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Lorisaur

Alternate Devon puzzle?

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I was pretty hyped for the Devon Puzzle as Marcello said that he couldn't solve it and neither could Ame. 

My idea was to figure out a solution on paper first and complete it in game afterwards. I'm a silly person or at least I paied too little attention as I did not read the words "exept the center" in the rules. This means I have been thinking at least an hour about how to make two magical squares with two different centers. Usually that should be impossible, but bringing more than one number to one side and vice versa could maybe be a solution. After a bit of reasoning I understood that the centers HAD to be 4 and 6, but I found no real solution: number 9 had to both be on the square 4 and not on 6, this means that also both numbers 2 had to be by the side of 4. On the other hand, 1 and 8 had to be on square 6. The numbers left were the 3s one 4 one 6 the 7s and the 5s. My idea was to use the 5s on the two different squares and complete them with the opposite centre number (like 5 4 6 and 5 6 4). The 3 and 7 caused a bit of trouble as the 7 HAD to be on 4 side because on the 6 side could be completed only by the 2, that were on side 4. This means that 3s had to be on side 3, but 3 + 6 left space for another 6 that was on the opposite square. I felt I was close, but I wasn't. After a while I realized I could have both 5s in the center and from there it was EXTREMELY easy, it took me skmethung like 3 minutes to figure out the solution on paper and at least 15 or just a bit more to realize it on the in-game puzzle. Now, after a couple days I was thinking... was it math possible to complete the puzzle as I was thinking of or it was just... not possible at all?

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To me the puzzle is like a mixture of the magic square and a Rubix cube. Its more of a pattern than anything to get them in the right place.

 

I forget who said what but I'm sure that person will find this topic eventually...

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1 hour ago, Lorisaur said:

I was pretty hyped for the Devon Puzzle as Marcello said that he couldn't solve it and neither could Ame. 

My idea was to figure out a solution on paper first and complete it in game afterwards. I'm a silly person or at least I paied too little attention as I did not read the words "exept the center" in the rules. This means I have been thinking at least an hour about how to make two magical squares with two different centers. Usually that should be impossible, but bringing more than one number to one side and vice versa could maybe be a solution. After a bit of reasoning I understood that the centers HAD to be 4 and 6, but I found no real solution: number 9 had to both be on the square 4 and not on 6, this means that also both numbers 2 had to be by the side of 4. On the other hand, 1 and 8 had to be on square 6. The numbers left were the 3s one 4 one 6 the 7s and the 5s. My idea was to use the 5s on the two different squares and complete them with the opposite centre number (like 5 4 6 and 5 6 4). The 3 and 7 caused a bit of trouble as the 7 HAD to be on 4 side because on the 6 side could be completed only by the 2, that were on side 4. This means that 3s had to be on side 3, but 3 + 6 left space for another 6 that was on the opposite square. I felt I was close, but I wasn't. After a while I realized I could have both 5s in the center and from there it was EXTREMELY easy, it took me skmethung like 3 minutes to figure out the solution on paper and at least 15 or just a bit more to realize it on the in-game puzzle. Now, after a couple days I was thinking... was it math possible to complete the puzzle as I was thinking of or it was just... not possible at all?

There seems to be one important property to a magic square (ms):

 the sum of all squares devided by the number of squares is equal to the center-number

 

 As such, your center-4 magic square would need a sum of 36 and your center-6 magic square would need a sum of 54. Added together this would be a sum of 90 - so far it seems possible. The 4-ms would need a row/column/diagonal sum of 12 and the 6-ms would need one of 18. It follows, that the 4-ms needs all the 1s and 2s, the 6-ms on the other hand needs all the 9s and 8s. Looking at the 4ms: 4 + 1 + x = 12 -> the 4ms needs a 7 for every 1 it has. Same goes for 4+2+x=12 -> it needs a 6 for every 2 it has. And this is the point where it stops working, since you only have 2 6s.

 

 

So as long as the mentioned property is mandatory, it doesn´t work. I wasn´t able to proof/find proof yet, that it is mandatory. It just seems intuitive by looking at proper magic squares. But should you be able to find proof that it is mandatory, it would equal proof that your initial idea doesn´t work.

