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| Author | Topic: Discontinuing research about ID | |||||||||||||||||||||||||||||
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
RAZD writes: So P.WeC- and P.WeC+ are not appearances -- does the person (P.WeC in this case) need to be present for +/- to occur? No. For example: P.Da: "I feel so sick"P.Ri: "Does this medicine help?" P.Da: "Yes, it helps a lot. I feel better now" P.Ri: "I will go to P.WeC. This medicine will also heal him." is: *P.Da, P.Da-, *P.Ri, *P.Da, P.Da+, *P.Ri, *P.WeC, P.WeC+(This row of appearances would break the pattern) RAZD writes: Can you have P.WeC without it being * P.WeC, P.WeC- or P.WeC+ ? I'm just trying to understand your marking system. There are only *P.WeC, P.WeC- or P.WeC+. P.WeC without *,+,- is not defined in the notation and only describes the person.
RAZD writes: M1, M2, M3, M4, M5, M6, M7, M10, M11, M12, M13, and M14 -- M8 and M9 seem to be missing from the "pattern" table 4 Yes, M8 and M9 are missing. There are only 12 M's.
RAZD writes: Would I be correct in thinking that anything not covered by this list is ignored in the "pattern" derivation and applications? Yes.
RAZD writes: The question you did not answer was why\how should I conclude that these three hypothetical episodes should be part of the same pattern: A data source of only three hypothetical episodes is to small to create a non-arbitrary pattern. Therefore it is not possible for only three episodes.
RAZD writes: and how do I correct D and E so they fit the pattern? Episode D:E2: Not *P.WeC, but *P.Ya E7: Not *P.Wo, but *P.Da E8: Not *P.Da, but M6 E9: Not *P.BW, but *P.Ya E12: Not M2, but *P.BW Episode E: E15: Not *P.BeC, but M4 RAZD writes: I plan to, but as an initial comment you say that the calculation is based on the observed incidents rather than on an accounting of the possibilities, yes? You can also approximate a probability by accounting possibilities of single appearances. For E1 there are 25 occurrences that fit with the pattern (P.Al, P.BW, P.Da, P.LF, P.Pi, P.Tr, P.WeC, P.Wo, P.WSA, M1, M2, M5, M6, M7, M13, P.Al-, P.BW+, P.Tr+, P.WeC-, P.BeC, P.Ri, P.Ya, M4, P.BW-, P.Da) and 26 occurrences that break the pattern (*P.En, M3, M10, M11, M12, M14, P.Al+, P.BeC+, P.BeC-, P.Da+, P.En+, P.En-, P.LF+, P.LF-. P.Pi+, P.Pi-, P.Ri+, P.Ri-, P.Tr-, P.WeC+, P.Wo+, P.Wo-, P.WSA+, P.WSA-, P.Ya+, P.Ya-). Altogether the probability would probably be about 0.5 that random data fits with the pattern. The test on random data resulted in a probability of 0.625. But at (00:00) the probability is 0.95. That is the big difference that can't be explained by chance.
RAZD writes: If I throw a di 10 times and only get numbers between 1 and 3, can I calculate with confidence the probability of what the next throw will be? Would you calculate that probability based on the number of 1's the number of 2's and the number of 3's in those 10 throws to predict the next toss? Certainly I can take the results of those 10 throws and put them through standard probability calculations while ignoring possibilities that did not occur during the data gathering phase, yes? If we assume that we don't know what shape the di was -- Tetrahedron (four faces), Cube or hexahedron (six faces), Octahedron (eight faces), Dodecahedron (twelve faces), Icosahedron (twenty faces) -- then the only evidence we have for the possibilities is what is observed during the data gathering phase, yes? Let's say we know the die is a Tetrahedron (four faces) or an Octahedron (eight faces) and if we throw it 10 times we only get numbers between 1 and 4. The probability that the di is an Octahedron can then be calculated with the probability mass function: The resource cannot be found. Probability of success: 0.5An unbiased Octahedron (eight faces) will only show numbers between 1 and 4 every second time (1-4, 5-8). Successes: 10Trials: 10 Result: 0.00097656 = 1:10^3 Therefore the next throw will also show a number between 1 and 4 with a residual uncertainty of 1:10^3. More throws are necessary to increase the certainty. If we throw the di 20 times and we only get numbers between 1 and 4, then the probability is 0.00000095 = 1:10^6. The found pattern was tested 47 times and the beginning was very different to the random data source (15/24,45/47).
RAZD writes: Essentially you are assuming that the three observed results are the only possibilities. Nor do we know if the di was weighted so we are assuming that the results are not biased. For the probability calculation was assumed that the two observed results are the only possibilities, the pattern fits or it doesn't fit.
Cat Sci writes: It is not the odds of getting the pattern, it is the odds of its existence. Yeah, that's what I thought. Your whole argument is flawed: You don't figure out why and how a pattern exists by calculating the odds of it existing. You are right about this. The probability calculation is solely about the patterns existence. But there is also a residual uncertainty of 1:10^3 about a triune God.
Cat Sci writes: Your pattern could just be a natural result of the TV series making process, you have not eliminated that possibility and that the chance of it happening is very low doesn't either. For the origin I will point back to something I already said. From [Msg=162]:
quote: I point back to that, because until now no one has answered these four question with for example "Yes, Yes, Yes, No" or "Yes, No, No, No".
Cat Sci writes: You also haven't refuted this counter argument: I did. From [Msg=120]:
quote: Cat Sci writes: TV shows follow rules and they are going to have patterns. Quantizing events in the shows and then looking for patterns in the notations is going to make more complicated patterns that are going to have lower odds of occurring. There can be actually patterns like this with a residual uncertainty of 1:10 or 1:10^2. For example there are at least three persons discussing with each other in the first 30 seconds. This could probably happen in 9 out of 10 times. The found pattern has a residual uncertainty of 1:10^7, not 1:10. And the found pattern is more complex and includes complex pattern on its own.
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
RAZD writes: Thus we could define elements as
It would be:
RAZD writes:
ie no, P.WeC- and P.WeC+ are not appearances, but yes the person (P.WeC in this case) need to be presenobserved for +/- to occur, is this correct?RAZD writes: Because of the way you define appearance\observation as including being named, it seems that it would not be possible to affect a person without identifying the person being affected and counting that as an appearance\observation, yes? Yes, a person must be identified first. But there can also be references with "he" or "she" that can create +/- without an appearance, if it is clear who "he" or "she" is.
RAZD writes: Would you be able to change that so each hypothetical episode matches your "pattern" and is unique? This is actually not possible. Episode A-D shall have a E3->E4 transition. That are 4 episodes, but there are only 3 possible transitions: P.Al-, M4, M10. *P.Wo and *P.Al doesn't trigger E4 at E3 in the episodes A and B. There are other transitions like this: E1->E2: 6 possible transitionsE2->E3: 13 possible transitions E3->E4: 3 possible transitions E5->E6: 6 possible transitions E10->E11: 11 possible transitions It is also nearly impossible that there is no element within a episode that repeats itself. In this example: *P.Da, P.Da-, *P.Ri, *P.Da, P.Da+, *P.Ri, *P.WeC, P.WeC+ appeared *P.Da and *P.Ri already repeatedly. *P.Da and *P.Ri would triggered new events repeatedly. An episode without repetitions would be rare to find. If you want to group elements yourself, then simultaneous appearances would be important. If there is a simultaneous appearance of P.BeC, P.Da, P.LF, P.Ri and P.WeC, then you will need a part for your pattern that contains all this appearances together. But I don't think all this is necessary. The pattern was created for the first 76 episodes (table 5 on page 8) and tested on a random data source and further 47 episodes. The test showed, that the pattern fits at (00:00), but not for random data. The answer to your question could be table 5 on page 8. It is shown there from where the elements were added to E1-E15.
NoNukes writes: the use of complicated pattern rules increases the probability of identifying a pattern. You can actually increase the number of alternatives for identifying a particular pattern until you insure that the pattern is matched. The pattern was created for the first three seasons of the data source and has no predictive power for them.
NoNukes writes: The individual items of the pattern are not independent. Some items cannot happen unless other items have happened. For example? There is no appearance that can't happen until an other appearance has happened, for example.
NoNukes writes: Before I even pick up the paper, my first question would be whether or not the calculated probabilities are meaningful and correct based on these considerations. To understand a paper about a theory you mostly have to read the paper. You would have to read it to understand it. But you don't want to read it, before you have found out that the content is correct. That's not how to acquire knowledge.
NoNukes writes: If in fact, there is something of value in the paper, it is not just of interest to mathematicians. The ideas here are potentially world transforming. In short they suggest the existence of a logical and inescapable argument that Jesus is Lord and that special creation is real. And as the author of the paper, you claim to have no time to translate this earth shattering, world transforming idea into English for mass consumption. I already spend some time to explain it here. For your other opinions: Add some reasons, references and/or examples and not your opinion solely. Otherwise there can't be a discussion.
