magic hypercubes

Returns TRUE if a hypercube is semimagic, magic, perfect

Usage

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)

Arguments

a: The hypercube (array) to be tested
give.answers: Boolean, with TRUE meaning to also return the sums
func: Function to be applied across each dimension
ndim: In is.alicehypercube(), dimensionality of subhypercube to take sums over. See the details section
boolean: Boolean, with TRUE meaning that the hypercube is deemed magic, semimagic, etc, if all applications of func evaluate to TRUE. If boolean is FALSE, the hypercube is magic etc if all applications of func are identical
...: Further arguments passed to func()

Details

(Although apparently non-standard, 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 \(\sum a[i_1,i_2,\ldots,i_{r-1},,i_{r+1}, \ldots,i_d]\) with \(1\leqslant r\leqslant d\) 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,d-1),d). The first d-1 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 \(n-1\)), not necessarily in that order. Each integer thus appears \(n^{d-1}\) times.
A magic hypercube is a semimagic hypercube with the additional requirement that all \(2^{d-1}\) long (ie extreme point-to-extreme 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 one-dimensional 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^{d-ndim} subhypercubes have the same sum.

For example, if a is four-dimensional with dimension \(5\times 5\times 5\times 5\) then is.alicehypercube(a,1) is TRUE if and only if a is a semimagic hypercube: all \({4\choose 1}5^3=500\) one-dimensional subhypercubes have the same sum. Then is.alicehypercube(a,2) is TRUE if all 2-dimensional subhypercubes (ie all \({4\choose 2}\times 5^2=150\) of the \(5\times 5\) 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 \({4\choose 3}\times 5^1=20\) of the \(5\times 5\times 5\) 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, \(q\times 2^d\) 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 \(d^{\rm th}\) 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[,3-i_1,3-i_2,...,3-i_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]}).

References

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).

Author

Robin K. S. Hankin

Note

On this page, “subhypercube” is restricted to rectangularly-oriented subarrays; see the note at subhypercubes.

Not all subhypercubes of a magic hypercube are necessarily magic! (for example, consider a 5-dimensional 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 \(5\times5\times 5\) (ie three dimensional), cube. Say a=magiccube.2np1(2). Then the square defined by sapply(1:n,function(i){a[,i,6-i]}, 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 point-to-point” 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 bad-ass requirement. I reckon that is a total of \(\frac{1}{2}\left( 3^d-1\right)\cdot n^{d-1}\) correct summations. This is not coded up yet; I can't see how to do it in anything like a vectorized manner.

Examples

library(abind)
is.semimagichypercube(magiccube.2np1(1))
#> [1] TRUE
is.semimagichypercube(magichypercube.4n(1,d=4))
#> [1] TRUE

is.perfect(magichypercube.4n(1,d=4))
#> [1] FALSE

# 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
#> [[1]]
#>      [,1] [,2] [,3]
#> [1,]   42   42   42
#> [2,]   42   42   42
#> [3,]   42   42   42
#> [4,]   42   42   42
#> [5,]   42   42   42
#> [6,]   42   42   42
#> 
#> [[2]]
#>      [,1] [,2] [,3]
#> [1,]   84   84   84
#> [2,]   84   84   84
#> [3,]   84   84   84
#> 
#> [[3]]
#>      [,1] [,2] [,3] [,4] [,5] [,6]
#> [1,]   42   42   42   42   42   42
#> [2,]   42   42   42   42   42   42
#> [3,]   42   42   42   42   42   42
#> 

# 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
#> [1] FALSE
is.semimagichypercube(not_mag,func=mymax,UP=81)  # TRUE
#> [1] TRUE


a2 <- magichypercube.4n(m=1,d=4)
is.diagonally.correct(a2)
#> [1] TRUE
is.diagonally.correct(a2,g=TRUE)$diag.sums
#> , , 1, 1
#> 
#>      [,1] [,2]
#> [1,]  514  514
#> [2,]  514  514
#> 
#> , , 2, 1
#> 
#>      [,1] [,2]
#> [1,]  514  514
#> [2,]  514  514
#> 
#> , , 1, 2
#> 
#>      [,1] [,2]
#> [1,]  514  514
#> [2,]  514  514
#> 
#> , , 2, 2
#> 
#>      [,1] [,2]
#> [1,]  514  514
#> [2,]  514  514
#> 

## To extract corner elements (note func(1:n) != func(n:1)):
is.diagonally.correct(a2,func=function(x){x[1]},g=TRUE)$diag.sums 
#> , , 1, 1
#> 
#>      [,1] [,2]
#> [1,]    1   13
#> [2,]    4   16
#> 
#> , , 2, 1
#> 
#>      [,1] [,2]
#> [1,]   49   61
#> [2,]   52   64
#> 
#> , , 1, 2
#> 
#>      [,1] [,2]
#> [1,]  193  205
#> [2,]  196  208
#> 
#> , , 2, 2
#> 
#>      [,1] [,2]
#> [1,]  241  253
#> [2,]  244  256
#> 


#Now for a subhypercube of a magic hypercube that is not semimagic:
is.magic(allsubhypercubes(magiccube.2np1(1))[[10]])
#> [1] FALSE

data(hendricks)
is.perfect(hendricks)
#> [1] TRUE


#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))
#> [1] TRUE

#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)
#> $answer
#> [1] TRUE
#> 
#> $alice.sums
#>        [,1]   [,2]   [,3]   [,4]   [,5]
#> [1,] 131200 131200 131200 131200 131200
#> [2,] 131200 131200 131200 131200 131200
#> [3,] 131200 131200 131200 131200 131200
#> [4,] 131200 131200 131200 131200 131200
#>