## I need help with this

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

### I need help with this

Hi

I'm stuck with this...

Can you tell me the next number with a logical explanation, please?

Thanks a lot
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sudokudumb

Posts: 9
Joined: 27 September 2018

### Re: I need help with this

Your pencil marks should show that cells I23 are occupied by a pair. Now consider where the next value of 4 can be placed.

HTH

Mike Metcalf

m_b_metcalf
2017 Supporter

Posts: 11218
Joined: 15 May 2006
Location: Berlin

### Re: I need help with this

sudokudumb wrote:Can you tell me the next number with a logical explanation, please?

1) Look at box 7 (south-west). Digits 5 and 9 can only go to cells I2 and I3 (hidden pair), which leaves no room for other digits in those cells.

2) Look at column 2. Digit 4 seems to have two possible rows (B and I), but because of the hidden pair it can't go to row I.

3) Thus B2=4 (hidden single); singles to the end.

-SpAce-: Show
Code: Select all
`   *             |    |               |    |    *        *        |=()=|    /  _  \    |=()=|               *            *    |    |   |-=( )=-|   |    |      *     *                     \  ¯  /                   *    `

"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."

SpAce

Posts: 2581
Joined: 22 May 2017

### Re: I need help with this

Thanks guys

I really appreciate your advice. Now looks easy (and obvious) but I didn't see the next number about 2 days

Cheers
sudokudumb

Posts: 9
Joined: 27 September 2018

### Re: I need help with this

Here we go again

Another 3-day enigma for me...

Please give me hint for next move

Thanks
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sudokudumb

Posts: 9
Joined: 27 September 2018

### Re: I need help with this

Look at box 5 and column 5. Digits 3 and 9 can't go to r56c5, so they get locked into r5c46 (and are forbidden in the rest of the row 5) and another pair of digits gets locked into r56c5 (forbidden in the rest of the column 5). The latter pair gives you a naked single in r8c5 (and a few other solved cells after that). It won't solve the puzzle, though. (You'll still need a hidden-pair/locked-triple.) Ask again if you need more help.

..4...853..78354..853.9....43.2769.5.7....6.2..25.83.774.....3..2.....6..8..51..4

lp.png (40.24 KiB) Viewed 456 times

PS. There's also a relatively easy UR Type 1 in r56c58 which would give you a few more solved cells, but it won't solve the puzzle.
Last edited by SpAce on Tue Nov 06, 2018 10:36 pm, edited 1 time in total.

SpAce

Posts: 2581
Joined: 22 May 2017

### Re: I need help with this

Unfortunately, I need more help

I 'found' a naked single in r8c5 and r7c9 and BIG wall
I don't see 'relatively easy UR Type 1 in r56c58'
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current situation
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sudokudumb

Posts: 9
Joined: 27 September 2018

### Re: I need help with this

sudokudumb wrote:Unfortunately, I need more help

I 'found' a naked single in r8c5 and r7c9 and BIG wall

Don't worry, I kind of anticipated you might hit another wall The next step is perhaps a bit harder to see. Look at box 8 and row 7. You need 1 and 5 for row 7, but you have them as solved cells in box 8, which leaves just two cells for them in row 7. Thus you have a Hidden Pair (15) in r7c37. It means that no other digits can go to those two cells. (It also leaves a Locked Triple in r7c456, but we don't really need that.) Then look at column 3. One of the needed digits there now has only one cell available: r9c3. Can you see which one? (Answer hidden below the image.)

hp.png (39.55 KiB) Viewed 433 times

Hidden Text: Show

You still need one more step before you're home free. Now you have to combine both of the previous deductions. First, remember the Locked Pair (39) in r5c46. Neither of those digits can exist elsewhere in row 5. Then remember the Hidden Pair (15) in r7c37, which blocks other digits from those cells. Lastly, the placement we got in the previous step blocks another cell in column 3 (r9c3). It leaves only one cell (r8c3) for one of the digits we need for column 3. (Answer hidden below the image.)

combo.png (39.6 KiB) Viewed 433 times

Hidden Text: Show

After those two placements, the rest is hidden singles and full houses.

