| Solve.block {limSolve} | R Documentation |
Solves the linear system A*X=B where A is an almost block diagonal matrix of the form:
TopBlock
... Array(1) ... ... ...
... ... Array(2) ... ...
...
... ... ... Array(Nblocks)...
... ... ... BotBlock
The method is based on Gauss elimination with alternate row and column elimination with partial pivoting, producing a stable decomposition of the matrix A without introducing fill-in.
uses FORTRAN subroutine colrow
Solve.block(Top, AR, Bot, B, overlap)
Top |
the first block of the almost block diagonal matrix |
AR |
intermediary blocks; |
Bot |
the last block of the almost block diagonal matrix |
B |
Right-hand side of the equations, a vector with length = number
of rows of |
overlap |
the number of columns in which successive blocks
overlap, and where |
matrix with the solution, X, of the block diagonal system of equations Ax=B, the number of columns of this matrix = number of columns of B.
A similar function but that requires a totally different input can now
also be found in the Matrix package
Karline Soetaert <karline.soetaert@nioz.nl>
J. C. Diaz , G. Fairweather , P. Keast, 1983. FORTRAN Packages for Solving Certain Almost Block Diagonal Linear Systems by Modified Alternate Row and Column Elimination, ACM Transactions on Mathematical Software (TOMS), v.9 n.3, p.358-375
Solve.tridiag to solve a tridiagonal system of linear equations.
Solve.banded to solve a banded system of linear equations.
Solve the generalised inverse solution,
solve the R default
# Solve the following system: Ax=B, where A is block diagonal, and
# 0.0 -0.98 -0.79 -0.15 Top
# -1.00 0.25 -0.87 0.35 Top
# 0.78 0.31 -0.85 0.89 -0.69 -0.98 -0.76 -0.82 blk1
# 0.12 -0.01 0.75 0.32 -1.00 -0.53 -0.83 -0.98
# -0.58 0.04 0.87 0.38 -1.00 -0.21 -0.93 -0.84
# -0.21 -0.91 -0.09 -0.62 -1.99 -1.12 -1.21 0.07
# 0.78 -0.93 -0.76 0.48 -0.87 -0.14 -1.00 -0.59 blk2
# -0.99 0.21 -0.73 -0.48 -0.93 -0.91 0.10 -0.89
# -0.68 -0.09 -0.58 -0.21 0.85 -0.39 0.79 -0.71
# 0.39 -0.99 -0.12 -0.75 -0.68 -0.99 0.50 -0.88
# 0.71 -0.64 0.0 0.48 Bot
# 0.08 100.0 50.00 15.00 Bot
B <- c(-1.92, -1.27, -2.12, -2.16, -2.27, -6.08,
-3.03, -4.62, -1.02, -3.52, 0.55, 165.08)
AA <- matrix (nrow = 12, ncol = 12, 0)
AA[1,1:4] <- c( 0.0, -0.98, -0.79, -0.15)
AA[2,1:4] <- c(-1.00, 0.25, -0.87, 0.35)
AA[3,1:8] <- c( 0.78, 0.31, -0.85, 0.89, -0.69, -0.98, -0.76, -0.82)
AA[4,1:8] <- c( 0.12, -0.01, 0.75, 0.32, -1.00, -0.53, -0.83, -0.98)
AA[5,1:8] <- c(-0.58, 0.04, 0.87, 0.38, -1.00, -0.21, -0.93, -0.84)
AA[6,1:8] <- c(-0.21, -0.91, -0.09, -0.62, -1.99, -1.12, -1.21, 0.07)
AA[7,5:12] <- c( 0.78, -0.93, -0.76, 0.48, -0.87, -0.14, -1.00, -0.59)
AA[8,5:12] <- c(-0.99, 0.21, -0.73, -0.48, -0.93, -0.91, 0.10, -0.89)
AA[9,5:12] <- c(-0.68, -0.09, -0.58, -0.21, 0.85, -0.39, 0.79, -0.71)
AA[10,5:12]<- c( 0.39, -0.99, -0.12, -0.75, -0.68, -0.99, 0.50, -0.88)
AA[11,9:12]<- c( 0.71, -0.64, 0.0, 0.48)
AA[12,9:12]<- c( 0.08, 100.0, 50.00, 15.00)
## Block diagonal input.
Top <- matrix(nrow = 2, ncol = 4, data = AA[1:2 , 1:4] )
Bot <- matrix(nrow = 2, ncol = 4, data = AA[11:12, 9:12])
Blk1 <- matrix(nrow = 4, ncol = 8, data = AA[3:6 , 1:8] )
Blk2 <- matrix(nrow = 4, ncol = 8, data = AA[7:10 , 5:12])
AR <- array(dim = c(4, 8, 2), data = c(Blk1, Blk2))
overlap <- 4
# answer = (1, 1,....1)
Solve.block(Top, AR, Bot, B, overlap = 4)
# Now with 3 different B values
B3 <- cbind(B, 2*B, 3*B)
Solve.block(Top, AR, Bot, B3, overlap = 4)