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[Mogulmeister]

Hopefully another absorbing puzzle:

[Mogulmeister]

After basics

[immpy]

Depending on how one travels, there are 2 String Kites, a Finned X-Wing, possibly a W-Wing here. Nice puzzle.

cheers...immp

[Mogulmeister]

A rather lovely solution if you like doing the spatial stuff.

ALS XZ

A = {1,2,4,6,7,8} (Pink)
B = {1,2,6,8} (Green)

Restricted Common x = 6 in row 6
Common Candidate z = 1 (Blue) so r8c6<>1 because it can see all the 1's in both A and B.

[Mogulmeister]

Finding two ALS that works can be time consuming but it gets better with practice. The strong link between the two <18> act as a focus as you try and construct two ALS that share enough candidates. Remember any ALS you construct must have its member cells in the same row, column or box.

You then look for a restricted common which is a number shared by both sets but will only end up in one or the other. This put me off for a while because usually a restricted common (in this case 6) only appears once in each set. I then realised that just because there are two sixes in set A at r6c45 the basic idea is unchanged - the 6 is either in A or B.

Constructing the ALS is always interesting and you have to remember the rules. If there are N squares then there must be N+1 candidates in those squares. So A has 5 squares and so 6 candidates. Set B has 3 squares and so 4 candidates.

I had got to the stage where I had 4 squares in A but needed a 5th and r6c4 was left but the extra 6 threw me for a while.

[Mogulmeister]

If you prefer chains, there is an ANP (almost naked pair) solve too giving the same result r8c6 <>1.

[Mogulmeister]

The ANP solve is often worth looking at. In this case, if you look again at the green cells r2c9 and r6c9. Now these two cells will either make a <28> naked pair (NP) or the 6 in r6c9 will prevent that.

So let's play it out. Both scenarios.


(1) ANP (6=28)r26c9-(8=1)r8c9-(1)r8c6 so r8c6<>1 (If NP True)

The above creates a naked pair that knocks out the 8 in r8c9 making it 1. This then knocks out the 1 in r8c6.

(2)ANP (28=6)r26c9-(6=478)r6c4|r46c5-(8=12)r45c6-(1)r8c6; r8c6<>1(IF NP False)

The above 6 in r6c9 knocks out the 6's in r6c45. This allows a triple to form <478> in r6c4, r46c5.

This kills the 8's in r45c6 which creates another NP <12> which ALSO knocks out the 1 in r8c6.

[Mogulmeister]

I showed the two sides of the ANP in this way to mirror the analysis of something that takes seconds by inspection but far longer to write down !

For more completeness there is, as always, a chain that covers it all.

ANP (28=6)r26c9-(6=478)r6c4|r46c5-(8=12)r45c6-(1=8)r8c6-(8)r8c9; r8c9<>8 and r8c6<>1

[Mogulmeister]

Which leads to the final observation. You can take advantage of the links above to prove a contradiction on 1.

Start by plugging 1 into r8c6. We can use our earlier chain but in reverse.

This creates a loop which insists on eliminating that same 1 in r8c6. This is a contradiction and the 1 can be removed.

(8=1)r8c6-(1=8)r8c9-(28=6)r26c9-(6=478)r6c4|r46c5-(8=12)r45c6-(1=8)r8c6; r8c6<> 1