Could someone please provide a definition of decision nodes, change nodes and end nodes.
I have view the decision tree interpretation on wikipedia and haven't found the clear definition about the three tree nodes.
Thanks!
The way I understand it is like this:
Decision Nodes:
The node where a there is a requirement set that determines the outcome: ex:Profit>50k.
Chance Node:
A node were there isn't a set requirement to determine where to go in the split but something that just has a probability of happening or not Ex: 50% chance of success or failure of a business.
End Nodes:
The end of a split, so something feeds into this but nothing comes out of it. Usually some result. Ex: Business is sucessful.
DECISION NODE: These are variable's included on an influence diagrams or decision tree, they are points where decisions would have to be made, they are usually depicted by square or rectangle. by Master E. Felix.
Related
(reposted from cs.stackexchange since I got no answers or comments)
I want to solve the problem of finding a shortest path on a directed weighted graph from a certain node to any of a specified set of destination nodes (preferably the closest one, but that's not that important). The standard (I believe) way to do this with the A* algorithm is to use a distance-to-closest-goal heuristic (which is admissable) and exit as soon as any of the goal nodes is reached.
However, in my scenario (which is game AI, if that matters) some (or all) of the goals might be unreachable; furthermore, the set of nodes reachable from such goals is typically quite small (or, at least, I want to optimize in that particular case). For the case of a single goal, bidirectional search sounds promising: the reverse search direction would quickly exhaust all reachable nodes and conclude that no path exists. These slides by Andrew Goldberg et al. describe the bidirectional A* algorithm with proper conditions on the heuristics, as well as stopping conditions.
My question is: is there a way to combine these two approaches, i.e. to perform bidirectional A* to find path to any of a specified set of goal nodes? I'm not sure what heuristic function to choose for the reverse search direction, what are the stopping conditions, etc. Googling for anything on this topic didn't get me anywhere either.
Trying to learn MCST using YouTube videos and papers like this one.
http://www0.cs.ucl.ac.uk/staff/D.Silver/web/Applications_files/grand-challenge.pdf
However I am not having much of a luck understanding the details beyond the high level theoretical explanations. Here are some quotes from the paper above and questions I have.
Selection Phase: MCTS iteratively selects the highest scoring child node of the current state. If the current state is the root node, where did these children come from in the first place? Wouldn't you have a tree with just a single root node to begin with? With just a single root node, do you get into Expansion and Simulation phase right away?
If MCTS selects the highest scoring child node in Selection phase, you never explore other children or possibly even a brand new child whilst going down the levels of the tree?
How does the Expansion phase happen for a node? In the diagram above, why did it not choose leaf node but decided to add a sibling to the leaf node?
During the Simulation phase, stochastic policy is used to select legal moves for both players until the game terminates. Is this stochastic policy a hard-coded behavior and you are basically rolling a dice in the simulation to choose one of the possible moves taking turns between each player until the end?
The way I understand this is you start at a single root node and by repeating the above phases you construct the tree to a certain depth. Then you choose the child with the best score at the second level as your next move. The size of the tree you are willing to construct is basically your hard AI responsiveness requirement right? Since while the tree is being constructed the game will stall and compute this tree.
Selection Phase: MCTS iteratively selects the highest scoring child node of the current state. If the current state is the root node, where did these children come from in the first place? Wouldn't you have a tree with just a single root node to begin with? With just a single root node, do you get into Expansion and Simulation phase right away?
The selection step is typically implemented not to actually choose among nodes which really exist in the tree (having been created through the Expansion step). It is typically ipmlemented to choose among all possible successor states of the game state matching your current node.
So, at the very beginning, when you have just a root node, you'll want your Selection step to still be able to select one out of all the possible successor game states (even if they don't have matching nodes in the tree yet). Typically you'll want a very high score (infinite, or some very large constant) for game states which have never been visited yet (which don't have nodes in the tree yet). This way, your Selection Step will always randomly select among any states that don't have a matching node yet, and only really use the exploration vs. exploitation trade-off in cases where all possible game states already have a matching node in the tree.
If MCTS selects the highest scoring child node in Selection phase, you never explore other children or possibly even a brand new child whilst going down the levels of the tree?
The ''score'' used by the Selection step should typically not just be the average of all outcomes of simulations going through that node. It should typically be a score consisting of two parts; an "exploration" part, which is high for nodes that have been visited relatively infrequently, and an "exploitation" part, which is high for nodes which appear to be good moves so far (where many simulations going through that node ended in a win for the player who's allowed to choose a move to make). This is described in Section 3.4 of the paper you linked. The W(s, a) / N(s, a) is the exploitation part (simply average score), and the B(s, a) is the exploration part.
