dailysudoku.com Forum Index dailysudoku.com
Discussion of Daily Sudoku puzzles
 
 FAQFAQ   SearchSearch   MemberlistMemberlist   UsergroupsUsergroups   RegisterRegister 
 ProfileProfile   Log in to check your private messagesLog in to check your private messages   Log inLog in 

Puzzle 9th March

 
Post new topic   Reply to topic    dailysudoku.com Forum Index -> Daily Sudoku puzzles
View previous topic :: View next topic  
Author Message
Bigdad



Joined: 02 Nov 2005
Posts: 2

PostPosted: Thu Mar 09, 2006 9:54 am    Post subject: Puzzle 9th March Reply with quote

Hi all,
I'm stuck on today's puzzle. I've got as far as
5?69?7??4
?9?5?467?
74????195
46??59??1
359???426
8??64??59
9?4???5?8
*8549???7
6??8?594?

The hint gives * as 2. I can't figure out why. Can anyone help?

Eric
[/code]
Back to top
View user's profile Send private message
Steve R



Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England

PostPosted: Thu Mar 09, 2006 1:27 pm    Post subject: Puzzle 9th March Reply with quote

Hi Eric

Note that the 1 in row 3 causes the 1 in box 2 to occupy column 5. Now row 9 shows that the 1 in box 7 lies in column 2 or 3, not column 1.

Steve
Back to top
View user's profile Send private message
Sarah
Guest





PostPosted: Fri Mar 10, 2006 5:58 pm    Post subject: Mar. 9 puzzle Reply with quote

Steve-
That doesn't make any sense! The 1 in box 2 can be in either row 1 or 2...
Help!
Back to top
TKiel



Joined: 22 Feb 2006
Posts: 292
Location: Kalamazoo, MI

PostPosted: Sat Mar 11, 2006 12:41 pm    Post subject: Reply with quote

Sarah,

Perhaps another way of explaining it. In box 2, all the cells with candidate 1's are in column 5, so all other candidate 1's in c5 can be excluded (r5c5, r7c5 & r9c5). That leaves all the cells with candidate 1's for row 9 in box 7 (r9c2&3). Since box 7 & row 9 must contain a 1, all other candidate 1's in box 7 can be excluded, leaving r8c1 with a single possibility.
Back to top
View user's profile Send private message
alanr555



Joined: 01 Aug 2005
Posts: 194
Location: Bideford Devon EX39

PostPosted: Sun Mar 12, 2006 11:01 pm    Post subject: Reply with quote

Code:

Starting from the position initially posted, I used Mandatory
Pairs to record additional information before proceeding to
derivation of the candidates. Normally I would have the M/P
data marked already but some had been eliminated on my
own solution which I completed on 10th.

The markings include

Box1- 8 in col3
Box2- 1 in col5, 6 in row3
Box3- 8 in row 1, 2 askew
Box4- 1 in row 6
Box5- Triple in row 5 (pairs for 1,7,8)
          Mutual Reception (23)
Box6- 8 in row4, 7 in col7.
Box7- No pairs at this stage
Box8- No pairs at this stage
Box9- Two Mutual Receptions (16) and (23)

To continue:

a) Box 8 must have '1' in row 7 or row 8
    (as r9c5 is excluded by pair in box 2)
b) Box8 AND Box 9 now have '1' restricted
    to row 7 or row 8. By one of the M/P
    technique rules the '1' in box 7 MUST be
    in row 9. M/P in r9c2 and r9c3.
c) Column 1 has seven cells resolved. The
    two missing values are '1' and '2'.
    From 'b' there cannot be a '1' in col 1
    on row 8 and so the '1' must be in row 2.
d) This leaves only one cell in column 1 and
    so r8c1 MUST have value '2'.

This leads to resolution of:

r1c5 (remove M/P from r2c5)
r8c7 (Mutual reception restricts to 23)
r9c9 (Other end of Mutual Reception)
r1c7 (remove partner in r2c9)
r1c8 (remove partner in r1c7)
r1c2 (last cell in row)
r2c9 (last cell in box3)
There is now a mutual reception on (28) in box 1
r6c2 (by slice/dice on '2')
r6c3 (remove partner in r6c2)
r4c3 (last cell in box 4)
r9c3 (sole candidate in col3 discounting M/R of 28)
r9c2 (remove partner in r9c3)
r7c2 (last cell in box7)
r9c5 (last cell in row 9)
r6c6 (remove partner in r6c2)
r4c4 (remove partner in r6c6 - M/R)
r6c7 (last cell in row)
r4c8 (slice/dice on 3)
r4c7 (last cell in box 6)
r5c4 (remove partner in r5c5)
r5c6 (remove partner in r5c4)
r5c5 (remove partner in r5c6)
r7c4 (block/col intersect on 1)
r8c8 (remove partner in r7c8)
r7c8 (last cell in column)
r8c6 (last cell in row)
r7c5 (sole position for 3)
r7c6 (last cell in row)
r3c5 (remove partner from r3c6)
r3c6 (slice/dice on 8)
r2c3 (remove partner in r3c3)
r3c3 (last cell in box 1)
remainder is trivial.

++++

When I solved it originally, I resorted to use of candidate profiles
but on this second look, it has been solved WITHOUT having to
use candidates at all - BUT this was a different puzzle in that
some cells not given in the original puzzle had been resolved
before the query was raised under this topic.

One of the paradoxes of SamGJ's grading system is that "V.Hard" is
often more amenable to Mandatory Pairs than are the plain "Hard"!

Alan Rayner  BS23 2QT
Back to top
View user's profile Send private message
jabejochke



Joined: 16 Mar 2006
Posts: 21
Location: Reading

PostPosted: Sat Mar 18, 2006 2:22 pm    Post subject: Reply with quote

First time responding so not fully in tune with the terminology and unsure how the words will translate (I am planning to create in Word, then paste into response field and then look at what shows up).

I had difficulty with this puzzle also. The way the solution came to me was when 4 fell into r2c6 then:

The 1s in box 2 at r1c5 and r2c5 were paired (hence no other 1s in c5)
Hence, 1 in r9c5 was eliminated creating paired set of 1s in row 9 in cells r9c2 and r9c3
Hence, eliminating the 1 in the 12 pair in box 7 at r8c1. This 2 in r8c1..
Eliminated all 2s in row 8 leaving the only remaining 2 in box 9 sitting at r9c9

That 2 in r9c9 broke it open for me.

Hope this helps.
Back to top
View user's profile Send private message
Display posts from previous:   
Post new topic   Reply to topic    dailysudoku.com Forum Index -> Daily Sudoku puzzles All times are GMT
Page 1 of 1

 
Jump to:  
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum


Powered by phpBB © 2001, 2005 phpBB Group