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Set C Puzzle 25

 
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Sat Nov 29, 2008 6:10 pm    Post subject: Set C Puzzle 25 Reply with quote

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

Play this puzzle online at the Daily Sudoku site
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storm_norm



Joined: 18 Oct 2007
Posts: 1741

PostPosted: Sat Nov 29, 2008 8:03 pm    Post subject: Reply with quote

Code:
.---------------------.---------------------.---------------------.
| 15     9      15    | 3      78     48    | 47     6      2     |
| 3      267    247   | 5      9      246   | 47     1      8     |
| 46     267    8     | 1      67     246   | 9      3      5     |
:---------------------+---------------------+---------------------:
| 2      678    17    | 9      168    68    | 3      5      4     |
| 14     3      9     | 24     5      7     | 12     8      6     |
| 14568  568    145   | 24     168    3     | 12     9      7     |
:---------------------+---------------------+---------------------:
| 58     258    25    | 7      3      1     | 6      4      9     |
| 9      4      3     | 6      2      5     | 8      7      1     |
| 7      1      6     | 8      4      9     | 5      2      3     |
'---------------------'---------------------'---------------------'

Quote:
xyz-wing {1,4,5} in box 1
x-wing 6
xy-wing {2,4,7}
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arkietech



Joined: 31 Jul 2008
Posts: 1717
Location: Northwest Arkansas USA

PostPosted: Sat Nov 29, 2008 9:05 pm    Post subject: Reply with quote

One stepper
Quote:
xy-chain solves it
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Sat Nov 29, 2008 10:42 pm    Post subject: Reply with quote

There is an interesting AIC that solves it in one step. First, note the non-matching digit "W-Wing" in r2c7 and r3c5 activated by the conjugate <7>s in r1. This creates the strong inference (6)r3c5=(4)r2c7. Think of it as a Medusa W-Wing.

Next, notice that this strong inference is joined up with the conjugate <4>s in b1 via the weakly linked <4>s in r2. Think of this as a Medusa Turbot Fish with the pincers (4)r3c1 and (6)r3c5. This eliminates ("traps") the <6> in r3c1.

In Eureka:
(6)r3c1 - (6=7)r3c5 - (7)r1c5=(7)r1c7 - (7=4)r2c7 - (4)r2c3=(4-6)r3c1; r3c1<>6
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Marty R.



Joined: 12 Feb 2006
Posts: 5160
Location: Rochester, NY, USA

PostPosted: Sun Nov 30, 2008 1:34 am    Post subject: Reply with quote

For what it's worth (exactly -0-), 14 can be removed from r6c1 to avoid the 12-24-14 DP.
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keith



Joined: 19 Sep 2005
Posts: 3165
Location: near Detroit, Michigan, USA

PostPosted: Mon Dec 01, 2008 12:12 am    Post subject: Reply with quote

What is this? (I have not read any messages in this thread.)

Starting this puzzle with pencil & paper, I got to here:
Code:
+---------+--------+--------+
| .  9 .  | 3 . .  | 47 6 2 |
| 3  . .  | . 9 .  | 47 1 8 |
| .  . 8  | . . .  | 9  3 5 |
+---------+--------+--------+
| 2  . 17 | 9 . .  | 3  5 4 |
| 14 3 9  | . 5 7  | 12 8 6 |
| .  . .  | . . 3  | 12 9 7 |
+---------+--------+--------+
| .  . .  | 7 3 .  | 6  4 9 |
| 9  4 3  | 6 2 58 | 58 7 1 |
| 7  . 6  | . 4 9  | 58 2 3 |
+---------+--------+--------+
Now,
Code:
+---------+--------+--------+
| .  9 .  | 3 . .  | 47 6 2 |
| 3  . .  | . 9 .  | 47 1 8 |
| .  . 8  | . . .  | 9  3 5 |
+---------+--------+--------+
| 2  . 17 | 9 . .  | 3  5 4 |
| 14 3 9  | . 5 7  | 12 8 6 |
| .  . .  | . . 3  | 12 9 7 |
+---------+--------+--------+
| .  . .  | 7 3 a  | 6  4 9 |
| 9  4 3  | 6 2 b  | c 7 1 |
| 7  a 6  | c 4 9  | b  2 3 |
+---------+--------+--------+

a and a must be the same, they have the same unsolved peers in R7B7. b and b must be the same, because of the <58> pairs in R8C7. So, it seems to me, c and c must be the same. b and c are not <1>, so a must be <1>.

I put this into Sudoku Susser. At this point in the puzzle, it has no deductions to make in these cells.