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
NoNukes writes: So the fact of an appearance increases the probability that some other things will occur like "becomes negatively affected" prior to the occurrence of the corresponding disappearance Yes, if someone appears, then there can be higher possibility that he becomes negatively affected then. This pattern could have a certainty of maybe 30%. This is a residual uncertainty of 7:10. It is not a question that patterns are existing. There can be a lot patterns with a certainty of 90% or even 99%. But this pattern has a certainty of 5.3 sigma or 99.99999%. Your arguments should be more mathematical to refute a pattern with a high certainty like the found pattern. Only the appearances are not noted. The next post could be helpful for the other answers you wrote.
RAZD writes: If it is clear who they are, then are they not identified\observed\appeared? Only if they are named, start to speak or appear. "he" or "she" is not a name.
RAZD writes: I suppose they could be in the immediate previous event for the reference to work. Yes. For example: P.Da: "I feel so sick"P.Ri: "Does this medicine help?" P.Da: "Yes, it helps a lot. I feel better now" P.Ri: "I will go to P.WeC. This medicine will also heal him." P.Da: "Go to P.Tr instead. It will help her, but it doesn't help humans. He will certainly remain sick." is: *P.Da, P.Da-, *P.Ri, *P.Da, P.Da+, *P.Ri, *P.WeC, P.WeC+, *P.Da, *P.Tr, P.Tr+, P.WeC- There can be a few events between * and +/-.
RAZD writes: Or verbally: Data appears, bad happens Riker appears, good happens, Wesley appears, good happens. This creates problems. For example: P.Da: "P.Wo, you just won all my chips in this poker game" It is good for P.Wo, but bad for P.Da. It would be difficult to objectively evaluate what is "good" and what is "bad". It can be evaluated more easily how it affects every person. An other example: P.anevilperson: "I feel so sick" This is P.anevilperson-, but is it M+ or M-? To feel sick could be a bad thing to happen, but it could be good that an evil person is affected by this.
RAZD writes: And this would mean
Only 27 elements is not much. The pattern would probably not be distinct and you would need M+ and M- at every event from E1-E15.
RAZD writes: With this system all possibilities are described and we can then calculate probabilities. I have calculated probabilities in the next post.
Cat Ski writes: Thanks for admitting it. That's a huge chunk of your argument. I only referred to the 1:10^7 probability. Not to the 1:10^3 probability or the arguments below.
Cat Ski writes: But I've already accepted that the pattern didn't come about by chance. I've also pointed out that this doesn't mean that it didn't occur naturally. Your only argument against that has been incredulity. The residual uncertainty of 1:10^7 showed that it didn't come about by chance. The four questions you are referring to, show that the involvement of chance precludes any pattern with a residual uncertainty of 1:10^7. Any naturally imprinted pattern, for example imprinted by writers, would be corrupted to a residual uncertainty below 1:10^2 through the involvement of chance. Your only argument against the four questions was to ignore them and to not answer them.
Cat Ski writes: That does not refute the couter argument. You refer to a question asked by RAZD 100 messages ago. If it would have not refuted his counter argument, then he would have mentioned it. Ask and formulate your own questions.
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
RAZD mentioned the idea to approximate the probability about the pattern by accounting possibilities of single appearances. I have done this now. The calculations below are not part of the paper, but they support the results.
For E1 there are 25 occurrences that fit with the pattern (P.Al, P.BW, P.Da, P.LF, P.Pi, P.Tr, P.WeC, P.Wo, P.WSA, M1, M2, M5, M6, M7, M13, P.Al-, P.BW+, P.Tr+, P.WeC-, P.BeC, P.Ri, P.Ya, M4, P.BW-, P.Da-) and 26 occurrences that break the pattern (*P.En, M3, M10, M11, M12, M14, P.Al+, P.BeC+, P.BeC-, P.Da+, P.En+, P.En-, P.LF+, P.LF-. P.Pi+, P.Pi-, P.Ri+, P.Ri-, P.Tr-, P.WeC+, P.Wo+, P.Wo-, P.WSA+, P.WSA-, P.Ya+, P.Ya-). Appearances occur about 10 times more often than affected persons or M's. Therefore the 25 occurrences that fit with the pattern will happen more often than the 26 occurrences that break the pattern. There are (*=appearances) *: 12; +/-: 6; M's: 7 that fit with the pattern and *: 1; +/-: 20; M's: 5 that break the pattern. * will have a value of 10 to occur and +/- and M's a value of 1. Because appearances occur about 10 times more often than affected persons or M's. This is summarised to the values 133 (12*10 + 6*1 + 7*1) and 35 (1*10 + 20*1 + 5*1). Therefore an occurrence at E1 has a probability of p=133/(133+35)=133/168=0.792 to fit with the pattern. The probability for all other events for single occurrences are: E1:fit with the pattern: *:12; +/-:6; M's:7 doesn't fit with the pattern: *:1; +/-:20: M's:5 probability for the next single occurrence to fit with the pattern: p=133/168=0.79 E2:fit with the pattern: *:13; +/-:5; M's:5 doesn't fit with the pattern: *:0; +/-:21: M's:7 probability for the next single occurrence to fit with the pattern: p=140/168=0.83 E3:fit with the pattern: *:13; +/-:11; M's:11 doesn't fit with the pattern: *:0; +/-:15: M's:1 probability for the next single occurrence to fit with the pattern: p=152/168=0.90 E4:fit with the pattern: *:10; +/-:4; M's:6 doesn't fit with the pattern: *:3; +/-:22: M's:6 probability for the next single occurrence to fit with the pattern: p=110/168=0.65 E5:fit with the pattern: *:11; +/-:5; M's:7 doesn't fit with the pattern: *:2; +/-:21: M's:5 probability for the next single occurrence to fit with the pattern: p=122/168=0.73 E6:fit with the pattern: *:12; +/-:5; M's:7 doesn't fit with the pattern: *:1; +/-:21: M's:5 probability for the next single occurrence to fit with the pattern: p=132/168=0.79 E7:fit with the pattern: *:12; +/-:12; M's:9 doesn't fit with the pattern: *:1; +/-:14: M's:3 probability for the next single occurrence to fit with the pattern: p=141/168=0.84 E8:fit with the pattern: *:12; +/-:16; M's:10 doesn't fit with the pattern: *:1; +/-:10: M's:2 probability for the next single occurrence to fit with the pattern: p=146/168=0.87 E9:fit with the pattern: *:13; +/-:18; M's:9 doesn't fit with the pattern: *:0; +/-:8: M's:3 probability for the next single occurrence to fit with the pattern: p=157/168=0.93 E10:fit with the pattern: *:13; +/-:14; M's:6 doesn't fit with the pattern: *:0; +/-:12: M's:6 probability for the next single occurrence to fit with the pattern: p=150/168=0.89 E11:fit with the pattern: *:13; +/-:12; M's:6 doesn't fit with the pattern: *:0; +/-:14: M's:6 probability for the next single occurrence to fit with the pattern: p=148/168=0.88 E12:fit with the pattern: *:13; +/-:14; M's:7 doesn't fit with the pattern: *:0; +/-:12: M's:5 probability for the next single occurrence to fit with the pattern: p=151/168=0.90 E13:fit with the pattern: *:12; +/-:17; M's:7 doesn't fit with the pattern: *:1; +/-:9: M's:5 probability for the next single occurrence to fit with the pattern: p=144/168=0.86 E14:fit with the pattern: *:13; +/-:21; M's:9 doesn't fit with the pattern: *:0; +/-:5: M's:3 probability for the next single occurrence to fit with the pattern: p=160/168=0.95 The shortest possible pattern is: E3->E9->E12->E13->E14->E15 The probability for it to fit would be: 0.90*0.93*0.90*0.86*0.95=0.615 But there is almost never a row of appearances that includes only one occurrence in every event. For 1x01-1x10 (Appendix A) there were 218 occurrences at 66 events, E15 not included. This are averagely 3.3 occurrences per event. The probability to fit with two occurrences at every event would be: 0.90^2*0.93^2*0.90^2*0.86^2*0.95^2=0.378The probability to fit with averagely 3.3 occurrences per event is: 0.90^3.3*0.93^3.3*0.90^3.3*0.86^3.3*0.95^3.3=0.201 The pattern can start at E1, E3, E4 and E5. The probabilities for the likeliest fits with the pattern are: E3->E9->E12->E13->E14->E15: p=0.201E3->E9->E10->E12->E13->E14->E15: p=0.139 E3->E9->E11->E12->E13->E14->E15: p=0.132 E3->E9->E10->E11->E12->E13->E14->E15: p=0.090 E1->E2->E3->E9->E12->E13->E14->E15: p=0.050E1->E2->E3->E9->E10->E12->E13->E14->E15: p=0.034 E1->E2->E3->E9->E11->E12->E13->E14->E15: p=0.033 E1->E2->E3->E9->E10->E11->E12->E13->E14->E15: p=0.022 E5->E6->E7->E8->E9->E12->E13->E14->E15: p=0.003E5->E6->E7->E8->E9->E10->E12->E13->E14->E15: p=0.002 E5->E6->E7->E8->E9->E11->E12->E13->E14->E15: p=0.002 E5->E6->E7->E8->E9->E10->E11->E12->E13->E14->E15: p=0.001 E4->E5->E6->E7->E8->E9->E12->E13->E14->E15: p=0.001E4->E5->E6->E7->E8->E9->E10->E12->E13->E14->E15: p=0.0005 E4->E5->E6->E7->E8->E9->E11->E12->E13->E14->E15: p=0.0005 E4->E5->E6->E7->E8->E9->E10->E11->E12->E13->E14->E15: p=0.0002 The other fits are negligible. The overall probability that the pattern fits is: p=0.201+0.139+0.132+0.090+0.050+...=0.711 This is the probability for the pattern to fit with single occurrences only. There can be simultaneous occurrences as well. For simultaneous appearances there are 12+11+10...=78 possibilities for two persons to appear simultaneous. If one of this two persons is P.Ya, then there are at E1 5 simultaneous appearances that fit: P.Ya with P.Al, P.BeC, P.LF, P.Ri, P.WeC and 7 simultaneous appearances that doesn't fit: P.Ya with P.BW, P.Da, P.En, P.Pi, P.Tr, P.Wo, P.WSA. For a single appearance the ratio of appearances was 12/1. For two simultaneous appearances the ratio is 5/7. The ratio to fit with the pattern will further decrease, if three or four simultaneous appearances are examined. Therefore the probability for single occurrences and simultaneous occurrences is: p<0.711 Approximately every 20th occurence is a simultaneous occurrences, therefore its only a small effect. The theoretical result fits well with the experimental result. The pattern was tested on a random data source and the probability that the pattern fits was 0.625. Only for the actual beginning the rules of chance are suspended. The result is a stable statistical anomaly. The involvement of chance would normally prevent a pattern with a residual uncertainty higher than 1:10^2. The pattern even includes a reference about a triune God with a residual uncertainty of 1:10^3. This results are really intruiging, if you only wanted to write a paper about how to quantise nontrivial patterns first. Edited by Dubreuil, : No reason given.
