User:Polygon/Page for testing

1RB1RD1LC_2LB1RB1LC_1RZ1LA1LD_2RB2RA2RD (bbch) is a pentational halting BB(4,3) TM. It was discovered in May 2024 by Pavel Kropitz as one of seven long running TMs and achieves a score of over $2{\uparrow }^{4}5<math>,makingittheBB(4,3)champion.==Analysis==<pre>Sisanytapeconfiguration1.SD>2^{a}S-->S{2}^{a}D>S[+asteps]2.SB>1^{a}S-->S{1}^{a}B>S[+asteps]3.SA>0^{2}S-->S<A{1}^{2}S[+5steps]4.SD>(11{)}^{a}S-->S(21{)}^{a}D>S[+2asteps]SA>(11{)}^{a}S-->S(12{)}^{a}A>S[+2asteps]5.S(21{)}^a<CS-->S<C(11{)}^{a}S[+2asteps]S(12{)}^a<AS-->S<A(11{)}^{a}S[+2asteps]6.S(12{)}^{a}A>0^{2}S-->S<A(11{)}^a+1S[+2a+5steps]7.SA>(11{)}^1{2}^{b}S-->S2A>(11{)}^1{2}^b-1S[+5steps]8.SA>(11{)}^1{2}^{b}S-->S{2}^{b}A>(11{)}^{1}S[+5bsteps]9.SD>0^{2}S-->S<B{2}^{2}S[+3steps]10.S2<D(11{)}^a{0}^{2}S-->S<D(11{)}^a+12S[+4a+7steps]11.S2<D(11{)}^{a}20^{2}S-->S<D(11{)}^a+12^{2}S[+4a+7steps]12.S{1}^a<A(11{)}^b{0}^{2}S-->S{1}^a-1<A(11{)}^b+12S[+4b+7steps]13.S{1}^a<A(11{)}^{b}20^{2}S-->S{1}^a-1<A(11{)}^b+12^{2}S[+4b+7steps]14.S(12{)}^{a}1<D(11{)}^b{0}^{2}S-->S(12{)}^a-11<D(11{)}^b+2[+4b+8steps]15.S(12{)}^{a}1<D(11{)}^b{0}^{i}nf-->S1<D(11{)}^b+2a{0}^{i}nf[+4a^2+4ba+4asteps]16.S(12{)}^{a}21<D(11{)}^b{0}^{i}nf-->S(12{)}^a-12(12{)}^b+21<D(11{)}^1{0}^{i}nf[+10b+28steps]17.S(12{)}^{a}21<D(11{)}^b{0}^{i}nf-->S(12{)}^a-121<D(11{)}^{2}b+50^{i}nf18.S(12{)}^{a}21<D(11{)}^b{0}^{i}nf-->S21<D(11{)}^{(}{2}^a)*b+(2^a)*5-50^{i}nf19.S(12{)}^{a}21<D(11{)}^{b}20^{i}nf-->S(12{)}^a{2}^{2}1<D(11{)}^{2}b-10^{i}nf20.S(12{)}^{a}1<D(11{)}^{b}20^{i}nf-->S(12{)}^{a}21<D(11{)}^{2}b-10^{i}nf21.S(12{)}^a{2}^{2}1<D(11{)}^b{0}^{i}nf-->S(12{)}^a-12^{2}1<D(11{)}^2{}^{(}b+4)*3-50^{i}nf22.S1<D(11{)}^b{2}^2{0}^{i}nf-->S2(12{)}^b-121<D(11{)}^1{0}^{i}nf23.S(11{)}^a{2}^{2}1<D(11{)}^b{0}^{i}nf-->S(11{)}^a-3(12{)}^{2}b+112^{2}1<D(11{)}^1{0}^{i}nf24.0^{i}nf{2}^{2}1<D(11{)}^c{0}^{i}nf-->0^{i}nf(11{)}^c+1(12{)}^3{2}^{2}1<D(11{)}^1{0}^{i}nf25.0^{i}nf(11{)}^2{2}^{2}1<D(11{)}^c{0}^{i}nf-->0^{i}nf1(11{)}^{2}c+8(12{)}^3{2}^{2}1<D(11{)}^1{0}^{i}nf26.0^{i}nf1(11{)}^1{2}^{2}1<D(11{)}^c{0}^{i}nf-->0^{i}nf1(11{)}^{2}c+7(12{)}^3{2}^{2}1<D(11{)}^1{0}^{i}nf27.0^{i}nf12^{2}1<D(11{)}^c{0}^{i}nf-->0^{i}nf(11{)}^{2}c+5(12{)}^3{2}^{2}1<D(11{)}^1{0}^{i}nf28.0^{i}nf(11{)}^1{2}^{2}1<D(11{)}^c{0}^{i}nf-->0^{i}nf1Z>1(11{)}^{2}c+80^{i}nf</pre>LetD(a,b,c)=0^{i}nf(11{)}^a(12{)}^b{2}^{2}1<D(11{)}^c{0}^{i}nfLet{D}_1(a,b,c)=0^{i}nf1(11{)}^a(12{)}^b{2}^{2}1<D(11{)}^c{0}^{i}nfLet<math>f_1(n)=2^{n+4}\times 3-5$

