Sudoku Players' Forum (clone)

* 3 4 * 7 * * * 5
* * * 2 * * * 7 6
7 * 9 5 * * * * *

4 1 5 * * * * * 7
9 7 * * 3 * * 6 *
* * 3 * * * 5 4 9

* * * * * 1 7 * 4
3 * * * * 2 6 * *
* * * * 9 * 2 5 *

Note there are *NO* 8's in the whole puzzle. That means 8's are candidates for every open square. I found half a dozen places where there was only one choice for a # (they are already filled in above) and a few #'s that could be eliminated because they would conflict with a value known to be on a row. (Example: r6c4, r6c5 and r6c6 can't contain a 6.)

I'm stumped at how to attack this mess.

Look closely at where the 5 can go in block 5. Then look where the 4 can go in column 4. Then look at candidates that can only be in one column or row, and from that make some eliminations in blocks. Then you will have to look for some pairs, and make yet more eliminations.

Have a good look at row 5. You should be able to fill in a couple of values in that row.

scrose wrote:Look closely at where the 5 can go in block 5. Then look where the 4 can go in column 4. Then look at candidates that can only be in one column or row, and from that make some eliminations in blocks. Then you will have to look for some pairs, and make yet more eliminations.

I see now. I saw 4,5,8 and 4,5,8 and didn't see that I could do anything with it.

SteveF wrote:Have a good look at row 5. You should be able to fill in a couple of values in that row.

I need new eyes I think. I had a 5 in r5c4.

Loren Pechtel wrote:I need new eyes I think. I had a 5 in r5c4.

If I had a dollar/pound/ruble/yen/etc. for every time I've done something like that...

Getting that 4 and 5 in box 5 didn't crack anything open. I'm as stuck as ever.

Look for a hidden triplet in row 2. That should break the puzzle wide open.

scrose wrote:Look for a hidden triplet in row 2. That should break the puzzle wide open.

I'm being dense somehow. There's no triplet there as far as I can tell.

Loren Pechtel wrote:I'm being dense somehow. There's no triplet there as far as I can tell.

Actually, now that I look at it closer, there is a hidden pair in row 2. Look closely at where the 3's and 9's can go in that row. Where does that let you put the 4 in that row?

Update: I just noticed that after you figure out where to put the 4 in row 2, you will be able to quickly place the remaining three 4's.

scrose wrote:

Loren Pechtel wrote:I'm being dense somehow. There's no triplet there as far as I can tell.

Actually, now that I look at it closer, there is a hidden pair in row 2. Look closely at where the 3's and 9's can go in that row. Where does that let you put the 4 in that row?

Update: I just noticed that after you figure out where to put the 4 in row 2, you will be able to quickly place the remaining three 4's.

I see said the blind man.

Great! The rest of the puzzle should fall like a house of cards.

The hidden triplet I saw was {349} but then I realized the pair {39} only occurred in r2c6 and r2c7. There is also a naked triplet of {158} in r2c1, r2c2, and r2c3. Any of these combinations will isolate where the 4 can go in row 2.

This situation is similar to the puzzle where BadCujo found one set and MCC found a corresponding set. Whenever you find a set of candidates in a row/column/block, there is usually a second (or even a third) set that is also present.

scrose wrote:Great! The rest of the puzzle should fall like a house of cards.

The hidden triplet I saw was {349} but then I realized the pair {39} only occurred in r2c6 and r2c7. There is also a naked triplet of {158} in r2c1, r2c2, and r2c3. Any of these combinations will isolate where the 4 can go in row 2.

I've already got that 3,9 pair but it doesn't help. I haven't found any way to make it fall.

Here are my candidates for row 2.

{158} {58} {18} [2] {148} {3489} {13489} [7] [6]

The {158} set in the three left cells lets you eliminate the candidate 1's and 8's from r2c5, r2c6, and r2c7. This leaves the following candidates.

{158} {58} {18} [2] {4} {349} {349} [7] [6]

So that lets you place a 4 in row 2. Now where can the 4's go in blocks 3 and 8? And then where can the 4 go in block 7?

scrose wrote:Here are my candidates for row 2.

{158} {58} {18} [2] {148} {3489} {13489} [7] [6]

The {158} set in the three left cells lets you eliminate the candidate 1's and 8's from r2c5, r2c6, and r2c7. This leaves the following candidates.

{158} {58} {18} [2] {4} {349} {349} [7] [6]

Got this far.

So that lets you place a 4 in row 2. Now where can the 4's go in blocks 3 and 8? And then where can the 4 go in block 7?

Ack! I forgot to remove the 4 from r9c2. The house of cards finally fell.