I've been looking for the difference of ExitRule and EnterRule in Listener in "the definitive antrl4 reference" book but I still don't understand the difference. What is the difference between these two? And how does Listener travel the tree?
simply said, these are auto generated events created by Antrl
to keep track of the walker, so to speak.
Imagine, you stand in the middle of a long floor with countless doors on either site. Which you have to open and something in the rooms.
To keep track on which doors you've already visited, you mark them with a X and an incrementing number behind.
enterRule == You opens the door of a room.
you get in and search
exitRule == leave the room and paint the X and the next number on the door.
Now you are able to tell exactly which rooms you have already visited, but further more you are able to go to a specific room again for another search, without having to go all the way back and start all over again.
more technically spoken.
Antrl creates an enter and exit method for each Rule that is defined.
These Methods, or also known as callbacks, being used by the walker to walk the given tree.
Using a *ParseTreeListener you provide an entry point which indicates the beginning of the tree. For example enterAssign.
The Walker looks for this event and triggers it.
It then looks for a sub enterRule to trigger this event, and so on...
It keeps walking for as long as it find further enter rules or triggers exitAssign and the walker stops its walk.
Keypoint here is the automated or indipendent walk behavior.
The ParseTreeVisitor on the other hand,
will not generating enter/exit Rules to walk a tree.
It will generate visitRule instead.
The visit methods have to be called explicitly!
That means, if you forget to invoke a visit all its children don't get visited at all.
Antrl-Doc --> Parse Tree Listeners
Antrl-Mega-Tutorial --> many Information, short and precise
Tree Walking
Walking starts from a Root-Node and goes down on it until it have found the very left-most nested item. Then goes back up until it reaches the first node and looks for a sub-tree on the right.
It enters the right tree if one is found.
Or, it gets futher up the chain to find the next node.
...
Until it reaches the Root-Node again.
short Picture:
parent
|
/ \
/ \
Child1 Child2
/
/
Grandchild
chain of Calls:
enter parent
enter Child1
enter Grandchild
exit Grandchild
exit Child1
enter Child2
exit Child2
exit parent
Related
Could anybody please clarify how (as I have not found any clear example anywhere) The MCTS algorithm iterates for the second player.
Everything I seem just seems to look like it is playing eg P1 move every time.
I understand the steps for one agent but I never find anything showing code where P2 places its counter, which surely must happen when growing the tree.
Essentially I would expect:
for each iter:
select node Player1
expand Player1
select node Player2
expand player 2
rollout
backpropogate
next iter
Is this right?? Could anybody please spell out some psuedocode showing that? Either iteratively or recursion i don't mind.
Thanks for any help.
The trick is in backpropagation part, where you update "wins" variable from the point of view of player whose move led into this position.
Code for MCTS
Notice under UCT function, specially the comments:
#Backpropagate
while node != None: # backpropagate from the expanded node and work back to the root node
node.Update(state.GetResult(node.playerJustMoved)) # state is terminal. Update node with result from POV of node.playerJustMoved
node = node.parentNode
IF you follow the function call, you would realize visit variable is always updated; wins however, is not.
I have one parent node ,based on the user input Iam setting a context variable at my application level eligibility:yes or no and passing back.And for my parent node I have two child nodes for conditions $eligibility=="yes" and $eligibility=="no".So once users input from parent node validation is done and context variable is passed back ,then I need to jump and look for condition of eligibility.If yes I need to go one node ,if no then to other.How can I do?
I tried putting true to node and added these two nodes to this and jump to true..But didnt worked..How can we achieve this?
what #data_henrik has mentioned is a good way to set context value and then switch to different flows depending upon the set value. But when you need to perform some logic before setting that value in the context from your application, it won't be a suitable way.
I had a requirement like this, so we used to send a dummy text from our application after we were done with setting the value in context after the parent node execution. Check out the images and explanation after that.
We didn't use Jump because we had to do some validation in the Conversation service after parent node before moving forward. Using jumps would've allowed the Conversation to move to next node before we could set value in context.
