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jomapil
Posted on: Tue Nov 01, 2011 5:46 pm
Posts: 246 Location: Lisbon, Portugal Joined: Sun Sep 18, 2011 5:40 pm

The difficult puzzle 6x6 of today ( 01NOV2011 )
I solved all the puzzles of the last 7 days except the 6x6 difficult today. It has " infinite " possibilities and the time to solve it can be so great. So I gave up. Even the explanation of Clm in his 9x9 difficult presented here doesn't apply because the sums in the corners are too great. So, how is it possible to solve it in 5 minutes? Is there a special treatment in this case? Or was a question of luck? If anyone has a special process I thank very much for the explanation.
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Last edited by jomapil on Tue Nov 01, 2011 8:21 pm, edited 1 time in total.




sneaklyfox
Posted on: Tue Nov 01, 2011 6:23 pm
Posts: 422 Location: Canada Joined: Fri May 13, 2011 2:43 am

Re: The difficult puzzle 9x9 of today ( 01NOV2011 )
The large cages in the corner are probably among the last to be solved. I did mostly everything else first. Sometimes you can know that a certain number may be in those cages but not the exact placement. Also, I think for this puzzle I used the product rule/method. After enough cells are filled, you can use that... I forget exactly where it's explained in the tips section of the forum though.




sneaklyfox
Posted on: Tue Nov 01, 2011 6:26 pm
Posts: 422 Location: Canada Joined: Fri May 13, 2011 2:43 am

Re: The difficult puzzle 9x9 of today ( 01NOV2011 )
Ok, I think it's here: viewtopic.php?f=3&t=26That one talks about addition but it also works for multiplication. For this 9x9 puzzle, I used it on the four leftmost columns and four bottommost rows.




pnm
Posted on: Tue Nov 01, 2011 6:44 pm
Posts: 2469 Joined: Thu May 12, 2011 11:58 pm

Re: The difficult puzzle 9x9 of today ( 01NOV2011 )
sneaklyfox wrote: Ok, I think it's here: viewtopic.php?f=3&t=26That one talks about addition but it also works for multiplication. For this 9x9 puzzle, I used it on the four leftmost columns and four bottommost rows. But is jomapil referring to the 9x9 puzzle (stated in the subject), or the 6x6 (mentioned once in the text, and he writes about solving it in 5 minutes)? Patrick




jomapil
Posted on: Tue Nov 01, 2011 8:16 pm
Posts: 246 Location: Lisbon, Portugal Joined: Sun Sep 18, 2011 5:40 pm

Re: The difficult puzzle 9x9 of today ( 01NOV2011 )
You are right, Patrick. Excuse me but it's the 6x6, of course!
_________________Visit http://www.calcudoku.org the most interesting and addictive site of puzzles.




clm
Posted on: Wed Nov 02, 2011 12:31 am
Posts: 735 Joined: Fri May 13, 2011 6:51 pm

Re: The difficult puzzle 6x6 of today ( 01NOV2011 )
jomapil wrote: I solved all the puzzles of the last 7 days except the 6x6 difficult today. It has " infinite " possibilities and the time to solve it can be so great. So I gave up. Even the explanation of Clm in his 9x9 difficult presented here doesn't apply because the sums in the corners are too great. So, how is it possible to solve it in 5 minutes? Is there a special treatment in this case? Or was a question of luck? If anyone has a special process I thank very much for the explanation. You refer to the stepbystep solution of the 6x6 difficult, of course, but that is a different type of 6x6 (Patrick uses 10 different types of 6x6's). It is true that the corners must sum 13 (55  42) being 55 the inner part and 42 the value of two lines, but this is not very useful to start solving this puzzle. I have not space for the full solution in this moment but try to follow this: Observe that you must have the pair 56 in e2e3 (either if you obtain the 8+ with 62 or with 53) ok?. Then e2e3e4e5 have a sum of 14 so c6d6 have a sum of 6 (18 + 16 + 14  42, being 42 te sum of columns e and f). It is easy to see that c6 = 5 and d6 =1 is not possible because when you arrive to b2 if b2 = 4 then e1, f1 and f2 must be simultaneously 1 (by the addition of the first two rows) or if b2 = 1, then e1 + f1 + f2 = 6 which is impossible because f3 + f4 = 12 (7 + 5). So c6d6 is the pair 42 and now it is easy to see that c6 = 4 and d6 = 2 is not valid since d3 = 1 d3 = 5 and the cage 8+ would be impossible, so c6 = 2, d6 = 4, e1 = 4, e6 = 3. The rest is not difficult.
Last edited by clm on Wed Nov 02, 2011 9:54 pm, edited 3 times in total.




