Conference rusure::math

One of my first reactions to the problem, other than it being basically
interesting, is that since we''re dealing with the squares of up to ten
digits, the squares don't fit in 32 bits, so it seems like it will be 
difficult to work on this problem in Bliss.

So use Bliss-36.
				/AHM

I originally solved the problem using a TI-58 calculator and some ingenuity.
An exhaustive search is going to chew up tons of CPU time, so some ideas to
trim the search are needed.

To the wise guy that said to use BLISS36:  I meant that we're dealing
with the SQUARES of 10-digit numbers, so we're dealing with 20-digit 
numbers, so BLISS36 probably isn't any better than BLISS32.

As far as TI-58, how many digits of exact integer arithmetic does that
calculater have ?

p.s.	Yes, I was starting to realize that brute force was going to
	be unavailable, so I was starting to think about methods that
	eliminate lots of things.  Now that you've admitted you actually
	SOLVED the problem, I'm more motivated.  Thanks.

The TI-58 has 10 display digits and two guard digits. I actually used it 
mostly to explore the 10 MOST significant digits after I had a pretty good
idea what the least significant digits would look like.

Re .4:

So you'd rather use triple precision on a Vax than double precision on
a PDP-10?  Perhaps you should use Bliss-10, then you could declare
REGISTER variables of type VECTOR[2].
				/AHM

re .4
	Is it assured that the "ideas to trim the search" won't also
	trim the sought-after maximum from the search?

I think we can all assume that anything that "trims the search" doesn't
throw out the baby with the bath water by definition.  Otherwise it's
useless.

For motivation, Lynn, please tell us:  Have you actually SOLVED this for THE
maximum answer ?

I'm still thinking about it during various leisuremind
every day.  I haven't gotten too far.  I've been trying to see what sorts
of inspections might dictate that I no longer have to look at some large
set of squares.  For instance, suppose I found this:

		x x x x x\x x x x x\x x x x x\x x x x x
		        y y       y y       y y

That is, suppose I've found one permutation such that there are four
five-digit squares plus three two-digit ones formed at the boundaries.
That permutation gets a score of 4*5+3*2 = 26.  Once I've found this
permutation, can I prove somehow that I don't need to try any squares
larger than some ??? number of digits in size since such squares can't
possibly help me do better than 26.  Or instead of "larger than some number
of digits", maybe some other large set.

Or maybe the score of 26 combined with the fact that we're dealing with a
permutation, so digits can "used up" in a sense, will help.

I'm sort of thinking out loud here, but maybe a company-wide joint solution
could be fun and productive ! So if you've "gotten anywhere at all" on this
problem even though not far, you might consider sharing even though you
don't have a complete solution yet.

Here's a hint: I found a very large solution (>100) which I am not certain
is the absolute maximum. The longest square in my solution was 16 digits.

It's been several months since the previous reply, so I'll record my solution.

The permutation 49497813631827562500 contains the following squares:
49 (twice)     		 4
81			 2
36			 2
25	       		 2
2500	       		 4
625	       		 3
62500	       		 5
5625			 4
562500			 6
75625	       		 5
7562500	       		 7
275625	       		 6
27562500       		 8
18275625       		 8
1827562500     		10
78136318275625 		14
7813631827562500	16
                          
Total		       106

I dimly recall hearing of a solution of 107, but I can't reproduce it. There
are other 106 solutions.

Lynn Yarbrough