Re: Does the number of nines increase?

On 7/10/2024 5:32 PM, Ben Bacarisse wrote:

"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
 

On 7/10/2024 4:29 PM, Moebius wrote:

Am 11.07.2024 um 01:02 schrieb Moebius:

Am 11.07.2024 um 01:00 schrieb Moebius:

Am 11.07.2024 um 00:55 schrieb Moebius:
>

Math SHOULD BE fun! (imho)

>
Hmmm... It's quite clear that WM doesn't have ANY fun with math, I'd
say.

>
Actually, he doesn't DO any math.

Most cranks don't to any (real) math, they just "criticizes" (sort of)
things they don't really understand.

>
If I am having a hard time understanding something, I ask/read around until
I can finally grasp the underlying meaning of the problem to a point where
I can code up a working solution. It has certainly tended worked for me in
the past.

 Here's and interesting problem to code up.  The input is a finite array
of n pairs of strings, so there are 2n strings in all.  They are often
thought of as strings on the top and bottom of a collection of tiles or
dominos, but that just help visualise the problem.  An example with n=3
might be
    aa     bb    abb
   aab    ba     b

Just making up an example:
ab   ab  ab = ababab
aba  ba   b = ababab
Would that work? Does each string need to be unique?

 The pairs (or tiles) are numbered from 1 to n.  The problem is to
determine, for any finite set of tiles, if there is a finite sequence of
numbers such that when those numbered tiles are laid out (we can use any
tile as often as we like) the concatenation of the top stings is the
same as the concatenation of the bottom strings.
 For example, with these tiles, the sequence 1, 2, 1, 3 produces
    aa    bb    aa    abb
   aab   ba    aab    b

      aa    bb    aab    b

 with top string aabbaaabb = bottom string aabbaaabb.  So for this input,
the program should print "yes".  You don't need to give the sequence
that gives the matching top and bottom strings (though that would be a
neat extra), you just have print yes or no depending on whether such a
sequence of tiles exists or not.