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[SOLVED] Coms w4701: artificial intelligence homework 1 problem 1: hodl (15 points) consider the problem of investing in cryptocurrency.

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Consider the problem of investing in cryptocurrency. We have many trader agents who must
decide when and how much to invest at any given time. Their decisions are based on individual
risk levels, current market conditions, and inferences about company events (e.g., IPO, earnings
reports, acquisition, etc.).(a) Give a state space description of this problem. What information should individual states
contain? What are the valid actions that an agent can take in each state?(b) Classify this task environment according to the six properties discussed in class, and include a
one- or two-sentence justification for each. For some of these properties, your reasoning may
determine the correctness of your choice.This is a variation of the famous Monty Hall problem. Let’s say we play the following game prior to
our final exam. You are given 50 mystery boxes and you have to choose one of them. One mystery
box grants you a free A on the final (U = 20). Another one will force you to retake the entire class
in the fall (U = −100). All other boxes give you a single bonus point on the final exam (U = 1).(a) Compute the expected utility of selecting a mystery box at random.
(b) Suppose you’ve chosen a box. Before you open it, we choose and open five other boxes granting
the single exam bonus point. What is the expected utility of switching to another box after
this occurs?(c) Is it better to stick with your original choice or switch to another one? Compute the value of
information of seeing the five boxes granting the single bonus point.(d) We will now slightly change the game rules. After you select a box, we will open it for you.
You can then either claim the revealed prize, or choose to open another mystery box and keep
the new prize (forgoing the first one). What is the expected utility of this game procedure?In the state space graph below, S is the start state and G is the goal state. Costs are shown along
edges and heuristic values are shown adjacent to each node. All edges are bidirectional. Assume
that search algorithms expand states in alphabetical order when ties are present.(a) List the ordering of the states expanded as well as the solution (as a state sequence) returned
by each of DFS and BFS, assuming that they use the early goal test (insertion into the frontier)
and a reached table.(b) List the ordering of the states expanded as well as the solution (as a state sequence) returned
by each of DFS and BFS, assuming that they use the late goal test (removal from the frontier)
and a reached table.(c) List the ordering of the states expanded as well as the solution (as a state sequence) returned
by each of UCS and A*, assuming that they use the late goal test (removal from the frontier)
and a reached table.(d) Change exactly one heuristic value so that A* returns a suboptimal solution. Give the value
of the heuristic that you changed and explain why it becomes inadmissible. List the ordering
of the states expanded as well as the solution (as a state sequence) returned by A*.
(e) A heuristic value has not been assigned to S. Give the upper bound on the value of h(S) so
that i) h is admissible (but possibly inconsistent), and ii) h is consistent (and admissible). Do
these values change how A* runs on this graph?In this problem you will implement and compare the performance of various search algorithms on
solving word ladder puzzles. Given two English words, the goal is to transform the first word into
the second word by changing one letter at a time. The catch is that each new word in the process
must also be an English (dictionary) word. For example, given a start word “fat” and a goal word
“cop”, a solution would be the word sequence “fat”, “cat”, “cot”, “cop”.We have provided a Python skeleton file for you to complete this problem. Note that the function
headers are type annotated to clearly indicate argument and return value types. You can optionally
use static type checkers like mypy to verify your code as you work on it.In our implementation, states will be equated with words (strings). Successor states are words
that differ from the current state by one letter. Using this idea, we provide a successors function
that returns a list of “actions” and successor states given a state. The action is simply the index
of the changed letter. This function uses the pyenchant library to perform dictionary checking.
