Very good! Non consecutive is such a weird constraint. Sometimes you can look at a cell and find it randomly only has one option. Rest assured, you hit every key deduction along the way.

Easy to Puzzle solving but reading comments it's very hard. 1. 349 XY wing gives 28,28 then puzzle solved. 2. 3s X wing , start with R7C2,R1C2,R2C8,R2C9, R5C9, R5C6, Eliminating 3 in R8C6. 3. The Medusa method... *See the row 7, (39), (238),( 29) connecting with the 3,2 and 9, conjugate pairs. *Col 9, (39) and (29) is 9s conjugate. *Col 6 conjugate with 3 4 and 9. We start with R8C3 (39) 3 is off 9 is on, R8C6 (238) 3 is on, 2 is off, R8C9 (29) 2 is on 9 is off, R2C9 (39) 9 is on 3 is off, R5C9 (23) 3 is on. Now come to col 9.., If R2C9 (39) 9 is on means R2C6 (49) 9 is off and 4 is on. If R5C9 (23) 3is on means R5C6 (34) 3 is off 4 is on. In col 7 two 4s is on, so any other 4 will be spotte so, 4 spotted in R1C6, puzzle solved.

Hi again! Used my only single # in all the BVCs strategy again. It was 8 in this puzzle. Using colouring I built out several Houses on 8s. Found r2c3 can not be a 2. Then, realized that r2c7 must be 28 BVC with known colours. IE, that cell can not be a 4. Found multiple eliminations from there on to solve. Regards. Keep up the good work!

Excellent puzzle. Had a lot of fun solving it. Solved it in 29:04. I used the following logic: The first step I noticed from box 9: 2 is either in row 8 or in r7c8. These put a 4 to either r8c1 or r6c8, which put a 2 to either r2c1 or r6c2, thus r1c2, which sees both, cannot be 2, leaving only 34 there. Now earlier I noticed a swordfish on 3s in columns 2, 5, and 8, and in each of these columns there are exactly two possible locations for 3. So in the final solution 3 is either in cells r1c2, r7c5, and r4c8, or 3 is in cells r7c2, r4c5, and r1c8. This means that exactly one of r1c2 and r4c5 will be 3, and since at that point both can only be a 34 pair, exactly one of them will be 4, which eliminates 4 from r4c2 and r1c5. This creates a 29 pair in cells r4c2 and r4c4, eliminating 2 from r4c8. Now both r1c2 and r4c8 can only be 3 or 4, and since in row 6 we have a 24 pair in r6c28, we cannot have r1c2 and r4c8 be both 4, so r1c8 cannot be 3. This places 9 to r1c8, after which the puzzle solves easily.

I went back to this great puzzle to try to solve it using BVC/colouring approach I outlined in a comment to your 19 July puzzle. It worked! 159 appear multiple times in BVCs. There is 38 BVC but I ignored that because is seems isolated and not related to anything in the grand scheme of things. 2 & 7 appear once and are attached to the 159s. Couldn't find a House that "sees" the 7 at one end of a strong link. Found one for 2 though! Box 3 has a strong link for 2s at r1c8 & r3c8 and one sees the 2 in r1c4. So Colour the 2s in Box 3, Green in r1, Blue in r3. Then the 5 in r1c4 is Green which makes the 1 in r1c7 also Green. But r3c8 is Blue so r3c8 can not be a 1! Puzzles solves easily from there. I thought that this approach was interesting and worth sharing it with you. Keep up the good work on your channel! Regards.

Great puzzle and solve. Solved it in 31:02 after eliminating some possibilities in a much more complicated manner.

Thanks for covering this puzzle of mine. I had already featured this combination of variants in 2022 Sudoku Grand Prix, round 4. It makes for very interesting geometrical deductions.

I used a simple chain to determine that R2C2 was always a 4, used the same claiming/ pointing logic as you did on 2s as well as a pointing sashimi x-wing, and then followed loops on BVCs with the 12479. It was tough! I was delighted you featured this puzzle, as I was intrigued to uncover what the intended solve path was when I first saw it on the discord. Thanks for your video 😊

A good challenging puzzle! Focusing on Bi-Value Cells, as mentioned at about 6:30, can give a lot of information. Collectively, the BVCs have multiple occurrences of 2,4,7&9s. Except r9c9 which has a 1. Further more, r9c9 sees a strong link on 1s in r1. Setting up colours with 1s on r1: lets say c1 is Blue and c9 is Green. If Green is the correct colour, then the 4 in r9c9 is Green which implies that the 2 in r9c1 is also Green. But, if Blue is the correct colour, r1c1 is a 1. Therefore r1c1 can not be a 2! Seems to solve from there. Trust that this is somewhat clear.

