | n ) ≤ log n and C t ( ω 1: n | n ) ≥ 1 4 n − log n for every n . (ii) The set in item (i) c ontains a c ountably infinite numb er of ( that is ℵ 0 ) r e cursive se quenc es ω = ω 1 ω 2 . . . such that C t ( ω 1: n | n ) ≥ 1 4 n − log n for every n . Pro of. (i) Let g ( n ) = 1 2 n − log n . Let c ≥ 2 b e a constant to b e c ho sen later, m i = c 2 i , B ( i ) , C ( i ) , D ( i ) ⊆ { 0 , 1 } m i for i = 0 , 1 , . . . , and C ( − 1) = { ǫ } . The C sets are constructed so that they contain the target strings in the form of a binary tree, where C ( i ) contains all t a rget strings of length m i . The B ( i ) sets corresp ond to fo rbidden prefixes of length m i . The D ( i ) sets consist of the set of strings o f length m i with prefixes in C ( i − 1) fr om which the strings in C ( i ) are selected . Algorithm C ( t, g ): for i := 0 , 1 , . . . do Step 1. Using the univ ersal reference T uring ma chine U , recursiv ely enume r- ate the finite list of a ll binary progr a ms p of length | p | < g ( m i ) with m i = c 2 i and the constant c defined b elo w. There a re at mo st 2 g ( m i ) − 1 suc h prog rams. Execute eac h of these prog r a ms o n all inputs m i + j with 0 ≤ j < m i . Con- sider the set o f all programs with input m i + j that halt with output x = y z within t ( | x | ) time with | x | = m i + j , y ∈ C ( i − 1) (then | y | = m i − 1 for i > 0 and | y | = 0 for i = 0), and z is a binary string suc h tha t x s atisfies m i ≤ | x | < m i +1 . There are a t most m i (2 g ( m i ) − 1) suc h x ’s. Let B ( i ) b e the set of the m i -length prefixes of these x ’s. Then, | B ( i ) | ≤ m i (2 g ( m i ) − 1) and it 7 can b e computed in time O ( m i 2 g ( m i ) t ( m i +1 )). Note that if u ∈ { 0 , 1 } m i \ B ( i ) then C t ( uw | | uw | ) ≥ g ( | u | ) for ev ery w suc h that | u w | < m i +1 . Step 2. Let C ( i − 1 ) = { x 1 , x 2 , . . . , x h } and D ( i ) = ( C ( i − 1) { 0 , 1 } ∗ T { 0 , 1 } m i ) \ B ( i ). for l := 1 , . . . , h do for k := 0 , 1 do put the k th string with initial segmen t x l , in the lexicographic order of D ( i ), in C ( i ). If there is no such string then halt with output ⊥ . o d o d o d E nd of A lgorithm Clearly , C ( i ) { 0 , 1 } ∗ ⊆ C ( i − 1) { 0 , 1 } ∗ for eve ry i = 0 , 1 , . . . . Therefore, if ∞ \ i =0 C ( i ) { 0 , 1 } ∞ 6 = ∅ , (5) then the elemen ts of this in tersection constitute the infinite sequences ω in the statemen t of the lemma. Claim 1 With g ( m i ) = 1 2 m i − log m i , w e ha v e | C ( i ) | = 2 i +1 for i = 0 , 1 , . . . . . Pro of. The pro of is b y induction. Recall that m i = c 2 i with the constan t c ≥ 2. Base c a s e : | C (0) | = 2 since C ( − 1) = { ǫ } and | D (0) | ≥ 2 m 0 − m 0 (2 g ( m 0 ) − 1) ≥ 2. Induction : Assume that the lemma is true for eve ry 0 ≤ j < i . Then, ev ery string in C ( i − 1) has tw o extensions in C ( i ), since for ev ery string in C ( i − 1) there a re 2 m i − m i − 1 extensions a v aila ble of whic h at most | B ( i ) | ≤ m i (2 g ( m i ) − 1) are fo rbidden. Namely , 2 m i − m i − 1 − | B ( i ) | ≥ 2 m i / 2 − 2 g ( m i )+log m i + m i ≥ 2. Hence it follow s that the binary k -choice can a lwa ys b e made in Step 2 o f t he algorithm for ev ery l . Therefore | C ( i ) | = 2 i +1 . ✷ Let a constan t c 1 accoun t for the constan t n um b er of bits to sp ecify the f unc- tions t, g , the algorithm C , and a reconstruction program that execute s the follo wing: W e can sp ecify ev ery initial m i -length segmen t of a particular ω in the set o n the lefthand side of (5) b y running the algorit hm C using the da ta represen ted b y the c 1 bits, m i , and the indexes k j ∈ { 0 , 1 } of the strings in D ( j ) with initial segmen t in C ( j − 1), 0 ≤ j ≤ i , that form a prefix of ω . Therefore, C ( ω 1: m i | m i ) ≤ c 1 + i + 1 . Setting c = 2 c 1 +1 yields C ( ω 1: m i | m i ) ≤ log c + i = log m i . By the choic e o f B ( i ) in the algorithm w e know that C t ( ω 1: m i + j | m i + j ) ≥ g ( m i ) for ev ery j satisfying 0 ≤ j < m i . Becaus e 2 m i = m i +1 , for ev ery n satisfying m i ≤ n < m i +1 , C t ( ω 1: n | n ) ≥ 1 2 m i − log m i ≥ 1 4 n − log n . Since this holds for ev ery i = 0 , 1 , . . . , item (i) is pro v en with C t ( ω 1: n | n ) ≥ 1 4 n − log n for ev ery n . The n um b er of 8 ω ’s concerned equals the n umber of pat hs in an infinite complete binary tree, that is, 2 ℵ 0 . (ii) This is the same as item (i) except that w e alwa ys tak e, for example, k i = 0 (no binar y c hoice) in Step 2 of the algor it hm. In f act, w e can sp ecify an arbitrary computable 0–1 v alued f unction to choose the k i ’s. There are a coun tably infinite n um b er of (that is ℵ 0 ) suc h functions. The sp ecification of ev ery such function φ take s C ( φ ) bits. Hence we do not ha v e to sp ecify the success iv e k i bits, and C ( ω 1: n | n ) = c 1 + 1 + C ( φ ) = O (1) with c 1 the constan t in the pro of of it em (i). T rivially , still C t ( ω 1: m i + j | m i + j ) ≥ g ( m i ) for ev ery j satisfying 0 ≤ j < m i . Since this holds f o r ev ery i = 0 , 1 , . . . , item (ii) is pro v en by item (i). ✷ 5 Conclusions W e ha v e pro v ed the items promised in the a bstract. In Lemma 5 w e iter- ated the pro o f metho d o f Lemma 2 t o prov e a result whic h is reminiscen t of Barzdins’s lemma 1, relating compressiblit y and time-b ounded incompress- iblit y of infinite sequences in another manner. Alternativ ely , we c ould ha ve studied space-b o unded incompressibilit y . It is easily v erified that the results also hold when the time-b ound t is replaced b y a space b ound s and the time- b ounded K o lmogorov complexit y is replaced by space-b ounded Kolmogoro v complexit y . Ac kno wledgemen t W e thank the referees fo r commen ts, references, p oin ting out an error in the original pro of of Lemma 2 and that the argumen t used there is b oth indep en- den t and close to that used to pro v e Theorem 3.2 in [1]. References [1] L. An tunes, L . F ortno w, D . v an Melk eb eek, and N. V. 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