Recently I've found that my old dimensional array notation can't get very far but it's very complex, so here I bring a new version of Dimensional array notation (DAN) - it's more simple. It's still the 4th part of R function series.
Separators
Now I bring you a separator, which comes after the comma - {1*} (here the * is at the left superscript of the "}"). It work as: nR{0{1*}1}=nR{0,0,...,0,1} with n 0's. Then nR{1{1*}1}=nR{0{1*}1}{0{1*}1}...{0{1*}1} with n {0{1*}1}'s. And then nR{{0,1{1*}1},0{1*}1}=nR{{...{0,0{1*}1}...,0{1*}1},0{1*}1} and so on.
nR{0{1*}2}=nR{0,0,...,0,1{1*}1} with n 0's, and generally nR{0{1*}a+1}=nR{0,0,...,0,1{1*}a} with n 0's. Now some definitions come in.
Something
Symbol * is called asterisk, with ASCII=42.
A brace with an asterisk at the left superscript of the rbrace is a separator, e.g. \(\{2 ^*\}\), and the comma is still a separator. The comma is a shorthand for {0*}.
And notice that things inside a separator can be any string! For example, {{0{1*}1}*} is a separator.
Planar and higher dimensional arrays
The comma, or {0*}, is a separator for entries. The {1*} is a separator for rows. So how does "the next row" work? The first entry in the next row decreases by 1 if rows before are all default, and fill the previous row with 0,0,...,0,1 with n 0's. It's very simple. What matters is that the new version doesn't use complex things like "brace rank".
3R{0{1*}3}=3R{0,0,0,1{1*}2}, noticing that dimensional arrays work just at the array itself, and doesn't use a lower leveled array like LAN.
3R{0{1*}0,2}=(S2)3R{0{1*}{0{1*}0,2},1}=(R5)3R{0{1*}{0{1*}{0{1*}0,1},1},1}, using LAN rules.
3R{0{1*}0{1*}2}=3R{0{1*}0,0,0,1{1*}1}=(S2)3R{0{1*}0,0,{0{1*}0,0,0,1{1*}1},0{1*}1}=(R5)3R{0{1*}0,0,{0{1*}0,0,{0{1*}0,0,0,0{1*}1},0{1*}1},0{1*}1}=(S1)3R{0{1*}0,0,{0{1*}0,0,{0{1*}0{1*}1}{1*}1}{1*}1}=3R{0{1*}0,0,{0{1*}0,0,{0{1*}0,0,0,1{1*}0}{1*}1}{1*}1}=(S1)3R{0{1*}0,0,{0{1*}0,0,{0{1*}0,0,0,1}{1*}1}{1*}1}=...
Then higher dimensional arrays. nR{0{2*}1}=nR{0{1*}0{1*}...0{1*}1} with n 0's, and nR{0{a+1*}1}=nR{0{a*}0{a*}...0{a*}1} with n 0's.
But what about {{0}*} ? For further extensions I've to change Rule 4 a little here -
- Rule 4: \(nR\oslash\{a+1\odot_1\}\odot_2=nR\oslash\{a\odot_1\}0\{a\odot_1\}0...0\{a\odot_1\}\odot_2\) with n \(\{a\odot_1\}\)'s.
Notice the \(\odot_1\) can be any string, so it can conclude an asterisk at the end. We can apply this on dimensional separators. And notice that Subrule 1 applies if we apply this Rule 4 on a brace not a separator, so it doesn't matter when we apply rules on Basic R funcrion, brace notation and LAN. Subrule 1 doesn't apply if we apply this Rule 4 on a separator, so the new Rule 4 works well.
But we made things weaker. e.g. 4R{0{1*}5}=(R4)4R{0,0,0,0,5} - really weaker. How to fix it?
Formal definition
After some work I come up with the rules. This is the last time I use pattern tests in my definitions.
- Rule 1: \(nR0=10^n\) ----- base rule
- Rule 2: \(nRa+1\odot=nRa\odot Ra\odot...Ra\odot\) with n R's.
- Rule 3: \(nR\oslash\{0\}\odot=nR\oslash n\odot\)
- Rule 4: \(nR\oslash\{a+1\odot_1\}\odot_2=nR\oslash\{a\odot_1\}0\{a\odot_1\}0...0\{a\odot_1\}\odot_2\) with n \(\{a\odot_1\}\)'s.
- Subrule 1: 0 is default, so \(0\{=\{\) if this 0 isn't immediately before a separator, and \(\diamond0\}=\}\), and \(\diamond_10\diamond_2=\diamond_2\) if \(\diamond_1\) has lower level than \(\diamond_2\), where the \(\diamond\) means separator.
