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 6x6 difficult, Sep. 4, analytical solution? 
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Posted on: Sat Sep 07, 2013 9:50 am




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Post 6x6 difficult, Sep. 4, analytical solution?
I got a question from a user asking if there is an analytical solution possible for this puzzle.

I'm thinking yes, and I'm also thinking, since so many people solved it by now, someone
can give a few hints to get him/her started?

thanks,

Patrick


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Posted on: Sat Sep 07, 2013 1:27 pm




Posts: 690
Joined: Fri May 13, 2011 6:51 pm
Post Re: 6x6 difficult, Sep. 4, analytical solution?
pnm wrote:
I got a question from a user asking if there is an analytical solution possible for this puzzle.

I'm thinking yes, and I'm also thinking, since so many people solved it by now, someone
can give a few hints to get him/her started?

thanks,

Patrick


First observe that both cages "5+" (at bottom right) cann´t simultaneously contain either [1,4] or [2,3] (this is elementary and very easy to see for several reasons) so one of them must be [1,4] and the other [2,3]. Now since b2 = 6, b3 = 5, b4 = 4 (c4 = 6), b5 = 1 (c5 = 5) >>> c1 + d1 = 17 + 15 + 16 - 42 (two columns) = 6.

Now "5+" (c2-c3) cann´t be [1,4] because c1 = 2, d1 = 4 and you would arrive to the impossibility of obtaining [1,4] for at least one of the cages "5+" in the bottom right area, so "5+" (c2-c3) = [2,3] with c2 = 3 and c3 = 2. Now observe that c1 = 4, d1 =2 is not possible because d2 = 1 and this would make impossible any combination for "5+" in d4-d5. Consequently c1 = 1, d1 = 5, c6 = 4.

Next, f3 + f4 = 6 here applying again the addition rule: 17 + 19 + 12 - 42 (two rows) = 6, so f3 =1, f4 = 5 and then a3 = 3, a2 = 5, f2 = 4.

Next: e1 + f1 = 9 >>> [3,6] so a1 = 4, b1 =2, b6 = 3. The 6 of column "a" does not go to a6 because >>> f5 = 8 (impossible) so a5 = 6.
The rest is easy with some simple logic.
I hope this helps.

(Perhaps, Patrick, it would be interesting to add the drawing for future reference).


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Posted on: Sat Sep 07, 2013 8:09 pm




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Joined: Tue May 24, 2011 4:55 pm
Post Re: 6x6 difficult, Sep. 4, analytical solution?
Thanks, clm, for your very helpful analytical solution! I was one of those who had to solve the puzzle by trial and error because even though I did realise this:
clm wrote:
First observe that both cages "5+" (at bottom right) cann´t simultaneously contain either [1,4] or [2,3] (this is elementary and very easy to see for several reasons) so one of them must be [1,4] and the other [2,3].

I completely missed this:
clm wrote:
Now since b2 = 6, b3 = 5, b4 = 4 (c4 = 6), b5 = 1 (c5 = 5) >>> c1 + d1 = 17 + 15 + 16 - 42 (two columns) = 6.

and therefore couldn't go on to see this:
clm wrote:
Now "5+" (c2-c3) cann´t be [1,4] because c1 = 2, d1 = 4 and you would arrive to the impossibility of obtaining [1,4] for at least one of the cages "5+" in the bottom right area, so "5+" (c2-c3) = [2,3] with c2 = 3 and c3 = 2. Now observe that c1 = 4, d1 =2 is not possible because d2 = 1 and this would make impossible any combination for "5+" in d4-d5. Consequently c1 = 1, d1 = 5, c6 = 4.

Regarding the latter bit of analysis, I noticed a similarity between d1 = 4 and d1 = 2 that means it could also be phrased like this:

    Now c1-d1 can't be [2,4] because d1 = 4, e3 = 4 would make obtaining [1,4] for at least one of the cages "5+" in the bottom right area impossible and, similarly, d1 = 2, e2 = 2 would mean that [2,3] could not be obtained for any of those cages. Consequently d1 = 5, c1 = 1, c2 = 3, c3 = 2, c6 = 4.