How does the Expansion phase happen for a node? In the diagram above, why did it not choose leaf node but decided to add a sibling to the leaf node?
The Expansion step is typically implemented to simply add a node corresponding to the final game state selected by the Selection Step (following what I answered to your first question, the Selection Step will always end in selecting one game state that has never been selected before).
During the Simulation phase, stochastic policy is used to select legal moves for both players until the game terminates. Is this stochastic policy a hard-coded behavior and you are basically rolling a dice in the simulation to choose one of the possible moves taking turns between each player until the end?
The most straightforward (and probably most common) implementation is indeed to play completely at random. It is also possible to do this differently though. You could for example use heuristics to create a bias towards certain actions. Typically, completely random play is faster, allowing you to run more simulations in the same amount of processing time. However, it typically also means every individual simulation is less informative, meaning you actually need to run more simulations for MCTS to play well.
The way I understand this is you start at a single root node and by repeating the above phases you construct the tree to a certain depth. Then you choose the child with the best score at the second level as your next move. The size of the tree you are willing to construct is basically your hard AI responsiveness requirement right? Since while the tree is being constructed the game will stall and compute this tree.
MCTS does not uniformly explore all parts of the tree to the same depth. It has a tendency to explore parts which appear to be interesting (strong moves) deeper than parts which appear to be uninteresting (weak moves). So, typically you wouldn't really use a depth limit. Instead, you would use a time limit (for example, keep running iterations until you've spent 1 second, or 5 seconds, or 1 minute, or whatever amount of processing time you allow), or an iteration count limit (for example, allow it to run 10K or 50K or any number of simulations you like).
Basically, Monte Carlo is : try randomly many times(*) and then keep the move that led to the best outcome most of the times.
(*) : the number of times and the depth depends on the speed of the decision you want to acheive.
So the root node is always the current game state with immediate children being your possible moves.
If you can do 2 moves (yes/no, left/right,...) then you have 2 sub-nodes.
If you cannot do any moves (it may happen depending on the game) then you do not have any decision to make, then Montec Carlo is useless for this move.
If you have X possible moves (chess game) then each possible move is a direct child node.
Then, (in a 2 player game), evey level is alternating "your moves", "opponent moves" and so on.
How to traverse the tree should be random (uniform).
Your move 1 (random move of sub-level 1)
His move 4 (random move of sub-level 2)
Your move 3 (random move of sub-level 3) -> win yay
Pick a reference maximum depth and evaluate how many times you win or lose (or have a sot of evaluation function if the game is not finished after X depth).
You repeat the operation Y times (being quite large) and you select the immediate child node (aka: your move) that leads to you winning most of the times.
This is to evaluate which move you should do now. After this, the opponent moves and it is your turn again. So you have to re-create a tree with the root node being the new current situation and redo the Monte Carlo technique to guess what is your best possible move. And so on.
I know how to implement union find in general, but I was thinking of whether there would be a way to utilize the set structure in python to achieve the same result.
For example, we can union sets pretty easily. But I'm not sure how to determine if two elements are in the same set using just sets.
So, I am wondering if there is a data structure in python that would support such operation, other than the usual implementation?
You could always solve this problem by visualizing it as a tree and its nodes connecting to each other via the root, and then looking up the tree if you want to know if two nodes are connected. If the two nodes you are comparing has the same root (they are in the same tree), than they are connected.
To connect two nodes, just go to the root of each tree they are in, and make one root become the parent of the other.
This video will give you a great intuition about it:
https://www.youtube.com/watch?v=YIFWCpquoS8&list=PLUX6FBiUa2g4YWs6HkkCpXL6ru02i7y3Q&index=1
The connection between the tree nodes can be made via pointers in a language which supports it, but if your language dont (python), than you can create your own pointers by storing positions and links via an array.
The array would be such that its positions would represent your nodes, and the values inside it represents the connection of the specific node to its root. On the beginning, the position in the array is filled with the node number because the nodes has initially no parent, but as you connect nodes, the roots changes, and the array has to represent this. Actually, the value stored there is the identificator of the root.
But try visualizing the problem visually first instead of thinking of arrays and too much mathematical artificats. Visually dealing with it makes the solution sound banal, and can be a good guidance while writing code.
I say this because I have watched the video from Robert Sedgewick I just posted, with a graphical simulation of the solution, and implemented myself without paying too much attention to the code on his book. The intuition the video gave me is much more valuable than any mathematics.
It will help you to encapsulate the nodes into a class, with the following methods:
climbTreeFromNodeUpToRoot
setNewParentToThisNodeAndUpdateHeights
The first method, as the name says, takes you from a node and goes up the tree until finding the root of it, which is then returned.
