An 8x8 magic square that BEATS Ben Franklin's!
jwa...@az.com
thought a magic square that *I* had made would be a welcome sight here
on rec.puzzles. I thought I'd show mine side-by-side with the well
known 8x8 square made by Benjamin Franklin, when it occured to me that
by leaving some of the numbers out of my square, I'd be posing a rather
challenging puzzle:
Mine (I've left out some of
Ben Franklin's: the numbers as a challenge):
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
|52|61| 4|13|20|29|36|45| | | | | | | | | |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
|14| 3|62|51|46|35|30|19| | | | | | | | | |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
|53|60| 5|12|21|28|37|44| | | | | | | | | |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
|11| 6|59|54|43|38|27|22| | | | | | | | | |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
|55|58| 7|10|23|26|39|42| | | | | | | | | |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
| 9| 8|57|56|41|40|25|24| | | | | | | | | |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
|50|63| 2|15|18|31|34|47| | | | | | | | | |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
|16| 1|64|49|48|33|32|17| | 1| | | | | | | |
+--+--+--+--+--+--+--+--+ +--+--+--+--+--+--+--+--+
Properties:
1. All rows and columns add to 1. All rows and columns add to
260 stopping halfway at 130. 260 stopping halfway at 130.
Main diagonals do NOT add to Main diagonals ALSO add to
260, however. 260 and ALSO stop halfway at
130.
2. All bent diagonals & parallel 2. All bent diagonals & parallel
bent diagonals add to 260. bent diagonals add to 260,
Ex: from 9 diagonally up to 12, but broken diagonals ALSO add
plus 21 diagonally down to 24. to 260!
Ben probably considered his So mine is THREE times as
square to be twice as magical magical as normal pandiagonal
as pandiagonal magic squares, magic squares and half again
since there are twice as many as magical as Ben's!
bent diagonals and parallel bent
diagonals as broken diagonals.
3. The shortend bent diagonal, 3. The same shortend bent
from 53 up to 4 and 29 down to diagonal, (now with different
44, with the top two corners numbers) with the top two
adds to 260. corners adds to 260.
You can slide this configuration You can also slide this
up and down and it remains magic. configuration up and down AND
ALSO ROTATE IT 90 degrees
left or right and it
remains magic!
4. The two-piece diagonals around 4. The two-piece diagonals
the corners: 14, 61, 36, 19, around the corners also add
47, 32, 1 and 50 add to 260 and to 260 and can be divided in
can be divided in half top and half top and bottom OR left
bottom to make 130. and right to make 130!
5. All 2x2 subsquares within Ben's 5. Same property.
magic square add to 130.
To help you fill in the numbers (after all, there ARE 2^6 * 6! magic
squares all having the properties I've listed; 2^3 * 6! if you don't
count rotations and reflections; just as there are 2^6 * 6! magic
squares all having the same properties of Ben Franklin's magic square;
2^4 * 6! if you don't count all the rotations and reflections of his
square -- and no, that's not a misprint; since some properties of Ben's
square no longer hold if for example, his square is rotated 90 degrees
clockwise) - *anyway* - to help you narrow it down:
1/4 of all the numbers 1-64 occur in the SAME POSITIONS in my square as
in Ben Franklin's square. *WHEW!* - That narrows it down to just 4
possible answers! And finally:
The number 1 is in the lower left corner in my magic square. In fact,
the entire ascending main diagonal consists of 1, 10, and those numbers
up to 64 whose digits ADD to 10. As a matter of fact, that last clue I
give you was an extra freebie, since once I told you that 1 was in the
bottom left corner, you now had EVERY CLUE you needed in order to
figure out MY magic square!
Enjoy!
-- Jim Waters <jwa...@az.com>
Sent via Deja.com http://www.deja.com/
Before you buy.
MR
[..............]
Give me a 4x4 magic square that has that properties and i'll construct the 8x8 one.
--
......___ _____..
___ _ .....| \/ __ \..
| |_| |_ ....| |\/| |_> >..
| | | |_ ...| | | /....
..| | | |\ \.....
.|_| |_| >_\.....
Jim Waters
"MR" <mr.(arroba)wanadoo.es@NOSPAM> wrote:
> <jwa...@az.com> ...
Properties:
>
> Give me a 4x4 magic square that has that properties and i'll
> construct the 8x8 one.
>
My 8x8 magic square DOES EXIST and DOES have all the properties I've
mentioned. As for a 4x4 square having even the half-way properties of
the rows and columns, common sense should suffice to show that for a
row of 4 to stop half-way at half the magic sum, then for a column of
4 intersecting the row somewhere, where the half-column adds to the
same half-sum there would be no escaping a duplicate number adjacent
either above or below.
