7 posts
• Page **1** of **1**

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.

- udosuk
**Posts:**2698**Joined:**17 July 2005

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.

- underquark
**Posts:**299**Joined:**06 September 2005

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.

- motris
**Posts:**71**Joined:**13 March 2006

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:

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...

Solution

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

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...

- udosuk
**Posts:**2698**Joined:**17 July 2005

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:

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):

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:

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...

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...

Mission accomplished...

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...

- udosuk
**Posts:**2698**Joined:**17 July 2005

7 posts
• Page **1** of **1**