模块 LUSolve
使用 LU 分解求解 a*x = b 中的 x。
公共实例方法
ludecomp(a,n,zero=0,one=1) 点击切换源代码
执行 n×n 矩阵 a 的 LU 分解。
# 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) 点击切换源代码
使用 LU 分解求解 a*x = b 中的 x。
a 是矩阵,b 是常数向量,x 是解向量。
ps 是主元,一个向量,指示在 LU 分解期间执行的行置换。
# 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