Shift Complexity and Depth

I. 3.1 (Partial) Recursive Functions and Sets

V.1.1 Shift Complexity and Depth

Unlike COMPLEX(δ, c), SC(δ, c) is a deep Π⁰₁ class for every computableδ∈ (0,1)and c ∈N. Our proof is essentially a more detailed presentation of the proof given by Rumyantsev in [23, Theorem 3, essentially]

plus the uniformity observation by Bienvenu & Porter given in [2].

Theorem V.1.3. [23, Theorem 3, essentially] SC(δ, c)is a deepΠ⁰₁class for all rationalδ∈ (0,1)andc∈N. To prove Theorem V.1.3, we make use of the following probabilistic lemma:

Lemma V.1.4. [23, Lemma 6]Supposeδ∈ (0,1)is rational. For every rational ε>0 andn0∈Nthere exist natural numbers n<N and random variables An,An+1, . . . ,AN such that

(i) Ai is a subset of{0,1}ⁱ of size at most 2^δi,

(ii) for every stringσ∈ {0,1}^N, the probability thatσ has no substring in⋃^Ni=nAi is less thanε, and (iii) n≥n₀.

Moreover, the natural numbersn, N ∈Nand the (probability distributions of the) random variablesAn,An+1, . . . ,AN can be found effectively as functions of εandn0.

Proof. Letm≥2 be such thatδ> _m¹. We define natural numbers n=n1<n2< ⋯ <nm=N satisfying the following properties:

(i) nk dividesnk+1 for allk∈ {1, . . . , m−1}.

(ii) δnk∈Nfor allk∈ {1, . . . , m}. (Assuming (i) holds, it suffices forδn to be a natural number.)

Wheniis not of the formnk fork∈ {1, . . . , m}, we will defineAi to take the constant value∅, leaving only An_k to define. Roughly speaking, we will define the random variablesAn_k to address stringsσwhich exhibit relatively few substrings of increasingly greater lengths.

Suppose σ is a string of length p, q divides p, and α∈ (0,1) satisfies αq ∈ N. If σ has less than 2^αq substrings of lengthq, then we may encodeσin the following way. Writingσ=σ ^⌢σ ^⌢⋯^⌢σ _/ where∣σ ∣ =q,

our hypothesis implies∣{σ1, σ2, . . . , σp/q}∣ <2^αq (our hypothesis is much stronger than this, but is the natural hypothesis for the remaining arguments in the proof). Letρbe a string encoding, in lexicographical order, the distinct elements of{σ₁, σ₂, . . . , σ_p/q}; for exactness, we may encode a finite sequenceν₁, . . . , ν_nof strings by

ν₁^0⌢⟨1,1⟩^⌢ν₂^0⌢⟨1,1⟩^⌢⋯^⌢ν_n^0⌢⟨1,1⟩

whereν_i⁰denotes the string⟨νi(0),0, νi(1),0, νi(2),0, . . . , νi(∣νi∣ −1),0⟩. Then we may encodeσby τ=τ1⌢τ2⌢⋯^⌢τ_p/q^⌢ρ

whereτi is the binary representation of the index ofσi inρ(regarding ρas a finite sequence of strings). It is convenient to ensureτ has a length depending only onpandq, so we appropriately pad eachτi by 0’s to ensure that∣τ_i∣ =αq (ρencodes at most 2^αq−1 strings and hence requires at most αq bits to describe) and appropriately padρby 1’s to ensure that∣ρ∣ =2(q+1)(2^αq−1)(the length ofρif it encodes the maximum number of 2^αq−1 strings). Thus, to each such stringσof lengthpwe associate with it a unique stringτ of lengthαp+2(q+1)(2^αq−1).

Supposing nk has already been defined and nk divides i, we say that a string σ of length i isk-sparse if it has less than 2^{m−km nk} substrings of lengthnk. As observed above, we may encode a k-sparseσ with a string of length ^m_m^−k∣σ∣ +2(nk+1)(2^{m−km nk}−1). For simplicity, we writeck+1∶=2(nk+1)(2^{m−km nk}−1).

We may now define nk and the associated random variableAn_k fork∈ {1, . . . , m}.

k=1. Letn1=nbe the least natural number greater than n0such thatδn∈Nand (1−

1 2^n/m)

2^δn

<ε.

Note that such annexists as

(1− 1 2^n/m)

2^δn

= (1− 1 2^n/m)

2^n/m⋅2^(δ−1/m)n

=

⎛

⎝ (1−

1 2^n/m)

2^n/m

⎞

⎠

2^(δ−1/m)n

≈ ( 1 e)

2^(δ−1/m)n

with both the approximation getting tighter and the final expression tending toward 0 asn→ ∞.

