User blog:P進大好きbot/New Difference Sequence System

Here is a new difference sequence system, which I came up with today through a greedy trial to imitate Y-sequence system originally created by a Japanese googologist Yukito. Although it expands in a different way from Y-sequence system, it looks not so bad.

WIP: I am trying to inventigate a way to avoid using BadRoot.

= Definition =

I denote by \(\mathbb{N}_+\) the set of positive integers. For a set \(X\), I denote by \(X^{< \omega}\) the set of arrays of elements of \(X\). I denote by \(S \subset \mathbb{N}_+^{< \omega}\) the subset of non-empty sequences whose leftmost entry is \(1\). We denote by \(\leq_S\) the lexicographic order on \(S\).

For an array \(a\), I denote by \(\textrm{Lng}(a)\) the length of \(a\). For an array \(a\) and an \(i \in \mathbb{N}\) smaller than \(\textrm{Lng}(a)\), I denote by \(a_i\) the \((1+i)\)-th entry of \(a\). An initial segment of an array \(a\) is an array of the form \((a_i)_{i=0}^{j-1}\) for some \(j \in \mathbb{N}\) smaller than \(\textrm{Lng}(a)-1\). In particular, the empty array is an initial segment of any array.

For arrays \(a\) and \(b\), I denote by \(a \frown b\) the concatenation of \(a\) and \(b\).

Difference Sequence
I denote by \begin{eqnarray*} \textrm{Parent} \colon \mathbb{N}^{< \omega} \times \mathbb{N} \to \mathbb{N} \end{eqnarray*} the partial computable map introduced here. Roughly speaking, \(\textrm{Parent}\) is the function which outputs the index of the parent of the index in the input. Since \(\textrm{Parent}\) forms a bad root searching rule in the sense of the terminology introduced here, it induces the total computable maps \begin{eqnarray*} \begin{array}{lcrcl} \textrm{Ancestor} & \colon & \mathbb{N}^{< \omega} & \to & \mathbb{N}^{< \omega} \\ \textrm{RightNodes} & \colon & \mathbb{N}^{< \omega} & \to & \mathbb{N}^{< \omega} \\ \textrm{Kaiser} & \colon & \mathbb{N}^{< \omega} \setminus \{\} & \to & \mathbb{N}^{< \omega} \end{array} \end{eqnarray*} introduced in the same article. Roughly speaking, \(\textrm{Ancestor}\) is the function which outputs the sequence of indices of ancestors of the rightmost entry of the input, \(\textrm{RightNodes}\) of the function which outputs the sequence of entries of ancestors of the rightmost entry of the input, \(\textrm{Kaiser}\) is the function which outputs the difference sequence of the value of \(\textrm{RightNodes}\).

I define a total computable map \begin{eqnarray*} \textrm{Royal} \colon \mathbb{N}_+^{< \omega} \times \mathbb{N}_+^{< \omega} & \to & \mathbb{N}_+^{< \omega} \times \mathbb{N}_+^{< \omega} \\ (a,b) & \mapsto & \textrm{Royal}(a,b) \end{eqnarray*} in the following recursive way: This map imitate the bad root searching rule in Y-sequence system, but I intensionally use only the first difference sequences.
 * 1) Suppose that \(a \frown b \in S\), \(b \neq \), and \(\textrm{Ancestor}(b)_0 = 0\).
 * 2) If \(\textrm{Lng}(b) = 1\), then set \(\textrm{Royal}(a,b) := (a,b)\).
 * 3) Suppose \(\textrm{Lng}(b) \neq 1\).
 * 4) Put \(i := \textrm{Ancestor}(\textrm{Kaiser}(b))_0\).
 * 5) Denote by \(c\) the initial segment of \(b\) of length \(\textrm{Ancestor}(b)_i+1\).
 * 6) Denote by \(d\) the sequence given by removing the first \(i\) entries from \(\textrm{Kaiser}(b)\).
 * 7) Set \(\textrm{Royal}(a,b) := (a \frown c,d)\).
 * 8) Otherwise, then set \(\textrm{Royal}(a,b) := (,)\).

Constructibility
A sequence \(a\) is said to be additive principal if \(\textrm{Ancestor}(a)_0 = 0\), and is said to be a decreasing sum if \(a\) can be expressed as a concatenation of an \(A \in S^{< \omega}\) satisfying that \(A_i\) is additive principal for any \(i \in \mathbb{N}\) smaller than \(N\) and \(A_{i+1} \leq_S A_i\) for any \(i \in \mathbb{N}\) smaller than \(N-1\). If \(a\) is a decreasing sum, such an \(A\) is unique, and will be denoted by \(\textrm{Principal}(a)\).

