## FANDOM

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Okay, here is a new (complete?) version of Dollar Function.

## Symbols

$$\bullet$$ can be anything, but it has a higher level than the thing your're expanding.

$$\circ$$ is a set of brackets

$$\diamond$$ is a bracket ( with content ) or a & with only lower level brackets surrounding it

$$\text{◆}$$ is a group of zeroes

$$\text{◈}$$ is a subarray with not any non-nested numbers. Example: [[0]2][0] Counterexample: [0]3

$$v(a)$$ is the subarray $$a$$ after applying a rule to it once. Note that $$v(a)$$ has a lower level than $$a$$ itself, but a higher level than any other subarray that has a lower level than $$a$$. $$v(a)$$ should be solved when you have to solve it.

1. If there is nothing after the $, use rule 0 2. If there are any non-nested non-subscript numbers, use rule 1 3. If there are any non-nested non-subscript [0]'s, use rule 2 4. If there are any non-nested non-subscript [b]'s, use rule 3 5. If the previous things doesn't apply but the lowest level bracket can be solved with normal bracket notation: 1. Search in the bracket for the least nested lowest level bracket or number 1. If it is a 0: 1. If the zero is the only content, use rule 2 2. Otherwise, use rule 5 2. If it is another number, use rule 3 3. If it is a bracket: Return to step 5 6. If the lowest level bracket can be solved with extended bracket notation: 1. If the level of the outermost bracket is 1, use rule 7 2. If the level of the outermost bracket contains a number larger than 1: 1. If the bracket contains only brackets with a larger subscript than the outermost bracket, use rule 8 2. If the content is a zero, use rule 6 3. Otherwise, use rule 9 3. If it is an bracket with no non-nested numbers 1. If the content is a 0, solve the bracket with one step and use rule 6 2. If the content is a number, use rule 10 7. If you have to solve it with linear array notation: 1. If the last entry is a zero, use rule 5b 2. If the first entry is non-zero and the level is 1, use rule 11 3. If the first entry is non-zero and the level isn't 1, use rule 12 4. Otherwise, use rule 1-10 to solve the first entry 8. If you have to solve it with extended array notation: 1. If there is an array with comma's before the first arrow, solve it using rules 1-12 2. If there is a non-zero number after the first arrow, use rule 13 3. If there is a bracket solve bracket until there is a number using rules 1-12 4. If there is an array, solve it using rules 1-14 9. If you have to solve it with ultimate array notation: 1. If there is 1 normal &, use rule 15 2. If there is an array of &, use rule 16 3. If there is an extended array of &, use rule 17 4. If there is more then 1 &, use rule 18 1. If there the bracket with one & less doesn't exist, use rule 19 5. If you have a structure of &, use rule 20, 21 and 22 6. If you have an {} array, use rule 23, 5c or 13-22 ## Bracket Notation 0. If there is nothing after the$, the array is solved. The value of the array is the number before the \$.

1. $$a\b\bullet=(a+b)\\bullet$$

2. $$a\\circ[0]\bullet\circ=a\\circ a\bullet\circ$$

3. $$a\\circ[\bullet+1]\bullet\circ=a\\circ[\bullet][\bullet]...[\bullet][\bullet]\bullet\circ$$ with a $$\bullet$$'s

4. There is only one dollar sign in the array, and that is the one after the first number.

5. If the bracket contains a zero and the bracket has other content, you can remove the zero.

## Extended Bracket Notation

6. $$a\[0]_{b\bullet}\bullet=a\[[...[[0]_{(b-1)\bullet}]_{(b-1)\bullet}...]_{(b-1)\bullet}]_{(b-1)\bullet}\bullet$$ with a nests

7. $$a\\circ[\diamond]\circ=a\\circ\diamond\diamond...\diamond\diamond\circ$$ with a $$\diamond$$s

8. $$a\\circ[\diamond]_{b\bullet}\circ=a\\circ[[...[[\diamond]_{(b-1)\bullet}]_{(b-1)\bullet}...]_{(b-1)\bullet}]_{(b-1)\bullet}\circ$$

9. $$a\[c\bullet]_{b\bullet}\bullet=a\[[...[[[c-1\bullet]_{b\bullet}]_{(b-1)\bullet}]_{(b-1)\bullet}]...]_{(b-1)\bullet}]_{(b-1)\bullet}\bullet$$