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
Cat Ski writes: No, it doesn't. As you've admitted, the odds of a pattern existing don't tell is how or why it occured. It doesn't tell where it come from, but it tells where it doesn't come from. The probability for the pattern to occur coincidental was 0.625 for the random data test and <0.711 for the calculation, although the pattern fit at (00:00). The probability that this is the result of chance was calculated to 1:10^7. Because of this it is almost impossible that the pattern occured because of coincidental effects.
Cat Ski writes: You said, along with the questions, that if the answers to all questions were 'yes' then the pattern could not have come about by chance. I told you that I already accepted that the pattern did not come about by chance so your questions are irrelevant. That it could not have come about by chance was shown with the 1:10^7 probability. The four questions show that any naturally imprinted pattern would be corrupted to a residual uncertainty below 1:10^2 through the involvement of chance.
Cat Ski writes: Any naturally imprinted pattern, for example imprinted by writers, would be corrupted to a residual uncertainty below 1:10^2 through the involvement of chance. No, that's not true. You've not proved this. Let's take a look to the four questions again:
quote: The pattern is really sensitive to coincidental effects. An example from [Msg=58]: A row of appearance that doesn't fit:*P.Al, *P.Pi, M14, P.Al-, P.Al+ But two rows of appearances that would fit:*P.Pi, *P.Al, M14, P.Al-, P.Al+ M14, *P.Al, *P.Pi, P.Al-, P.Al+ Only the first three appearances were mixed here and it has already changed whether the pattern fits or not. Every coincidental contribution can change the row of appearances or simultaneous appearances and affects whether the pattern fits or not. Although coincidental contributions should have highly influenced any naturally imprinted pattern, the found pattern has a high likelihood to appear and a high residual uncertainty, against the coincidental contributions. If you want to keep discussing about this, then I insist that you answer all four question. That will simplify the discussion.
Cat Ski writes: And again, the odds of the pattern existing tell us nothing about how or why it happened. Yes, it is only an indication for Intelligent Design. Intelligent Design claims there is a intelligent cause in evolution and this pattern shows a signal in evolution-like processes. The residual uncertainty of 1:10^3 is an other reference and the four questions above can preclude an other natural origin.
Cat Ski writes: quote: Its left unrefuted. The distinctness of the pattern was actually created this way. From [Msg=166]: "The elements therefore have to be grouped in a way to create a pattern of the most possible distinctness". The pattern was created to fit with the first three seasons of the data source and was then tested on a data source it was not created for. The argument about this is that there is such a complex pattern at all. From [Msg=166]: "If you are interested about this, then the pages 11 to 13 of the paper could be revealing to you. It was tried there to add actual random data from episodes to the pattern. To make the pattern fit with this random data, it became a random pattern itself. Large gaps were removed and the patterns within the pattern had to be removed too. The pattern also didn't fit anymore with previous episode, for example 4x08 as shown at the end. To make the pattern fit again with 4x08, M3 must be added to E12 and M14 to E13. This would again remove large gaps and would make the pattern even more random.". For the random data source it was not possible to create a distinct pattern that fit with the additional 9 episodes. For the first 76 episodes it was possible to create a distinct pattern, but not for the 9 episodes out of the random data source. This shows a difference between the actual beginning and random data. But for the residual uncertainties of 1:10^3 and 1:10^7 is the actual complexity of the pattern unimportant. This additional complexity (E11=E13,...) is mainly a side note, not part of any calculation.
RAZD writes: There are other transitions like this: E1->E2: 6 possible transitions What are those 6 possible transitions? I get 13 to 15 *P.BeC, P.BW-, P.Da-, *P.Ri, *P.Ya, M4 Every element that is present in E2 but not in E1 will cause a transition. An element that is present in E1 and E2 will not cause a transition. An element that is not present in E1 and E2 will break the pattern at E1. E3->E4: P.Al-, M4, M10E5->E6: *P.Al, P.Pi-, P.Ri-, *P.Tr, M1, M6 RAZD writes: Caveat: +/- cannot be observed without * appearance of individual. It can. For Example:
E5: *P.Pi /E6: P.Pi- /E7: P.Pi+, *P.Pi /E8: P.Pi-
RAZD writes: But you know who they are to assign +/- and therefore they are identified\observed. Yes, they are identified but it is not an appearance. An appearance for this pattern is defined as someone being named, start to speak or appear. And "he" or "she" is not a name. You try to rewrite the pattern rules. For example that every affected person includes an appearance of the person. If you look at table 4 on page 5, then you will see there a lot - or + without *. If they are all replaced from "+" and "-" to "*, +" and "*, -", then an reappearance of this person would not cause a transition. a row of appearances:*P.Da, P.Da-, *P.Da, *P.WeC, *P.Da, P.Da-, *P.Da With the actual rules:
E1: *P.Da /E2: P.Da- /E3: *P.Da /E9: *P.WeC /E11: *P.Da /E12: P.Da- /E13: *P.Da For your rules (probably):
E1: *P.Da /E2: P.Da-, *P.Da, *P.WeC, *P.Da, P.Da-, *P.Da You can test your revised system A and B, but they are completely different to the pattern introduced in the paper. Didn't you said you wanted to reproduce the work and not to create your own?
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
RAZD writes: it seems to me that I can make B15 be P.Da and then E is unique from A-C: No. I will explain it again: An element that is present at E1 and present at E2 will not cause a transition.An element that is present at E1 and not present at E2 will not cause a transition. An element that is not present at E1 and present at E2 will cause a transition. An element that is not present at E1 and not present at E2 will break the pattern. *P.Da will not cause a transition at E14. *P.Da is part of E14 and E15. This is detailed explained on page 6:E11: P.Da appears. *P.Da is not part of E9, but part of E11, therefore E11 is triggered. E12: The crew is mentioned. The crew in ST:TNG consist of 1000 persons. *P.Al is not part of E11, but part of E12, therefore E12 is triggered E13: The first officer is named. *P.Ri is not part of E12, but part of E13, therefore E13 is triggered. E14: P.Pi appears. *P.Pi is not part of E13, but part of E14, therefore E14 is triggered. E15: Commander William Riker is commended to be a highly experienced man. P.Ri+ is not part of E14, but part of E15, therefore E15 is triggered Other examples in your A-E table:Episode A E4: *P.Wo is part of E3 and E4. *P.Wo doesn't cause a transition at E3. Episode A E9: *P.Ri is part of E8 and E9. *P.Ri doesn't cause a transition at E8. Episode B E2: M5 is part of E1 and E2. M5 doesn't cause a transition at E1. Episode B E4: *P.Al is part of E3 and E4. *P.Al doesn't cause a transition at E3. Episode B E6: *P.Ri is not part of E6. Episode B E7: *P.Tr is part of E6 and E7. *P.Tr doesn't cause a transition at E6. Episode B E8: *P.Wo is part of E7 and E8. *P.Wo doesn't cause a transition at E7. Episode B E11: *P.LF is part of E9 and E11. *P.LF doesn't cause a transition at E9. Episode B E15: *P.Da is part of E14 and E15. *P.Da doesn't cause a transition at E14. Episode C E2: *P.LF is part of E1 and E2. *P.LF doesn't cause a transition at E1. Episode C E3: *P.Ri is part of E2 and E3. *P.Ri doesn't cause a transition at E2. Episode C E5: *P.Wo is part of E4 and E5. *P.Wo doesn't cause a transition at E4. Episode C E8: *P.BW is part of E7 and E8. *P.BW doesn't cause a transition at E7. Episode C E10: *P.Pi is part of E9 and E10. *P.Pi doesn't cause a transition at E9. Episode D E2: *P.Al is part of E1 and E2. *P.Al doesn't cause a transition at E1. Episode D E8: *P.Ri is part of E7 and E8. *P.Ri doesn't cause a transition at E7. RAZD writes: There is no pattern here that I can see: what should I be seeing that I am missing? You use a random data source. For the random data test it was not possible to create a distinct pattern that fits with the random data.
