module LUSolve

Solves a*x = b for x, using LU decomposition.

Public Instance Methods

ludecomp(a,n,zero=0,one=1) click to toggle source

Performs LU decomposition of the n by n matrix a.

# File ext/bigdecimal/lib/bigdecimal/ludcmp.rb, line 11
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

lusolve(a,b,ps,zero=0.0) click to toggle source

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.

# File ext/bigdecimal/lib/bigdecimal/ludcmp.rb, line 67
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