# HG changeset patch
# User athos
# Date 1161952753 0
# Node ID 90c66dc93ca4f2d88031cd3e1abe618e0f9a214a
# Parent  9273fe7d850c23c9753fb257b1b7875168684a55
A little test was born for Expr::simplify().

diff -r 9273fe7d850c -r 90c66dc93ca4 test/lp_test.cc
--- a/test/lp_test.cc	Thu Oct 26 14:20:17 2006 +0000
+++ b/test/lp_test.cc	Fri Oct 27 12:39:13 2006 +0000
@@ -19,7 +19,7 @@
 #include <sstream>
 #include <lemon/lp_skeleton.h>
 #include "test_tools.h"
-
+#include <lemon/tolerance.h>
 
 #ifdef HAVE_CONFIG_H
 #include <config.h>
@@ -175,6 +175,37 @@
     lp.addRow(x[1]+x[3]<=x[5]-3);
     lp.addRow(-7<=x[1]+x[3]-12<=3);
     lp.addRow(x[1]<=x[5]);
+
+    std::ostringstream buf;
+
+
+    //Checking the simplify function
+
+//     //How to check the simplify function? A map gives no information
+//     //on the question whether a given key is or is not stored in it, or
+//     //it does?
+//   Yes, it does, using the find() function.
+    e=((p1+p2)+(p1-p2));
+    e.simplify();
+    buf << "Coeff. of p2 should be 0";
+    //    std::cout<<e[p1]<<e[p2]<<e[p3]<<std::endl;
+    check(e.find(p2)==e.end(), buf.str());
+
+     
+
+
+    e=((p1+p2)+(p1-0.99*p2));
+    double tolerance=0.001;
+    e.simplify(tolerance);
+    buf << "Coeff. of p2 should be 0.01";
+    check(e[p2]>0, buf.str());
+    
+    tolerance=0.02;
+    e.simplify(tolerance);
+    buf << "Coeff. of p2 should be 0";
+    check(e.find(p2)==e.end(), buf.str());
+    
+
   }
   
   {