Let $f_2(a,b)=f_1^{2\times f_2(a-1,b)+11}(1)$, where$f_2(0,b)=b$

Rule 21 becomes:

• $D(a,b,c)-->D(a,b-1,2^{b+4}\times 3-5)$
• $D_1(a,b,c)-->D_1(a,b-1,2^{b+4}\times 3-5)$

Rule 23 becomes:

• $D(a,0,c)-->D(a-3,2c+11,1)$
• $D_1(a,0,c)-->D_1(a-3,2c+11,1)$

Rule 24 becomes:

• $D(0,0,c)-->D(c+1,3,1)$

Rule 25 becomes:

• $D(2,0,c)-->D(2c+8,3,1)$

Rule 26 becomes:

• $D_1(1,0,c)-->D_1(2c+7,3,1)$

Rule 27 becomes:

• $D_1(0,0,c)-->D(2c+5,3,1)$

Rule 28 becomes:

• D(1, 0, c) --> halt with score 4c + 18

Rule 29 becomes:

• $D_1(2,0,c)-->D(2c+10,2,1)$

By repeating rule 21, a stronger rule can be constructed:

• $D(a,b,c)-->D(a,0,f_1^b(c))$
• $D_1(a,b,c)-->D1(a,0,f_1^b(c))$

If a is greater than or equal to 3: $D(a,0,c)-->D(a-3,2c+11,1)-->D(a-3,0,f_1^{2c+11}(1))$ =$D(a-3,0,f_2(1,c))$

• $D(a,0,c)-->D(a-3,0,f_1^{2c+11}(1))$

This rule can also be repeated, also note that $f_1^{2c+11}(1)=f_2(1,c)$ and $f_1^{2\times f_2(a,b)+11}(1)=f_2(a+1,b)$:

• $D(3k+d,0,c)-->D(d,0,f_2(k,c))$
• $D_1(3k+d,0,c)-->D_1(d,0,f_2(k,c))$

The TM starts in configuration D(2, 2, 1).

D(2, 2, 1) -->

$D(2,0,f_1^2(1))=D(2,0,f_1(91))$

$e_1=f_1(91)=2^{95}\times 3-5$

f_1(n) = 2^(n+4)*3 - 5
Note that the times three means that this expression of of the form 3k - 5 which can be rewritten as 3(k-1)-2 which can again be rewritten as 3(k-2)+1.
Next, 3k+1 mod 3 = 1
So f_1(n) mod 3 = 1
Thus f_1^a(n) mod 3 = 1
f_2(a,b) = f_1^(2*f_2(a-1,b)+11)(1)
Note that f_1^(2*f_2(a-1,b)+11)(1) is also of the form f_1^a(n)
Thus f_2(a,b) mod 3 = 1

$D(2,0,e_1)$

-->$D_1(2e_1+8,3,1)-->D_1(2e_1+8,0,f_1^2(91))$

e_1 mod 3 = 1; 2*1 + 8 = 10 --> 10 mod 3 = 1

$D_1(2e_1+8,0,f_1^2(91))$

--> $D_1(1,0,f_2(\frac{2e_1+7}{3},f_1^2(91)))$

$e_2=f_2(\frac{2e_1+7}{3},f_1^2(91))$

$D_1(1,0,e_3)$

$e_{2}mod3=1$

--> $D_1(2e_2+7,3,1)-->D_1(2e_2+7,0,f_1^2(91))$

2e_3 + 7

Modulus: 2 + 7 --> 9 mod 3 = 0

--> $D_1(0,0,f_2(\frac{2e_2+7}{3},f_1^2(91)))$

$e_3=f_2(\frac{2e_2+7}{3},f_1^2(91))$

$D_1(0,0,e_3)$

--> $D(2e_3+5,3,1)-->D(2e_3+5,0,f_1^2(91))$

e_3 mod 3 = 1; 2*1+5 = 7 --> 7 mod 3 = 1

--> $D(1,0,f_2(\frac{2e_3+4}{3},f_1^2(91)))$

$e_4=f_2(\frac{2e_3+4}{3},f_1^2(91))$

$D(1,0,e_4)$

--> halts with score $4e_4+18$.

This can be bounded by:

$2{\uparrow }^{4}5<2\uparrow \uparrow \uparrow 2\uparrow \uparrow \uparrow 2\uparrow \uparrow \uparrow (7.92\times 10^{28})<e_4<\sigma <S<2\uparrow \uparrow \uparrow 2\uparrow \uparrow \uparrow 2\uparrow \uparrow \uparrow (7.93\times 10^{28})$