RAZD writes: When you assemble mirrors and bits of colored plastic in a kaleidoscope you can create the appearance of colored patterns, where any one would be highly unlikely to occur, but the pattern is an artifact caused by the mirrors, not the colored plastic bits. It's not an unlikely pattern that occurred once. It is an unlikely pattern with a high complexity, but it didn't occurred only once, it occurred 45 out of 47 times. It's like you use your kaleidoscope and 45 out of 47 times the same unlikely pattern occurs. And a kaleidoscope is not an evolution-like process.
RAZD writes: Well isn't that a problem that applies to other instances -- your subjective evaluation of good/evil, based on your worldview? It's based on the worldview of the persons. For example: P.evilperson1, P.evilperson2, P.otherperson1, P.otherperson2 P.ep1: "I feel so sick."P.ep2: "Where is P.op2? That's terrible." P.op1: "I will go to P.op2. I will make sure he won't help them." Is: *P.ep1, P.ep1-, *P.ep2, *P.op2, P.ep2-, *P.op1, *P.op2, {P.ep1-, P.ep2-} The worldview of the persons is defined. An objective worldview is not always defined.
RAZD writes: Curiously I think the pattern would be more universal: what I am trying to do is figure out is how you develop the pattern, so I figure a simpler system would be easier to start with. The pattern was created for the first 76 episode. It was revised until it did fit with the first 76 episode. Table 5 on page 8 shows the origin of the pattern. If you want to develop a pattern on your own, then you will need a data source with a similar size. Write down all quantisations and create a distinct pattern for it. Then test it on a random data source and one or more data sources it was not created for. That will take a few years. It took that long for this paper with 60 pages. You can then tell about your results in a few years.
RAZD writes: So I get your 6 "begin" transitions but it seems you missed the 14 "end" transitions -- elements that must exit before Event #2 can begin (and before any of the "begin" transitions occur). There are no "end transitions" with appearances. Only appearances:* and affected person:+/- were quantised. It would be an other pattern, if they are added.
RAZD writes: Pedantic hair splitting -- you know who is identified by "he" or "she" as plainly as if they were named because you assign +/- to them. The paper defined that an appearance is if a person is named, becomes visible or starts to speak. That are three possible appearances and "being identified" is not part of the possible appearances. "he", "she" or "I" are not appearances with this rules.
RAZD writes: Now I note that you previously said that P.(A)+ and P.(A)- were not elements, but you are treating them as elements now. Can you clarify this? *P.(A), P.(A)+, P.(A)- are different independent elements. They appear independent in the pattern. Sometimes solely as "+", "-", "*", sometimes together as "*, -", "*, +" at E1-E15.
RAZD writes: No, I would get (system A with 52 independent elements):E1: *P.Da‘ /E2: *P.Da- /E3: *P.Da‘ /E9: *P.WeC‘ /E11: *P.Da‘ /E12: *P.Da- /E13: *P.Da‘ Is *P.Da- = {*P.Da, P.Da-}?If so, then P.Da can't appear at E2. *P.Da is not part of E2 and E12. P.Da: "I feel very sick"Then P.Da walks offscreen and onscreen again P.Da: "I still feel very sick" Then P.Da walks offscreen and onscreen again P.Da: "I still feel very sick" Then P.Da walks offscreen and onscreen again P.Da: "I still feel very sick" Then P.Da walks offscreen and onscreen again P.Da: "I still feel very sick" is: *P.Da, P.Da-, *P.Da, P.Da-, *P.Da, P.Da-, *P.Da, P.Da-, *P.Da, P.Da- There is no P.Da‘ defined. And in this situation it would never happen.
RAZD writes: Curiously I see them as being the same pattern. Perhaps there is something missing that is not in your verbal descriptions of how the pattern works, or is applied. They are quantised differently. You count "he" or "she" as appearance and added end transitions. The transitions themselves are also not correctly applied yet.
Cat Sci writes: It doesn't tell where it come from, but it tells where it doesn't come from. No, it really doesn't.
That it could not have come about by chance was shown with the 1:10^7 probability. You said that number was the odds of the pattern existing. The odds of something existing tells us nothing about how it came about. I said this number is the odds of the pattern existing out of chance. 1:10^7 is a very low probability, therefore it's very unlikely that chance created the pattern.
Cat Sci writes: The four questions show that any naturally imprinted pattern would be corrupted to a residual uncertainty below 1:10^2 through the involvement of chance. Not really. A coincidental contribution changing the rows of appearances so that the pattern doesn't fit as well doesn't have anything to do with whether or not the pattern occurs naturally. It does. The pattern fits 45 out of 47 times. If chance would have been involved in the usual way, then the pattern would not fit that well and it would have only fit for example 35 out of 47 times. This is a residual uncertainty of 1:10.
Cat Sci writes: You can put me down for 4 yes's. I've already accepted that the pattern did not come about by chance. The fourth question maybe was not exactly formulated. I will specify the question to: 4. Do you agree that if the pattern doesn't fit that often, then any pattern will have only a low residual uncertainty like 1:10^2? The fourth question referred to any possible pattern. The involvement of chance would normally preclude any pattern with a residual residual uncertainty of 1:10^7 because of chance. The revised questions: 1. Do you agree there is an coincidental contribution?2. Do you agree that a coincidental contribution will change the row of appearances? 3. Do you agree that a change in the row of appearances will cause the pattern to not fit sometimes? 4. Do you agree that if the pattern doesn't fit that often, then any pattern will have only a low residual uncertainty like 1:10^2? If all this questions are answered with Yes, then the involvement of chance precludes any pattern with a residual uncertainty of 1:10^7 because: 1.->2.->3.->4. Do you still agree with all four question? If so, how could a pattern with a high residual uncertainty occur, although chance was involved? The paper suggests a pattern or a bias in chance itself.
Cat Sci writes: Yes, it is only an indication for Intelligent Design. I don't think you'll find anyone to disagree with the fact that Star Trek episodes were designed by intelligent people. Yes, and they all had preferences which can be retrieved as patterns with a certainty of 10% or 90% or even 99%. It's not about the existence of patterns, it's about their certainty.
Cat Sci writes: The fact that if you deviate from the pattern then your pattern doesn't fit anymore doesn't tell us anything about how the patterns emerged. The paper doesn't say anything how the pattern exactly emerged, there are only indications for the origin named. First: The similarity with the claims of ID about evolution and evolution-like processes.Second: The calculated residual uncertainty of 1:10^3 about a triune God Cat Sci writes: Intelligent Design claims there is a intelligent cause in evolution and this pattern shows a signal in evolution-like processes. No offense, but the real actual Intelligent Design movement is just a bunch of unscientific malarkey to disguise creationism. I agree. The story about the dispute with the ID proponents: Petition Petition for academic freedom
Cat Sci writes: That sounds like the thing that you are making the pattern for has an effect on the pattern itself. When its random you get randomness and when its not you don't. So, if your making your pattern on a TV show that already follows patterns, then you're just going to get a more complicated pattern. The random data source was not really "random". The starting time was randomised. The random data source was also a quantisation like the actual data source only with randomised starting times. Already present patterns would affect both data sources in the same way and there would be no difference in randomness about this already present patterns.
Cat Sci writes: How much stuff did you leave out, for instance? You're M# observations are just arbitrary. I added recurring appearances that are not persons as M's. There are possibly more M's that also fit with pattern but wasn't added yet. On page 15 a possible M15 is mentioned that could appear at E16. In five episodes followed right after E15 speechlessness as a possible M15. It didn't appeared at E1-E15 for the whole data source. I assume that there are not only 15 events. There are possibly a lot more events like 20, 30 or more. The residual uncertainty would drastically increase then. But 1:10^7 is also a good residual uncertainty which is accepted as a declaration of a discovery in science.
