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Offline JWinslow23

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How I wish I could calculate pi...
« on: May 04, 2015, 01:27:55 pm »
...in vanilla TI-BASIC. To a precision of more than 15 places.

Well, with the help of Jeremy Gibbons and myself, now there's hope!

EDIT: Now there's MORE hope...click here

This program is as fast as I could make it (though not too fast...this ain't Asm), and I welcome any improvements on size and speed. It uses simulated arbitrary precision arithmetic with lists. I have verified it for 100 digits, and the algorithm itself has been verified for 1,000 digits in Haskell.

Download link coming soon attached.

Code is as follows:

PICALC.8xp
Code: [Select]
{1→ʟQ
{180→ʟR
{60→ʟT
2→I
ClrHome
501→D
"CHANGE THE ABOVE NUMBER TO CALCULATE MORE OR LESS DIGITS
"?→Str1
For(J,1,D
Output(1,1,Str1
Output(8,1,J
"ABOVE FOR TESTING
3(3I+1)(3I+2→U
5ʟT→L1
prgmAADJUST
L1→L3
5ʟR→L1
prgmAADJUST
L1→L2
(27I-12)ʟQ→L1
prgmAADJUST
L1+L2→L1
prgmAADJUST
L3→L2
prgmADIVIDE
θ→Y
Str1+sub("0123456789",Y+1,1→Str1
(5I-2)ʟQ→L1
prgmAADJUST
L1→L2
YʟT→L1
prgmAADJUST
L2-L1+ʟR→L1
prgmAADJUST
10UL1→L1
prgmAADJUST
L1→ʟR
10I(2I-1)ʟQ→L1
prgmAADJUST
L1→ʟQ
UʟT→L1
prgmAADJUST
L1→ʟT
I+1→I
EndAADJUST.8xp
Code: [Select]
DelVar BDelVar C1→A
Repeat B or A>dim(L1
C+L1(A→L1(A
If not(Ans
A=dim(L1)+1-sum(int(1/(1+cumSum(abs(seq(L1(Z),Z,dim(L1),1,-1→B
int(10^(7)⁻¹Ans→C
If 0>L1(A
-10^(7)Ans+L1(A→L1(A
10^(7)fPart(10^(7)⁻¹L1(A→L1(A
A+1→A
If C and Ans>dim(L1
Then
0→L1(A
1+dim(ʟQ→dim(ʟQ
Ans→dim(ʟR
Ans→dim(ʟT
Ans→dim(L2
Ans→dim(L3
End
EndACOMPARE.8xp
Code: [Select]
DelVar Gdim(L1→A
Repeat G or A<1
(L1(A)>L2(A))-(L1(A)<L2(A→G
A-1→A
EndADIVIDE.8xp
Code: [Select]
DelVar θprgmACOMPARE
While G≥0
Output(6,1,"
Output(6,1,θ
"ABOVE FOR TESTING
L1-L2→L1
prgmAADJUST
θ+1→θ
prgmACOMPARE
End
« Last Edit: May 06, 2015, 08:18:38 pm by JWinslow23 »
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Offline Sorunome

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Re: How I wish I could calculate pi...
« Reply #1 on: May 04, 2015, 01:56:51 pm »
This sound awesome!
Would you mind explaining some logic behind it / posting the Haskel program?
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Offline JWinslow23

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Re: How I wish I could calculate pi...
« Reply #2 on: May 04, 2015, 02:14:58 pm »
Quote from: Sorunome on May 04, 2015, 01:56:51 pm
This sound awesome!
Would you mind explaining some logic behind it / posting the Haskel program?
All the logic is explained in the paper linked at the top. The program I am referring to is the last program listed in the paper:
Code: [Select]
> piG3 = g(1,180,60,2) where
> g(q,r,t,i) = let (u,y)=(3*(3*i+1)*(3*i+2),div(q*(27*i-12)+5*r)(5*t))
> in y : g(10*q*i*(2*i-1),10*u*(q*(5*i-2)+r-y*t),t*u,i+1)I'm not a math wizard or anything :P , but I think it's based off of this series for pi:

And it views that as an infinite composition of linear fractional transforms. The program also has an optimization based off a conjecture that has not yet been fully proven, but it has been verified to hold for calculation of the first 1,000 digits.

Hope I helped! Check the paper for more: this guy knows more about the subject than I ever would hope to tell you :P .

