Description Usage Arguments Details Note Author(s) References See Also Examples
Returns TRUE
if a hypercube is semimagic, magic, perfect
1 2 3 4 5 6  is.semimagichypercube(a, give.answers=FALSE, func=sum, boolean=FALSE, ...)
is.diagonally.correct(a, give.answers = FALSE, func=sum, boolean=FALSE, ...)
is.magichypercube(a, give.answers = FALSE, func=sum, boolean=FALSE, ...)
is.perfect(a, give.answers = FALSE, func=sum, boolean=FALSE)
is.latinhypercube(a, give.answers=FALSE)
is.alicehypercube(a,ndim,give.answers=FALSE, func=sum, boolean=FALSE)

a 
The hypercube (array) to be tested 
give.answers 
Boolean, with 
func 
Function to be applied across each dimension 
ndim 
In 
boolean 
Boolean, with 
... 
Further arguments passed to 
(Although apparently nonstandard, here a hypercube is defined to have dimension d and order n—and thus has n^d elements).
A semimagic hypercube has all “rook's move” sums
equal to the magic constant (that is, each sum(a[i_1,i_2,
…,i_{r1},,i_{r+1},…,i_d]) with 1 <=
r <= d is equal to the magic constant for all values of the
i's). In is.semimagichypercube()
, if
give.answers
is TRUE
, the sums returned are in the
form of an array of dimension c(rep(n,d1),d)
. The first
d1
dimensions are the coordinates of the projection of the
summed elements onto the surface hypercube. The last dimension
indicates the dimension along which the sum was taken over.
Optional argument func
, defaulting to sum()
, indicates
the function to be taken over each of the d
dimensions.
Currently requires func
to return a scalar.
A Latin hypercube is one in which each line of elements whose coordinates differ in only one dimension comprises the numbers 1 to n (or 0 to n1), not necessarily in that order. Each integer thus appears n^{d1} times.
A magic hypercube is a semimagic hypercube with the
additional requirement that all 2^(d1) long (ie
extreme pointtoextreme point) diagonals sum correctly. Correct
diagonal summation is tested by is.diagonally.correct()
; by
specifying a function other than sum()
, criteria other than
the diagonals returning the correct sum may be tested.
An Alice hypercube is a different generalization of a semimagic square to higher dimensions. It is named for A. M. Hankin (“Alice”), who originally suggested it.
A semimagic hypercube has all onedimensional subhypercubes (ie
lines) summing correctly. An Alice hypercube is one in which all
ndim
dimensional subhypercubes have the same sum, where
ndim
is a fixed integer argument. Thus, if a
is a
hypercube of size n^d, is.alicehypercube(a,ndim)
returns TRUE
if all n^{dndim}
subhypercubes have the
same sum.
For example, if a
is fourdimensional with dimension
5x5x5x5 then
is.alicehypercube(a,1)
is TRUE
if and only if a
is a semimagic hypercube: all 4*5^3=500
onedimensional subhypercubes have the same sum. Then
is.alicehypercube(a,2)
is TRUE
if all 2dimensional
subhypercubes (ie all 6x5^2=150 of
the 5x5 squares, for example a[,2,4,]
and
a[1,1,,]
) have the same sum. Then
is.alicehypercube(a,3)
means that all 3d subhypercubes (ie
all 4x5=20 of the 5x5x5 cubes, for example a[,,1,]
and
a[4,,,]
) have the same sum. For any hypercube a
,
is.alicehypercube(a,dim(a))
returns TRUE
.
A semimagic hypercube is an Alice hypercube for any value of
ndim
.
A perfect magic hypercube (use is.perfect()
) is
a magic hypercube with all nonbroken diagonals summing correctly.
This is a seriously restrictive requirement for high dimensional
hypercubes. As yet, this function does not take a
give.answers
argument.
A pandiagonal magic hypercube, also Nasik hypercube (or sometimes just a perfect hypercube) is a semimagic hypercube with all diagonals, including broken diagonals, summing correctly. This is not implemented.
The terminology in this area is pretty confusing.
In is.magichypercube()
, if argument give.answers=TRUE
then a list is returned. The first element of this list is Boolean
with TRUE
if the array is a magic hypercube. The second
element and third elements are answers
fromis.semimagichypercube()
and is.diagonally.correct()
respectively.
