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BenMac
Joined: 24 Jun 2007
Posts: 5
Location: Adelaide, South Australia
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 Posted: Sun Jun 24, 2007 4:51 am Post subject: New reader - understanding terminology. |
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Hi fellow Su Dokans,
I have been doing these puzzles for quite some time using my own spread sheet to help solve them by hilighting a single digit across the grid.
I have been reading your discussions for a short time, and would like to know where I can find an apendix of all of your descriptive terms which are confusing to new readers. e.g. xy-chain, y-chains, y-wing chains, xy-chain, xy-ring, forcing chain etc.
I have been doing today's puzzle (Sun 24-Jun-2007), and have arrived at the following position.
+------------+----------------+------------+
| 14 58 24 | 28 9 158 | 6 3 7 |
| 58 36 7 | 4 368 358 | 2 1 9 |
| 19 36 29 | 236 1236 7 | 5 8 4 |
+------------+----------------+------------+
| 7 4 6 | 38 38 2 | 9 5 1 |
| 589 58 589 | 1 7 6 | 34 24 23 |
| 2 1 3 | 9 5 4 | 8 7 6 |
+------------+----------------+------------+
| [b]468[/b] 7 48 | 5 12368 138 | 34 9 238 |
| 3 2 458 | 7 68 9 | 1 46 58 |
| 568 9 1 | 2368 4 38 | 7 26 2358 |
+------------+----------------+------------+
[code][url=http://www.dailysudoku.com/sudoku/play.shtml?
p=14:58:24:28:9:158:6:3:7:
58:36:7:4:368:358:2:1:9:
19:36:29:236:1236:7:5:8:4:
7:4:6:38:38:2:9:5:1:
589:58:589:1:7:6:34:24:23:
2:1:3:9:5:4:8:7:6:
468:7:48:5:12368:138:34:9:238:
3:2:458:7:68:9:1:46:58:
568:9:1:2368:4:38:7:26:2358:][/code]
My question is 'when I ask for a hint I get R7C1=6', which I can't reconcile, as R9C1=6 & R7C5=6, and I can't see how they are eliminated.
Generally when I get to this position I save a copy, then with a pair I try one digit, and if it doesn't work go back to the saved copy and try the other, but this doesn't seem quite logical.
Thanks,
BenMac. |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Sun Jun 24, 2007 11:20 am Post subject: |
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Hello, Ben
| Code: | +------------+----------------+------------+
| 14 58 24 | 28 9 158 | 6 3 7 |
| 58 36 7 | 4 368 358 | 2 1 9 |
| 19 36 29 | 236 1236 7 | 5 8 4 |
+------------+----------------+------------+
| 7 4 6 | 38 38 2 | 9 5 1 |
| 589 58 589 | 1 7 6 | 34 24 23 |
| 2 1 3 | 9 5 4 | 8 7 6 |
+------------+----------------+------------+
| 468 7 48 | 5 12368 138 | 34 9 238 |
| 3 2 458 | 7 68 9 | 1 46 58 |
| 568 9 1 | 2368 4 38 | 7 26 2358 |
+------------+----------------+------------+ |
If you look at the fifth column, you will see that 1 and 2 can be placed in only two cells, r3c5 and r7c5. Between them 1 and 2 must therefore occupy these cells and any other candidates can be eliminated from them. Allowing for the pair (12) in this way leaves r7c1 as the only the cell in the seventh row with room for a 6.
You can read about pairs and x-wings at SadMan or Sudopedia.
Steve |
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BenMac
Joined: 24 Jun 2007
Posts: 5
Location: Adelaide, South Australia
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| Posted: Sun Jun 24, 2007 11:51 am Post subject: Makes sense |
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Thanks Steve,
I can see your logic, I have been using 'clean pairs' only. I will now look for pairs with other candidates in the way.
I shall read Sadman & Sudopedia.
