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author | Jari Vetoniemi <jari.vetoniemi@indooratlas.com> | 2020-03-16 18:49:26 +0900 |
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committer | Jari Vetoniemi <jari.vetoniemi@indooratlas.com> | 2020-03-30 00:39:06 +0900 |
commit | fcbf63e62c627deae76c1b8cb8c0876c536ed811 (patch) | |
tree | 64cb17de3f41a2b6fef2368028fbd00349946994 /jni/ruby/ext/bigdecimal/sample/linear.rb |
Fresh start
Diffstat (limited to 'jni/ruby/ext/bigdecimal/sample/linear.rb')
-rw-r--r-- | jni/ruby/ext/bigdecimal/sample/linear.rb | 72 |
1 files changed, 72 insertions, 0 deletions
diff --git a/jni/ruby/ext/bigdecimal/sample/linear.rb b/jni/ruby/ext/bigdecimal/sample/linear.rb new file mode 100644 index 0000000..a33255f --- /dev/null +++ b/jni/ruby/ext/bigdecimal/sample/linear.rb @@ -0,0 +1,72 @@ +#!/usr/local/bin/ruby + +# +# linear.rb +# +# Solves linear equation system(A*x = b) by LU decomposition method. +# where A is a coefficient matrix,x is an answer vector,b is a constant vector. +# +# USAGE: +# ruby linear.rb [input file solved] +# + +# :stopdoc: +require "bigdecimal" +require "bigdecimal/ludcmp" + +# +# NOTE: +# Change following BigDecimal.limit() if needed. +BigDecimal.limit(100) +# + +include LUSolve +def rd_order(na) + printf("Number of equations ?") if(na <= 0) + n = ARGF.gets().to_i +end + +na = ARGV.size +zero = BigDecimal.new("0.0") +one = BigDecimal.new("1.0") + +while (n=rd_order(na))>0 + a = [] + as= [] + b = [] + if na <= 0 + # Read data from console. + printf("\nEnter coefficient matrix element A[i,j]\n"); + for i in 0...n do + for j in 0...n do + printf("A[%d,%d]? ",i,j); s = ARGF.gets + a << BigDecimal.new(s); + as << BigDecimal.new(s); + end + printf("Contatant vector element b[%d] ? ",i); b << BigDecimal.new(ARGF.gets); + end + else + # Read data from specified file. + printf("Coefficient matrix and constant vector.\n"); + for i in 0...n do + s = ARGF.gets + printf("%d) %s",i,s) + s = s.split + for j in 0...n do + a << BigDecimal.new(s[j]); + as << BigDecimal.new(s[j]); + end + b << BigDecimal.new(s[n]); + end + end + x = lusolve(a,b,ludecomp(a,n,zero,one),zero) + printf("Answer(x[i] & (A*x-b)[i]) follows\n") + for i in 0...n do + printf("x[%d]=%s ",i,x[i].to_s) + s = zero + for j in 0...n do + s = s + as[i*n+j]*x[j] + end + printf(" & %s\n",(s-b[i]).to_s) + end +end |