| View previous topic :: View next topic |
| Author |
Message |
daj95376
Joined: 23 Aug 2008
Posts: 3854
|
 Posted: Sat Nov 29, 2008 6:10 pm Post subject: Set C Puzzle 25 |
 |
|
| 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 |
|
| Back to top |
|
 |
storm_norm
Joined: 18 Oct 2007
Posts: 1741
|
| Posted: Sat Nov 29, 2008 8:03 pm Post subject: |
|
|
| 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} |
|
|
| Back to top |
|
|
arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
|
| Posted: Sat Nov 29, 2008 9:05 pm Post subject: |
|
|
One stepper
| Quote: | | xy-chain solves it |
|
|
| Back to top |
|
|
Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
|
| Posted: Sat Nov 29, 2008 10:42 pm Post subject: |
|
|
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 |
|
| Back to top |
|
|
Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
|
| Posted: Sun Nov 30, 2008 1:34 am Post subject: |
|
|
| For what it's worth (exactly -0-), 14 can be removed from r6c1 to avoid the 12-24-14 DP. |
|
| Back to top |
|
|
keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
|
| Posted: Mon Dec 01, 2008 12:12 am Post subject: |
|
|
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 |
|
| Back to top |
|
|
daj95376
Joined: 23 Aug 2008
Posts: 3854
|
| Posted: Mon Dec 01, 2008 2:16 am Post subject: |
|
|
| 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. |
|
| Back to top |
|
|
keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
|
| Posted: Mon Dec 01, 2008 3:40 am Post subject: |
|
|
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 |
|
| Back to top |
|
|
daj95376
Joined: 23 Aug 2008
Posts: 3854
|
| Posted: Mon Dec 01, 2008 8:30 am Post subject: |
|
|
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 |
|
| Back to top |
|
|
Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
|
| Posted: Mon Dec 01, 2008 1:13 pm Post subject: |
|
|
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 |
|
| Back to top |
|
|
daj95376
Joined: 23 Aug 2008
Posts: 3854
|
| Posted: Mon Dec 01, 2008 4:19 pm Post subject: |
|
|
| 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. |
|
| Back to top |
|
|
keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
|
| Posted: Mon Dec 01, 2008 7:30 pm Post subject: |
|
|
The jury's verdict seems to be I was lucky.
Guilty as charged!
Keith |
|
| Back to top |
|
|
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
