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Apr0122
  al wazir: <johnlspouge>: I assume that you found those sets by a computer search. (That would have been my first move.) But is there a formula or criterion for them?
I think of this problem as a generalization of the formula for integer solutions of the Pythagorean relation a^2 + b^2 = c^2: a = 2mn; b = m^2 – n^2; c = m^2 + n^2, where m and n are integers. 

Apr0122
  al wazir: I'm having no luck finding "a more systematic approach" to the problem you posted. At this point I've found two examples (153 and 407) in addition to the ones you gave and an upper bound on the range within which others can occur (2,916). Based on my trialanderror searching, I doubt that there are any more threedigits examples, but I expect that there are some with four. Trialanderror isn't a very elegant way of looking for them, and I'm not going to bother writing a code. I'll give myself half credit and move on. 

Apr0322
  al wazir: I suspect that a prescription that generates some or all triples of positive integers whose cubes add up to a fourth cube must exist, though I have no idea of how to go about finding it. I googled "sums of three cubes," but all I found was a discussion of a different problem: https://en.wikipedia.org/wiki/Sums_.... I await enlightenment from other sources.
Here's a different problem, one that I *have* managed to solve: <On a lonely desert island, there are 17 yellow chameleons, 15 green chameleons, and 13 blue chameleons. It's a big island, but occasionally a pair of chameleons meet. If they are the same color, nothing happens, but if they are of differing colors, then both of them, out of politeness, change to the remaining color. If this is the only way that color changes can occur, then is it possible that after an appropriate sequence of meetings, all the chameleons could be the same color?> https://people.sc.fsu.edu/~jburkard... 

Apr0322   johnlspouge: The original configuration of chameleons is (17, 15, 13) has remainders (2, 0, 1) mod 3. By enumeration of cases, the given operation on (2, 0, 1) mod 3 just permutes (2, 0, 1) mod 3, so iterations of the operation always leave at least 2 types of chameleon nonzero. 

Apr0422   FM David H. Levin: <al wazir: Supposedly the QWERTYUIOP layout of the keyboard on a standard typewriter was chosen so that salesmen could quickly spell out "typewriter." (This story may be apocryphal, but I believe it. Why else would anyone who thought for five minutes about it have chosen such an inefficient design?)> I read long ago that the QWERTY layout was designed to slow down typists because they kept jamming the mechanism on typewriters that employed the original layout (called Dvorak, as I recall). One time, when I visited a telephone operating company in 1985 as part of my work, I saw Dvorak keyboards at a data center, but I don't recall seeing them elsewhere. 

Apr0422
  al wazir: <FM David H. Levin: I read long ago that the QWERTY layout was designed to slow down typists because they kept jamming the mechanism on typewriters that employed the original layout (called Dvorak, as I recall).> I think you've got your facts mixed up. The Dvorak keyboard is a later optimized design intended to remedy the defects and inefficiency of the QWERTY layout. https://en.wikipedia.org/wiki/Dvora... I too have heard that the QWERTY layout was chosen to prevent jamming of the keys. Pray tell, why would they be less likely to jam in that configuration than in another, randomly chosen one? Jamming of the keys in the original mechanical typewriters occurred whenever two or more keys were struck at once. That can happen with any layout. So I don't believe it. 

Apr0422
  al wazir: <johnlspouge>: Thanks for posting your solution. It saves me from posting my own, which involved solving a set of three simultaneous equations three times, once for each color. I'm sure <nok> would have sneered and called it ugly. 

Apr0422   FM David H. Levin: <al wazir: The Dvorak keyboard is a later optimized design intended to remedy the defects and inefficiency of the QWERTY layout.> Thanks for the information and the link.
<al wazir: I too have heard that the QWERTY layout was chosen to prevent jamming of the keys. Pray tell, why would they be less likely to jam in that configuration than in another, randomly chosen one?> I thought we were discussing not a randomly chosen alternative to the QWERTY layout, but rather an alternative layout that has been claimed to increase typing speed. 