If you don't believe that such an 8x8 really exists, consider the
following square of 16:
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
|238| 20| 13|243| 46|212|205| 51| 77|179|174| 84|141|115|110|148|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| 17|239|242| 16|209| 47| 50|208|178| 80| 81|175|114|144|145|111|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
|244| 14| 19|237| 52|206|211| 45| 83|173|180| 78|147|109|116|142|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| 15|241|240| 18|207| 49| 48|210|176| 82| 79|177|112|146|143|113|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
|246| 12| 21|235| 54|204|213| 43| 85|171|182| 76|149|107|118|140|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| 9 |247|234| 24|201| 55| 42|216|170| 88| 73|183|106|152|137|119|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
|236| 22| 11|245| 44|214|203| 53| 75|181|172| 86|139|117|108|150|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| 23|233|248| 10|215| 41| 56|202|184| 74| 87|169|120|138|151|105|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
|252| 6 | 27|229| 60|198|219| 37| 91|165|188| 70|155|101|124|134|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| 7 |249|232| 26|199| 57| 40|218|168| 90| 71|185|104|154|135|121|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
|230| 28| 5 |251| 38|220|197| 59| 69|187|166| 92|133|123|102|156|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| 25|231|250| 8 |217| 39| 58|200|186| 72| 89|167|122|136|153|103|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
|228| 30| 3 |253| 36|222|195| 61| 67|189|164| 94|131|125|100|158|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| 31|225|256| 2 |223| 33| 64|194|192| 66| 95|161|128|130|159| 97|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
|254| 4 | 29|227| 62|196|221| 35| 93|163|190| 68|157| 99|126|132|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
| 1 |255|226| 32|193| 63| 34|224|162| 96| 65|191| 98|160|129|127|
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
Properties:
1. All rows, columns, and diagonals add to 2056, EVERY QUARTER of
which adds to 514.
2. Not only do all diagonals and broken diagonals add to 2056, but
if any one of these has its cells colored in 4 alternating colors,
then the 4 cells of any particular color will add to 514.
3. Not only do all bent diagonals and parallel bent diagonals add to
2056 as with Ben Franklin's 16x16, but all four QUADRANTS have all
their bent diagonals and parallel bent diagonals adding to 1028 as
with Ben's 8x8, only with a different magic sum.
4. The shortend bent diagonal, from 236 up to 205 plus 179 down to
150, plus the 2 top corners add to 2056; as does the further
shortend parallel bent diagonal from 9 up to 212 plus 174 down to
119, plus the 2-cell diagonals about the top corners, 17, 20, 110,
and 111; as also the still further shortend parallel bent diagonal
from 246 up to 46 plus 84 down to 140 plus 244 up to 13 plus 115
down to 142. All these configurations are magic as with Ben's
16x16, yet they can also be ROTATED 90 DEGREES AS WELL and they
retain their magic properties!
5. As with Ben's 16x16 and his 8x8, all 2x2 subsquares add to the
same total - in this case 514.
I believe my 16x16 compares VERY well against Ben's, and yet he wrote
that he considered his square of 16 to be "the most magically magical
of any magic square ever made by any magician."
Perhaps my claim regarding the existence of my 8x8 now has a bit more
credibility!
MR
> "MR" :
> > Jim Waters...
> > > With the recent thread involving the construction of magic squares,
> > > I thought a magic square that *I* had made would be a welcome sight
> > > here on rec.puzzles. I thought I'd show mine side-by-side with the
> > > well known 8x8 square made by Benjamin Franklin, when it occured to
> > > me that by leaving some of the numbers out of my square, I'd be
> > > posing a rather challenging puzzle:
[blah blah blah]
> > Give me a 4x4 magic square that has that properties and i'll
> > construct the 8x8 one.
>
> My 8x8 magic square DOES EXIST and DOES have all the properties I've
> mentioned. As for a 4x4 square having even the half-way properties of
> the rows and columns, common sense should suffice to show that for a
> row of 4 to stop half-way at half the magic sum, then for a column of
> 4 intersecting the row somewhere, where the half-column adds to the
> same half-sum there would be no escaping a duplicate number adjacent
> either above or below.
I didn't say that the 8x8 magic square with your properties doesn't exist... I said that if a 4x4 magis square with that properties
exist, then i can apply my method of duplicating size in order to make a 8x8 sqare with all that properties, as my method remains
most of the properties of the half-size magic sqare.
I didn't say either that a 4x4 magic square with that properties MUST exist in order to the 8x8 to be possible... I DIDN'T
[..............]
Be sure, with more spare time i would be able to analyze the 16x16 square and extract ideas to construct the 8x8 one, or even the
32x32 and 64x64 ones...
But, dude, i'm on holidays!! In a couple of days i'm leaving the city!!
HINT to any person who wants to solve the problem:
Take the 16x16 square, substract 1 from every cell, and work with HEXADECIMAL BASE. The properties of the 16x16 square with
hexadecimal base WILL be very clear!!!
In order to make the 8x8 one, work with OCTAL BASE (and then add 1 to every cell)
See you all in september!
dus...@gate.sinica.edu.tw
> I didn't say that the 8x8 magic square with your properties doesn't
> exist... I said that if a 4x4 magis square with that properties
> exist, then i can apply my method of duplicating size in order to
> make a 8x8 sqare with all that properties, as my method remains
> most of the properties of the half-size magic sqare.
Here is a 1x1 magic square with all of the desired properties:
1
You seem to be saying that you have a method that can convert this into
a 2x2 magic square, using the digits 1, 2, 3 and 4. But this is clearly
impossible. Or else, what are you saying? I don't think you simply
intend to use the numbers in the 4x4 over and over again, as that seems
silly.
Jonathan Dushoff
Kendel
> jwa...@az.com wrote:
>
> >I thought I'd show mine side-by-side with the well
> >known 8x8 square made by Benjamin Franklin,
>
> Can you offer any evidence whatsoever that Ben Franklin was in any way
> involved with magic squares? I didn't think so.
Well, let's see: who cares? The "well known “Benjamin Franklin 8x8 Magic
Square”" does indeed seem to be fairly well known regardless of the
authenticity of the credited creatorship (?).
