[VA] SRC#006- Pi Day 2020 Special: A New Fast Way to Compute Pi

[Valentin Albillo]

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Hi, EdS2:

(03-21-2020 07:21 PM)EdS2 Wrote:  I hadn't, but now I have! (I even found how to run at full speed.) And indeed, of course, your two programs both produce the hoped-for results. No typo.

Of course indeed, I told you as much.

Quote:There remains the question of my understanding, which is still missing a piece.

Let's see ...

Quote:What's very curious to me is that you have two quite different final expressions in the two versions of the HP-71B rnd/rnd programs that you've posted.

I didn't remember the exact earlier version, the one-liner I posted many years ago, but as I know the underlying theory I simply coded it again from scratch and came up with the second version, the 3-liner, which of course produces the exact same results. That explains why I posted two different versions.

Quote:In the one-liner, you use the same form as my Basic port - you counted negatively and therefore perform a subtraction instead of an addition, but otherwise the same. In the multi-line version which you posted most recently, the final expression is quite different - and I don't (yet) understand why it should be.

Your earlier program: 1-4*N/K
My Basic port: 1+4*S%/N%
Your recent program: 5-P*4

It's quite simple, actually. My recent program is this:

      1   DESTROY ALL @ RANDOMIZE 1 @ FOR K=1 TO 5 @ N=10^K @ S=0
      2   FOR I=1 TO N @ IF NOT MOD(IROUND(RND/RND),2) THEN S=S+1
      3   NEXT I @ P=S/N @ STD @ DISP N, @ FIX 3 @ DISP 5-P*4 @ NEXT K

which is computing the probability that the closest integer to A/B is even, where A and B are uniformly distributed random numbers in [0,1), as produced by the RND keyword. Each time the rounded value is even (i.e., it's 0 modulo 2) the number of favorable outcomes (S) is incremented by one (see line 2). After N tries have been sampled, the probability P for the even case will be the number of favorables outcomes (S) divided by the number of tries (N), thus we have the estimated probability P = S/N.

But I know from theory that in the limit, for N -> Infinity, the exact probability P = (5-Pi)/4, so isolating Pi we have Pi = 5-P*4, which is displayed by the program in line 3 above.

Now, my earlier program, the one-liner, namely:

      10   INPUT K @ N=0 @ FOR I=1 TO K @ N=N-MOD(IROUND(RND/RND),2) @ NEXT I @ DISP 1-4*N/K

is computing the probability that the closest integer to A/B is odd, where A and B are uniformly distributed random numbers in [0,1), as produced by the RND keyword. Each time the rounded value is odd (i.e., isn't 0 modulo 2) the number of favorable outcomes (N) is decremented by one, and after K tries have been sampled, the probability for the odd case will be the number of favorables outcomes (-N) divided by the number of tries (K), thus we have the estimated probability P = -N/K.

As the probability of the rounded division being either even or odd is 1 (certainty), the probability for the odd case is 1 minus the probability for the even case, thus it's P = 1-(5-Pi)/4 = (Pi-1)/4, so isolating Pi we have Pi = 1+4*P = 1+4*(-N/K) = 1-4*N/K, which is then displayed by the one-line program.

As you can see, the source of your confusion (sorry for it !) was due to the fact that I didn't remember my previous one-liner program posted many years ago and thus I simply created code anew based on the same theory, but as it happened the new code was a three-liner, used different variables for the number of favorable outcomes and the number of tries, and further it was using the probability for the even case to compute Pi instead of the probability for the odd case, as the one-liner did.

I hope this explains the whole affair to you, and thanks for your interest.

Regards.
V.