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Jedd
1980 Pong World Champion


Elite


Joined: 18 Nov 2003
Posts: 823

Posted: 01 Dec 2003 11:01:04 pm    Post subject:

I made a program to solve for pi using the method:

pi / 4 = 1 - 1/3 + 1/5 - 1/7 + 1/9.....

and it only lets you see 9 decimal places. Does anyone know of a way to let the variable contain more than 9 decimal places? I know in C there's something like "float" but im not to good with computer programming. btw I wouldnt reccomend making this program; its best uses are battery waster and ram clearer.
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Fr0sty


Member


Joined: 27 Nov 2003
Posts: 202

Posted: 01 Dec 2003 11:16:18 pm    Post subject:

Uhh... you can try multiplying out your number by like 10000000000000000 then tell them that the decimal is supposed to be after the first digit (in this case, a 3)
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NETWizz
Byte by bit


Bandwidth Hog


Joined: 20 May 2003
Posts: 2369

Posted: 01 Dec 2003 11:53:10 pm    Post subject:

Yes,

you use that method, but you write your own calculation routinees using strings. It is slower and more difficult, but you can get more accuracy.
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Jedd
1980 Pong World Champion


Elite


Joined: 18 Nov 2003
Posts: 823

Posted: 02 Dec 2003 01:04:19 am    Post subject:

How would i do that with strings? here's the program as it is now...


ClrHome
0->P
-1->B
For(A,1,1000001,2
-B->B
P+B(1/A0->P
Output(4,1,P4
End


Last edited by Guest on 02 Dec 2003 09:05:20 am; edited 1 time in total
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DarkerLine
ceci n'est pas une |


Super Elite (Last Title)


Joined: 04 Nov 2003
Posts: 8328

Posted: 02 Dec 2003 05:30:46 pm    Post subject:

You could use lists to get up to 999 digits. That would be faster than using strings.

Code:
{0 -> L1
999 -> dim(L1
-1 -> B
For(A,1,1000001,2
-B -> B
prgmINVERSE
L1 + BL2 -> L1
For(I,999,2,-1
L1(I-1)-(L1(I)<0)+(L1(I)>9 -> L1(I-1
L1(I)+10((L1(I)<0)-(L1(I)>9 -> L1(I
End
End
-------------------------------------------
prgmINVERSE
{int(10/A -> L2
AfPart(10/A -> R
{1,R -> L3
2 -> I
Repeat 1-sum(R=L3) or I=999 or not(R
int(10R/A -> L2(I
I+1 -> I
AfPart(10R/A -> R
R -> L3(I
End
If R:Then
dim(L3 -> R
DelVar L3
seq(L2(RfPart(I/R)),I,1,999 -> L2
End

I think this will work.


Last edited by Guest on 03 Dec 2003 04:43:23 pm; edited 1 time in total
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