I can solve most of them eventually, but the very hardest ones defeat me. I need some new “logic tools” to apply to the more difficult ones. I already know basic scanning (going through rows, columns and 3x3 regions looking for gotta-be’s) and marking up (eliminating everything a space can’t be to see if clues emerge). But when that isn’t enough, I need some analysis methods to help me out. I only know three: (1.) If in a particular 3x3 region a number has to be in one row or one column, then that number is eliminated for the rest of the row or column. (2.) If a pair of spaces are the only two places that two numbers can be, then all other alternatives for those two spaces are eliminated. (3.) Very rarely, I can find a reductio ad absurdum situation where the chain of logic is brief and simple enough for me to follow.

Can anyone guide me through “Multiple Contingencies for Dummies”?

After you have done the quick scanning you need to go though each empty box and eliminate any numbers in that row that column and that 3 by 3 box. Once that is done look in each row/column/box to see if any of them there is only one box that can have a specific number.

Also if two boxes can have only the same two numbers those numbers cannot be anywhere else in the row/column/box. This tends to help some.

After that you have to start guessing. It helps at that point to have a notation system so you can back out your guesses to the poit where you are sure.

Similarly, if two digits can only be in the same two boxes of a 9x9, no other digit can be in those boxes.

Also look out for the common following situation…

— -9- —
123 — —

where you have a complete row (or column) of 3 within a 9x9. In this example you know that a ‘9’ must occur on the bottom row of the first 9x9 and the middle row of the last.

Remember, narrowing down in which row/column of a 9x9 a digit lies gives the same amount of information in one dimension as does the exact position. On difficult puzzles I write tiny reminder digits in the corners of the squares, though as I get better I find I can complete harder and harder puzzles without doing so - an extra challenge!

Here we disagree. With practice you shouldn’t need to guess even on the hardest puzzles. If you find you are making no progress, take a break. Often I have solved quite easily one which seemed impossible the day before.

The key to completing Sudoku for me is to use a dot notation in each square that lets me keep track of what numbers are possible. Simply writing the numbers in creates a mess and confusion, putting small dots lets you see whats going on. Just doing this in every square makes a surprising difference, you never miss any numbers through marking up and basic scanning.

After that, its the second method you’re talking about that usually breaks a fiendish SuDoku. I think it needs expanding though, its not just recognising that a pair of spaces is reduced to two numbers, its also recognising that those two numbers are then not available for any other square in the row (or column). This also works for a pair of spaces in a 3x3 grid, and can be very helpful.

I completed all of the fiendish Sudokus in the most recent Times book using this approach, it got to be quite easy. I heard recently that one of the UK papers had introduced an even more difficult Sudoku, one that had extra symbols as well as the numbers 1-9, sounds good!

Lastly, ticker is right, you never have to guess to complete a Sudoku.

Take a look at this site. It explains it a lot better than I could.
I found this to be one of the clearest explanations for sudoku solving with good examples of each solving hint.
On the other hand if you need to resort to either the"x-wing" or “swordfish” methods then you could be in trouble.
I would also agree that in a well designed puzzle you should not have to guess.

Here is all I could finish for the Sudoku from last Sunday in my newspaper. The solution was published so I know that I was right as far as I got. But can anyone explain how you can deduce any more clues?

row 3, cols 4 & 6 can only be digits 4 & 8 since both digits are already in row 2 and col 5. Therefore row 1, cols 7…9 must also contain 4 & 8. Furthermore row 1, cols 7…9 must also contain a 2 since rows 2 & 3 already have a 2. That is all three digits for the top row of the top-right 9x9 cell, even if the exact order is not known yet.

There must be a 1 in col 7, rows 3 or 4 since there is already a 1 in col 9 and the only other free box in that 9x9 is limited to 2, 4 or 8 as above. That means in the bottom right 9x9 the only position left for a 1 is between the 9 & 7.

It is then easy to place a 1 at row 9, col 5. In the same column it is then easy to place the 9 (top) and 3 (middle).

Thanks everyone, especially for the link to the “Simple Sudoku” site. I finally solved last week’s puzzle with the help of “naked triples”. Just the sort of advanced tool I needed.