Help - Stuck - Not sure how to proceed from here.

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Help - Stuck - Not sure how to proceed from here.

Postby sanchin » Wed Feb 15, 2006 8:04 pm

Hi,

I'm relatively new to this andhaving lots of fun. I'm not familiar with advance techniques, and I'm guessing that is probably what is needed at this point (?). Can someone help me out with how to proceed from here?

Thanks!

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

5       6       7     |  {24}    {24}    9     |  3       {18}    {18}
8       2       9     |  6       1       3     |  4       5       7
{14}    3       {14}  |  5       7       8     |  9       6       2
----------------------+------------------------+-------------------------
9       8       5     |  {1247}  {24}    {147} |  {167}   3       {146}
2       {17}    3     |  9       8       6     |  5       {147}   {14}
{167}   4       {16}  |  {17}    3       5     |  8       2       9
----------------------+------------------------+-------------------------
3       5       {146} |  {147}   9       2     |  {167}   {1478}  {1468}
{147}   9       8     |  3       6       {147} |  2       {147}   5
{1467}  {17}    2     |  8       5       {147} |  {167}   9       3
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Postby Animator » Wed Feb 15, 2006 8:21 pm

First note that you can remove 4 as a candidate from r4c9. (4 is locked on row 4 in box 5.)

A possible solution: the uniqueness pattern.

If r4c4 is 2 or 4 then you change the 2's and the 4's in rows 1 and 4.

(If you decide not to use the uniqness pattern then you can remove some candidates by colouring the 7. But I'm not yet sure if you that takes you to the solution.)
Last edited by Animator on Wed Feb 15, 2006 5:21 pm, edited 1 time in total.
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Postby tarek » Wed Feb 15, 2006 8:27 pm

The uniqueness pattern can solve it in one step
Code: Select all
*--------------------------------------------------------*
| 5     6     7    |*24   *24    9    | 3     18    18   |
| 8     2     9    | 6     1     3    | 4     5     7    |
| 14    3     14   | 5     7     8    | 9     6     2    |
|------------------+------------------+------------------|
| 9     8     5    |-1247 *24    147  | 167   3     146  |
| 2     17    3    | 9     8     6    | 5     147   14   |
| 167   4     16   | 17    3     5    | 8     2     9    |
|------------------+------------------+------------------|
| 3     5     146  | 147   9     2    | 167   1478  1468 |
| 147   9     8    | 3     6     147  | 2     147   5    |
| 1467  17    2    | 8     5     147  | 167   9     3    |
*--------------------------------------------------------*
r4c4 must only contain 17 (Candidates 24 in r1c4,r1c5,r4c4 & r4c5 form a type-1 unique quadrangle)


However you may decide either to postpone it until last (as there are several other interesting eliminations), or for a challenge try not using it at all.

[Edit: I just noticed Animator's similar response]

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Postby QBasicMac » Wed Feb 15, 2006 8:40 pm

Restating what Animator said:

Box 5 has 1247,24,147 in row 4. Note that 4 appears only on this line in that box. Thus 4 MUST be placed in one of those 3 cells. Hence it cannot go in row 4 column 9 (r4c9) where you see 146. Erase 4.

Now, as he indicated, you need advanced techniques.

If you are not interested in advanced techniques, do what I do:

Make two more copies of your puzzle so far.
Place 1, 6, and 7 in cell r9c7 to make three puzzles you can work on.

You will find 1 and 7 are invalid. Good.

You will be able to solve 6 using easy techniques, proving the puzzle is valid and has only one solution!

Three puzzles for the price of one!

(Most, if not everybody but me, on this forum specialize in very advanced techniques, striving to solve any puzzle however difficult without resorting to the multiple-puzzle technique. You might find that enjoyable, too. I think it converts a jolly fun activity to advanced calculus type brain strain. To each his/her own!)

Mac
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Postby sanchin » Wed Feb 15, 2006 10:21 pm

Thanks to all of the suggestions. I totally missed the 4 in r4c9.
QBasicMac wrote:If you are not interested in advanced techniques, do what I do:
Make two more copies of your puzzle so far.
Place 1, 6, and 7 in cell r9c7 to make three puzzles you can work on.