I was just curious though: can you offer any reason as to why you refute
his ability to provide evidence in the time it took you to hit the space
bar once and before you'd actually even asked him the question?
*Shrug*
—Kendel
“A polar bear is a rectangular bear after a coordinate transform.”
—Geek Joke
MR
> MR :
>
> [......]
>
> Here is a 1x1 magic square with all of the desired properties:
>
> 1
>
> You seem to be saying that you have a method that can convert this into
> a 2x2 magic square, using the digits 1, 2, 3 and 4. But this is clearly
> impossible. Or else, what are you saying? I don't think you simply
> intend to use the numbers in the 4x4 over and over again, as that seems
> silly.
No, no, no, no...
My method is only useful for "large" squares... it doesn't work for sizes 4,6 or 10 because of some special properties of
transposing elements. I'm not asking to use the numbers on the 4x4 over and over, but working with them to make a real 8x8 sqare.
In fact, I believe that the 8x8 square that we're trying to make has some regularities with a 4x4 one. Regularities, but it is not
equal.
Rene ALBERT
>
>
> My method is only useful for "large" squares... it doesn't work for sizes 4,6 or 10 because of some special properties of
> transposing elements. I'm not asking to use the numbers on the 4x4 over and over, but working with them to make a real 8x8 sqare.
>
> In fact, I believe that the 8x8 square that we're trying to make has some regularities with a 4x4 one. Regularities, but it is not
> equal.
If you are interested in, I can forward to you an Excel file (182 ko)
who contains my general "cloning" method to construct magic squares of
any even size.
If I understand correctly what you say, you are on the way I have found.
Following 2 examples (size 8 and 12)
8x8:
49 63 17 31 14 4 46 36
60 54 28 22 7 9 39 41
1 15 33 47 62 52 30 20
12 6 44 38 55 57 23 25
56 58 24 26 11 5 43 37
61 51 29 19 2 16 34 48
8 10 40 42 59 53 27 21
13 3 45 35 50 64 18 32
12x12
4 9 2 130 135 128 121 126 119 31 36 29
3 5 7 129 131 133 120 122 124 30 32 34
8 1 6 134 127 132 125 118 123 35 28 33
103 108 101 49 54 47 58 63 56 76 81 74
102 104 106 48 50 52 57 59 61 75 77 79
107 100 105 53 46 51 62 55 60 80 73 78
67 72 65 85 90 83 94 99 92 40 45 38
66 68 70 84 86 88 93 95 97 39 41 43
71 64 69 89 82 87 98 91 96 44 37 42
112 117 110 22 27 20 13 18 11 139 144 137
111 113 115 21 23 25 12 14 16 138 140 142
116 109 114 26 19 24 17 10 15 143 136 141
--
Amicalement.
mailto:r_al...@club-internet.fr
Jim Waters
I don't know how helpful your excel file may be, but I don't see your
example squares as being in any way helpful toward solving the puzzle
I've posed. Aside from the rows, columns, and main diagonals adding to
the proper total, your squares have *no* other extraordinary properties
whatsoever. Neither of them are even pandiagonal!
I was considering posting examples of an 8x8 and a 12x12 of my own, but
that would spoil a hint that I'm considering giving if nobody comes up
with the answer after a week. (My examples would both be pandiagonal
and my 12x12 would have its rows, columns and main diagonals not only
adding to 870, but dividing into even *thirds*; that is, every block of
4 consecutive squares along a row, column, or main diagonal would add
to 290.) -- Hmmm, perhaps *that* hint might spark a few ideas!?
Rene ALBERT
>
> In article <39867E55...@club-internet.fr>,
> Rene ALBERT <r_al...@club-internet.fr> wrote:
> > MR a écrit :
>
> I don't know how helpful your excel file may be, but I don't see your
> example squares as being in any way helpful toward solving the puzzle
> I've posed. Aside from the rows, columns, and main diagonals adding to
> the proper total, your squares have *no* other extraordinary properties
> whatsoever. Neither of them are even pandiagonal!
Sorry, I did'nt get your post, just seen "MR" one, who said he was still
looking for even magic square solution if I did'nt misunderstood.
I have never looked for what you call pandiagonal, too bad.
--
Amicalement.
mailto:r_al...@club-internet.fr
Jim Waters
Rene ALBERT <r_al...@club-internet.fr> wrote:
> Sorry, I did'nt get your post, just seen "MR" one, who said he
> was still looking for even magic square solution if I did'nt
> misunderstood.
>
> I have never looked for what you call pandiagonal, too bad.
>
For the benefit of everyone else who didn't catch my post which
began this thread, I'm reposting it now. I'll also add a hint to
help out.
With the recent thread involving the construction of magic squares, I
thought a magic square that *I* had made would be a welcome sight here
on rec.puzzles. I thought I'd show mine side-by-side with the well
known 8x8 square made by Benjamin Franklin, when it occured to me that
by leaving some of the numbers out of my square, I'd be posing a rather
challenging puzzle:
Mine (I've left out some of
Ben Franklin's: the numbers as a challenge):
52 61 4 13 20 29 36 45 | | | | | | | | |
14 3 62 51 46 35 30 19 | | | | | | | | |
53 60 5 12 21 28 37 44 | | | | | | | | |
11 6 59 54 43 38 27 22 | | | | | | | | |
55 58 7 10 23 26 39 42 | | | | | | | | |
9 8 57 56 41 40 25 24 | | | | | | | | |
50 63 2 15 18 31 34 47 | | | | | | | | |
16 1 64 49 48 33 32 17 | 1| | | | | | | |
Properties:
1. All rows and columns add to 1. All rows and columns add to
260 stopping halfway at 130. 260 stopping halfway at 130.
