The New Sudoku Players' Forum

[packerjg10]

I was given this puzzle, and I don't even know where to begin... any help would be appreciated.....

Thanks!


73X X15 X9X
6XX XXX XXX
81X 92X 4XX

XX6 XX9 X18
XX1 XXX XXX
X8X 14X 25X

5XX X7X XXX
XXX XX1 3XX
1X7 XXX X8X


I have the 1's set, but I'm just lost. It has been three days... thanks a lot!

[vidarino]

There are quite a few cells there that can only be one particular number. The other candidates are eliminated by their surroundings, so to speak. Look closer in cells R3C3 and R4C7 for starters.

Vidar

[Cec]

"I was given this puzzle, and I don't even know where to begin... "

Hi packerjg10,

I gather that you are just getting started. Firstly, I suggest you click on HERE to understand the recommended sudoku terminology. Then you could click on the following links which provide lots of basic to advanced techniques for sudoku puzzle solving.
http://www.paulspages.co.uk/sudoku/howtosolve/index.htm
http://www.angusj.com/sudoku/hints.php
http://www.simes.clara.co.uk/programs/sudokutechniques.htm

Make sure you fully understand the basic techniques such as "singles, pairs, triples and locked candidates" before you venture into the more advanced techniques. You've certainly got some homework now

Cec

[QBasicMac]

I was given this puzzle, and I don't even know where to begin.

Here's what you do:

Look at row 3, column 3.

Can it be a 1? No: 1 is already in row 5, column3 and you can't have two 1's in the same column.

Can it be a 2? No: 2 is at row 3, column 5 (r3c5)

Can it be a 3? No: There is already a 3 in the first box (It has 7,3,6,8,and 1)

Can it be a 4? No: r3c7
Can it be a 5? Yes, it could be.
Can it be a 6? No: r4c3
Can it be a 7? No: r1c1 (Same box)
Can it be a 8? No: r3c1 (Same box)
Can it be a 9? No: r3c4

OK, there is only one number that can go in r3c3: 5. So put a 5 there.

See? One cell solved already!

Study r4c7 the same way. You will find it must be a 7.
Another cell solved!

Now another way to find placements:

There is a 7 in r1c1 and r9c3. Now the middle box (called box 4 - has cells r4c1 and r6c3 at the corners) needs a 7. But it can't go in columns 1 or 3 as there are already 7's there. And r6c2 is already a 8. So the only cells that the 7 of box 4 can go into are r4c2 and r5c2. But row 4 already has a 7 at r4c7. So you must put the 7 in r5c2.

Another cell solved.

While looking at 7's I see that row 4 has a 7 and row 5 has a 7. The cell at r6c6 therefore must be a 7.

Good Luck. This is all you need. No "jargon" and other complexities. This puzzle can be solved as I outlined.

Mac

[packerjg10]

It took me awhile, but i figured it out!

734 615 892
692 784 531
815 923 467
456 129 718
271 856 943
389 147 256
563 478 129
928 561 374
147 392 685

YEAH! I know it might be easy to some, but it took me awhile!