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ib002(graphs): convert pseudocode to Ada

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cause why not…

Signed-off-by: Matej Focko <mfocko@redhat.com>
This commit is contained in:
Matej Focko 2023-02-04 18:06:49 +01:00
parent a4a9bcfcd2
commit 2c0d77b689
Signed by: mfocko
GPG key ID: 7C47D46246790496
1 changed files with 119 additions and 75 deletions
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194
ib002/10-graphs/iterative-and-iterators.md
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@ -21,81 +21,108 @@ On the other hand, we have seen iterative implementation in the exercises and I
### Recursive DFS implementation from exercises without colors
```
FUNCTION VisitedDFS(u, Visited) IS
Visited <- Union(Visited, { u })
FOR v IN u.successors DO
IF v NOT IN Visited THEN
Visited <- VisitedDFS(v, Visited)
FI
OD
```ada showLineNumbers
function VisitedDFS(u: Vertex, visited: VertexSet) return VertexSet is
v: Vertex;
begin
visited.Union(To_Set(u));
RETURN Visited
END
for v in u.successors loop
if not Contains(visited, v) then
visited := visitedDFS(v, Visited);
end if;
end loop;
return visited;
end VisitedDFS;
```
This implementation is correct, does the DFS traversal as it should, however it has one „smallish“ downside and that is the time complexity. The usage of set raises the time complexity, of course it is implementation dependant. However in case of either RB-tree or hash-table implementation, we get look-up in time $\mathcal{O}(n)$ for hash-table in worst-case or $\mathcal{O}(\log n)$ for the other in the worst-case. Both are not ideal compared to checking color on vertex.
### Iterative DFS from the exercises
```
PROCEDURE IterDFS(u)
stack <- empty stack
Push(stack, (u, 0))
u.color <- gray
time <- 1
u.d <- time
```ada showLineNumbers
procedure IterDFS(u: Vertex) is
stack: StateVector;
i, time: Integer;
v: Vertex;
begin
stack.Append(VertexState(u, 0));
u.color := Gray;
time := 1;
u.d := time;
WHILE NOT Empty(stack) DO
(u, k) <- Pop(stack)
while not stack.Is_Empty loop
u := stack.Last_Element.Vertex;
i := stack.Last_Element.NextIndex;
stack.Delete_Last;
IF k < Length(u.successors) THEN
// search is not finished, is pushed back to stack
Push(stack, (u, k + 1))
if i < u.successors.Length then
-- search is not finished, is pushed back to stack
stack.Append(VertexState(u, k + 1));
v <- u.successors[k + 1]
IF v.color = white THEN
Push(stack, (v, 0))
v.color <- gray
time <- time + 1
v.d <- time
FI
ELSE
// u has no other successors, we can finish the search
time <- time + 1
u.f <- time
u.color <- black
FI
OD
END
v := u.successors.Element(i);
if v.color = White then
stack.Append(VertexState(v, 0));
v.color := Gray;
time := time + 1;
v.d := time;
end if;
else
-- u has no other successors, we can finish the search
time := time + 1;
u.f := time;
u.color := Black;
end if;
end loop;
end IterDFS;
```
As we can see, there is some ordering in which we search through the successors. Time complexity is OK, stack holds at most all vertices (they must be on the current path).
### My iterative with path in stack
```
PROCEDURE DFS(G, start) IS
path <- [ start ]
time <- 1
start.d, start.color <- time, gray
```ada showLineNumbers
procedure DFS(start: Vertex) is
path: VertexVector;
time: Integer;
hasSuccessor: Bool;
successor: Vertex;
begin
path.Append(start);
time := 1;
WHILE NOT Empty(path) DO
hasSuccessor <- false
FOR successor IN path[-1].successors DO
IF successor.color = white THEN
hasSuccessor <- true
successor.d, successor.color <- ++time, gray
path <- Append(path, successor)
BREAK
FI
OD
IF NOT hasSuccessor THEN
lastVertex <- Pop(path)
lastVertex.f, lastVertex.color <- ++time, black
FI
OD
END
start.d := time;
start.color := Gray;
while not path.Is_Empty loop
hasSuccessor := false;
for successor in path.Last_Element.successors loop
if successor.color = White then
hasSuccessor := true;
successor.d := time + 1;
successor.color := Gray;
time := time + 1;
path.Append(successor);
exit;
end if;
end loop;
if not hasSuccessor then
path.Last_Element.f := time + 1;
path.Last_Element.color := Black;
time := time + 1;
path.Delete_Last;
end if;
end loop;
end DFS;
```
This approach is similar to the iterative solution from the exercises, but it does not keep the index of the next successor, therefore it always iterates through all of them, which raises the time complexity.
@ -104,27 +131,44 @@ This approach is similar to the iterative solution from the exercises, but it do
On the other hand, we do not actually have to depend on the representation of the graph. In this case, we just _somehow_ obtain the iterator (which yields all of the succesors) and keep it in the stack.
```
PROCEDURE DFS(G, start) IS
path <- [ (start, Iterator(start.successors)) ]
time <- 1
start.d, start.color <- time, gray
```ada showLineNumbers
procedure DFS(start: Vertex) is
path: StateVector;
time: Integer;
current: State;
nextVertex: Vertex;
begin
path.Append(State(start));
time := 1;
WHILE NOT Empty(path) DO
lastVertex, successors <- path[-1]
start.d := time;
start.color := Gray;
IF NOT MoveNext(successors) THEN
Pop(path)
lastVertex.f, lastVertex.color <- ++time, black
ELSE IF successors.Current.color = white THEN
nextVertex <- successors.Current
nextVertex.d, nextVertex.color <- ++time, gray
path <- Append(path, (nextVertex, Iterator(nextVertex.successors)))
FI
OD
END
while not path.Is_Empty loop
current := path.Last_Element;
if not Move_Next(current.successors) then
path.Delete_Last;
time := time + 1;
current.vertex.f := time;
current.vertex.color := Black;
else if current.successors.Value.color = white then
nextVertex := current.successors.Value;
time := time + 1;
nextVertex.d := time;
nextVertex.color := Gray;
path.Append(State(nextVertex));
end if;
end loop;
end DFS;
```
( The way we manipulate with the iterators is closest to the C# implementation. Apart from the `Iterator` thing :) In case you tried to implement it in C++, you would more than likely need to change the check, since you would get first successor right at the beginning )
So here we don't keep indices, but the iterators. We can also check existence of other successors easily: by the iterator moving after the last successor.