Of course, if you don't mind a previous commitment to the number of digits calculated, I have one that fits entirely within the precision of the calculator's numbers... that one fails before the first 0 and I don't know why
« Last Edit: May 04, 2015, 07:25:46 pm by JWinslow23 »
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Offline JWinslow23

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Re: How I wish I could calculate pi...
« Reply #3 on: May 06, 2015, 08:16:57 pm »
I have new code! With this one, I tested that the maximum amount of digits it can supply on a fully RAM cleared calculator is 377, and this is done so in 3 hours 45 minutes. (don't worry, it doesn't go at the speed of 1.675 minutes per digit...it just gets slower with digits calculated, as expected)

Download links attached. You will be amazed or else :P

Code is as follows:

PICALC.8xp
Code: [Select]
SetUpEditor Q,T,R
{1->|LQ
{60->|LT
3Ans->|LR
ClrHome
377->D
"?->Str1
For(J,2,D+1
Output(1,1,Str1
Output(6,1,"
Output(8,1,J-1
5|LT->L1
prgmAADJUST
L1->L2
(27J-12)|LQ+5|LR->L1
prgmAADJUST
prgmADIVIDE
Str1+sub("0123456789",theta+1,1->Str1
30((5J-2)|LQ-theta|LT+|LR->L1
prgmAADJUST
(3J+1)L1->L1
prgmAADJUST
(3J+2)L1->L1
prgmAADJUST
L1->|LR
10J(2J-1)|LQ->L1
prgmAADJUST
L1->|LQ
3(3J+1)|LT->L1
prgmAADJUST
(3J+2)L1->L1
prgmAADJUST
L1->|LT
EndAADJUST.8xp
Code: [Select]
DelVar C
For([recursiven],1,dim(L1
C+L1([recursiven]->|N
int(Ans|E~7->C
|N-Ans|E7->L1([recursiven]
End
If C
C->L1([recursiven]
dim(L1->dim(L2
Ans->dim(|LQ
Ans->dim(|LR
Ans->dim(|LTACOMPARE.8xp
Code: [Select]
1+dim(L1->A
Repeat Ans or A<2
A-1->A
(L1(A)>L2(A))-(L1(A)<L2(A
EndADIVIDE.8xp
Code: [Select]
DelVar thetaprgmACOMPARE
While Ans>=0
Output(6,1,"
Output(6,1,theta
L1-L2->L1
prgmAADJUST
theta+1->theta
prgmACOMPARE
End
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Offline JWinslow23

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Re: How I wish I could calculate pi...
« Reply #4 on: June 02, 2015, 06:12:49 pm »
Bump.

Even newer code. Now, no divide program. Also a bit faster.

Download links attached, here is the code to type in:

PICALC.8xp
Code: [Select]
SetUpEditor Q,T,R
{1->|LQ
{60->|LT
3Ans->|LR
ClrHome
Output(7,1,"CLEAR TO EXIT.  #    NEXT: 0...
"?->Str1
2->I%
Repeat getKey=45
If 1<length(Str1
Then
"   
If not(fPart((length(Str1)-2)/16
Output(6,1,Ans+Ans+Ans+Ans
2+16int((length(Str1)-82)/16)(98<=length(Str1
Output(1,1,sub(Str1,Ans,length(Str1)+1-Ans
End
Output(8,2,I%-2
Output(8,12,0
5|LT->L1
prgmAADJUST
L1->L2
(9I%-4)|LQ->L1
prgmAADJUST
3L1+5|LR->L1
prgmAADJUST
0->PV
prgmACOMPARE
While Ans>=0
PV+1->PV
Output(8,12,PV
L1-L2->L1
prgmAADJUST
prgmACOMPARE
End
sub("0123456789",PV+1,1
If I%=2
Ans+".
Str1+Ans->Str1
(5I%-2)|LQ+|LR-PV|LT->L1
prgmAADJUST
30L1->L1
prgmAADJUST
If I%<33
Then
9I%^^2+9I%+2
Else
(3I%+1)L1->L1
prgmAADJUST
3I%+2
End
AnsL1->L1
prgmAADJUST
L1->|LR
10I%|LQ->L1
prgmAADJUST
(2I%-1)L1->L1
prgmAADJUST
L1->|LQ
3|LT->L1
prgmAADJUST
If I%<33
Then
9I%^^2+9I%+2
Else
(3I%+1)L1->L1
prgmAADJUST
3I%+2
End
AnsL1->L1
prgmAADJUST
L1->|LT
I%+1->I%
End
ClrHome
sub(Str1,2,length(Str1)-1->Str1
Disp "# OF DIGITS:","DIGITS STORED IN","Str1
Output(1,14,I%-2AADJUST.8xp
Code: [Select]
If min(L1)>0 and |E10>=max(L1
Return
0->PMT
For([recursiven],1,dim(L1
PMT+L1([recursiven]->|N
int(|N|E~10->PMT
|N-PMT|E10->L1([recursiven]
End
If not(PMT
Return
[recursiven]->dim(L2
Ans->dim(|LQ
Ans->dim(|LR
Ans->dim(|LT
PMT->L1(AnsACOMPARE.8xp
Code: [Select]
1+dim(L1->|N
Repeat Ans or |N=1
|N-1->|N
L1(|N)-L2(|N
End
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