Help!!!

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

Help!!!

Postby Kent » Wed Mar 01, 2006 3:10 am

I need help for this!!! Thanks!!

__1!9_7!__8
6__!185!73_
__7!46_!1__
_34!_9_!___
___!5_4!___
___!_1_!42_
__5!_71!9__
_1_!84_!__7
7__!_59!2__

By the way.Do we need to pay if we register here??? Or is it free to use this forum???I'm new.
Kent
 
Posts: 98
Joined: 28 February 2006

Postby bennys » Wed Mar 01, 2006 10:11 am

Code: Select all
+-------------------+-------------------+-------------------+
| 2345 ^245   1     | 9     23    7     |^56   ^456   8     |
| 6    *249  *29    | 1     8     5     | 7     3     249   |
| 23589 2589  7     | 4     6     23    | 1     59    259   |
+-------------------+-------------------+-------------------+
| 1258  3     4     | 267   9     268   | 568   1567  156   |
| 1289  26789 2689  | 5     23    4     | 368   1679  1369  |
| 589   56789 689   | 367   1     368   | 4     2     3569  |
+-------------------+-------------------+-------------------+
| 248   2468  5     | 236   7     1     | 9     468   346   |
| 29    1     2369  | 8     4     236   | 356   56    7     |
| 7     468   368   | 36    5     9     | 2     1468  1346  |
+-------------------+-------------------+-------------------+

First step from ALS xz rule we get

R1C1<>4

and we get



+-------------------+-------------------+-------------------+
| 235   245   1     | 9     23    7     | 56    456   8     |
| 6     249   29    | 1     8     5     | 7     3     249   |
| 23589 2589  7     | 4     6     23    | 1     59    259   |
+-------------------+-------------------+-------------------+
| 1258  3     4     | 267   9     268   | 568   1567  156   |
| 1289  26789 2689  | 5     23    4     | 368   1679  1369  |
| 589   56789 689   | 367   1     368   | 4     2     3569  |
+-------------------+-------------------+-------------------+
| 4     26    5     | 236   7     1     | 9     8     36    |
| 29    1     2369  | 8     4     236   | 356   56    7     |
| 7     68    368   | 36    5     9     | 2     14    14    |
+-------------------+-------------------+-------------------+
here

R9C2=8=>R3C1=8

R9C2=6=>R7C2=2=>R8C6=2=>R3C6=3=>R3C1<>3

so we get R3C1<>3 and it solve the puzzle.


You dont need to pay
bennys
 
Posts: 156
Joined: 28 September 2005

Postby emm » Wed Mar 01, 2006 10:47 am

Kent - this is a pretty advanced puzzle if you're new. If this is where you're stuck then Havard’s thread on strong links could be just what you need.

Code: Select all
*--------------------------------------------------------------------*
 | 2345A  245    1      | 9      23     7      | 56     456    8      |
 | 6      249B   29     | 1      8      5      | 7      3      249b   |
 | 23589  2589   7      | 4      6      23     | 1      59     259    |
 |----------------------+----------------------+----------------------|
 | 1258   3      4      | 267    9      268    | 568    1567   156    |
 | 1289   26789  2689   | 5      23     4      | 368    1679   1369   |
 | 589    56789  689    | 367    1      368    | 4      2      3569   |
 |----------------------+----------------------+----------------------|
 | 248a   2468   5      | 236    7      1      | 9      468    346*   |
 | 29     1      2369   | 8      4      236    | 356    56     7      |
 | 7      468    368    | 36     5      9      | 2      1468   1346   |
 *--------------------------------------------------------------------*


The conjugate pairs of 4s r1c1, r7c1 (Aa) and r2c2, r2c9 (Bb) eliminate the 4 from r7c9 marked *
emm
 
Posts: 987
Joined: 02 July 2005

Help!!!

Postby Cec » Wed Mar 01, 2006 10:53 am

Hi Kent,
I'll answer your second question first. Regarding payment to register in this forum to my knowledge it is free.
You haven't specifically asked what help you want - some people want only the next step others want a "bit" more whilst some (usually not many) might only want the solution. If you have just started to learn about Sudoku then you have picked a fairly difficult puzzle which, after minor elimination of some candidates, requires an advanced technique to solve it.

If I'm "reading" your enquiry correctly and I could be wrong, I would suggest you read up on the following links and later submit your puzzle, including the "pencilmarks" (also known as candidates) in the manner fully explained by clicking on HERE .

The following links provide many solving techniques ranging from basic to advanced and these should provide plenty for you to study up on - just click on these:
www.SuDoku.funURL.com
http://www.angusj.com/sudoku/hints.php
http://www.simes.clara.co.uk/programs/sudokutechniques.htm

PS. The above posts have just preceded mine which I'll still let stand.

