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Two Puzzles in Recursion Theory: Verbose Sets and Terse Sets

Let Image may be NSFW.
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K
be the set of all Image may be NSFW.
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n
such that the Image may be NSFW.
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n
th Turing machine halts. (For these puzzles, we will assume that Turing machines are always run on a blank initial state, i.e., they take no input.) Recall that Image may be NSFW.
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K
is computably enumerable, but not decidable.

Puzzle #1. Describe a winning strategy for the following game: You are given three numbers Image may be NSFW.
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x_1, x_2, x_3
. You must correctly say for each number whether or not it is in Image may be NSFW.
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K
. You are allowed to ask (and receive a truthful answer to) two questions of the form “Is Image may be NSFW.
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n
in K?” for any Image may be NSFW.
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n
.

Puzzle #2. Show that there is no winning strategy for the game which is the same as that in Puzzle #1 except you are given two numbers and may ask only one question. (Even stronger, show that if Image may be NSFW.
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S\subset \mathbb{N}
is a set such that you can win that game, then Image may be NSFW.
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S
must be decidable.)

These puzzles are special cases of more general questions answered in Terse sets, superterse sets, and verbose sets by Richard Beigel, William Gasarch, John Gill, and James Owings. I also suggest looking at Richard Beigel’s page of online papers, which has a lot of interesting stuff.

I read about this first in Piergiorgio Odifreddi’s book “Classical Recursion Theory, Volume 1.”

Answer #1. The first thing to do is to find out how many of Image may be NSFW.
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x_1, x_2, x_3
are in Image may be NSFW.
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K
. To do this, let Image may be NSFW.
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n_1
be such that the Image may be NSFW.
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n_1
th Turing machine simulates the Image may be NSFW.
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x_1
th, Image may be NSFW.
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x_2
th and Image may be NSFW.
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x_3
th Turing machines in parallel, and halts after any two of them halt.

By asking if Image may be NSFW.
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n_1
is in Image may be NSFW.
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K
, you will find out either that there are two or more or that there are one or less of Image may be NSFW.
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x_1, x_2, x_3
in Image may be NSFW.
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K
. Use a similar second question to find out exactly how many of Image may be NSFW.
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x_1, x_2, x_3
are in Image may be NSFW.
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K
.

Once you know how many of Image may be NSFW.
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x_1, x_2, x_3
are in Image may be NSFW.
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K
, just run the Image may be NSFW.
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x_1
th, Image may be NSFW.
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x_2
th, and Image may be NSFW.
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x_3
th Turing machines in parallel, and wait until all the ones that are going to halt do halt. Then you will know which of them are in Image may be NSFW.
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K
.

Answer #2. Suppose that you have a winning strategy, and I’ll show how to compute Image may be NSFW.
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S
. Let Image may be NSFW.
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f^x_{\text{no}}(x,y)
be the function which, given Image may be NSFW.
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x
and Image may be NSFW.
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y
, returns the guess for whether or not Image may be NSFW.
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x
is in Image may be NSFW.
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S
, supposing that you received a “no” answer to whatever question your strategy decided to ask. Similarly define Image may be NSFW.
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f^y_{\text{no}}
, Image may be NSFW.
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f^x_{\text{yes}}
, and Image may be NSFW.
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f^y_{\text{yes}}
.

Either it is the case that for all Image may be NSFW.
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x
there is a Image may be NSFW.
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y
such that Image may be NSFW.
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f_{\text{no}}^x(x,y) = f_{\text{yes}}^x(x,y)
or it isn’t. Suppose first that it is. Then given Image may be NSFW.
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x
, we may compute whether or not is in Image may be NSFW.
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S
by searching for such a Image may be NSFW.
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y
and computing Image may be NSFW.
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f_{\text{no}}^x(x,y)
(which equals Image may be NSFW.
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f_{\text{yes}}^x(x,y)
). This must be the correct guess since either “yes” or “no” must be the correct answer to whatever question the strategy asked.

Now suppose that there is an Image may be NSFW.
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x
such that for all Image may be NSFW.
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y
, Image may be NSFW.
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f_{\text{no}}^x(x,y)\ne f_{\text{yes}}^x(x,y)
. Since Image may be NSFW.
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x
is just a single number, we can assume that we know whether or not it’s in Image may be NSFW.
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S
. Then, given any Image may be NSFW.
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y
, since Image may be NSFW.
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f_{\text{no}}^x(x,y)\ne f_{\text{yes}}^x(x,y)
we know which of them is correct, hence we know whether “yes” or “no” is the correct answer to the question that the strategy would pose. Hence we can compute its correct guess as to whether or not Image may be NSFW.
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y
is in Image may be NSFW.
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S
.

Additional Results. The paper linked above gives generalizations of both puzzles: You can find out whether any set of Image may be NSFW.
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2^n - 1
numbers is in Image may be NSFW.
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K
by asking Image may be NSFW.
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n
questions (by the same strategy of determining how many of them are in Image may be NSFW.
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K
), and, for any Image may be NSFW.
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S
, if you can determine whether any set of Image may be NSFW.
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2^n
numbers is in Image may be NSFW.
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S
by asking Image may be NSFW.
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n
questions then Image may be NSFW.
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S
is decidable.

The authors call a set verbose if it is such that you can determine whether any set of Image may be NSFW.
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2^n -1
numbers is in that set by asking Image may be NSFW.
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n
questions. They call it terse if you need Image may be NSFW.
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n
questions to determine whether or not Image may be NSFW.
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n
numbers are in the set. They are show a number of interesting results about these, mainly along the lines that (very roughly) lots of both kinds exist, and in lots of different places.


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