Keith
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Mon Dec 01, 2008 2:16 am    Post subject: Reply with quote

Code:
+---------+--------+--------+
| .  9 .  | 3 . .  | 47 6 2 |
| 3  . .  | . 9 .  | 47 1 8 |
| .  . 8  | . . .  | 9  3 5 |
+---------+--------+--------+
| 2  . 17 | 9 . .  | 3  5 4 |
| 14 3 9  | . 5 7  | 12 8 6 |
| .  . .  | . . 3  | 12 9 7 |
+---------+--------+--------+
| .  . .  | 7 3 a  | 6  4 9 |
| 9  4 3  | 6 2 b  | c  7 1 |
| 7  a 6  | C 4 9  | b  2 3 |
+---------+--------+--------+

Okay on "b = c" in [r8] and [c7].

Okay on "a = a" logic.

However, your logic is incorrect for "c = C".

"abC" form a Naked Triple, so "C" can have a third candidate in common with "a" ... and it does!

In fact, "a = C" is correct.
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keith



Joined: 19 Sep 2005
Posts: 3165
Location: near Detroit, Michigan, USA

PostPosted: Mon Dec 01, 2008 3:40 am    Post subject: Reply with quote

Danny,

I do not understand your comments. (When you say "=", I presume you mean "the same value in the solution".)

I said:
keith wrote:
Code:
+---------+--------+--------+
| .  9 .  | 3 . .  | 47 6 2 |
| 3  . .  | . 9 .  | 47 1 8 |
| .  . 8  | . . .  | 9  3 5 |
+---------+--------+--------+
| 2  . 17 | 9 . .  | 3  5 4 |
| 14 3 9  | . 5 7  | 12 8 6 |
| .  . .  | . . 3  | 12 9 7 |
+---------+--------+--------+
| .  . .  | 7 3 a  | 6  4 9 |
| 9  4 3  | 6 2 b  | c 7 1 |
| 7  a 6  | c 4 9  | b  2 3 |
+---------+--------+--------+

a and a must be the same, they have the same unsolved peers in R7B7. b and b must be the same, because of the <58> pairs in R8C7. So, it seems to me, c and c must be the same. b and c are not <1>, so a must be <1>.Keith

The solution is:
Code:
+-------+-------+-------+
| 5 9 1 | 3 7 8 | 4 6 2 |
| 3 6 2 | 5 9 4 | 7 1 8 |
| 4 7 8 | 1 6 2 | 9 3 5 |
+-------+-------+-------+
| 2 8 7 | 9 1 6 | 3 5 4 |
| 1 3 9 | 4 5 7 | 2 8 6 |
| 6 5 4 | 2 8 3 | 1 9 7 |
+-------+-------+-------+
| 8 2 5 | 7 3 1 | 6 4 9 |
| 9 4 3 | 6 2 5 | 8 7 1 |
| 7 1 6 | 8 4 9 | 5 2 3 |
+-------+-------+-------+

I am quite prepared to discuss the logic, but my deductions are correct!

My question is, perhaps, by what logic can we conclude c R9C4 is not <1>? With only the information I posted.

Keith
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Mon Dec 01, 2008 8:30 am    Post subject: Reply with quote

My apologies. I mistakenly thought you were talking about "sameness" in cell candidates instead of cell solutions.

When it comes to solutions. Here's what I see from the information available.

Code:
 your logic
 +---------+--------+--------+
 | .  9 .  | 3 . .  | 47 6 2 |
 | 3  . .  | . 9 .  | 47 1 8 |
 | .  . 8  | . . .  | 9  3 5 |
 +---------+--------+--------+
 | 2  . 17 | 9 . .  | 3  5 4 |
 | 14 3 9  | . 5 7  | 12 8 6 |
 | .  . .  | . . 3  | 12 9 7 |
 +---------+--------+--------+
 | .  . .  | 7 3 a  | 6  4 9 |
 | 9  4 3  | 6 2 b  | c  7 1 |
 | 7  a 6  | c 4 9  | b  2 3 |
 +---------+--------+--------+

Code:
 my logic
 +---------+--------+--------+
 | .  9 .  | 3 . .  | 47 6 2 |
 | 3  . .  | . 9 .  | 47 1 8 |
 | .  . 8  | . . .  | 9  3 5 |
 +---------+--------+--------+
 | 2  . 17 | 9 . .  | 3  5 4 |
 | 14 3 9  | . 5 7  | 12 8 6 |
 | .  . .  | . . 3  | 12 9 7 |
 +---------+--------+--------+
 | .  . .  | 7 3 c  | 6  4 9 |
 | 9  4 3  | 6 2 b  | c  7 1 |
 | 7  c 6  | a 4 9  | b  2 3 |
 +---------+--------+--------+

[Addendum using information from below] In both cases, a=1 must follow.


Last edited by daj95376 on Mon Dec 01, 2008 5:08 pm; edited 2 times in total
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Steve R



Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England

PostPosted: Mon Dec 01, 2008 1:13 pm    Post subject: Reply with quote

I suspect that Keithís argument is flawed though I canít be sure because some steps are missing and, as he says, it produces the right answer. Letís try to follow it through.

