Magic hexagons
William Rex Marshall
It has long been known that the only possible nontrivial magic hexagon
(using a regular hexagonal array) is the unique solution for an order 3
regular array, using the integers from 1 to 19, with a magic constant
of 38:
3 17 18
19 7 1 11
16 2 5 6 9
12 4 8 14
10 13 15
If the hexagonal array is NOT a regular hexagon but a semiregular
hexagon (in which the number of cells along each border edge alternates
between a and b) are any other magic hexagons possible? Remember, for
an array with N cells, all the positive integers from 1 to N must each
be used once, and the sum of the integers in each row must be the same
for every row.
William Rex Marshall
New Zealand
williamrexmarshall (at) yahoo (dot) com
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D & A Klinkenberg
&&&
Hello William,
Do you mean something like:
> xx xx xx
> xx xx xx xx
> xx xx xx xx xx
> zz xx xx xx xx zz
> zz xx xx xx zz
> zz zz zz zz
This looks difficult to solve. It has 27 numbers and with a little
calculation, I guess that the 'magic number' is 63. I don't know what
else can be said about it with only a little calculation.
It took me a program that considered many possibilities before I could
solve the 19 number one with the solution you showed. Once the
solution is found, it is easy to prove correct but (I believe) hard to
prove unique. With the zz'z added, I can imagine that it would have
even more possibilities to check.
Of course, other regular hexagons are possible that don't have a side
of 'length' three.
Dan in NY
Colin Barker
D & A Klinkenberg a écrit dans le message <3A3D4FE5...@hvc.rr.com>...
Many years ago I wrote a Prolog program to solve this puzzle. And it
"proved" that, if we ignore rotations and reflections, there is only one
solution.
>
>Of course, other regular hexagons are possible that don't have a side
>of 'length' three.
>
Yes, but none of these can be made into "magic" hexagons (except for the
trivial case of side 1). I also wrote an Excel spreadsheet to demonstrate
that, up to side 100, there are no other "magic" hexagons.
>Dan in NY
Colin
William Rex Marshall
Yes! Of course, it's not the only possibility.
> >
> >This looks difficult to solve. It has 27 numbers and with a little
> >calculation, I guess that the 'magic number' is 63. I don't know
what
> >else can be said about it with only a little calculation.
> >
> >It took me a program that considered many possibilities before I
could
> >solve the 19 number one with the solution you showed. Once the
> >solution is found, it is easy to prove correct but (I believe) hard
to
> >prove unique. With the zz'z added, I can imagine that it would have
> >even more possibilities to check.
>
> Many years ago I wrote a Prolog program to solve this puzzle. And it
> "proved" that, if we ignore rotations and reflections, there is only
one
> solution.
>
> >
> >Of course, other regular hexagons are possible that don't have a side
> >of 'length' three.
> >
>
> Yes, but none of these can be made into "magic" hexagons (except for
the
> trivial case of side 1). I also wrote an Excel spreadsheet to
demonstrate
> that, up to side 100, there are no other "magic" hexagons.
>
> >Dan in NY
>
> Colin
>
>
There are certainly no other magic hexagons for *regular* hexagonal
arrays, as I had already mentioned in my original posting. It can be
shown algebraically that the magic constant for regular magic hexagons
is only an exact integer when the side is either 1 (trivial) or 3
(first proven unique by Charles W. Trigg in the early 1960s).
But the big question is: what about *semiregular* hexagonal arrays of
the type as described above? Can any new magic hexagons of *this* type
be constructed? I have been unable to prove it impossible.
William Rex Marshall
New Zealand
williamrexmarshall (at) yahoo (dot) com
D & A Klinkenberg
> Yes! Of course, it's not the only possibility.
> arrays, as I had already mentioned in my original posting. It can be
> shown algebraically that the magic constant for regular magic hexagons
> is only an exact integer when the side is either 1 (trivial) or 3
> (first proven unique by Charles W. Trigg in the early 1960s).
>
> But the big question is: what about *semiregular* hexagonal arrays of
> the type as described above? Can any new magic hexagons of *this* type
> be constructed? I have been unable to prove it impossible.
>
> William Rex Marshall
> New Zealand
> williamrexmarshall (at) yahoo (dot) com
&&&
Hello William,
You say, "as I had already mentioned in my original posting" and
indeed you had. I had missed that and am sorry I brought it up. I am
looking into the case of 4 and 3 that I posted. I hope others also
look into it. I am looking into all the 'rules' I can state about
this 'figure'. I haven't tried to draw any other. Generalizing it to
a and b would come later for me. Maybe I should try smaller ones
first. Here it is with a better lineup than my first posting:
xx xx xx
xx xx xx xx
xx xx xx xx xx
zz xx xx xx xx zz
zz xx xx xx zz
zz zz zz zz
Dan in NY
D & A Klinkenberg
&&&
Hello William, [my 3rd reply]
I have been working on this problem, off and on. I don't have an
answer to the question about whether any semiregular hexagon is magic,
but I have a formula for the magic constant.
This post is to tell you some details of my interim results, and to
pose some problems that I have. After the previous posts, I looked at
why the regular hexagon is magic only when a =3. I don't know how
that was proved but I wondered if the proof used the magic constant.
