... by: jem

for Gnomeface: there's a popular misconception about Sudoku puzzle solving regarding what constitues "logic". The fact of the matter is that "trial and error" is iron-clad logic when used to eliminate candidates - the trial of a particular candidate that leads to an error (i.e., an unsolvable puzzle) is a 100% logical proof that the particular candidate can be eliminated.

... by: Andrew Stuart

Miss numbered number 42 - there were two 43s. Sorry about that! Fixed now and you can go back to 42

... by: X.

I manually solved this using an extension of 3D Medusa after a few simple initial steps. Spent a couple of hours on it.

... by: Gnomeface

ps. I've just seen the earlier comments and haven't a clue what they are talking about... so I guess that I'm out of my depth and my comment below is irrelevant! Apologies - I've just come across this site.

... by: Gnomeface

My solver can't solve this with direct logic, but its next level is to try each remaining possible number using its normal logical rules to see if the choice results in another unsolved cell having no possible valid number. If so, the selected original number is invalid. OK, this is part logic, part trial and error, but it does find that 4 of the possible numbers remaining (after exhausting the standard logic levels) would give at least one impossible cell - and when these 4 are removed the solution drops out.

... by: iwl

sorry, this is last weeks sudoku nr. 43, can we get a new one `?

... by: Jean-Claude Casari

Well done, Scott, but what are your logical steps ????

... by: Scott Bailey

This was very simple with my solver.

Solved in: 0.75 seconds

... by: Cenoman

A side observation : Andrew, you skipped #42 in your "unsolveable" puzzle numbering...

Solution for this #43, got from the "full tagging" method, with results transcripted in your notations.

After the first three eliminations with AICs (-7B3, -2J1, -1J5, and subsequent -1H7) the puzzle is unlocked by the following two steps :

#1 -7J2, using a forcing net.

As it will appear several times, the derived SIS (8)G9=G2-(3)G2=J2 is replaced by the shortcut 8G9=3J2

6J4-J79=(6-8)G9=3J2
||
6C4-(4)C4=C8-G8|=2G8-(2-8)H7-G9=3J2
_______________|=(3)G8-G2=J2
||
6B4-B9|=(7)B9-E9=E12-F3=G3
______|=(8)B9-G9=3J2

=> derived strong link 7G3=3J2 =>-7J2

For those who prefer TM tables, another proof of the same :

7G3_7F3
____7E12_7E9
3J2__________3G2
_____________8G2_8G9
_________7B9_____8B9_6B9
_________________8H7_____2H7
______________3G8________2G8_4G8
_____________________________4C8_4C4
_________________6G9_________________6J79
_____________________6B4_________6C4_6J4

#2 -2J2, also using a forcing net.

4A9-A6=C4-J4|=2J4
____________|=(6)J4-J79=(6-8)G9=3J2
4J9-J4|=2J4
______|=(6)J4-J79=(6-8)G9=3J2
4G9-8G9=3J2

=> derived strong link 2J4=3J2 =>-2J2

TM table :

3J2_3G2
____8G2_8G9
________6G9_6J79
2J4_________6J4__4J4
_________________4C4_4A6
________4G9______4J9_4A9

#3 The solver is now able to get the solution with AICs only

... by: simonetta

Un pò Impegnativo ma con mezz'ora tutto risolto!!