Use case flow - once user enters text for the parent node intent, for my case "#send-mail" intent, I show the parent response and do some functional validation in my app after that and add a value to the context. Now we send a dummy text "valid" which satisfies the intent "#Valid" and hence move to the next node in flow. In this node we check for the value in context (which is already set by now) and show appropriate response to user.
You can set within your first two test nodes, $testMe==true and $testMe==false a temp output variable within the output json packet, i.e. output{"temp":"true"} or "false". Then you can jump to a new set of nodes and test for the output.temp value, i.e. output.temp == 'true' then do something, or output.temp == 'false' then do something.
The nice side effect of this action is that the output.temp variable only has a life of that current conversation input. Unlike context variables which need to be removed / deleted.
I have to solve the "rush hour puzzle" by iterative deepening algorithm. I have read a lot of topics here on stackoverflow and also on the internet. I think that I understand the iterative deepening algorithm. Basically you just go deeper into the tree and try to find the solution.
I figured that I need to create a graph or a tree from the puzzle, but I really don't have an idea how. Also, if I would have the tree, then how would I tell if something is a valid move or a final state?
There were answers that the nodes should be possible moves and the edges are between the nodes that can be reached in one move. I can imagine this, but somehow I'm getting trouble in see how this can be useful or better yet how can this solve the problem.
Please help me, I'm not asking for complete solution or code sample, I just need some easy explanation of the problem.
There is a reason you need to use the deepening algorithm. Imagine you name each car A, B, C, D... The root node of your tree is the initial board state. Now, move car A. You go down one node in the tree. Move car A back. You are at the initial state, but you made two moves to get here, so you are two nodes down the tree. Repeat over and over. You will never hit a final state.
The root node of your tree is the initial board state. Given that node, add a child node to it for every possible valid move. So, each child node will be what the initial tree looks like after one move. Now, for each of those child nodes, do the same thing: make a child node where each node is one move off the original child node.
Eventually, you will hit a solution to the puzzle. When that happens, you print the moves from the root node to the solution child node and quit. This algorithm ensures that you find a solution with the least number of moves.
I'm currently working on some code for a project. The question is about targeting child with a unique parent. For example. Let's say I have a box people can move called box_mc and 3 platforms it can jump on called:
Platform_1
Platform_2
Platform_3
All of these platforms have a child element called hit.
Platform_1
Hit
Platform_2
Hit
Platform_3
Hit
I use an array and a for each statement to detect if box_mc hits one of the platforms childs.
var obj_arr:Array = [Platform_1, Platform_2, Platform_3];
for each(obj in obj_arr){
if(box_mc.hitTestObject(obj.hit)){
trace(obj + " " + obj.hit);
box_mc.y = obj.hit.y - box_mc.height;
}
}
obj seems to output the unique parent it is hitting but obj.hit ouputs hit so my theory in this is that it is applying the change of y to all the childs called hit in the stage. WOuld it be possible to only detect the child of that specific parent?
Thanks in advance again guys!
Question: how can we display tree nodes level by level ?. could you please give me time and space efficient solution .
Example :
A
/ \
B C
/ \ / \
D E F G
void PrintTree(struct tree *root);
Output:
You have to print tree nodes level by level
A
B C
D E F G
If you're feeling brutish, and want to think very simply about the level you are at...
You will need:
Two queues
A slight twist on Jack's approach
So, start with root.
Tack its children onto the first queue.
Step through them, tacking their children onto the second queue as you go.
Switch to the second queue, step through, pushing their children onto the first queue.
Wax on, wax off.
Really it's just a slight expansion of the same idea, the breadth first search or sweep, which is worth thinking about as a pattern, since it applies to a variety of data structures. Almost anything that's a tree or trie, and a few things that aren't, in fact!
To save space and time on SO:
http://thecodecracker.com/c-programming/bfs-and-dfs/
This kind of visit is called Breadth-first or Level Order. You can see additional infos here.
Basically you
first visit the current node
then all the children of that node
then all the children of every children and so on
This should be achieved easily with a FIFO structure:
push the root
until queue is empty
take first element, visit it, and push all its children to the end of the queue
repeat