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I am currently stuck at this point in the below Sudoku puzzle. I have tried to apply various tricks such as XY wing, swordfish etc. but the preconditions are not fulfilled for any number.

What is my next move and why?

Thanks!

enter image description here

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Note the chain of five marked (yellow) cells the image below.

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Start with either of the marked cells in the top row, and move in a loop through the chain of marked cells back around to the other marked cell in the top row (that is, start at the top left cell going counter-clockwise, or the top right cell going clockwise). Choose a value in your starting cell that forces only one choice for the next cell in the chain. Step through each cell in the chain setting the forced values, and when you reach the opposite end (back in the top row), there will be a contradiction. This rules out the value you started with, and lets you determine the values of both cells in the top row. For example, if starting with the top right yellow cell, choose 9. This forces 4 - 1 - 5 - 9 for the values of the other cells in the chain, which produces a contradiction in the top row, of both cells needing to be 9. So you know that the top right marked cell cannot be 9. It must be 5.

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  • $\begingroup$ Dear Lanny, thanks for the answer. How did you come up with the chain of five marked yellows? Is it like I pick a pair and then pick all the cells in which either of the two digits are occurring in a loop? $\endgroup$ – Varun Gupta May 17 at 5:35
  • $\begingroup$ Yes, exactly. But it doesn't work anywhere in the grid, only in certain places. It takes trial and error, and some guesswork, too, to identify places where you might be able to find a short chain. It's possible to do the same thing with longer chains, too, but it's much harder to keep everything straight in your head when dealing with longer chains. $\endgroup$ – Lanny Strack May 17 at 10:32
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If r4c4 is a $1$, then r2c4 is a $5$. Also r4c8 is $9$, therefore r3c8 is a $5$. But this cancels both $5$'s in the top-left block. So r4c4 is a $4$.

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  • $\begingroup$ Dear JMP, I understood your answer and agree with the logic. But is it like completely hit and trial to pick a cell to start from. For example I tried applying this technique to find other similar pairs and wasn't reaching to a solution that quickly. Thanks. $\endgroup$ – Varun Gupta May 17 at 6:11

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