In is.diagonally.correct()
, if argument
give.answers=TRUE
, the function also returns an array of
dimension c(q,rep(2,d))
(that is, q*2^d
elements), where q is the length of func()
applied to a
long diagonal of a
(if q=1, the first dimension is
dropped). If q=1, then in dimension d
having index 1
means func()
is applied to elements of a
with the
dth dimension running over 1:n
; index 2
means to run over n:1
. If q>1, the index of the first
dimension gives the index of func()
, and subsequent dimensions
have indices of 1 or 2 as above and are interpreted in the same way.
An example of a function for which these two are not identical is given below.
If func=f
where f
is a function returning a vector of
length i
, is.diagonally.correct()
returns an array
out
of dimension c(i,rep(2,d))
, with
out[,i_1,i_2,...,i_d]
being f(x)
where x
is the
appropriate long diagonal. Thus the 2^d equalities
out[,i_1,i_2,...,i_d]==out[,3i_1,3i_2,...,3i_d]
hold if and
only if identical(f(x),f(rev(x)))
is TRUE
for each long
diagonal (a condition met, for example, by sum()
but not by the
identity function or function(x){x[1]}
).
On this page, “subhypercube” is restricted to
rectangularlyoriented subarrays; see the note at subhypercubes
.
Not all subhypercubes of a magic hypercube are necessarily magic! (for
example, consider a 5dimensional magic hypercube a
. The square
b
defined by a[1,1,1,,]
might not be magic: the diagonals
of b
are not covered by the definition of a magic hypercube).
Some subhypercubes of a magic hypercube are not even semimagic: see
below for an example.
Even in three dimensions, being perfect is pretty bad. Consider a
5x5x5 (ie three dimensional), cube. Say
a=magiccube.2np1(2)
. Then the square defined by
sapply(1:n,function(i){a[,i,6i]}, simplify=TRUE)
, which is a
subhypercube of a
, is not even semimagic: the rowsums are
incorrect (the colsums must sum correctly because a
is magic).
Note that the diagonals of this square are two of the “extreme
pointtopoint” diagonals of a
.
A pandiagonal magic hypercube (or sometimes just a perfect hypercube) is semimagic and in addition the sums of all diagonals, including broken diagonals, are correct. This is one seriously badass requirement. I reckon that is a total of (3^d1)n^(d1)/2 correct summations. This is not coded up yet; I can't see how to do it in anything like a vectorized manner.
Robin K. S. Hankin
R. K. S. Hankin 2005. “Recreational mathematics with R: introducing the magic package”. R news, 5(1)
Richards 1980. “Generalized magic cubes”. Mathematics Magazine, volume 53, number 2, (March).
is.magic
, allsubhypercubes
, hendricks
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47  library(abind)
is.semimagichypercube(magiccube.2np1(1))
is.semimagichypercube(magichypercube.4n(1,d=4))
is.perfect(magichypercube.4n(1,d=4))
# Now try an array with minmax(dim(a))==FALSE:
a < abind(magiccube.2np1(1),magiccube.2np1(1),along=2)
is.semimagichypercube(a,g=TRUE)$rook.sums
# is.semimagichypercube() takes further arguments:
mymax < function(x,UP){max(c(x,UP))}
not_mag < array(1:81,rep(3,4))
is.semimagichypercube(not_mag,func=mymax,UP=80) # FALSE
is.semimagichypercube(not_mag,func=mymax,UP=81) # TRUE
a2 < magichypercube.4n(m=1,d=4)
is.diagonally.correct(a2)
is.diagonally.correct(a2,g=TRUE)$diag.sums
## To extract corner elements (note func(1:n) != func(n:1)):
is.diagonally.correct(a2,func=function(x){x[1]},g=TRUE)$diag.sums
#Now for a subhypercube of a magic hypercube that is not semimagic:
is.magic(allsubhypercubes(magiccube.2np1(1))[[10]])
data(hendricks)
is.perfect(hendricks)
#note that Hendricks's magic cube also has many broken diagonals summing
#correctly:
a < allsubhypercubes(hendricks)
ld < function(a){length(dim(a))}
jj < unlist(lapply(a,ld))
f < function(i){is.perfect(a[[which(jj==2)[i]]])}
all(sapply(1:sum(jj==2),f))
#but this is NOT enough to ensure that it is pandiagonal (but I
#think hendricks is pandiagonal).
is.alicehypercube(magichypercube.4n(1,d=5),4,give.answers=TRUE)

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