Initial revision

buddy/examples/money/makefile

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+# --------------------------------
+# Makefile for BVEC example
+# --------------------------------
+
+# --- Compiler flags
+CFLAGS = -O3 -pedantic -Wall -ansi -L../../src -I../../src
+
+# --- C++ compiler
+CPP = g++
+
+# --- C compiler
+CC = gcc
+
+
+# --- Do not touch ---
+
+.SUFFIXES: .cxx .c
+
+.cxx.o:
+	$(CPP) $(CFLAGS) -c $<
+
+.c.o:
+	$(CC) $(CFLAGS) -c $<
+
+money:	money.o bddlib
+	$(CPP) $(CFLAGS) money.o -o money -lbdd -lm
+
+bddlib:
+	cd ../../src; make
+
+clean:
+	rm -f *~
+	rm -f *.o
+	rm -f money
+
+money.o:	../../src/bvec.h ../../src/fdd.h ../../src/bdd.h

buddy/examples/money/money.cxx

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+#include "bvec.h"
+
+/* Find a solution to the send-more-money example
+ * The problem is to assign values for the digits s,e,n,d,m,o,r,y
+ * in such a way that the following equation is true:
+ *
+ *     s e n d +
+ *     m o r e =
+ *   m o n e y
+ *
+ * with the additional constraints that all digits must have different values
+ * and s>0 and m>0.
+ */
+
+int main(void)
+{
+      // Allocate 11 domains with room for up to 3*10
+   static int dom[11] = {30,30,30,30,30,30,30,30,30,30,30};
+   
+   bdd_init(10000,10000);
+   fdd_extdomain(dom,11);
+
+      // Assign binary vectors (expressions) to the digits
+   
+   bvec s = bvec_varfdd(0);  // The 's' digit
+   bvec e = bvec_varfdd(1);  // The 'e' digit
+   bvec n = bvec_varfdd(2);  // ...
+   bvec d = bvec_varfdd(3);
+   bvec m = bvec_varfdd(4);
+   bvec o = bvec_varfdd(5);
+   bvec r = bvec_varfdd(6);
+   bvec y = bvec_varfdd(7);
+   bvec m1 = bvec_varfdd(8);  // Carry out 1
+   bvec m2 = bvec_varfdd(9);  // Carry out 2
+   bvec m3 = bvec_varfdd(10); // Carry out 3
+
+      // Create a few constants of the right bit number (5)
+   bvec c10 = bvec_con(5,10);
+   bvec c2 = bvec_con(5,2);
+   bvec c0 = bvec_con(5,0);
+
+      // Create constraints
+
+      // Constraint 1:  addition of the last digits and constraints on
+      // the max. value of the involved variables and carry-out
+   bdd  t1 = (d + e       ==  y + m1*10) & d<c10 & e<c10 & y<c10 & m1<c2;
+
+   // The use of "m1*10" instead of "m1*c10" avoids a bitnum mismatch since
+   // "m1*10" results in 5 bits but "m1*c10" results in 10 bits!
+   
+      // And so on ...
+   bdd  t2 = (n + r + m1  ==  e + m2*10) & n<c10 & r<c10 & m2<c2;
+   bdd  t3 = (e + o + m2  ==  n + m3*10) & o<c10 & m3<c2;
+   bdd  t4 = (s + m + m3  ==  o +  m*10) & s<c10 & m<c2;
+   bdd  t5 = (m > c0  &  s > c0);
+
+      // Make sure all digits are different
+   bdd  t6 = (s!=e  &  s!=n  &  s!=d  &  s!=m  &  s!=o  &  s!=r  &  s!=y);
+   bdd  t7 = (e!=n  &  e!=d  &  e!=m  &  e!=o  &  e!=r  &  e!=y);
+   bdd  t8 = (n!=d  &  n!=m  &  n!=o  &  n!=r  &  n!=y);
+   bdd  t9 = (d!=m  &  d!=o  &  d!=r  &  d!=y);
+   bdd t10 = (m!=o  &  m!=r  &  m!=y);
+   bdd t11 = (o!=r  &  o!=y);
+   bdd t12 = (r!=y);
+
+      // Join all constraints
+   bdd t = t1 & t2 & t3 & t4 & t5 & t6 & t7 & t8 & t9 & t10 & t11 & t12;
+
+      // Print result
+   cout << fddset << t << endl;
+   
+   return 0;
+}