Minizinc: How to apply this constraint for scheduling model? -

i working on scheduling model , i'd figure out how use constraint: using minizinc 2.0.2 version & minizincide-0.9.7-linux , g12-mip & gecode solvers.

array[1..2,1..48] of var 0..1: table; array[1..48] of int:demand; array[1..2] of int: cost; constraint forall(j in 1..48)(sum(i in 1..2)(table[i,j]) >= demand[j] ); solve minimize sum(i in 1..2)(sum(j in 1..48)(table[i,j])*cost[i]); output [show(table)]; 

sample data.dzn file:

demand=[0,1,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]; cost=[3,5]; 

output table array using g12-mip solver gives result:

[0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 

this model 2 points , 48 hours(i.e 2 days).the constraint wanted add each employee have shift each day without break if alloted any. in desired output :

[0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 

approach tried:

var 1..5: k; array[1..2,1..2] of var 1..48: key2; array[1..2,1..2] of var 1..48: key1;  % aim store first , last index table[i,j]=1 each point (i) each day (k) constraint forall(i in 1..2,j in 1..48 k= ceil(j/24))(if          table[i,j]==1 key2[i,k]=j else true endif)           /\ forall(i in 1..2,j in 48..1 k= ceil(j/24))(if          table[i,j]==1 key1[i,k]=j else true endif);  % make table[i,j]=1 between first index , last index table[i,j]=1 constraint forall(i in 1..2,k in 1..2)(forall(t in key1[i,k]..key2[i,k])(table[i,t]=1));    

when ran through linux terminal using command: mzn-g12mip test.mzn data.dzn gave same previous result. when ran through minizincide-0.9.7-linux gave error :

minizinc: type error: type error in operator application `..'. no matching operator found left-hand side type `var int' , right-hand side type `var int' 

is there problem code or there other way satisfy constraint?

two notes:

i got same result you: running via command line works. using minizincide give same error, after resetting mzn_stdlib_dir in terminal run minizincide.sh works. have mzn_stdlib_dir set? try reset command.

     export mzn_stdlib_dir= 

regarding shift constraint, can use global constraint "regular". see minizinc tutorial example. specific constraint want here "contiguity" constraint requires 1s collected. it's not built-in in minizinc, can see model example: http://hakank.org/minizinc/contiguity_regular.mzn

here's version of model, including definition of contiguity.

include "globals.mzn";   array[1..2,1..48] of var 0..1: table; array[1..48] of int:demand; array[1..2] of int: cost;  var int: z = sum(i in 1..2)(sum(j in 1..48)(table[i,j])*cost[i]);  constraint forall(j in 1..48)(sum(i in 1..2)(table[i,j]) >= demand[j] );  constraint   forall(i in 1..2) (     contiguity([table[i,j] | j in 1..48])   ) ;  predicate contiguity(array[int] of var int: a) =   let {         int: n_states = 3,         int: input_max = 2,         int: initial_state = 1,         set of int: accepting_states = {1,2,3},         % transition matrix         array[1..n_states, 1..input_max] of int: transition_fn =     array2d(1..n_states, 1..input_max,     [        % note:all states accepting states       1,2, % state 1 (start): input 0 -> state 1, input 1 -> state 2 i.e. 0*       3,2, % state 2: 1*        3,0, % state 3: 0*      ]),       int: len = length(a),       % note: regular cannot handle 0 values, must > 0       array[1..len] of var 1..2: reg_input   } in    forall(i in 1..len) (      reg_input[i] = a[i] + 1    )    /\    regular(reg_input, n_states, input_max, transition_fn,                     initial_state, accepting_states) ;   solve minimize z;  output [   "table: ", show(table), "\n",   "z: ", show(z), "\n", ] ++ ["\ntable:"] ++ [   if j = 1 "\n" else " " endif ++     show(table[i,j])   | in 1..2, j in 1..48 ] ;  demand=[0,1,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]; cost=[3,5]; 

the optimal solution using model not same solution since tasks without demand costs if person 1 must take "empty" tasks between mandatory tasks. if have other constraint don't allow separation of tasks, have add in model.

table: 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0