- Rule 5: For \(nR\oslash_2\{\oslash_10,a+1\odot_1\}\odot_2\)(the \(\odot_1\) doesn't have an asterisk at the end, i.e. the \(\{\oslash_10,a+1\odot_1\}\) isn't a separator), find the nearest encloser \(\{\oslash_1\) and \(,a\odot_1\}\) enclosing the \(\{\oslash_10,a+1\odot_1\}\) (you can use Subrule 2 now), then \(nR\oslash_2\{\oslash_1\oslash_3\{\oslash_10,a+1\odot_1\}\odot_3,a\odot_1\}\odot_2=\) \(nR\oslash_2\{\oslash_1\oslash_3\{\oslash_1\oslash_3...\{\oslash_1\oslash_30\odot_3,a\odot_1\}...\odot_3,a\odot_1\}\odot_3,a\odot_1\}\odot_2\) with n \(\{\oslash_1\oslash_3\)'s and n \(\odot_3,a\odot_1\}\)'s.
- Subrule 2: If you cannot find an encloser \(\{\oslash_1\) and \(,a\odot_1\}\), add one to immediately enclose the whole string after R.
- Rule 6: \(nR\oslash_1\{{\odot_2}^*\}a+1\odot_1=nR\oslash_10\{{\odot_2}^*\}1\{{\odot_2}^*\}a\odot_1\) and then apply rules on the first \(\{{\odot_2}^*\}\), but return to the main process once Rule 3,4 or 5 apply.
Subrule 1 can apply any time, but subrule 2 applies only be called by Rule 5. Rule 4 on separators can only apply when called by Rule 6. The \(\odot\) can be any string or empty, the \(\oslash\) can be any string or empty excluding non-0 numbers on the same nested level.
That's it! These definitions are much clearer now. So we get
3R{0{{0}*}2}=(R6)3R{0{{0}*}1{{0}*}1}=(R3)3R{0{3*}1{{0}*}1}=(R6)3R{0{3*}1{3*}0{{0}*}1}=(R4)3R{0{2*}0{2*}0{2*}1{3*}0{{0}*}1}=(S1)3R{0{2*}0{2*}0{2*}1{{0}*}1}=3R{0{2*}0{2*}0{1*}0{1*}0{1*}1{{0}*}1} =3R{0{2*}0{2*}0{1*}0{1*}0,0,0,1{{0}*}1}=...
3R{0{{{{0,2},1}}*}3}=(R6)3R{0{{{{0,2},1}}*}1{{{{0,2},1}}*}2}=(R5)3R{0{{{{{0,1},1},1}}*}1{{{{0,2},1}}*}2}=...
Trimensional arrays and higher
Trimensional Array Notation (TAN) is the "\(4\frac{1}{2}\)-th part" of R function. It's not the 5th part because it still uses the DAN rules. The limit of DAN is nR{0{{0{...{0{0*}1}...*}1}*}1} with n nests of {0{ and *}1}. In TAN it's just nR{0{0,1*}1} - Now we let commas appear immediately inside a separator.
Notice the slightly change of Rule 5 - I add "the \(\odot_1\) doesn't have an asterisk at the end, i.e. the \(\{\oslash_10,a+1\odot_1\}\) isn't a separator". That's for the TAN extension.(i.e. the only difference between DAN and TAN is just here.) So 4R{0{0,1*}1}=(R6)4R{0{0,1*}1{0,1*}0}=(S2)4R{0{{0{0,1*}1{0,1*}0},0*}1{0,1*}0}=(R5)4R{0{{0{{0{{0{0,0*}1{0,1*}0},0*}1{0,1*}0},0*}1{0,1*}0},0*}1{0,1*}0}=(S1)4R{0{{0{{0{{0,1}*}1}*}1}*}1}=...
Then 3R{0{{0{0,1*}1}*}2}=(R6)3R{0{{0{0,1*}1}*}1{{0{0,1*}1}*}1}=(R6)3R{0{{0{0,1*}1{0,1*}0}*}1{{0{0,1*}1}*}1}=(S2)3R{0{{0{{0{0,1*}1{0,1*}0}*}1{{0{0,1*}1}*}1},0*}1{0,1*}0}=(R5)3R{0{{0{{0{{0{{0{{0{0*}1{{0{0,1*}1}*}1},0*}1{0,1*}0}*}1{{0{0,1*}1}*}1},0*}1{0,1*}0}*}1{{0{0,1*}1}*}1},0*}1{0,1*}0}=(S1)3R{0{{0{{0{{0{{0{{0,1{{0{0,1*}1}*}1}*}1}*}1{{0{0,1*}1}*}1}*}1}*}1{{0{0,1*}1}*}1}*}1}=...
Then nR{0{{0{0,1*}1}*}0,1}, nR{0{{0{0,1*}1}*}0{{0{0,1*}1}*}1}, nR{0{1{0{0,1*}1}*}1}, nR{0{{0{{0{0,1*}1}*}1}{0{0,1*}1}*}1}, nR{0{{0{0,1*}1}{0{0,1*}1}*}1}, nR{1{0,1*}1}, nR{0{{0{0,1*}1}*}1{0,1*}1}, nR{0{0,1*}2}, nR{0{0,1*}0{0,1*}1}, nR{0{1,1*}1}, nR{0{{0{0,1*}1},1*}1}, nR{0{0,2*}1}, nR{0{0,{0{0,1*}1}*}1}, nR{0{0,0,1*}1}, nR{0{0{1*}1*}1}, nR{0{0{0,1*}1*}1}, nR{0{0{0{1*}1*}1*}1}, etc. etc. etc.