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
Cat Sci writes: But you kept revising the pattern until you got something that fit. You kept all the positives and disregarded all the negatives. But only for the first 76 episodes. As stated on page 6: "The pattern was created to fit with season 1, 3 and 4 at the actual start of the episode (00:00). Afterwards it was tested on season 5 and 6 and a random data source". You will find the origin of the pattern in table 5 on page 8. You will see there that this was done only for the first part of the data source. For the second part, season 5 and 6, all positives and negatives are listed in table 5. Three elements didn't fit with the pattern created form the first part of the data source and they are listed in table 5.
Cat Sci writes: And you're not looking at everything, you're only looking at things that you selected to look at. It was looked at the complete data source. Only season 2 was skipped because of an other main cast. Season 7 was not looked at, because a residual uncertainty of 1:10^7 was already reached. But if it reassures you, the pattern was also tested for season 7 some time ago and it did fit again 22 out of 23 times.
Cat Sci writes: What is the usual way that chance is involved in making episodes for a TV series? The usual way is the normal way if there is no bias or cause or pattern in chance itself.
Cat Sci writes: Any pattern? No. The patter: "Scene opens and later scene closes" would fit every single time. Then I will specify again: Any nontrivial pattern. There are a lot patterns like this: There is always a person appearingThere is always someone affected There is always someone talking A nontrivial pattern is a pattern far beyond this.
Cat Sci writes: But they have restraints. As I said before, they're not going to do this:
quote: Yes, but the pattern quantises rows of appearances. I doubt there is a restraint which always defines the row of appearances. And there were 4 different series with different restraints examined, but the same pattern was found.
Cat Sci writes: Yeah, well ID's claims about evolution are bullshit. I think I just found out why you are not open-minded about this paper.
Cat Sci writes: The opening scenes are constrained even further than the rest of the episode. And a random start could start you in the middle of an ongoing scene, so of course it would not have the same pattern as starting at the very beginning of a scene, and especially starting at the very beginning of the opening scene. The opening scenes have less constraints than the rest of the episode. You will find 18 examples (3x01-4x23) in [Msg=120]. The random start never starts in the middle of an ongoing scene. From page 6: "For the random data source it was assumed for the first season, that all episodes start at an other time (03:00-07:00) right after the opening credits.". You could read a few pages of the paper, maybe you will like it.
Cat Sci writes: It doesn't matter how low your uncertainty is if you just looking at something that you created yourself. Table 5 on page 8 will show you that this was only done for the first part of the data source, not for the second part. The residual uncertainty was calculated for the second part of the data source, not for the first part.
Cat Sci writes: And there are also M's that would totally destroy the pattern. But you don't include those. Can you name an example or is that just your opinion? If there would be an M that would totally destroy the pattern, then just a "*" has to be added for every event and it could appear everywhere. The pattern would then lose distinctness and would have a lower residual uncertainty. The residual uncertainty is a measure for the quality of a pattern. If you want to keep arguing against it, then you should start to get familiar with the mathematics about it. Science is refuted with science, not with opinions.
RAZD writes: you are now saying that anything left over from E1 can be in E2 even when it is not listed as a part of E2. The pattern is used on the quantisations, not on the original visual information. For Example: P.Wo, P.Ya, P.WeC walk onscreenP.Da walks onscreen P.Pi, P.Ri, P.WeC have a conversation Is: *P.Wo, *P.Ya, *P.WeC, *P.Da, *P.Wo, *P.Ya, *P.WeC, *P.Wo, *P.Ya, *P.WeC, *P.Wo, *P.Ya, *P.WeC, *P.Wo, *P.Ya, *P.WeC A possible fit is:
E1: *P.Wo /E2: *P.Ya, *P.WeC /E3: *P.Da, *P.Wo, *P.Ya /E9: *P.WeC /E11: *P.Wo /E12: *P.Ya /E13: *P.WeC, *P.Wo /E14: *P.Ya, *P.WeC, *P.Wo, *P.Ya, *P.WeC P.Da remains visible in the original visual information, but in the quantisation he doesn't appear again. The pattern is used on the quantisations, not on the original visual information.
RAZD writes: That makes the possibilities even more open than before when you have infinite possibilities due to multiple ad nauseum appearances, while maintaining a finite list of pattern breaking possibilities. In essence you have n/∞ → 0 probability of not fitting the "pattern" ... There are infinite possibilities to fit and not fit with the pattern. For Example:
E1: *P.LF, *P.Tr, M10??
E1: *P.LF, *P.Tr, *P.LF, *P.Tr, M10?? E1: *P.LF, *P.Tr, *P.LF, *P.Tr, *P.LF, *P.Tr, M10?? E1: *P.LF, *P.Tr, *P.LF, *P.Tr, *P.LF, *P.Tr, *P.LF, *P.Tr, M10?? and so on. *P.LF is part of E1 and E2, *P.Tr is part of E1. Both appearances will never cause a transition at E1. If there would be a 0 probability of not fitting the pattern, then the pattern would have always fit in the random data test, wouldn't it?
RAZD writes: No, I used selected elements from your "pattern" description in Message 166 "Elements are observed, either singly or in combinations and all with possible repeated appearances" ... I meant you didn't retrieved your quantisations from an actual audible and visual data source. If you want to reproduce the work, then you must retrieve your quantisations from an evolution-like process. Arbitrary arranged appearances could contain any arbitrary pattern.
RAZD writes: That depends on how the "pattern" is defined. If I define the "pattern" as a symmetrical arrangement of colored elements then anything the kaleidoscope shows will fit that definition of the "pattern" ... This pattern will have a probability of 1 to fit. But the E1-E15 pattern has only a probability of 0.625 to fit. The 0.625 probability resulted from an experimental test and a theoretical calculation in [Msg=190] resulted in a probability of p<0.711. To create a comparable pattern, you would have to define a pattern that also fits only about every second time.
RAZD writes: Now suddenly, where I had 5 episodes that did not show a discernible "pattern" there is a completely perfect fit for each of these hypothetical episodes to the "pattern" You would have to test the pattern created from the first 5 episodes on further 5 or 50 episodes. A pattern has no predictive power for a data source it was created. This wasn't done for the E1-E15 pattern as shown in table 5 on page 8.
RAZD writes: The question that you haven't answered yet is why I should make a pattern this way. You can also try other ways if you want to create a pattern yourself.
RAZD writes: add these transitions (and who\what can stay from the previous event/s and who\what would be an invalidating elements) between each event: A possible transition is every element that is not part of the current event but part of a next event.An element that is part of the current event will not cause a transition. An element that is not part of the current event or a next event will invalidate the pattern. There is nothing like "can stay" or "can not stay". The pattern is not used on the original visual information, it is used on the quantisations. For E1:transitions to E2: *P.BeC, *P.Ri, *P.Ya, M4, P.BW-, P.Da- invalidating: *P.En, M3, M10, M11, M12, M14, P.Al+, P.BeC+, P.BeC-, P.Da+, P.En+, P.En-, P.LF+, P.LF-. P.Pi+, P.Pi-, P.Ri+, P.Ri-, P.Tr-, P.WeC+, P.Wo+, P.Wo-, P.WSA+, P.WSA-, P.Ya+, P.Ya- For E2:transitions to E3: *P.BW, *P.Da, *P.En, *P.Pi, *P.Tr, *P.Wo, *P.WSA, M1, M3, M6, P.BW+, P.Pi-, P.Wo+ invalidating: M2, M7, M10, M11, M12, M13, M14, P.Al+, P.Al-, P.BeC+, P.BeC-, P.Da+, P.En+, P.En-, P.LF+, P.LF-. P.Pi+, P.Ri+, P.Ri-, P.Tr+, P.Tr-, P.WeC+, P.WeC-, P.Wo-, P.WSA+, P.WSA-, P.Ya+, P.Ya- The other transitions and invalidations can be found with the already explained rules too.
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
Well, I'm quiet sure I don't want to comment this anymore. The tone of the conversation becomes increasingly hateful, and I don't want to read such hate messages. The knowledge about the paper is still minimal. I have corrected Cat Scis posts often enough to know he didn't read the paper and RAZD just found out 12 hours later that he had to rethink his fit vs fail and that his 26/∞ argument was just wrong. There is also obviously on one familiar with information science. Nearly all questions didn't had to be answered if there would be some knowledge about these sciences or the paper. I probably could continue writing here for months, and the knowledge about the respective sciences and the paper would still be insufficient to write a substantiated comment about it. I really don't need this comments. The paper was already looked at by people with enough knowledge about the respective sciences. It doesn't matter to me if this paper will never be published too, I don't even like the most of the ID proponents. I really don't want to read all this hateful comments anymore. If anyone knows the proceedings in science, then he knows that it is not usual to be offensive and hateful until a different opinion was exterminated. This thread can be closed. You hate me? I hate you too.
