Busy Beaver - Proof For Uncomputability of S(n) and Σ(n)

Proof For Uncomputability of S(n) and Σ(n)

Suppose that S(n) is a computable function and let EvalS denote a TM, evaluating S(n). Given a tape with n 1s it will produce S(n) 1s on the tape and then halt. Let Clean denote a Turing machine cleaning the sequence of 1s initially written on the tape. Let Double denote a Turing machine evaluating function n + n. Given a tape with n 1s it will produce 2n 1s on the tape and then halt. Let us create the composition Double | EvalS | Clean and let n₀ be the number of states of this machine. Let Create_n₀ denote a Turing machine creating n₀ 1s on an initially blank tape. This machine may be constructed in a trivial manner to have n₀ states (the state i writes 1, moves the head right and switches to state i + 1, except the state n₀, which halts). Let N denote the sum n₀ + n₀.

Let BadS denote the composition Create_n₀ | Double | EvalS | Clean. Notice that this machine has N states. Starting with an initially blank tape it first creates a sequence of n₀ 1s and then doubles it, producing a sequence of N 1s. Then BadS will produce S(N) 1s on tape, and at last it will clear all 1s and then halt. But the phase of cleaning will continue at least S(N) steps, so the time of working of BadS is strictly greater than S(N), which contradicts to the definition of the function S(n).

The uncomputability of Σ(n) may be proved in a similar way. In the above proof, one must exchange the machine EvalS with EvalΣ and Clean with Increment - a simple TM, searching for a first 0 on the tape and replacing it with 1.

The uncomputability of S(n) can also be established by reference to the blank tape halting problem. The blank tape halting problem is the problem of deciding for any Turing machine whether or not it will halt when started on an empty tape. The blank tape halting problem is equivalent to the standard halting problem and so it is also uncomputable. If S(n) was computable, then we could solve the blank tape halting problem simply by running any given Turing machine with n states for S(n) steps; if it has still not halted, it never will. So, since the blank tape halting problem is not computable, it follows that S(n) must likewise be uncomputable.