Then An₁ = An is defined to be randomly chosen uniformly among all subsets of {0,1}ⁿ of size 2^δn (i.e., each such subset of {0,1}ⁿ of size 2^δn has an equal probability(²

n

2^δn)of being the value ofAn).

1<k<m. Supposenk−1 has been defined. Thennk is the least multiple ofnk−1 such that

(1− 1 2^nk/m+ck)

2^δnk

=

⎛

⎝ (1−

1 2^nk/m+ck)

2^nk/m+ck

⎞

⎠

2^{(δ−1/m)nk−ck}

<ε.

(That such ann_k exists is analogous to the case wherek=1.)

ThenAn_k is defined to be randomly chosen uniformly among all subsets of{0,1}^nk of size 2^δnk which consist only of(k−1)-sparse strings.

k=m. Letn_k=n_m=N to be the least multiple ofn_m−1 such that _m¹N+c_m<δN (by hypothesis, 1/m<δ andc_mis constant with respect toN, so there is such anN).

Then AN is defined to be constantly equal to the set of all(m−1)-sparse strings of length N (note that an(m−1)-sparse string is described uniquely by a string of length ^m−(_m^m−1)N+cm= _m¹N+cm, so

∣AN∣ ≤2^N/m+cm <2^δN).

Finally, we show that for everyσ∈ {0,1}^N,

Prob(Ai has no substring ofσfor alli∈ {n, . . . , N}) <ε.

It suffices to show that Prob(A_i has no substring ofσ) <εfor at least onei∈ {n, . . . , N}.

Case 1: Suppose σ is not 1-sparse, so that σ has at least 2^{m−1m n} substrings of length n. Because of the definition of the output distribution of An, the probability that σ has no substring in An is at most the probability that σ has no substring among 2^δn strings chosen uniformly and indepe- dently at random (the latter probability may be higher because we allow duplicates). The inde- pendence and uniformity of those 2^δn random choices means that the latter probability is equal to Prob(τ∈ {0,1}ⁿ is not a substring ofσ)^2δn. Prob(τ∈ {0,1}ⁿ is not a substring ofσ) is at most the probability that a randomτ∈ {0,1}ⁿ (chosen uniformly) is not in a set of size 2^{m−1m n}. Thus,

Prob(A_n has no substring ofσ)

≤Prob(2^δn random strings of lengthnare not substrings ofσ)

≤Prob(random string of lengthnis not substring ofσ)^2δn

≤ (1− 2^mm−1n

2ⁿ )

2^δn

= (1− 1 2^n/m)

2^δn

<ε.

Case 2: Suppose σ is 1-sparse but is not k-sparse for some k ∈ {2, . . . , m−1}. Assume k is minimal with that property, so σ is not k-sparse (and hence has at least 2^{m−km nk} substrings of length nk) but is (k−1)-sparse. As in Case 1, we have

Prob(An_k has no substring ofσ)

≤Prob(2^δnk random(k−1)-sparse strings of lengthnk are not substrings ofσ)

≤Prob(random(k−1)-sparse string of lengthnk is not substring ofσ)^2δnk.

The probability Prob(random(k−1)-sparse string of lengthnk is not substring ofσ)is of the form Prob(E∖F), whereE= {τ∈ {0,1}^nk∣τ is(k−1)-sparse}andF = {τ∈ {0,1}^nk∣τ is a substring ofσ}.

Observe that F ⊆ E: if a substring of σ is not (k−1)-sparse, then it contains at least 2^{m−(k−1)m nk−1} substrings of lengthnk−1, and henceσdoes as well, contrary to the hypothesis thatσis(k−1)-sparse.

BecauseE is finite, we have Prob(E∖F) =1−^∣_∣^F_E^∣_∣. To get an upper bound on Prob(E∖F), it suffices to have an upper bound on∣E∣and a lower bound on∣F∣. Thus,

Prob(random(k−1)-sparse string of length nk is not substring ofσ)^2δnk

≤ (1−

2^{m−km nk} 2^{m−k+1m nk+ck})

2^δnk

= (1− 1 2^nk/m+ck)

2^δnk

<ε.

Case 3: Supposeσisk-sparse for allk∈ {1, . . . , m−1}. In particular,σis(m−1)-sparse and so an element ofAN. Thus, Prob(AN has no substring ofσ) =0<ε.

Proof of Theorem V.1.3. WithMa universal left r.e. continuous semimeasure on{0,1}^∗, by Proposition II.4.3 there is a partial recursive functional Ψ such that for everyσ∈ {0,1}^∗,

M(σ) =λ(Ψ⁻¹(σ)) =λ({X∈ {0,1}^N∣Ψ^X⊇σ}).

