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* GNU Lesser General Public License for more details.
*
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* If not, see http://www.gnu.org/licenses/lgpl-3.0.en.html
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package oscar.examples.cp.hakank
import oscar.cp.modeling._
import oscar.cp.core._
import scala.io.Source._
import scala.math._
/*
Traffic lights problem (CSPLib #16) in Oscar.
CSPLib problem 16
http://www.cs.st-andrews.ac.uk/~ianm/CSPLib/prob/prob016/index.html
"""
Specification:
Consider a four way traffic junction with eight traffic lights. Four of the traffic
lights are for the vehicles and can be represented by the variables V1 to V4 with domains
{r,ry,g,y} (for red, red-yellow, green and yellow). The other four traffic lights are
for the pedestrians and can be represented by the variables P1 to P4 with domains {r,g}.
The constraints on these variables can be modelled by quaternary constraints on
(Vi, Pi, Vj, Pj ) for 1<=i<=4, j=(1+i)mod 4 which allow just the tuples
{(r,r,g,g), (ry,r,y,r), (g,g,r,r), (y,r,ry,r)}.
It would be interesting to consider other types of junction (e.g. five roads
intersecting) as well as modelling the evolution over time of the traffic light sequence.
...
Results
Only 2^2 out of the 2^12 possible assignments are solutions.
(V1,P1,V2,P2,V3,P3,V4,P4) =
{(r,r,g,g,r,r,g,g), (ry,r,y,r,ry,r,y,r), (g,g,r,r,g,g,r,r), (y,r,ry,r,y,r,ry,r)}
[(1,1,3,3,1,1,3,3), ( 2,1,4,1, 2,1,4,1), (3,3,1,1,3,3,1,1), (4,1, 2,1,4,1, 2,1)}
The problem has relative few constraints, but each is very
tight. Local propagation appears to be rather ineffective on this
problem.
"""
Here are the four solutions from this model
Solution:
V: 0 2 0 2
P: 0 2 0 2
r r g g r r g g
Solution:
V: 1 3 1 3
P: 0 0 0 0
ry r y r ry r y r
Solution:
V: 2 0 2 0
P: 2 0 2 0
g g r r g g r r
Solution:
V: 3 1 3 1
P: 0 0 0 0
y r ry r y r ry r
@author Hakan Kjellerstrand hakank@gmail.com
http://www.hakank.org/oscar/
*/
object TrafficLights {
def main(args: Array[String]) {
val cp = CPSolver()
//
// data
//
val n = 4
val r = 0;
val ry = 1;
val g = 2;
val y = 3;
val lights = Array("r", "ry", "g", "y")
// The allowed combinations
val allowed = Array(Array(r,r,g,g),
Array(ry,r,y,r),
Array(g,g,r,r),
Array(y,r,ry,r))
//
// variables
//
val V = Array.fill(n)(CPIntVar(0 to n-1)(cp))
val P = Array.fill(n)(CPIntVar(0 to n-1)(cp))
//
// constraints
//
var numSols = 0
cp.solve subjectTo {
for(i <- 0 until n) {
val j = (1+i) % n
cp.add( table(Array(V(i),P(i),V(j),P(j)), allowed), Strong)
}
} search {
binaryStatic( V ++ P)
} onSolution {
println("\nSolution:")
println("V: " + V.mkString(""))
println("P: " + P.mkString(""))
for(i <- 0 until n) {
print(lights(V(i).value) + " " + lights(P(i).value) + " ")
}
println()
numSols += 1
}
println(cp.start())
}
}