Comments - New Stock Trading and Lottery Game Rooted in Deep Math - Data Science Central2019-08-19T20:33:27Zhttps://www.datasciencecentral.com/profiles/comment/feed?attachedTo=6448529%3ABlogPost%3A816772&xn_auth=noYes Victor, the answer is pos…tag:www.datasciencecentral.com,2019-04-23:6448529:Comment:8201152019-04-23T20:57:17.413ZVincent Granvillehttps://www.datasciencecentral.com/profile/VincentGranville
<p>Yes Victor, the answer is positive in both cases. I actually started the process with a pair of seeds (a, b) each being a positive integer smaller than 1000, run a fairly large number of iterations and arrived at the published (x(0), y(0)) with 250,000 digits, and then if you continue for many more millions of iterations (but less than 30,000,000 of them) you end up with the 2,000 past winning numbers that I published. Any subsequent number after these 2,000 numbers is a future winning…</p>
<p>Yes Victor, the answer is positive in both cases. I actually started the process with a pair of seeds (a, b) each being a positive integer smaller than 1000, run a fairly large number of iterations and arrived at the published (x(0), y(0)) with 250,000 digits, and then if you continue for many more millions of iterations (but less than 30,000,000 of them) you end up with the 2,000 past winning numbers that I published. Any subsequent number after these 2,000 numbers is a future winning number. I'll be presenting this system at the INFORMS annual meeting in Seattle, in October 2019.</p>
<p>I also formally proved (mathematical proof) today that the auto-correlations among the <em>x</em>(<em>t)</em> [and also among the winning numbers] are all equal to zero (lag-1, lag-2, lag-3 and so on, all of them.) The proof applies to a much wider class of sequences. </p> May I clarify a couple of poi…tag:www.datasciencecentral.com,2019-04-23:6448529:Comment:8201042019-04-23T17:23:31.076Zvictor zurkowskihttps://www.datasciencecentral.com/profile/victorzurkowski
<p>May I clarify a couple of points?</p>
<p>- is it true that the sequence of 2,000 winning numbers provided above is generated with a seed (a,b) consisting of a pair of positive integers less than 1,000 (after running the dynamics for over 30 million steps)?</p>
<p>- is it true that the "valid seeds" or "public seeds" x(0), y(0) provided in section 2.3 can be reached starting from (a,b)?</p>
<p>May I clarify a couple of points?</p>
<p>- is it true that the sequence of 2,000 winning numbers provided above is generated with a seed (a,b) consisting of a pair of positive integers less than 1,000 (after running the dynamics for over 30 million steps)?</p>
<p>- is it true that the "valid seeds" or "public seeds" x(0), y(0) provided in section 2.3 can be reached starting from (a,b)?</p> Victor, you need to have some…tag:www.datasciencecentral.com,2019-04-19:6448529:Comment:8190132019-04-19T21:48:33.224ZVincent Granvillehttps://www.datasciencecentral.com/profile/VincentGranville
<p>Victor, you need to have somewhere in your sequence x(t), starting at some iteration (smaller than 3 trillion), all the 2,000 successive values, in the right order, listed at <a href="http://www.datashaping.com/winningNumbers2000.txt">www.datashaping.com/winningNumbers2000.txt</a> (these are the past winning numbers).</p>
<p>Only one set of seeds (among the one million or so suggested) will meet this requirement. Yes, plenty of seeds produce a sequence x(t) being a constant number equal to…</p>
<p>Victor, you need to have somewhere in your sequence x(t), starting at some iteration (smaller than 3 trillion), all the 2,000 successive values, in the right order, listed at <a href="http://www.datashaping.com/winningNumbers2000.txt">www.datashaping.com/winningNumbers2000.txt</a> (these are the past winning numbers).</p>
<p>Only one set of seeds (among the one million or so suggested) will meet this requirement. Yes, plenty of seeds produce a sequence x(t) being a constant number equal to 255, and they are easy to eliminate. In section 2.3, I published seeds x(0) and y(0) that work (they will get you to the winning number) but these seeds are integers with 250,000 digits. </p>
<p>I'm glad to see the interest in this problem!</p> I have a question about the s…tag:www.datasciencecentral.com,2019-04-19:6448529:Comment:8190102019-04-19T21:17:12.960Zvictor zurkowskihttps://www.datasciencecentral.com/profile/victorzurkowski
<p>I have a question about the statement "<span>Then you known for sure that your next number will be a winning one." If y(0) > 4 x(0) + 2 ( which is larger than 2 x(0) + 1/2), then y(t) > 4 x(t) + 2 for all t, and so x(t) - 256 x(t-8) = 255 for all t >=8, i.e.: any seed that satisfies y(0) > 4 x(0) + 2 produces exactly the same sequence of winning numbers. Clearly, seed with y(0) > 4 x(0) + 2 are bad seeds (for the purpose of generating pseudo random numbers), but they…</span></p>