Say that Y ∈ {0,1}^∗∪ {0,1}^N avoids a k-tuple of finite sets of strings ⟨A₁, A₂, . . . , A_k⟩ if∣Y∣ ≥max{∣σ∣ ∣ σ∈ ⋃^k_i=1A_i} and no substring ofY is an element of⋃^k_i=1A_i. Finite sets of strings in {0,1}^∗ are implicitly G¨odel numbered by some recursive bijectionPfin({0,1}^∗) →N.

Our approach, roughly, involves us finding natural numbers n<N and setsAn, An+1, . . . , AN satisfying the following conditions:

(i) Ai is a subset of{0,1}ⁱ of size at most 2^δi for eachi∈ {n, n+1, . . . , N}and (ii) λ({X∈ {0,1}^N∣Ψ(X)avoids An, An+1, . . . , AN}) <ε.

The claim is then that each elementτof⋃^N_i=nA_isatisfies KP(τ) <δ∣τ∣ −c, so Ψ(X)being⟨δ, c⟩-shift complex

implies that Ψ(X)avoids the setsAn, An+1, . . . , AN. Then

λ({X∈ {0,1}^N∣Ψ(X) ∈SC(δ, c)}) ≤λ({X∈ {0,1}^N∣Ψ(X)avoidsA_n, A_n+1, . . . , A_N}) <ε.

There are two issues that we must work around, the first being that Ψ(X)need not be an element of {0,1}^N, and the second of which is that an elementτof⋃^N_j=nAj need not necessarily satisfy KP(τ) <δ∣τ∣ −c.

We start with addressing the first issue. Our use of avoidance only requires that∣Ψ^X∣ ≥N. Lemma V.1.4 shows that as a recursive function of⟨n0, m⟩ ∈N², we can find natural numbersn<N and random variables An,An+1, . . . ,AN such that:

(i) n≥n0,

(ii) Ak is a subset of{0,1}^k of size at most 2^δk/3 (the use ofδ/3 instead ofδ will become apparent when dealing with the second issue) for eachk∈ {n, n+1, . . . , N}, and

(iii) for every stringσ∈ {0,1}^N, the probability thatσhas no substring in⋃^N_i=nAi is less than 2^−m. Fixn₀andm and letn<N andA_n,A_n+1, . . . ,A_N be as above.

Let S ∶= {X ∈ {0,1}^N ∣ ∣Ψ^X∣ ≥ N}. S is Σ⁰₁, so there exists a recursive sequence ⟨σ_n⟩_n∈N of pairwise incompatible strings such that S = ⋃n∈NJσnK2. Let α∶= λ(S), so that ⟨λ(⋃_k≤nJσkK2)⟩n∈N is a recursive sequence converging monotonically toαfrom below. Fori∈N, letαi∶=i⋅2^−m/3, and leti0 be the largesti for whichαi<α, so thatα−αi₀ <2^−(m+1). Finally, define

S˜∶= ⋃

k≤p

JρkK2

where pis the smallest natural number for which λ(S) ≥˜ α_i0. ˜S is a recursive subset of S and λ(S∖S) <˜ 2^−(m+1). By virtue of being a subset ofS,∣Ψ^X∣ ≥N for allX ∈S. ˜˜ S is recursive and an index for ˜S can be computed fromn₀,m, andi₀. Define a probability measureµon{0,1}^N by setting

µ({σ}) ∶=λ(S)˜ ⁻¹⋅λ({X ∈ {0,1}^N∣X∈S˜and Ψ^X⊇σ})

for eachσ∈ {0,1}^N. µis a computable measure and an index forµcan be computed fromn0,m, andi0. Letν be the probability measure onP ({0,1}ⁿ) × P ({0,1}ⁿ⁺¹) × ⋯ × P ({0,1}^N)defined by

ν({⟨An, An+1, . . . , AN⟩}) ∶=Prob(Ai=Ai for eachi∈ {n, n+1, . . . , N})

(i.e., the joint probability distribution made up of the output distributions of the random variablesA_n,A_n+1, . . . ,A_N). Write

E∶= {⟨X,⟨A_n, . . . , A_N⟩⟩ ∈ {0,1}^N×

N

∏P ({0,1}ⁱ) ∣X∈S˜and Φ^X avoids A_n, . . . , A_N}.

By Fubini’s Theorem,

∫ (∫ χ_E(X,(A_n, . . . , A_N))dλ)dν= (λ×ν)(E)

= ∫ (∫ χ_E(X,(A_n, . . . , A_N))dν)dλ

≤ ∫ 2^−(m+1)dλ

=2^−(m+1).

