tag:blogger.com,1999:blog-2031071943178978556.post8142156518449726645..comments2022-10-25T16:43:39.384+01:00Comments on detuned radio: The MSLS: RevolutionsUnknownnoreply@blogger.comBlogger45125tag:blogger.com,1999:blog-2031071943178978556.post-14963117901952787252020-10-22T22:53:03.126+01:002020-10-22T22:53:03.126+01:00Sorry: 2014, not 2010...Sorry: 2014, not 2010...Fred76https://www.blogger.com/profile/03805064563109466105noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-26112925067567714082020-10-22T19:49:50.386+01:002020-10-22T19:49:50.386+01:00Tom, you've got me experimenting with more exo...Tom, you've got me experimenting with more exotic partitions now... :)<br /><br />For example, the one I'm playing with at the moment has the red cells as the intersections of r1379 with b1379, and the blue cells as the intersections of r2468 with b24568. Each set has 24 cells, and must contain the same digits. What's especially interesting with this arrangement as it relates to the challenge puzzle above is that boxes 13579 all have six cells of one color, so with a 5/4 split in the digits each of the boxes *must* haso you can immediately get away with only 22 givens matching the partition (ve at least one (or two) of the corresponding digits, even if the are not given.<br /><br />How long it will take to find an interesting puzzle with something like this, I have no idea...Philip Newmanhttps://www.blogger.com/profile/11212211922590240781noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-29520870967911146802020-10-22T11:09:27.657+01:002020-10-22T11:09:27.657+01:00This is not the first time that interesting and de...This is not the first time that interesting and deep discussions occur on your blog ;). I remember the fundamental discussion about WSC when you were in charge of it in 2010.Fred76https://www.blogger.com/profile/03805064563109466105noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-71890255528693164972020-10-22T01:24:54.078+01:002020-10-22T01:24:54.078+01:00correction 247 should read 279 which is this
+-...correction 247 should read 279 which is this <br /><br />+------------------------+---------------------------+------------------------+<br />| 1456-27 12367 13456 | 1345-2 12379 1345-279 | 13468-9 12489 13468 |<br />| 1456-2 1236 9 | 8 123 1345-2 | 1346 124 7 |<br />| -14(27) 8 134 | 134(2) 6 -134(79-2) | 134(9) 5 134 |<br />+------------------------+---------------------------+------------------------+<br />| -168(29) 5 168 | 16(-2) 4 -18(27) | -168(79) 3 168 |<br />| 1468 136 7 | 9 138 1358 | 14568 148 2 |<br />| 1468-29 12369 13468 | 1356-2 12378 1358-27 | 14568-79 14789 14568 |<br />+------------------------+---------------------------+------------------------+<br />| 1568 16 2 | 7 138 1348 | 13458 148 9 |<br />| (79) 4 18 | 13(2) 5 -138(29) | 138(7) 6 138 |<br />| 3 79 158 | 14 189 6 | 2 1478 1458 |<br />+------------------------+---------------------------+------------------------+<br />{3 digit muti fish} <br /><br />3 digits: 3 fish 3 base & 4 covers <br /><br />wont completely clear it all but does a lot of the work of a larger msls <br /> <br /><br /><br />strmckrhttp://forum.enjoysudoku.com/member2491.htmlnoreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-54683354741755008562020-10-22T01:13:55.187+01:002020-10-22T01:13:55.187+01:00I thought at this stage I'd add some further r...I thought at this stage I'd add some further reflections.<br /><br />1. I think we're finding our limits with blogger comments unfortunately!<br />2. It's amazing to have so many of the protagonists I've been following all here on the same page! I think at this stage we're only missing Phistomefel, 胡蒙汀(Hu Meng Ting), and whoever it was at nikoli that came up with that puzzle (Tea.M maybe suggests a collaboration)<br />3. There is so much gold here in these comments I don't really know where to start. I think Sam's recasting of intersection theory as some kind of multiplicity theory I think is more or less where I was hoping this would all end up at, without ever quite knowing myself where I was hoping this would end up.<br />4. It's probably worth acknowledging that in one form or another there isn't really anything here that comes as news to the real experts here, if it is dressed up in the appropriate language and notation.<br />5. I think the real appeal for me with intersection/multiplicity theory is that it very much focuses on cells in the grid, rather than candidates within cells. Once you know that the contents of one group of cells somewhere in the grid is equal to another group of cells, it feels very natural to scan the contents of these two groups and make deductions that way. Sam's diagram [ https://ibb.co/MZtXsMs ] demonstrates this beautifully!<br />6. I think the contrast with the established theory from where I'm sitting is that it is much more focused on candidates within cells, which for solvers like me is basically not much fun!