60 arithmetic/digits/equations/123456789.p

In how many ways can "." be replaced with "+", "-", or "" (concatenate) in
.1.2.3.4.5.6.7.8.9=1 to form a correct equation?

arithmetic/digits/equations/123456789.s

``` 1-2 3+4 5+6 7-8 9 = 1
+1-2 3+4 5+6 7-8 9 = 1
1+2 3+4-5+6 7-8 9 = 1
+1+2 3+4-5+6 7-8 9 = 1
-1+2 3-4+5+6 7-8 9 = 1
1+2 3-4 5-6 7+8 9 = 1
+1+2 3-4 5-6 7+8 9 = 1
1-2 3-4+5-6 7+8 9 = 1
+1-2 3-4+5-6 7+8 9 = 1
1-2-3-4 5+6 7-8-9 = 1
+1-2-3-4 5+6 7-8-9 = 1
1+2-3 4+5 6-7-8-9 = 1
+1+2-3 4+5 6-7-8-9 = 1
-1+2 3+4+5-6-7-8-9 = 1
-1 2+3 4-5-6+7-8-9 = 1
1+2+3+4-5+6+7-8-9 = 1
+1+2+3+4-5+6+7-8-9 = 1
-1+2+3-4+5+6+7-8-9 = 1
1-2-3+4+5+6+7-8-9 = 1
+1-2-3+4+5+6+7-8-9 = 1
1+2 3+4 5-6 7+8-9 = 1
+1+2 3+4 5-6 7+8-9 = 1
1+2 3-4-5-6-7+8-9 = 1
+1+2 3-4-5-6-7+8-9 = 1
1+2+3+4+5-6-7+8-9 = 1
+1+2+3+4+5-6-7+8-9 = 1
-1+2+3+4-5+6-7+8-9 = 1
1-2+3-4+5+6-7+8-9 = 1
+1-2+3-4+5+6-7+8-9 = 1
-1-2-3+4+5+6-7+8-9 = 1
1-2+3+4-5-6+7+8-9 = 1
+1-2+3+4-5-6+7+8-9 = 1
1+2-3-4+5-6+7+8-9 = 1
+1+2-3-4+5-6+7+8-9 = 1
-1-2+3-4+5-6+7+8-9 = 1
-1+2-3-4-5+6+7+8-9 = 1
-1+2 3+4 5-6 7-8+9 = 1
1-2 3-4 5+6 7-8+9 = 1
+1-2 3-4 5+6 7-8+9 = 1
-1+2 3-4-5-6-7-8+9 = 1
-1+2+3+4+5-6-7-8+9 = 1
1-2+3+4-5+6-7-8+9 = 1
+1-2+3+4-5+6-7-8+9 = 1
1+2-3-4+5+6-7-8+9 = 1
+1+2-3-4+5+6-7-8+9 = 1
-1-2+3-4+5+6-7-8+9 = 1
1+2-3+4-5-6+7-8+9 = 1
+1+2-3+4-5-6+7-8+9 = 1
-1-2+3+4-5-6+7-8+9 = 1
-1+2-3-4+5-6+7-8+9 = 1
1-2-3-4-5+6+7-8+9 = 1
+1-2-3-4-5+6+7-8+9 = 1
1-2 3+4+5+6+7-8+9 = 1
+1-2 3+4+5+6+7-8+9 = 1
1+2+3+4 5-6 7+8+9 = 1
+1+2+3+4 5-6 7+8+9 = 1
1 2+3 4+5-6 7+8+9 = 1
+1 2+3 4+5-6 7+8+9 = 1
1+2+3-4-5-6-7+8+9 = 1
+1+2+3-4-5-6-7+8+9 = 1
-1+2-3+4-5-6-7+8+9 = 1
1-2-3-4+5-6-7+8+9 = 1
+1-2-3-4+5-6-7+8+9 = 1
-1-2-3-4-5+6-7+8+9 = 1
-1-2 3+4+5+6-7+8+9 = 1
1-2+3 4-5 6+7+8+9 = 1
+1-2+3 4-5 6+7+8+9 = 1
1 2-3 4+5-6+7+8+9 = 1
+1 2-3 4+5-6+7+8+9 = 1

Total solutions  = 69 (26 of which have a leading "+", which is redundant)

69/19683 = 0.35 %

for those who care (it's not very elegant but it did the trick):

#include <stdio.h>
#include <math.h>

main (argc,argv)
int argc;
char *argv[];
{
int sresult, result, operator[10],thisop;
char buf[256],ops[3];
int i,j,tot=0,temp;

ops[0] = ' ';
ops[1] = '-';
ops[2] = '+';

for (i=1; i<10; i++) operator[i] = 0;

for (j=0; j < 19683; j++) {
result = 0;
sresult = 0;
thisop = 1;
for (i=1; i<10; i++) {
switch (operator[i]) {
case 0:
sresult = sresult * 10 + i;
break;
case 1:
result = result + sresult * thisop;
sresult = i;
thisop = -1;
break;
case 2:
result = result + sresult * thisop;
sresult = i;
thisop = 1;
break;
}
}

result  = result + sresult * thisop;
if (result == 1) {
tot++;
for  (i=1;i<10;i++)
printf("%c%d",ops[operator[i]],i);
printf(" = %d\n",result);
}
temp = 0;
operator[1] += 1;
for (i=1;i<10;i++) {
operator[i] += temp;
if (operator[i] > 2) { operator[i] = 0; temp = 1;}
else temp = 0;
}

}

printf("Total solutions  = %d\n" , tot);
}

cwren@media.mit.edu (Christopher Wren) Continue to:prev: 59 arithmetic/digits/divisible.p Indexnext: 61 arithmetic/digits/equations/1992.p