I agree that sometimes the simpler approach is best. However, I'm not sure I would not have known which of the many possible places to try. I did as you suggested (r9c7) and was able to finish it immediately. It would be good to also know which of the advanced approaches would yield solutions (or maybe hints).

Thanks again!
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Postby Animator » Sat Feb 18, 2006 1:10 pm

There were some extra posts in this thread...

The most important contribution: (it's not the original text... it's how I remember it) (It was original posted by Neunmalneun)


There is another uniqueness pattern in 1, 8 in the cells r1c8, r1c9, r7c8, r7c9 which allows you to remove 1 from r7c8 and r7c9.


It might not be so easy to see it so here's an example:
  • r7c8 = 1 ==> r7c9 is 8.
  • r7c9 = 1 ==> r7c8 is 8.
In both cases you end up with r1c8, r1c9 having the numbers 1, 8 and r7c8, r7c9 having the numbers 1, 8 (in some order). (Which basically means you end up with 2 solutions, not 1.)
Last edited by Animator on Sun Feb 19, 2006 9:37 am, edited 1 time in total.
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Postby MCC » Sat Feb 18, 2006 5:16 pm

I believe Ronk may have been the contributor on the uniqueness pattern of the 1,8 in cells r1c8, r1c9, r7c8, r7c9.

The 8's are locked in r7c89 so one of these must be an 8, if the other cell in r7c89 is a 1 then this leads to a UR, so neither can be a 1.

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Help - Stuck - Not sure how to proceed from here

Postby Cec » Sun Feb 19, 2006 12:52 am

MCC wrote:"I believe Ronk may have been the contributor on the uniqueness pattern of the 1,8 in cells r1c8, r1c9, r7c8, r7c9...."

Hi MCC. No - it was actually Neunmalneun whose post (Fri Feb 17) read as follows:
"There is another UR in this puzzle (18 in R1C89/R7C89). As the 8 is "needed" for row 7 in these cells, you can eliminate the 1 in R7C89".

Before the hacking incident, just by chance, I happened to print some pages of this interesting thread, particularly tso's excellent description of the simple Unique Rectangle in this puzzle where the naked pairs [24] in r1c45 and also the [24] in r4c5 enabled the [24] to be excluded from r4c4.

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Postby MCC » Sun Feb 19, 2006 12:17 pm

Thanks Cec.

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Re: Help - Stuck - Not sure how to proceed from here.

Postby windbell » Tue Feb 21, 2006 4:49 pm

sanchin wrote:Hi,

I'm relatively new to this andhaving lots of fun. I'm not familiar with advance techniques, and I'm guessing that is probably what is needed at this point (?). Can someone help me out with how to proceed from here?

Thanks!

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

5       6       7     |  {24}    {24}    9     |  3       {18}    {18}
8       2       9     |  6       1       3     |  4       5       7
{14}    3       {14}  |  5       7       8     |  9       6       2
----------------------+------------------------+-------------------------
9       8       5     |  {1247}  {24}    {147} |  {167}   3       {146}
2       {17}    3     |  9       8       6     |  5       {147}   {14}
{167}   4       {16}  |  {17}    3       5     |  8       2       9
----------------------+------------------------+-------------------------
3       5       {146} |  {147}   9       2     |  {167}   {1478}  {1468}
{147}   9       8     |  3       6       {147} |  2       {147}   5
{1467}  {17}    2     |  8       5       {147} |  {167}   9       3


Code: Select all
5   6   7   2   4   9   3   8   1
8   2   9   6   1   3   4   5   7
1   3   4   5   7   8   9   6   2

9   8   5   1   2   4   7   3   6
2   7   3   9   8   6   5   1   4
6   4   1   7   3   5   8   2   9

3   5   6   4   9   2   1   7   8
7   9   8   3   6   1   2   4   5
4   1   2   8   5   7   6   9   3


What I do is try-&-error~:D
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Postby CathyW » Tue Feb 21, 2006 4:54 pm

You may be happy with that approach Windbell but now that you've posted complete solutions a few times I think I should say that most of us when seeking assistance with a puzzle are asking for the next logical step to help us solve the puzzle. Simply giving the solution doesn't actually help very much without details of how you got there.
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