Please help a newbie

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Please help a newbie

Postby RobA » Thu Apr 20, 2006 2:58 am

I was hoping to get some help with this puzzle. I am new to this so don't know any advanced tricks - or how to post my puzzle well.

**2 |*** |1**
*3* |629 |*8*
4** |*** |**9
----------------
**5 |4*8 |2**
9** |*** |**6
**4 |7*3 |9**
----------------
8** |*** |**4
*1* |846 |*3*
**6 |*** |8**

I made it as far as shown below and then solved it by guessing R1C4 to be a 5 (it could only be 3 or 5) but know there must be a better way.

*92 |*84 |1**
*31 |629 |48*
4** |*** |*29
----------------
**5 |498 |2**
9** |*** |*46
124 |763 |958
----------------
8** |*** |**4
*19 |846 |*3*
*46 |*** |8**[/code]
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Postby re'born » Thu Apr 20, 2006 5:02 am

RobA,

Here is a candidate grid for your puzzle:

Code: Select all
  *--------------------------------------------------------------------*
 |*567    9      2      |-35     8      4      | 1      67    *357    |
 |*57     3      1      | 6      2      9      | 4      8     *57     |
 | 4      56     8      | 135    1357   157    | 356    2      9      |
 |----------------------+----------------------+----------------------|
 | 36     67     5      | 4      9      8      | 2      17     137    |
 | 9      8      37     | 125    15     125    | 37     4      6      |
 | 1      2      4      | 7      6      3      | 9      5      8      |
 |----------------------+----------------------+----------------------|
 | 8      57     37     | 12359  135    125    | 56     169    4      |
 | 25     1      9      | 8      4      6      | 57     3      257    |
 | 235    4      6      | 1359   1357   157    | 8      19     125    |
 *--------------------------------------------------------------------*


It turns out that there is a very nice explanation for why your guess works. If you set r1c4=3, then the only remaining 3 in row 3 will be in r3c7. Now the only remaining 6 in box 3 will be in r1c8. Look closely now at the 4 cells r1c1, r2c1, r1c9, r2c9. Starting with the assumption r1c4=3, we get that all 4 of the above cells can only be 5 or 7. This is what is known as a deadly pattern and must be avoided at all costs. Therefore, r1c4 cannot be a 3 and hence must be a 5. As you pointed out, this solves the puzzle.

For more info on this idea, search for "unique rectangles".

Hope this helps.
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Postby re'born » Thu Apr 20, 2006 5:11 am

Oh, and if you want a bit of homework, try using the the cells in r8c1, r9c1, r8c9, r9c9 to deduce r9c8 = 9.

Code: Select all
 *--------------------------------------------------------------------*
 | 567    9      2      | 35     8      4      | 1      67     357    |
 | 57     3      1      | 6      2      9      | 4      8      57     |
 | 4      56     8      | 135    1357   157    | 356    2      9      |
 |----------------------+----------------------+----------------------|
 | 36     67     5      | 4      9      8      | 2      17     137    |
 | 9      8      37     | 125    15     125    | 37     4      6      |
 | 1      2      4      | 7      6      3      | 9      5      8      |
 |----------------------+----------------------+----------------------|
 | 8      57     37     | 12359  135    125    | 56     169    4      |
 |*25     1      9      | 8      4      6      | 57     3     *257    |
 |*235    4      6      | 1359   1357   157    | 8     -19    *125    |
 *--------------------------------------------------------------------*


This won't solve the puzzle like the other one did, but it is fun to look for nonetheless.
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Another Solution

Postby Carcul » Thu Apr 20, 2006 8:21 am

Hi RobA.

You can also use Almost Unique Rectangles (in this case, the one in cells r12c19):

Code: Select all
 *--------------------------------------------------------------------*
 | 567    9      2      | 35     8      4      | 1      67     357    |
 | 57     3      1      | 6      2      9      | 4      8      57     |
 | 4      56     8      | 135    1357   157    | 356    2      9      |
 |----------------------+----------------------+----------------------|
 | 36     67     5      | 4      9      8      | 2      17     137    |
 | 9      8      37     | 125    15     125    | 37     4      6      |
 | 1      2      4      | 7      6      3      | 9      5      8      |
 |----------------------+----------------------+----------------------|
 | 8      57     37     | 12359  135    125    | 56     169    4      |
 | 25     1      9      | 8      4      6      | 57     3      257    |
 | 235    4      6      | 1359   1357   157    | 8      19     125    |
 *--------------------------------------------------------------------*


[r4c1]=3=[r4c9]-3-[r1c9]=3|6=[r1c1]-6-[r4c1],

which implies that r4c1 cannot be "6" and that solves the puzzle.
Check this post for an explanation of the notation used, and read this post for a description of almost unique rectangles.

Regards, Carcul
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Re: Another Solution

Postby ronk » Thu Apr 20, 2006 10:40 am

Carcul wrote:You can also use Almost Unique Rectangles (in this case, the one in cells r12c19) ...

Even when used in a loop, I think it's still a Type 3 Unique Rectangle to most of us.
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Postby Carcul » Thu Apr 20, 2006 11:02 am

Ronk wrote:Even when used in a loop, I think it's still a Type 3 Unique Rectangle to most of us.


I don't see a bivalue cell with candidates "3,6" in row 1 in order for it to be a Type-3 Unique Rectangle. Do you Ronk? Well, in spite of the apparent fact that it looks an UR for most of "you", it is still an AUR for me. Good try Ronk, but you didn't convince me, sorry.:D
Also, I think what is really important is the logic behind a deduction, not its correct name.

