User:Polygon/Page for testing: Difference between revisions

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Undo revision 4160 by Polygon (talk) uncommented the starting configuration again
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Polygon (talk | contribs)
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Added a calculation
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={{TM|1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD|halt}}=
{| class="wikitable"
|+
!
!0
!1
!2
|-
|A
|1RB
|1RD
|1LC
|-
|B
|2LB
|1RB
|1LC
|-
|C
|1RZ
|1LA
|1LD
|-
|D
|2RB
|2RA
|2RD
|}
<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
by:
S (12)^a 2 (12)^b A> 0^2 S
--> S (12)^a 2 (12)^b 1 B> 0 S
--> S (12)^a 2 (12)^b <A (11) S
--> S (12)^a 2 <A (11)^b+1 S
--> S (12)^a <C 1 (11)^b+1 S
--> S (12)^a-1 1 <D (11)^b+2 S
--> S (12)^a-1 2 A> (11)^b+2 S
--> S (12)^a-1 2 (12)^b+2 A> S
 
8. S (12)^a 2 (12)^b A> 0^inf --> S 2 (12)^b+2a A> 0^inf
Obtained by repeating rule 8.
 
9. S (12)^a <D (11)^b 0^inf --> S (12)^a-1 <D (11)^2b+3 0^inf
by:
S (12)^a <D (11)^b 0^inf
--> S (12)^a D> (11)^b 0^inf
--> S (12)^a (21)^b D> 0^inf
--> S (12)^a (21)^b 2 B> 0^inf
--> S (12)^a (21)^b 2 <B 2 0^inf
--> S (12)^a (21)^b <C 1 2 0^inf
--> S (12)^a <C (11)^b 1 2 0^inf
--> S (12)^a-1 1 <D (11)^b+1 2 0^inf
--> S (12)^a-1 2 A> (11)^b+1 2 0^inf
--> S (12)^a-1 2 (12)^b+1 A> 2 0^inf
--> S (12)^a-1 2 (12)^b+1 <C 1 0^inf
--> S (12)^a-1 2 (12)^b 1 <D 11 0^inf
--> S (12)^a-1 2 (12)^b 2 A> (11)^1 0^inf
--> S (12)^a-1 2 (12)^b 2 (12)^1 A> 0^inf
--> S (12)^a-1 2 2 (12)^2b+1 A> 0^inf
--> S (12)^a-1 2^2 <A (11)^2b+2 0^inf
--> S (12)^a-1 2 <C 1 (11)^2b+2 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
Obtained by repeating rule 9.
 
11. S (11)^a <D (11)^b 0^inf --> S (11)^a-2 (12)^b+3 <D (11)^3 0^inf
by:
S (11)^a <D (11)^b 0^inf
--> S (11)^a-1 1 2 A> (11)^b 0^inf
--> S (11)^a-1 (12)^b+1 A> 0^inf
--> S (11)^a-1 <A (11)^b+2 0^inf
--> S (11)^a-1 D> (11)^b+2 0^inf
--> S (11)^a-1 (21)^b+2 D> 0^inf
--> S (11)^a-1 (21)^b+2 2 B> 0^inf
--> S (11)^a-1 (21)^b+2 2 <B 2 0^inf
--> S (11)^a-1 (21)^b+2 <C (12)^1 0^inf
--> S (11)^a-1 <C (11)^b+2 1 2 0^inf
--> S (11)^a-2 1 <A (11)^b+3 2 0^inf
--> S (11)^a-2 1 D> (11)^b+3 2 0^inf
--> S (11)^a-2 1 (21)^b+3 D> 2 0^inf
--> S (11)^a-2 1 (21)^b+3 2 D> 0^inf
--> S (11)^a-2 1 (21)^b+3 2^2 B> 0^inf
--> S (11)^a-2 1 (21)^b+3 2^2 <B 2 0^inf
--> S (11)^a-2 1 (21)^b+3 2 <C (12)^1 0^inf
--> S (11)^a-2 1 (21)^b+3 <D 1 1 2 0^inf
Note that 1 (21)^k = (12)^k 1
= S (11)^a-2 (12)^b+3 1 <D (11)^1 2 0^inf
--> S (11)^a-2 (12)^b+3 2 A> (11)^1 2 0^inf
--> S (11)^a-2 (12)^b+3 2 (12)^1 A> 2 0^inf
--> S (11)^a-2 (12)^b+3 2 (12)^1 <C 1 0^inf
--> S (11)^a-2 (12)^b+3 2 1<D (11)^1 0^inf
--> S (11)^a-2 (12)^b+3 2^2 A> (11)^1 0^inf
--> S (11)^a-2 (12)^b+3 2^2 (12)^1 A> 0^inf
--> S (11)^a-2 (12)^b+3 2^2 (12)^1 1 B> 0^inf
--> S (11)^a-2 (12)^b+3 2^2 (12)^1 <A (11)^1 0^inf
--> S (11)^a-2 (12)^b+3 2^2 1 <C 1 (11)^1 0^inf
--> S (11)^a-2 (12)^b+3 2^2 <A (11)^2 0^inf
--> S (11)^a-2 (12)^b+3 2 <C 1 (11)^2 0^inf
--> S (11)^a-2 (12)^b+3 <D (11)^3 0^inf
</pre>
Let A(a,b,c) = S (11)^a (12)^b <D (11)^c 0^inf
* Rule 9: A(a, b, c) --> A(a, b - 1, 2c + 3)
* Rule 10: A(a, b, c) --> <math>A(a,0,2^{b} \times c + 2^{b} \times 3 - 3)</math> which becomes <math>A(a,0,2^{b+1} \times 3 - 3)</math> if c = 3.
* Rule 11: A(a, 0, c) --> A(a - 2, c + 3, 3)
 
Further: let <math>f(n) = 2^{n+1} \times 3</math>
* If c = 3: A(a, b, 3) --> A(a, 0, f(c) - 3) --> A(a - 2, f(c), 3)
 
* A(2k + d, 0, c) --> <math>A(d, f^{k-1}(c+3), 3)</math>
<pre>
The TM enters configuration A(19, 2, 3) after 799 steps with tape:
0^inf 2 1 (11)^19 (12)^2 <D (11)^3 0^inf
</pre>
==Trajectory==
A(19, 2, 3) --> A(19, 0, 21) --> <math>A(1, f^{8}(24), 3)</math>
 
--> <math>A(1, 0, f^{9}(24) - 3)</math>
 
Let's have <math> f^{9}(24) - 3</math> = m.
<pre>
Final trajectory:
0^inf 2 1 (11)^1 <D (11)^m 0^inf
--> 0^inf 2 1 1 2 A> (11)^m 0^inf
--> 0^inf 2 1 (12)^m+1 A> 0^inf
--> 0^inf 2 1 <A (11)^m+2 0^inf
--> 0^inf 2 1 D> (11)^m+2 0^inf
--> 0^inf (21)^m+3 D> 0^inf
--> 0^inf (21)^m+3 2 B> 0^inf
--> 0^inf (21)^m+3 2 <B 2 0^inf
--> 0^inf (21)^m+3 <C (12)^1 0^inf
--> 0^inf <C (11)^m+3 (12)^1 0^inf
--> 0^inf 1 Z> (11)^m+3 (12)^1 0^inf
Score = 2m + 9
 
Score calculated in HyperCalc:
(10^)^8 30,302,671.815163
Or in tetration: 10^^10.873987 (truncated)
</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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