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Huh?
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Huh?
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 Posted: Thu Feb 16, 2006 9:24 pm Post subject: |
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| I beg your pardon for a double post, but I guess the answer I should ask: why can the red 1 not be a 2? |
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dotdot
Joined: 07 Dec 2005
Posts: 29
Location: oberseen
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| Posted: Thu Feb 16, 2006 10:59 pm Post subject: two ways of explaining why not 2 |
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Hallo Huh
The conventional explanation is that the two empty cells at the bottom of col3 can both take only the values 2 or 7, i.e. their candidate lists are {2,7}. Thus between them they must contain a 2, and thus prevent a 2 occurring elsewhere in col3.
But this {2,7} pair is not at all obvious unless you systematically maintain the candidate lists.
An alternative explanation involves the Glassman pan based at the shared triplet c3b7:
the rest of col3 must have the same values as the rest of box7.
So the {1,3,8} of box7 must somehow fit in the r1..6c3, i.e. into the empty cells r3..5c3.
This is an exact fit, leaving no room for any other values. |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Thu Feb 16, 2006 11:13 pm Post subject: It looks obvious ... |
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| dotdot wrote: | | But this {2,7} pair is not at all obvious unless you systematically maintain the candidate lists. |
Hi, dotdot!
This "hidden" pair {2, 7} in r7c3 & r9c3 looks pretty obvious to me.
-- The only values missing in column 3 are {1, 2, 3, 7, 8}
-- The triplet {1, 3, 8} is already visible in the bottom left 3x3 box
-- Therefore {1, 3, 8} must appear at r3c3, r4c3, & r5c3; and {2, 7} must lie in r7c3 & r9c3.
You can also look at it the other way round:
-- The only values missing in the bottom left 3x3 box are {2, 4, 5, 6, 7}
-- The triplet {4, 5, 6} already appears in column 3 outside that box
-- Therefore {4, 5, 6} can't appear in r7c3 or r9c3, leaving the {2, 7} pair as the only possible values in those two cells. dcb |
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emily hodges
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| Posted: Thu Feb 16, 2006 11:20 pm Post subject: Feb 15 puzzle |
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 I got stuck after finding only 9 numbers that fit. I don't understand why the hint (#5 in the third box down of the first vertical row) should have been apparent, since I had found the following combinations in that row
359
13
135 (the box in question)
7 (given)
1369
2 (found)
8 (given)
456
45
Sorry I don't yet know the nomenclature for identifying specific boxes...
Can anyone help with this one?
Emily |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Fri Feb 17, 2006 12:07 am Post subject: Are you sure? |
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Hi, Emily! It's nice to meet you.
I'm a little confused by your question, because row3, column 1 ("the box in question") must contain a "1" and not a "5". The possibilities you've listed look good, though.
Are you sure you keyed your partial solution into the "Draw" program correctly?
Oh -- you asked about nomenclature. People usually refer to the horizontal lines in a sudoku puzzle as rows, and we call the vertical lines columns. We number the rows from top to bottom, and the columns from left to right. So your "box in question" is row 3, column 1, usually abbreviated r3c1.
Over to you ... dcb |
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Guest
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| Posted: Fri Feb 17, 2006 2:22 pm Post subject: |
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THanks, David, you are right. I reentered them and got the red 1 in r4c4, as mentioned in the discussion above, BUT - I don't understand why it has to be the 1 and not the 3, without just guessing.
i also got stuck again right after I entered the 3 in r4c8, and couldn't make any more entries in the test puzzle. Is there a way to reload it without making all the entries individually, so that I can continue to get hints?
I really do appreciate your help. Thank you again.
Emily |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Fri Feb 17, 2006 8:15 pm Post subject: Here are some hints |
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Hi, Emily!
Let's take the easy question first. You can use the "copy to draw" button from the archived puzzle on this site to get all the original boxes filled in. From there I'm afraid you'll have to reenter the items you've figured out already ... but as long as you haven't gotten too far along this will only be a few values to be rekeyed.
OK, now for the trickier part. I'm not too sure exactly where you are on this puzzle, so have taken it up at a point where only the most obvious values have been entered.
| Code: | 359 579 6 237 247 1 2345 8 2345
13 2 4 5 8 9 13 7 136
135 578 138 237 2467 2467 12345 13456 9
7 4 13 6 9 5 8 13 123
369 689 38 127 127 27 3459 34569 3456
2 69 5 4 3 8 19 169 7
8 56 27 127 124567 2467 13459 13459 1345
456 1 9 8 456 3 7 2 45
45 3 27 9 12457 247 6 145 8 |
The first thing you'll notice is that I've listed the possibilities at r1c5 as {3, 6, 9} and not as the {1, 3, 6, 9} shown in your earlier post. This is true because the values {1, 2, 7} must lie in r5c4, r5c5, & r5c6 -- all but three of the cells in the middle center 3x3 box being already filled in. So there can't be a "1" at r5c1.
So there's only one place in the middle left 3x3 box where a "1" can fit -- at r4c3. And this implies that r4c8 = 3, and also that r4c9 = 2. Now we have completed row 4, and the matrix looks like this.
| Code: | 359 579 6 237 247 1 2345 8 345
13 2 4 5 8 9 13 7 136
135 578 38 237 2467 2467 12345 1456 9
7 4 1 6 9 5 8 3 2
369 689 38 127 127 27 459 4569 456
2 69 5 4 3 8 19 169 7
8 56 27 127 124567 2467 13459 1459 1345
456 1 9 8 456 3 7 2 45
45 3 27 9 12457 247 6 145 8 |
Now it's pretty clear that r2c9 = 6 (it's the only place a "6" can fit in row 2) -- this creates the pair {4, 5} in column 9, implying that r1c9 = 3 and that r7c9 = 1.
| Code: | 359 579 6 27 247 1 245 8 3
13 2 4 5 8 9 1 7 6
135 578 38 3 2467 2467 1245 456 9
7 4 1 6 9 5 8 3 2
369 689 38 127 127 27 459 4569 45
2 69 5 4 3 8 19 169 7
8 56 27 27 2456 246 3459 459 1
456 1 9 8 456 3 7 2 45
45 3 27 9 12457 247 6 45 8 |
Now all sorts of good things start happening. The "1" at r7c9 has uncovered the {2, 7} pair in row 7, and also in column 4 (because of the "3" at r1c9), allowing us to set a "3" at r3c4, which in turn forces an "8" at r3c3. The "3" at r1c9 also forces a "1" at r2c7, which in turn forces a "3" at r2c1. At this point the matrix looks like this.
| Code: | 59 579 6 27 247 1 245 8 3
3 2 4 5 8 9 1 7 6
15 57 8 3 2467 2467 245 456 9
7 4 1 6 9 5 8 3 2
69 689 3 127 127 27 459 4569 45
2 6 5 4 3 8 9 16 7
8 56 27 27 2456 246 3459 459 1
456 1 9 8 456 3 7 2 45
45 3 27 9 12457 247 6 45 8 |
Now the puzzle is falling apart -- you can stick a fork in it. Clearly we must have r3c1 = 1 because it's the only spot for a "1" in the top left 3x3 box. And a great number of additional implications are also visible in this matrix -- I hope I haven't spoiled the puzzle for you! dcb |
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