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
I have rethought my decision about stop commenting here. I will give it second chance, but only with clear rules I will observe for myself. I will ignore every comment that includes:
1. sarcasmor 2. insults or 3. expletives (BS, Bullshit) RAZD showed that it is possible to write comments like this. I will also ignore all posts from: 1. Coyote: He suggested to ignore all portions of the paper, although there was still a discussion about it. Therefore I will also ignore all posts of him if he wants to join a discussion one day.2. Dr Adequate: He has more likely a Ph.D in insulting than in maths. Hopefully this rules I will observe for myself will change the currently destructive conversation into a constructive conversation. Science is not war. And I don't want to be attacked for the opinions and arguments. We will see if this new rules I will observe for myself are sufficient to allow a civil discussion.
RAZD writes: But if it reassures you, the pattern was also tested for season 7 some time ago and it did fit again 22 out of 23 times. Curiously that is about what I would expect from the way your pattern is constructed. I assume you mean that it would always fit with every data source? The probability was tested to 0.625 to fit with random data and calculated to <0.711 in [Msg=190]. If you are agreeing with this calculation and the test, then you would expect only a 14 out of 23 fit.
RAZD writes: The pattern is used on the quantisations, not on the original visual information. For Example: P.Wo, P.Ya, P.WeC walk onscreenP.Da walks onscreen P.Pi, P.Ri, P.WeC have a conversation Is: *P.Wo, *P.Ya, *P.WeC, *P.Da, *P.Wo, *P.Ya, *P.WeC, *P.Wo, *P.Ya, *P.WeC, *P.Wo, *P.Ya, *P.WeC, *P.Wo, *P.Ya, *P.WeC A possible fit is: E1: *P.Wo /E2: *P.Ya, *P.WeC /E3: *P.Da, *P.Wo, *P.Ya /E9: *P.WeC /E11: *P.Wo /E12: *P.Ya /E13: *P.WeC, *P.Wo /E14: *P.Ya, *P.WeC, *P.Wo, *P.Ya, *P.WeC P.Da remains visible in the original visual information, but in the quantisation he doesn't appear again. The pattern is used on the quantisations, not on the original visual information. So in one episode he stays and in another he doesn't and you think these are the same. With your rules the quantisation would be: *P.Wo, {*P.Wo, *P.Ya}, {*P.Wo, *P.Ya, *P.WeC}, {*P.Wo, *P.Ya, *P.WeC. *P.Da}, {*P.Wo, *P.Ya, *P.WeC. *P.Da} ... If someone is named or affected, then it continues: {*P.Wo, *P.Ya, *P.WeC. *P.Da, *P.Pi}, {*P.Wo, *P.Ya, *P.WeC. *P.Da, P.Da-}, {*P.Wo, *P.Ya, *P.WeC. *P.Da, P.Da+} You would create a pattern with *, +, - for every part of the pattern and it would always fit. You could not create a distinct pattern then, if you also quantise all stays.
RAZD writes: With 52 elements (counting +/- and adding 1 for "something else") you have 26/52 = 0.50 for E1 and 28/52 = 0.54 for E2 probability to fail. Your 26/52 is better than your 26/infinite argument but still incomplete. Appearances occur about 10 times more often than +/- and M's. Therefore the probability to fit with E1 is higher than 0.50. The probability to fit with E1 and E2 in a row would be E1*E2=0.50*0.54=0.27. The probability to fit with 15 elements in a row would maybe be 0.0001. This contradicts the experimental result of a probability of 0.625 to fit. I already calculated the probability in [Msg=190]. I suggest you read and refer to this calculation first.
RAZD writes: P.Ya+ and P.Ya- cannot occur because *P.Ya does not exist in E1 and the appearance of P.Ya means you are in E2. Likewise P.BeC+ and P.BeC- cannot occur without *P.BeC causing a transition to E2, *P.En is not listed in E1 or E2 or in your invalidating list, so P.En+ and P.En- also cannot occur without *P.En (which presumably would be an invalidating element), and similarly P.Ri+ and P.Ri- need to be preceded by *P.Ri (which presumably would be an invalidating element), and so we see that the actual possible invalidation elements are reduced to 20 elements. You are actually right about this. But that's only for E1. The later events have a possible appearance before.
RAZD writes: They are take from your quantisations from actual audible and visual data. And you still haven't said why your elements are grouped the way they are when there is no discernible common thread within each group. The pattern was created to fit with the first 76 episodes. If you remove *P.Da from E7, then the pattern wouldn't fit anymore with 1x03 (table 5 page 8). If you add *P.Da to E6, then it wouldn't fit anymore with 1x03 too. The pattern is the simplest pattern known to me, that fits with this whole first part and can describe all *, + and - over a time of nearly a few minutes.
RAZD writes: Curiously something is missing that would de facto cause an event change: a scene change. Only elements were quantised within the episodes. A scene change can cause an event change. For example if a new person appears.
RAZD writes: My overall impression is that you have not tried to simplify the pattern (equation), preferring to add elements to get a 100% fit, and as a result it is unwieldy and clunky. The reason for the different events is to describe the happenings over time. 7 Events would maybe only describe 30 seconds. 15 events already describe a minute. 100 events would describe the happenings for maybe 5 minutes. The more events, the more predictive power has the pattern.
RAZD writes: You might find that a system with half of your elements would be accurate 95% of the time ... which would still be a strong signal of pattern. Do you refer to the M's or to ...? It is actually possible, that the pattern without M's could be accurate 95% of the time. But I found that humour (M3) only appeared at E3 and E14. It only appeared there, therefore I added M3 to the pattern. But it probably could be removed too without affecting the patterns fit.
RAZD writes: First - because the "pattern is used on the quantizations, not on the original visual information" the "pattern" depends on what you chose to quantisize Yes. The paper was based on previous studies that found law-like patterns. According to a previous study, every time a person is avoided, every time a person gets a positive benefit and every time a person just appears was observed. With this basic outline, there were M's that always appeared at the same places, therefore I added them to the pattern.
RAZD writes: Second - the "possible fit" is based on how elements are grouped, which appears arbitrary and has not been explained. It was explained in table 5 on page 8. If you remove *P.Da from E7, then the pattern wouldn't fit anymore with 1x03. If you add *P.Da to E6, then it wouldn't fit anymore with 1x03 too.
RAZD writes: Now obviously if you included every possible element into the Event #1 cast there could only be one event, so the question is how and why are the elements divided into different events? To describe the happenings over time. 100 events could describe the happenings for maybe 5 minutes.
RAZD writes: Well an easy first approximation would be to start with those elements that start the episodes - ie start with the element that is at the start of episode 1 and then add the element that is at the start of episode 2 (if it is not the same element) and so on, until all 76 of the "data set" episodes is included. That's how it was done. There are 76 episode. The pattern is the simplest pattern known to me, that fits with this whole first part and can describe all *, + and - over a time of nearly a few minutes.
RAZD writes: And we can continue this way until the Event #n cast would include all the elements, there would be none left to cause a transition to a next event. True. It wouldn't describe the happenings over time anymore.
RAZD writes: It would also be different from the published one, and likely it would have fewer total events. The less events, the less predictive power over time.
RAZD writes: This process is dependent on what is chosen to observe as an element that triggers an event: is the picture of a sun (from a planet) different from a picture of a star field? Do I count an empty room (view without people) as an element? Do I then count the room as an element when other elements (people) are included? Only *, + and - for persons were observed first. All M's that appeared always at the same events were then added to the pattern. There are possibly more M's that also always appear at the same events.
RAZD writes: Back the bus up, young fella ... where are events 1, 2, 4, 5, 6, 7, 8, and 10? AND you only count elements that trigger the next event in the latter seasons??? From page 6 and [Msg=190]: "At the events 1, 3, 4 and 5 the pattern is allowed to start". For 1x01 the pattern starts at E3. The optional events 10 and/or 11 and/or group {4,5,6,7,8} can be omitted. E{4,5,6,7,8} and E10 were omitted. "there are only appearances and affected persons noted that trigger the next event." means that the notation is shortened from:
E1: *P.LF, *P.Da, *P.Tr, *P.Da, *P.LF /E2: *P.Ri to
E1: *P.LF /E2: *P.Ri
RAZD writes: So the "pattern" must have been built from the first two seasons in order to know the triggers. Your text talks about adding new elements for later episodes to make them fit the "pattern" ... did you go back and review the first two seasons for these additional elements to see if they changed the documentation for those seasons? That doesn't make sense. The pattern was built from the first three seasons of the data source (table 5 page 8).
