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Bill Denholm
Joined: 24 Mar 2006
Posts: 17
Location: Mountain View, California
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 Posted: Sun Jul 09, 2006 9:46 pm Post subject: July 8 puzzle Very Hard |
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I played 16 moves and then got stuck, so I asked for a hint. The program gave me 9 at C4R1. I was hoping that I would see the logic to the hint, but in this case, I don't. Can anybody explain it?
Bill Denholm |
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Chuck B
Joined: 24 Jun 2006
Posts: 24
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| Posted: Sun Jul 09, 2006 10:37 pm Post subject: |
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Bill,
If your 16 moves correspond to the hints, there is an X-Wing in '7' at R19C37. This eliminates 7 as a candidate from R1C4, leaving 9 as the only possibility.
HTH,
Chuck |
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Bill Denholm
Joined: 24 Mar 2006
Posts: 17
Location: Mountain View, California
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| Posted: Mon Jul 10, 2006 1:23 pm Post subject: |
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Thanks, Chuck. I see it when you point it out, but they (X-Wings) are not something I usually look for and I'm not good at spotting them.
Bill |
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oaxen
Joined: 10 Jul 2006
Posts: 96
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| Posted: Tue Jul 11, 2006 4:54 pm Post subject: july 8 |
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No, I am not so sure or do not understand the X-wing technic so well.
There is still a possibility for a "7" in r9c8 |
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Chuck B
Joined: 24 Jun 2006
Posts: 24
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| Posted: Tue Jul 11, 2006 6:15 pm Post subject: Re: july 8 |
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| oaxen wrote: | No, I am not so sure or do not understand the X-wing technic so well.
There is still a possibility for a "7" in r9c8 |
Here's the candidate list after the 16 moves: | Code: | *-----------------------------------------------------------*
| 1 2456 2467 | 79 489 48 | 467 3 456 |
| 47 3 9 | 6 5 1 | 8 47 2 |
| 467 456 8 | 37 34 2 | 9 1 456 |
|-------------------+-------------------+-------------------|
| 234 8 234 | 5 234 9 | 1 6 7 |
| 5 9 346 | 1 346 7 | 34 2 8 |
| 2346 7 1 | 8 2346 34 | 34 5 9 |
|-------------------+-------------------+-------------------|
| 367 16 5 | 4 389 38 | 2 79 16 |
| 9 124 234 | 23 7 6 | 5 8 14 |
| 8 246 2467 | 29 1 5 | 467 479 3 |
*-----------------------------------------------------------*
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Notice that the 7 for columns 3 and 7 are constrained to rows 1 and 9.
Since these columns' 7's cannot be in the same row, one must be in row 1 and the other in row 9, i.e., on opposing corners of the R19C37 rectangle.
This fact blocks 7 from occupying any other cell in row 1 or row 9. In this particular case, it eliminates 7 as a candidate in R1C4 and R9C8.
That's the X-wing logic applied to this particular situation.
HTH,
Chuck |
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kronik
Joined: 11 Jul 2006
Posts: 1
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| Posted: Tue Jul 11, 2006 7:01 pm Post subject: |
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Oaxen,
I just recently understood the X-Wing.... and yes there are links to explain it but here's how I interpreted it...
I think of it more as a rectangle where the X-Wing creates 4 corners (A, B, C, D) of the rectangle and a number must go in opposite corners of that rectangle (AD or BC), creating only a pair of options. Those two options eliminate all other possibilites, similar to a naked pair.
A--------B
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C--------D
In this case 7's can still go in C1, C3, C7 and C8....here's how I create the X-Wing/Rectangle...
C8 has possibilities for 7's but at more than two rows, that's out immediately because you can only have two possibilities in any unit.
C1 has only two options, but no matching rows in C3 or C7... It share's rows with C8, but there are no other possibilities for 7's in those rows.
It's important to see the pair of 7's in C3 and C7 because they share R1 and R9...
What creates the X-Wing/Rectangle is that the 7's in C3 and C7 also only occur in R1 and R9.
The 7's must be placed at either (C3R1 and C7R9) or (C3R9 and C7R1). Why? Because a 7 at C3R1 eliminates obviously any other 7 on R1 more specifically the 7 in C7R1. If that's the case the only place left for a 7 in C7 is R9 eliminating all other possibilities on that row. Follow the same theory for a 7 at C7R1, or any of the other remaining corners.
The X-Wing creates only two options for where a 7 can go in R1 and R9 at C3 and C7. Therefore any other 7 on R1 or R9 gets eliminated.
X-wings can also occur in rows eliminating possibilities in columns in the same manner.
Wow reading back I don't know if that will be helpful or more confusing, but I guess that's how my crazy brain works...
HTH,
..kev |
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pedro
Joined: 16 Jul 2006
Posts: 1
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| Posted: Sun Jul 16, 2006 10:13 pm Post subject: |
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Hello everyone-
I'm relatively new to Sudoku. I've been struggling with this puzzle and can't see how the 1 ends up in R2C6. That corresponds to hint #9 if one plays this puzzle online.
