示例#1
0
func removeExactEdges(n *coupling.Node, exact [][]bool) {
	n.Visited = true

	for _, row := range n.Adj {
		for _, edge := range row {
			// if the edge node adj matrix is nill, we do not have to recursivevly call it,
			// and just just delete n from its successor slice
			if edge.To.Adj == nil {
				coupling.DeleteNodeInSlice(n, &edge.To.Succ)
				continue
			}

			// if the edge has already been visited, we do not have have to recursively call
			if edge.To.Visited {
				coupling.DeleteNodeInSlice(n, &edge.To.Succ)
				continue
			}

			// recursively removes edges and successor node bottom up
			removeExactEdges(edge.To, exact)
			coupling.DeleteNodeInSlice(n, &edge.To.Succ)
		}
	}

	// here we do the actual removing of edges
	exact[n.S][n.T] = true
	exact[n.T][n.S] = true
	n.Adj = nil
	n.Visited = false

	return
}
示例#2
0
func preparenode() coupling.Node {
	var n coupling.Node

	e1 := coupling.Edge{&coupling.Node{S: 0, T: 0}, 0.1, true}
	e2 := coupling.Edge{&coupling.Node{S: 0, T: 1}, 0.5, true}
	e3 := coupling.Edge{&coupling.Node{S: 1, T: 0}, 0, false}
	e4 := coupling.Edge{&coupling.Node{S: 1, T: 1}, 0.4, true}

	n.Adj = [][]*coupling.Edge{
		[]*coupling.Edge{&e1, &e2},
		[]*coupling.Edge{&e3, &e4}}

	return n
}
示例#3
0
func findReverseReachable(node *coupling.Node, reachables []*coupling.Node) []*coupling.Node {
	node.Visited = true

	for _, succ := range node.Succ {
		if !coupling.IsNodeInSlice(succ, reachables) {
			reachables = append(reachables, succ)
		}

		if len(succ.Succ) == 0 || succ.Visited {
			continue
		}

		reachables = findReverseReachable(succ, reachables)
	}

	node.Visited = false

	return reachables
}
示例#4
0
func setUpCoupling() coupling.Coupling {
	c := coupling.New()
	n1 := coupling.Node{S: 0, T: 0}
	n2 := coupling.Node{S: 0, T: 1}
	n3 := coupling.Node{S: 1, T: 0}
	n4 := coupling.Node{S: 1, T: 1}
	e1 := coupling.Edge{&n1, 0.5, true}
	e2 := coupling.Edge{&n2, 0.2, true}
	e3 := coupling.Edge{&n3, 0, false}
	e4 := coupling.Edge{&n4, 0.3, true}
	n1.Succ = []*coupling.Node{&n1, &n2, &n4}
	n2.Succ = []*coupling.Node{&n2, &n4, &n1}
	n4.Succ = []*coupling.Node{&n2, &n1, &n4}
	n2.Adj = [][]*coupling.Edge{[]*coupling.Edge{&e1, &e2}, []*coupling.Edge{&e3, &e4}}
	c.Nodes = []*coupling.Node{&n1, &n2, &n3, &n4}

	return c
}
示例#5
0
func setUpLinearEquations(n *coupling.Node, exact [][]bool, d [][]float64, a *[][]float64, b *[]float64, i int, index *[]*coupling.Node, lambda float64) {
	n.Visited = true

	for _, row := range n.Adj {
		for _, edge := range row {
			// if the node is non-basic, we do not have to look at it
			if !edge.Basic {
				continue
			}

			// if the node is exact, we add its probablity times its distance to the i'th row in the b vector
			if exact[edge.To.S][edge.To.T] {
				(*b)[i] += d[edge.To.S][edge.To.T] * edge.Prob * lambda
				continue
			}

			// if the node has already been visited, we do not have to add it as a new linear equation
			// and subtract its probability from its corresponding place in the a matrix
			if edge.To.Visited {
				rowindex := findRowIndex(index, edge.To)
				(*a)[i][rowindex] -= edge.Prob * lambda
				continue
			}

			// we have to add a new linear equation, so we update a, b, and index
			addLinearEquation(a, b, index, edge.To)

			(*a)[i][len(*a)-1] -= edge.Prob * lambda

			// recursively sets up the linear equation
			setUpLinearEquations(edge.To, exact, d, a, b, len(*a)-1, index, lambda)
		}
	}

	return
}