I define a recursive subset \(T \subset \mathbb{N}_+^{< \omega}\) in the following recursive way: The set \(T\) is an analogue of the set of Cantor normal forms.
 * 1) For any \(i \in \mathbb{N}_+\), \((i) \in T\).
 * 2) For any sequence \(a\), if \(\textrm{Principal}(a) \in T^{< \omega}\), then \(a \in T\).
 * 3) For any additive principal sequence \(a\), if \(a = (i) \frown b\) for some \((i,b) \in \mathbb{N}_+ \times (T \setminus )\) and the concatenation of \(\textrm{Royal}(,a)\) belongs to \(T\), then \(a \in T\).

WIP

I define a total computable map \begin{eqnarray*} \textrm{IsConstructibleFromBelow} \colon S^{< \omega} \times S & \to & \{0,1\} \\ (A,a) & \mapsto & \textrm{IsConstructibleFromBelow}(A,a) \end{eqnarray*} in the following recursive way: Namely, \(\textrm{IsConstructibleFromBelow}\) is a function which tells us whether every entry of \(A\) is constructible from below \(a\) in some sense. I will use this in order to define a "diagonal" expansion.
 * 1) If every entry of \(A\) is \((1)\), then set \(\textrm{IsConstructibleFromBelow}(a,A) := 1\).
 * 2) If \(A \notin T^{< \omega}\), then set \(\textrm{IsConstructibleFromBelow}(a,A) := 0\).
 * 3) If some entry of \(A\) is not \((1)\), \(A \in T^{< \omega}\), and there is an entry \(b\) of \(A\) such that \(a \leq_S b\), then set \(\textrm{IsConstructibleFromBelow}(a,A) := 0\).
 * 4) Suppose that some entry of \(A\) is not \((1)\), every entry of \(A\) is a decreasing sum, and there is no entry \(b\) of \(A\) such that \(a \leq_S b\).
 * 5) Denote by \(B_0\) by the array given by enumerating by \(\leq_S\) the finite set of sequences which are entries of \(\textrm{Principal}(b)\) for some entry \(b\) of \(A\).
 * 6) Denote by \(B_1\) by the array given by enumerating by \(\leq_S\) the finite set of sequences which are concatenations of \(\textrm{Royal}(,b)\) for some entry \(b\) of \(A\).
 * 7) Denote by \(B_2\) by the array given by enumerating by \(\leq_S\) the finite set of sequences given as non-empty initial segments of some entry \(b\) of \(B_1\).
 * 8) Set \(\textrm{IsConstructibleFromBelow}(A,a) := \textrm{IsConstructibleFromBelow}(B_2,a)\).