This alternative phrasing is not as reflective as yours of the actual steps we as solvers are likely to take in determining which numbers go where. It is more of an afterthought that illustrates how your initial observation about the cages "5+" at bottom right can be combined with our knowledge of the content of the cages "3+" and "10+" above them to show that neither 2 nor 4 can go into any of the cages d1, e1, d6 and e6 (and, given your second observation that c1 + d1 = 6, that c1-d1 then can't be [2,4] but must be [1,5] etc.).


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Posted on: Sat Sep 07, 2013 10:21 pm




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Joined: Fri May 13, 2011 6:51 pm
Post Re: 6x6 difficult, Sep. 4, analytical solution?
bram wrote:
...

Regarding the latter bit of analysis, I noticed a similarity between d1 = 4 and d1 = 2 that means it could also be phrased like this:

    Now c1-d1 can't be [2,4] because d1 = 4, e3 = 4 would make obtaining [1,4] for at least one of the cages "5+" in the bottom right area impossible and, similarly, d1 = 2, e2 = 2 would mean that [2,3] could not be obtained for any of those cages. Consequently d1 = 5, c1 = 1, c2 = 3, c3 = 2, c6 = 4.

This alternative phrasing is not as reflective as yours of the actual steps we as solvers are likely to take in determining which numbers go where. It is more of an afterthought that illustrates how your initial observation about the cages "5+" at bottom right can be combined with our knowledge of the content of the cages "3+" and "10+" above them to show that neither 2 nor 4 can go into any of the cages d1, e1, d6 and e6 (and, given your second observation that c1 + d1 = 6, that c1-d1 then can't be [2,4] but must be [1,5] etc.).


Thanks. Sure, bram, your additional observations are perfect and show the beauty of the structure of this type of 6x6's; the alternative phrasing is also more concise, I wrote the notes too quickly I think, I probably was thinking more of focusing or underlining the relevance and use of the "addition rule" (horizontally or vertically) to solve this particular structure of the 6x6's.


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Posted on: Sun Sep 08, 2013 12:22 pm




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Joined: Mon Jul 29, 2013 3:05 am
Post Re: 6x6 difficult, Sep. 4, analytical solution?
clm wrote:
Now since b2 = 6, b3 = 5, b4 = 4 (c4 = 6), b5 = 1 (c5 = 5) >>> c1 + d1 = 17 + 15 + 16 - 42 (two columns) = 6 ...........

Next, f3 + f4 = 6 here applying again the addition rule: 17 + 19 + 12 - 42 (two rows) = 6, so f3 =1, f4 = 5 and then a3 = 3, a2 = 5, f2 = 4.


What is this addition rule you speak of? Where do the numbers you use (17+15+16) come from, and how do they relate to the cells previously mentioned?
I get that in a 6x6 each column or row adds up to 21, so the number 42 is being used for two columns combined.

Thanks,
Jermain


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Posted on: Sun Sep 08, 2013 1:21 pm




Posts: 690
Joined: Fri May 13, 2011 6:51 pm
Post Re: 6x6 difficult, Sep. 4, analytical solution?
main wrote:
clm wrote:
Now since b2 = 6, b3 = 5, b4 = 4 (c4 = 6), b5 = 1 (c5 = 5) >>> c1 + d1 = 17 + 15 + 16 - 42 (two columns) = 6 ...........

Next, f3 + f4 = 6 here applying again the addition rule: 17 + 19 + 12 - 42 (two rows) = 6, so f3 =1, f4 = 5 and then a3 = 3, a2 = 5, f2 = 4.


What is this addition rule you speak of? Where do the numbers you use (17+15+16) come from, and how do they relate to the cells previously mentioned?
I get that in a 6x6 each column or row adds up to 21, so the number 42 is being used for two columns combined.

