
MAGIC
SQUARE PROBLEMS.
"By magic numbers." CONGREVE,
The Mourning Bride.
This is a very ancient branch of mathematical puzzledom, and it has an
immense, though scattered, literature of its own. In their simple form of
consecutive whole numbers arranged in a square so that every column, every row,
and each of the two long diagonals shall add up alike, these magic squares offer
three main lines of investigation: Construction, Enumeration, and
Classification. Of recent years many ingenious methods have been devised for the
construction of magics, and the law of their formation is so well understood
that all the ancient mystery has evaporated and there is no longer any
difficulty in making squares of any dimensions. Almost the last word has been
said on this subject. The question of the enumeration of all the possible
squares of a given order stands just where it did over two hundred years ago.
Everybody knows that there is only one solution for the third order, three cells
by three; and Frénicle published in 1693 diagrams of all the arrangements of the
fourth order—880 in number—and his results have been verified over and over
again. I may here refer to the general solution for this order, for numbers not
necessarily consecutive, by E. Bergholt in Nature, May 26, 1910, as it is
of the greatest importance to students of this subject. The enumeration of the
examples of any higher order is a completely unsolved problem.
As to classification, it is largely a matter of individual taste—perhaps an
æsthetic question, for there is beauty in the law and order of numbers. A man
once said that he divided the human race into two great classes: those who take
snuff and those who do not. I am not sure that some of our classifications of magic squares
are not almost as valueless. However, lovers of these things seem somewhat
agreed that Nasik magic squares (so named by Mr. Frost, a student of them, after
the town in India where he lived, and also called Diabolique and Pandiagonal)
and Associated magic squares are of special interest, so I will just explain
what these are for the benefit of the novice.
I published in The Queen for January 15, 1910, an article that would
enable the reader to write out, if he so desired, all the 880 magics of the
fourth order, and the following is the complete classification that I gave. The
first example is that of a Simple square that fulfils the simple conditions and
no more. The second example is a SemiNasik, which has the additional property
that the opposite short diagonals of two cells each together sum to 34. Thus,
14 + 4 + 11 + 5 = 34 and 12 +
6 + 13 + 3 = 34. The third example is not only
SemiNasik but also Associated, because in it every number, if added
to the number that is equidistant, in a straight line, from the centre gives 17.
Thus, 1 + 16, 2 + 15, 3 + 14, etc. The fourth example,
considered the most "perfect" of all, is a Nasik. Here all the broken diagonals
sum to 34. Thus, for example, 15 + 14 + 2 + 3, and
10 + 4 + 7 + 13, and 15 + 5 +
2 + 12. As a consequence, its properties are such that if you repeat
the square in all directions you may mark off a square, 4 × 4,
wherever you please, and it will be magic.
The following table not only gives a complete enumeration under the four
forms described, but also a classification under the twelve graphic types
indicated in the diagrams. The dots at the end of each line represent the
relative positions of those complementary pairs, 1 + 16, 2 + 15, etc.,
which sum to 17. For example, it will be seen that the first and second magic
squares given are of Type VI., that the third square is of Type III., and that
the fourth is of Type I. Edouard Lucas indicated these types, but he dropped
exactly half of them and did not attempt the classification.
NASIK 
(Type I.) 


48 
SEMINASIK 
(Type II., Transpositions of Nasik) 

48 

" 
(Type III., Associated) 

48 

" 
(Type IV.) 
96 


" 
(Type V.) 
96 
192 

" 
(Type VI.) 

96 
384 
SIMPLE. 
(Type VI.) 