Apr0422   johnlspouge: < <al wazir> wrote: <johnlspouge>: Thanks for posting your solution. It saves me from posting my own, which involved solving a set of three simultaneous equations three times, once for each color. > My pleasure. I am reminded
"That would be a butcher's way, not an artist's" [ Search Kibitzing ] Still,
#36. You gotta go with what works.
[ https://memoryalpha.fandom.com/wik... ] 

Apr0422
  al wazir: <FM David H. Levin: I thought we were discussing not a randomly chosen alternative to the QWERTY layout, but rather an alternative layout that has been claimed to increase typing speed.> My post was poorly phrased.
Keys can jam with any configuration, whether chosen with a view to speed in typing or randomly. But if the aim was to minimize jamming, putting all of the mostfrequently used keys (ETAIONSHRD) on one row would be the way to go. QWERTY doesn't do that. Linotype does. I think you are saying that the QWERTY layout was deliberately chosen to be inefficient, to slow down typists. What QWERTY does do is cause the left hand to make about 75% of the keystrokes. For righthanded people (which is most of us), this is indeed inefficient. It would have made more sense to reverse the arrangement, giving the right hand the majority of keystrokes. This would have been just as effective in reducing jamming. To my mind, that is evidence that the design was not thought out carefully or with any consideration of typing speed. 

Apr0422   FM David H. Levin: <al wazir: ...snip...What QWERTY does do is cause the left hand to make about 75% of the keystrokes. For righthanded people (which is most of us), this is indeed inefficient. It would have made more sense to reverse the arrangement, giving the right hand the majority of keystrokes. This would have been just as effective in reducing jamming. To my mind, that is evidence that the design was not thought out carefully or with any consideration of typing speed.> One does get the impression that the adoption of the QWERTY layout was a decision by Sales or Marketing rather than Engineering or Design. (As a former systems engineer, I occasionally take a swipe at the less technical business functions.) 

Apr0422
  harrylime: Oh Clever Boys here who have discovered the Universe and beyond ! lol lol 

Apr1722
  al wazir: Where I live  and maybe everywhere  the National Weather Service expresses its forecasts in terms of probabilities. For example, today at https://forecast.weather.gov/MapCli... I see this prediction for Monday night, April 18: <Rain likely, mainly after midnight. Mostly cloudy, with a low around 51. South wind 9 to 14 mph. Chance of precipitation is 60%. New precipitation amounts of less than a tenth of an inch possible.> And for Tuesday, I see this: <A 40 percent chance of rain before noon. Mostly cloudy, with a high near 57. New precipitation amounts of less than a tenth of an inch possible.> I once asked an NWS employee what these percentages mean. His answer, as I recall it, is that if the probability is given as 60%, then six times out of ten someone living *anywhere* within the area covered by the forecast will get rain *sometime* during the period in question. Of course the higher the percentage, the more likely it is to rain. But if the chance of rain next Wednesday is given as 90% and I don't get any rain, that doesn't prove that the Weather Service goofed. A 90% probability means I can expect that one time out of ten, it *won't* rain. For a long time I've wondered how, at least in principle, I can check the accuracy of these probabilities. How can I tell if, e.g., they average too high or too low? If they were always the same percentage it would be easy. For example, if the probability were always stated as 50%, then I would keep a record of the times rain is predicted and find the fraction in which I actually got some rain. How well that fraction approximates 1/2 would be a measure of the accuracy of the predictions. Over a long time it should be pretty close. But the stated probabilities vary. Sometimes they're only 20% or 30%, sometimes as high as 90%, and very rarely, even 100%. How can I determine the accuracy for all these predictions taken together? I want a single figure of merit that says how closely the NWS predictions approximate reality. I've found a simple solution, and what's more, I think it's the only solution. 

Apr1722
  beatgiant: <al wazir>
Did you try doing a literature survey on "calibrating predictive models"? With the rise of machine learning, we have more and more such models and there are a lot of papers on the subject.Meanwhile, xkcd has this bit: https://xkcd.com/2370/ 

Apr1822
  al wazir: <beatgiant: Did you try doing a literature survey on "calibrating predictive models"?> No. I didn't know the right buzz words to use. But I'm going to do it now. Thanks. 