Start by calling the entries missing from box 8 a, b and c:

Code:
+---------+--------+--------+
| .  9 .  | 3 . .  | 47 6 2 |
| 3  . .  | . 9 .  | 47 1 8 |
| .  . 8  | . . .  | 9  3 5 |
+---------+--------+--------+
| 2  . 17 | 9 . .  | 3  5 4 |
| 14 3 9  | . 5 7  | 12 8 6 |
| .  . .  | . . 3  | 12 9 7 |
+---------+--------+--------+
| .  . .  | 7 3 a  | 6  4 9 |
| 9  4 3  | 6 2 b  | .  7 1 |
| 7  . 6  | c 4 9  | .  2 3 |
+---------+--------+--------+


Now, r9c2 contains a because a cannot be placed in the top row of box 7 and it cannot be 7, 3, 6, 4 or 9 by virtue of the entries in row 7. This is just as Keith says. So is the placement of b in r9c7 because of the (58) pairs. This brings us to:

Code:
+---------+--------+--------+
| .  9 .  | 3 . .  | 47 6 2 |
| 3  . .  | . 9 .  | 47 1 8 |
| .  . 8  | . . .  | 9  3 5 |
+---------+--------+--------+
| 2  . 17 | 9 . .  | 3  5 4 |
| 14 3 9  | . 5 7  | 12 8 6 |
| .  . .  | . . 3  | 12 9 7 |
+---------+--------+--------+
| .  . .  | 7 3 a  | 6  4 9 |
| 9  4 3  | 6 2 b  | .  7 1 |
| 7  a 6  | c 4 9  | b  2 3 |
+---------+--------+--------+

We know from box 8 that c is not 3, 4 or 9, so it cannot fall in the middle row of box 7. Accordingly it falls in the middle row of box 9.

Conclusion: c is to be placed in r8c7 or it is 1 or it is 7. The case 7 is excluded by the 7 already present in box 8. The only way I can see to exclude the possibility that c = 1 is by using the rest of the puzzle and I think this case has to be excluded in order to place c in r8c7. If it is in fact placed there, Keithís inference that a = 1 follows.

Dannyís logic must have gone awry because r8c7 and r9c2 contain different entries in the solution.

Steve
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Mon Dec 01, 2008 4:19 pm    Post subject: Reply with quote

Steve R wrote:
Conclusion: c is to be placed in r8c7 or it is 1 or it is 7. The case 7 is excluded by the 7 already present in box 8. The only way I can see to exclude the possibility that c = 1 is by using the rest of the puzzle and I think this case has to be excluded in order to place c in r8c7. If it is in fact placed there, Keithís inference that a = 1 follows.

I agree with you completely to here.

Quote:
Dannyís logic must have gone awry because r8c7 and r9c2 contain different entries in the solution.

I'm not talking about the final solution. I'm talking about the information Keith provided.

Code:
 reality vs. a-b-c for band 3
 |----------------------+----------------------+----------------------|
 | 158    1258   125    | 7      3      158    | 6      4      9      |
 | 9      4      3      | 6      2      58     | 58     7      1      |
 | 7      158    6      | 158    4      9      | 58     2      3      |
 *--------------------------------------------------------------------*

Code:
 scenario 1: assuming [r8c6]=5 and Keith's conclusion that [r7c6]=1
 |----------------------+----------------------+----------------------|
 | 58     258    25     | 7      3      1      | 6      4      9      |
 | 9      4      3      | 6      2      5      | 8      7      1      |
 | 7      1      6      | 8      4      9      | 5      2      3      |
 *--------------------------------------------------------------------*

 scenario 2: assuming [r8c6]=8 and Keith's conclusion that [r7c6]=1
 |----------------------+----------------------+----------------------|
 | 58     258    25     | 7      3      1      | 6      4      9      |
 | 9      4      3      | 6      2      8      | 5      7      1      |
 | 7      1      6      | 5      4      9      | 8      2      3      |
 *--------------------------------------------------------------------*

Code:
 scenario 3: assuming [r8c6]=5 and Danny's premise that [r9c4]=1
 |----------------------+----------------------+----------------------|
 | 15     125    125    | 7      3      8      | 6      4      9      |
 | 9      4      3      | 6      2      5      | 8      7      1      |
 | 7      8      6      | 1      4      9      | 5      2      3      |
 *--------------------------------------------------------------------*

 scenario 4: assuming [r8c6]=8 and Danny's premise that [r9c4]=1
 |----------------------+----------------------+----------------------|
 | 18     128    12     | 7      3      5      | 6      4      9      |
 | 9      4      3      | 6      2      8      | 5      7      1      |
 | 7      5      6      | 1      4      9      | 8      2      3      |
 *--------------------------------------------------------------------*

None of these scenarios can be excluded with the information Keith provided.

This also explains why Susser failed to provide any assignments/eliminations in band 3 for Keith.
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keith



Joined: 19 Sep 2005
Posts: 3165
Location: near Detroit, Michigan, USA

PostPosted: Mon Dec 01, 2008 7:30 pm    Post subject: Reply with quote

The jury's verdict seems to be I was lucky.

Guilty as charged!

Keith
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