tag:blogger.com,1999:blog-7729093380675162051.comments2016-09-25T02:28:01.728+10:00moyhuNick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comBlogger6875125tag:blogger.com,1999:blog-7729093380675162051.post-8732573254209737012016-09-25T02:28:01.728+10:002016-09-25T02:28:01.728+10:00Balancer looks fine now!Balancer looks fine now!HaroldWnoreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-23431451130975674202016-09-25T01:55:12.132+10:002016-09-25T01:55:12.132+10:00I hope it's OK now.I hope it's OK now.Nick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-90407537617168204822016-09-25T01:33:15.135+10:002016-09-25T01:33:15.135+10:00Harold,
Yes, on checking I see that the balancing ...Harold,<br />Yes, on checking I see that the balancing coins arrangement is not working. It was - I must have messed it up somehow. Working on it.Nick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-9127231186536788342016-09-25T01:27:11.342+10:002016-09-25T01:27:11.342+10:00HaroldW,
Thanks. The need for prior good coins to ...HaroldW,<br />Thanks. The need for prior good coins to balance is only pressing for the maximal case of (3^N-1)/2 - 13 coins in 3 weighings, for example, or 4 in 2. If there is 1 coin less, (like the common 12) it's not needed. Of course, you can make use of the bonus coins if you want - by putting 8 on one side with good coins to balance, for example. But you can do the first weighing without that, and later there are always enough known coins from that weighing.<br /><br />Of course the program just tips the balance based on its knowledge of where the fake is - arranging to add the good coins to balance numbers is for appearance, and it may not always get it right. I'll check.<br /><br />Nick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-8544089673520949912016-09-24T23:09:54.646+10:002016-09-24T23:09:54.646+10:00Nick -
Cool Java program. But when the script &quo...Nick -<br />Cool Java program. But when the script "imagines" the existence of known good coins, it's not true to the original problem statement. And that's significant: with N=4 coins (2*N=8 possibilities), the script allows a solution in 2 steps. [That's consistent with the ceil(log3(2*N)) formula based on information. One can start by weighing #1&#2 vs. #3, for example.] But without a known good coin, one can only start by comparing 1-v-1 or 2-v-2, either of which allow a path with 4 possibilities remaining, requiring two more steps for a guaranteed solution. <br /><br />Another example with N=12 occurs when the initial 4-v-4 comparison is unequal; one can put 3 from the heavier side & 3 from the lighter side on the left pan and weigh. This is a good strategy -- that is, one stays on the max-3-weighings path -- but without positing additional coins beyond the 12, there are only 4 known good coins at that point in the tree.<br /><br />[Also, a nitpick - the balance code isn't quite right in what it shows, displaying 2 known-good coins offsetting one unknown. At least some of the time.]HaroldWnoreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-85859769948141183402016-09-24T19:27:18.035+10:002016-09-24T19:27:18.035+10:00I thought more about the version of the problem wh...I thought more about the version of the problem where you don't have to say whether the fake is heavy or light. It isn't much harder - I think it extends the number you can identify by just 1. Think of the case of 3 weighings and N coins. If N≤13 you can just follow the same method; you'll identify the fake and whether heavy or light. But what if N>13?<br /><br />The max you can put on the balance at first is 9 (with good coin to even up). If it tilts, you have the same remaining problem as above - 9 coins where you have ruled out heavy or light, and the N-9 you left off are all good. If no tilt, then at the next weighing you can put at most 3 on, since that's all you can resolve if tilt. But if no tilt again, then in the third weighing you can only put 1 on; with 2 unknown coins you can't say anything if it tilts and no more weighing possible. So that is 9+3+1=13 coins put on. But at the last stage, you could also have a coin off balance. If tilt, then it's good, and if not, then it is the fake, even though it has never been on the balance, so you don't know if it is heavy or light. So N=14, but not more.<br />Nick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-30393136586608711962016-09-24T15:27:37.215+10:002016-09-24T15:27:37.215+10:00"sorry to waste your time"
That's w...<i>"sorry to waste your time"</i><br /><br />That's what this problem is famous for :)<br />Besides, I wasn't clear - hope it's better now.<br />Nick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-70772030028145754222016-09-24T15:17:11.206+10:002016-09-24T15:17:11.206+10:00I can now see why it wasn't as easy as I thoug...I can now see why it wasn't as easy as I thought. sorry to waste your time.j fergusonhttp://www.blogger.com/profile/05188939085463962922noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-74759390674373593742016-09-24T12:31:28.099+10:002016-09-24T12:31:28.099+10:00ps Kevin,
Sorry I didn't make that extra requi...ps Kevin,<br />Sorry I didn't make that extra requirement clear. I see how it was causing a difficulty. It's actually a much harder problem if you don't have that requirement. Nick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-9342894303333406632016-09-24T12:18:01.349+10:002016-09-24T12:18:01.349+10:00I've added above to specify that the answer ha...I've added above to specify that the answer has to include whether the fake is heavy or light. Nick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-67343798049583076122016-09-24T12:13:41.924+10:002016-09-24T12:13:41.924+10:00Kevin.
