can read data from file.

git-svn-id: svn+ssh://ci.ruby-lang.org/ruby/trunk@4818 b2dd03c8-39d4-4d8f-98ff-823fe69b080e

ext/bigdecimal/sample/linear.rb

@@ -6,43 +6,66 @@
 # 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]
+#
+
 require "bigdecimal"
 require "ludcmp"
 
-include LUSolve
+#
+# NOTE:
+#   Change following BigDecimal::limit() if needed.
+BigDecimal::limit(100)
+#
 
-def rd_order
-   printf("Number of equations ?")
-   n = gets().chomp.to_i
+include LUSolve
+def rd_order(na)
+   printf("Number of equations ?") if(na <= 0)
+   n = ARGF.gets().to_i
 end
 
-BigDecimal::limit(100)
-
+na   = ARGV.size
 zero = BigDecimal::new("0.0")
 one  = BigDecimal::new("1.0")
 
-while (n=rd_order())>0
+while (n=rd_order(na))>0
   a = []
   as= []
   b = []
-  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 = gets
-      a <<=BigDecimal::new(s);
-      as<<=BigDecimal::new(s);
-    end
-    printf("Contatant vector element b[%d] ? ",i);b<<=BigDecimal::new(gets);
+  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
-  printf "ANS="
   x = lusolve(a,b,ludecomp(a,n,zero,one),zero)
-  p x
-  printf "A*x-b\n"
+  printf("Answer(x[i] & (A*x-b)[i]) follows\n")
   for i in 0...n do
-    s = zero
-    for j 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
-    p s-b[i]
+     end
+     printf(" & %s\n",(s-b[i]).to_s)
   end
 end