Cec
Cec
 
Posts: 1039
Joined: 16 June 2005

Postby absolute beginner » Wed Mar 01, 2006 9:42 pm

I think "ALS xz shows r1c1 <> 4" is sort of short.
I am just trying to understand ALS and so the long description
is like:

If r2c2 is 4
r1c1 cannot be, too, of course.
If r2c2 is not 4,
then r2c2 and r2c3 are both 2 or 9.
So no cell in this block except these two can be 2 or 9.
In particular r1c2 <> 2.
In the first row r1c2, r1c7 and r1c8 are 4 and 5 and 6.
And no cell else can have one of these digits.
Especially r1c1 <> 4.
So r1c1 <> 4 in both cases.

Is that ok?
absolute beginner
 
Posts: 22
Joined: 26 February 2006

thanks cecbevwr

Postby Kent » Thu Mar 02, 2006 7:10 am

cecbevwr what i mean is that do we pay when we use the forum.thats all.the help asked was what was the following step and how u derive the step.1 more question.What do you mean by r1c1<>4?? Does that mean r1c1=4?Thanks
Kent
 
Posts: 98
Joined: 28 February 2006

question

Postby Kent » Thu Mar 02, 2006 7:15 am

bennys, how do u get R9C2=8 and whats ALZ ? Thanks
Kent
 
Posts: 98
Joined: 28 February 2006

Re: Help!!!

Postby Cec » Thu Mar 02, 2006 11:48 am

Kent wrote:"cecbevwr what i mean is that do we pay when we use the forum..."

Still the same answer - NO - unless you'd like to pay:)

Kent wrote:"..thats all.the help asked was..."

Well, that's not quite correct because in your first post you asked "I need help for this!!! Thanks!!" which presumably related to the puzzle you submitted followed by your next two questions:
"..Do we need to pay if we register here??? Or use this forum???I'm new"
As I see it you actually asked three questions which I understand have been answered.

Kent wrote:".. what was the following step and how u derive the step..."

Your initial puzzle and candidate grid (as existed prior to Benny's above grid) would have looked like this:
Code: Select all
 *-----------*
 |..1|9.7|..8|
 |6..|185|73.|
 |..7|46.|1..|
 |---+---+---|
 |.34|.9.|...|
 |...|5.4|...|
 |...|.1.|42.|
 |---+---+---|
 |..5|.71|9..|
 |.1.|84.|..7|
 |7..|.59|2..|
 *-----------*

 
 *--------------------------------------------------------------------*
 | 2345   245    1      | 9      23     7      | 56     456    8      |
 | 6      249    29     | 1      8      5      | 7      3      249    |
 | 23589  2589   7      | 4      6      23     | 1      59     259    |
 |----------------------+----------------------+----------------------|
 | 1258   3      4      | 267    9      268    | 568    15678  156    |
 | 1289   26789  2689   | 5      23     4      | 368    16789  1369   |
 | 589    56789  689    | 367    1      368    | 4      2      3569   |
 |----------------------+----------------------+----------------------|
 | 2348   2468   5      | 236    7      1      | 9      468    346    |
 | 239    1      2369   | 8      4      236    | 356    56     7      |
 | 7      468    368    | 36     5      9      | 2      1468   1346   |
 *--------------------------------------------------------------------*


In box7 (the lower left 3X3 box) the candidate 3's must be in column3 (c3) as no other 3's exist in this column outside this box. This enables the remaining 3's in this box7 (ie r7c1 and r8c1) to be excluded. This elimination doesn't immediately place a (big) number in a cell owing to necessary further eliminations. This technique is known as "Locked Candidates(1)" and is one of the many other techniques which you can read about in the links suggested above.
A similar "Locked Candidates" situation exists in box9 (lower right box) where the candidate 8's are "locked" in c8 being the only column in this box where an 8 can be placed. This enables the remaining 8's in c8 outside of this box9 (ie. r4c8 and r5c8) to be excluded.
Kent wrote:"..1 more question.What do you mean by r1c1<>4?? Does that mean r1c1=4?Thanks"

This query relates to Benny's above post who hopefully will post a reply to you. Looks like plenty of homework eh!:)

Cec
Cec
 
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Re: thanks cecbevwr

Postby Havard » Thu Mar 02, 2006 11:59 am

Kent wrote:What do you mean by r1c1<>4?? Does that mean r1c1=4?Thanks

<> means "not like" or "different"

r1c1<>4 means that in r1c1 there can be no 4. (hence you can kill the candidate 4 in that cell)

ALS you can read about here:
http://forum.enjoysudoku.com/viewtopic.php?t=2510
Havard
 