It is not always an integer as a changes. When it is not an integer,
there is no magic hexagon. Maybe when it is an integer, a magic
hexagon is possible.
I studied hexagons when 13>a>b>2, (a small sample) where a and b are
the sides of a hexagon. I believe that for any a and b, it is
possible to draw the figure.
I have derived these formulas:
S =a+b-1, where S is the longest diagonal in the hexagon.
N =((a^2-3a+b^2-3b)/2)+2ab+1, where N is the number of cells in the
hexagon. With this formula, N is always an integer.
M =N(N+1)/(2S), where M is the Magic constant. Sometimes M is an
integer and sometimes not. If M is not an integer, no magic hexagon
can be found for that combination of a and b.
Problem 1) if a>b is there a method for finding all and only the
values of a and b such that M is an integer. In the small range I
gave, 12 to 3, I found 10 have integer M, and 35 do not. I will post
list on request.
I don't have a method which decides whether M is an integer except by
using this formula. If M is an integer, I don't have any method to
determine whether a magic hexagon is possible.
It has been stated (and I believe it is true) that there is no magic
hexagon if a =b except when a =b=3. When a =b,
S =2a-1
N =3a^2-3a+1
M =N(N+1)/(2S).
Problem 2) If a = b>3, is M ever an integer? See the new thread,
"Integer solution?"
When a =3 and b =4, M is an integer. The hexagon for this (as already
shown) has N =27 and M =63 and is:
xx xx xx
xx xx xx xx
xx xx xx xx xx
zz xx xx xx xx zz
zz xx xx xx zz
zz zz zz zz
I have tried to determine whether there is a magic hexagon for this
figure and I do not have an answer. While studying it I came to the
equations listed in the thread, "six eq. six unk" but I don't yet know
whether a solution will help me with this question.
If someone finds a typo in the above, I will try to delete it and post
a correction.
Dan in NY
William Rex Marshall
days ago.
"D & A Klinkenberg" <dklink...@hvc.rr.com> wrote in message
news:3A5B7EC4...@hvc.rr.com...
In my second posting to rec.puzzles on this problem I noted that it did,
though I didn't give an actual proof. However, Geoff Bailey has posted
one in the "Integer Solution?" thread.
To end some of the misery, the answer for a = 3, b = 4 is given below.
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A simple impossibility proof for a = 3, b = 4:
A A A
D D D D
E - - - F
B E - - F C
B E - F C
B E F C
The numbers in the six rows each marked in the above diagram by one of
the letters from A to F must add to six times the magic constant (6 x
63), or 378. But the maximum sum that can possibly be achieved by
adding the largest numbers (from 27 down to 7) is only 357. Therefore a
magic hexagon for a = 3, b = 4 is impossible.
This proof can be extended to many larger semiregular hexagonal arrays,
but there seem to be difficulties dealing adequately with large "nearly
regular" hexagons. Can a rigorous proof be found that covers every
case, or are there semiregular magic hexagons after all?
Here's a table for all non-trivial cases with integer magic constants
and <500 cells (including all those with sides <14). Here, a and b are
the alternate sides, N is the total number of cells in the hexagonal
array, and M is the magic constant:
a b N M
3 3 19 38 1 solution
3 4 27 63 no solutions
4 5 48 147 no solutions
5 6 75 285 ?
4 8 87 348 no solutions
3 10 96 388 no solutions
3 13 144 696 no solutions
7 8 147 777 ?
6 10 165 913 no solutions
8 9 192 1158 ?
9 10 243 1647 ?
5 15 246 1599 no solutions
6 15 280 1967 no solutions
4 19 307 2149 no solutions
7 15 315 2370 no solutions
8 15 351 2808 no solutions
11 12 363 3003 ?
12 13 432 3897 ?
William Rex Marshall
New Zealand
williamrexmarshall (at) yahoo (dot) com
D & A Klinkenberg
> Things have been happening since I last checked the newsgroups a few
> days ago.
> though I didn't give an actual proof. However, Geoff Bailey has posted
> one in the "Integer Solution?" thread.
> William Rex Marshall
> New Zealand
> williamrexmarshall (at) yahoo (dot) com
Hello William,
Thank you for your reply to my comments. I have no solutions, only
observations about the part I skipped above. However, I agree that
the post by Geoff can be used to prove that there is no solution for
a=b when a>3.
This is my post 1 (of 3?) to this thread today. I hope to post my
observations soon but I have some irrelevant things to do before I
write them. I hope that at least one of them may be helpful.
Dan in NY
D & A Klinkenberg
> Things have been happening since I last checked the newsgroups a few
> days ago.
> To end some of the misery, the answer for a = 3, b = 4 is given below.
> p
> a
> r
> t
> i
> a
> l
> .
> s
> p
> o
> i
> l
> e
> r
> A simple impossibility proof for a = 3, b = 4:
>
> A A A
> D D D D
> E - - - F
> B E - - F C
> B E - F C
> B E F C
>
> The numbers in the six rows each marked in the above diagram by one of
> the letters from A to F must add to six times the magic constant (6 x
> 63), or 378. But the maximum sum that can possibly be achieved by
> adding the largest numbers (from 27 down to 7) is only 357. Therefore a
> magic hexagon for a = 3, b = 4 is impossible.
