Great puzzle and solve. Solved it in 33:48. Immediately after recognizing the 4567 quadruple in row 5 one can deduce that one of r5c346 must be 4, so at least one of r4c35 must be 3, placing 3 to r3c4.

Thank you for featuring my puzzle! I'm glad you enjoyed it. It's named Siphonophorae because the shape of the larger thermo reminded me of a siphonophore, which is a deep sea creature.

Beautiful puzzle. Solved it in 54:13. Nice coloring exercise.

That was an amazing solve Ashish! Thank you so much for this very nice feature! This puzzle was really fun to construct actually. A large part of the puzzle can be colored, using the geometry of the cages. Then I noticed indeed that I can disambiguate the puzzle using only the digits 1 to 9 once each. After that, I observed that I can swap the numbers in any way I like, so I could make sure that the break in is on the 1 and also that the central box is a magic square. That was a thematic choice for me, because Mjolnir is a magic hammer. It’s the hammer of Thor in Norse mythology. There’s also roping in the middle three rows and the middle three columns.

Puzzle solved in 8 minutes. Though its with pencil marks

मुझे तो R3C1& R3C8 में 16 का hidden pair मिल गया। क्योंकि R3C2 , R3C4, R3C6 & R3C9 में 2,4,5&9 का quartz आ रहा है।इसके बाद R1C8 में 2 का hidden single आ गया।

I spotted it was rotationally symmetrical so 5 goes in the middle and 1-9, 2-3, 4-7, 6-8 and 5-5 are all rotationally opposite. It was quite easy after that.

I spotted it was rotationally symmetrical so 5 goes in the middle and 1-9, 2-3, 4-7 and 6-8 are rotationally opposite. Makes it quite easy to solve once you spot it.

Excellent puzzle and great solve. Made a mistake near the end, which took me a while to find, so solved it in 37:47. The break-in is really beautiful. In box 1 there are only two places for 1, and each will force a 5 in boxes 3 or 7. Similarly, there are only two places for 9 in box 9, and each will force a 5 in boxes 3 or 7. Since boxes 3 and 7 contain only two 5s, these must be the 5s in those boxes, and we can eliminate 5 from the other cells in these boxes. This also means that the 5s must be the ones forced by the 1s and the 9s in boxes 1 and 9. So in box 3 either r3c9 is 9, or if it is 5, it must have been forced by the 9 in box 9, so r7c9 is 9. Therefore we can eliminate 9 from r2c9, placing 3 there. After this the puzzle collapses.

Greetings Chaotix here. Apologies for the late reply. Thanks for the feature and the live solve of the puzzle. The remote pairs is the intended way on cracking the puzzle. Coincidentally I wanted to feel the solver, on using remote pairs.

oooo, a new puzzle type! thank you so much for sharing!

I’ve been completing these puzzles for a few weeks, they are a lot of fun. Often Friday’s puzzles are a little more tricky.

Great video!

fun one. puzzle solved in 48 seconds

Too hard to be enjoyable.

It looks so easy when you do it! What a great Logic and Quick identification of patterns 😊

Excellent puzzle and an equally excellent video. Please stick to this formula of very good classic sudokus.

Excellent puzzle and nice solve. Solved it in 24:03 using the same methods.

Is Zatu ko aata hi nahi Sudoku solve karna … Fake trick bata raha he

Nice puzzle and solve. Solved it in 12:54. The break-in is a variant of the Phistomefel ring. It says that the cells in r5678c2345 contain exactly the same digits as the cells r123c16 and r49c16789 (to see this you can either permute the rows and columns of the Phistomefel ring, or color columns 2,3,4,5 blue, then color rows 4, 9 and boxes 1 and 2 red, then remove the cells with both colors; the remaining blue cells must contain exactly the same digits as the remaining red cells). This immediately gives that cells r58c25 must contain two 1s, a 2, and a 3. The interaction between r5c2 and r8c5 was really nice.

Amazing solve! You found the intended logic very quickly. And you're defnitely not the first person to get stuck on the simple simple afterwards (because I was the first person to get stuck on the simple step 🤪).

Excellent puzzle and great solve. I took a while to notice some of the logic, so took me 45:48 to solve.

Tq Nice sir ji

Excellent puzzle, quite difficult. Solved it in 42:30.

Oh my God!! What a superb video. I just couldnt understand how to solve this variant, so didnt even look at it at the SM. But you explained it so beautifully. Thanks very much for this. And extra special kudos to young Nityant. I understood the beauty of his construction, thanks to you. Super sudoku+explanation combo.

Wow, what a beautiful construction and the crucial deduction to blow the puzzle wide open was quite amazing! I also missed 1 of the 2 negative constraint deductions in c1; failing to spot these 2 deductions is what killed my progress in solving this amazing puzzle which up to the point I had reached was correct.

Thanks for the video! I enjoyed watching you solve

Very enjoyable video. Thanks for covering this. A very well made puzzle.