 

 

 

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1 minute ago, LilyX said:

There seem to be two properties to a magic square (ms):

1- the sum of all squares devided by the number of squares is equal to the center-number

2- center-number x3 equals the sum of the rows/columns/diagonals

 As such, your center-4 magic square would need a sum of 36 and your center-6 magic square would need a sum of 54. Added together this would be a sum of 90 - so far it seems possible. The 4-ms would need a row/column/diagonal sum of 12 and the 6-ms would need one of 18. It follows, that the 4-ms needs all the 1s and 2s, the 6-ms on the other hand needs all the 9s and 8s. Looking at the 4ms: 4 + 1 + x = 12 -> the 4ms needs a 7 for every 1 it has. Same goes for 4+2+x=12 -> it needs a 6 for every 2 it has. And this is the point where it stops working, since you only have 2 6s.

 

 

So as long as properties 1 and 2 are mandatory, it doesn´t work. I wasn´t able to proof/find proof yet, that they are mandatory. It just seems this way by looking at proper magic squares. But should you be able to find proof that they are mandatory, it would equal proof that your initial idea doesn´t work.

 

 

 

This is a bit like what I was thinking... but I hoped to be missing some other options I did not see to solve the puzzle. Looks like there are not. Thanks for your help

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56 minutes ago, Lorisaur said:

This is a bit like what I was thinking... but I hoped to be missing some other options I did not see to solve the puzzle. Looks like there are not. Thanks for your help

I mean, there´s always the possibility that I made a logical error. If you are really interested in the topic, I´m sure there´s a lot of literature regarding the maths behind the Magic Square (or special versions of it, in your case).

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You cannot have six, or indeed any number above 5 as the center of a 3*3 magic square because then you couldn't place 9 anywhere on it as you note. You can't use any number BELOW 5 either, because that would leave nowhere you can put one. Its really that simple. Magic squares really aren't a math puzzle at all, its a logic puzzle.

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29 minutes ago, wcv said:

You cannot have six, or indeed any number above 5 as the center of a 3*3 magic square because then you couldn't place 9 anywhere on it as you note. You can't use any number BELOW 5 either, because that would leave nowhere you can put one. Its really that simple. Magic squares really aren't a math puzzle at all, its a logic puzzle.

Your first point is just not correct in this case because he/she wasn´t talking about standard magic squares, but rather two special magic squares that between each other have to use the numbers from 1-9 twice. This includes using the 9s only in the square with the 6-center and the 1s only in the square with the 4-center.

And your second point is just semantics. Arguing that something is a logic puzzle and not a math puzzle doesn´t make sense.

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24 minutes ago, LilyX said:

Your first point is just not correct in this case because he/she wasn´t talking about standard magic squares, but rather two special magic squares that between each other have to use the numbers from 1-9 twice. This includes using the 9s only in the square with the 6-center and the 1s only in the square with the 4-center.

And your second point is just semantics. Arguing that something is a logic puzzle and not a math puzzle doesn´t make sense.

I took the puzzle to clearly be saying that you would use the numbers 1-9 on each side.

 

The puzzle is explicitly asking for the horizontals, verticals, and diagonals to add to 15, not to find any general magic square using two sets of numbers from 1-9.

 

It also wouldn't work anyway, with 4 in the middle you also need both 2's on that side. Leaving you with something along the lines of:

 

92x

x4x

x92

 

In order to finish the top and bottom row you need two more 4's, which isn't there.

 

So you already can't have both 9's and both 2's on the same side, but you have to have it that way since the 9's CANNOT be on the same side as the 6 given the parameters of the puzzle.

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30 minutes ago, wcv said:

I took the puzzle to clearly be saying that you would use the numbers 1-9 on each side.

 

The puzzle is explicitly asking for the horizontals, verticals, and diagonals to add to 15, not to find any general magic square using two sets of numbers from 1-9.

 

It also wouldn't work anyway, with 4 in the middle you also need both 2's on that side. Leaving you with something along the lines of:

 

92x

x4x

x92

 

In order to finish the top and bottom row you need two more 4's, which isn't there.

 

So you already can't have both 9's and both 2's on the same side, but you have to have it that way since the 9's CANNOT be on the same side as the 6 given the parameters of the puzzle.

As I stated above, a square with 4 in the center would not have sums of 15 but rather 12. So 9+2+1. But yes, as stated above also: It doesn´t work. I understood the puzzle the same way you did, but Lorisaur apparently tried an alternative - and I think it´s perfectly fine to explore other possibilities.

To make it short: You are right, it is not possible - especially if you follow the instructions closely (Lorisaur: " I paied [sic] too little attention "). BUT: It´s not "really that simple" as you stated.

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