Zen & the Art of the Guardian Sudoku is a chat and sudoku channel promoting sudoku learning and anything that's currently in the news or seems good to discuss. Come for the sudoku, stay for the chat. Come for the chat, stay for the sudoku. If you want to send me a (classic) sudoku to try, don't bother with RU-vid as they will remove any links so please try twitter: twitter.com/ZenSudoku or Facebook: @ZenArtOfGuardianSudoku
Fascinating in the latter stages of the puzzle to assist to a furious placement of candidates when all that was necessary to finish was to play sudoku for beginners ! Like the 3 that has been locked in r1 c8 for ages now, the 7 you finally got in r3c9 and, so on acting upon the consequences of each new easy placement you make.
An interesting way to start this puzzle was to enter a 7 in r1c3 and r9c2. Why? In block 1, we have a 9 and 4 in the middle column. In addition, there is a 6 pointing from the right in row 1, that will have to be located somwhere in col 1 or 2. So, an obvious 469 triple in rows 6,8 and 9 of column 3, and 7 completed the column in row 1. This puzzle was full of nice restrictions of this kind, particularly enjoyable.
The 9 has been restricted to r9c1 for a long time, ever since the 2 was placed in r2c1. Once you have the 139 triple in the lower row of block 7, the puzzle becomes quite easy.
Nice puzzle. A little bit different trick than we are used to with the NYT. In the situation you have when you get stuck around minute 15, what is the most interesting restrictionÉ It actually stands in the not too inviting column 8 where you have only one given digit. One thing Zen has already established is that there is a 12 bi-value cell in row 8. Where else can there be a 1 and a 2 in column 8? Well the only other place for a 1 is in row 5, and, for the 2, in row 4. Next question: what else can there be in rows 4 and 5? Pretty clear there must be an 8 in either of those spots. So, broken triple 128 of which r4c8 is a 28 and r5c8 is a 18. Eureka ! We already know there is a 18 in r5c3 ! Hence no other 8s possible in row 5, crucially, no 8 possible in r5c4. So now in that central vertical chute, the 8s are confined to either columns 5 or 6 in blocks 5 and 8, and theredore, there must be an 8 on the left side of block 2, where we already know we have a 3 and 6. So 68 pair in rows 1 and 3, and the 3 in row 2, confirming r3c3 must be a 4. Puzzle done. Thank you NYT for giving us something different.
Using a technique called a Unique Rectangle (UR). If r2c7 contained either a 6 or 7, the puzzle would have more than one possible solution (not allowed). Therefore, the only number possible digit there was 3. This old video from this channel might help: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-MD0ntsNMJj4.html
Fun puzzle. Column 6 can be filled out in the first few seconds. Series of digits easy to fill from the newly created restrictions, and then quickly row 4 becomes occupied by 47 and 39 pairs with the 6 now limited to r4c3, r6c2 and r9c9, and the rest is basic easy sudoku.
Almost from the first few moments of the puzzle, rows 4 and 9 of column 6 were occupied by a 28 pair, so r5c6 was an easy 7, and finishing the game from there was without difficulty.
Minute 17: rather plain in row 2 that 4 can only be in column 4, and therefore 6 in col 3 and 2 in col 8 AND of course column 1 is a 34 in block 1 and a 2 in row 7. Then the last straw for the puzzle was that, in column 7, row 1 was a 349 and row 3 was a 34. A 34 pair was already established in the same rows 1 and 3 in column 1. So, potential UR avoided by placing the 9 in r1c7. Adieu puzzle.
Strange. The most difficult digit to spot was the 9 in r3c2. Which you did. Then the rest is barely above sudoku for advanced beginners. Of course a 9 is now in r9c7, r7c9 etc. 12 minutes notation free.
@@johnmiller5629 I never had to see a 25 pair in column 9.. Going back to Zen's position at minute 16 after finding the 9, as Zen spotted, we know the 9 is restricted to row 8 in block 7. At that time, it was well establiched, and Zen knew, that column 8 was 79 in the upper rows, and a 56 pair in rows 4 and 9. With this, the only available spot for a 9 was in r9c7, r9c8 being already known to be a 56 bi-value cell. From there, you can place all the 9s, the 3s, the 8s, the 4s, etc. by just tracking them in a quick merry-go-round through the grid.
@@johnmiller5629 In the position Zen had at 16 minutes after finding the 9 in block 1, the 9 was now restricted to row 8 in block 7 as Zen saw. At that time, column 8 was already known to be 79 in rows 1 and 2 and 56 in rows 4 and 9. So in block 9, the only place for a 9 was in r9c7. With this, the next 9 as already established is in r7c9, and everything unfolds eaily with 3 in 77 (r7c7), and in 93 and 11, the 8s now in 29 and 33, the 4s in 31 and 43, the 1 in 17 and 22, 7 in 41, 2 in 12 and 51, etc. With such a sequence, we always have an easy to find locked-in digit available and no necessity of looking for pairs.
