User:Polygon/Page for testing: Difference between revisions

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Polygon (talk | contribs)
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Expanded rules
Polygon (talk | contribs)
955 edits
Added a calculation
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Placeholder
{{TM|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD|halt}}
<pre>
<pre>
S is any tape configuration
Let's rewrite the first rule as:
1. S D> 2^a S --> S 2^a D> S
A(a, b) --> A(2a + 4, b - a - 3) if b >= a + 3
2. S B> 1^a S --> S 1^a B> S
Then, after n applications of rule 1 (assuming b is large enough):
3. S 1 B> 0 S --> S <A 1^2 S
A(a, b) --> A(2^n * a + 2^(n + 2) - 4, b - 2^n * a + a - 2^(n + 2) + n + 4)
4. S D> (11)^a S --> S (21)^a D> S
Considering that a is always set to 4 when leaving B, this simplifies to:
  S A> (11)^a S --> S (12)^a A> S
A(4, b) --> A(2^(n + 3) - 4, b - 2^(n + 3) + n + 8)
5. S (21)^a <C S --> S <C (11)^a S
Now consider the halting configuration: A(a, a + 1):
  S (12)^a <A S --> S <A (11)^a S
So, for this to be in a halting configuration, the following must be true:
6. S (12)^a A> 0^2 S --> S <A (11)^a+1 S
2^(n+3)-4 + 1 = b-2^(n+3)+n+8
 
This simplifies to:
7. S (12)^a 2 (12)^b A> 0^2 S --> S (12)^a-1 2 (12)^b+2 A> S
b = 2^(n+4)-n-11
8. S (12)^a 2 (12)^b A> 0^inf --> S 2 (12)^b+2a A> 0^inf
 
9. S (12)^a <D (11)^b 0^inf --> S (12)^a-1 <D (11)^2b+3 0^inf
10. S (12)^a <D (11)^b 0^inf --> S <D (11)^((2^(a))*b+(2^(a))*3-3) 0^inf
11. S (11)^a <D (11)^b 0^inf --> S (11)^a-2 (12)^b+3 <D (11)^3 0^inf
</pre>
</pre>

Revision as of 20:33, 25 October 2025

Let's rewrite the first rule as:
A(a, b) --> A(2a + 4, b - a - 3) if b >= a + 3
Then, after n applications of rule 1 (assuming b is large enough):
A(a, b) --> A(2^n * a + 2^(n + 2) - 4, b - 2^n * a + a - 2^(n + 2) + n + 4)
Considering that a is always set to 4 when leaving B, this simplifies to:
A(4, b) --> A(2^(n + 3) - 4, b - 2^(n + 3) + n + 8)
Now consider the halting configuration: A(a, a + 1):
So, for this to be in a halting configuration, the following must be true:
2^(n+3)-4 + 1 = b-2^(n+3)+n+8
This simplifies to:
b = 2^(n+4)-n-11
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