topic Datatype problem in project euler in LabVIEW

Datatype problem in project euler

frostie — Wed, 13 May 2009 13:51:31 GMT

Today I had nothing to do, so I participated in Project Euler http://projecteuler.net with Labview. I'm at question 10 now. The question itself isn't much of a challenge, but the problem I'm having is that the datatypes are to small for the value I would like to put in there. The value first goes up, then changes from + to - and then goes back to zero. This all happens while I'm just adding up to the value.

I guess there is a sollution to this problem but I don't seem to find it. Is there someone who can and will help me please?

Re: Datatype problem in project euler

RavensFan — Thu, 14 May 2009 03:32:42 GMT

What datatypes are you trying to use? Are they integer or floating point? How many bytes?

Re: Datatype problem in project euler

frostie — Thu, 14 May 2009 09:31:31 GMT

For example:

Calculate the full value of 2^1000. Even if I use the 64bit datatype I come in (very) short.

Thank you very much for your response!

Re: Datatype problem in project euler

frostie — Thu, 14 May 2009 11:07:33 GMT

by the way, I am using integers because the result has to be accurate. In the end I need the sum of all numbers of the result.

(example: 456789 becomes 4+5+6+7+8+9 = 39 and the 39 is what I need)

Re: Datatype problem in project euler

RavensFan — Thu, 14 May 2009 17:33:33 GMT

Well, the link you gave said problem 10 was "Find the sum of all the primes below two million", a little different problem than 2^1000.

In order represent 2^1000 at full resolution, you would need a 1000 bit number! Of course a U64 won't be able to hold it. A double will hold it and it comes to about 1.07...e301, so 301 decimal digits. But a double won't store the full resolution.

In order to do what you want, you will have to set up your own math functions using 125+ bytes. You will have to do the multiplication and do all of your own carries. That why they made it a problem, to challenge you to figure out the algorhitm to do it. Otherwise going 2^1000 in any programming language is a trivial exercise.