Posts: 377
Joined: 25 December 2005

Postby vidarino » Thu Mar 02, 2006 12:05 pm

There's a Grouped Turbot Fish there as well:

R1C1-2-R1C5=2=R5C5-2-[R5C123]=2=R4C1-2-R1C1 -> R1C1 <> 2

(Edit:) And on second glance, we can also apply the Hinge / Empty Rectangle;

Code: Select all
 2  2  .  |  .  2  .  |  .  .  .
 .  2  2  |  .  .  .  |  .  .  2
 2  2  .  |  .  .  2  |  .  .  2
----------+-----------+----------
 2  .  .  |  2  .  2  |  .  .  .
 2  2  2  |  .  2  .  |  .  .  .
 .  .  .  |  .  .  .  |  .  2  .
----------+-----------+----------
 2  2  .  |  2  .  .  |  .  .  .
 2  .  2  |  .  .  2  |  .  .  .
 .  .  .  |  .  .  .  |  2  .  .


"Pivot point" at R5C1, "partner" conjugate pair in C5.


Yes, the ALS rules are very powerful, but incredibly hard to spot, IMHO.:)

Vidar
vidarino
 
Posts: 295
Joined: 02 January 2006

Re: thanks cecbevwr

Postby Cec » Thu Mar 02, 2006 2:28 pm

Havard wrote:"..r1c1<>4 means that in r1c1 there can be no 4.."

Thanks Havard for answering this question.
Cec
Cec
 
Posts: 1039
Joined: 16 June 2005

Postby absolute beginner » Fri Mar 03, 2006 12:57 am

Kent,

you asked "bennys, how do u get R9C2=8 "
Benny wrote before
"R9C2=8=>R3C1=8
R9C2=6=>R7C2=2=>R8C6=2=>R3C6=3=>R3C1<>3
"

benny didnt have r9c2=8. He ment:
"if r9c2 were 8 then r3c1 would be 8
if r9c2 were 6 then r7c2 were 2 and therefore r8c6 2
and therefore r3c6 were 3 and so r3c1 were not 3"
So in both cases (r9c2=8 or r9c2=6) r3c1 is not 3,
so you can delete 3 from the candidate list of r3c1.
Benny wrote that in a very short way ;-)

You wrote "What is ALZ"
I asked the same question and benny answered with the link
http://forum.enjoysudoku.com/viewtopic.php?t=2033#16664
absolute beginner
 
Posts: 22
Joined: 26 February 2006

Postby niznadia » Fri Mar 03, 2006 5:54 am

i've tried this puzzle..it's kind of hard since this is my 3rd puzzle..but i managed to solve it..

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

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

4/2/5 3/7/1 9/8/6
9/1/6 8/4/2 3/5/7
7/8/3 6/5/9 2/4/1
niznadia
 
Posts: 1
Joined: 01 March 2006

Postby Carcul » Fri Mar 03, 2006 11:28 am

Hi Niznadia.

Welcome to this forum. Just a comment on your first post: in this forum, we are not interested in the solution of a puzzle, but in the logical methods that can be used to reach that solution. So, why don't you tell us how you have reached your solution?

Regards, Carcul
Last edited by Carcul on Fri Mar 03, 2006 1:37 pm, edited 1 time in total.
Carcul
 
Posts: 724
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Alternative Solution

Postby Carcul » Fri Mar 03, 2006 11:29 am

Code: Select all
 2345   245    1     | 9    23  7    | 56     456    8     
 6      249    29    | 1    8   5    | 7      3      249   
 23589  2589   7     | 4    6   23   | 1      59     259   
---------------------+---------------+---------------------
 1258   3      4     | 267  9   268  | 568    1567   156   
 1289   26789  2689  | 5    23  4    | 368    1679   1369   
 589    56789  689   | 367  1   368  | 4      2      3569   
---------------------+---------------+---------------------
 248    2468   5     | 236  7   1    | 9      468    346   
 29     1      2369  | 8    4   236  | 356    56     7     
 7      468    368   | 36   5   9    | 2      1468   1346   

[r8c7]{=(AUR:r1c7/r1c8/r8c7/r8c8)=3|4=[r1c8]-4-[r1c1]=4=[r7c1]-4-[r9c2]}=3=[r5c7](=8=[r4c7]-8-[r4c6])-3-[r5c5](-2-[r4c6]-6-[r8c6])-2-[r1c5](-3-[r3c6]-2-[r8c6])-3-[r1c1]=3=[r3c1]=8=[r3c2]-8-[r9c2]-6-[r9c4]-3-[r8c6],

(AUR=Almost Unique Rectangle) which means that, if r8c7 is not "3" then r8c6 turns out to be an Empty Cell - a contradiction. So, we must have r8c7=3 and that solve the puzzle.

Carcul
Carcul
 
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