Time complexity of an enumeration of SUBSET SUM instances

Question

An instance of the SUBSET SUM problem (given $y$ and $A = \{x_1,...,x_n\}$ is there a non-empty subset of $A$ whose sum is $y$) can be represented on a one-tape Turing Machine with a list of comma separated numbers in binary format. If $\Sigma = \{0,1,\#\}$ a reasonable format could be:

$( 1 \; (0|1)^* \; \#)^* \#$

Where the first required argument is the value $y$ and $\#\#$ encodes the end of the input. For example:

 1  0  0  #  1  0  #  1  #  #
^^^^^^^^     ^^^^     ^
   y          x1     x2
Instance: y=4, A={2,1}

I would like to enumerate the SUBSET SUM instances.

Question: What is the (best) time complexity that can be achieved by a Turing Machine $TM_{Enum}$ that on input $m$ (which can be represented on the tape with a string of size $\log m + 1$) - outputs the $m$-th SUBSET SUM instance in the above format?

EDIT:

Yuval's answer is fine, this is only a longer explanation.

Without loss of generality we set that $y > 0$ and $0 < x_1 \leq x_2 \leq ... \leq x_n$, $n \geq 0$

And we can represent an instance of subset sum using this encoding:

$y \# x_1\# d_2\# ...\# d_{n} \#\#$ where $d_i \geq 1, x_i = x_{i-1} + d_i - 1 \; , i \geq 2$

Using a binary representation for $y,x_1, d_2, d_3, ...$ we have the following representation:

$1 \; ((0|1)^* \# 1)^* \; \#\#$

Equivalent to $1 \; (0|1|\#1)^* \; \#\#$. There is always a leading 1 and a trailing ## so we can consider only the $(0|1|\#1)^*$ part.

So the decoder TM on input $m$ in binary format should:

• output the leading 1
• convert $m$ to base 3 mapping digit 2 to $\#1$
• when outputing the i-th intermediate $\#$ calculate $x_i = d_i + x_{i-1}-1$
• output the trailing $\#\#$

No duplicate instances are generated.

Answer 1

SUBSET-SUM instances can be encoded in base 3. We have codes for $0,1,\#$. Some codings are invalid, but in that case we can just immediately output $\#\#$ (or $\#$, if we have just written $\#$). Every SUBSET-SUM problem has infinitely many encodings, I hope that's not a problem.

If the input has length $\ell$, then (assuming the tape alphabet has at least 4 symbols) we can do the conversion in time $O(\ell^2)$. I don't know whether this is the "best" time complexity achievable.

Edit: Here is a better encoding. We still have only three input codes, $0,1,\#$. The output string always starts with $1$ and ends with $\#\#$. Further, $\#$ is output as $\#1$. Now each output string is generated once, though several output strings could correspond to the same instance.

As an example, your instance is encoded by "00#0#".