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Trial and error
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Posted on: Fri Sep 23, 2011 11:29 am

Posts: 2342
Joined: Thu May 12, 2011 11:58 pm
Re: Trial and error
pnm wrote:
I think the 5x5 difficult is almost always solvable with analysis
(someone correct me if I'm wrong..), simply because it's such
a small puzzle.

Just did the 5x5, and yes, definitely doable...

For example, to get started, by considering the combinations for the 1- cage
at the bottom left, and the 16x cage at the bottom right, you can figure out
quite easily that there should be a 3 in C5. Which gives you the whole 11+ cage.

Then try to infer what has to go into E4.

Then consider what the options are for the 8+ cage, and for the 12+ cage
in the same row, and figure out what the solution is for C3.

etc. etc...

Posted on: Sun Oct 02, 2011 8:18 pm

Posts: 116
Joined: Sat May 14, 2011 3:18 am
Re: Trial and error
pnm wrote:
For example, to get started, by considering the combinations for the 1- cage
at the bottom left, and the 16x cage at the bottom right, you can figure out
quite easily that there should be a 3 in C5. Which gives you the whole 11+ cage.

Given that the 16x could be either 4x1x4 or 2x4x2 and that 1- could be either 5-3-1, 5-2-2, or 4-2-1 (each in various combinations within the cage), I fail to see how "you can figure out quite easily that there should be a 3 in C5." To me, it looks like a lot of trial and error just to get to that point.

Also, earlier in this thread a couple of people mention "parity checks." What is that, and how do you use it to solve calcudoku?

Posted on: Sun Oct 02, 2011 9:32 pm

Posts: 2342
Joined: Thu May 12, 2011 11:58 pm
Re: Trial and error
pharosian wrote:
Given that the 16x could be either 4x1x4 or 2x4x2 and that 1- could be either 5-3-1, 5-2-2, or 4-2-1 (each in various combinations within the cage), I fail to see how "you can figure out quite easily that there should be a 3 in C5." To me, it looks like a lot of trial and error just to get to that point.

Well, yes, you 're right, you would have to try those combinations.
pharosian wrote:
Also, earlier in this thread a couple of people mention "parity checks." What is that, and how do you use it to solve calcudoku?

Check out

Patrick

Posted on: Mon Oct 03, 2011 12:34 am

Posts: 720
Joined: Fri May 13, 2011 6:51 pm
Re: Trial and error
pnm wrote:
pharosian wrote:
Given that the 16x could be either 4x1x4 or 2x4x2 and that 1- could be either 5-3-1, 5-2-2, or 4-2-1 (each in various combinations within the cage), I fail to see how "you can figure out quite easily that there should be a 3 in C5." To me, it looks like a lot of trial and error just to get to that point.

Well, yes, you 're right, you would have to try those combinations.
pharosian wrote:
Also, earlier in this thread a couple of people mention "parity checks." What is that, and how do you use it to solve calcudoku?

Check out

Patrick

A quick way to see this:
Although the combination 3-1-1 is also possible for the cage 1-, it is rapidly supressed since 16x should be 2-4-2 and now the total known sum would be 5 + 11 + 8 = 24 so a 6 (30 - 24 to complete 2 rows) would go to E4.
Then, if a 3 is inside the cage -1, the other two numbers must be 1 and 5 necessarily and now 9 + 11 = 20 and the sum of 16x and the cell E4 equals 10, but 2-4-2 and a 2 in E4 is not possible and 4-4-1 and a 1 in E4 go against the hypothesis (because in this case two 1's have already been considered present in rows 4 and 5).
So the 3 must go to C5.
(and, inmediately, B4 = 3, C4 = 5). Also we see that 2-4-2 is not valid in 16x since 5-1 in A5-B5 produce a 3 in A4,... , concluding that 16x = 4-1-4 and then 5-2-2 for the cage 1-, etc.).

But perhaps the fastest way is:
To consider only 3-1-1 (as before), supressing this combination, and since 4-2-1 is not possible (two 4's in case of 4-1-4 or two 2's in case of 2-4-2), the other possibilities 5-3-1 or 5-2-2 have a sum of 9, so the sum of 16x and E4 equals 10, then 2-4-2 is not valid (E4 = 2) so 16x = 4-1-4 and E4 = 1, and now two 1's in rows 4 and 5 drive to 1- = 5-2-2 (and not 5-3-1). Rows 4 and 5 are totally defined.

For the question of the parity, and using this same diagram: The sum of the whole puzzle is odd (5 x 15 = 75). Every two cages odd produce a sum of all numbers inside them even (for instance the cages 1- and 11+); 2- we know is an even cage, and since 3- is odd, finally the addition of the cages 2: and 16x must be even, but now we know that 16x is odd (4-1-4) so 2: must be odd too and then 1-2 is the only valid combination (not 2-4); we can write the pair 1-2 as the final candidates in B1-C1 inmediately after solving the cage 16x.

