Calcudoku puzzle forum
http://www.calcudoku.org/forum/

Puzzle 9x9 20NOV11
http://www.calcudoku.org/forum/viewtopic.php?f=16&t=113
Page 1 of 1

Author:  jomapil  [ Mon Nov 21, 2011 6:50 pm ]
Post subject:  Puzzle 9x9 20NOV11

Here it is the puzzle where I arrived.
I thank anyone who tell me if there are any continuations without the use of trial and error.



Image

Uploaded with [url=http://www.calcudoku.org/im/forum/img_f16_t113_p923_i1.png]ImageShack.us[/url]


Besides I have the following multiple candidates:

b2=3,6,9; d2=1,3,6,9; e2=1,3,6,9; i2=1,3,6,9; h7=2,4,8; h8=1,4,8; g6=g7=g8=3,4,5,6,9;
g9=3,5,6,9. ( I begin to use the terminology of Clm ).

Of course I indicate the multiple candidates of each cell on account of the impossibility to include them in the puzzle.

Author:  clm  [ Mon Nov 21, 2011 11:05 pm ]
Post subject:  Re: Puzzle 9x9 20NOV11

jomapil wrote:
Here it is the puzzle where I arrived.
I thank anyone who tell me if there are any continuations without the use of trial and error.
...
Besides I have the following multiple candidates:

b2 = [3, 6, 9]; d2 = [1, 3, 6, 9]; e2 = [1, 3, 6, 9]; i2 = [1, 3, 6, 9]; h7 = [2, 4, 8]; h8 = [1, 4, 8]; g6-g7-g8 = [3, 4, 5, 6, 9]; g9 = [3, 5, 6, 9] ( I begin to use the terminology of Clm ).

Of course I indicate the multiple candidates of each cell on account of the impossibility to include them in the puzzle.


Thank you, and apologize for making small adjustments on the terminology introducing some spaces, square brackets, etc., also g6 = g7 = g8 would not be possible, see? ...
Now, the puzzle: To assign candidates you must depart from correct conclusions but this is not so in this case, your bolded diagram is wrong. Why did you take the decision of doing a5 = 3; a3-a4 = [5, 7]; a1 = 4; b1 = 1; d1 = 7? (in fact this assignment is wrong, see the solution). Why not a3-a4 = [3, 5]; a1 = 7; b1 = 4; d1 = 1; a5 = 4 which is the correct one?, I think you should revise your initial inferences; it's impossible to continue solving the puzzle and arrive to the valid solution if you depart from a wrong initial setting, right?.

Author:  jomapil  [ Tue Nov 22, 2011 9:01 am ]
Post subject:  Re: Puzzle 9x9 20NOV11

I'm sorry with my negligence. I abandoned the puzzle without to verify if it was correct. Even so the question maintains: IF, by hypothesis, those values were correct, what will be the next step?

With the terminology that I accept as very important as a uniformity between us, one must begin with the simplest.

I began with b2=3,6,9 ... according

" 6. If a cell may have several possibilities we write: b1 = 3, 5, 7 "

"11. If a cage must contain a already well defined numbers we say: “75x” (e1-f1-f2) = [3, 5, 5]"

you used the brackets with cages and not with cells.

So "112X" (g3-g4-g5)=[1,2,8] and I introduced g6=g7=g8=3,4,5,6,9. Because g3-g4-g5 form a cage but g6 g7 g8 are cells dispersed by two different cages.



I have two suggestions about the terminology that I'll write in the other thread within a little time.

Author:  clm  [ Tue Nov 22, 2011 1:02 pm ]
Post subject:  Re: Puzzle 9x9 20NOV11

jomapil wrote:
I'm sorry with my negligence. I abandoned the puzzle without to verify if it was correct. Even so the question maintains: IF, by hypothesis, those values were correct, what will be the next step?

With the terminology that I accept as very important as a uniformity between us, one must begin with the simplest.