A way to think about it - could you get 2 c...Kevin.<br />A way to think about it - could you get 2 coins in 1 weighing? You could weigh them against each other, but if it tilts left, you don't know whether the left is heavy or the right light. And if it balances, you know it's the off one, but not whether heavy or light. Actually, the last is probably OK within the rules; the method I'm describing always not only identifies the fake, but gives that info. I'm not sure whether not having to know that makes a huge difference.<br /><br />INT(log3(2))+1=1, but INT(log3(2*2+1))=2, and I think 1 weighing isn't enough in all cases.<br />Nick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-65471453194065792312016-09-24T12:03:09.462+10:002016-09-24T12:03:09.462+10:00Kevin,
"And with 21 coins you can find the fa...Kevin,<br /><i>"And with 21 coins you can find the false coin in 3 weighings."</i><br />Only if you know whether it is heavy or light. Otherwise, with 21 coins there are 21*2=42 possibilities to distinguish. And with 3 weighings, there are only 3*3*3=27 possible outcomes to discriminate them.<br />Nick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-31619083589199414732016-09-24T11:59:35.141+10:002016-09-24T11:59:35.141+10:00jf,
Try out the gadget. The final weighing can res...jf,<br />Try out the gadget. The final weighing can resolve three coins, as long as you by know know that they can only be one kind (heavy or light, H or L in my terms). Say you have 2 H, 1 L. Put an H each side, and leave one off. If it tilts, you'll know which is heavy; if not, then it's the one left off. You always have to learn as much as possible from each result, including level, which tells you the fake is not on the balance.<br /><br /><i>"Maybe what you are interested in here isn't solving the thing but establishing the limits within which it can or cannot be solved."</i><br />The limits are in the earlier arithmetic; to do n coins in M weighings, 2*n≤3^M. With this method, you can actually achieve that, which isn't too hard to demonstrate.<br /><br />Remember, we don't know if the fake is heavy or light. But if the first weighing tilts left then we know:<br />1. The left coins can't be light<br />2. The right coins can't be heavy<br />3. The coins off balance must be good.<br />Nick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-89454869975205721142016-09-24T11:48:38.716+10:002016-09-24T11:48:38.716+10:00Nick, I think using n*2 adds an extra step. E.g., ...Nick, I think using n*2 adds an extra step. E.g., with 21 coins: INT[Log3(n*2)]+1 = 4 while INT[Log3(n)] + 1 = 3. And with 21 coins you can find the false coin in 3 weighings.Kevin O'Neillhttp://www.blogger.com/profile/15751040367339659805noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-59407720213030798532016-09-24T11:44:11.905+10:002016-09-24T11:44:11.905+10:00You don't know the bias up or downYou don't know the bias up or downAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-84851981255489211302016-09-24T11:15:08.864+10:002016-09-24T11:15:08.864+10:00I must be missing something. FWIW, this was intui...I must be missing something. FWIW, this was intuitively obvious or I didn't understand it. The third weighing has to be one coin against another. the second weighing would have to be two coins on each side and the only way to get there would be to qualify the group of four - this can be done by weighing 8 of the coins first, if they balance the off-coin isn't in that group and you now can look at the four coin remainder.<br /><br />Maybe what you are interested in here isn't solving the thing but establishing the limits within which it can or cannot be solved.<br /><br />it's too bad I've never been able to do anything useful with this talent (if that's what it is) at this sort of thing.j fergusonhttp://www.blogger.com/profile/05188939085463962922noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-31289089144572021392016-09-24T09:33:35.290+10:002016-09-24T09:33:35.290+10:00"I then realized that when n is greater than ...<i>"I then realized that when n is greater than 15 using 1/3 vs 1/3 is the best weighing strategy. "</i><br />I think you should always get as cloase as you can. That is related to maximising uncertainty - 1/3 chance left, 1/3 R and 1/3 balance.<br /><br /><i>"INT[Log3(n)] +1"</i><br />I think it is INT[Log3(n*2)]+1, if the coins could be heavy or light (hence the *2). That determines the first weighing. You should kep off the max number you could resolve in one less weighing, which you'll have to do if it balances.<br />Nick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-84989754218064525752016-09-24T09:06:44.788+10:002016-09-24T09:06:44.788+10:00Intuitively I thought:
Weight one half of the coin...Intuitively I thought:<br />Weight one half of the coins against the other half (if odd, leave one out)<br />Take the half that reads lower and repeat. (if they're equal, then you must have left one out and it's the guilty party).<br /><br />This essentially then just becomes a matter of converting a decimal (# of coins) to binary. INT[Log2(n)] where the integer portion will tell you how many weightings you'll need to reach a conclusion.<br /><br />I then realized that when n is greater than 15 using 1/3 vs 1/3 is the best weighing strategy. Not knowing exactly how to formalize this I experimented a little in a spreadsheet and it looks like when n > 15, then INT[Log3(n)] +1, and if n < 15, then INT[Log2(n)] gives what I think are the minimum number of weighings for any number (n) of coins where one coin is different than the rest.<br /><br />Kevin O'Neillhttp://www.blogger.com/profile/15751040367339659805noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-5105311691140014252016-09-23T04:32:20.929+10:002016-09-23T04:32:20.929+10:00Bryan, when you write "...my subjective feel ...Bryan, when you write "<i>...my subjective feel for the uncertainty is more like at least about +/- 0.3C and maybe as much as 0.5C..."</i> do you believe they left out important uncertainties or disagree with the values they place on individual uncertainties?<br /><br />To accept uncertainties of +/- 0.5 C we'd pretty much have to ignore all of the phenological evidence we've accrued. Either that or make the argument that most earth systems are much more sensitive to small changes than currently accepted. For biological systems that might make some sense, but for lake ice, glaciers, sea-level and other non-biological systems one has a hard time escaping the physics. One can ignore thermometer or temperature readings and simply look at phenological data and arrive at the same conclusion.<br /><br /><br /><br />Kevin O'Neillhttp://www.blogger.com/profile/15751040367339659805noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-26819957252519468082016-09-22T08:56:01.271+10:002016-09-22T08:56:01.271+10:00Bryan,
Hadcrut has been committed to the ensemble ...Bryan,<br />Hadcrut has been committed to the ensemble approach used also in forecasting, and I expect that is where th confidence comes from. As with any approach, the answer depends on what you vary going into the ensemble (or equivalent). When GISS or NOAA quote (larger) uncertainty, my understanding is that the largest component is station spread - ie how much different would it be if we measured in different places. I don't know if HADCRUT includes this. But they have written quite a lot about it - I should look it up.<br />Nick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-63398337918884463492016-09-22T07:03:53.543+10:002016-09-22T07:03:53.543+10:00Nick, your explanation of the changes in HadCRUT4 ...Nick, your explanation of the changes in HadCRUT4 makes good sense to me. However, I still have trouble with their "95% confidence limits", which average at +/- 0.15C for January through July of 2016. Based on my extensive experience in working with surface air temperature measurements and data for over 40 years now, my subjective feel for the uncertainty is more like at least about +/- 0.3C and maybe as much as 0.5C and increasing the farther back you go in the estimates (probably more like +/- 0.5C to 1.0C for the 1850-1900 period where they show an average of +/- 0.3C).Bryan - oz4casterhttp://www.blogger.com/profile/18027990322659101002noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-90422433761637175372016-09-21T07:19:53.587+10:002016-09-21T07:19:53.587+10:00Actually, there is a wrinkle that I should have me...Actually, there is a wrinkle that I should have mentioned; they average by hemispheres, so in effect the hemisphere average is used there. But yes, and the Arctic too. That is basically what C&W were doing - replacing the imputed NH average of those cells by something more locally appropriate.Nick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-84978724391701111932016-09-21T03:08:49.174+10:002016-09-21T03:08:49.174+10:00"HADCRUT is an average by grid (area-weighted..."HADCRUT is an average by grid (area-weighted), in which cells without data are simply not included. That has the effect of assigning to them the global average, which is dominated by sea temp."<br /><br />I'm embarrassed to say that despite reading through some of the relevant publications and documentation I didn't realize that was the case. So 1/3 of Africa and 1/5 of South America are simply omitted... that is something I'll keep in mind in future.Magmanoreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-73997437849399456392016-09-17T06:13:36.577+10:002016-09-17T06:13:36.577+10:00Bryan,
Well, I'm not smoothing them; the numb...Bryan, <br />Well, I'm not smoothing them; the numbers as plotted are tabulated and plotted as reported. But it may be because the SH winter ice has open ocean frontage, while NH ice has all sorts of land features and currents - bays that melt (or freeze) together, straits where a wind can wash ice away. Or possibly even that the land just causes some measurement confusion.<br />Nick Stokeshttp://www.blogger.com/profile/06377413236983002873noreply@blogger.comtag:blogger.com,1999:blog-7729093380675162051.post-57495019316066850662016-09-17T04:36:56.793+10:002016-09-17T04:36:56.793+10:00Ice extent is an insufficient proxy for assessing ...Ice extent is an insufficient proxy for assessing the short-term dynamics of the arctic ice. If you examine satellite imagery of the arctic (MODIS, for example) over the past few weeks, you can see that the real culprit in extent reversal was not re-freezing (though there was likely some), but rather the dramatic fracturing of the ice just north of the Canadian Archipelago. Those massive cracks suddenly increased the extent in the Central Arctic Basin due to expansion of ice extent via fissuring; and sent ice floes into the CA and Greenland Sea, via transport (also increasing extent). More of that ice will meet its demise if the weather encourages further transport to warmer waters. Long story short, though the extent may be increasing, the ice has still been getting hammered.Robert Bitmanhttp://www.blogger.com/profile/15996926846675777693noreply@blogger.com