208 

" 
(Type VII.) 
56 


" 
(Type VIII.) 
56 


" 
(Type IX.) 
56 


" 
(Type X.) 
56 
224 

" 
(Type XI.) 
8 


" 
(Type XII.) 
8 
16 
448 




880 
It is hardly necessary to say that every one of these squares will produce
seven others by mere reversals and reflections, which we do not count as
different. So that there are 7,040 squares of this order, 880 of which are
fundamentally different.
An infinite variety of puzzles may be made introducing new conditions into
the magic square. In The Canterbury Puzzles I have given examples of such
squares with coins, with postage stamps, with cuttingout conditions, and other
tricks. I will now give a few variants involving further novel conditions.
399.—THE
TROUBLESOME EIGHT.
Nearly everybody knows that a "magic square" is an arrangement of numbers in
the form of a square so that every row, every column, and each of the two long
diagonals adds up alike. For example, you would find little difficulty in merely
placing a different number in each of the nine cells in the illustration so that
the rows, columns, and diagonals shall all add up 15. And at your first attempt
you will probably find that you have an 8 in one of the corners. The puzzle is
to construct the magic square, under the same conditions, with the 8 in the
position shown.
Solution 400.—THE
MAGIC STRIPS.
I happened to have lying on my table a number of strips of cardboard, with
numbers printed on them from 1 upwards in numerical order. The idea suddenly
came to me, as ideas have a way of unexpectedly coming, to make a little puzzle
of this. I wonder whether many readers will arrive at the same solution that I
did.
Take seven strips of cardboard and lay them together as above. Then write on
each of them the numbers 1, 2, 3, 4, 5, 6, 7, as shown, so that the numbers
shall form seven rows and seven columns.
Now, the puzzle is to cut these strips into the fewest possible pieces so
that they may be placed together and form a magic square, the seven rows, seven
columns, and two diagonals adding up the same number. No figures may be turned upside
down or placed on their sides—that is, all the strips must lie in their original
direction.
Of course you could cut each strip into seven separate pieces, each piece
containing a number, and the puzzle would then be very easy, but I need hardly
say that fortynine pieces is a long way from being the fewest possible.
Solution 401.—EIGHT
JOLLY GAOL BIRDS.
The illustration shows the plan of a prison of nine cells all communicating
with one another by doorways. The eight prisoners have their numbers on their
backs, and any one of them is allowed to exercise himself in whichever cell may
happen to be vacant, subject to the rule that at no time shall two prisoners be
in the same cell. The merry monarch in whose dominions the prison was situated
offered them special comforts one Christmas Eve if, without breaking that rule,
they could so place themselves that their numbers should form a magic
square.
Now, prisoner No. 7 happened to know a good deal about magic squares, so he
worked out a scheme and naturally selected the method that was most
expeditious—that is, one involving the fewest possible moves from cell to cell.
But one man was a surly, obstinate fellow (quite unfit for the society of his
jovial companions), and he refused to move out of his cell or take any part in
the proceedings. But No. 7 was quite equal to the emergency, and found that he
could still do what was required in the fewest possible moves without troubling
the brute to leave his cell. The puzzle is to show how he did it and,
incidentally, to discover which prisoner was so stupidly obstinate. Can you find
the fellow?
Solution 402.—NINE
JOLLY GAOL BIRDS.
Shortly after the episode recorded in the last puzzle occurred, a ninth
prisoner was placed in the vacant cell, and the merry monarch then offered them
all complete liberty on the following strange conditions. They were required so
to rearrange themselves in the cells that their numbers formed a magic square
without their movements causing any two of them ever to be in the same cell
together, except that at the start one man was allowed to be placed on the
shoulders of another man, and thus add their numbers together, and move as one
man. For example, No. 8 might be placed on the shoulders of No. 2, and then they
would move about together as 10. The reader should seek first to solve the
puzzle in the fewest possible moves, and then see that the man who is burdened
has the least possible amount of work to do.
Solution 403.—THE
SPANISH DUNGEON.
Not fifty miles from Cadiz stood in the middle ages a castle, all traces of
which have for centuries disappeared. Among other interesting features, this
castle contained a particularly unpleasant dungeon divided into sixteen cells,
all communicating with one another, as shown in the illustration.
Now, the governor was a merry wight, and very fond of puzzles withal. One day
he went to the dungeon and said to the prisoners, "By my halidame!" (or its
equivalent in Spanish) "you shall all be set free if you can solve this puzzle.
You must so arrange yourselves in the sixteen cells that the numbers on your
backs shall form a magic square in which every column, every row, and each of
the two diagonals shall add up the same. Only remember this: that in no case may
two of you ever be together in the same cell."
One of the prisoners, after working at the problem for two or three days,
with a piece of chalk, undertook to obtain the liberty of himself and his
fellowprisoners if they would follow his directions and move through the
doorway from
cell to cell in the order in which he should call out their numbers.
He succeeded in his attempt, and, what is more remarkable, it would seem from
the account of his method recorded in the ancient manuscript lying before me,
that he did so in the fewest possible moves. The reader is asked to show what
these moves were.
Solution 404.—THE
SIBERIAN DUNGEONS.
The above is a trustworthy plan of a certain Russian prison in Siberia. All
the cells are numbered, and the prisoners are numbered the same as the cells
they occupy. The prison diet is so fattening that these political prisoners are
in perpetual fear lest, should their pardon arrive, they might not be able to
squeeze themselves through the narrow doorways and get out. And of course it
would be an unreasonable thing to ask any government to pull down the walls of a
prison just to liberate the prisoners, however innocent they might be. Therefore
these men take all the healthy exercise they can in order to retard their
increasing obesity, and one of their recreations will serve to furnish us with
the following puzzle.
Show, in the fewest possible moves, how the sixteen men may form themselves
into a magic square, so that the numbers on their backs shall add up the same in
each of the four columns, four rows, and two diagonals without two prisoners
having been at any time in the same cell together. I had better say, for the
information of those who have not yet been made acquainted with these places,
that it is a peculiarity of prisons that you are not allowed to go outside their
walls. Any prisoner may go any distance that is possible in a single move.
Solution 405.—CARD
MAGIC SQUARES.
Take an ordinary pack of cards and throw out the twelve court cards. Now,
with nine of the remainder (different suits are of no consequence) form the
above magic square. It will be seen that the pips add up fifteen in every row in
every column, and in each of the two long diagonals. The puzzle is with the
remaining cards (without disturbing this arrangement) to form three more such
magic squares, so that each of the four shall add up to a different sum. There
will, of course, be four cards in the reduced pack that will not be used. These
four may be any that you choose. It is not a difficult puzzle, but requires just
a little thought.
Solution 406.—THE
EIGHTEEN DOMINOES.
The illustration shows eighteen dominoes arranged in the form of a square so
that the pips in every one of the six columns, six rows, and two long diagonals
add up 13. This is the smallest summation possible with any selection of
dominoes from an ordinary box of twentyeight. The greatest possible summation
is 23, and a solution for this number may be easily obtained by substituting for
every number its complement to 6. Thus for every blank substitute a 6, for every
1 a 5, for every 2 a 4, for 3 a 3, for 4 a 2, for 5 a 1, and for 6 a blank. But the
puzzle is to make a selection of eighteen dominoes and arrange them (in exactly
the form shown) so that the summations shall be 18 in all the fourteen
directions mentioned.
Solution
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