Can't even start on this 'easy' grid

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Can't even start on this 'easy' grid

Postby delt » Wed Jun 07, 2006 6:39 pm

Hello everyone, i'm new here, glad to be onboard.

This one is from a book with about 400 sudoku grids, it's in the "easy" section but i can't even put a first number in it. After eliminating all the "already taken" in each row, column, and box, there's no giveaway anywhere (cell with only 1 possible number left)

Code: Select all
- - -   - - 4   - - 5
- - 6   - 5 -   - - -
- 1 5   - 8 2   7 - -

- - ?   - - 9   3 - 6
- 7 3   - - -   8 9 -
1 - 9   7 - -   - - -

- - 2   3 4 -   9 8 -
- - -   - 9 -   6 - -
4 - -   2 - -   - - -


Based on the row, i -think- where i put the "?" (middle left box) is a 4, but i'm not even sure about that. I don't know about advanced stuff like "X or Y wing" (yet) so i stick to easy puzzles for now, but... what approach would you take to start solving this grid?

Knowing this will also help me in other cases where i might be stuck in the same way.
delt
 
Posts: 1
Joined: 07 June 2006

Postby tso » Wed Jun 07, 2006 6:51 pm

Try to place a 4 in box one. (the 3x3 box on the upper left.) See how the 4 on the top row and the 4 in the left column narrow your choice? After this, you should be able to place a few more 4's by similar methods.
The is how a "hidden single" is found.

Try to place an 8 in box three or a 9 in the bottom row by similar means.
tso
 
Posts: 798
Joined: 22 June 2005

Not exactly my concept of "easy".

Postby QBasicMac » Thu Jun 08, 2006 9:55 pm

Not exactly my concept of "easy".

Anyway, look at box 1. It needs 234789, right?
Well, test each cell for each value, starting with 2.
2) There are a couple of cells that 2 can fit in. Forget it.
3) Same.
4) It can only go in r2c2 (row 2 column 2). It can't go into any other cell because there are already 4's in row 1 and column 1. Great!! Put a 4 there. One cell solved!
789) Not unique. Forget them

Now that we have placed a 4 in column 2 and considering the 4 in column 1, we see that the 4 in column 3 must go into box 4. But 3 and 9 are already there. There is only one place! Put a 4 there. Another cell solved.

But no more progress here:(

For example with the new 4 in box one on row 2 and the existing 4 on row 1, we conclude a 4 must go into row 3 at box 3. But there are two cells where it would fit. Rats!

But wait! Look at columns 5 and 6. So a 4 must go into r4c4 or r5c4. But there is already a 4 in row 4! Yea! Put a 4 at r5c4!

You are here:
Code: Select all
---  --4  --5
-46  -5-  ---
-15  -82  7--

--4  --9  3-6
-73  4--  89-
1-9  7--  ---

--2  34-  98-
---  -9-  6--
4--  2--  ---


See if you can see why r4c8=7, r2c9=8, r9c2=9

Fill those in.

Now you can note that as r7c7=9 and r5c8=9 then r3c9=9. Fill in another one!

There may be enough cells filled into row 3 to look for a placement there. Missing is 346.

Look at r3c3. There is a 4 in the column and a 6 in the box. Has to be 3! Fill it in.

What about r3c4? There is a 4 in the box so r3c4 must be 6! Fill it in.

Last missing digit: Put a 4 in r3c8.

One whole row done:

Code: Select all
---  --4  --5
-46  -5-  --8
315  682  749

--4  --9  376
-73  4--  89-
1-9  7--  ---

--2  34-  98-
---  -9-  6--
49-  2--  ---


Well, just keep doing the same thing. It is tough, but you can get there without advanced techniques. Just patience and accuracy.

Mac
QBasicMac
 
Posts: 441
Joined: 13 July 2005


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