Could anyone enlighten me as to how one deduces this move?
Thanks,
pedro |
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Chuck B
Joined: 24 Jun 2006
Posts: 24
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| Posted: Sun Jul 16, 2006 11:11 pm Post subject: |
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| Code: | *-----------------------------------------------------------------------------*
| 1 2456 2467 | 379 3489 348 | 345678 347 4568 |
| 47 3 9 | 6 5 148 | 478 147 2 |
| 467 456 8 | 37 134 2 | 9 1347 1456 |
|-------------------------+-------------------------+-------------------------|
| 234 8 234 | 5 234 9 | 1 6 7 |
| 5 9 346 | 1 346 7 | 348 2 48 |
| 2346 7 1 | 8 2346 34 | 34 5 9 |
|-------------------------+-------------------------+-------------------------|
| 3678 16* 5 | 4 1389 >138< | 2 179 16* |
| 9 124 234 | 23 7 6 | 45 8 145 |
| 24678 1246 2467 | 29 1289 5 | 467 1479 3 |
*-----------------------------------------------------------------------------* |
Pedro, there's a naked pair '16' in R7C2/C9 , which eliminates '1' as a candidate in the other cells of row 7 -- in particular, the '1' in R7C6.
That leaves R2C6 as the only cell in column 6 that can hold a '1'.
- cb
edited to correct row/column # error (swapped) |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Sun Jul 16, 2006 11:20 pm Post subject: Naked pair {1, 6} in row 7 |
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Hi, Pedro! Welcome to the forum.
After the first 8 (elementary) moves, the puzzle looks like this.
| Code: | 1........
.3965...2
..8..29..
.8.5.9167
59.1.7.2.
.718...59
..54..2..
9...76.8.
.....5..3 |
Now examine row 7 closely. At r7c2 you will notice that the only values that can possibly be entered are {1, 6} -- {2, 4, 5} are already in row 7, and {3, 7, 8, 9} are in column 2.
At r7c9 you will also find that the only possible values are {1, 6} -- {2, 4, 5} are in row 7, {3, 7, 9} are in column 9, and "8" is in the bottom right 3x3 box.
These two cells -- r7c2 and r7c9 -- must contain the "naked pair" {1, 6}. We don't know the order in which these appear, but it's clear that if one of them is "1", the other one is "6". And it's also clear that neither "1" nor "6" can appear anywhere else in row 7. In particular, there cannot possibly be a "1" at r7c6.
Now if you look at column 6 you will see that "1" must appear at r2c6 -- r1c6 and r6c6 cannot contain a "1" because of "1"s already entered in the puzzle, and r7c6 can't be a "1", either, because of the {1, 6} pair described above. dcb |
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Mimi
Joined: 20 Jul 2006
Posts: 2
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| Posted: Thu Jul 20, 2006 8:05 pm Post subject: to Chuck B about July 8 puzzle |
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Chuck, i ALMOST got your explanation but am so confused that i can't figure out what you mean by the R19C37 rectangle in this puzzle.
Would youplease illustrate this?
Mimi |
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Chuck B
Joined: 24 Jun 2006
Posts: 24
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| Posted: Thu Jul 20, 2006 8:32 pm Post subject: Re: to Chuck B about July 8 puzzle |
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| Mimi wrote: | Chuck, i ALMOST got your explanation but am so confused that i can't figure out what you mean by the R19C37 rectangle in this puzzle.
Would youplease illustrate this?
Mimi |
My apologies, Mimi! Guess my shorthand was too short. 
'R19C37' refers to the rectangle with corners at R1C3, R1C7, R9C3 and R9C7 - which are flagged with '+' and '-', below: | Code: | *-----------------------------------------------------------*
| 1 2456 2467+ | 79 489 48 | 467- 3 456 |
| 47 3 9 | 6 5 1 | 8 47 2 |
| 467 456 8 | 37 34 2 | 9 1 456 |
|-------------------+-------------------+-------------------|
| 234 8 234 | 5 234 9 | 1 6 7 |
| 5 9 346 | 1 346 7 | 34 2 8 |
| 2346 7 1 | 8 2346 34 | 34 5 9 |
|-------------------+-------------------+-------------------|
| 367 16 5 | 4 389 38 | 2 79 16 |
| 9 124 234 | 23 7 6 | 5 8 14 |
| 8 246 2467- | 29 1 5 | 467+ 479 3 |
*-----------------------------------------------------------* |
The 7's for columns 3 and 7 cannot be in the same row, so they must be on diagonally opposite corners of the rectangle, either in the '+' cells or in the '-' cells.
You might find it helpful to consider the flip side and visualize what would happen to column 3 or column 7 if R1C4 or R9C8 were allowed to hold a 7: the result would be that one of the columns (C3 or C7) would then have no legal place to put a 7.
Hope that helps!
- cb |
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Mimi
Joined: 20 Jul 2006
Posts: 2
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| Posted: Thu Jul 20, 2006 9:34 pm Post subject: |
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| Thanx, Chuck. got THIS one now! |
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