Main diagonals do NOT add to Main diagonals ALSO make 260
260, however. ALSO stopping halfway at 130.
2. All bent diagonals & parallel 2. All the same bent diagonals &
bent diagonals add to 260. parallel bent diagonals add
Ex: from 9 diagonally up to 12, to 260, but broken diagonals
plus 21 diagonally down to 24. ALSO add to 260!
Ben probably considered his So mine is THREE times as
square to be twice as magical magical as normal pandiagonal
as pandiagonal magic squares, magic squares and HALF AGAIN
since there are twice as many as magical as Ben's!
bent diagonals and parallel bent
diagonals as broken diagonals.
3. The shortend bent diagonal, 3. The same shortend bent
from 53 up to 4 and 29 down to diagonal, (now with different
44, with the top two corners numbers) with the top two
adds to 260. corners adds to 260.
You can slide this configuration You can also slide this
up and down and it remains magic; configuration up and down AND
that is, the numbers within the ALSO ROTATE IT 90 degrees
configuration still add to 260. left or right and IT STILL
remains magic!
4. The two-piece diagonals around 4. The two-piece diagonals
the corners: 14, 61, 36, 19, around the corners also add
47, 32, 1 and 50 add to 260 and to 260 and can be divided in
can be divided in half top and half top and bottom OR left
bottom to make 130. and right to make 130!
5. All 2x2 subsquares within Ben's 5. Same property.
magic square add to 130.
To help you fill in the numbers (after all, there ARE 2^6 * 6! magic
squares all having the properties I've listed; 2^3 * 6! if you don't
count rotations and reflections; just as there are 2^6 * 6! magic
squares all having the same properties of Ben Franklin's magic square;
2^4 * 6! if you don't count all the rotations and reflections of his
square -- and no, that's not a misprint; since some properties of Ben's
square no longer hold if for example, his square is rotated 90 degrees
clockwise) - *anyway* - to help you narrow it down:
1/4 of all the numbers 1-64 occur in the SAME POSITIONS in my square as
in Ben Franklin's square. *WHEW!* - That narrows it down to just 4
possible answers! And finally:
The number 1 is in the lower left corner in my magic square. In fact,
the entire ascending main diagonal consists of 1, 10, and those numbers
up to 64 whose digits ADD to 10. As a matter of fact, that last clue I
give you was an extra freebie, since once I told you that 1 was in the
bottom left corner, you now had EVERY CLUE you needed in order to
figure out MY magic square!
Now,
1st new hint since my original post:
Consider properties 1, 2, and 5 which I claim my own magic square to
have: What property can you conclude that ALL 4 QUADRANTS of my square
MUST have in order for my square to be able to have those 3 properties?
(It's a property which the quadrants of Ben's square DO NOT have!)
If you know what that property is, you'll know where 64 must go!
Enjoy!
Jim Waters
*SIGH!* One more time, with a hint to that hint!
In article <8mdcok$tok$1...@nnrp1.deja.com>,
Now a hint to the above hint:
Property #5, about all 2x2 subsquares throughout the square adding to
the same total, is actually a pretty powerful property having some
consequences:
Consequence #1: All subsquares of even by even dimension have their
4 corners adding to the same total as the 2x2
subsquares.
Consequence #2: All subsquares of even by odd dimension have the sum
of the 2 corners on one even side equal to the sum of
the 2 corners on the other even side.
Consequence #3: All subsquares of odd by odd dimension have the sums
of the two pairs of opposite corners equal to each
other.
+---+---+---+---+ Consequence #4: In any 4x4 subsquare, the sum of
| | a | b | | one broken diagonal added to the
+---+---+---+---+ sum of the same broken diagonal
| | b | a | | rotated 90 degrees adds to twice
+---+---+---+---+ the sum of any 2x2 square.
| b | | | a |
+---+---+---+---+ Consequence #5: Likewise, the sum of one broken
| a | | | b | diagonal equals itself rotated
+---+---+---+---+ 180 degrees.
Now for any parallel bent diagonal in my 8x8 square, what conclusion
can you make regarding the sum of the two closest fitting broken
diagonals in 2 of the quadrants?
For any broken diagonal in the 8x8, can you make the same conclusion
in *its* two closest fitting quadrants broken diagonals?
Can you force a conclusion regarding ALL broken diagonals within each
quadrant of my 8x8?
Adrian Smith
: Aw, C'mon! No replies since I posted my last hint?
