The New Sudoku Players' Forum

[Pat]

1..2....3.4..5......6..12..5..7..6...8.....7...1..8..2..98..4......7..5.8....3..1

[ play ]

24 cells given
(layout 198 of the Patterns Game)

Code: Select all


 1 . . | 2 . . | . . 3
 . 4 . | . 5 . | . . .
 . . 6 | . . 1 | 2 . .
-------+-------+------
 5 . . | 7 . . | 6 . .
 . 8 . | . . . | . 7 .
 . . 1 | . . 8 | . . 2
-------+-------+------
 . . 9 | 8 . . | 4 . .
 . . . | . 7 . | . 5 .
 8 . . | . . 3 | . . 1



[7b53]

this puzzle can be solved by basic techniques. it doesn't means it is easy.
many players are not used to spotting hidden subsets for whatever reasons.
and supposed the challenge is to solved it without any pencil-marked.

[Marty R.]

I'll readily call anyone a genius with a photographic memory if they can solve this without PMs.

Yes, basics, but one must look extremely carefully for the quads.

[JasonLion]

Quads? I saw just three triples, a couple of locked candidates, and singles.

[Marty R.]

Quads? I saw just three triples, a couple of locked candidates, and singles.

Not a shock. I constantly miss stuff that's there right in front of my nose.

[Pat]

three triples

yes, 3 heavy moves
— i thought that's high enough to be unusual and interesting

[7b53]

It is highly unusual needing three triples.
Plus the way those hidden triples are form makes it harder to spot.
If you noticed the first two are form pretty much in an identical way.
And yes, is a very interesting one.

[Pat]

did a little digging,
found we already had a better puzzle —
joel64 # 163 # 40

[JasonLion]

joel64 # 163 # 40

That uses hidden triples as well as naked triples.

<withdrawn>

[7b53]

one of the greatest enemy for non pencil-marker is spotting naked subsets.
just a naked pair can takes forever. not to mention naked triple,quad. or maybe is just for me.

joel64 # 163 # 40

for some reason, i like the previous one better.

[keith]

I'll readily call anyone a genius with a photographic memory if they can solve this without PMs.

Well, I don't know about that, but I did solve it without PMs. I must admit to staring at it for a long time!

If you don't try to solve without PMs, it's kind of hard to explain. It works by finding the hidden subsets rather than the naked subsets. First, take a look at C3. Note the cells which are not 578; they must be 234. That solves a couple of cells, R2C14.

Now, look at R3 for the cells that are not 357. They must be 489. That solves R1C2 as 9 and, more importantly, fixes 4 in B6C9.

Now, look at B4 for the cells that are not 679. They must be 234. This fixes 4 in B4C3. Looking at which squares can be 4 solves the puzzle.

Keith

[Marty R.]

Well, I don't know about that, but I did solve it without PMs. I must admit to staring at it for a long time!

I'll stand by my statement. It's very impressive as far as I'm concerned.

[eleven]

Maybe it helps, when you see them marked.

Code: Select all

 1 . . | 2 . . | . . 3
 . 4 . | . 5 . | . . .
 . . 6 | . . 1 | 2 . .
-------+-------+------
*5 . x |*7 . . | 6 . .
 .*8 x | . . . | .*7 .
 . . 1 | . . 8 | . . 2
-------+-------+------
 . . 9 | 8 . . | 4 . .
 . . x | .*7 . | .*5 .
*8 . . | . . 3 | . . 1

The 578 all see the cells marked with x. As Keith says, it needs some time to spot it. But it's the same for patterns in pencilmark grids.

[keith]

OK,

I'll take eleven's cue and spell it out.

Code: Select all

 1 . y | 2 . . | . . 3
 . 4 y | . 5 . | . . .
 . . 6 | . . 1 | 2 . .
-------+-------+------
*5 . x |*7 . . | 6 . .
 .*8 x | . . . | .*7 .
 . . 1 | . . 8 | . . 2
-------+-------+------
 . . 9 | 8 . . | 4 . .
 . . x | .*7 . | .*5 .
*8 . y | . . 3 | . . 1

The cells y are 578. The cells x are 234. Solving 2 and 3 in R2.

Code: Select all

 1 . . | 2 . . | . .*3
 2 4 . |*3*5 . | . . .
 y y 6 | x x 1 | 2 x y
-------+-------+------
 5 . . |*7 . . | 6 . @
 . 8 . | . . . | .*7 @
 . . 1 | . . 8 | . . 2
-------+-------+------
 . . 9 | 8 . . | 4 . .
 . . . | .*7 . | .*5 .
 8 . . | . . 3 | . . 1

The cells y are 357. The cells x are 498. @ is where 4 must be in B6C9. Solving 9 in R1C2.

Code: Select all

 1 9 . | 2 . . | . . 3
 2 4 . |*3*5 . | . . .
 . .*6 | . . 1 | 2 . y
-------+-------+------
 5 x x@|*7 . . |*6 . @
 . 8 x@| . . . | .*7 @
 . . 1 | # # 8 | . . 2
-------+-------+------
 . .*9 | 8 . . | 4 . .
 . . . | . 7 . | . 5 .
 8 . . | # # 3 | . . 1

The cells x are not 679. They are 234. The cells @ must contain 4 in B4C3 and B6C9. You can solve 4 in B7. Then look at the pattern for 4 in B58 (#) and solve 4 in B2. The puzzle is done.

Keith

[7b53]

I must admit to staring at it for a long time!

you're not the only one Keith. maybe I was staring it longer than you.
trying to find pair at the beginning..... none !