Regards,
Ben. |
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PouLeeps
Joined: 19 Jun 2007
Posts: 6
Location: Lismore, Australia
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| Posted: Sun Jun 24, 2007 12:31 pm Post subject: Re: New reader - understanding terminology. |
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[code]+------------+----------------+------------+
| 14 58 24 | 28 9 158 | 6 3 7 |
| 58 36 7 | 4 368 358 | 2 1 9 |
| 19 36 29 | 236 1236 7 | 5 8 4 |
+------------+----------------+------------+
| 7 4 6 | 38 38 2 | 9 5 1 |
| 589 58 589 | 1 7 6 | 34 24 23 |
| 2 1 3 | 9 5 4 | 8 7 6 |
+------------+----------------+------------+
| 468 7 48 | 5 12368 138 | 34 9 238 |
| 3 2 458 | 7 68 9 | 1 46 58 |
| 568 9 1 | 2368 4 38 | 7 26 2358 |
+------------+----------------+------------+
Hi Ben,
I'm pretty new to this and I'm not sure if this was a proper application but I think I used a swordfish on the 6s in columns 2,4 and 8. This should remove 6s from rows 3 and 9. This leaves only one possible 6 in block 7.
I also have struggled trying to learn the terms and have just googled each term I come across. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Jun 24, 2007 3:50 pm Post subject: |
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| Quote: | | I can see your logic, I have been using 'clean pairs' only. I will now look for pairs with other candidates in the way. |
Ben,
What Steve described is a "hidden pair", a perfectly valid and useful technique. I have trouble with hidden pairs, but they are complemented by other subsets. In this case, note that column 5 contains a 368-38-68 triple, or what you might call a "clean triple." Eliminating the 368 candidates from other cells in the column exposes the 12 pair.
Just two ways of looking at the same thing. |
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jLo
Joined: 30 Apr 2007
Posts: 55
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| Posted: Sun Jun 24, 2007 7:55 pm Post subject: |
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| Quote: |
I have trouble with hidden pairs, but they are complemented by other subsets.
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I agree that the complementing subsets are usually easier to spot than the hidden
pairs, but there are rare occations when a hidden pair exists without a complementing
subset. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Jun 24, 2007 8:53 pm Post subject: |
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| Quote: | | but there are rare occations when a hidden pair exists without a complementing subset. |
That I didn't know. Next time you come across one, I'd appreciate it if you'd post it or PM me. Thanks. |
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jLo
Joined: 30 Apr 2007
Posts: 55
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| Posted: Sun Jun 24, 2007 9:28 pm Post subject: |
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| Quote: |
That I didn't know. Next time you come across one, I'd appreciate it if you'd post it or PM me. Thanks.
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This is from the 6/5 very hard. I was pretty surprised this could happen to. There
is a hidden pair in row 8. If there is a corresponding subset, I can't find it.
| Code: |
+-------------+-----------------+-------------+
| 17 378 | 1478 48 | 49 39 |
| 18 238 | 146 2468 | 146 36 |
| 27 | 167 26 | 16 |
+-------------+-----------------+-------------+
| | | |
| | 46 46 | |
| | | 26 68 268 |
+-------------+-----------------+-------------+
| 24 38 | 468 346 | 246 |
| 247 138 | 478 1347 | 249 48 289 |
| 147 78 | 14678 468 | 68 |
+-------------+-----------------+-------------+
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Sun Jun 24, 2007 10:08 pm Post subject: |
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The complementary set is made up of the other candidates, here (24789).
Steve |
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jLo
Joined: 30 Apr 2007
Posts: 55
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| Posted: Mon Jun 25, 2007 2:56 am Post subject: |
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| Quote: |
The complementary set is made up of the other candidates, here (24789).
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I stand corrected.
I thought at one time I had proven to myself that hidden pairs/triples ... ran
in parallel with set theory eliminations. I just couldn't get it to work out in this
particular case. |
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BenMac
Joined: 24 Jun 2007
Posts: 5
Location: Adelaide, South Australia
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| Posted: Mon Jun 25, 2007 6:32 am Post subject: |
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Sorry PouLeeps, I haven't got to the paragraph on swordfish yet, so I will try to decypher your answer when I get to read about it. As I said in my original, I'm not used to your terminology. SadMan is in complexity order, and Sudopedia is in Alpha order.
Marty, I understood Steve's description of the 'hidden pair', and up till now was only using obvious pairs, but will now look for hidden ones. I will have to identify triples, and practice with them for a while. I can see both logics. Thanks.
jLo,
| Quote: | | I agree that the complementing subsets are usually easier to spot than the hidden pairs, but there are rare occations when a hidden pair exists without a complementing subset. |
In 5/6VH why do we need to identify the complementing subset, when we can see the 13 hidden pair? Don't we just eliminate the 8 & 47?
Thanks again Guys. |
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jLo
Joined: 30 Apr 2007
Posts: 55
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| Posted: Mon Jun 25, 2007 1:28 pm Post subject: |
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| Quote: |
In 5/6VH why do we need to identify the complementing subset, when we can see the 13 hidden pair? Don't we just eliminate the 8 & 47?
