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kragzy
Joined: 01 May 2007
Posts: 112
Location: Australia
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 Posted: Fri Oct 26, 2007 11:42 pm Post subject: Oct 27 NVH |
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It's happened again. A VH that was solved in a few minutes with basic skills. I'll do it again just to make sure that I didn't jag a lucky mistake.
Cheers |
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kragzy
Joined: 01 May 2007
Posts: 112
Location: Australia
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| Posted: Sat Oct 27, 2007 3:28 am Post subject: |
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Well, I had a go from another direction and ended up with a very difficult puzzle! Still not solved - have to get back to work so will try it again tomorrow.
First go must have been a fluke. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Oct 27, 2007 5:11 am Post subject: |
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| After the basics, I started looking for XY-Wings and quickly found one that didn't eliminate anything. However, coloring on one of the pincers broke it wide open. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Oct 27, 2007 11:47 am Post subject: |
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Seeems to be to be a standard "Very Hard".
The basics get you here:
| Code: | +----------------+----------------+----------------+
| 9 6 2 | 1 4 5 | 8 7 3 |
| 7 5 4 | 3 2 8 | 19 6 19 |
| 1 8 3 | 6 7 9 | 5 4 2 |
+----------------+----------------+----------------+
| 6 2 57 | 8 35 1 | 4 39 79 |
| 8 137 157 | 45 9 34 | 6 2 17 |
| 4 13 9 | 2 6 7 | 13 8 5 |
+----------------+----------------+----------------+
| 35 79 8 | 457 1 34 | 2 359 6 |
| 35 179 6 | 57 8 2 | 379 1359 4 |
| 2 4 17 | 9 35 6 | 37 135 8 |
+----------------+----------------+----------------+
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There are an X-wing, three XY-wings and two XYZ-wings in this grid.
Keith |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Sat Oct 27, 2007 1:03 pm Post subject: |
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| Certainly not an easy VH. I used an xy and xyz wing - then got stuck. But then I noticed there was only one triple cell left - a 457 in R7C4. There was also a 45 and a 57 in Col 4, so I used BUG+1 logic and stuck the 5 in R7C4. Was that just lucky? I thought BUG+1s applied to boxes not rows and columns. |
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sdq_pete
Joined: 30 Apr 2007
Posts: 119
Location: Rotterdam, NL
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| Posted: Sat Oct 27, 2007 2:12 pm Post subject: |
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The following did it for me:
| Code: | XY 357 R4C5
XYZ 137 R5C2
X 3 R49
XYZ 359 R8C7 |
Peter |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Sat Oct 27, 2007 3:42 pm Post subject: |
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| Coloring on '3' got rid of 3 in r78c8 and r9c7, from there an xy-wing (35-59-39 pivot r7c8). Not too easy for me. |
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Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
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| Posted: Sat Oct 27, 2007 10:18 pm Post subject: |
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| I used two xy-wings(pivot in R4C5, and R8C7) an x-wing on <3> in R49 and xyz-wing [137] pivot in R5C2 |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Oct 28, 2007 12:33 am Post subject: |
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The key is those <3>s, it seems to me. Folks eliminated them in various ways. But, here's another way to see it:
Finned X-Wing in C58: Fin is R78C8. Eliminates <3> in R9C7. This creates...
X-Wing in R49: Eliminates the two "fin" <3>s from above.
After that, there is just that BUG+1.
cgordon,
I'm not sure if I'm understanding your BUG+1 question. However, this may answer it: In a BUG+1, one of the trivalue cell digits occurs 3 times in all 3 of the houses to which the trivalue cell belongs. In this case, there are 3 <5>s in B8, 3 <5>s in R7, and 3 <5>s in C4. You only need to spot any one of these to place the <5> in the trivalue cell.
[Edit to correct name.] |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Sun Oct 28, 2007 12:05 pm Post subject: |
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| Asellus: Yeah that answers it. I didn't know if BUG+1's applied to columns and rows as well as boxes. I do things by rote: so henceforth if there is only one xyz triple cell left in the puzzle , I will look for the xy and xz - the stick the x in the triple cell - without question. |
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DennyOR
Joined: 12 Sep 2007
Posts: 33
Location: Portland, Oregon
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| Posted: Sun Oct 28, 2007 8:23 pm Post subject: |
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From Keith's position:
XY Wing: (r8c4) 75-53-37 (r9c7)
- eliminates 7 from r8c7
- r9c7 -> 7
- r9c3 -> 1
XY Chain: (r9c8) 53-39-97-75 (r8c4)
- r9c5 -> 3 |
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re'born
Joined: 28 Oct 2007
Posts: 80
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| Posted: Sun Oct 28, 2007 11:00 pm Post subject: |
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This puzzle solves at once with xyz-transport.