RAZD writes: Now I notice that disappearances are mentioned even though you tell me they are not part of the "pattern" ... From page 3: "If a person gets interrupted while speaking through someone else and then starts to speak again it counts as an-other appearance. Equally if a person walks away and becomes visible again after this disappearance." Every appearance precedes a disappearance. If disappearances would be observed too, then every time a person is interrupted while speaking would be {appearance: P.personwhostartedtospeak, disappearance: P.personwhostoppedtospeak}. It would complicate the pattern a lot to add all this simultaneous appearances. It is nice that you finally started to read a few more parts of the paper. This discussion here will probably go one for a few months until a few persons have acquired the knowledge about the text and content of the paper. There will probably be more unknown sentences no one has read until now. The new rules I introduced in this post will make the discussion hopefully more civil than before. @Cat Sci: I will answer your last post in one of the next comments. You still don't know how the pattern works ([Msg=219]) and you haven't read the paper yet. It will take more time to answer your posts to explain all this to you. Expect your posts to be answered maybe every third day. I will answer your last post, although it contains a lot of bullshit. You have to remove this word in future posts, if you want me to answer to them.
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
Cat Sci writes: Cat Sci writes: But you kept revising the pattern until you got something that fit.You kept all the positives and disregarded all the negatives. But only for the first 76 episodes. Even worse. Your criteria for your pattern is a hodgepodge of arbitrary observations for a portion of the episodes in the series. It matches all the other episodes because you defined it into place by basing the criteria on the observations that you made. It was not created for the whole data source, only for the first 76 episodes. You will see this in the big table 5 on page 8. You will also see there, that it didn't matched all the other episodes, how you claim. It only matched 45 out of 47 times.
Cat Sci writes:
No, I mean that your M#'s are just based on what you noticed and selected for or against based on whether it could fit a pattern or not. You don't have M#'s for things you didn't notice and you don't include M#'s that don't fit the pattern you are creating.Cat Sci writes: If instead you used: M1 closed window, color pink/greenM2 musical instrument, "How many?" M3 sadness, crying and so on Obviously, you wouldn't have found a pattern, right? As far as I know: Nowhere appeared a closed windowNowhere appeared a color pink/green One time appeared a musical instrument Nowhere appeared a "How many? Nowhere appeared a crying Recurring appearances that are not persons were added as M's. All M's appeared recurringly. The pattern was found for *, + and -. Afterwards the M's were added. For the M's it was not selected against anything. All recurring appearances that appeared recurringly at specific events in the first part of the data source was added to the pattern.
Cat Sci writes: You're diluting the criteria for your pattern into one that can fit a lot of possibilities. The pattern fits with a lot of possibilities and doesn't fit with a lot of possibilities too. If the pattern would fit almost ever, then there wouldn't be 9 out of 24 episodes that didn't fit in the test with random starting times. If the pattern would fit almost ever, then the probability for the pattern to fit wouldn't have been calculated to <0.711 in [Msg=190].
Cat Sci writes: The opening scene is like a summation of the episode, or even a mini-episode, in its form. This is no surprise. Have you read this?
quote: Nothing of this is a summation.
Cat Sci writes: Cat Sci writes: What is the usual way that chance is involved in making episodes for a TV series? The usual way is the normal way if there is no bias or cause or pattern in chance itself. Pssh. You can't use the definition of the word to define the word you're using. The usual way is if there is no bias or cause or pattern in chance itself, the existence of chance.
Cat Sci writes: Then I will specify again: Any nontrivial pattern. So then, it is up to you what counts as trivial or not? Triviality (mathematics) - Wikipedia For example, consider the differential equation y'=y where y = f(x) is a function whose derivative is y′. The trivial solution is y = 0, the zero function while a nontrivial solution is y(x) = e^x, the exponential function. Your example "Scene opens and later scene closes" is a trivial pattern, it has only one part and fit always. The E1-E15 pattern has 15 parts and describes *, + and - for about a minute.
Cat Sci writes: Yes, but the pattern quantises rows of appearances. I doubt there is a restraint which always defines the row of appearances. Is incredulity your only argument? Because that's a logical fallacy... I will give you an example in words what the pattern predicts. The pattern predicts there is no P.Ya+ from E10 to E14. In words: 1. If P.Tr and P.Ri appeared simultaneous and if then P.LF appeared and if then P.Da appeared, then P.Ya can't be positively affected until for example P.Tr has appeared and then P.Da has appeared and and then P.Tr has appeared again.2. If P.LF appeared and if then P.Pi was negatively affected and if then P.Da appeared and if then P.Pi was negatively affected again and if then P.BeC appeared and if then P.Wo appeared, then P.Ya can't be positively affected until for example P.LF was negatively affected and then P.Wo appeared and then P.BeC appeared again. 3. If P.Ya, P.Pi and P.Tr appeared simultaneous and if then P.Pi was positively affected and if then P.Pi was negatively affected, then P.Ya can't be positively affected until for example P.Tr has appeared again and then was negatively affected and then P.Pi appeared again. This list is far from being complete for no P.Ya+ from E10 to E14. There are more predictions made for P.Ya and there are 12 more persons. Maybe you see now that there is no restraint that can create this complexity in every episode. It is also only a "restraint", something that can't happen. You need to assume a "force", that creates this complexity. The pattern is about something that is happening. For example the reference about a triune God. No one has yet disagreed with this. This reference was proved in the appendices D-I. There are three persons "God", "Jesus" and "Bible". It was shown that there are 12 persons the pattern would be broken with and only 1 person that fits. It was shown that P.Go=P.Je=P.Bi=P.Ya. Any other person for this three persons would break the pattern, only P.Ya fits. For 1x09 Bsfk (Appendix I) the persons Abraham, Sarai and God must appear as P.BW, P.Wo and P.Ya. Any other fit than P.Ab=P.BW, P.Sa=P.Wo and P.Go=P.Ya would break the pattern. There are 13 persons "God", "Jesus" and "Bible" can appear as, but they all appeared as P.Ya every time. This pattern is about something that is happening. The residual uncertainty is (1/13)*(1/13)*(1/13)≈1:2.2*10^3.
Admin writes: I'm still waiting for an answer for how you could start a thread about a paper with "Triune God" in the title and then in discussion completely disavow any discussion of God. A theologist could answer: "God is omniscient. What you call bizarre is part of a greater plan which you can't understand because you are only a inferior human being". An theologist will answer this to every question, including evolution, the question about evil and death in the world and why innocent babies die in Africa. Evidence is the opposite of religion. Any religious belief could claim: "That does not concur with my beliefs about God" and any religious belief could claim: "But it does concur with my beliefs about God". Discussions about religion are the more arbitrary, the more religious beliefs start to discuss. This is a statistical paper, not a religious. I don't claim it proves the cristian God, I only present my finding that the pattern supports a triune God with a residual uncertainty of 1:10^3. This is a mathematical argument, not a theological.
Admin writes: or that your "patterns" are so ill defined as to match almost anything. If the pattern would fit almost ever, then there wouldn't be 9 out of 24 episodes that didn't fit in the test with random starting times. If the pattern would fit almost ever, then the probability for the pattern to fit wouldn't have been calculated to <0.711 in [Msg=190].
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
RAZD writes: Not really. I'm just saying that there is a difference to people remaining and people leaving a scene, and that this muddles the dependent P.(A)+/- assignments. It can be talked about a person no matter if this person is visible or not. A person can be also commended if the person is not present.
RAZD writes:
That still does not explain the method used to determine the event groups (or how events are defined) and this makes it difficult to duplicate.RAZD writes: That's a start, but as noted, I ran into a couple of problems doing that. First a simple pattern was created for a few episodes and the more episodes were added, the more complex it became to fit with all episodes. But at about 50 episodes the pattern stopped to become more complex. The other parts of the data source fit then with it. Of course it will be difficult to duplicate it, it will take you maybe a year to create a pattern from nothing from a big data source, which describes *, + and - for about a minute.
RAZD writes: The way I saw it was that the probability of a deal breaker decreased with each event if you don't have a mechanism (and a reason for it) for removing elements from the event cast, so you have increasing probability of pattern fit when it should be constant or increasing. You have for example to remove elements from certain events to allow further transitions.
RAZD writes: Curiously this makes the pattern less restrictive rather than more ... adding optional alternatives when you don't need it to define the pattern. It makes it more restrictive. For example: Row of appearances:*P.Al, {*P.Tr, *P.Ri}, *P.Pi, M13 E1: *P.Al, {*P.Tr, *P.Ri??}E3: *P.Al, {*P.Tr, *P.Ri}, *P.Pi, M13?? E4: *P.Al, {*P.Tr??, *P.Ri} E5: *P.Al?? You were right with "the invalidations would have to work in ALL RAZD writes: So the pattern has ...
E2 is not a possible start. The pattern has
for a total of 24 different pattern variations. Only E1, E3, E4 or E5 can be arbitrary chosen. Which events are omitted is decided by the elements that occur. If at E3 an element of E9 occurs which is not part of E3, then E4-E8 are skipped. If at E3 an element of E4 occurs which is not part of E3, then E4-E8 are not skipped. If at E9 an element of E12 occurs which is not part of E9, then E10-E11 are skipped. And so on.
