# Model file for the Google sum problem

var x{i in 1..9, j in 0..9} binary; 

# So, x[2,3]=1 means second letter (D) is equal to digit '3'

minimize objective: 0;

subject to Subtraction
: 
 100000*sum{j in 0..9}(x[1,j]*j) #w
+ 10000*sum{j in 0..9}(x[1,j]*j) #w
+  1000*sum{j in 0..9}(x[1,j]*j) #w
+   100*sum{j in 0..9}(x[2,j]*j) #d
+    10*sum{j in 0..9}(x[3,j]*j) #o
+       sum{j in 0..9}(x[4,j]*j) #t

-100000*sum{j in 0..9}(x[5,j]*j) #g
- 10000*sum{j in 0..9}(x[3,j]*j) #o
-  1000*sum{j in 0..9}(x[3,j]*j) #o
-   100*sum{j in 0..9}(x[5,j]*j) #g
-    10*sum{j in 0..9}(x[6,j]*j) #l
-     1*sum{j in 0..9}(x[7,j]*j) #e
=
 100000*sum{j in 0..9}(x[2,j]*j) #d
+ 10000*sum{j in 0..9}(x[3,j]*j) #o
+  1000*sum{j in 0..9}(x[4,j]*j) #t
+   100*sum{j in 0..9}(x[8,j]*j) #c
+    10*sum{j in 0..9}(x[3,j]*j) #o
+     1*sum{j in 0..9}(x[9,j]*j) #m
;

subject to assign1{i in 1..9}: sum{j in 0..9}x[i,j]=1;
subject to assign2{j in 0..9}: sum{i in 1..9}x[i,j]<=1;