Expansion
I define total computable maps \begin{eqnarray*} \textrm{BadRoot} \colon \mathbb{N}_+^{< \omega} \times \mathbb{N}_+^{< \omega} & \to & \mathbb{N} \\ (a,b) & \mapsto & \textrm{BadRoot}(a,b) \end{eqnarray*} and \begin{eqnarray*} [ \ ] \colon S \times \mathbb{N} & \to & S \\ (a,n) & \mapsto & a[n] \end{eqnarray*} simultaneously in the following recursive way: Roughly speaking, the parallel expansion is supposed to be the same one as Y-sequence system, but the diagonal expansion is supposed to be as strong as possible following the restriction on the constructibility.
 * Definition of \(\textrm{BadRoot}(a,b)\):
 * 1) If \(\textrm{Royal}(a,b) = (,)\), then set \(\textrm{BadRoot}(a,b) := 0\).
 * 2) If \(\textrm{Royal}(a,b) \neq (,)\) and \(b = (1)\), then set \(\textrm{BadRoot}(a,b) := \textrm{Lng}(a) + \textrm{Lng}(b) - 2\).
 * 3) Suppose that \(\textrm{Royal}(a,b) \neq (,)\), \(b \neq (1)\), and \(\textrm{Lng}(b) = 1\), then \(\textrm{BadRoot}(a,b)\) is the second rightmost entry of \(\textrm{Ancestor}(a \frown b)\).
 * 4) Suppose that \(\textrm{Royal}(a,b) \neq (,)\) and \(\textrm{Lng}(b) > 1\).
 * 5) Put \((c,d) := \textrm{Royal}(a,b)\).
 * 6) Put \(i := \textrm{Lng}(c)\).
 * 7) Put \(r := \textrm{BadRoot}(\textrm{Royal}(a,b))\).
 * 8) If \(r < i\), then set \(\textrm{BadRoot}(a,b) := r\).
 * 9) If \(r \geq i\), then set \(\textrm{BadRoot}(a,b) := \textrm{Lng}(a) + \textrm{Ancestor}(b)_{r-i+1}\).
 * Definition of \(a[n]\):
 * 1) If \(a = (1)\), then set \(a[n] := (1)\).
 * 2) If \(a \neq (1)\) and the rightmost entry of \(a\) is \(1\), then \(a[n]\) is the sequence given by removing the rightmost entry from \(a\).
 * 3) Suppose \(a \neq (1)\) and the rightmost entry of \(a\) is not \(1\).
 * 4) Denote by \(s\) the initial segment of \(a\) of length \(\textrm{Ancestor}(a)_0\).
 * 5) Denote by \(t\) the sequence given by removing the first \(\textrm{Ancestor}(a)_0\) entries from \(a\).
 * 6) Put \((c,d) := \textrm{Royal}(s,t)\)
 * 7) Put \(r := \textrm{BadRoot}(c,d)\).
 * 8) Denote by \(b\) the sequence given by removing the first \(r\) entries and the rightmost entry from \(a\).
 * 9) Put \(L_0 := \textrm{Lng}(b)\).
 * 10) If \(n = 0\), then \(a[n]\) is the sequence given by removing the rightmost entry from \(a\).
 * 11) If \(n \neq 0\) and the rightmost entry of \(d\) is \(1\), then set := \(a[n] \frown b\).
 * 12) Suppose \(n \neq 0\) and the rightmost entry of \(d\) is not \(1\).
 * 13) Suppose \(\textrm{Lng}(c) \leq r\)
 * 14) Put \(L_1 := \textrm{Lng}(c) + \textrm{Lng}(d) - r\)
 * 15) Denote by \(\delta\) the sequence characterised by the following conditions:
 * 16) \(\textrm{Lng}(\delta) = (n+1)L_1\).
 * 17) \(\delta_0 = 0\).
 * 18) \(\delta_{i+1} - \delta_i = (c \frown d)[n]_{r+i}\) for any \(i \in \mathbb{N}\) smaller than \((n+1)L_1-1\).
 * 19) Denote by \(b^{(n)}\) the sequence characterised by the following conditions:
 * 20) \(\textrm{Lng}(b^{(n)}) = L_0\).
 * 21) \(b^{(n)}_i = b_i\) for any \(i \in \mathbb{N}\) satisfying that \(i < L_0\) and there is no \(j \in \mathbb{N}\) such that \(j < L_1\) and \(i + r\) is the \((L_1-j)\)-th rightmost entry of \(\textrm{Ancestor}(a)\).
 * 22) \(b^{(n)}_i = b_0 + \delta_{nL_1+j}\) for any \((i,j) \in \mathbb{N}^2\) satisfying that \(j < L_1\) and \(i + r\) is the \((L_1-j)\)-th rightmost entry of \(\textrm{Ancestor}(a)\).
 * 23) Set \(a[n]] := a[n-1] \frown b^{(n)}\).
 * 24) Suppose \(\textrm{Lng}(c) > r\)
 * 25) Denote by \(b^{(n)}\) the sequence characterised by the following conditions:
 * 26) \(\textrm{Lng}(b^{(n)}) = L_0\).
 * 27) For any \(i \in \mathbb{N}\) smaller than \(L_0\), \(b^{(n)}_i\) is the maximum of a \(k \in \mathbb{N}\) satisfying \(\textrm{IsConstructibleFromBelow}((a[n-1] \frown e \frown (k)),a) = 1\), where \(e\) is the initial segment of \(b^{(n)}\) of length \(i\).
 * 28) Set \(a[n] := a[n-1] \frown b^{(n)}\).

Large Function
I define a partial computable map \begin{eqnarray*} \langle \cdot, \cdot \rangle \colon S^{< \omega} \times \mathbb{N} & \to & \mathbb{N} \\ (A,n) & \mapsto & \langle A, n \rangle \end{eqnarray*} in the following recursive way: This is just a computable variant of FGH. Although I do not know whether it is total or not, \(\langle ((1,10)), 10 \rangle\) will be a computable large number if it actually terminates.
 * 1) If \(A = \), then set \(\langle A, n \rangle := n\).
 * 2) Suppose \(A = (a)\) for some \(a \in S\).
 * 3) If \(a = (1)\), then set \(\langle A, n \rangle := n+1\).
 * 4) If \(a \neq (1)\), then set \(\langle A, n \rangle := \langle ((\underbrace{a[n+1],\ldots,a[n+1]}_{n+1}), n+1 \rangle\).
 * 5) Suppose that the length of \(A\) is greater than \(1\).
 * 6) Denote by \(B\) the array given by removing the rightmost entry from \(A\).
 * 7) Denote by \(a\) the rightmost entry of\(A\).
 * 8) Set \(\langle A, n \rangle := \langle B, \langle (a), n \rangle \rangle\).