The limit of TAN is nR{0{0{0...{0,1*}...1*}1*}1} with n nests of {0 and 1*}.
Analysis
The equvalent ordinal of {0{1*}1} is \(I(\omega,0)\), {0{1*}2} is \(I(\omega,1)\), and {0{1*}0,1} is \(I(\omega+1,0)\). Further,
{0{1*}0{1*}1} has level of \(I(\omega2,0)\)
{0{2*}1} has level of \(I(\omega^2,0)\)
{0{{0}*}1} has level of \(I(\omega^\omega,0)\)
{0{{1}*}1} has level of \(I(\omega^{\omega^2},0)\)
{0{Template:0,1*}1} has level of \(I(\varepsilon_0,0)\)
{0{{{{{0,1},1},1}}*}1} has level of \(I(\Gamma_0,0)\)
{0{{{{0,2},1}}*}1} has level of \(I(\vartheta(\varepsilon_{\Omega+1}),0)\)
{0{{{{{0,3},2},1}}*}1} has level of \(I(\vartheta(\varepsilon_{\Omega_2+1}),0)\)
But {0{Template:0,2*}1} or {0{{0,1}*}1} will suddenly jump to about \(\psi(\psi_{I(1,0,0)}(0))\), and the limit of DAN (not including TAN) may be slightly above (or equal to) \(\psi(\psi_{I(1,0,0)}(0))\).
I'm not sure about the growth rate of TAN. It's at least \(\psi(\psi_{\chi(\varepsilon_{M+1})}(0))\), which I think it is at one point. But recently I've found that it may be far beyond the reach of stage function, even the limit of \(\alpha\)-perstage functions.
P.S.
(About stage functions)
Think of these: \(\Omega\) is a cardinal.
\(\Omega_2\) is next cardinal
\(\Omega_3\) is next next cardinal
\(\Omega_{\Omega_2}\) is next cardinal-ex-next cardinal
\(psi_I(0)\) is (1,0)-ex-next cardinal
I is inaccessible cardinal (stage-1 cardinal)
\(I_2\) is next inaccessible cardinal
\(I_I\) is inaccessible cardinal-ex-next inaccessible cardinal
\(psi_{I(1,0)}(0)\) is (1,0)-ex-next inaccessible cardinal
\(I(1,0)\) is inaccessible inaccessible cardinal
\(I(1,1)\) is next inaccessible inaccessible cardinal
\(I(2,0)\) is inaccessible inaccessible inaccessible cardinal
\(I(1,0,0)\) is (1,0)-ex-inaccessible cardinal
\(M\) is mahlo cardinal (stage-2 cardinal)
\(M_2\) is next mahlo cardinal
\(M(1,0)\) is inaccessible mahlo cardinal (stage-1 stage-2 cardinal)
\(M(1,0,0)\) is (1,0)-ex-inaccessible mahlo cardinal
\(\Xi(3,0)\) is mahlo mahlo cardinal
\(\Xi(3,1)\) is next mahlo mahlo cardinal
\(\Xi(4,0)\) is mahlo mahlo mahlo cardinal
\(\Xi(K,0)\) is (1,0)-ex-mahlo cardinal
\(K\) is compact cardinal (stage-3 cardinal)
\(K_2\) is next compact cardinal
\(\Xi_2(1,0)\) is inaccessible compact cardinal
\(\Xi_2(2,0)\) is mahlo compact cardinal (stage-2 stage-3 cardinal)
\(\Xi_2(3,0)\) is mahlo mahlo compact cardinal
Diagonalizer of \(\Xi_2()\) is compact compact cardinal (stage-3 stage-3 cardinal)
Diagonalizer of \(\Xi_3()\) is compact compact compact cardinal
Diagonalizer of \(\Xi_\alpha(\beta,\gamma)\) in subscript is ultimate cardinal...
Inaccessible is stage 1, mahlo is stage 2, compact is stage 3, and ultimate is stage 4... If we use a \(S(\alpha)\) function for stage \(\alpha\), and use T for a diagonalizer, we'll call this T "stage cardinal".
Then next stage cardinal, inaccessible stage cardinal, mahlo stage cardinal,... and stage stage cardinal.
Then stage stage stage cardinal, (1,0)-ex-stage cardinal, and superstage-1 cardinal for diagonalizer of ()-ex-stage cardinals...
Then stage superstage-1 cardinal, superstage-1 superstage-1 cardinal, superstage-2 cardinal, and superstage cardinal...
Then superstage superstage cardinal, duperstage-1 cardinal, duperstage-2 cardinal, and duperstage cardinal...
Then truperstage cardinal, quadruperstage cardinal, \(\omega\)-perstage cardinal, and (1,0)-perstage cardinal...
But these're just imaginations without definitions...