If ∫ χE(X,(An, . . . , AN))dλ≥2^−(m+1) for every ⟨An, . . . , AN⟩ ∈ ∏^Nj=nP ({0,1}^j), we reach a contradiction.

Thus, there is a least one tuple⟨An, . . . , AN⟩ ∈ ∏^Ni=nP ({0,1}ⁿ)with the desired property, i.e., that λ({X∈ {0,1}^N∣X∈S˜ and Ψ^X avoidsA_n, . . . , A_N}) <2^−(m+1),

and hence

λ({X∈ {0,1}^N∣X∈S and Ψ^X avoids An, . . . , AN}) <2^−(m+1)+2^−(m+1)=2^−m.

Letc1∈Nbe such that KP(n) ≤2 log₂n+c1 for alln∈N. The recursiveness of ˜S allows us to effectively find such a tuple ⟨An, . . . , AN⟩ for whichλ({X ∈ {0,1}^N ∣X∈S˜and Ψ^X avoids ⟨An, . . . , AN⟩}) <2^−(m+1). As noted before, an index for ˜S can be found effectively fromn0,m, andi. As such, there isc^′₁ such that

KP(An, . . . , AN) ≤KP(n0) +KP(m) +KP(i) +c^′₂.

Although icannot be found recursively from n₀ andm in general, we regardless have the bound have the boundi≤ (2^−m/3)⁻¹=3⋅2^m. Thus,

KP(i) ≤ max

0≤k≤⌊3⋅2^m⌋KP(k) ≤ max

0≤k≤⌊3/ε⌋(2 log₂k+c1) ≤2m+2 log₂3+c1. Letc2=2 log₂3+c1+c^′₂, so that for every⟨n0, m⟩we have

KP(An, . . . , AN) ≤KP(n0) +KP(m) +2m+c2.

Now we address the second issue. Letc3∈Nbe such that, where τi is thei-th element of A_∣τi∣ ordered lexicographically,

KP(τ_i) ≤KP(∣τ_i∣) +KP(i) +KP(A_n, A_n+1, . . . , A_N) +c₃. Then for anyτ∈ ⋃^N_j=nA_j we have

KP(τ) ≤KP(∣τ∣) +KP(index of τ inA_∣τ∣) +KP(An, . . . , AN) +c2

≤ (2 log₂∣τ∣ +c₁) + (2 log₂(index ofτ inA_∣τ∣) +c₁) + (KP(n₀) +KP(m) +2m+c₂) +c₃

≤ (2 log₂∣τ∣ +c1) + ((2/3)δ∣τ∣ +c1) + (2 log₂n0+2 log₂m+2m+2c1+c2) +c3

= 2

3δ∣τ∣ +2 log₂∣τ∣ +2 log₂n0+2 log₂m+2m+ (4c1+c2+c3).

Note thatd∶=4c1+c2+c3 is independent of⟨n0, m⟩. For∣τ∣ sufficiently large, ²₃δ∣τ∣ +2 log₂∣τ∣ +2 log₂n0+ 2 log₂m+2m+d<δ∣τ∣−c. Thus, definen0=n0(m)to be the leastnsuch that ²₃δn+4 log₂n+2 log₂m+2m+d<

δn−c. Finally, definer∶N→Nbyr(m) ∶=N(n0(m), m).

If Ψ(X)is⟨δ, c⟩-shift complex, then Ψ(X) =Ψ^X∈ {0,1}^N(thereforeX ∈S) and KP(τ) ≥δ∣τ∣ −cfor every substringτ of Ψ(X). Then

M(SC(δ, c)↾r(m)) =λ({X∈ {0,1}^N∣ ∣Ψ^X∣ ≥r(m)and Ψ^X↾r(m)is⟨δ, c⟩-shift complex})

≤λ({X∈ {0,1}^N∣ ∣Ψ^X∣ ≥r(m)and Ψ^X↾r(m)avoidsAn, . . . , AN})

<

1 2^m. Hence, SC(δ, c)is deep.

Corollary V.1.5. No difference random computes a shift complex sequence. Consequently,SC≰wMLR.

Remark V.1.6. X ∈ {0,1}^N is Kurtz random if X ∉ P for any Π⁰₁ class P with λ(P) =0. [15, Theorem 6.7] shows that for every δ ∈ (0,1)there is a Y ∈SC(δ) such that Y computes no Kurtz random. Every Martin-L¨of random sequence is Kurtz random, so this showsMLR≰_wSC(δ)for every δ∈ (0,1).