shell bypass 403
require 'bigdecimal'
#
# Solves a*x = b for x, using LU decomposition.
#
module LUSolve
module_function
# Performs LU decomposition of the n by n matrix a.
def ludecomp(a,n,zero=0,one=1)
prec = BigDecimal.limit(nil)
ps = []
scales = []
for i in 0...n do # pick up largest(abs. val.) element in each row.
ps <<= i
nrmrow = zero
ixn = i*n
for j in 0...n do
biggst = a[ixn+j].abs
nrmrow = biggst if biggst>nrmrow
end
if nrmrow>zero then
scales <<= one.div(nrmrow,prec)
else
raise "Singular matrix"
end
end
n1 = n - 1
for k in 0...n1 do # Gaussian elimination with partial pivoting.
biggst = zero;
for i in k...n do
size = a[ps[i]*n+k].abs*scales[ps[i]]
if size>biggst then
biggst = size
pividx = i
end
end
raise "Singular matrix" if biggst<=zero
if pividx!=k then
j = ps[k]
ps[k] = ps[pividx]
ps[pividx] = j
end
pivot = a[ps[k]*n+k]
for i in (k+1)...n do
psin = ps[i]*n
a[psin+k] = mult = a[psin+k].div(pivot,prec)
if mult!=zero then
pskn = ps[k]*n
for j in (k+1)...n do
a[psin+j] -= mult.mult(a[pskn+j],prec)
end
end
end
end
raise "Singular matrix" if a[ps[n1]*n+n1] == zero
ps
end
# Solves a*x = b for x, using LU decomposition.
#
# a is a matrix, b is a constant vector, x is the solution vector.
#
# ps is the pivot, a vector which indicates the permutation of rows performed
# during LU decomposition.
def lusolve(a,b,ps,zero=0.0)
prec = BigDecimal.limit(nil)
n = ps.size
x = []
for i in 0...n do
dot = zero
psin = ps[i]*n
for j in 0...i do
dot = a[psin+j].mult(x[j],prec) + dot
end
x <<= b[ps[i]] - dot
end
(n-1).downto(0) do |i|
dot = zero
psin = ps[i]*n
for j in (i+1)...n do
dot = a[psin+j].mult(x[j],prec) + dot
end
x[i] = (x[i]-dot).div(a[psin+i],prec)
end
x
end
end