<br />7. I wonder how many other techniques that I'd previously considered to be impenetrable might be seen through this more regional lens?Tom Collyerhttps://www.blogger.com/profile/03663579809785456634noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-10963533604727573272020-10-22T00:44:09.040+01:002020-10-22T00:44:09.040+01:00My answer to Philip's challenge is to place a ...My answer to Philip's challenge is to place a 6, but I won't add any further spoilers for nowTom Collyerhttps://www.blogger.com/profile/03663579809785456634noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-52714283866855802302020-10-21T23:05:13.942+01:002020-10-21T23:05:13.942+01:00Thanks for that. It's certainly a lovely diagr...Thanks for that. It's certainly a lovely diagram with the coloring. :)<br /><br />Interested what you think of the above challenge as well. It's a little bit trickier of an application; I'm definitely going to have to get it working in a puzzle that solves cleanly afterward.Philip Newmanhttps://www.blogger.com/profile/11212211922590240781noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-38834102060241911812020-10-21T22:37:12.785+01:002020-10-21T22:37:12.785+01:00It is possible to solve the tatooine sunset sudoku...It is possible to solve the tatooine sunset sudoku with intersection theorem, which gives an equivalent solving path than Trevor's one. I explained it there: https://tcollyer.blogspot.com/2020/10/tatooine-sunsets-and-partitions-of.html?showComment=1603315915175#c2274367086054215763Fred76https://www.blogger.com/profile/03805064563109466105noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-35513805921513809132020-10-21T21:34:22.495+01:002020-10-21T21:34:22.495+01:00You're welcome. :) I'm actually pretty new...You're welcome. :) I'm actually pretty new to sudoku myself, but I kinda got thrown in the deep end on the player's forum and had to get familiar very quickly with the crazy techniques they effortlessly use. So a relatively tame AIC/CNL is quickly becoming familiar.Philip Newmanhttps://www.blogger.com/profile/11212211922590240781noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-34415841268520690322020-10-21T21:31:47.929+01:002020-10-21T21:31:47.929+01:00Oof. Definitely not what I had, but I follow the l...Oof. Definitely not what I had, but I follow the logic so looking into chains with you guys is definitely paying off!<br /><br />If 6 is not in r1c7, 6r1c2 (immediately) and:<br />2r1c4, 4r1c6, 5r4c2, 1r4c5, 9r2c5, 6r6c5<br /><br />a. 1r4c5, 9r2c5, 6r6c5 => 8r8c5<br />b. 6r1c2, 6r6c5 => 6r7c3 => 8r8c3<br /><br />Contradiction, so 6r1c7.Philip Newmanhttps://www.blogger.com/profile/11212211922590240781noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-16433173184502895452020-10-21T20:36:34.387+01:002020-10-21T20:36:34.387+01:00+------------------------+-----------------------+...+------------------------+-----------------------+--------------------+<br />| 7 4(56) 1 | (28) 3 (48) | -8(6) 245 9 |<br />| 245 8 249 | 7 (1249) 6 | 2345 12345 123 |<br />| 246 469 3 | 1289 5 1489 | 7 124 68 |<br />+------------------------+-----------------------+--------------------+<br />| 3568 367(15) 678 | 4 68(1) 2 | 36 9 367 |<br />| 23468 3469 24689 | 3689 7 389 | 1 234 5 |<br />| 2346 134679 24679 | 1369 (169) 5 | 2346 8 2367 |<br />+------------------------+-----------------------+--------------------+<br />| 1 347(6) 47(68) | 5 2489(6) 34789 | 2389 237 238 |<br />| 36(8) 2 67(8) | 13689 (1689) 13789 | 3589 1357 4 |<br />| 9 347 5 | 1238 1248 13478 | 238 6 1238 |<br />+------------------------+-----------------------+--------------------+<br /><br /><br /><br /><br />challenge accepted: however i wont let that 1 digit easily solve it. <br />strmckrhttp://forum.enjoysudoku.com/member2491.htmlnoreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-66524565669983036262020-10-21T20:32:11.900+01:002020-10-21T20:32:11.900+01:00Phillip, thank you so much for your patient explan...Phillip, thank you so much for your patient explanations, you are making all this stuff in the terms of these old techniques much easier for a layman to understand!<br /><br />I expect my next blog pop-culture reference is going to have to be Butch Cassidy. Which I guess makes Strmckr the sundance kid...Tom Collyerhttps://www.blogger.com/profile/03663579809785456634noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-2161288650889392672020-10-21T20:23:18.597+01:002020-10-21T20:23:18.597+01:00In fact, if you go back to my first post on the su...In fact, if you go back to my first post on the subject where I expressed some doubts about the detail of the sk loop - phistomefel correspondence, which were precisely because somehow we had boxes involved on the one side but not the other. That’s another itch which will probably scratched if I ever get around to making myself comfortable with the notations. Tom Collyerhttps://www.blogger.com/profile/03663579809785456634noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-8591330478430873592020-10-21T20:17:57.844+01:002020-10-21T20:17:57.844+01:00Sam, I think this is exactly how I felt like it mi...Sam, I think this is exactly how I felt like it might be possible to explain things without ever really getting there myself. Absolutely fabulous stuff, and I was kind of hoping might come out of some of the discussion and a few weeks thought!