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Set C Puzzle 25

 
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daj95376



Joined: 23 Aug 2008
Posts: 3854

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: 1834
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: 5770
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: 3355
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: 3854

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: 3355
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: 3854

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 Keith’s argument is flawed though I can’t be sure because some steps are missing and, as he says, it produces the right answer. Let’s 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, Keith’s inference that a = 1 follows.

Danny’s 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: 3854

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, Keith’s inference that a = 1 follows.

I agree with you completely to here.

Quote:
Danny’s 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: 3355
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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