/**
* Test debugging till given DPU instance in a more complex graph. ! Scenario: E0 -> T1 -> T2 ->
* L3 E4 -> T5 ---^ E6 -^
*/
@Test
public void testDebugNodeInComplexGraph() {
PipelineGraph graph = buildGraph(7);
graph.addEdge(nodes[0], nodes[1]);
graph.addEdge(nodes[1], nodes[2]);
graph.addEdge(nodes[2], nodes[3]);
graph.addEdge(nodes[4], nodes[5]);
graph.addEdge(nodes[5], nodes[1]);
graph.addEdge(nodes[6], nodes[2]);
DependencyGraph dGraph = new DependencyGraph(graph, nodes[1]);
GraphIterator iter = dGraph.iterator();
Set<Node> nodesToRun = new HashSet<>();
nodesToRun.add(nodes[0]);
nodesToRun.add(nodes[1]);
nodesToRun.add(nodes[4]);
nodesToRun.add(nodes[5]);
for (int i = 0; i < 4; i++) {
assertTrue(iter.hasNext());
assertTrue(nodesToRun.contains(iter.next()));
}
assertFalse(iter.hasNext());
assertNull(iter.next());
}
/** Test debugging till given DPU instance in a graph with a single node. ! Scenario: E -> L */
@Test
public void testDebugNodeInTrivialGraph() {
PipelineGraph graph = buildGraph(5);
graph.addEdge(nodes[0], nodes[1]);
DependencyGraph dGraph = new DependencyGraph(graph, nodes[0]);
GraphIterator iter = dGraph.iterator();
assertTrue(iter.hasNext());
assertEquals(nodes[0], iter.next());
assertFalse(iter.hasNext());
assertNull(iter.next());
}
/** Test dependency resolution for serialized DPURecord setup. Scenario: E -> T -> T -> T -> L */
@Test
public void testInlineDependencyResolution() {
PipelineGraph graph = buildGraph(5);
graph.addEdge(nodes[2], nodes[3]);
graph.addEdge(nodes[0], nodes[1]);
graph.addEdge(nodes[1], nodes[2]);
graph.addEdge(nodes[3], nodes[4]);
DependencyGraph dGraph = new DependencyGraph(graph);
GraphIterator iter = dGraph.iterator();
// check correct order
for (int i = 0; i < 5; i++) {
assertTrue(iter.hasNext());
assertSame(nodes[i], iter.next());
}
// no more nodes
assertFalse(iter.hasNext());
assertNull(iter.next());
}
/** Test more complex setup of DPUs. Scenario: E0 -> T1 -> T2 -> L3 E4 ---^ */
@Test
public void testComplexDependencyResolution() {
PipelineGraph graph = buildGraph(5);
graph.addEdge(nodes[0], nodes[1]);
graph.addEdge(nodes[1], nodes[2]);
graph.addEdge(nodes[2], nodes[3]);
graph.addEdge(nodes[4], nodes[2]);
DependencyGraph dGraph = new DependencyGraph(graph);
GraphIterator iter = dGraph.iterator();
// first must be E0 or E4
Node n = iter.next();
assertTrue(n == nodes[0] || n == nodes[4]);
// second may be any of E0, E4, T1
n = iter.next();
assertTrue(n == nodes[0] || n == nodes[1] || n == nodes[4]);
// third may be E4 or T1
n = iter.next();
assertTrue(n == nodes[1] || n == nodes[4]);
// fourth is always T2
n = iter.next();
assertSame(nodes[2], n);
// last is always L3
n = iter.next();
assertSame(nodes[3], n);
// no more nodes
assertFalse(iter.hasNext());
assertNull(iter.next());
}
/** Test circular dependency resolution. Scenario: E -> T -> T -> T -> T ^---------' */
@Test
public void testCircularDependencyResolution() {
PipelineGraph graph = buildGraph(5);
graph.addEdge(nodes[0], nodes[1]);
graph.addEdge(nodes[1], nodes[2]);
graph.addEdge(nodes[2], nodes[3]);
graph.addEdge(nodes[3], nodes[4]);
graph.addEdge(nodes[3], nodes[1]);
DependencyGraph dGraph = new DependencyGraph(graph);
GraphIterator iter = dGraph.iterator();
// first node is not in the circle
assertTrue(iter.hasNext());
assertSame(nodes[0], iter.next());
// second node is in the circle
// graph still says to have more nodes and so return true for hasNext,
// however does not return any node, because it is impossible to tell
// which node is next
assertTrue(iter.hasNext());
assertNull(iter.next());
}