The New Sudoku Players' Forum

[cwgannon]

It began as:

Code: Select all

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

and has come to:

Code: Select all

23  347 247 | 5 1 8 | 6  9   47
168 16  5   | 7 9 4 | 3  2   18
18  9   478 | 6 3 2 | 14 145 14578
------------+-------+--------------
4   2   6   | 1 7 5 | 8  3   9
39  37  79  | 2 8 6 | 14 145 145
5   8   1   | 3 4 9 | 7  6   2
------------+-------+--------------
169 146 49  | 8 5 3 | 2  7   14
28  14  28  | 9 6 7 | 5  14  3
7   5   3   | 4 2 1 | 9  8   6


It's really quite possible that I've already made a mistake, but if I haven't, could somebody please point me in the right direction? I'm looking more to understand how to progress from here, rather than just how to fill in the numbers.

Thanks.

[vidarino]

It looks good so far.

You have a "naked triple" of 146 in column 2 (16, 146 and 14). This means you can eliminate a 4 in the top row. Then, look closely at the remaining 4s in the top box.

Vidar

[tarek]

hi cwgannon & vidar,

you can also achievethe same result with the triple if you look for a hidden double in column 2......


Tarek

[cwgannon]

I hope I don't come off as an idiot, but here's where the suggestions thus far have gotten me:

Code: Select all

23  37  247 | 5 1 8 | 6  9   47
168 16  5   | 7 9 4 | 3  2   18
18  9   47  | 6 3 2 | 14 145 14578
------------+-------+--------------
4   2   6   | 1 7 5 | 8  3   9
39  37  79  | 2 8 6 | 14 145 145
5   8   1   | 3 4 9 | 7  6   2
------------+-------+--------------
169 146 49  | 8 5 3 | 2  7   14
28  14  28  | 9 6 7 | 5  14  3
7   5   3   | 4 2 1 | 9  8   6

I'm not seeing anything that can be done with the 7's in the top row. What am I missing?

And thanks, already, for the help.

[tarek]

if you look at column 2

All the 4s are in box 7

As Column 2 must have a 4, you can safely eliminate other 4s in box 7, here we haveonly one elimination r7c3

Tarek

[tarek]

Anyway,

i just noticed the original puzzle, I don't think it's moderate at all..

on the contrary, It is extremely difficult to solve by newspaper standardsespecially when there are multiple solutions..................

I'm curious, how did you make some of the elimination along the way cwgannon ???

Or could you check the starting grid again........

Tarek

[tarek]

Don't worry,

From your pencilmarks...

r3c6 must have been 2 & was dropped out....


Tarek

[cwgannon]

if you look at column 2

All the 4s are in box 7

As Column 2 must have a 4, you can safely eliminate other 4s in box 7, here we haveonly one elimination r7c3

Tarek

That hint was all it took for me to solve the rest. I just wish I'd spotted it earlier myself.

As it turns out, this puzzle was actually rated as "Hard. Like Diamonds." by the Daily Cardinal in Madison. I figured I'd just label it as a moderate one, as it seems to be compared to some of those I've seen here.

Thanks again for the help. I appreciate it.

[Cec]

"..you can also achieve the same result with the triple if you look for a hidden double in column 2...."

Adding to your correct comment, when four clues ('given' numbers) appear, as in this case, in a unit (row, column or box) and a naked triple exists in that unit, then after eliminating these three candidates if existing in other cells of that unit, the remaining candidates in that unit will always form a naked pair .

I normally find it "easier" to spot naked triples than hidden pairs but that's just me. You are correct but thought I'd throw this in.

Cec

[ronk]

I normally find it "easier" to spot naked triples than hidden pairs but that's just me.

I find it easier to even spot naked quads than a hidden pair, so I certainly don't think it's "just you".

Ron

[QBasicMac]

After removing 4 in r1c2 because of the hidden pair 37 in col 2, the locked candidate 4 in r1c3 f3c3 eliminate the 4 in r7c3.

Nothing but singles after that.

Mac

[ronk]

Looking at the wrong grid, so comment deleted. Sorry, Ron

[Myth Jellies]

Code: Select all

23  347 247 | 5 1 8 | 6  9   47
168 16  5   | 7 9 4 | 3  2   18
18  9   478 | 6 3 2 |*14*145*145+78
------------+-------+--------------
4   2   6   | 1 7 5 | 8  3   9
39  37  79  | 2 8 6 |*14*145*145
5   8   1   | 3 4 9 | 7  6   2
------------+-------+--------------
169 146 49  | 8 5 3 | 2  7   14
28  14  28  | 9 6 7 | 5  14  3
7   5   3   | 4 2 1 | 9  8   6


You may think this is funny, but the first thing I saw when looking at this grid was the very obvious MUG formed by the 145 in the starred cells. This forces r3c9 = 78 which solved the 5's. It didn't exactly crack the puzzle, but it did get rid of some clutter.

[MCC]

Code: Select all

23  347 247 | 5 1 8 | 6  9   47
168 16  5   | 7 9 4 | 3  2   18
18  9   478 | 6 3 2 | 14 145*145*+78
------------+-------+--------------
4   2   6   | 1 7 5 | 8  3   9
39  37  79  | 2 8 6 | 14 145*145*
5   8   1   | 3 4 9 | 7  6   2
------------+-------+--------------
169 146 49  | 8 5 3 | 2  7   14
28  14  28  | 9 6 7 | 5  14  3
7   5   3   | 4 2 1 | 9  8   6

In a similar vein to Myth Jellies I saw the uniqueness rectangle of 145's which implied that r3c9=78.

MCC

[ronk]

In a similar vein to Myth Jellies I saw the uniqueness rectangle of 145's which implied that r3c9=78.

I don't believe that's a valid deduction, since ...

Code: Select all

1    4

5    1

... is one of several possible outcomes not containing 78.

Ron