Perfect couples number place (6 more added)

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Perfect couples number place (6 more added)

Postby udosuk » Mon Aug 07, 2006 7:28 am

Image

The rules are simple:

Fill in the white boundary cells with the numbers 1 to 32, so that all adjacent cells (couples) sum to a perfect square number... You could mark the sums in the yellow cells and cross out the blue candidates in the middle as you use each of them... 3 of the white cells are already filled for you...

The solution is unique and you can find it logically... Enjoy!:)

Edit: Added solution and 6 more puzzles. Just scroll down the page...
Last edited by udosuk on Tue Aug 08, 2006 2:52 pm, edited 1 time in total.
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Postby underquark » Mon Aug 07, 2006 3:22 pm

I got as far as a chain of 25 and then the smile on my face faded. 25's a perfect square - can I jut stop there? Nice one.
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Postby udosuk » Mon Aug 07, 2006 3:56 pm

Thanks underquark... For giving me a chance to picture the fading of the smile on your face...:D

Seriously... Thank you for trying it out... I'm a sure a player of your calibre will conquer this soon... Hope it gives you some fun in the process!:)
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Postby motris » Mon Aug 07, 2006 8:10 pm

Thanks for this challenge. While the graphic image of the puzzle is not really needed to describe what is intended, I liked how focused attack on what digits must be connected works really well to give the whole cycle. I believe I ended up with about 4 reasonably sized "chains" and 5 leftover numbers from just logic and then I made a good intuition that ended up piecing them all together to get the solution.
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Postby udosuk » Tue Aug 08, 2006 5:22 am

Well done Thomas... You're right the pic just serves as a tool to tackle this problem, which is a bit more like maths...

It's also proved that no shorter chain exists for this property (32 is the minumum you need to form such a chain)...:idea:
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Postby udosuk » Tue Aug 08, 2006 6:44 pm

Here is the solution for the original puzzle...

Solution

BTW, if we change the candidate range from 1..32 to 0..31, we get 6 more puzzles:

Image

Each of them has a different unique solution...

PS: Thomas, I tried your approach, and with just logic I got 6 reasonably sized chains and 1 single leftover number from which I easily linked all the chains together with more logical reasoning... No "intuition" moves or guesswork were necessary...:idea:
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Postby udosuk » Sat Aug 12, 2006 11:02 am

Here is the solution of the 6 new ones...

Solution

And this is the "chains and links" approach suggested by motris...

First, we need to construct a table of all possible "spouses" for each candidate:
Code: Select all
 1   3,8,15,24
 2   7,14,23
 3   1,6,13,22
 4   5,12,21,32
 5   4,11,20,31
 6   3,10,19,30
 7   2,9,18,29
 8   1,17,28
 9   7,16,27
10   6,15,26
11   5,14,25
12   4,13,24
13   3,12,23
14   2,11,22
15   1,10,21
16   9,20
17   8,19,32
18   7,31
19   6,17,30
20   5,16,29
21   4,15,28
22   3,14,27
23   2,13,26
24   1,12,25
25   11,24
26   10,23
27   9,22
28   8,21
29   7,20
30   6,19
31   5,18
32   4,17

Those with only 2 possible "spouses" are automatically hooked together and form a "ménage à trois"... The ones hooked on both sides are no longer available to all other candidates... As a result more "forced relationships" will appear and eventually we will have a single loop linking all the candidates...

At the very first stage, we can deduce these chains easily (with possible spouses of each end listed):
Code: Select all
4|11 [5,31,18,7,29,20,16,9,27,22] 3|14
6|15 [10,26,23,2,14] 11|22
5|12|21 [4,32,17] 8|19
3|10|19 [6,30,19] 6|17
1|17 [8,28,21] 4|15
5|14 [11,25,24] 1|12
1|6|22 [3,13,12] 4|24
3|8|15|24 [1] 3|8|15|24
1|10|21 [15] 1|10|21

Looking at [6,30,19], 19 could be linked with both 6 and 17, but if we link 19 and 6 together, it will form a "closed" triangular loop and the 3 candidates no longer could be linked with the others... Therefore 19 must be linked with 17 which is no longer available for 8... So the 8 in [8,28,21] must be linked with 1... Hence we have this situation:
Code: Select all
4|11 [5,31,18,7,29,20,16,9,27,22] 3|14
5|12|21 [4,32,17,19,30,6] 3|10
6|15 [10,26,23,2,14] 11|22
3|15|24 [1,8,28,21] 4|15
5|14 [11,25,24] 1|12
1|6|22 [3,13,12] 4|24
1|10|21 [15] 1|10|21

Next, notice that 1 and 21 are opposite ends of a chain. 15 must be linked with 10, otherwise it would be "forced" to have 1 and 21 on both sides and thus forming a "closed" pentagonal loop... Therefore 10 is no longer available for 6 which must now be linked with 3, forming a 9-candidate chain... Hence 22 must be linked with 14 and then 11 must be linked with 5...
Code: Select all
1|12 [24,25,11,5,31,18,7,29,20,16,9,27,22,14,2,23,26,10,15] 1|21
12|21 [4,32,17,19,30,6,3,13,12] 4|24
15|24 [1,8,28,21] 4|15

Lastly, 12 cannot be linked with 4 as it would form a 9-candidate closed loop... Thus 12 and 24 are linked together and 4 is forced to be linked with 21... Finally 1 and 15 are linked together and we have the big loop formed...
Code: Select all
[1,8,28,21,4,32,17,19,30,6,3,13,12,24,25,11,5,31,18,7,29,20,16,9,27,22,14,2,23,26,10,15]

Mission accomplished...:)
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