Hi rjamil,

rjamil wrote:Well, it's not to be perfect but,I see the latest

yzfwsf's program YZF_Sudoku finds BUG+3 move as per

RW's logic as follows:

BUG + 3 : => r1c7<>9; (but the move not produces STTE solution.

Yes, RW's logic works here. In fact, its elimination could be included (due to a subchain) in Cenoman's solution, though it makes no difference:

(9)b3p34 == (8)r1c7 - (8=39)r17c9 => -9

r1c7,r3c9; ste

Similarly, its counter Bivalue Universal Grave + 3 not detected.)

I don't know what you mean by that. What is a "counter BUG+3"?

Note: no further discussion needed as I have difficulty in identifying the BUG+candidates in this case, especially 8 @ r1c7.

I would rather think it specifically warrants further discussion! Or don't you want to learn?

You're right that the guardian cell r1c7 is the trickiest of the three. The reason is that it sees at least one other guardian cell in each of its houses (row, col, box) and shares digits with them. The process is easier if you start with the other two cells with three candidates, i.e. r1c9 and r2c7, and identify their guardians first. They're easier because they both have a house in which they're the sole guardian cells.

First an optional step. If you notice that all three (potential) guardian cells (r1c7, r1c9, r2c7) are in box 3, you can easily count what the guardian digits must be, without yet worrying about their exact cells. In box 3 there are exactly two 2s and two 3s, so neither of those digits can be a guardian. On the other hand, there are three 8s and four 9s. That tells us directly that if we have a BUG+ situation (which we don't yet know) the guardians must be two 9s and one 8, because that would leave us with two of every digit in box 3 (if the guardians were removed).

Next you should figure out which guardian digits go to which cell. There are three possible configurations with three cells and two digits, but only one is correct (if any) -- it's the one that leaves exactly two instances of each digit in each house if the guardians were removed. If you did the previous step, you could simply figure out where the 8 must be, and the 9s would necessarily go to the other two cells. However, that's actually a bit harder route here. Like I said, it's easier to start with r1c9 or r2c7. Let's start with the latter, because it's the simplest of them all.

If you look at r2c7 you should notice that it's the only guardian cell in its row, as the other two are bivalue cells. Next you should just check which digit exists three times in that row. The answer is 9, so it must be the guardian. If 9r2c7 were removed, there would be exactly two of each digit (2,8,9) in that row, as per BUG requirements. Thus 9r2c7 can be locked as a guardian. Note that it also removes the possibility that 8r2c7 could be a guardian.

Similarly, if you look at r1c9 you should notice that it's the only guardian cell in its column. Again, if you count the digit instances in that column you notice that 9 is the only one that exists three times, so it must be the guardian digit. Thus 9r1c9 can be locked as a guardian. It also removes the possibility that 8r1c9 is a guardian.

Whether you did the optional step or not, you now know that the third guardian must be 8r1c7 because r1c7 is the only bivalue cell left, and 8 is the only digit in box 3 that has three candidates left. There are also three 8s in every house r1c7 sees: r1,c7,b4. Thus 8r1c7 can be locked as a guardian.

The last step is to double-check that if those three guardians were removed you'd actually have a BUG+3 (all bivalue cells, every digit exactly twice in every house). It holds, so we have a BUG+3 with guardians 8r1c7, 9r1c9, 9r2c7.

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Edit. Fixed a typo (box 4 -> box 3).