1. RE: Defeating Murphy's laws and a new challenge
- Posted by rforno at tutopia.com
Apr 19, 2003
Pete:
What is the value for 'a' in Data[Lendata]+=2*a+1 ?
----- Original Message -----
From: Pete Lomax <petelomax at blueyonder.co.uk>
Subject: Re: Defeating Murphy's laws and a new challenge
On Thu, 17 Apr 2003 20:03:05 -0300, rforno at tutopia.com wrote:
>What perplexes me is that this algorithm defeats Murphy's laws. In fact, it
>is nearly impossible to find a data set to make it run slow.
>This is the new challenge for all: find this kind of data set.
Bored minds think alike....
I was thinking about odd numbers before but realised you'd caught at
least a few obvious cases, so I said nowt. A quick experiment or two
gave me this nasty little spanner:
Data*=2
Data = reverse(sort(Data))
Data[1]+=1
Data[Lendata]+=2*a+1
Not unexpectedly that gives appalling times for sets as low as 30-40.
Multiple search threads with random sort patterns?
>As this problem is closely related to what is called the knapsack problem,
I
>wonder if this solution can be applied to it. When I have some more spare
>time, I'll give it a try.
If you ever do, the program should have both volume and weight limits,
with higher volume a better solution for equal weights.
> Years ago I wrote a program in APL for filling to
>brim a number of diskettes. I'm thinking of translating it into Euphoria.
LOL guess what my 15-year old cobol program did?
Pete
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2. RE: Defeating Murphy's laws and a new challenge
- Posted by rforno at tutopia.com
Apr 19, 2003
Pete:
OK. But anyway, using your approach, I found number sequences that force the
program to work really hard. Now I'm analyzing them and perhaps I'll find a
way to defeat Murphy once more...
Regards.
----- Original Message -----
From: Pete Lomax <petelomax at blueyonder.co.uk>
Subject: Re: Defeating Murphy's laws and a new challenge
On Fri, 18 Apr 2003 23:00:17 -0300, rforno at tutopia.com wrote:
>Pete:
>What is the value for 'a' in Data[Lendata]+=2*a+1 ?
My apologies. I was updating the wrong ends of the array which is why
it defeated all the optimisations. Sorry. I take it all back, I can't
find a data set which messes it up either 
One small point: if the highest value in the array is a partition all
by itself, eg {111,9,9,9,9,9,5,1,1,1}, then the partitions are
correctly printed but BestVal is zero.
Pete
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