FANDOM


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.

Which rule should you use?

  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

I've added 2 rules.

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.

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