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 March 13 9x9 cages 25+, 27+, 27+ 
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Posted on: Wed Mar 14, 2012 5:10 pm




Posts: 65
Joined: Mon Mar 05, 2012 6:17 pm
Post March 13 9x9 cages 25+, 27+, 27+
Need a solving strategy for determination of cells in the 25+ and 27+ cages in the 9x9 of March 13.....i seem to be stumped or not seeing something crucial. Much Thanks!!! BTW -- new to this site about 9 days ago, and I am HOOKED!!! Great puzzles and website. Also appreciate the new SHIFT key feature that holds guesses..Got here just in time to use it. Saves printing paper etc!

THANKS AGAIN!


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Posted on: Wed Mar 14, 2012 6:08 pm




Posts: 47
Joined: Sun Oct 16, 2011 8:58 am
Post Re: March 13 9x9 cages 25+, 27+, 27+
Same here, I was not able to find the correct numbers in the two 27+ cages.


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Posted on: Wed Mar 14, 2012 11:36 pm




Posts: 274
Location: Ladysmith, BC, Canada
Joined: Thu May 12, 2011 11:37 pm
Post Re: March 13 9x9 cages 25+, 27+, 27+
Not sure this will help, but check out clm's help on solving a 6x6 and Tuesday's 9x9 on the heading of "Solving Strategy and Tips" and sub topic of ".....show an analytical solution for 6x6"


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Posted on: Thu Mar 15, 2012 12:09 am




Posts: 274
Location: Ladysmith, BC, Canada
Joined: Thu May 12, 2011 11:37 pm
Post Re: March 13 9x9 cages 25+, 27+, 27+
Also check out Imc's post on "Calcudoku helpers".


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Posted on: Thu Mar 15, 2012 12:55 am




Posts: 575
Joined: Fri May 13, 2011 4:51 pm
Post Re: March 13 9x9 cages 25+, 27+, 27+
beaker wrote:
Not sure this will help, but check out clm's help on solving a 6x6 and Tuesday's 9x9 on the heading of "Solving Strategy and Tips" and sub topic of ".....show an analytical solution for 6x6"


The 9x9 discussed (as an example for some techniques) under the topic "Can anyone show an analytical solution for the 10 March 6x6?" corresponds to monday, march 12, 2012 (Puzzle id: 433452) and additional info for this same puzzle (continuing the discussion) is under the topic "march 12 9x9 cages of 27+ and 31+"; perhaps I should have said "The monday's 9x9 ... " instead of saying " ... using the yesterday's 9x9 (Puzzle id: 433452) ... "; the misunderstanding comes from the time tag of the post, but when I sent the comment the new date and puzzles for march 13 had already appeared in the site, of course, the time tag should show March 13... .

Anyway, when I have time I will open a new post with the full solution for this 433452 (monday's 9x9 march 12, 2012) basically following the same procedures and then compacting the comments actually divided in the two "subtopics", and in order to have another full solution for a 9x9 (besides the one provided by sneaklyfox's in the youtube's video and the one written in the Forum by jomapil some time ago).

In this moment, I understand that what rickt and arjen are asking for is a technique, strategy or similar to solve the tuesday's 9x9 march 13, 2012 (Puzzle id: 433483, solver rating 105.6). This is something totally different because this type of puzzle (usually the tuesday's 9x9 is the most difficult of the untimed, regular, puzzles along the week) requires a more detailed and extended analysis (and certainly several graphics so it will take time and possibly a few days to show the full process). More specifically they ask about how to find the candidates for the cages "25+", "27+"(h1) and "27+"(i2), I understand. But an important part of the puzzle must be solved before we are prepared to define first which are those candidates and secondly where to place them in the proper cells. I hope to have some extra time in the next days in order to cooperate in the discussion to clarify all the steps that we must follow until the final solution.


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Posted on: Thu Mar 15, 2012 2:48 am




Posts: 16
Joined: Fri May 13, 2011 12:12 am
Post Re: March 13 9x9 cages 25+, 27+, 27+
A quick explanation of how I approach this style puzzle is by filling in the known numbers. Than divide puzzle into top, bottom, right & left sections.
When you add all the known numbers & subtract from column totals or section totals, it will give you an available total for the x cells. once I arrive at
the available total, I then evaluate the available combinations that fit the x cells. (x cells are mutiplication cells) I hope this quick explanation helps


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Posted on: Thu Mar 15, 2012 6:18 am




Posts: 47
Joined: Sun Oct 16, 2011 8:58 am
Post Re: March 13 9x9 cages 25+, 27+, 27+
Image

This is what I achieved.
Too many unknown correct positions and running out of time.