Absolutely majestic 👍

Nice puzzle and solve. Solved it in 23:17. I also found a finned X-wing on 5s in rows 3 and 7, which eliminated 5 from r9c2. Then I was able to eliminate 6 from r7c2 (gave a contradiction for box 1 after a relatively long chain), which placed 6 to r7c3 in box 7. After that the puzzle solved easily.

I did a -79 on r3c6 by uniqueness. Then a -5 on r9c2 by finned X-Wing. Basics there after. Please check.

DONE Used an AIC TYPE 2 . R6C1(45)--->R6C3(57)-->R4C3(67)-->R7C3(56)-->R9C2(45)-->R8C1(14) Starts with a strong link frm 4 to 5 in the cell R6C1 and ends with a strong link frm 4 to 1 in the terminating cell R8C1 .Therefore the start digit 4 can be eliminated frm the end cell R8C1(14) to give us a single 1. That's all! Let's watch ur solve now.

I used the same 28/ 89 logic, but then saw the second logic as remote pairs on 29 in boxes 2 and 6. Fantastic puzzle, very pleased to have cracked this very tricky puzzle without any hints first!

Whatever u explained abt 28 pairs in boxes 4 and 8 is clear to me , one of them has to be a 2 and the other an 8 . The formation of 18 pair in box 5 ???. Now talking abt 89 pairs in boxes 4 and 8. Here both of them can't be 8 , that's fine , but why both of them can't be 9? The logic related to 12 pair at R5C2 and R8C5 was enough to solve the grid completely.

Oh my God! Awesome deductions. I would never have thought of these. Too good.

Another nice classic sudoku constructed by your good self :) I love the 2 hidden singles 1 after the other; these deductions are the ones you need to train yourself to spot by carefully scanning the grid. I spotted the hidden 7 in r9c1 a few minutes before you, but then you instantly saw the subsequent hidden 3 in r9c4 where I didn't! I whole-heartedly agree that the reason sudokus, especially some classics, can take someone (quite) a bit longer to solve is the hidden singles/pairs, especially those involving box reduction. The more sudokus I solve like this one - only needing good careful scanning without advanced techniques - the better I become at solving sudokus, especially during a contest where time is of the essence.

What an unbelievable set-up! To use the logic that is seen at the start encompassing as many arrows as it did is truly amazing; no wonder it took Akash so long to handcraft this gem of a puzzle.

Beautiful construction with a great start! I especially love the sharp deductions involving firstly b9, and then soon after with b6/4 - it's these sort of deductions that LK puzzles have due to the key cells that see multiple cells along a particular LK diagonal.

Superb construction and lovely solve path! Thinking outside the box is very useful for puzzles like this one, so thankyou for giving this puzzle the public recognition and appreciation it most certainly needs :)

Excellent puzzle, quite difficult. Took me quite a while to notice the steps. Solved it in 72:31.

Hi Ashish! I came up with the same elimination as you but I used finned X-wing logic. Looking for a X-wings on 4s starting with c1. Almost have on in c5 except for the possibility r8c5 being a 4. Either way, though, r7c6 can't be a 4!

Bhai tu isko hindi mai bolega to jyada se jyada samjh payenge 😌

Nice puzzle and solve. Solved it in 13:27. The first logic was also a skyscraper on 4s in many different ways (rows 5 and 8 or 6 and 8, or columns 1 and 6 or 2 and 6).

An AIC(type1) on 4's in boxes 4,5,7 & 8 gets the 4 placed in box 7 ( ofcourse after placing the single 4 in box 9). Precisely R8C2-->R8C5-->R6C5-->R6C1 ( Strong/Weak/Strong) hence 4 gets deleted frm R7C1. Then used 29 remote pairs coloring to place 1 in box 6. DONE. Now it's time to watch ur solve.

Hey Ashish! Thanks for giving this a go! The kite-like chain followed by double skyscraper is the exact intention, well done. It is not well telegraphed, I'll admit, but you spotted it in the end. :) Oh, and nice thumbnail!

This is an excellent puzzle. Solved it in 67:57. This stumped me for quite some time. For some reason I always included the arrow in box 9, so I always had a freedom of 2. Then I figured out that all the freedom must come from box 9, and everywhere else the bounds must be tight. After this it became relatively easy. At 18:40 there are quite a few easy deductions one can make. 1 and 2 can be placed in box 7, 5 can be placed in r4c6 (because of a quintuple 56789 in column 6). In box 6 we can see that 2 cannot be on r56c9, because then r9c6 would have to be 6. Thus 2 is in column 7 in box 6, making r3c7 3.

Using sword fish helped me to some extent. Very good puzzle. Thanks

I hope I can reach this level of knowledge in sudoku.... It was really, really difficult

प्रारम्भ में 8 & 9 को हम empty cell तकनीक का प्रयोग करके हटा सकते है।

Amazing!