For me, the key at the 13:18 mark was the unique rectangle of 29 pairs in Blocks 7 and 8 which gave me 29 pairs at r7c1 and r9c3 in Block 7. That then made the placement of the 3 at r8c3. Fairly easy after that.
I found it tough, spent 38 minutes on it. The key thing I discovered stemmed from the 3 & 8 pair in C4 and the 8 placement. If R3C4 is 8, R1C9 becomes 8 and R9C9 is 5. If R9C4 is 8, then again R9C9 is 5, so 5 is a write in there regardless of where the 8 is in C4.
Just before you went into adding all the candidates. You can see a Phistomefel Ring on 3, forcing 1:9 to be 3 . The puzzle unwinds from there without the need for all the candidates being filled in.
A few months ago, I watched a video on the that technique on CTC and I remember thinking it was pretty neat. However, it's not something I've ever used and subsequently, I've forgotten exactly what it entails. In fact, playing these newspaper games, I've forgotten a lot of the techniques I used to use playing the harder level games, simply because of lack of practice. Thanks for the comment.
I know the said technique, but wouldn't have thought to apply it when having 9 of the 16 numbers in the ring itself, and 7 out of 16 in the corner 2 x 2 cells. I can't say how by that means I'd be able to eliminate 1, 5 or 8 from the cell.
It’s not that complex. The rings says there must be a 3 in one of the four corner boxes and must be in the top right 4 boxes of block 3. This leaves only one cell 1:9 open for the 3 block 1,7 and 9 can not meet the criteria for the 3
But then R2C2 is a 3 and inside one of the 2 x 2s. For R1C9 to be a 3, there would need to be two 3s in the ring - having solved the puzzle I know there are two. As I said i a comment, positioning the 8 in either position in C4 produces a 5 in R9C9, and that set me on a solution path.
Around 11th minute re UR of 29s: Zen, it was so obvious if 29 was excluded from cell r9c1, then there was a 29 in r9c3 ! Just look. You know in that block you cannot have a 2 or 9 in the middle row.. For those who like to play with candidates, the first lesson is that the use of candidates is meant to accelerate the findings of actual solutions, not to entering a maze of candidates in cells where they are actually excluded. One example here is that when you find out in block 9 that the 7s are in column 7, the next move is not to contunue entering false candidates, but rather to act on the meaning of the newly found location of those 7s and immediately enter the locked-in 7 in r1c8. From there, you continue movin towards true solutions because your newly acquired knowledge is that you now have a 67 pair in r3c1 and r3c2, and can place actual digits thanks to this newly found restriction, such as the 2 in r3c5, etc. Likewise, entering 2 and 9 as candidates in r9c1, knowing that these candidates are excluded from that cell, seems srange. Fun puzzle, a little bit more difficult for me than the recent editions, but quite enjoyable to do notation free. Finding that naked 4 in block 6 was most productive.
The power of the 17 combos... It takes a minute or so to get to the point where we are in the video at minute 8. At that point, we know we have in block 4 a 456 triple in column 3 (with cell r5c3 being a 45. Then we have the greatly positionned 17 in block 6 forcing a 17 pair along row 5 in col 2 and 6 (and also, incidentally, along column 9 in rows 2 and 8) Now in row 5, in columns 7 and 8, we have a 23 (due to 2 and 3 both already intercepting from above in column 9). Therefore there only remains 45 as possibility to complete row 5 in column 9. A 4 is already in column 9, so now, 5 in r5c9 and 4 in r5c3. The rest is very basic sudoku, at best medium level difficulty. 11m25s notation free.
I guess this is a puzzle that particuloarly suits my style of playing. I never hesitated before finding the next restrictions. So playing at a slow pace as usual: 5m47s One example of what I was saying comes up in the situation you have when Zen gets blind sided in the 8th minute. I look at the most congested area of the puzzle, block 3,(the geometry of the area provides a 100% chance of getting at least 1 digit) and 5 seconds later, I have all 5 remaining 2s placed in the puzzle just doing a merry go round dance starting with the 2 in r3c6, then down in block 9, left in block 7, up in blocks 4 and 1.
I can't crow too much about my attempts this time as I went wrong on my first one. I got it out the second time though, and found limited rather than full notation worked. Block 9 broke up into 6 & 8 and 3 & 9 pairs, with 7 placed, and that can be seen at the 15 minute mark on the video.
Another super easy, medium level, puzzle with loads of digits clearly restricted as we progress through it. Like at the time you start pencil marks, block 9 with the 19 completing the left side and the 34 squeezed in the remaining cells of row 9, which quickly places a series of 3s and then all the other digits at a sprint pace.