| Code: | *-----------------------------------------------------------*
| 9 6 2 | 1 4 5 | 8 7 3 |
| 7 5 4 | 3 2 8 | 19 6 19 |
| 1 8 3 | 6 7 9 | 5 4 2 |
|-------------------+-------------------+-------------------|
| 6 2 57 | 8 35a 1 | 4 39* 79 |
| 8 137 157 | 45 9 34 | 6 2 17 |
| 4 13 9 | 2 6 7 | 13 8 5 |
|-------------------+-------------------+-------------------|
| 35* 79 8 | 457 1 34- | 2 359* 6 |
| 35 179 6 | 57 8 2 | 379 1359 4 |
| 2 4 17 | 9 35A 6 | 37 135 8 |
*-----------------------------------------------------------*
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Note the seemingly useless xyz-wing in r4c8, r7c18. The implication of such an xyz-wing is that one of the cells must be a 3. We now transport these 3's around the grid to create a deduction.
The simplest way to do so is to note the weak link r4c8 -3- r4c5 and then the strong link r4c5 =3= r9c5. The combination implies that if r9c5<>3, then r4c8 <> 3. So we now know that r7c1 = 3, r7c8 = 3 or r9c5 = 3. This immediately implies that r7c6 <> 3, solving the puzzle.
If you want to try it some more, you can show that each of the 3's in the xyz-wing above can be transported to r9c5. In particular, this shows that r9c5 = 3 (of course, that already followed from the previous deduction, but it's still pretty cool). |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Mon Oct 29, 2007 1:05 pm Post subject: |
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| I followed Re'borns logic but it seems like an awful lot of "what ifs" to prove the strong/weak links. Looks like "logical trial and error". Why not just go to the first available pair - a 19 - then assume it's a 1. Less headache. |
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re'born
Joined: 28 Oct 2007
Posts: 80
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| Posted: Mon Oct 29, 2007 3:26 pm Post subject: |
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| cgordon wrote: | | I followed Re'borns logic but it seems like an awful lot of "what ifs" to prove the strong/weak links. Looks like "logical trial and error". Why not just go to the first available pair - a 19 - then assume it's a 1. Less headache. |
There are no more what ifs than in any other technique. A strong link exists if there only two places in a unit the candidate can go (hence the strong link on 3 in column 5) and a weak links exists if there are at least two places in a unit the candidate can go (hence the weak link on 3 in row 4). From there I only have to note that a strong link followed by a weak link preserves negation of a candidate, i.e., rAcB =x= rA'cB' -x- rA''cB'' tells us that rAcB <x> rA''cB'' <> x, or in this puzzle, if 3 is not in r9c5 then 3 is not in r4c8.
But...to each his own. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Oct 29, 2007 4:37 pm Post subject: |
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| Quote: | | A strong link exists if there only two places in a unit the candidate can go (hence the strong link on 3 in column 5) and a weak links exists if there are at least two places in a unit the candidate can go (hence the weak link on 3 in row 4). |
I have a persistent problem with this concept. A strong link, as I understood it, is where there are only two possibilities for a number. I can't see a difference between c5 and r4, why one is strong and the other weak. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Oct 29, 2007 5:50 pm Post subject: |
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I hope this is what we all can agree on (if not pls correct me, I'm certainly no expert):
"strong link" is eqivalent with "if cell x is NOT a then cell y is b"
"strong link" always implies "weak link", which is less demanding because weak link is equivalent to "if cell x IS a then cell y is not b".
strong link might be between the only 2 candidates in a single cell (x=y) or between two cells sharing the same unit OR between two cells in different units.
(nataraj's personal shortcut:)
strong links are "nicer to have" because they can stand in for weak links in a pinch.
With this interpretation in mind, I have no problem with the quote:
| Quote: | | weak link r4c8 -3- r4c5 and then the strong link r4c5 =3= r9c5. The combination implies that if r9c5<>3, then r4c8 <> 3 |
In the puzzle in question, there is a strong link standing in for a weak link.