<br /><br />I think it gives an alternative view of phistomefel as well, the one I was trying to get my around when I first saw it. That is very satisfying as it scratches another itch I’d forgotten I even had!<br /><br />Ie: r1289 and c1288<br /><br />Is the same thing as<br /><br />B1379, r37, c37Tom Collyerhttps://www.blogger.com/profile/03663579809785456634noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-7337345515340370722020-10-21T18:04:26.351+01:002020-10-21T18:04:26.351+01:00Ok, here's a challenge puzzle derived from Sam...Ok, here's a challenge puzzle derived from Sam's which makes use of the above forumla as an inequality:<br /><br />7 . 1 . 3 . . . 9 <br />. 8 . 7 . 6 . . . <br />. . 3 . 5 . 7 . . <br />. . . 4 . 2 . 9 . <br />. . . . 7 . 1 . 5 <br />. . . . . 5 . 8 . <br />1 . . 5 . . . . . <br />. 2 . . . . . . 4 <br />9 . 5 . . . . 6 . <br /><br />The goal is not to solve (though I'm sure strmckr will probably post a solution quickly!) - the goal is to place one digit. (Andrew's solver cannot place a digit, but if given that digit is able to solve the puzzle, though it's still Extreme.)Philip Newmanhttps://www.blogger.com/profile/11212211922590240781noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-39343709193580040152020-10-21T17:41:58.287+01:002020-10-21T17:41:58.287+01:00Tom, this is a very insightful comment. I had some...Tom, this is a very insightful comment. I had some thinking about it, and you're exactly right, this deduction *can* be phrased in terms of partitions / "Fred theory", albeit generalised to include boxes (of course!).<br /><br />First, it's helpful to take a slightly different view of Fred's original insight. Fred's exposition divided the grid into four sets of cells, which he labelled Red, Green, Yellow and Blue. Well, really there's no difference between the Red/Green cells and the Yellow/Blue cells - if we take a different Fred-style partition using the same rows but the complementary columns (or vice-versa), then the Yellow/Blue cells take on the role of the Red/Green cells in the new partition.<br /><br />Now we'll prove Fred's conclusion about the Red and Green sets using a different argument. It's all about adding and subtracting sets of cells that we know contain a complete 1-9, and so whatever we're left with must be a whole number of 1-9 sets. Add all of the rows that define the Fred-style partition, and subtract away the *complementary* columns from the partition. Some cells cancel out, some cells are included once, and some cells are included -1 (yes, minus one) times, and the result must be a whole number of 1-9 sets.<br /><br />For example, if we *add* rows 1,2,8 and 9 and *subtract* columns 3,4,5,6 and 7, then we get the following picture, where the Green cells are included once each, and the Red cells are included -1 times each. The conclusion is exactly what Fred originally deduced - that the digits in the Red set must be the same as those in the Green set, plus a copy of 1-9: https://ibb.co/4NBFDv8<br /><br />But with this new proof, it's clear how we can generalise Fred's "partitions" to include boxes too. The word partitions, though, is probably no longer appropriate once we've done away with Yellow/Blue as we are no longer insisting that the whole grid is covered. Nonetheless, here's how to deal with the Botsu Baku puzzle using this methodology.<br /><br />*Add* Column 1 and Row 9, and *subtract* Box 1, Box 7 and Box 9. Some cells cancel, some are included once (or -1 times), and r9c1 is even included (2 - 1) times. The picture we're left with is as follows - and this time, I've put the givens in the grid: https://ibb.co/MZtXsMs<br /><br />The deduction is obvious. Red is equal to Green plus a 1-9 set. But Red already contains two each of 1 and 2, whereas Green has none. Therefore Green must contain a 1 and a 2, and Red cannot have any additional 1s or 2s.Sam Cappleman-Lyneshttps://www.blogger.com/profile/09809527140050302213noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-72824053082657094072020-10-21T17:31:36.194+01:002020-10-21T17:31:36.194+01:00And to emphasize: None of this diminishes the beau...And to emphasize: None of this diminishes the beauty of Tom's pattern! It being a special case of Phistomefel makes it no less useful or elegant. :)Philip Newmanhttps://www.blogger.com/profile/11212211922590240781noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-61818782735175358192020-10-21T17:28:15.624+01:002020-10-21T17:28:15.624+01:00Small correction.
