User blog:King2218/BAL Notation Part 4

Let's make the numbering system of both the ARRAY and the LEVEL the same.

It's a bit complicated right now.

Intro
In Part 3, googleaarex mentioned something called a Mega n-bal which is:

$$\text{Mega n-bal} = n(\underbrace{n, n, \cdots, n}_{n})\{0\}$$

It gave me the basis on a multidimensional LEVEL. (I can't really think of anything to make the ARRAY multidimensional but meh)

I tried to make multiple LEVELs but ended up just making them compact by storing them inside a multidimensional LEVEL that contains SEPARATORS.

Things got complicated and I decided to shift the numbering system of the ARRAY and the SEPARATORS into something more reasonable.

Here is an example of how to evaluate:

$$2(0,1[1]1)\{0\}$$
 * $$= 2(2,0[1]1)\{0\}$$
 * $$= 2(2[1]1)\{0\}$$
 * $$= 2(1[1]1)\{2,2\}$$
 * $$= 4(1[1]1)\{1,2\}$$
 * $$= 4\uparrow\uparrow4(1[1]1)\{0,2\}$$
 * $$= 4\uparrow\uparrow4(0[1]1)\{4\uparrow\uparrow4,1\}$$
 * $$= 4\uparrow\uparrow4(\underbrace{4\uparrow\uparrow4, 4\uparrow\uparrow4, \cdots}_{4\uparrow\uparrow4}[1]0)\{4\uparrow\uparrow4,1\}$$
 * $$= 4\uparrow\uparrow4(\underbrace{4\uparrow\uparrow4, 4\uparrow\uparrow4, \cdots}_{4\uparrow\uparrow4})\{4\uparrow\uparrow4,1\}$$

Indeed, this is very large.

If you didn't get the process, read along.

After the last step shown above, the BASE becomes $$(4\uparrow\uparrow4)\uparrow^{(4\uparrow\uparrow4) + 1}(4\uparrow\uparrow4)$$.

If you're curious about the +1 on the number of arrows, it's because I changed the numbering system so I have to add one to get the right amount of arrows.

Let's look at the rules.

Even More Rules
Let $$\#$$ and $$\#^*$$ represent rest of ARRAY/LEVEL. For $$\#$$, there MUST be a rest of ARRAY/LEVEL but for $$\#^*$$, it's optional.

Let $$a$$, $$b$$, $$c$$, and $$d$$ be COUNTING NUMBERS.

Let '$$[0]$$' be '$$,$$'.


 * $$a(0)\{0\} = a$$ (You'll notice the shift in the numbering system here)
 * $$a(\#)\{\#, 0\} = a(\#)\{\#\}$$
 * $$a(\#)\{\underbrace{0, 0, \cdots, 0}_{n}, b, \#^*\} = a(\#)\{\underbrace{a, a, \cdots, a}_{n}, b - 1, \#^*\}$$
 * $$a(c \#^*)\{0\} = a(c-1 \#^*)\{\underbrace{a, a, \cdots, a}_{a}\}$$
 * $$a(\underbrace{0, 0, \cdots, 0}_{n}, c \#)\{\#\} = a(\underbrace{a, a, \cdots, a}_{n}, c - 1 \#^*)\{\#\}$$
 * $$a(c \#^*)\{b, \#^*\} = a\uparrow^{c+1}a(c \#^*)\{b - 1, \#^*\}$$
 * $$a(\# [d - 1] 0)\{\#\} = a(\#)\{\#\}$$
 * $$a(\#, 0 [d] \#)\{\#\} = a(\# [d] \#)\{\#\}$$
 * $$a(0 [d_1] \cdots 0 [d_n] c, \#^*)\{\#\} = a(\underbrace{a [d_1 - 1] \cdots [d_1 - 1] a}_{a} [d_1] \cdots$$
 * $$ \cdots \underbrace{a [d_n - 1] \cdots [d_n - 1] a}_{a} [d_n] c - 1, \#^*)\{\#\}$$

Three new rules.

The last two bullets are just one rule.

You can see how fast this grows by examining the example earlier. LARGE, right? And that's just $$[1]$$.