The implication uses the "strong" property (if not, then) and subsequently the "weak" property "if, then not" to arrive at "if not, then not".
just my 2 (Euro-)cents' worth ...  |
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re'born
Joined: 28 Oct 2007
Posts: 80
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| Posted: Mon Oct 29, 2007 6:06 pm Post subject: |
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| Marty R. wrote: | | Quote: | | A strong link exists if there only two places in a unit the candidate can go (hence the strong link on 3 in column 5) and a weak links exists if there are at least two places in a unit the candidate can go (hence the weak link on 3 in row 4). |
I have a persistent problem with this concept. A strong link, as I understood it, is where there are only two possibilities for a number. I can't see a difference between c5 and r4, why one is strong and the other weak. |
If there are only two places to put a number in a unit, there is a strong link, but there can also between strong links between candidates that are not in the same unit! (see unique rectangles) The way I think about these ideas is with the logical operators OR and NAND. A strong link represents OR, a weak link NAND. In the case where there are exactly two places for a candidate to go in a unit (list these possibilities as A and B), then A OR B and A NAND B are both true, hence we can treat it as either a strong link or weak link.
In the original language of nice loops, the way they talked about things was using strong and weak links and inferences (see here for more on this point). So in my situation, they would say both are strong links, and to get from r4c8 to r4c5 I'm using the weak inference (associated to every strong link) while to get from r9c5 to r4c5, I'm using the strong inference (associated to every strong link).
I've forgone the nice loop language and just equate strong link with OR and weak link with NAND. |
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eddieg
Joined: 12 Jan 2006
Posts: 47
Location: San Diego, CA USA
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| Posted: Tue Oct 30, 2007 2:58 pm Post subject: Still have not solved |
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I have a feeling that I am one step away from solving last Saturday's puzzle. Note that I am not to "up" on coloring and some other advanced technigues (but willing to learn). Any help appreciated.
| Code: |
+----------+-----------+-----------+
| 9 6 2 | 1 4 5 | 8 7 3 |
| 7 5 4 | 3 2 8 | 19 6 19 |
| 1 8 3 | 6 7 9 | 5 4 2 |
+----------+-----------+-----------+
| 6 2 57 | 8 35 1 | 4 39 79 |
| 8 13 57 | 45 9 34 | 6 2 17 |
| 4 13 9 | 2 6 7 | 13 8 5 |
+----------+-----------+-----------+
| 35 79 8 | 457 1 34 | 2 359 6 |
| 35 79 6 | 57 8 2 | 39 1 4 |
| 2 4 1 | 9 35 6 | 7 35 8 |
+----------+-----------+-----------+
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Play this puzzle online at the Daily Sudoku site |
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re'born
Joined: 28 Oct 2007
Posts: 80
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| Posted: Tue Oct 30, 2007 3:54 pm Post subject: Re: Still have not solved |
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| eddieg wrote: | I have a feeling that I am one step away from solving last Saturday's puzzle. Note that I am not to "up" on coloring and some other advanced technigues (but willing to learn). Any help appreciated.
| Code: |
+----------+-----------+-----------+
| 9 6 2 | 1 4 5 | 8 7 3 |
| 7 5 4 | 3 2 8 | 19 6 19 |
| 1 8 3 | 6 7 9 | 5 4 2 |
+----------+-----------+-----------+
| 6 2 57 | 8 35 1 | 4 39 79 |
| 8 13 57 | 45 9 34 | 6 2 17 |
| 4 13 9 | 2 6 7 | 13 8 5 |
+----------+-----------+-----------+
| 35 79 8 | 457 1 34 | 2 359 6 |
| 35 79 6 | 57 8 2 | 39 1 4 |
| 2 4 1 | 9 35 6 | 7 35 8 |
+----------+-----------+-----------+
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Play this puzzle online at the Daily Sudoku site |
A fast way to solve it is to notice that it is a BUG+2 grid, i.e., every unsolved candidate occurs exactly twice in each row, column, and block, except for 2 of them, namely there is an extra 3 in r7c8 and an extra 5 in r7c4. The rule is that one of these two extra candidates must be true. Now note that r7c8=5 would simultaneously make both extra candidates false. Hence r7c8 <> 5, which reduces the puzzle to singles.
In case you don't like uniqueness stlye answers, there is an x-wing on the number 3 which eliminates 3 from r7c8. After this, there is an xy-wing that eliminates 3 from r8c1, solving the puzzle. |
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eddieg
Joined: 12 Jan 2006
Posts: 47
Location: San Diego, CA USA
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| Posted: Tue Oct 30, 2007 4:16 pm Post subject: Thanks re'born |
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I will study the unqiueness style response when time permits. In the meantime, the silly X-wing on 3's to eliminate the 3 in R7C8, then I saw an XY-wing involving cells R7C1, R7C8, and R8C7 that eliminated the 3 from R8C1, which totally opened up the puzzle.
Thanks for the time you put into the response. Appreciated. |
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