Thanks,
Jermain


The addition rule is exactly what you explain: in a 6x6 each line sums 21 (1 + 2 + ... + 6), in a 7x7 is 28, in an 8x8 is 36, etc. (the multiplication rule would refer to the fact that the product of all numbers in a line is a well known number, factorial of n, being n the size of the puzzle: 6! = 720 = 6 x 5 x 4 x 3 x 2 x 1, 7! = 5040, etc. (you may use this property when you have a large number of "product" cages in some area).

With respect to this particular puzzle (below) observe that the two leftmost columns (sum of 42) plus the cells c1 and d1 represent the same area than: 17 (the cage "17+") + 15 (the cage "15+") + the numbers you have already defined (positions b2, b3, b4 and b5, which sum is 16 = 6 + 5 + 4 + 1), so c1 + d1 = 17 + 15 + 16 - 42 = 6.

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Posted on: Sun Sep 08, 2013 3:32 pm




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Joined: Mon Jul 29, 2013 3:05 am
 Re: 6x6 difficult, Sep. 4, analytical solution?
Thanks for taking the time to explain further clm. I get it now!

It feels good learning these advanced techniques for getting at the more difficult puzzles. This will serve me well in the future.


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Posted on: Sun Sep 08, 2013 4:01 pm




Posts: 690
Joined: Fri May 13, 2011 6:51 pm
Post Re: 6x6 difficult, Sep. 4, analytical solution?
main wrote:
Thanks for taking the time to explain further clm. I get it now!

It feels good learning these advanced techniques for getting at the more difficult puzzles. This will serve me well in the future.


Welcome, the same idea is used in this puzzle to obtain f3 + f4 = 17 + 19 + 12 (that is, 6 + 3 + 1+ 2, in other words, b2 + c2 + d2 + e2, once we have demonstrated that c2 = 3 ...) - 42 (the two upmost rows) = 6.
Sometimes playing with sums and subtractions of cages and/or some individual cells (and taking into account the addition rule or even, in some very special cases, the value of the full puzzle which is 126 in a 6x6) we arrive to relationships between cells that are very useful.

Edit: Something additional for this puzzle since I see that you are interested in the analysis (and also the person who originally asked Patrick for some analytical solution, ... , as well as all those puzzlers who like the analysis).

In my first post in this thread I said: “ … The 6 of column “a” does not go to a6 because >>> f5 = 8 (impossible) so a5 = 6 …”.). This is an initial approach but now I am going to underline a complementary and more advanced technique to quickly finish this 6x6. We depart from the graphic below (where we have entered the numbers found) and I am going to explain why I have suddenly and simultaneously placed f1 = 6, a6 = 1 and f6 = 2 (in blue colour). The reasons are:

First the sum of the corners is 13, why?: because C = I - (p - 4) x s (in this equation C is the sum of the corners, I is the sum of the 4x4 inner area, 55 in this case, p is the size of the puzzle, 6 in this case, and s is the sum of a line, 21 in this case; see the demonstration of this equation in the original post of the thread viewtopic.php?f=3&t=41 , so C = 55 – 2 x 21 = 13.

Secondly , if we observe the inner area, we quickly see that the numbers 1, 2, 4 and 6 are repeated three times while the numbers 3 and 5 are repeated only twice, so the corners are 1, 2, 4 and 6 and the 3 and the 5 cann’t be in the corners (for the demonstration see also the topic “Segmentation” in the section “Solving strategies and tips”: viewtopic.php?f=3&t=239 ). We know that in a 6x6 puzzle the inner area, I (the internal 4x4 box), is made with the four corners plus twice the full set of numbers in the puzzle so the full set 1 thru 6.

Consequently from the pair [3,6] in e1-f1 the 6 goes to the corner f1 and the 3 obviously not because the 3 cannot be a corner. The only place for a 1 as a corner is a6 and the 2 goes to the corner f6 and from this moment to finish the puzzle is a question of seconds.

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