I began with b2=3,6,9 ... according

" 6. If a cell may have several possibilities we write: b1 = 3, 5, 7 "

"11. If a cage must contain a already well defined numbers we say: “75x” (e1-f1-f2) = [3, 5, 5]"

you used the brackets with cages and not with cells.

So "112X" (g3-g4-g5)=[1,2,8] and I introduced g6=g7=g8=3,4,5,6,9. Because g3-g4-g5 form a cage but g6 g7 g8 are cells dispersed by two different cages.



I have two suggestions about the terminology that I'll write in the other thread within a little time.


Thank you for paying so good attention to the terminology, you are absolutely right with respect to the individual cells (I don't know what I was thinking about, "there's non worse shod than the shoemaker's wife", our "en casa del herrero, cuchillo de palo", sorry for making you loosing time), though we must show the spaces, b2 = 3, 6, 9, ... to be consistent (for clarity and also thinking in the possibilitie of creating a compiler); same thing inside the brackets "112x" (g3-g4-g5) = [1, 2, 8] and small "x" for the multiplication factor.

With respect to g6 = g7 = g8 = 3, 4, 5, 6, 9 I have to think in something alternative (certainly there is a lack and the terminology does not include this situation, and g6 = 3, 4, 5, 6, 9; g7 = 3, 4, 5, 6, 9; g8 = 3, 4, 5, 6, 9 looks too long), I don't like the first part because the equal signs create confusion since two or more cells in the same column can not be equal. Perhaps I have to introduce in the paragraph 6 something like (g6, g7, g8) = 3, 4, 5, 6, 9 now it would not be confusion because the colons mean they do not form a cage as in the case of the dashes... and that "any" of the g6, g7 and g8 may take "any" of the values 3, 4, 5, 6 and 9 ... will think, this has been a good opportunity to discuss about the terminology, this must be improved.

I will go to the other section and thread for your suggestions on the terminology.

Wrt the 9x9 puzzle itself I'll give you my comments in a separate reply.

Author:  clm  [ Tue Nov 22, 2011 5:33 pm ]
Post subject:  Re: Puzzle 9x9 20NOV11

jomapil wrote:
I'm sorry with my negligence. I abandoned the puzzle without to verify if it was correct. Even so the question maintains: IF, by hypothesis, those values were correct, what will be the next step?

...


For this discussion I provide two graphics, in the first I show why “14+” (b7-b8) <> [5, 9] and in the second I advance a little bit the puzzle once determined that “14+” (b7-b8) = [6, 8]. I keep the good part of your initial setting (I believe you will not need more help to finish the puzzle, otherwise tell me). The first three steps are for the first graphic, the oher two steps for the second graphic. From now on I use only the “international terminology”

C = 9

1. “4:” = [2, 8]; “23+” = [6, 8, 9] ---> c1 = 2; c2 = 8

2. a6 = 8 ---> “15+” = [6, 9]; “2-“ = [3, 5], [5, 7] ---> "2-" = [x, 5]; “8+” = [1, 7] ---> a1 <> 1, 2, 3 *b1 = 6*, 4 *b1 = 1, b1= 7*, 5, 6, 7, 8, 9 = [000] ---> “8+” = [3, 5] ---> h1= 3 ---> i1 = 5

3. “14+” (b7-b8) = [5, 9] ---> “16+” (a5-b4-b5) <> [4, 4, 8] *d1 = 4 ---> a1-b1 = [1, 7] = [000]*;
“14+” (b7-b8) = [5, 9] ---> “16+” (a5-b4-b5) <> [1, 7, 8] *b4 = 1 = 7 = [000]*;
“14+” (b7-b8) = [5, 9] ---> “16+” (a5-b4-b5) <> [2, 6, 8] *a2 = 2; b6 = 2*;
“14+” (b7-b8) = [5, 9] ---> “16+” (a5-b4-b5) <> [3, 5, 8] *(aaaa) [] 5!; (bbbb) [] 5!*;
“14+” (b7-b8) = [5, 9] ---> “16+” (a5-b4-b5) <<>> 8 ---> b9 = 8 *"972x" <<>> 8* ---> c8-c9 = [3, 5] ---> b8 = 9 ---> a8 = 6 ---> f9 = 6 ---> e8-f8 = [5, 7] ---> c8 = 3 ---> c9 = 5 ---> g9-h9-i9 = [3, 7, 9] ---> g8 = 6 = [000] ---> “14+” (b7-b8) = [6, 8]