If you compare two nodes with this method (actually, the roots returned by it), you know easily if they are connected by just comparing their roots.
Once you want to connected them, you go up the trees of both nodes, and ask one root to take the other one as its parent.
The trees can grow very big in height (sorry I dont use the official nomeclature, but this is the one that makes sense to me), so this simple approach will get very slow when you have to climb the tree at a later time.
To prevent trees from becoming to high, dont just set one root as the parent to another without criterium, but attach the smallest tree (in terms of height, not quantity of elements) to the highest one.
For this, you need to know the heights of each tree, and this information you can store on their respective root (via an extra array in your case, or an extra pointer from each node in other languages). This information should be updated everytime another tree connects to it.
It is not possible for a tree to know that she just got a new tree attached to it, so its important that every tree attaching to a second one informs the second as to update its height.
This information can be sent to the root of the second tree, and later used to judge (as writen before) which tree is the smallest. Remember, attaching a small tree to a big one instead of the opposite will save you incredible amounts of time.
Do you want something like this?
myset = ...
all(elt in myset for elt in (a,b))
Say we have a graph that is similar to a linked list (or a directed acyclic graph). An independent set consists of nodes that don't share edges with any other node in the set. If each node is weighted, how can we calculate the max possible value of the independent set of nodes? I understand we have to use Dynamic Programming so I have a slight clue but I'm hoping someone could explain how they would approach it. Thank you!
I believe that this problem is NP-hard for arbitrary directed acyclic graphs. The corresponding problem for undirected graphs is known to be NP-hard, and that problem can be converted into the directed version of the problem by directing all of the edges in a way that makes the resulting graph a DAG. Any independent set in the original graph will be an independent set in the directed graph and vice-versa, so any solution to the directed case will solve the undirected case.
Your question talks about solving this problem on a linked list. If you're solving the problem just for linked lists, there is a polynomial-time solution using dynamic programming. As a hint, if you choose one node in the linked list, you have to skip the next node, then should maximize what remains. If you don't choose the node, you just maximize the value of the rest of the list. Taking the better of these two options and evaluating this bottom-up will give you a really fast DP algorithm.
Hope this helps!
Basically my decision tree can't classify a value using the normal algorithm.
I get to a node, and there are two options (say, sunny and windy), but at this node my value is different (for example, rainy).
Are there any methods to deal with this, e.g. change the tree or just estimate based on other data?
I was thinking of assigning the most common value at that node but this is just a guess.
Have you considered fuzzy logic for the rich/poor continuum? As for things that can't be expressed as a continuum, I can't think of a way it can be done. Rainy weather, for example, is so fundamentally different from sunny and windy weather in how we experience and react to it, I'm not sure how you expect a computer (or whatever it is you're writing your decision tree for) to figure out what to do. (Aside from simply having an "I don't know what to do" output state, but I'm assuming you wanted something more meaningful than that.)
The whole point in decision trees is that the options are complete and (hopefully) mutual exclusive.
If it is not you'll get into trouble. Redefine poor and rich to cover everything. (all incomes, all states of mind...)
But honestly, interpret such weather examples as what they are: just examples for a concept, not the holy grail of meteorology.
The issue here is that you've learned a decision from different data as you are using to classify it. More specific, your decision tree knows only two values (i.e., sunny and windy) for the attribute Weather. But your data for classification also allows the value rainy.
Since your decision tree has no observation when the weather was rainy, this value turns useless. In other words, you have to eliminate this value from your classification.
The only solution is to do data cleaning before using the decision tree as classifier.
You have two options:
1. Remove all observations/instances with Weather="rainy" from your data set because you can't classify them. The disadvantage is that all instances with Weather="rainy" are not classified.
2. For all observations/instances with Weather="rainy", remove the value or rather set it to unknown/null. In case that your decision tree can handle null values, it can classify all of your data set. If not, you still have a problem. In that case you should go for option 3.
3. Relearn your decision tree with Weather={sunny, windy, rainy}
(4). In your case the following is not an option. Replace "rainy" with either "sunny" or "rainy. There are different heuristics for that.
You are talking about the "normal algorithm", which is a quite blurry statement. I assume you are using a strictly-binary rooted decision tree, where the each internal node makes a binary split of the data. Thus, the condition evaluation at each internal node outputs a Boolean variable, which splits the data into the left node (true) and right node (false). In your case, you can have a categorical variable weather with two possible values in the training data, which makes only two possible node: weather==sunny or weather==windy. Hence, the rainy samples will be always on the right node, as it is not sunny and not windy.
In the following picture, the rainy samples will be classified as not sunny, not windy.