: In article <8mdcok$tok$1...@nnrp1.deja.com>,
: Jim Waters <jwa...@az.com> wrote:
: >
<snip>
: > With the recent thread involving the construction of magic squares,
: > I thought a magic square that *I* had made would be a welcome sight
: > here on rec.puzzles. I thought I'd show mine side-by-side with the
: > well known 8x8 square made by Benjamin Franklin, when it occured to
: > me that by leaving some of the numbers out of my square, I'd be
: > posing a rather challenging puzzle:
I missed that thread because I've only just readded rec.puzzles to my
newsfeed, but back in 1985 I found a method of making 2n by 2n magic
squares from n by n magic squares, and 2n by 2n by 2n magic cubes from
n by n by n magic cubes. Perhaps it is of interest. If so, a postscript
description can be found at http://www.snaffles.demon.co.uk/mcubes.ps
: >
: > Mine (I've left out some of
: > Ben Franklin's: the numbers as a challenge):
: >
: > 52 61 4 13 20 29 36 45 | | | | | | | | |
: > 14 3 62 51 46 35 30 19 | | | | | | | | |
: > 53 60 5 12 21 28 37 44 | | | | | | | | |
: > 11 6 59 54 43 38 27 22 | | | | | | | | |
: > 55 58 7 10 23 26 39 42 | | | | | | | | |
: > 9 8 57 56 41 40 25 24 | | | | | | | | |
: > 50 63 2 15 18 31 34 47 | | | | | | | | |
: > 16 1 64 49 48 33 32 17 | 1| | | | | | | |
Well, the square on the left can be constructed from the following two
squares by multiplying a value in the first square by 16 and adding
1 + the value in the second square.
3 3 0 0 1 1 2 2 3 12 3 12 3 12 3 12
0 0 3 3 2 2 1 1 13 2 13 2 13 2 13 2
3 3 0 0 1 1 2 2 4 11 4 11 4 11 4 11
0 0 3 3 2 2 1 1 10 5 10 5 10 5 10 5
3 3 0 0 1 1 2 2 6 9 6 9 6 9 6 9
0 0 3 3 2 2 1 1 8 7 8 7 8 7 8 7
3 3 0 0 1 1 2 2 1 14 1 14 1 14 1 14
0 0 3 3 2 2 1 1 15 0 15 0 15 0 15 0
You could use the same method but with the first square rotated 90
degrees anti clockwise to get a different square with the 1 in the same
place.
Adrian Smith
http://www.snaffles.demon.co.uk
QSCGZ
Adrian Smith wrote:
> I missed that thread because I've only just readded rec.puzzles to my
> newsfeed, but back in 1985 I found a method of making 2n by 2n magic
> squares from n by n magic squares, and 2n by 2n by 2n magic cubes from
> n by n by n magic cubes. Perhaps it is of interest. If so, a postscript
> description can be found at http://www.snaffles.demon.co.uk/mcubes.ps
is there an ASCII desciption available ?
>: > Ben Franklin's: the numbers as a challenge):
>: >
>: > 52 61 4 13 20 29 36 45 | | | | | | | | |
>: > 14 3 62 51 46 35 30 19 | | | | | | | | |
>: > 53 60 5 12 21 28 37 44 | | | | | | | | |
>: > 11 6 59 54 43 38 27 22 | | | | | | | | |
>: > 55 58 7 10 23 26 39 42 | | | | | | | | |
>: > 9 8 57 56 41 40 25 24 | | | | | | | | |
>: > 50 63 2 15 18 31 34 47 | | | | | | | | |
>: > 16 1 64 49 48 33 32 17 | 1| | | | | | | |
>
>Well, the square on the left can be constructed from the following two
>squares by multiplying a value in the first square by 16 and adding
>1 + the value in the second square.
>
> 3 3 0 0 1 1 2 2 3 12 3 12 3 12 3 12
> 0 0 3 3 2 2 1 1 13 2 13 2 13 2 13 2
> 3 3 0 0 1 1 2 2 4 11 4 11 4 11 4 11
> 0 0 3 3 2 2 1 1 10 5 10 5 10 5 10 5
> 3 3 0 0 1 1 2 2 6 9 6 9 6 9 6 9
> 0 0 3 3 2 2 1 1 8 7 8 7 8 7 8 7
> 3 3 0 0 1 1 2 2 1 14 1 14 1 14 1 14
> 0 0 3 3 2 2 1 1 15 0 15 0 15 0 15 0
>
> You could use the same method but with the first square rotated 90
>degrees anti clockwise to get a different square with the 1 in the same
>place.
why did you split it 4*16 and not 8*8 ?
52 61 4 13 20 29 36 45 67012345 34343434
14 3 62 51 46 35 30 19 10765432 52525252
53 60 5 12 21 28 37 44 67012345 43434343
11 6 59 54 43 38 27 22 10765432 25252525
55 58 7 10 23 26 39 42 = 8* 67012345 + 61616161 + (1)
9 8 57 56 41 40 25 24 10765432 07070707
50 63 2 15 18 31 34 47 67012345 16161616
16 1 64 49 48 33 32 17 10765432 70707070
then rearrange the rows a bit to get:
0 7 0 7 0 7 0 7
4 3 4 3 4 3 4 3
7 0 7 0 7 0 7 0
let A:= 3 4 3 4 3 4 3 4
6 1 6 1 6 1 6 1
2 5 2 5 2 5 2 5
1 6 1 6 1 6 1 6
5 2 5 2 5 2 5 2
A already satisfies all Jim Water's sums.
Now let B:=8*A+transpose(A)+(1).
Rotate B by 90' and you have a magic square with the desired properties.
--qscgz
Adrian Smith
: Adrian Smith wrote:
: > I missed that thread because I've only just readded rec.puzzles to my
: > newsfeed, but back in 1985 I found a method of making 2n by 2n magic
: > squares from n by n magic squares, and 2n by 2n by 2n magic cubes from
: > n by n by n magic cubes. Perhaps it is of interest. If so, a postscript
: > description can be found at http://www.snaffles.demon.co.uk/mcubes.ps
: is there an ASCII desciption available ?
