User blog:Wythagoras/Proof that S(n+1)-S(n) is unbounded

Proof that S(n+1) > S(n)
You probably all know this proof.

Here it is:
 * 1) Simulate the machine for S(n). Instead of halting, go to the new state.
 * 2) Halt using the new state.

Proof that S(n+1) > S(n) + 1
Peng said in chat that he had never seen a proof of this one before. So, after a while I came up with this:
 * 1) Simulate the machine for S(n). Instead of halting, go to the new state and to the right. Leave the symbol that IS NOT to the right.
 * 2) In the new state go to the left and stay in the new state.
 * 3) Now the head reads opposite symbol as before, as we left it like that in step 1. So now we can use this to halt.

Proof that S(n+1) > S(n) + 2
Let R be the symbol to the right, and H is the other symbol. Here is now the ruleset: oldhaltingstate oldhaltingsymbol * r newstate newstate       R                H l oldhaltingstate newstate       H                * r halt So, yeah, 3 extra steps.
 * 1) Simulate the machine for S(n). Instead of halting, go to the new state and to the right. Do not change symbol.
 * 2) In the new state go to the left, leave opposite symbol and go to the old halting state.
 * 3) Now the head reads same symbol as before, as we left it like that in step 1. So now we can use this go to the right again.
 * 4) Now there is opposite symbol at this place and we are in the new state, so we halt.

Proof that S(n+2) > S(n) + 7
Let R1 be the symbol to the first right, and H1 is the other symbol. Let R2 be the symbol to the second right, and H2 is the other symbol. So the tape looks like this: oldhaltingsymbol R1 R2

Here is now the ruleset: oldhaltingstate oldhaltingsymbol * r newstate newstate       R1               H1 l oldhaltingstate newstate       H1               R1 r newstate2 newstate2      R2               H2 l newstate newstate2      H2               *  r halt

Proof that \(S(n+k) > S(n) + 3\cdot2^k-k-3\)
Let Rk be the symbol to the kth right, and Hk is the other symbol. oldhaltingsymbol R1 R2 R3 .... Rn Here is now the ruleset: oldhaltingstate oldhaltingsymbol * r newstate newstate       R1               H1 l oldhaltingstate newstate       H1               R1 r newstate_2 newstate_i     Ri               Hi l newstate_(i-1) for all 1<i<k newstate_i     Hi               Ri r newstate_(i+1) for all 1<i<k newstate_k     Rk               Hk l newstate_(k-1) newstate_k     Hk               *  r halt Now we notice that the number of additional steps \(u_n\) can be defined by the following recurrence relation:

\(u_{n+1}=2u_n+n+1\).

With a bit of higher math we find the following direct formula of \(u_n\):

\(u_n=3\cdot2^n-n-2\).

To make it sharp inequality, we need to subtract one.

Proof that S(n+1)-S(n) is unbounded

 * Feel free to edit this section for clarity improving.

For this we use the pigeonhole principle. We see with the pigeonhole principle that at least one difference S(n+1)-S(n) is \(\rfloor\frac{3\cdot2^k-k-3}{k}\lfloor\). As \(\rfloor\frac{3\cdot2^k-k-3}{k}\lfloor\) is unbounded (I will not go into the details), S(n+1)-S(n) is unbounded.