Apr1822
  scutigera: <al wazir>:
<If "prototype" can be verbed, then someone who prototypes is a prototyper, unless he's a prototypist. > The word "prototype" was verbed decades ago: the formal name for 3D printing is "rapid prototyping", which is not only a verb form, but one that, as a gerund, has been made back into a different noun. English is a hot pervy slut of a tongue, Lord bless it. 

Apr1922
  beatgiant: I rarely have any success explaining probability and statistics on chessgames.com. This time, I'll just post a very simple example of the model calibration problem. Suppose an insurance company is providing insurance to the shipping industry. Suppose they have a model that says 10% of shipments through a certain region of the South Pacific are lost, and also 10% of shipments through a certain region of the Caribbean are lost. They set their insurance rates accordingly. In the following year, suppose they insure 9,000 shipments through the dangerous region in the South Pacific, of which 500 are lost, and 1,000 shipments through the dangerous region in the Caribbean, of which 500 are lost. So, out of 10,000 shipments through the dangerous regions, 1,000 are lost, which is indeed 10% as their model provided. Success! The insurance company makes money. But how accurately did their model perform? Based on this sample, what would you predict will happen if, in the year after, they insure 9,000 shipments through the dangerous part of the Caribbean and 1,000 shipments through the dangerous part of the South Pacific? 

Apr1922
  al wazir: <beatgiant: But how accurately did their model perform? Based on this sample, what would you predict will happen if, in the year after, they insure 9,000 shipments through the dangerous part of the Caribbean and 1,000 shipments through the dangerous part of the South Pacific?> That's not quite the same problem. Presumably the NWS makes its forecasts based on models that simulate the objective features of the ocean/atmosphere environment, not on its previous record of success or failure. 

Apr1922
  beatgiant: <al wazir>
The NWS has a model that takes the myriad available observations (temperature, pressure, imagery, local geography, etc.) as inputs to a complex system to estimate the probability of a weather event. The insurance company in the toy example has a model that takes one observation (the route of the shipment) and applies two simple rules (through Caribbean, through South Seas) to estimate the probability of a shipment loss event. If you like, those rules could be based on a very simple simulation of the activities of the pirates of the Caribbean and South Pacific storms. In both the cases, they need a way to check how well the models perform so they can make adjustments. I'm not understanding what you view as the crucial difference between the two systems in principle. Yes, the NWS system uses many more inputs and a much more complex function, but those are differences in scale, not in principle. But as I said above, I'm never very successful explaining these things on chessgames.com, even with savvy interlocutors such as yourself, so feel free to move on if I'm not making sense to you. 

