设计非Sipser版识别{0^(2^n);n>0}的确定性图灵机

Awesome question! Let's work through designing this deterministic Turing machine step by step—since you already have a non-deterministic version sorted, the key here is replacing numeric count storage with clever use of an extended alphabet to track our cross-out totals.

Core Idea: Use Marker Symbols Instead of Storing Counts #

Since we’re allowed an arbitrary-sized alphabet, we can use unique symbols (like A, B, C, D, ...) to represent "count groups". The number of symbols of a given type will directly equal the number of 0s we crossed out in that pass. This lets us avoid ever storing a numeric value—we just use the quantity of markers to guide our next steps.

Step-by-Step Algorithm & State Design #

Let’s break this down into concrete states and transitions that follow your required cross-out rules (1, 1, 2, 4, ... each pass equals the sum of all prior passes):

1. First Two Passes (Cross Out 1 Zero Each) #

• State S₀ (Start):
  • Scan right until you hit the first 0. Replace it with A (this is our first cross-out). Slide back to the left end of the tape, then switch to State S₁.
• State S₁:
  • Scan right again to find the next unmarked 0. Replace it with B (second cross-out). Slide back to the left end, switch to State S₂.
  • If you can’t find an unmarked 0 here? Reject immediately—n>0 means we need at least 2 zeros, so a single zero is invalid.

2. General Passes (Cross Out Sum of All Prior Passes) #

For each pass after the second, we’ll use a new unique symbol (e.g., C for pass 3, D for pass 4, etc.):

• State S_k (Prepare for Pass k):
  • Instead of counting the total number of prior markers, we’ll process them one by one:
    1. Find a marker from the previous pass(es) (e.g., A or B for pass 3). Replace it with the new symbol.
    2. Scan right to find an unmarked 0. Replace it with the same new symbol (this is one cross-out for this pass).
    3. Repeat until all old markers are replaced and we’ve crossed out exactly the sum of all prior passes (since each old marker maps to one cross-out in this pass).
  • Once all old markers are replaced and cross-outs are done, slide back to the left end, switch to State S_check.

3. Termination Check State #

• State S_check:
  • Scan the entire tape for unmarked 0s:
    1. If there are no unmarked 0s left? Accept! We’ve crossed out exactly 2ⁿ zeros (1+1=2, 2+2=4, 4+4=8, etc.—each total is a power of two).
    2. If there are unmarked 0s, but the number of new markers (which equals how many we need to cross out next) is more than the remaining 0s? Reject—we can’t complete the next pass, so the input isn’t a valid 2ⁿ length.
    3. If there are enough 0s left? Move to the state for the next pass (using the next unique symbol in our alphabet).

Why This Is Fully Deterministic #

Every state transition is defined exactly by the symbol under the tape head—no guesswork or non-deterministic choices. We never store numbers; the count is implicit in how many of each marker symbol we have on the tape. Since we can use an infinite alphabet, we’ll never run out of new symbols for each pass.

Example Run for Input 0000 (4 zeros = 2²) #

1. S₀: Replace first 0 with A → A000
2. S₁: Replace next 0 with B → AB00
3. S₂ (Pass 3):
   • Replace A with C, cross out next 0 → CB00 → CC00
   • Replace B with C, cross out next 0 → CCC0 → CCCC
4. S_check: No unmarked 0s left → accept!

内容的提问来源于stack exchange，提问作者Alyssa Buchthal