Can't even start on this 'easy' grid

Post the puzzle or solving technique that's causing you trouble and someone will help

Can't even start on this 'easy' grid

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)

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`- - -   - - 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

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".

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:
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`---  --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:

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`---  --4  --5-46  -5-  --8315  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