Not quickly, but if the original TeX version is any good to you, I've just
put it at:
http://www.snaffles.demon.co.uk/mcubes.tex
: >: > Ben Franklin's: the numbers as a challenge):
: >: >
: >: > 52 61 4 13 20 29 36 45 | | | | | | | | |
: >: > 14 3 62 51 46 35 30 19 | | | | | | | | |
: >: > 53 60 5 12 21 28 37 44 | | | | | | | | |
: >: > 11 6 59 54 43 38 27 22 | | | | | | | | |
: >: > 55 58 7 10 23 26 39 42 | | | | | | | | |
: >: > 9 8 57 56 41 40 25 24 | | | | | | | | |
: >: > 50 63 2 15 18 31 34 47 | | | | | | | | |
: >: > 16 1 64 49 48 33 32 17 | 1| | | | | | | |
: >
: >Well, the square on the left can be constructed from the following two
: >squares by multiplying a value in the first square by 16 and adding
: >1 + the value in the second square.
: >
: > 3 3 0 0 1 1 2 2 3 12 3 12 3 12 3 12
: > 0 0 3 3 2 2 1 1 13 2 13 2 13 2 13 2
: > 3 3 0 0 1 1 2 2 4 11 4 11 4 11 4 11
: > 0 0 3 3 2 2 1 1 10 5 10 5 10 5 10 5
: > 3 3 0 0 1 1 2 2 6 9 6 9 6 9 6 9
: > 0 0 3 3 2 2 1 1 8 7 8 7 8 7 8 7
: > 3 3 0 0 1 1 2 2 1 14 1 14 1 14 1 14
: > 0 0 3 3 2 2 1 1 15 0 15 0 15 0 15 0
: >
: > You could use the same method but with the first square rotated 90
: >degrees anti clockwise to get a different square with the 1 in the same
: >place.
: why did you split it 4*16 and not 8*8 ?
I was expecting the square to be derived in some obvious way from a 4x4
square, but I can't see how if it is.
I got it wrong anyway because you can't just rotate the first square.
It matters that no two occurrences of a number in the second square get
paired with the same number in the first square. You can manipulate them
though, as long as you preserve that property. You could rotate the
second one clockwise 90 degrees and reflect the first one abou x=y to
give a square with a 1 in the bottom left position, as was asked for.
Adrian Smith
http://www.snaffles.demon.co.uk
Jim Waters
qs...@aol.com (QSCGZ) wrote:
[snip]
>
> 0 7 0 7 0 7 0 7
> 4 3 4 3 4 3 4 3
> 7 0 7 0 7 0 7 0
> let A:= 3 4 3 4 3 4 3 4
> 6 1 6 1 6 1 6 1
> 2 5 2 5 2 5 2 5
> 1 6 1 6 1 6 1 6
> 5 2 5 2 5 2 5 2
>
> A already satisfies all Jim Water's sums.
> Now let B:=8*A+transpose(A)+(1).
> Rotate B by 90' and you have a magic square with the desired
> properties.
>
> --qscgz
>
You appear to be on the right track.
The square thus made is:
62 27 6 35 14 43 54 19
2 39 58 31 50 23 10 47
59 30 3 38 11 46 51 22
7 34 63 26 55 18 15 42
60 29 4 37 12 45 52 21
8 33 64 25 56 17 16 41
61 28 5 36 13 44 53 20
1 40 57 32 49 24 9 48
And it solves every requirement but one. In my square, one fourth
of all the numbers are in the same positions as in Ben Franklin's.
Now believe it or not, there are 2^6 * 6! magic squares ALL having
THE SAME additive properties! (That IS counting all rotations and
reflections.) 6! = 720 of these all have 1 in the lower left corner.
Hint to speed up the process of tracking down my square:
In QSCGZ's square above, try highlighting the cells whose numbers are
equal to a power of 2 plus 1. Now set up a fresh 8x8 grid with 1 in
the lower left corner and THE SAME CELLS highlighted. Can you find
a different way to arrange the six powers of 2 plus 1, filling the
rest of the cells accordingly, to maximize the number of cells having
the same numbers with Ben's square?
QSCGZ
4 3 4 3 4 3 4 3
A already satisfies all Jim Water's sums.
Now let B:=8*A+transpose(A)+(1).
Rotate B by 90' and you have a magic square with the desired
properties.
>...Now believe it or not, there are 2^6 * 6! magic squares ALL having
>THE SAME additive properties! (That IS counting all rotations and
>reflections.) 6! = 720 of these all have 1 in the lower left corner.
I found six 8-reduced matrices suitable for composing to make such a
magic square :
07070707 07070707 07070707 07070707 07070707 07070707
43434343 43434343 16161616 16161616 25252525 25252525
70707070 70707070 70707070 70707070 70707070 70707070
34343434 34343434 61616161 61616161 52525252 52525252
61616161 52525252 52525252 34343434 61616161 34343434
25252525 16161616 43434343 25252525 43434343 16161616
16161616 25252525 25252525 43434343 16161616 43434343
52525252 61616161 34343434 52525252 34343434 61616161
Choose two of them , transpose one of them , multiply one of them by 8
and add them up. That's 60 possibilities. Which am I missing ?