Thanks, Patrick, for the reference to my original post, in that moment I did not provide a diagram showing a good example of the use of this property, but anyway the target is understanding the nuclear idea.

Posted on: Tue Oct 04, 2011 3:56 pm

Posts: 116
Joined: Sat May 14, 2011 3:18 am
Re: Trial and error
Thanks, clm!

I have since read your article on parity. I'm still trying to get used to thinking about cages in these terms, but I see how it could be helpful.

Of more immediate assistance, however, is your comment about keeping the sum of the rows (or columns) in mind. I use a sum on each row and column when I'm using Excel to solve book puzzles, but it's always been more of a "sanity check" to ensure that my solution doesn't have dupes in a row or column because my Excel spreadsheet doesn't have Patrick's handy "continuous checking" property to flag errors.

I see now, though, that if I keep the sum in mind as you describe in your first sentence, I could rule out possibilities more quickly.

Posted on: Tue Oct 11, 2011 7:23 pm

Posts: 246
Location: Lisbon, Portugal
Joined: Sun Sep 18, 2011 5:40 pm
Re: Trial and error
Patrick, the 5x5 difficult of today (11OCT11) I only "analysed" the row 5 :

34,34,125,125,125

After that, it is impossible ( for me ) to continue the solution without the Trial and Error. Did I fail anything or are there any of those little puzzles only solved with Trial and Error?

_________________
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Posted on: Tue Oct 11, 2011 8:08 pm

Posts: 2342
Joined: Thu May 12, 2011 11:58 pm
Re: Trial and error
jomapil wrote:
After that, it is impossible ( for me ) to continue the solution without the Trial and Error. Did I fail anything or are there any of those little puzzles only solved with Trial and Error?

The way I'd continue would be along these lines:
- the 1- in e1,e2,e3 can only be 4,1,2 or 5,1,3
- so the number in e5 has to be a 2 or a 5
- so the number in e4 has to be a 3 or a 4
- so the 8+ cage must have a 3 somewhere

then:
- the 2- cage in a3,a4,b4 can only be 4,1,1 or 5,1,2
- the 11+ cage can only be 2,4,5 or 5,1,5 (3,5,3 not possible because of the 3 in the 8+ cage)
- but 5,1,5 also impossible, because then the 2- cage should be 4,1,1
- so the 11+ cage must have 2,4,5

then:
- the 2- cage cannot be 4,1,1 because then the left column would have 4,1,3, leaving
2,5 for the 2- cage at the top left

so:
- the 8+ cage cannot have 2,3 in d3 and d4
- so a 1 has to go into c5

etc. etc.

I'll leave it to others to finish the story

Patrick

Posted on: Tue Oct 11, 2011 10:50 pm

Posts: 422
Joined: Fri May 13, 2011 2:43 am
Re: Trial and error
I did it more like this:
- after 3,4 and 1,2,5 in row 5...
- a3,a4,b4 has to be 1,1,4 or 1,2,5
- it cannot be 1,1,4 because then a5,b5 would be 3,4 which only leaves 2 and 5 to go into the 2- cage at a1,a2
- so a3,a4,b5 must be 1,2,5
- then sum row 3 and 4 = 1+2+5+11+8+e3 = 30 so e3=3
- e1,e2,e3 can only be 1,3,5 with e3=3
- then e5=2, e4=4
- then d3, d4 must be 1,3 with d3=1, d4=3
- then d5=5, c5=1
- then d1,d2 must be 4,2 so c1=3
- since solutions are unique, a1,a2 cannot be 2,4 so there must be a2=3
- then a5=4, b5=3
- there can only be 5,2,2 or 3,1,1 or 4,1,2 or 5,1,3 in 1- L-shaped cage so b1,b2,c2 must be 4,1,2 or 5,1,3 (must have a 1)
- so a1=5, e1=1, e2=5, a3=2, a4=1, b4=5
- then b1,b2,c2 must be 4,1,2
- since solutions are unique, b2=1
- then b1=2, b3=4, c2=4, c3=5, c4=2
DONE

Lots of other ways to arrive at solution...

Posted on: Tue Oct 11, 2011 10:56 pm

Posts: 2342
Joined: Thu May 12, 2011 11:58 pm
Re: Trial and error
sneaklyfox wrote:
I did it more like this:
- after 3,4 and 1,2,5 in row 5...
...
- then b1=2, b3=4, c2=4, c3=5, c4=2
DONE

This shows why you're always faster than me

Posted on: Tue Oct 11, 2011 11:11 pm

Posts: 246
Location: Lisbon, Portugal
Joined: Sun Sep 18, 2011 5:40 pm
Re: Trial and error
I never performed an analysis in that way. It looks like it is appropriate for little puzzles. With 3 weeks as a calcudolic I conclude the resolution of a 5x5 puzzle is a little different that a 9x9 or a 12x12, though the principles are the same.

Thank you Patrick and sneaklyfox for the lesson.

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