For all you pi nerds:

[Jorge]

Then you can be the first one who can tell me how Johnny could verify Alice's abilities. If you don't have the answer by your next post I won't mind supplying what you can't figure out on your own. Its not a trick question, Johnny has a way of testing Alice without himself counting all the leaves and buds on the tree.

Whatever you have on your mind (God only knows what that may be),
one way or another you are going to 'smuggle' information into the picture.

And how would you know that a proof was correct Jorge? Read through all the steps carefully and do double checking with other peers? Even then there's a chance that somewhere there's a subtle error hiding that escaped attention, though the more eyes and error-checking used the less the probability of this happening. The proof of any computation of pi being correct is simple that the algorithms used converge to pi, and that they have been correctly implemented. The programs are complex, but not overwhelmingly so and they can be bug tested, which they have been.

The 'correctness' of a proof is established in many ways. Of course, if you take this
to a Cartesian level of skepticism then we cannot be certain of just about anything.

The BBP algorithm can quickly and efficiently be used to calculate small stretches of pi which can used to verify the results. Small stretches at any arbitary location say between the 1,244,513,153,111th and 1,244,514,153,113th decimal point. Any error would have a cumulative effect rendering the sequence after the error basically random. The odds of any random stretch of numbers accidentally matching a correct segment when doing an error check would be exponentially low depending on how long stretches that are being compared. And each time you run the error checking algorithm on a new segment the odds of the sequence being wrong would be decreased by a large factor.

How do you know that you have a "correct" segment when it's the FIRST time?
It is true that a 'random' error is highly unlikely to be reproduced the same
way each time and so seeing the same number time after time corroborates its
validity. However, this requires multiple runs of the algorithm - multiple runs
generating trillions of digits and then cross-checking.

Very few things are known with absolute certainty, though I'd make an exception with your idiocy, Jorge. If this is correct then even mathematical proofs don't count as mathematical certainty rendering the concept meaningless. You're free to use this restricted definition of 'mathematical certainty' but mathematicians would shy away from it.

I wouldn't keep pointing out my "idiocy" if I were you ... I mean, if an "idiot" is able
to make jackasses out of all you folk then what does that say about you?

Did you know that we know the quadrillionth bit of pi in hexadecimals using the BBP algorithm? It happens to be 0 if you're curious. The same method can be used to verify just about any hexadecimal of pi. The hexadecimal format is what it will have immediately after computation, and that's the stage where its being verified. Calculating all the digits of pi using this method is terribly slow, but calculating small stretches is very fast.

My earlier comments still stand.

Question: How do we verify with 'absolute certainty' that the 31st decimal place of pi is 5? I can write down an algorithm that extracts this result, and then you can question whether the algorithm is correct. I can write down a proof that the algorithm gives the right result and then you can question whether the proof is correct. I can carefully label out all the steps, and have other people (including yourself) verify them. Would that give 'absolute certainty'. No. It would still be possible that some error had been done somewhere. Its even possible that everytime someone has done a computation of pi people all over the world have made an error because of an accidental mental fluke, and that all our calculators have had a cosmic ray interfere with memory registers in just such a way as to produce '5' as the 31st digit of pi. Even assuming that the algorithms would have produced the right result. If there's even a shadow of a chance of error then you're not dealing with absolute certainty. It seems you're making a mistake in equivocating between absolute certainty and mathematical certainty.

I am making no such equivocation, thank you.

Also, as I said earlier, if you go into a Cartesian level of skepticism,
then we cannot be certain of just about anything. [That line of thought
is what eventually led Descartes to his "cogito, ergo sum" remark.]

Mathematical certainty, to me, merely means that we have analytic means of verifying a result to arbitrarily high certainty and that it has been checked to such a high degree that withholding assent is obscene. For example, a proof whose validity can be independently verified by mathematicians. And for this calculation of pi we have several methods: Verifying the implementations of the algorithms by studying the source code (though nobody has gotten the source code as of yet), checking it with the previous values, using the BBP-algorithm to check stretches and so forth.. We have the means to rule out even creative proposals that this two-man japanese team just copy pasted the past record and used the BBP-algorithm to calculate some small stretches here and there later used in the verification. All it takes is just to pick some areas at random, calculate them using BBP and check them. It would be extremely unlikely to hit the pre-calculated regions and not just random digits. I can do that, but I think you see the point.

I see your point - I always have.

My point is that just because independent methods agree on the first N digits, this
is in no way a rigorous mathematical proof that they will agree on the (N+1) digit.
There are several historical examples of this (which I'm sure you know).

Propose a way that this result could be wrong and slip past their proposed testing regime no more unlikely than a one in a billion chance of occurring.

