How to "restore" a 3-dimensional (2 x 2 x 2) array (cube) of three two-dimensional matrices (cube faces)

Question

I have this array:

T <- array(c(.25,.1,.1,.1,.05,.1,.1,.2),c(2,2,2)) # , , 1 # [,1] [,2] # [1,] 0.25 0.1 # [2,] 0.10 0.1 # , , 2 # [,1] [,2] # [1,] 0.05 0.1 # [2,] 0.10 0.2 

I suppose this can be understood as a kind of "cube" cut in the third dimension. It has rows (size 1), columns (dim 2) and "height" (dim 3), so to speak ...

Now I can summarize its values ​​in one of these dimensions. 3 combinations are possible:

  Tm1 <- apply(T0,c(1,2),sum) Tm2 <- apply(T0,c(1,3),sum) Tm3 <- apply(T0,c(2,3),sum) 

Now I have this:

  #> Tm1 # [,1] [,2] #[1,] 0.3 0.2 #[2,] 0.2 0.3 #> Tm2 # [,1] [,2] #[1,] 0.35 0.15 #[2,] 0.20 0.30 #> Tm3 # [,1] [,2] #[1,] 0.35 0.15 #[2,] 0.20 0.30 

These are cubic "faces."

Is it possible to restore the original array from these 3 matrices? . In other words, is it possible to find out the distribution inside this “cube” simply by looking at its “faces”?

If so, how to do it? (I mean, the "algebra path" and the algorithm R ...)

Answer 1

This is how I came up with a solution to your question. First, construct the system of equations so that A %*% x = b (where x are the values ​​that need to be solved for those inside T0 ):

 n <- prod(dim(T0)) b <- c(Tm1, Tm2, Tm3) m <- length(b) Ti <- array(seq_along(T0), dim(T0)) Ti1 <- unlist(apply(Ti, c(1,2), list)) Ti2 <- unlist(apply(Ti, c(1,3), list)) Ti3 <- unlist(apply(Ti, c(2,3), list)) A <- matrix(0, nrow = m, ncol = n) A[cbind(rep(1:m, each = 2), c(Ti1, Ti2, Ti3))] <- 1 cbind(A, b) # b # [1,] 1 0 0 0 1 0 0 0 0.30 # [2,] 0 1 0 0 0 1 0 0 0.20 # [3,] 0 0 1 0 0 0 1 0 0.20 # [4,] 0 0 0 1 0 0 0 1 0.30 # [5,] 1 0 1 0 0 0 0 0 0.35 # [6,] 0 1 0 1 0 0 0 0 0.20 # [7,] 0 0 0 0 1 0 1 0 0.15 # [8,] 0 0 0 0 0 1 0 1 0.30 # [9,] 1 1 0 0 0 0 0 0 0.35 # [10,] 0 0 1 1 0 0 0 0 0.20 # [11,] 0 0 0 0 1 1 0 0 0.15 # [12,] 0 0 0 0 0 0 1 1 0.30 

A is a non-square matrix, so I used the generalized inverse to solve for x :

 library(MASS) xsol <- ginv(A) %*% b Tsol <- array(xsol, dim(T0)) Tsol # , , 1 # # [,1] [,2] # [1,] 0.2375 0.1125 # [2,] 0.1125 0.0875 # # , , 2 # # [,1] [,2] # [1,] 0.0625 0.0875 # [2,] 0.0875 0.2125 

This solution does not match your initial T0, however you can verify that

 apply(Tsol, c(1,2), sum) # [,1] [,2] # [1,] 0.3 0.2 # [2,] 0.2 0.3 apply(Tsol, c(1,3), sum) # [,1] [,2] # [1,] 0.35 0.15 # [2,] 0.20 0.30 apply(Tsol, c(2,3), sum) # [,1] [,2] # [1,] 0.35 0.15 # [2,] 0.20 0.30 

Output? No, it is impossible to restore the original matrix. Another way to show this is that the rank qr(A)$rank matrix A is 7 , while you have 8 unknowns. So you need one extra bit of information, for example. that T[1, 1] is 0.25 to restore the original array:

 A <- rbind(A, c(1, rep(0, n - 1))) b <- c(b, 0.25) qr(A)$rank # [1] 8 xsol <- ginv(A) %*% b Tsol <- array(xsol, dim(T0)) Tsol # , , 1 # [,1] [,2] # [1,] 0.25 0.1 # [2,] 0.10 0.1 # , , 2 # [,1] [,2] # [1,] 0.05 0.1 # [2,] 0.10 0.2 

Answer 2

Here is an algebraic explanation in a more general case with a continuous variable. This can help get the root reason why you cannot do this. The problem is that you cannot build a reverse map. Below you can replace the integration sign with a summation, try to find the inverse matrix, and they will achieve the result that the flannel showed. So, suppose that f is integrable in the domains x, y, and z. Your source table

$$ w = f (x, y, z) $$

your conversion

$$ t (x) = \ int_x f (x, y, z) dx = g (y, z) $$

You want to have a reverse card from t (x) to w. This card will be

$$ \ frac {\ partial t (x)} {\ partial x} = \ frac {\ partial} {\ partial x} \ left (\ int_x f (x, y, z) dx \ right) = \ frac { \ partial} {\ partial x} g (y, z) = 0 $$

That is, once you integrate x, you cannot restore it from g (y, z).