813554 wrote:I've stuck here for a while and then I saw 4s can help me. In box 1, r2c2 and r3c1 could be 4 so first, I assumed r2c2 is 4. In that case, r3c9 would be 4 aswell in box 3. Then I assumed r3c1 is 4. In that case, r1c9 would be 4. So, there was no way that r1c8 could be 4 in box 3. I've tried that and it was right, I can definitely eliminate 4 from r1c8 and probably solve this but obviously trial and error is not my favorite method.
I wouldn't call your original method trial and error. Trial and error would be if you tried to place 4 in r8c1 and then found a contradiction (i.e. "error") which would tell you that r8c1 can't be 4. Your method works like our chains in that it proves r8c1<>4 without ever assuming that value in that cell. It's basically a verity type forcing chain (or actually a forcing net, because it seems to use branching) which does exactly what an AIC does. Leren's 2-String Kite is an AIC, just like my longer chain. A non-branching forcing chain can be easily converted into an AIC as well, so they're basically equivalent techniques (just different points of view).
Your logic could be expressed as a forcing net like this:
(4)r2c2 - r3c1 = (4)r3c9
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(4*)r3c1 - r2c2 = r8c2 - r9c3 = r9c8 - (r1c8|*r3c9) = (4)r1c9
=> -4 r1c8, r8c9
It proves that no matter which 4 in box 1 is true (and one must be), r1c8 (or r8c9) can't be 4 because the chains force either r3c9 or r1c9 to be 4 (and they both see the victims). It's perfectly valid logic, and it's not t&e or "assumptive" any more than an AIC is (even though some might say otherwise). Its only problem is that it's a bit too complicated for this situation, especially because the second chain uses branching. There's no reason to prove (4)r3c1 -> (4)r1c9 because (4)r3c1 -> (4)r9c8 does the trick just as well more simply:
(4)r2c2 - r3c1 = (4)r3c9
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(4)r3c1 - r2c2 = r8c2 - r9c3 = (4)r9c8
=> -4 r1c8, r8c9
Again, both end points see the victims. That forcing chain is equivalent to the X-Chain (a single-digit AIC) I presented in the previous post. To turn it into an AIC you can just "pull" one of the end points to straighten it into a single chain:
(4)r3c9 = r3c1 - (4)r2c2 = (4)r3c1 - r2c2 = r8c2 - r9c3 = (4)r9c8
Once the duplicate nodes are removed we have my X-Chain:
(4)r3c9 = r3c1 - r2c2 = r8c2 - r9c3 = (4)r9c8 => -4 r1c8, r8c9
Similarly Leren's 2-String Kite could be expressed as a forcing chain:
(4)r2c2 - r3c1 = (4)r3c9
||
(4)r3c1 - r2c2 = (4)r8c2
=> -4 r8c9
But the AIC format is preferable:
(4)r3c9 = r3c1 - r2c2 = (4)r8c2 => -4 r8c9
I just wanted to say that you were on the right track, so don't beat yourself too hard!