For the upper 27+:
1 4 5 8 9
2 4 5 7 9
2 4 5 8 8
3 4 5 7 8

And the lower 27+:
1 2 7 8 9
1 3 7 7 9
1 3 7 8 8
1 5 5 7 9
2 3 7 7 8


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Posted on: Thu Mar 15, 2012 1:31 pm




Posts: 98
Joined: Thu May 12, 2011 10:48 pm
Post Re: March 13 9x9 cages 25+, 27+, 27+
[quote="arjen"]Image

It might help if you bolded a few more things. The double 6's in the 23+ cage can be bolded
as they can not be in any other position.
You can also look at rows 6 and 7. You can figure out that the 5 and 1 in row 7 can only be
in 2 spots, hence they are fixed that makes row 7 columnn 5 be 8, thus r8c1=8. I think there is
more...


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Posted on: Thu Mar 15, 2012 7:34 pm




Posts: 575
Joined: Fri May 13, 2011 4:51 pm
Post Re: March 13 9x9 cages 25+, 27+, 27+
arjen wrote:
...

This is what I achieved.
Too many unknown correct positions and running out of time.

For the upper 27+:
1 4 5 8 9
2 4 5 7 9
2 4 5 8 8
3 4 5 7 8

And the lower 27+:
1 2 7 8 9
1 3 7 7 9
1 3 7 8 8
1 5 5 7 9
2 3 7 7 8


The cells coloured in green you have found are correct and also the content of the cages in white (regardless of the “position of the numbers”). But there is a little mistake in the combinations you have found for the cages “27+”, for instance, the upper “27+” initially would additionally admit the 14778; also for the lower “27+” the 15579 is not correct, two 5’s are not possible (there are two 5’s already in the three rightmost columns, in “6+” and “630x”) and missing the 23589. Think that the combinations must be “paired” (so the same number of them), but we will see all those details later.

Rickt, do you agree with the result found by arjen to this point?: the content of “25+” is [1,3,5,7,9].

Anyway, let’s resume a little. First, we write the candidates (and the final numbers) in the three central columns and in the central row (as much as we can).

“10080x” = [4,5,7,8,9], [5,6,6,7,8] but the second is not possible (two 6’s and only one could go to b2 in such a case); “10080x” = [4,5,7,8,9]; “10080x” (+) = 33. Now “128x” (+) = 17 (calculating the difference to 135, which is the sum of the three leftmost columns) and then “128x” = [1,2,2,4,8] (the only valid among the five initial possibilities, [1,1,2,8,8] (20), [1,1,4,4,8] (18), [1,2,2,4,8] (17), [1,2,4,4,4] (15) and [2,2,2,4,4] (14)), apart of the fact that three of these combinations have two or three 4’s which is impossible once determined that a 4 is inside “10080x”.

“11664x” = [2,8,9,9,9], [3,6,8,9,9], [4,4,9,9,9], [4,6,6,9,9]; three 9’s are not possible (should go to g6, h7 and i8, but there is a 9 in e6) so “11664x” = [3,6,8,9,9], [4,6,6,9,9]; the two 9’s now must go to h7, i8 (as in the arjen’s graphic) and then [4,6,6,9,9] is not valid so “11664x” = [3,6,8,9,9], which sum is 35; consequently “630x” (+) = 22 (calculating the difference to 135, the sum of the three righmost columns). Then “630x” = [1,3,5,6,7] (the only valid among the three initial possibilities, [1,2,5,7,9] (24), (1,3,5,6,7] (22) and [2,3,3,5,7] (20)). Additionally, the cage “2:” must be even so [4,8] (not [3,6]) and “1-” = [6,7] (“1-“ (+) + “2:”(+) = 25, which is the difference between 180, the sum of the four bottom rows, and the sum of the rest of the cages). Now, i.e., counting the numbers, we stablish that “25+” = [1,3,5,7,9] (and all candidates in column f are defined too).

Once we have “25+” = [1,3,5,7,9], counting the numbers in the three leftmost columns, we find “23+” = [1,2,6,6,8].

Image

The two 6’s of “23+” must go necessarily to a3, b4 (due to d2 = 6 and c5 = 6) (two more green cells to the graphic).

Let’s observe that the 2 of column c must go to c6 (it is not inside “10080x” or “25+” and c4 is not allowed due to i4 = 2) (yellow colour). Observe that we need a 1 in row 6 so b6 = 1. Observe that the 1’s in b6 and d9 force the 1 of cage “25+” to be in c7 or c8, then c4 = [1,2,8] must be 8 (yellow).

In brown colour: a 2 is not inside “10080x” so the 2 in column a goes to a7. This forces a8 = 8, b7 = 4 (completing the final positions in cage “128x”). Also a1 = 4. And e7 = 8 (brown).

… to be continued …


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Posted on: Thu Mar 15, 2012 7:52 pm




Posts: 65
Joined: Mon Mar 05, 2012 6:17 pm
Post Re: March 13 9x9 cages 25+, 27+, 27+
yes I do agree with the contents of the 25+ cage 57193 (top to bottom)


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