45 minutes, I solved it on my own....I think this was the toughest XV sudoku I have faced till now

Beautiful puzzle, lot of fun to solve it. Solved it in 25:05 using same ideas.

Beautiful set-up and 2 gorgeous deductions to break the solve wide open; I enjoyed both deductiona very much :)

Absolutely fantastic 👏

Nice to see you posting regular videos on puzzle solving.

Eliminating 2 is solution for puzzle. By bifurcation we can eliminate 2 from R1C1. R1C9 (179) spott 1 or 7 or 9 eliminated 2 from R1C1. R9C1 (24) spott 2 or 4 eliminated 2 from R1C1. R9C5 (124) spotte 1 or 2 or 4 eliminated 2 from R1C1. R9C9 (14) spotte 1 or 4 eliminated 2 from R1C1. These four cells targetting to eliminating 2 from R1C1. If 2 eliminate from R1C1 then puzzle solved.

I finished in 42 mins, without single mistake or guess . Did not use any of your method. very confusing. Did directly, Synder Notation, and simple logic.

R3C4 &R3C6 दोनों जगह से 2 को हटा देंगे। क्योंकि इन दोनों जगह पर 2 होने से R 1 में कहीं पर भी डिजिट 1 नहीं आयेगा... फिर इसके बाद तो यह पहेली आसानी से हल हो जायेगी।

Nice puzzle. Solved it in 32:57. While checking the possible location of each digit in the grid to see if any easy technique could be used, I was able to eliminate 7 from r1c9 (it forces 7 to r8c8 in box 9, to r7c1 in box 7, to r2c5 in box 2, to r6c4 in box 5, to r5c3 in box 4, no place for a 7 in box 6). This didn't help much, I could only eliminate 7 from r2c5. However, it left only 1 and 9 as candidates for r1c9. Since I noticed that 1 can only go to r1c1 or r1c9 in row 1, a 9 in r1c9 would force 1 into r1c1. On the other hand, if r1c9 is 1, that puts 4 to r9c9 and 2 in r9c1. This then eliminated 2 from r1c1, after which the puzzle solved easily.

beautiful sudoku! and explained it so well! thank you

This puzzle is so incredibly beautiful and just keeps on giving until the very end. Great solve - There were some quite challenging spots in there that you got immediately.

Thank you for this wonderful feature, Unshackling Sudokus & Puzzles! Hearing you speak in such appreciating words about our puzzle made my day. I also was astonished how differently the puzzle can be approached. In our testing, we tended to use colouring techniques. You did not use any of those and still solved it in such a smooth, effortless style. As a setter, it always is interesting to see what solvers find in the puzzle. You made use of a few deductions that I was not aware of (gaining the 26 pair early on, using the geometry of the Box 4/Box 7 zipper to limit positions for 8s and 9s). Thus, I myself learned a lot watching your solve path and I enjoyed it immensely. Thanks again for this amazing video solve! I am looking forward to your upcoming content and wish you all the best for the future. Kind regards, Nils/sujoyku

Excellent puzzle and great solve. Solved it in 30:49.

Wow, a wonderful construction and great deductions! New variant for me.

Thank Ashish for the feature. I myself did not realize while setting that the digits in r6c2 and r8c4 can never be the same and also that they cannot be 4. Also, I did not realize that the digits at r3c5 and r5c7 cannot both be 8. These two steps made the sudoku a bit easier (although the puzzle can still be solved without using these observations).

What a marvellous sudoku! I solved this puzzle after the contest and actually started with significant progress with the top right, but I errenously ruled out the possibility of double 8 = 16. Happily, the puzzle didn't break. You took everyone through the exact solve path by starting in the intended spot and from there, you can write many 678 triples and 78 pairs. I did spot the b1/b9 min-max logic but I had real trouble logically understanding the placment of 1 in r9c1 when I attempted to solve it on my twitch stream last night. Now, I absolutely understand after your perfect explanation. You can quickly resolve the puzzle by realising that there's a given 3 in c9, giving a 14 pair in r67c9 which quickly finishes the b3 arrows. As an aside, I also want to give you massive props once again for your simply stunning set-up for your inclusion sudoku. I've set quite a few quadruple/inclusion sudokus as it's 1 of my favourite variants and for me, your puzzle was absolutely the best inclusion sudoku I've ever seen and solved from here to Pluto; simply superb! :)))

Fantastic

Great!