"Note, the Schrodinger Cel...Small correction.<br /><br />"Note, the Schrodinger Cell technique does *not* require the 12 in box 9. Box 9 is required to make it bidirectional, though - 12 must be locked out of r9c789, though that could be done in other ways."<br /><br />Box 9 is required whether it is bidirectional or not (which should mean that all SC setups are bidirectional). What isn't required is that 1 and 2 appear as givens in r78c789.<br /><br />It's not as clear, but Phistomefel can still get to the result either way:<br /><br />Say 1 and 2 do not appear in box 9, and instead r9c789 are other given digits (which will probably result in some other basic deduction, but anyway). We have one of two cases:<br /><br />a. The digit 1 is in r78c78, in which case the logic is identical to above.<br />b. The digit 1 is in r78c9. In this case, r456c9 are ruled out. Now, Phistomefel only guarantees a single 1 in our remaining target cells (r19c456+r456c1); however, if 1 appears in r1c456, that again forces a 1 into r23c78, so we need a second 1 and it must appear in r9c456+r456c1.<br /><br />The same logic holds for 2, so again the result is the same as Tom's approach.Philip Newmanhttps://www.blogger.com/profile/11212211922590240781noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-77297541391699470362020-10-21T16:52:50.045+01:002020-10-21T16:52:50.045+01:00The 247 (digit) muti-fish crushs alot on the puzzl...The 247 (digit) muti-fish crushs alot on the puzzle. <br /><br />Nxn+k fish cover set mathematics used by algorthiums is what is being described here.<br />n base sectors with n cover sectors(+exfra if not all bases are covered)<br />where by n base cells are covered by n cover sectors exactly n times. Thus all cells in the cover not in the base are excluded. See the ultimate fish guide or obiwans mathmatics on the players forum for refrence or ask and ill link it. Strmckrhttp://forum.enjoysudoku.com/member2491.htmlnoreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-41139796582790021752020-10-21T16:48:02.404+01:002020-10-21T16:48:02.404+01:00I suspected this might also be related to the SK L...I suspected this might also be related to the SK Loop, and a forum discussion concurred - this is basically a baby SK Loop, on two digits instead of four.<br /><br />Apparently, Steve K's original name for what came to be called an SK Loop was a "Hidden Pair Loop", which seems particularly appropriate here.Philip Newmanhttps://www.blogger.com/profile/11212211922590240781noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-80680767938273120382020-10-21T16:37:18.587+01:002020-10-21T16:37:18.587+01:00I get Patrick occasionally for some reason, Paul i...I get Patrick occasionally for some reason, Paul is a new one. :)Philip Newmanhttps://www.blogger.com/profile/11212211922590240781noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-59677098701511483382020-10-21T16:36:04.363+01:002020-10-21T16:36:04.363+01:00(Partitioning three ways in each case - as in the ...(Partitioning three ways in each case - as in the Tatooine Sunset puzzle - makes Fred's much messier, which is why the wonkyfish visual really shines there.)Philip Newmanhttps://www.blogger.com/profile/11212211922590240781noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-74174949929035374102020-10-21T16:34:35.951+01:002020-10-21T16:34:35.951+01:00Sorry phillp Sorry phillp Strmckrnoreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-44706463451709727732020-10-21T16:34:17.245+01:002020-10-21T16:34:17.245+01:00Trevor, Fred's intersection theory is basicall...Trevor, Fred's intersection theory is basically what you are talking about with two sets being congruent (mod sets of 1-9) and dissonant.<br /><br />This should work as a general formula?<br /><br />Given the following partitions:<br /><br />Split the 9 rows into sets of size R1 and R2.<br />Split the 9 columns into sets of size C1 and C2.<br />Split the 9 digits into sets of size D1 and D2.<br /><br />Say R1xC1 contains X1 digits from R1. Likewise, R2xC2 contains X2 digits from R2. Then, if X1+X2 = R1*C1*D1/9 + R2*C2*D2/9, the remaining digits of R1xC1 must be from D2, and the remaining digits of R2xC2 must be from D1. (Or more generally: that formula must hold in the solution to the puzzle; if X1+X2 is already equal to rhs in the current state of the puzzle, we can conclude what the remaining digits in the intersections must be.)<br /><br />So for the specific case of odd digits in 25 red squares and even digits in 16 blue squares, we need 21 total digits to conclude that the remainder must be of the opposite parity.Philip Newmanhttps://www.blogger.com/profile/11212211922590240781noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-29829745762599545212020-10-21T16:33:50.303+01:002020-10-21T16:33:50.303+01:00This comment has been removed by the author.Philip Newmanhttps://www.blogger.com/profile/11212211922590240781noreply@blogger.com