tag:blogger.com,1999:blog-2031071943178978556.post1985717234721409786..comments2023-12-03T22:29:11.592+00:00Comments on detuned radio: Portrait of a flying fishUnknownnoreply@blogger.comBlogger2125tag:blogger.com,1999:blog-2031071943178978556.post-24669400976260743222021-07-28T16:56:41.084+01:002021-07-28T16:56:41.084+01:00Hi Unknown, no need to apologise for your English,...Hi Unknown, no need to apologise for your English, I can understand everything you wrote well. Thanks for your explanation of the technique from yet another perspective and another intersection - I think it's very helpful and is something I missed!<br /><br />On Fred's theorem - yes this is exactly what I was trying to get at. I got into the habit of calling it intersection theory, and I think there's a comment on this blog from Sam CL explaining the nicer way of thinking about it. Others have also called it Set Equivalence Theory ("SET").<br /><br />My understanding of this puzzle is that it was specifically designed to show-off the exocet technique; what I was trying to explore is the extent to which different solving techniques can be interpreted through the lens of Fred/intersection theory/SETTom Collyerhttps://www.blogger.com/profile/03663579809785456634noreply@blogger.comtag:blogger.com,1999:blog-2031071943178978556.post-29543321718800246902021-07-26T15:09:30.520+01:002021-07-26T15:09:30.520+01:00Hi, Tom. Great observation
Just a few comments:
1....Hi, Tom. Great observation<br />Just a few comments:<br />1. This could be explained through the Fred's theorem. The red cells (intersections of rows 2, 4, 6 & 8 and columns 3, 5 & 7) have the same contents as the blue cells (could be defined as the cells not included in any of rows 2, 4, 6 8 or columns 3, 5, 7) minus 2 full sets 1-9.<br /><br />2. Your final conclusion should be even stronger: none of the other blue cells may contain ANY of 2, 3 or 4 (regardless of what's in the double-weighted blue cells). This immediately solves the puzzle (no strategies other than the basic ones are then required)<br />3. This sudoku can be easily solved by using MSLS on 2,3 & 4 in rows 3, 5 & 7: <br />- 3 must be either in r1c3 or r1c5 in raw 1<br />- 4 must be either in r3c5 or r3c7 in raw 3<br />- r4c7 and r6c3 must both contain 2, 3 or 4<br />- r7c3, r7c5 & r7c7 are a 2-3-4 triple<br />- 2 in raw 9 must be either in r9c3, r9c5 or r9c7<br />Then r4c9 is a 6, r1c7 & r3c7 are a 6-8 pair, the only place for 4 in raw 3 is r3c5, etc. The rest is trivial<br />And sorry for my English - I am not an English speaker Anonymoushttps://www.blogger.com/profile/00531176406066043364noreply@blogger.com