Apr1922
  al wazir: <Beatgiant>: I'm not having much luck relating the weather forecasting problem to what I found by googling “calibrating predictive models.” Most of the sites seem to involve the success or failure of medical diagnoses, which is not a blackandwhite yes/no sort of question like whether or not it will rain. So I'll go ahead and post my solution and hope that someone will comment on it. Suppose the odds that something will happen are stated as n':n", where n' and n" are positive integers. This means that if you bet n' dollars at these odds and it happens, you win n'' dollars. The odds that it won't happen are n":n'. A person betting against you puts up n" dollars and wins your n' dollars if it doesn't happen. This is equivalent to saying that the chance of its happening is given by a percentage p = n'/n, and the chance of its not happening is 1 – p = n"/n, where n = n' + n". (Apparently the “chances” stated by the NWS are always multiples of 10%, so we can assume that n = 10, but this isn't essential.) Let's suppose you repeatedly bet that rain will occur assuming that the forecast is right, i.e., you always bet at the equivalent odds, and that the stakes S (the total of the amounts on the two sides of the bet) are the same each time. (I'll come back to that assumption.) Then you stand to win (n"/n)S or lose (n'/n)S. If the forecast always said that the chance is the same unchanging percentage p, and if the forecast of rain turns out to be correct m' times and wrong m" times, where m' + m" = m, the number of bets, then your total winnings or losses T would be given by T = S(m' x n"/n – m" x n'/n) = mS[(m'/m)(1 – n'/n) – (1 – m'/m)(n'/n)] = mS(q – p), where q = m'/m.
If q = p, i.e., if the observed frequency of rain equals the predicted chance of rain, then T = 0. This would confirm that the predictions are accurate. T may not exactly vanish, but the closer it is to zero the better the agreement between predictions and observations. But the predicted chance of rain varies from one forecast to the next: the forecasts employ different values of the probability p. If we label them p_1, p_2, etc., the corresponding numbers of bets m_1, m_2 , etc., and the observed frequencies q_1, q_2, etc., the formula can be generalized to T = SΣ m_i(q_i – p_i),
where the summation is over the index i.
If all the terms are close to 0, then T will again be close to 0. But T can also be close to 0 if some of the terms are positive and some are negative and they cancel. (This is the situation in your insurance example.) Saying that this confirms the accuracy of the predictions would be very misleading. The way to avoid that is to rewrite the formula using the *absolute value* of each term: T = SΣ m_iq_i – p_i.
Of course, since each term is nonnegative, the more forecasts that are included in the sample, the bigger T becomes. Thus defined, T is more likely to be close to 0 for a small sample than for a large one, which makes it a misleading figure of merit. The way to correct for this is to normalize by dividing by m, the total number of cases in the sample. And since the absolute size of the “bet” is irrelevant, we can set S = 1: T = (1/m)Σ m_iq_i – p_i,
where m = Σ m_i.
This is my answer.
But is it realistic? If you were trying to maximize your gains, would you vary the size of your bets at different odds n':n" in proportion to n'/n, i.e., would you bet more when both your potential loss and the chance of winning are larger? Maybe you would bet the same amount each time, or bet more when the potential payoff is bigger. My answer is that the purpose of this experiment is not to make money but to test the accuracy of the “chances” in the NWS predictions. For this reason I think the size of the stakes should be the same for every bet, i.e, each observation should be given the same weight. 

Apr2022   nok: <The word "prototype" was verbed decades ago: the formal name for 3D printing is "rapid prototyping", which is not only a verb form, but one that, as a gerund, has been made back into a different noun.> Verbed and gerunded, then. 

Apr2022
  beatgiant: <al wazir>
I have time to post a few comments only.
I don't really agree that "rain or no rain" is more absolute than something like "stomach cancer or no stomach cancer." Over a given region at a given time, there could be anything from heavy rain over the whole region to 1mm of rain in one small patch, and an arbitrary decision is made where to draw the line. The typical buzzword for these functions to measure the difference between the model's distribution and the real world's distribution is "loss function," and the one you derived is "absolute loss." Another common one is "quadratic loss" where the differences are squared instead of placed under absolute value. There are many loss functions. Again I recommend a literature survey. 

Apr2022
  al wazir: <beatgiant: I have time to post a few comments only.> Don't waste any more time on this until I have another go at the literature. <Over a given region at a given time, there could be anything from heavy rain over the whole region to 1mm of rain in one small patch, and an arbitrary decision is made where to draw the line.> Of course. If you look at the examples I posted you'll see that the bar is set pretty low, e.g., "new precipitation amounts of less than a tenth of an inch possible." What does "less than a tenth of an inch" mean? One hundredth of an inch is less than a tenth of an inch. A light fog drip is less than a tenth of an inch. When the forecast is for half an inch or more (which happens, alas, all too seldom in California), I am much more confident that there will be actual, measurable amounts of precipitation. My formula doesn't distinguish between a diffident notice of an impending drizzle and near certainty that a deluge of biblical proportions is coming. They both count as forecasts of rain. 

Jan2023   Cheapo by the Dozen: I recognized John Wayne, Nick Cage, Cary Grant, Lady Gaga, Charlie Sheen, Woody Allen (barely) and Madonna. 

Jan2023   Cheapo by the Dozen: The only one I'm sure I could have gotten in the other direction was John Wayne. 



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