RAZD writes: Curiously I think your calculations are flawed by not properly accounting for the multiple (32) patterns within your overall pattern, any one of which can be fit by the new season episodes. The probability was calculated for the multiple possible patterns. It was calculated for the 16 most likely possibilities. The other fits were negligible. For Example: From [Msg=190]: E3->E9->E12->E13->E14->E15: p=0.201E3->E9->E10->E12->E13->E14->E15: p=0.139 E3->E9->E11->E12->E13->E14->E15: p=0.132 E3->E9->E10->E11->E12->E13->E14->E15: p=0.090 and so on.
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
Cat Sci writes: Anyways, what about "bullspit", can I say bullspit? No. Your tone is still somehow sarcastic. I will give you an other example: The pattern predicts there is no simultaneous appearance of P.Da and P.Tr from E10 to E13. In words: 1. If P.Tr and P.Ri appeared simultaneous and if then P.LF appeared and if then P.Da appeared, then P.Da and P.Tr can't appear simultaneous until for example P.Tr has appeared and then P.Wo has appeared.2. If P.LF appeared and if then P.Tr appeared and if then P.Ya and P.Wo appeared simultaneous and if then M1 (an open door) appeared and if then P.BeC appeared and if then P.Wo appeared and if then P.Ri appeared, then P.Da and P.Tr can't appear simultaneous until for example P.Ya appeared and then P.Da appeared. 3. If P.Ya, P.Pi and P.Tr appeared simultaneous and if then P.LF appeared and if then P.Wo appeared, then P.Da and P.Tr can't appear simultaneous until for example P.BeC has appeared and then P.LF appeared.
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
NoNukes writes: quote: Proof... Uh, you are aware that Star Trek is science fiction written and directed by someone whose religious position is not documented? The "Proof" shows that the three persons "God", "Jesus" and "Bible" can only appear as P.Ya. It is also shown that all other 12 person would break the pattern. This proof was done on two independent series.
NoNukes writes: How are you separating God's actions from man's actions in your analysis? The pattern is a "nontrivial" pattern. Everyone could include a sentence "A triune God exists". But the examination made in appendices D-I is different. The result that P.Go=P.Je=P.Bi=P.Ya was created through simultaneous appearances that would break the pattern and similar occurrences. Any person who wanted to intentionally include this data, would have needed a broad knowledge about the pattern to design every episode in a way which creates this result. But you can maybe assume that God made the man doing this. For this purpose the residual uncertainty was calculated. You would expect that if the persons P.Go, P.Je and P.Bi can only appear as one person, then it would be a random person with a probability of 1/13 for every single person. They all appeared as P.Ya resulting in a residual uncertainty of 1:10^3. A person who wanted to intentionally include this data, would have needed a broad knowledge about how to design all episodes properly.
NoNukes writes: Why is it reasonable to assume (and in fact to assume conclusively) that your patterns are not artifacts of deliberate choices made by writers and directors working on the show? Because of the residual uncertainty of 1:10^7. The pattern did fit 45 out of 47 times for a data source it was not created for. This is only a 95% certainty. The high residual uncertainty results from the behaviour contrary to chance. The probability to fit with a data source the pattern was not created for was tested to 0.625 and calculated to <0.711 in [Msg=190]. The probability for the pattern to fit behaves contrary to the tested and calculated expectations for the two seasons it wasn't created for and also for the other 3 series that were examined. This behaviour contrary to chance can be calculated to a residual uncertainty of 1:10^7. You can verify it yourself: [Msg=171]. And chance would normally preclude any nontrivial pattern with a high residual uncertainty of 1:10^7. I also formulated four questions about this:
quote: A nontrivial pattern is a pattern a lot more complex than: "Scene opens and later scene closes". The involvement of chance would normally corrupt this pattern to a residual uncertainty below 1:10^2.
NoNukes writes: So should we ignore the explicitly religious parts of your paper? Why is your paper in the Religion and Spiritualism section of the archive? There are no religious parts. The residual uncertainties are based on statistics, not on beliefs. It is in the Religion section because ID is classified as Religion.
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
RAZD writes: The problem with having 24 optional patterns is that it is difficult to use for predictions -- you would have 24 different predictions ... which each would need to be tested. There are only 4 real different predictions. The sub-variations can't be chosen arbitrary. Only in this cases are more than one sub-variation possible: E4-E8:M4 appears at E3. M4 is part of E4 and E9, but not part of E3. This has never happened. E10/E11:Sometimes E10 OR E11 can be triggered. If so, then a next occurrence (for example: *P.Ri, P.Da+ or *P.Tr) will remove the possibility to choose in the most cases. An additional choice exists almost never. In all other cases there are no optional sub-variations. Every row of appearances has only 4 variations it can fit with. There is almost no arbitrariness beyond this. For example the 9 examples in Appendix B that didn't fit, there were always only one sub-variation possible. There is for example no possibility to not skip E4-E8, if P.LF appears at E3. There are only 4 possible variations for every row of appearances. A fifth possible variation occurs almost never.
RAZD writes: So both the 10 event model and the 11 event model have higher predictive value than the 15 event model, and the 10 event model has fewer parts, so it is the better model. Yes, you have accounted 616 errors for your 15 event model and 136 errors for your 10 event model and 11 event model. Therefore your 10 event model and 11 event model have higher predictive value than your 15 event model. For the multi-pattern model there were 3 errors accounted, therefore the multi-pattern model has a higher predictive value then your 10 event model and 11 event model.
RAZD writes: The problem with having This was done in appendix B.
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Dubreuil Member (Idle past 3069 days) Posts: 84 Joined: |
NoNukes writes: The "Proof" shows that the three persons "God", "Jesus" and "Bible" can only appear as P.Ya. The item you claimed to be testing was the proposition that the Christian God is the intelligent designer of the ID hypothesis. Where did you test that proposition? From page 14: "If there is a triune God as designer that wants to be known, then a person called "God" could always appear as P.Ya. For this purpose it was looked for series that include "God" at the beginning. If God appears always as P.Ya, then this would strongly indicate the existence of a triune God as the designer of intelligent design." If there is a triune God as designer that wants to be known, then he could add a reference about the existence of a triune God into this design. The pattern already included the number 3 as part of P.Ya. An easy way to add a reference about the existence of a triune God into this design would be to let all persons like "God", "Jesus" and "Bible" appear as P.Ya. This was tested and verified.
NoNukes writes: And you did not actually, "Prove" that even the proposition that you claim. What you have showed, at best, is that a particular appearance did not occur in a very few television series episodes. You have not established that such an appearance could not occur. This was shown.Appendix G shows that P.Bi can't appear as: P.Al, P.BeC, P.BW, P.Da, P.En, P.LF, P.Pi, P.Ri, P.Tr, P.WeC, P.Wo, P.WSA. Any of these persons as P.Bi would break the pattern. Appendix H shows that P.Je can't appear as: P.Al, P.BeC, P.BW, P.Da, P.En, P.LF, P.Pi, P.Ri, P.Tr, P.WeC, P.Wo, P.WSA. Any of these persons as P.Je would break the pattern. Appendix I shows that P.Go can't appear as: P.Al, P.BeC, P.BW, P.Da, P.En, P.LF, P.Pi, P.Ri, P.Tr, P.WeC, P.Wo, P.WSA. Any of these persons as P.Go would break the pattern. Appendix D shows that P.Bi can appear as P.Ya.Appendix E shows that P.Je and P.Go can appear as P.Ya. Appendix E shows that the number 3 is a part of a person at all and that the number 3 can't be part of P.Al, P.BeC, P.BW, P.Da, P.En, P.LF, P.Pi, P.Ri, P.Tr, P.WeC, P.Wo, P.WSA. Any of these persons would break the pattern. Appendix A and E shows that the number 3 can be part of P.Ya. It was not shown that an other appearance than P.Go=P.Je=P.Bi=P.Ya could never ever occur. Because of this there is a residual uncertainty of 1:10^3.
NoNukes writes: But more importantly, the fact remains that the writers do not insert events randomly, and they are not, in fact attempting and failing to generate random events. They are trying to tell a story that makes some sense. I already answered this in the last comment. A coincidental contribution would preclude any nontrivial pattern with a residual uncertainty of 1:10^7 because of 1.->2.->3.->4.. I suggest you answer first the four questions in [Msg=239] with for example "Yes, No, No, No" or "Yes, Yes, Yes, No" or 4 times Yes. This will simplify the discussion.
NoNukes writes: In short, you are nowhere near making a convincing argument that you are not simply drawing your bull's eyes around whatever your shooting happens to hit. And every time you force an element or provide rules that allow alternatives, all you are doing is force fitting data because the pattern did not work without the alternatives. The pattern and the rules were completely outlined before the test with the result P.Go=P.Je=P.Bi=P.Ya was done. @Cat Sci: Your comment will be ignored. It contains:1. expletives: bullshit, pissed 2. sarcasm: "that pissed the Captain off gets chastised. That actually makes perfect sense." You can keep discussing with other persons here, but I will not respond to your comment until all expletives and all sarcasm were removed. If you did so, post an other comment that you have removed the expletives and the sarcasm and that you now want to have an answer to your comment.This is done as outlined in [Msg=231] about the rules I observe for myself.
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