SpAce

Posts: 2581
Joined: 22 May 2017

### Re: I need help with this

sudokudumb wrote:I don't see 'relatively easy UR Type 1 in r56c58'

Well, in this case it doesn't help much and it's not very obvious without candidates either. I just mentioned it because URs are often available and can be more effective and easier to spot than some basic techniques (but neither is really true here). In any case, you can find it here:

ur1.png (40.78 KiB) Viewed 431 times

You have the floor pair (14) in box 5 because of the first (39) step. The ceiling pair is in box 6. Note that you only have three digits missing in box 6 so they're easy to check: 1, 4 and 8. You have an 8 in row 6 which leaves just 14 in r6c8. Thus three corners of the UR(14) are already naked which makes it a UR Type 1, and 14 can be removed from the fourth corner leaving a naked single of 8. Easier to see with candidates:

ur2.png (58.11 KiB) Viewed 431 times

SpAce

Posts: 2581
Joined: 22 May 2017

### Re: I need help with this

Solved without looking in Hidden Text

Thank you SpAce for help and for having the patience and for detailed explanations for me.

(Bonus question:

''In box 8 and row 7. You need 1 and 5 for row 7, but you have them as solved cells in box 8, which leaves just two cells for them in row 7''

Why-how we know that 1 and 5 go in row 7? Why not in row 8? It disturbed me for a while )
sudokudumb

Posts: 9
Joined: 27 September 2018

### Re: I need help with this

sudokudumb wrote:Solved without looking in Hidden Text

Thank you SpAce for help and for having the patience and for detailed explanations for me.

Good job, and you're welcome!

(Bonus question:

''In box 8 and row 7. You need 1 and 5 for row 7, but you have them as solved cells in box 8, which leaves just two cells for them in row 7''

Why-how we know that 1 and 5 go in row 7? Why not in row 8? It disturbed me for a while )

We need 1 and 5 for both rows, of course, but row 7 has only two cells where they can go, which is valuable information because no other digits can fit there (and it also tells us that 2, 6 and 9 are locked in r7c456). On the other hand, row 8 has four empty cells outside of box 8, and while we similarly know that 1 and 5 must be somehow distributed in them it's not much to work with. Even though 5 is blocked from r8c9, it still has three possible cells left in row 8, and 1 has all four, so there's no identifiable subset pattern available.

SpAce

Posts: 2581
Joined: 22 May 2017

### Re: I need help with this

Again, a big THANK YOU

to the next mystery
sudokudumb

Posts: 9
Joined: 27 September 2018

### Re: I need help with this

It's been a long time
And just when I think that there is no mystery for me... voila
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sudokudumb

Posts: 9
Joined: 27 September 2018

### Re: I need help with this

sudokudumb wrote:It's been a long time
And just when I think that there is no mystery for me... voila

The first thing that jumps out to my eye, even without candidates, is the Unique Rectangle (19)r13c29. We've talked about those before. It solves r3c2, after which it's singles.

I use URs a lot even with basic puzzles such as this, because they're often much easier to see than some basic patterns -- especially without candidates. For example, in this case I find it much harder to see (without candidates) the naked pair on row 5, which also solves the puzzle.

SpAce

Posts: 2581
Joined: 22 May 2017

### Re: I need help with this

SpAce wrote:The first thing that jumps out to my eye, even without candidates, is the Unique Rectangle (19)r13c29. We've talked about those before. It solves r3c2, after which it's singles.

Without candidates, i would say, it solves r3c3:
You have the 19 pair in r13c9, and 19 missing in box 1 (or column 2) -> possible UR 19 with r13c2.
So (at least one of) 1 or 9 has to be elsewhere in the box (column). Outside the UR cells r13c2 there is only one free cell in the box, and it cannot be 9, so it must be 1.
(Same in the column: only 1 can be in r56c2, which leaves 5 in r5c3)

There is also a naked pair in row 5, which - also without candidates - is easier to spot for me than the hidden triple in that row.
eleven

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Joined: 10 February 2008

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