Megabal Group

 * $$\text{Mega Unibal} = 1(0 [1] 1)\{0\}$$
 * $$\text{Mega Bibal} = 2(0 [1] 1)\{0\}$$
 * $$\text{Mega Tribal} = 3(0 [1] 1)\{0\}$$
 * $$\text{Mega Quadribal} = 4(0 [1] 1)\{0\}$$
 * $$\text{Mega Decabal} = 10(0 [1] 1)\{0\}$$
 * $$\text{Mega King Bal} = 2218(0 [1] 1)\{0\}$$
 * $$\text{Mega Balgong} = 100000(0 [1] 1)\{0\}$$

Megablex Group

 * $$\text{Unimegablex} = \text{Mega Unibal}(0 [1] 1)\{0\}$$
 * $$\text{Bimegablex} = \text{Mega Bibal}(0 [1] 1)\{0\}$$
 * $$\text{Trimegablex} = \text{Mega Tribal}(0 [1] 1)\{0\}$$
 * $$\text{Quadrimegablex} = \text{Mega Quadribal}(0 [1] 1)\{0\}$$
 * $$\text{Decamegablex} = \text{Mega Decabal}(0 [1] 1)\{0\}$$
 * $$\text{King Megablex} = \text{Mega King Bal}(0 [1] 1)\{0\}$$
 * $$\text{Megablexigong} = \text{Mega Balgong}(0 [1] 1)\{0\}$$

Megadublex Group

 * $$\text{Unimegadublex} = \text{Unimegablex}(0 [1] 1)\{0\}$$
 * $$\text{Bimegadublex} = \text{Bimegablex}(0 [1] 1)\{0\}$$
 * $$\text{Trimegadublex} = \text{Trimegablex}(0 [1] 1)\{0\}$$
 * $$\text{Quadrimegadublex} = \text{Quadrimegablex}(0 [1] 1)\{0\}$$
 * $$\text{Decamegadublex} = \text{Decamegablex}(0 [1] 1)\{0\}$$
 * $$\text{King Megadublex} = \text{King Megablex}(0 [1] 1)\{0\}$$
 * $$\text{Megadublexigong} = \text{Megablexigong}(0 [1] 1)\{0\}$$

Hyperbal Group

 * $$\text{Hyper Unibal} = 1(1 [1] 1)\{0\}$$
 * $$\text{Hyper Bibal} = 2(0 [2] 2)\{0\}$$
 * $$\text{Hyper Tribal} = 3(0 [3] 3)\{0\}$$
 * $$\text{Hyper Quadribal} = 4(0 [4] 4)\{0\}$$
 * $$\text{Hyper Decabal} = 10(0 [10] 10)\{0\}$$
 * $$\text{Hyperior Bal} = 2218(0 [2218] 2218)\{0\}$$
 * $$\text{Hyper Balgong} = 100000(0 [100000] 100000)\{0\}$$

Hyperblex Group

 * $$\text{Unihyperblex} = \text{Hyper Unibal}(0 [\text{Hyper Unibal}] \text{Hyper Unibal})\{0\}$$
 * $$\text{Bihyperblex} = \text{Hyper Bibal}(0 [\text{Hyper Bibal}] \text{Hyper Bibal})\{0\}$$
 * $$\text{Trihyperblex} = \text{Hyper Tribal}(0 [\text{Hyper Tribal}] \text{Hyper Tribal})\{0\}$$
 * $$\text{Quadrihyperblex} = \text{Hyper Quadribal}(0 [\text{Hyper Quadribal}] \text{Hyper Quadribal})\{0\}$$
 * $$\text{Decahyperblex} = \text{Hyper Decabal}(0 [\text{Hyper Decabal}] \text{Hyper Decabal})\{0\}$$
 * $$\text{Hyperior Blex} = \text{Hyperior Bal}(0 [\text{Hyperior Bal}] \text{Hyperior Bal})\{0\}$$
 * $$\text{Hyperblexigong} = \text{Hyper Balgong}(0 [\text{Hyper Balgong}] \text{Hyper Balgong})\{0\}$$

Hyperdublex Group

 * $$\text{Unihyperdublex} = \text{Unihyperblex}(0 [\text{Unihyperblex}] \text{Unihyperblex})\{0\}$$
 * $$\text{Bihyperdublex} = \text{Bihyperblex}(0 [\text{Bihyperblex}] \text{Bihyperblex})\{0\}$$
 * $$\text{Trihyperdublex} = \text{Trihyperblex}(0 [\text{Trihyperblex}] \text{Trihyperblex})\{0\}$$
 * $$\text{Quadrihyperdublex} = \text{Quadrihyperblex}(0 [\text{Quadrihyperblex}] \text{Quadrihyperblex})\{0\}$$
 * $$\text{Decahyperdublex} = \text{Decahyperblex}(0 [\text{Decahyperblex}] \text{Decahyperblex})\{0\}$$
 * $$\text{Hyperior Dublex} = \text{Hyperior Blex}(0 [\text{Hyperior Blex}] \text{Hyperior Blex})\{0\}$$
 * $$\text{Hyperdublexigong} = \text{Hyperblexigong}(0 [\text{Hyperblexigong}] \text{Hyperblexigong})\{0\}$$

More Extensions?
Of course there are but I'm going to post them when I feel like it.

For now, don't expect a new part to come out!

Oh and...
The rules were hard...

Did I miss a rule?

Is a rule wrong?

Comment about it!