Beautiful puzzle, quite difficult. Solved it in 84:33. The rules as written DO NOT say that a cage cannot be both a killer and a quadruple clue, and this is not necessary to solve the puzzle. There was a small error at 14:43: The quad 5679 adds to 27, not 26 (like in box 8), so if there is no 8 in the quad in box 5, it would need to be 4679, so the 5s do not give you a contradiction. Instead, the 7s do.

Absolutely majestic 👍

thank you,i learn a vary important skill in this video

FYI: Y-WING EXPLANATION STARTS @21:55

FYI: X-Y CHAIN EXPLANATION STARTS @14:44

Really great to see someone do the solve! Fantastic logic used throughout, really enjoyed the video! Thanks for the feature!

This was a head scratching to me I have started from top to bottom and really went smooth This was not hard Struggled with renban arrows aswell

Nice puzzle and solve. Solved it in 30:47.

Wow! I managed to complete although it took me very long. Now let me watch ur solve. Yeah, i noticed that u used contradiction method a couple of times to eliminate the unwanted candidate frm a col/ box which was nice . I didn't use any in my solve, nice learning for me. Well this was the second puzzle on clones which i tried , the first one being " Clone of the attacks" which i viewed for the first time on ' 'GAS ' Channel solved by Clover dated 6th july '24. Thanks for the tough version on clones variant! Thank you Richard stolk!

Nice video! Well done Ashish, you found all the steps I put into it, and even a few nicer ones. 😀

Awesome! New type of learning for me . Thanks!

Thank you! I appreciate you showing another of my puzzles!

Great! There is so much to learn. Thanks for a diff variant puzzle which I viewed for the first time. God bless!

What a great explanation to this fiendishly beautiful puzzle

14:33 XY wing 135 in box 4 and 7 solved the puzzle.

Done using coloring for remote pairs.

Puzzle solved in 5:03. The remote pairs does the trick.

18 min . Find no's and two sting kite of remote pair gives 3 puzzle solved. Solve with Skyscraper, turbot fish, and single chains start any where targetting only one cell very easy puzzle.Thanks.

Thankyou for solving another puzzle of mine, and a very tough one at that - I'm really glad you persisted with this one; it was a real joy to see those lightbulb moments! There are many narrow deductions here: the 16 pair, seeing that 1 is limited to r7c89 & that r7c9 doesn't work because r7c8 can ONLY be 1 or 9 (amazingly enough), the 2 in b2, and there's lots of action with 3s in r6/b4. I seem to remember there being more traction with b4 but it's been a long while since I test solved this puzzle. The Y-wing is a happy coincidence but as has been pointed out by Laszlo, 3 can only go in r8c46 (not possible in r8c1 as it would be adjacent to a 2) so you can eliminate 3 from the bottom of c5. The skyscraper was not intended at all. I like XV (-) a lot and as luck would have it, this video has come at the perfect time for a real hard practice puzzle before the WPF Sudoku GP Rd 6 where the highest pointer is an XV (-) sudoku which is well over 100 points. I will look into doing another XV (-) sudoku in the future with a few given X and V clues as there is lots of potential for nice deductions in these puzzles but hopefully it can be a bit friendlier with a few given clues.

Wow! Great puzzle and fantastic solve. Learnt the diff types of reasoning in XV pairs variant. Thanks a lot!

Beautiful puzzle, extremely hard. Solved it in 50:54. I used a lot more pencilmarking, basically entering all candidates in each cell eventually, slowly eliminating possibilities. This allowed me to notice some of the deductions much easier. For example, already at 13:35, r7c8 can only be 1 or 9, which eliminates 1 from r7c9, placing 1 to r7c8. In box 8 I was able to eliminate 2 from r9c4, because a 2 there would put 3 to r8c6, 7 to r8c4 and 8 to r9c5, putting 2 and 8 next to each other (this didn't help much, though). Then I found a skyscraper on 6 in columns 3 and 8 (6 can only go to r29c3 and r39c8), eliminating 6 from r2c79 and r3c1. This restricted 6 to r3c89 in box 3, removing 4 from those cells, so now 4 could only go to r3c2 in row 3. Then I was able to eliminate 3 from r6c2, because a 3 there would place 8 to r6c5, making r6c3 either 2 or 7, neither of which can be next to a 3. This then limited 3 to r4c12 and r5c1 in box, eliminating 2 from r4c1, leaving only r6c3 for 2 in the box. After this the puzzle collapsed. At 48:09 you can already limit 3 to row 8 in box 8, so the pretty xy-wing on 238 was not actually needed :) Pity.

Best one!

A very hard sudoku indeed. Thankyou author and thankyou Kishore for solving this logically and patiently. I too didn't expect a Y wing in this variant. Nice