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The question was not about the "need" for a complementing set, but
rather whether one exists or not. In the general case it is this -
1. If we have a group of cells with a hidden pair/triple/quad/...
can we always find a corresponding closed set within that
group of cells.
2. If we have a group of cells with a closed set,
can we always find a corresponding hidden pair/triple/quad/...
within that group of cells.
In other words, are closed sets and hidden pairs/triples/quads/...
just different ways to view the same solution.
With hidden pairs/triples/quads/... we look for a set of N candidate
values that occur in no more than N different cells within a row/column/block.
If we find them, we can eliminate any non-candidate value from these
N cells.
With closed sets we look for a set of N candidate cells that are made up
of no more that N different values within a row/column/block.
If we find them, we can eliminate any of these values from any non-candidate
cell.
As it turns out, the cells that contain the hidden values with the hidden
pair/triple/quad/... strategy are the non-candidate cells when we
employ the closed set strategy. Also, the values that are found in our
candidate cells with the closed set strategy are the non-candidate
values when we employ the hidden pair/triple/quad/... strategy.
In this case -
1. hidden pair
We have a set of 2 values {1, 3} that occur in 2 different cells
{R8C2, R8C5} within row 8. The non-candidate values are
{2, 4, 7, 8, 9}. We can eliminate 8 from R8C2 and
{4, 7} from R8C5.
2. closed set
We have a set of 5 cells {R8C1, R8C4, R8C7, R8C8, R8C9} that are made
up of 5 different values {2, 4, 7, 8, 9} within row 8. Our non-candidate
cells are {R8C2, R8C5}. We can eliminate 8 from R8C2, and {4, 7}
from R8C5.
Same result. Different view. You say tomato, I say . . .uhhh tomato.
In general I find the closed set easier to spot. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Jun 25, 2007 3:03 pm Post subject: |
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| Quote: | | In 5/6VH why do we need to identify the complementing subset, when we can see the 13 hidden pair? Don't we just eliminate the 8 & 47? |
There is no need, it's just a matter of how an individual sees things. As jlo said, you say toh-MAY-toh, I say toh-MAH-toh. I have difficulty spotting hidden pairs and triples, but do pretty well at spotting naked triples, quads and quints. Both lead to the same result, so each of us does what works for us. If someone's good as spotting the hiddens, then by all means that's what they should look for. |
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BenMac
Joined: 24 Jun 2007
Posts: 5
Location: Adelaide, South Australia
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| Posted: Mon Jun 25, 2007 3:27 pm Post subject: |
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Okay,
24/6
| Code: |
+----------------+---------------+------------+
| 14 58 24 | 28 9 158 | 6 3 7 |
| 58 36 7 | 4 368 358 | 2 1 9 |
| 19 36 29 | 36 12 7 | 5 8 4 |
+----------------+---------------+------------+
| 7 4 6 | 38 38 2 | 9 5 1 |
| 589 58 589 | 1 7 6 | 34 24 23 |
| 2 1 3 | 9 5 4 | 8 7 6 |
+----------------+---------------+------------+
| 6 7 48 | 5 12 138 | 34 9 238 |
| 3 2 458 | 7 68 9 | 1 46 58 |
| 58 9 1 | 2368 4 38 | 7 26 2358 |
+----------------+---------------+------------+
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Can I remove 8 from R2C4 and 3 from R2C5 as that will leave 2 naked pairs?
Or am I still missing something?
e.g the 3 is in the 36 and the 8 is in the 58.
Cheers,
Ben. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Jun 25, 2007 4:45 pm Post subject: |
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| BenMac wrote: | Okay,
24/6
| Code: |
+----------------+---------------+------------+
| 14 58 24 | 28 9 158 | 6 3 7 |
| 58 36 7 | 4 368 358 | 2 1 9 |
| 19 36 29 | 36 12 7 | 5 8 4 |
+----------------+---------------+------------+
| 7 4 6 | 38 38 2 | 9 5 1 |
| 589 58 589 | 1 7 6 | 34 24 23 |
| 2 1 3 | 9 5 4 | 8 7 6 |
+----------------+---------------+------------+
| 6 7 48 | 5 12 138 | 34 9 238 |
| 3 2 458 | 7 68 9 | 1 46 58 |
| 58 9 1 | 2368 4 38 | 7 26 2358 |
+----------------+---------------+------------+
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Can I remove 8 from R2C4 and 3 from R2C5 as that will leave 2 naked pairs?
