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Not the End Of Year Puzzle :)
http://www.calcudoku.org/forum/viewtopic.php?f=5&t=518
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Author:  pnm  [ Thu Dec 26, 2013 6:40 pm ]
Post subject:  Not the End Of Year Puzzle :)

If you prefer a smaller End Of Year Puzzle, here's one:

Image

[tongue]

Author:  clm  [ Thu Dec 26, 2013 10:29 pm ]
Post subject:  Re: Not the End Of Year Puzzle :)

pnm wrote:
If you prefer a smaller End Of Year Puzzle, here's one:

...


Very curious, a typical 4x4 "difficult" (and not valid for 4x4 timed puzzles since it would probably increase a bit the solution time :-)).

Would it be possible to generate a 4x4 subtraction only without individual cells (solution unique, of course)?.
--------------------------------------------------------

Edit: And BTW, what solution would you give to this other puzzle?

Image

Author:  jaek  [ Sat Dec 28, 2013 5:02 pm ]
Post subject:  Re: Not the End Of Year Puzzle :)

clm wrote:
Edit: And BTW, what solution would you give to this other puzzle?


Very clever indeed. My natural inclination was to assume that when the 1+ cage couldn't be {0,1} that it would be {-1,2}. But then the -2x cage above that threw in yet another 'plot twist'.

For the record, I came up with

Code:
   0   1  -2   3
   3   0   1  -2
  -2   3   0   1
   1  -2   3   0

Author:  clm  [ Sat Dec 28, 2013 8:41 pm ]
Post subject:  Re: Not the End Of Year Puzzle :)

jaek wrote:
... My natural inclination was to assume that when the 1+ cage couldn't be {0,1} that it would be {-1,2}. But then the -2x cage above that threw in yet another 'plot twist'.

For the record, I came up with

Code:
   0   1  -2   3
   3   0   1  -2
  -2   3   0   1
   1  -2   3   0


Thank you for your interest and comments. Correct! [thumbup]. Another way: Since the 0's must be in the main diagonal a1-c4 >>> "2-" cann't be [-1,1,2] because d1 + d2 + d3 = -1 + 1 + 2 = 2 >>> c4 = 3 but we could not introduce a fifth number so necessarily "2-" = [1,1,-2] and then we find the 3 as the fifth number though now everything goes in its place: for instance, if the unknown quantity c4 is named "a", in the cage "5+": a + 0 + 0 + (-2) + 1 + a = 5 >>> 2a = 6 >>> a = 3.

What I tried to show is that it's possible to build puzzles where initially you do not know the components, you must determine them until arriving to the solution (which, of course, must be unique for consistency).

Author:  nicow  [ Mon Dec 30, 2013 12:49 pm ]
Post subject:  Re: Not the End Of Year Puzzle :)

Thank you, CLM, very nice puzzle!

Author:  clm  [ Mon Dec 30, 2013 8:35 pm ]
Post subject:  Re: Not the End Of Year Puzzle :)

nicow wrote:
Thank you, CLM, very nice puzzle!


Welcome, nicow, probably a 5x5 with, let's say, -3, 1, 0, 2, 5, or similar, could be curious and harder (if, for instance, the "5" is not visible at first sight, etc.).

Author:  nicow  [ Tue Dec 31, 2013 1:48 pm ]
Post subject:  Re: Not the End Of Year Puzzle :)

Yes, clm, it is harder.
In my model I began with 1,2,3,... Then a new one with 0,1,2...
When -1 was introduced I was forced to redesign; I added an extra layer with indexes and both models were integrated.
So the puzzle 'with only prime numbers' yielded no problem.

But this one is a step further. I have to add variables. So this is a new challenge!

Author:  beaker  [ Wed Jan 01, 2014 3:09 am ]
Post subject:  Re: Not the End Of Year Puzzle :)

This is not a reply to the previous post but instead a wish for a Happy New Year to all puzzlers .........Happy New year to those who have not reached 24:00 and wish for a great puzzling year for those who are into Jan.1 already........and from me, because of where I am, Hau'oli Makahaki Hou......to all........son's wedding was yesterday and what a beautiful setting it was.........aloha

Author:  pnm  [ Wed Jan 08, 2014 7:40 pm ]
Post subject:  Re: Not the End Of Year Puzzle :)

clm wrote:
Would it be possible to generate a 4x4 subtraction only without individual cells (solution unique, of course)?.


I think so :-)

Image

Author:  clm  [ Wed Jan 08, 2014 11:12 pm ]
Post subject:  Re: Not the End Of Year Puzzle :)

pnm wrote:
clm wrote:
Would it be possible to generate a 4x4 subtraction only without individual cells (solution unique, of course)?.


I think so :-)

...


Thanks [thumbup], undoubtedly very interesting (yes, solution unique), this type of puzzle requires a little more "thinking", at least to apply the parity rule, i.e., ...

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