Need help with this one

Post the puzzle or solving technique that's causing you trouble and someone will help

Need help with this one

This is what I have so far:
Code: Select all
` *-----------* |..3|.46|18.| |174|853|296| |6.8|19.|..3| |---+---+---| |3.1|.6.|..8| |.6.|581|.3.| |8..|32.|6.1| |---+---+---| |43.|615|8.9| |.86|.3.|.1.| |.1.|478|36.| *-----------*`

What is the next step? Thanks.
shurd12

Posts: 21
Joined: 07 March 2006

Hi Shurd12.

In row 5, candidate "4" is restricted to box 6, so the others "4s" in that box can be eliminated. This allows you to put a "4" in column 8. Next you can eliminate one more candidate in box 6 due to locked candidates. Hope this help.

Regards, Carcul
Carcul

Posts: 724
Joined: 04 November 2005

Thanks. I should have seen that.
shurd12

Posts: 21
Joined: 07 March 2006

Re: Need help with this one

Stuck again
shurd12 wrote:
Code: Select all
` *-----------* |.93|.46|18.| |174|853|296| |6.8|19.|.43| |---+---+---| |3.1|.64|..8| |.6.|581|.3.| |84.|32.|6.1| |---+---+---| |43.|615|8.9| |.86|.3.|.1.| |.1.|478|36.| *-----------*`

shurd12

Posts: 21
Joined: 07 March 2006

Shurd12,

I'm assuming your candidate list looks like this:
Code: Select all
` *-----------------------------------------------------------* | 25    9     3     | 27    4     6     | 1     8     57    | | 1     7     4     | 8     5     3     | 2     9     6     | | 6     25    8     | 1     9     27    | 57    4     3     | |-------------------+-------------------+-------------------| | 3     25    1     | 79    6     4     | 579   257   8     | | 279   6     279   | 5     8     1     | 479   3     247   | | 8     4     579   | 3     2     79    | 6     57    1     | |-------------------+-------------------+-------------------| | 4     3     27    | 6     1     5     | 8     27    9     | | 2579  8     6     | 29    3     29    | 457   1     2457  | | 259   1     259   | 4     7     8     | 3     6     25    | *-----------------------------------------------------------*`

Colouring on digit 5 leads to exclusion of 5 in r9c9 and placement of 2 there.

Tracy
TKiel

Posts: 209
Joined: 05 January 2006

I solved it with 2 xy wings and 1 locked candidate.
First xy wing eliminated 2 at R9C1
Locked 2 in box 7 eliminated 2 at R5C3
Second xy wing eliminated 2 at R9C3
Hud

Posts: 570
Joined: 29 October 2005

Thanks.
shurd12

Posts: 21
Joined: 07 March 2006

Hud: Could you please spell out the xy wings you used? I don't see it. Thanks
shurd12

Posts: 21
Joined: 07 March 2006

shurd12 wrote:Hud: Could you please spell out the xy wings you used? I don't see it. Thanks

I don't see the xy-wings either. Note that you also have a naked pair of candidates 2 and 9 in r8c4,6.
lb2064

Posts: 19
Joined: 29 March 2006

I was afraid I'd have to do this, but here goes:

Code: Select all
`+------------+----------+--------------+ | 25  9  3   | 27  4 6  | 1   8   57   |  | 1   7  4   | 8   5 3  | 2   9   6    | | 6   25 8   | 1   9 27 | 57  4   3    | +------------+----------+--------------+ | 3   25 1   | 279 6 4  | 579 257 8    | | 279 6  279 | 5   8 1  | 479 3   247  | | 8   4  579 | 3   2 79 | 6   57  1    | +------------+----------+--------------+ | 4   3  27  | 6   1 5  | 8   27  9    | | 57  8  6   | 29  3 29 | 457 1   457  | | 259 1  259 | 4   7 8  | 3   6   25   | +------------+----------+--------------+`

This is my candidate list after some minor eliminations.

My first xy wing was 57 at R8C1, 25 at R1C1, and 27 at R7C3. There are no elims in C3 but in C1 the 2 can be eliminated at R9C1 (I believe).

That creates the locked 2s in box 7 eliminating 2 at R5C3.

I'm having trouble reconstructing the last xy wing but when and if I locate it, I'll add it to this. If I've made an invalid assumption, please let me know.

OK, I believe I found it. After making the previous elims, the next xy wing is formed by:
57 at R8C1
25 at R1C1
27 at R7C3
Eliminate the 2 at R9C3 in similar fashion to the first xy wing.
Last edited by Hud on Thu Mar 30, 2006 1:25 am, edited 1 time in total.
Hud

Posts: 570
Joined: 29 October 2005

Hud wrote:...My first xy wing was 57 at R8C1, 25 at R1C1, and 27 at R7C3. There are no elims in C3 but in C1 the 2 can be eliminated at R9C1 (I believe).....

That was neat! I missed your first xy altogether. There actually another xy on the other side of the board formed by r9c9, with the branches r7c8 and r1c9. This eliminate a 7 at r8c9.
lb2064

Posts: 19
Joined: 29 March 2006

Nicely done, I didn't spot that one and believe me, I looked for it. I located my other xy wing and edited my previous post. Time for bed, but that was a nice exercise.
Hud

Posts: 570
Joined: 29 October 2005

Hud wrote:.....
OK, I believe I found it. After making the previous elims, the next xy wing is formed by:
57 at R8C1
25 at R1C1
27 at R7C3
Eliminate the 2 at R9C3 in similar fashion to the first xy wing.

Don't think that works. Both branches must be able to eliminate the target candidate. If r8c1 is 5 then r1c1 becomes a 2 but it cannot eliminate r9c3.

Yeah.. time to go to bed on my end as well. Nice discussion. Helps me to understand things.
Thanks!
lb2064

Posts: 19
Joined: 29 March 2006

My candidates have a 79 in c7r4 and a 25 in c8r4. But, on the main point, I am relatively new at this and I don't understand the logic of the xy-wing you describe. Maybe I'm thinking of an x-wing where everything has to line up vertically? If you would be so kind to explain this to me. Or, give me a link to an explanaion. Thanks.
shurd12

Posts: 21
Joined: 07 March 2006

An xy-wing is a 3-cell chain where each cell has two candidates in it. The chain looks something like ab-bc-ca, where the bc cell sees both the ab and the ca cells. It is easy to prove in a chain like this, that one end or the other must equal 'a'. Therefore, any cell which sees both endpoints of the chain cannot be an 'a'.
Myth Jellies

Posts: 593
Joined: 19 September 2005

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