Jim Waters
Actually that's 72 possibilities. Since you'll be transposing one
of them anyway, you CAN pick the same one both times -- as you did
yourself in your earlier post. My square IS amongst these.
Other possiblities exist by the two sets exchanging bits. But to
simplify the task of counting them all, just take ONE example square
having all the desired additive properties. Now imagine you're in
front of a classroom of 64 students seated at their desks which are
all arranged in an 8x8 array. You've handed out numbers 1 through
64 on cards, each student holds a card with a number; now if you
wanted to, you could stand on top of your desk with your camera, have
each student hold up their number, and snap a picture of your initial
square.
Now you give your class the following instructions:
1. Subtract 1 from your number.
2. Convert your number into base 2. Express your number as a 6-digit
binary number left-justified by zeros.
Before we move on to step 3, consider some of your favorite additive
properties. Pick a few specific examples. Notice that the binary
numbers involved, if listed in a column for adding, have all columns
of digits evenly divided between 0's and 1's.
3. Flip a coin 6 times. Let's say that the 3rd and 5th times were
tails, the rest heads. You're about to ask all your students to
CHANGE their 3rd and 5th digits when you're told, "Hey, we want
to keep the number 1 right where it is". Ok, that *would* have
been 2^6 = 64 possibilities.
4. Now write the numbers 1 - 6 on scraps of paper, toss them into a
hat, and randomly draw them out. Let's say they come out 5, 2, 4,
1, 6, 3. So you ask all your students to rearrange the digits of
their binary number accordingly. Each student's new binary number
now starts with what was their 5th digit and ends with what was
their 3rd digit. There are 6! = 720 possibilities.
Notice how in all the additive property checking, all of the columns
of binary digits are STILL half filled with 0's and half with 1's.
This would still be the case if some of the 6 digit positions had
first all been changed.
5. Now have each student convert their new 6-digit binary number back
to decimal and add 1. You now have a new arrangement of the numbers
1 - 64 with all the same additive properties. The process that was
done to each number is reversible, so every source had a target and
every target number has a source. This transformation is one-to-one
and onto.
Adrian Smith
: Adrian Smith wrote:
: > I missed that thread because I've only just readded rec.puzzles to my
: > newsfeed, but back in 1985 I found a method of making 2n by 2n magic
: > squares from n by n magic squares, and 2n by 2n by 2n magic cubes from
: > n by n by n magic cubes. Perhaps it is of interest. If so, a postscript
: > description can be found at http://www.snaffles.demon.co.uk/mcubes.ps
: is there an ASCII desciption available ?
I've now put one at:
http://www.snaffles.demon.co.uk/mcubes.txt
Adrian Smith
http://www.snaffles.demon.co.uk
Jim Waters
was a terrific hint that practically gives it away, I thought for
sure that I'd be congratulating someone by now for figuring out my
magic square. I'm reposting that reply of mine, with a little
improvement in some of the wording:
In article <8n33c3$3j9$1...@nnrp1.deja.com>,
Jim Waters <jwa...@az.com> wrote:
> In article <20000811054145...@ng-fz1.aol.com>,
> qs...@aol.com (QSCGZ) wrote:
> [snip]
> >
> > 0 7 0 7 0 7 0 7
> > 4 3 4 3 4 3 4 3
> > 7 0 7 0 7 0 7 0
> > let A:= 3 4 3 4 3 4 3 4
> > 6 1 6 1 6 1 6 1
> > 2 5 2 5 2 5 2 5
> > 1 6 1 6 1 6 1 6
> > 5 2 5 2 5 2 5 2
> >
> > A already satisfies all Jim Water's sums.
> > Now let B:=8*A+transpose(A)+(1).
> > Rotate B by 90' and you have a magic square with the desired
> > properties.
> >
> > --qscgz
> >
>
> You appear to be on the right track.
>
> The square thus made is:
>
> 62 27 6 35 14 43 54 19
> 2 39 58 31 50 23 10 47
> 59 30 3 38 11 46 51 22
> 7 34 63 26 55 18 15 42
> 60 29 4 37 12 45 52 21
> 8 33 64 25 56 17 16 41
> 61 28 5 36 13 44 53 20
> 1 40 57 32 49 24 9 48
>
> And it solves every requirement but one. In my square, one fourth
> of all the numbers are in the same positions as in Ben Franklin's.
In the square made from QSCGZ's directions above, NONE of the numbers
are in the same positions as Ben's, although a few of the numbers are
are horizontaly adjacent to their positions in Ben's.
> Now believe it or not, there are 2^6 * 6! magic squares ALL having
> THE SAME additive properties! (That IS counting all rotations and
> reflections.) 6! = 720 of these all have 1 in the lower left corner.
>
> Hint to speed up the process of tracking down my square:
In QSCGZ's square above, try highlighting the cells whose numbers are
equal to a power of 2 plus 1. Now set up a fresh 8x8 grid with 1 in
the lower left corner and THE SAME CELLS highlighted. Can you find
a different way to arrange the six powers of 2 plus 1 into the six
highlighted squares, then filling the rest of the cells accordingly,
so as to maximize the number of cells having numbers in the same cell
position as Ben's square? You get a NICE magic square as a result!