Regards, Carcul
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Postby ronk » Thu Apr 20, 2006 11:18 am

Carcul wrote:
Ronk wrote:Even when used in a loop, I think it's still a Type 3 Unique Rectangle to most of us.

I don't see a bivalue cell with candidates "3,6" in row 1 in order for it to be a Type-3 Unique Rectangle. Do you Ronk? Well, in spite of the apparent fact that it looks an UR for most of "you", it is still an AUR for me. Good try Ronk, but you didn't convince me, sorry.:D

If a naked pair doesn't cause an exclusion in its unit (row, col, or box), but does cause an exclusion as part of an xy-chain, should we then call it an Almost Naked Pair?:D

Carcul wrote:Also, I think what is really important is the logic behind a deduction, not its correct name.

Yes, even though improperly named, it was a fine deduction.
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Postby tarek » Thu Apr 20, 2006 2:25 pm

As an alternative:
Code: Select all
*-----------------------------------------------------------------*
| 567    9      2     | 35     8      4     | 1      67    %357   |
| 57     3      1     | 6      2      9     | 4      8     %57    |
| 4      56     8     | 135    1357   157   |*356    2      9     |
|---------------------+---------------------+---------------------|
| 36     67     5     | 4      9      8     | 2      17     137   |
| 9      8      37    | 125    15     125   | 37     4      6     |
| 1      2      4     | 7      6      3     | 9      5      8     |
|---------------------+---------------------+---------------------|
| 8      57     37    | 12359  135    125   |*56     169    4     |
| 25     1      9     | 8      4      6     |^57     3     -257   |
| 235    4      6     | 1359   1357   157   | 8      19     125   |
*-----------------------------------------------------------------*
Eliminating 7 from r8c9(ALS-XY  A=356 in r7c7,r3c7   B=357 in r2c9,r1c9   C=57 in r8c7   x=3 y=5 z=7)

solving the puzzle

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Postby ronk » Thu Apr 20, 2006 4:10 pm

tarek wrote:Eliminating 7 from r8c9(ALS-XY A=356 in r7c7,r3c7 B=357 in r2c9,r1c9 C=57 in r8c7 x=3 y=5 z=7)

By combining sets A and C, the ALS-xz rule may be applied instead, i.e., ...

ALS-XZ A=3567 in r8c7,r7c7,r3c7 B=357 in r2c9,r1c9 x=3 z=7

Is your solver missing this case of the xz-rule ... or are you just giving higher priority to the xy-rule?
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Postby tarek » Thu Apr 20, 2006 4:43 pm

No... but it does give preference to smaller sets.....Th xy rule in my example uses triple while the xz in yours has quads......

I'm not sure if my POV is correct, because I'm not that good in spotting either ...........

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Thanks

Postby RobA » Fri Apr 21, 2006 10:10 pm

I appreciate the help - especially all the useful links.

rep'nA, I did have your candidate grid except I somehow missed all the 7 eliminations. I do see how whether r8c1 is a 2 or 5 you end up with r9c8=9 and r9c9=1.
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Postby lb2064 » Sat Apr 22, 2006 5:34 am

rep'nA wrote:Oh, and if you want a bit of homework, try using the the cells in r8c1, r9c1, r8c9, r9c9 to deduce r9c8 = 9.

Code: Select all
 *--------------------------------------------------------------------*
 | 567    9      2      | 35     8      4      | 1      67     357    |
 | 57     3      1      | 6      2      9      | 4      8      57     |
 | 4      56     8      | 135    1357   157    | 356    2      9      |
 |----------------------+----------------------+----------------------|
 | 36     67     5      | 4      9      8      | 2      17     137    |
 | 9      8      37     | 125    15     125    | 37     4      6      |
 | 1      2      4      | 7      6      3      | 9      5      8      |
 |----------------------+----------------------+----------------------|
 | 8      57     37     | 12359  135    125    | 56     169    4      |
 |*25     1      9      | 8      4      6      | 57     3     *257    |
 |*235    4      6      | 1359   1357   157    | 8     -19    *125    |
 *--------------------------------------------------------------------*


This won't solve the puzzle like the other one did, but it is fun to look for nonetheless.


I presume we have to show that [r9c8]=1 causes the deadly pattern of {25} in [r8c1], [r9c1], [r8c9] and [r9c9]; then [r9c8]<>1 => [r9c8]=9.

a) If [r9c8]=1 => [r9c9]<>1; => [r9c9] = {25}
b) If [r9c8]=1 => [r4c8]=7 =>[r5c7]<>7 =>[r8c7] = 7 => [r8c9]<>7; => [r8c9] = {25}
c) If [r9c8]=1 => [r4c8]=7 =>[r4c2]=6 => [r4c1]=3 => [r9c1] <> 3; => [r9c1] = {25}
Therefore, we get the deadly pattern so [r9c8]<>1 => [r9c8]=9.

Pretty neat exercise rep'nA. Creating the deadly pattern to eliminate a candidate is a new technique for me. Thanks for the initial explanation.

LB
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Postby re'born » Sat Apr 22, 2006 5:51 am

lb2064 wrote:Pretty neat exercise rep'nA. Creating the deadly pattern to eliminate a candidate is a new technique for me. Thanks for the initial explanation.
LB


LB,

Your welcome. If you want to see more examples, check out Carcul's use of Almost Unique Rectangles (which he cited above). Though Carcul often uses them in a slightly different fashion, many of his deductions can be translated into the form:

Set rXcY = a, then... and we get a deadly pattern. This is how I use AUR's for the most part. Also, I recommend checking out the thread on unique rectangles and x-wings. For me, the moral of the story is that there are many overlooked ways of forcing deadly patterns.
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