Image

4. (a7-a8 = [6, 9]; b7-b8 = [6, 8]) ---> f9 = 6 ---> f1 = 8 ---> “5-“ (f5-f6) = [2, 7] ---> f5 = 2 ---> f6 = 7

5. f9 = 6 ---> e8-f8 = [5, 7]; f6 = 7 ---> f8 = 5 ---> e8 = 7

… … …

Image

Reedited Nov 23, 2011, to correct an observed typo error, the *because of* comment *"972x" <> 8*, almost at the end of step three, it must say *"972x" <<>> 8* (*because of cage "972x" can not contain an 8*).

Author:  jomapil  [ Tue Nov 22, 2011 8:03 pm ]
Post subject:  Re: Puzzle 9x9 20NOV11

clm wrote:
“14+” (b7-b8) = [5, 9] ---> “16+” (a5-b4-b5) <<>> 8 ---> b9 = 8 *"972x" <> 8* ---> c8-c9 = [3, 5] ---> b8 = 9 ---> a8 = 6 ---> f9 = 6 ---> e8-f8 = [5, 7] ---> c8 = 3 ---> c9 = 5 ---> g9-h9-i9 = [3, 7, 9] ---> g8 = 6 = [000] ---> “14+” (b7-b8) = [6, 8]


I didn't understand this step ---> b8 = 9. Why not ---> b8 = 5, 9 and then to analyse, also, the hypothesis b8 = 5 ?

Author:  clm  [ Wed Nov 23, 2011 1:18 am ]
Post subject:  Re: Puzzle 9x9 20NOV11

jomapil wrote:
clm wrote:
“14+” (b7-b8) = [5, 9] ---> “16+” (a5-b4-b5) <<>> 8 ---> b9 = 8 *"972x" <> 8* ---> c8-c9 = [3, 5] ---> b8 = 9 ---> a8 = 6 ---> f9 = 6 ---> e8-f8 = [5, 7] ---> c8 = 3 ---> c9 = 5 ---> g9-h9-i9 = [3, 7, 9] ---> g8 = 6 = [000] ---> “14+” (b7-b8) = [6, 8]


I didn't understand this step ---> b8 = 9. Why not ---> b8 = 5, 9 and then to analyse, also, the hypothesis b8 = 5 ?


Because, for the hypothesis we are considering in this step 3. (the full hypohesis is shown in red in graphic 1), that is, the pair 5-9 in the cage "14+" (b7-b8), once one we have arrived to the conclusion that b9 = 8 and that the pair c8-c9 would be 3-5, since you now have two 5's in rows 8 and 9 (there is another 5 inside "210x") both 5's would "erase" both rows 8 and 9 simultaneously so the 9 of the cage "14+" would go to b8 (and the 5 necessarily to b7, as in the graphic). Later, logically, the 9 of b8 forces a 6 in a8, and this 6 forces the 6 of cage "210x" to f9, etc., until we arrive to those 6's in a8 and g8 (encircled) which is an impossibilitie.

Author:  jomapil  [ Wed Nov 23, 2011 12:36 pm ]
Post subject:  Re: Puzzle 9x9 20NOV11

Of course, Clm. *5 in row 8 * --> b8 <> 5.
It's so evident and so easy!
We sometimes are so blind we can't see anything.

Page 1 of 1 All times are UTC + 1 hour [ DST ]
Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Group
http://www.phpbb.com/