That is far too difficult a problem for a lowly "pathetic diploma-mill-phd clown" like me.

Seriously, earlier in this post I hinted at how I would test up to 'M' digits.
How to generalize for any number of digits is a much harder problem, I think.

Jorge

[phank]

So we can generalize. Any problem Jorge considers too hard for him, is therefore too hard for anyone. And this is consistent with his conviction that those who know more than he does, know things he doesn't know, which makes them automatically wrong.

[Leonhard]

Whatever you have on your mind (God only knows what that may be),
one way or another you are going to 'smuggle' information into the picture.

I don't know what you mean with 'smuggle information into the picture'. Certainly I won't need to tell Johnny the correct answer in advance if that's what you're suggesting. Though I doubt you have any clue what you're suggesting half the time.

Lets recap: Alice says she has a method for counting a large number of objects accurately. Johnny is skeptical and asks her to count the number of leaves and buds on a tree. She tells him that there's exactly 182.575 leaves and 112.333 buds on the tree, but Johnny wants to verify it. How can he do that without counting all the leaves and buds himself?

Here's the answer: Send Alice away, have Johnny gather a small number of leaves and buds, have him keep track of how many he's removed and have her do a recount. Say he removes 134 leaves and 65 buds. She comes back and lo and behold she counts 182.441 leaves and 112.270 buds, which matches how many he removed. Alright maybe she was lucky, she had figured he would pick a number of leaves and buds between say one and 200. The odds of her guessing correctly if he choose randomly would be 1 out of 40000 in that case. So he performs the thing again. The odds of her guessing correctly both times is then 1 8*10^8, after three times its 1 out 1.6*10^13, after four times its 2.2*10^21, etc... He performs a task he can do, removing a small number of leaves and buds, which can be used to verify her ability. You don't have to do this very many times until the certainty you get at the level where it becomes obscene to withhold assent.

This is one example of verifying an algorithm without having access to its inner workings. Gotta head off for tonight its 1am here.

[ChemMJW]

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(2) The probability that it is not a 3 is 0.90.

Jorge

One caveat:
The probability that the specified digit is not 3 is only 0.9 under the assumption that the digits of π are randomly distributed in base 10. While this is generally believed by mathematicians to be true, it has certainly not been proven in the formal mathematical sense.

[Jorge]

I don't know what you mean with 'smuggle information into the picture'. Certainly I won't need to tell Johnny the correct answer in advance if that's what you're suggesting. Though I doubt you have any clue what you're suggesting half the time.

Lets recap: Alice says she has a method for counting a large number of objects accurately. Johnny is skeptical and asks her to count the number of leaves and buds on a tree. She tells him that there's exactly 182.575 leaves and 112.333 buds on the tree, but Johnny wants to verify it. How can he do that without counting all the leaves and buds himself?

Here's the answer: Send Alice away, have Johnny gather a small number of leaves and buds, have him keep track of how many he's removed and have her do a recount. Say he removes 134 leaves and 65 buds. She comes back and lo and behold she counts 182.441 leaves and 112.270 buds, which matches how many he removed. Alright maybe she was lucky, she had figured he would pick a number of leaves and buds between say one and 200. The odds of her guessing correctly if he choose randomly would be 1 out of 40000 in that case. So he performs the thing again. The odds of her guessing correctly both times is then 1 8*10^8, after three times its 1 out 1.6*10^13, after four times its 2.2*10^21, etc... He performs a task he can do, removing a small number of leaves and buds, which can be used to verify her ability. You don't have to do this very many times until the certainty you get at the level where it becomes obscene to withhold assent.

This is one example of verifying an algorithm without having access to its inner workings. Gotta head off for tonight its 1am here.

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I was right, of course ... you smuggled information into this scenario.
That's fine, it is necessary to do so. Problem with you people is that
you typically fail to notice when you do this (granted, it is often subtle).

Suppose L is the true number of leaves and B the true number of buds.
A first guess is L1 = 182,575 and B1 = 112,333.
Remove 134 from L1 and 65 from B1 giving you L2 = 182,441 and B2 = 112,268
By the way, your arithmetic is off. 112,333 - 65 = 112,268 NOT 112,270 as you report above.
The point is that 134 and 65 are VERIFIED KNOWN figures - a.k.a. information.
Information has been smuggled into the picture.
Without doing this the problem is unsolvable given your conditions, parameters and constraints.
Q.E.D.

Jorge

[Jorge]

One caveat:
The probability that the specified digit is not 3 is only 0.9 under the assumption that the digits of π are randomly distributed in base 10. While this is generally believed by mathematicians to be true, it has certainly not been proven in the formal mathematical sense.

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True.

I should have written : "(2) The probability that it is not a 3 is, statistically, 0.90.

Jorge