QSCGZ
>Good Grief! After QSCGZ came so close and I gave what I believe
>was a terrific hint that practically gives it away, I thought for
>sure that I'd be congratulating someone by now for figuring out my
>magic square. I'm reposting that reply of mine, with a little
>improvement in some of the wording:
>
>In article <8n33c3$3j9$1...@nnrp1.deja.com>,
> Jim Waters <jwa...@az.com> wrote:
>> In article <20000811054145...@ng-fz1.aol.com>,
>> qs...@aol.com (QSCGZ) wrote:
>> [snip]
>> >
>> > 0 7 0 7 0 7 0 7
>> > 4 3 4 3 4 3 4 3
>> > 7 0 7 0 7 0 7 0
>> > let A:= 3 4 3 4 3 4 3 4
>> > 6 1 6 1 6 1 6 1
>> > 2 5 2 5 2 5 2 5
>> > 1 6 1 6 1 6 1 6
>> > 5 2 5 2 5 2 5 2
>> >
>> > A already satisfies all Jim Water's sums.
>> > Now let B:=8*A+transpose(A)+(1).
>> > Rotate B by 90' and you have a magic square with the desired
>> > properties.
>> >
>> > --qscgz
Jim Waters wrote on 13.August:
>In article <20000812141036...@ng-ba1.aol.com>,
> qs...@aol.com (QSCGZ) wrote:
>> 0 7 0 7 0 7 0 7
>> 1 6 1 6 1 6 1 6
>> 7 0 7 0 7 0 7 0
>> 6 1 6 1 6 1 6 1
>> let A:= 5 2 5 2 5 2 5 2
>> 4 3 4 3 4 3 4 3
>> 2 5 2 5 2 5 2 5
>> 3 4 3 4 3 4 3 4
>>
>> A already satisfies all Jim Water's sums.
>> Now let B:=8*A+transpose(A)+(1).
>> Rotate B by 90' and you have a magic square with the desired
>> properties.
didn't you notice , that these 2 squares were different , and that
I claimed that my second one did satisfy all your requirements ?
In decimal it's :
60 53 4 13 20 29 44 37
3 14 59 54 43 38 19 30
61 52 5 12 21 28 45 36
6 11 62 51 46 35 22 27
B = 63 50 7 10 23 26 47 34
8 9 64 49 48 33 24 25
58 55 2 15 18 31 42 39
1 16 57 56 41 40 17 32
--qscgz
Jim Waters
qs...@aol.com (QSCGZ) wrote:
>
> Jim Waters wrote on 13.August:
>
> >In article <20000812141036...@ng-ba1.aol.com>,
> > qs...@aol.com (QSCGZ) wrote:
> >> 0 7 0 7 0 7 0 7
> >> 1 6 1 6 1 6 1 6
> >> 7 0 7 0 7 0 7 0
> >> 6 1 6 1 6 1 6 1
> >> let A:= 5 2 5 2 5 2 5 2
> >> 4 3 4 3 4 3 4 3
> >> 2 5 2 5 2 5 2 5
> >> 3 4 3 4 3 4 3 4
> >>
> >> A already satisfies all Jim Water's sums.
> >> Now let B:=8*A+transpose(A)+(1).
> >> Rotate B by 90' and you have a magic square with the desired
> >> properties.
>
> didn't you notice , that these 2 squares were different , and that
> I claimed that my second one did satisfy all your requirements ?
Subtle! No, I actually *hadn't* noticed that you had changed that
first displayed square. I thought that post to be just a question
about how did I get 720 possibilities, and I *did* notice the new
A square amongst your six. And BTW, you had made NO CHANGE AT ALL
to the verbal portion of your quote.
> In decimal it's :
>
> 60 53 4 13 20 29 44 37
> 3 14 59 54 43 38 19 30
> 61 52 5 12 21 28 45 36
> 6 11 62 51 46 35 22 27
> B = 63 50 7 10 23 26 47 34
> 8 9 64 49 48 33 24 25
> 58 55 2 15 18 31 42 39
> 1 16 57 56 41 40 17 32
>
> --qscgz
>
I *was* under the assumption that whoever would first figure out my
square would actually form it and proudly display it, as you finally
have done.
This may well end this thread, although I leave behind the following
challenge: There exists a 16x16 magic square having ALL of the
following properties:
1. All the rows, columns, and main diagonals not only all add to 2056,
and not only stop halfway at 1028, they divide into equal quarters all
adding to 514.
2. Not only do the bent diagonals and parallel bent diagonals all add
to 2056, but so do diagonals bent after their first four and bent again
before their last four -- whether or not their was yet another bend in
the middle.
3. Not only do the broken diagonals add to 2056, but if the cells of
any are colored in 4 alternating colors, the 4 of any particular color
will add to 514. Getting 2 alternating colors to have their colors
adding to half would be almost automatic, but the 4 alternating colors
property is very tricky!
4. At this point, some readers of this thread may remember a 16x16
square I posted near the beginning of this thread that has the all the
properties I've mentioned thus far. Here's one property it doesn't
have: Not only do the shortend bent diagonals at the top plus the
top corners add to 2056, but this property occurs also in the four
QUADRANTS of the square! The configuration there adds to 1028.
As usual, these configurations may be rotated 90 degrees and their
magic is retained.
There are of course 2^8 * 8! possible squares which have this property.
This time I'm not comparing the square I have in mind to any other
published square, even though Ben Franklin DID ALSO make a 16x16.
So any square with the above properties would be acceptable, although
I WOULD appreciate keeping 1 in the lower left corner, as I can then
verify an answer very quickly by seeing where the powers of 2 plus 1
have been placed!