[skip]
> William Rex Marshall
> New Zealand
> williamrexmarshall (at) yahoo (dot) com
&&&
Hello William,
(I hope someone beside me is working on this question.) Thank you for
all this information. This is indeed a simple demonstration. I was
having fun but had (almost) given up on the figure for a=3, b=4. It
is even better to go on to the 'next step'. I only got as far as A,
B, C )and showed that they all had to be at least 10) before I gave up
on this approach. My thread about 'six eq. with six unk' was related
to an attempt to find a solution.
I still wander whether there is any relation between the fact that my
six equations do not have a solution and your proof that there is no
solution. Since I doubt such a relation, I do not intend to pursue
it.
I also expect to post observations about the part I skipped, the
general solution. This is my post 2 (of 3?) to this thread today.
Dan in NY
D & A Klinkenberg
> Things have been happening since I last checked the newsgroups a few
> days ago.
> p
> a
> r
> t
> i
> a
> l
> .
> s
> p
> o
> i
> l
> e
> r
> This proof can be extended to many larger semiregular hexagonal arrays,
&&&
Hello William,
Thank you for posting information and some of your results and
observations. I want to write more of my observations about the
general case when 3<a<b but I have no solutions to your problem.
I started by feeling my way, deciding that that the primary way to
eliminate choices for a and b is to determine whether the magic
constant is an integer. It still seems that is a good place to
start. I wrote some equations for determining the values of N and M.
Did you agree with my equations?
Using those equations, Geoff made some remarks which apply to the case
when b=a+1. (Is this what you mean by "nearly regular"?) Using his
results, I observe that the value of M is an integer >except< when b+1
is divisible by 4.
From this I observe that for b=a+1, exactly three of four values of a,
result in an integer M. (This agrees with my sample results and with
your posted sample results.) It means there are an infinite number of
ways to choose a and b resulting in an integer M >and< and infinite
number of ways to get a non integer M.
I would like to find a way to characterize all (or some) of the
remaining ways (a>b+1) to get non integer M. Is their number finite
or infinite?
This is my post 3 (of 3?) in this thread today and I hope I haven't
made errors or typos. I don't have time today but sometime, I also
want to post my observation about how a hexagon is related to a square
with two opposite angles truncated. Some of these are not
'semiregular'. I used this to derive my formulas for N and M.
Dan in NY
William Rex Marshall
> William Rex Marshall wrote:
> > Things have been happening since I last checked the newsgroups a few
> > days ago.
> [skip a proof]
>
arrays,
> > but there seem to be difficulties dealing adequately with
large "nearly
> > regular" hexagons. Can a rigorous proof be found that covers every
> > case, or are there semiregular magic hexagons after all?
> >
[skip table]
> >
> > William Rex Marshall
> > New Zealand
> > williamrexmarshall (at) yahoo (dot) com
>
> &&&
> Hello William,
>
> Thank you for posting information and some of your results and
> observations. I want to write more of my observations about the
> general case when 3<a<b but I have no solutions to your problem.
>
> I started by feeling my way, deciding that that the primary way to
> eliminate choices for a and b is to determine whether the magic
> constant is an integer. It still seems that is a good place to
> start. I wrote some equations for determining the values of N and M.
> Did you agree with my equations?
Yes.
>
> Using those equations, Geoff made some remarks which apply to the case
> when b=a+1. (Is this what you mean by "nearly regular"?)
It includes them, but they not the only cases. First case which isn't is
a=15, b=19, (with 813 cells and a magic constant of 10,027 I haven't
seriously spent much time on it) though it can probably be proved
impossible easily; conditions are very restrictive. It seems difficult
to characterise exactly what is meant by "nearly regular", but it is
those cases for which a generalisation of the impossibility proof I
used for a=3,b=4 didn't work. There may however be a way of patching
the proof to cover at least some of these cases.
I should think the case for b=a+1 would be easiest to work with.
> Using his
> results, I observe that the value of M is an integer >except< when b+1
> is divisible by 4.
>
> From this I observe that for b=a+1, exactly three of four values of a,
> result in an integer M. (This agrees with my sample results and with
> your posted sample results.) It means there are an infinite number of
> ways to choose a and b resulting in an integer M >and< and infinite
> number of ways to get a non integer M.
>
> I would like to find a way to characterize all (or some) of the
> remaining ways (a>b+1) to get non integer M. Is their number finite
> or infinite?
Infinite for both non-integer M and infinite M. Example: b = 2*a-2
gives integer M when a is either a multiple of 6, or congruent to 3
modulo 24; non-integer M otherwise.
>
> This is my post 3 (of 3?) in this thread today and I hope I haven't
> made errors or typos. I don't have time today but sometime, I also
> want to post my observation about how a hexagon is related to a square
> with two opposite angles truncated. Some of these are not
> 'semiregular'. I used this to derive my formulas for N and M.
>
> Dan in NY
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