10. $$a\[c\bullet]_{\text{◈}}\bullet=a\[[...[[[c-1\bullet]_{\text{◈}}]_{v(\text{◈})}]_{v(\text{◈})}]...]_{v(\text{◈})}]_{v(\text{◈})}\bullet$$

Normal brackets have level 1

## Linear Array Notation

11. $$a\\bullet[\text{◆},0,b,\bullet]\bullet = a\\bullet[\text{◆},[\text{◆},0,b-1,\bullet]_{[\text{◆},0,b-1,\bullet]_{[\text{◆},0,b-1,\bullet]_{...}}},b-1,\bullet]\bullet$$ with a nests

12. $$a\\bullet[\text{◆},0,b,\bullet]_{c+1}\bullet = a\\bullet[\text{◆},[\text{◆},[\text{◆},...,b,\bullet]_{c},b,\bullet]_{c},b,\bullet]_{c}\bullet$$ with a nests

5b. $$\bullet[\bullet0]_\bullet\bullet = \bullet[\bullet]_\bullet\bullet$$

## Extended Array Notation

13. $$a\[0 \rightarrow_{b\bullet}c\bullet]_d = a\[a\rightarrow_{(b-1)\bullet}a...a\rightarrow_{(b-1)\bullet}a \rightarrow_{b\bullet}c-1\bullet]_d$$

14. $$a\[\text{◆},0 \rightarrow_{\bullet} \text{◆},0,1] = a\[\text{◆},0 \rightarrow_{\bullet} \text{◆},[\text{◆},0 \rightarrow_{\bullet} \text{◆},1]_{[\text{◆},0 \rightarrow_{\bullet} \text{◆},1]_{...}}]$$

Commas have arrow level 0 and untyped arrows have arrow level 1.

## Ultimate Array Notation

15. $$[0 \rightarrow_{[0]\text{&}b\bullet} 1] = [0 \rightarrow_{[0 \rightarrow_{[0 \rightarrow_{[0 \rightarrow_{...} 1]\text{&}b-1\bullet} 1]\text{&}b-1\bullet} 1]\text{&}(b-1)\bullet} 1]$$

16. $$[0]\text{&}0...\text{&}0\text{&}b = [0]\text{&}0...\text{&}([0]\text{&}0...\text{&}([0]\text{&}0...\text{&}(...)\text{&}b-1)\text{&}b-1)\text{&}b-1$$

17. $$[0] \text{&...&}\rightarrow_{b\bullet}c\bullet = [0]\text{&...&}\rightarrow_{(b-1)\bullet}a...a\text{&...&}\rightarrow_{(b-1)\bullet}a \text{&...&}\rightarrow_{b\bullet}c-1\bullet$$

18. $$[0\text{&&...&}\rightarrow_{[0]\text{&&...&&}1\bullet}1] = [0\text{&&...&}\rightarrow_{[0\text{&&...&&}\rightarrow_{[0\text{&&...&}\rightarrow_{...}1]\bullet}1]\bullet}1]$$

19. If the outermost bracket with one & less doesn't exist, add them.

20. $$[0][\bullet\text{&}]1 = [0][\bullet+[0][\bullet+[0][\bullet+...]1]1]1$$

21. $$\{[_{\text{&}}0],\bullet\} = \{[\text{&}],\bullet\}$$ and the other rules work the same

22. $$[0]\{[_{\text{&&...&&}}0],\bullet\}1= [0]\{[_{\text{&&...&}}0]\{[_{\text{&&...&}}0]\{...\}1,\bullet\}1,\bullet\}1$$

23. $$\{\text{◆},0,n,\bullet\} =\{\text{◆},[_{\{\text{◆},[_{\{\text{◆},[_{...}],n-1,\bullet\}}],n-1,\bullet\}}],n-1,\bullet\}$$ and extended and ultimate arrays work the same.

5c. Zeroes at the end of {} should be removed

15b. In {} arrays, you use the arrow of the $$\text{&}\rightarrow$$ to nest.

Normal brackets have level &0 and the outermost brackets can't have a higher & level than 0. The normal rules also apply to higher level brackets.