<p>I have a question about the statement "<span>Then you known for sure that your next number will be a winning one." If y(0) > 4 x(0) + 2 ( which is larger than 2 x(0) + 1/2), then y(t) > 4 x(t) + 2 for all t, and so x(t) - 256 x(t-8) = 255 for all t >=8, i.e.: any seed that satisfies y(0) > 4 x(0) + 2 produces exactly the same sequence of winning numbers. Clearly, seed with y(0) > 4 x(0) + 2 are bad seeds (for the purpose of generating pseudo random numbers), but they illustrate the fact that the winning numbers are not uniquely determined by the seed (assuming all starting pair of natural numbers are seeds). </span></p> From the public algorithm, th…tag:www.datasciencecentral.com,2019-04-17:6448529:Comment:8182142019-04-17T17:06:46.428Zvictor zurkowskihttps://www.datasciencecentral.com/profile/victorzurkowski
<p>From the public algorithm, the transition from x(t) to x(t+1) has 2 cases (one when 4x + 1 < 2y, and the other when not), so either x(t+1) = 2x(t) + 0, or x(t+1) = 2x(1) + 1.</p>
<p>From the public algorithm, the transition from x(t) to x(t+1) has 2 cases (one when 4x + 1 < 2y, and the other when not), so either x(t+1) = 2x(t) + 0, or x(t+1) = 2x(1) + 1.</p> Hi Victor, yes you are correc…tag:www.datasciencecentral.com,2019-04-17:6448529:Comment:8179922019-04-17T13:02:00.581ZVincent Granvillehttps://www.datasciencecentral.com/profile/VincentGranville
<p>Hi Victor, yes you are correct. I came up to that conclusion using the exact same line of reasoning. The part that is a bit more tricky (in my opinion) is to prove that r(t) is either 0 or 1. I think you might be on the right track to win our next competition when we announce it. At least, if I was a contestant with the goal of finding the future winning numbers, that's how I would start. Did you hear anything about the paper written regarding the past competition?</p>
<p>Hi Victor, yes you are correct. I came up to that conclusion using the exact same line of reasoning. The part that is a bit more tricky (in my opinion) is to prove that r(t) is either 0 or 1. I think you might be on the right track to win our next competition when we announce it. At least, if I was a contestant with the goal of finding the future winning numbers, that's how I would start. Did you hear anything about the paper written regarding the past competition?</p> Hi Vincent. As often, your ab…tag:www.datasciencecentral.com,2019-04-17:6448529:Comment:8177922019-04-17T06:40:18.020Zvictor zurkowskihttps://www.datasciencecentral.com/profile/victorzurkowski
<p>Hi Vincent. As often, your abstraction from numerical experiments is correct. We have x(t+1) = 2 x(t) + r(t), with r(t)=0 or 1 depending on y(t), therefore: x(t+8) = 2^8 x(t) + [2^7 r(t) + 2^6 r(t+1) + ...+ r(t+7)]. From here, x(t+8) - 256 x(t) is an integer between 0 and 2^7 + 2^6 + ..+ 1 = 2^8 - 1 = 255</p>
<p>Hi Vincent. As often, your abstraction from numerical experiments is correct. We have x(t+1) = 2 x(t) + r(t), with r(t)=0 or 1 depending on y(t), therefore: x(t+8) = 2^8 x(t) + [2^7 r(t) + 2^6 r(t+1) + ...+ r(t+7)]. From here, x(t+8) - 256 x(t) is an integer between 0 and 2^7 + 2^6 + ..+ 1 = 2^8 - 1 = 255</p> Hi Victor,
While there is no…tag:www.datasciencecentral.com,2019-04-16:6448529:Comment:8180382019-04-16T23:46:48.619ZVincent Granvillehttps://www.datasciencecentral.com/profile/VincentGranville
<p>Hi Victor,</p>
<p>While there is no explicit "mod 256" in the formula implemented, it does some kind of implicit "mod 256". I'd be curious to see someone proving this fact. I don't think it is hard to prove, but I stumbled upon it by chance (as an accidental, by-product result of some theoretical investigations, and later confirmed by actual computations.) Indeed, this is why I eventually stuck to <span>x(t) - 256 x(t-8) rather than a more complex formula that achieves the same business…</span></p>
<p>Hi Victor,</p>
<p>While there is no explicit "mod 256" in the formula implemented, it does some kind of implicit "mod 256". I'd be curious to see someone proving this fact. I don't think it is hard to prove, but I stumbled upon it by chance (as an accidental, by-product result of some theoretical investigations, and later confirmed by actual computations.) Indeed, this is why I eventually stuck to <span>x(t) - 256 x(t-8) rather than a more complex formula that achieves the same business goal. To put it differently, I was going to use a more intricate but more "obvious" formula that guarantees numbers between 0 and 255 (there is also some reasons for choosing 256 rather than 128 or 512), but it eventually (unexpectedly) resulted in cascading cancellations, finally simplifying to x(t) - 256 x(t-8). </span></p>
<p><span>Best,</span></p>
<p><span>Vincent</span></p> If the value of the winning n…tag:www.datasciencecentral.com,2019-04-16:6448529:Comment:8180282019-04-16T21:42:12.894Zvictor zurkowskihttps://www.datasciencecentral.com/profile/victorzurkowski
<p>If the value of the winning numbers are of the form x(t) - 256 x(t-8) for some suitable iteration t, how is it that the result is a number between 0 and 255? Are these integers or integers mod 256? </p>
<p>If the value of the winning numbers are of the form x(t) - 256 x(t-8) for some suitable iteration t, how is it that the result is a number between 0 and 255? Are these integers or integers mod 256? </p>