Finally, the cost of each action can be treated uniformly (e.g., 1).To perform search, we represent a search tree node as a Python dictionary containing three
components: state, parent, and cumulative cost. For example, the root node may be defined as
{‘state’:start, ‘parent’:None, ‘cost’:0}. The frontier can be implemented as a list, while
the implementation of the reached set will differ in each part below.4.1: Depth-Limited Depth-First Search (15 points)
We will start by implementing depth-limited DFS. First, write the expand function, which takes
in a node and returns a list of nodes, one per successor state to the state in the given node. You
should use the provided successors function here.The depth limited dfs function takes in a start state, a goal state, and a depth limit. Nodes at
depth will be considered to be leaves. In this procedure, you can implement the reached set as a
list of states (as opposed to the frontier, which is a list of nodes). Since this is DFS, you should
perform the early goal test for time efficiency by checking for the goal upon a node’s insertion
into, rather than removal from, the frontier.You should also continually update two local variables: nodes expanded and frontier size. The
first is an integer that is incremented every time that a node is expanded. The second is a list
that contains the size of the frontier at the beginning of each search iteration, updated before a
node is popped. Once the goal node has been found, the procedure returns that node, along with
nodes expanded and frontier size. Alternatively, if the frontier becomes empty, you should return None along with the latter two values.If you would like to test your solution before moving on, you can run the code in the main function
and also add your own (see 4.4). Investigate the return values of depth limited dfs. You can
also pass in the goal node into the provided sequence function to retrieve the sequence of words
from start to goal. Think about how changing the depth limit affects the solution.4.2: Iterative Deepening (5 points)
Since we have a DFS implementation that can account for depth limits, a natural extension
would be an iterative deepening wrapper around it. Recall that this algorithm repeatedly calls
depth limited dfs with a larger depth parameter each time, starting from 0. To prevent this
from potentially searching forever, iterative deepening will stop and return no solution if one is
not found by max depth.As with depth limited dfs, iterative deepening should also update and return quantities indicating nodes expanded and frontier size. The former should be the total number of expanded
nodes over all search iterations. The latter should be a single flat list containing the frontier sizes
over all iterations.4.3: A* Search (10 points)
Your iterative deepening implementation, like all uninformed search approaches, uses no information about the goal word. But we should use this knowledge to our advantage. It makes sense
to “favor” successor words that look more like the goal. We can do this using A* search, and a
suitable heuristic would be the Hamming distance between a current word and the goal, or the
number of indices where the corresponding letters are different.A* search will mostly follow the implementation of depth-first search with a few key changes. To
simulate priority queue behavior of the frontier, the heapq module, and in particular the heappush
and heappop functions, can be used to efficiently treat regular Python lists using a priority function.In order to “sort” the nodes in the frontier, you can first place each node within a data structure
like a tuple, so that the first element captures the f-cost and the last element is the node itself. For
example, (1, node1) would be ordered before (2, node2). If the first elements of two tuples are
equal, they are then compared according to their second elements, followed by their third and so
on. For this problem, you should break ties (and thus include an additional element between the
f-cost and node) by alphabetical order of the states.Another difference is that the reached structure is now a dictionary with each key as a state and
the value as the cheapest node reaching that state. Even if a state already exists in reached, a
cheaper path to it may be discovered later, and so a cost comparison should be done when determining whether a child node should be added to the frontier. Finally, remember to conduct the
goal test only when popping a node from the frontier—the “early” version should no longer be used.Implement the astar search function following the specifications above. The return values are
the same as those of your other implementations.4.4: Analysis (20 points)
You should now be able to run each search algorithm implementation and solve different word
ladder puzzle instances. We provide a sequence function that takes the goal node output from
best-first search and returns the entire sequence of words from start to goal. Consider the puzzles
below:
• Start: “fat”; goal: “cop”
• Start: “cold”; goal: “warm”
• Start: “small”; goal: “large”(a) Let’s investigate A* first as the results will help us better understand the other two algorithms.
Run A* on each of the puzzles above and report the solution length as well as number of nodes
expanded. Also generate three line plots (e.g., using matplotlib), one per puzzle, showing the
size of the frontier per iteration.(b) Now let’s look at iterative deepening search. From what you saw with A*, what is the range
of maximum allowable depth values that would yield solutions for the first two puzzles but no
solution for the third? Explain whether these solutions would be identical to those of A*.(c) Pick two different integer values in the range you found above and perform two runs of IDS on
the three puzzles with these max depths (we recommend you choose smaller values for faster
runs). Report the solution lengths and number of nodes expanded for each, and compare these
values with those of A* for each puzzle.(d) Do the same experiment using the max depth values you used above for depth-limited DFS.
Report the solution lengths and number of nodes expanded. Compare these values with those
of IDS for each puzzle. Why might the results be different even though the maximum depths
are all identical?Submission
You should have one PDF document containing your solutions for problems 1-3, as well as the
information and plots for 4.4. You should also have the completed code file implementing the
search algorithms for the word ladder problem in a .py file. Submit the document and code file to
the respective assignment bins on Gradescope. For full credit, you must tag your pages for
each given problem on the former.