Or am I still missing something?
e.g the 3 is in the 36 and the 8 is in the 58.
Cheers,
Ben. |
I don't think I can answer directly because there may be a typo, since r2c4 is already solved. In column 1 there is a naked pair 58 which solves the rest of the column, but I don't offhand see any other eliminations based on basic moves like subsets. |
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BenMac
Joined: 24 Jun 2007
Posts: 5
Location: Adelaide, South Australia
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| Posted: Tue Jun 26, 2007 3:32 am Post subject: |
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Sorry Marty,
| Quote: | | I don't think I can answer directly because there may be a typo, since r2c4 is already solved. In column 1 there is a naked pair 58 which solves the rest of the column, but I don't offhand see any other eliminations based on basic moves like subsets. |
Can't count, I meant r2c5=368 remove 8, and r2c6=358 remove 3.
Cheers,
Ben. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Jun 26, 2007 3:58 am Post subject: |
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| Quote: | | Can't count, I meant r2c5=368 remove 8, and r2c6=358 remove 3. |
There are moves that can be made in box 2 from the 58 pair in column 1, but I don't see a way to directly remove that 8 and 3. |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Wed Jun 27, 2007 8:18 pm Post subject: |
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Hi,
glad to see, that this forum is still living (again) 
| BenMac wrote: |
| Code: |
+----------------+---------------+------------+
| 14 58 24 | 28 9 158 | 6 3 7 |
| 58 36 7 | 4 368 358 | 2 1 9 |
| 19 36 29 | 36 12 7 | 5 8 4 |
+----------------+---------------+------------+
| 7 4 6 | 38 38 2 | 9 5 1 |
| 589 58 589 | 1 7 6 | 34 24 23 |
| 2 1 3 | 9 5 4 | 8 7 6 |
+----------------+---------------+------------+
| 6 7 48 | 5 12 138 | 34 9 238 |
| 3 2 458 | 7 68 9 | 1 46 58 |
| 58 9 1 | 2368 4 38 | 7 26 2358 |
+----------------+---------------+------------+
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Can I remove 8 from R2C4 and 3 from R2C5 as that will leave 2 naked pairs?
Or am I still missing something? |
No, you cant. What naked pairs ?
What you also missed is that with the 6 in r7c1 (however you got it - with Steve's hidden pair?) the 4 is a single in column 1.
To repeat an old post: When solving without pencilmarks, hidden pairs are often easier to spot than naked singles.
Look at this one:
+-------+-------+-------+
| x x x | x 9 x | 6 3 7 |
| x x 7 | 4 x x | 2 1 9 |
| x x x | x x 7 | 5 8 4 |
+-------+-------+-------+
| 7 4 6 | x x 2 | 9 5 1 |
| x x x | 1 7 6 | x x x |
| 2 1 3 | 9 5 4 | 8 7 6 |
+-------+-------+-------+
| x 7 x | 5 x x | x 9 x |
| 3 2 x | 7 x 9 | 1 x x |
| x 9 1 | x 4 x | 7 x x |
+-------+-------+-------+
In column 5 1 and 2 only can (and must) be in rows 3 and 7. Because in row 7 the 6 can only be in column 1 or 5, it must be in r7c1.
Btw, i dont solve without pencilmarks, but i just note cells with only 2 candidates (as long as i can). |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Wed Jun 27, 2007 10:55 pm Post subject: |
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PouLeeps,
I don't believe that your Swordfish exists. For Swordfish, a candidate digit must be limited to 3 rows in 3 columns (or vice versa). The 3 columns you mention (2, 4,  have <6> in 4 rows (2, 3, 8, 9).
I don't see a Swordfish on 6 in the puzzle in any of the rows or columns.
BenMac,
Just to be sure it's clear: You can only remove digits to reveal a Naked Pair if it was a Hidden Pair ("disrobe" it, in other words). You can't remove digits to create a Naked Pair where no Hidden Pair existed. |
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PouLeeps
Joined: 19 Jun 2007
Posts: 6
Location: Lismore, Australia
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| Posted: Thu Jun 28, 2007 12:23 am Post subject: |
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Thanks Asellus,
I didn't really understand what a swordfish was I realize now. I better stick to more basic techniques me thinks.  |
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