sneaklyfox
Posted on: Wed Nov 02, 2011 12:47 am
Posts: 422 Location: Canada Joined: Fri May 13, 2011 2:43 am

Re: The difficult puzzle 6x6 of today ( 01NOV2011 )
Ok, sorry I wasn't much help! Wasn't sure if it was 9x9 as in the topic (though now it's corrected) or 6x6 as mentioned in the text so I assumed it was the bigger puzzle.




jomapil
Posted on: Wed Nov 02, 2011 1:05 pm
Posts: 246 Location: Lisbon, Portugal Joined: Sun Sep 18, 2011 5:40 pm

Re: The difficult puzzle 6x6 of today ( 01NOV2011 )
Even so, thanks, Sneaklyfox, for your availability.
Clm, your explanation opened more a degree in my knowledges. The cells c6+d6 are the key for the correct solution of this puzzle. I never would attain the solution! As is said by our iberian compatriots " Yo no creo em brujas, pero que las hay las hay ". Thanks very much.
_________________Visit http://www.calcudoku.org the most interesting and addictive site of puzzles.




clm
Posted on: Wed Nov 02, 2011 10:52 pm
Posts: 735 Joined: Fri May 13, 2011 6:51 pm

Re: The difficult puzzle 6x6 of today ( 01NOV2011 )
jomapil wrote: Even so, thanks, Sneaklyfox, for your availability.
Clm, your explanation opened more a degree in my knowledges. The cells c6+d6 are the key for the correct solution of this puzzle. I never would attain the solution! As is said by our iberian compatriots " Yo no creo em brujas, pero que las hay las hay ". Thanks very much. Welcome. You probably observed that there was an slight error in "e1 + e2 + f1 = 6" that I have reedited to the correct "e1 + f1 + f2 = 6" (sorry). Yes... "No creo en las meigas ... pero haberlas haylas". The rough translation should be something like this: "I do not believe in witches but I am sure they exist". I am particularly fascinated with observing how the numbers in some part of the puzzle affect other parts very "far away" (specially in the 9x9's, 12x2's, etc.), it must be a question of witches. Now observe this other point of view: In puzzles like the yesterday's 6x6 difficult (id: 377321) we may "cut" the puzzle in different ways, for instance, using the addition of the first three rows: a3 + "4" + 5 + 9 + 8 + 21 + 18  f4 = 63 (where "4" represents the sum of the numbers in the cage "4"), that is: a3 + "4" = f4 + 2; but the maximum value of the cell f4 is 6, so the maximum value of "4" is 7 (since a3 values at least 1) so directly "4" is the pair 15 (it can never be 26, we could have started here).




arjen
Posted on: Thu Nov 03, 2011 11:09 am
Posts: 49 Joined: Sun Oct 16, 2011 10:58 am

Re: The difficult puzzle 6x6 of today ( 01NOV2011 )
My 'solution' was: Total of 4, 2 and 1 = 19 (6*21  sum(+cages)) By 26 in 4 cage: 11 left over for sum '2' and '1' cage Possible values: 13 / 34 (invalid because of +5 cage) 24 / 23 (invalid because of +3 cage) 35 / 12 (invalid because of +3 cage)
@clm: thanks for explaning an easier one.
@patrick: 'Show coordinates' doesnt work by solution view.




