AI midterm prep. DSStore .gitignore

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+.DS_Store

OMSCS/Courses/AI/Midterm_Prep/Midterm Prep.md

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+---
+tags:
+  - OMSCS
+  - AI
+---
+# Midterm Prep
+
+## Problem 1: Game Playing
+- Player 1 is "X"
+- Player 2 is "O"
+- Both players have a single piece, and this piece can move to adjacent squares, but cannot move diagonally.
+- Players cannot visit an occupied space, nor a previously visited space
+- The winner is the last one to move.
+
+![[Pasted image 20240225123322.png]]
+
+> Use the Minimax algorithm to determine the value of this first move for Player 1. Does Player 1 have a guaranteed win?
+
+- Transition from red layers to blue layers are "min" layers
+- Transition from blue layers to red layers are "min" layers
+- We can see the whole game tree, so we'll just propagate +1 for win and -1 from loss from the bottom of the tree to the top
+
+|  |  |  |  |  |  |  |  |  |
+| ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- |
+|  |  | +1 min |  |  |  |  |  |  |
+|  |  | +1 max |  | +1 max |  |  |  |  |
+|  | -1 min | +1 |  | +1 | +1 min |  |  |  |
+| -1 | +1 | +1 max |  | +1 | +1 max |  | +1 max |  |
+| -1 | +1 | +1 | -1 | +1 | +1 | +1 | +1 | -1 |
+| -1 | +1 win | +1 | -1 | +1 | +1 win | +1 | +1 | -1 |
+| -1 loss |  | +1 | -1 loss | +1 |  | +1 | +1 | -1 |
+|  |  | +1 |  | +1 |  | +1 win | +1 win | -1 |
+|  |  | +1 |  | +1 |  |  |  | -1 loss |
+|  |  | +1 win |  | +1 win |  |  |  |  |
+
+- X has a guaranteed win. No matter what O chooses as its first move, X can ensure a win.
+
+> Apply alpha-beta pruning to the tree (traversing branches from left to right). How many leaves are pruned?
+
+|  |  |  |  |  |  |  |  |  |
+| ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- | ---- |
+|  |  | +1 min |  |  |  |  |  |  |
+|  |  | +1 max |  | +1 max |  |  |  |  |
+|  | -1 min | +1 |  | +1 | ~~+1 min~~ |  |  |  |
+| -1 | ~~+1~~ | +1 max |  | +1 | ~~+1 max~~ |  | ~~+1 max~~ |  |
+| -1 | ~~+1~~ | +1 | ~~-1~~ | +1 | ~~+1~~ | ~~+1~~ | ~~+1~~ | ~~-1~~ |
+| -1 | ~~+1 win~~ | +1 | ~~-1~~ | +1 | ~~+1 win~~ | ~~+1~~ | ~~+1~~ | ~~-1~~ |
+| -1 loss |  | +1 | ~~-1 loss~~ | +1 |  | ~~+1~~ | ~~+1~~ | ~~-1~~ |
+|  |  | +1 |  | +1 |  | ~~+1 win~~ | ~~+1 win~~ | ~~-1~~ |
+|  |  | +1 |  | +1 |  |  |  | ~~-1 loss~~ |
+|  |  | +1 win |  | +1 win |  |  |  |  |
+
+- 25 nodes are never explored.
+- Those 25 nodes contain 6 leaves.
+
+> Search Depth of 6. Heuristic function of (num my moves - num their moves). Killer moves (win: $+\infty$, loss: $-\infty$). Alpha-Beta pruning.
+> 
+> How many leaves are pruned?
+
+![[Pasted image 20240225141302.png]]
+
+|           |                               |                               |          |                         |                         |     |          |     |
+| --------- | ----------------------------- | ----------------------------- | -------- | ----------------------- | ----------------------- | --- | -------- | --- |
+|           |                               | $[-\infty, +\infty]$ (min, B) |          |                         |                         |     |          |     |
+|           |                               | $[0, +\infty]$ (max, A)       |          | $[0, +\infty]$ (max, A) |                         |     |          |     |
+|           | $[-\infty, -\infty]$ (min, B) | $[0, +\infty]$                |          | 0                       | $[0, +\infty]$ (min, B) |     |          |     |
+| $-\infty$ | (pruned)                      | $[0, +\infty]$ (max, A)       |          | 0                       | $[+\infty, +\infty]$ (max, A)                |     | $[0, +\infty]$ (max, A) |     |
+| $-\infty$ | (pruned)                      | 0                             | (pruned) | 0                       | $+\infty$               | (pruned)    | 0         |     |
+| $-\infty$ | (pruned)                      | 0                             | (pruned) | 0                       | $+\infty$               | (pruned)    | 0         |     |
+
+- We can prune the 2nd column branch pretty obviously. It's a min node, the left branch found $-\infty$, nothing in the 2nd column branch will ever be better for the opponent.
+- We can prune the 4th column branch a little less obviously.
+	- We know from looking at the first 2 columns that $\alpha$ is "at least" $-\infty$. 
+	- By exploring the branch in the 3rd column, we find a 0, which is higher than $-\infty$.
+	- This is enough information to know that the agent should pick the column 3&4 branch over the column 1&2 branch, given the opportunity.
+	- With that information in hand, the search doesn't need to explore the branch further, pruning column 4.
+- Nothing else really interesting happens until column 6. In column 6, we find a $+\infty$ right before a (max, A) decision. This allows the search to prune the 7th column.
+- I don't freaking know man.
+
+## Problem 2: Search
+### Part 1: Search Knowledge
+> You are given an admissible heuristic $h_1(n)$ for A∗, and that $h_2(n) = 2h_1(n)$.
+
+- $h_2(n)$ is not always admissible.
+- A* using $h_2(n)$ would not be guaranteed to find the optimal solution.
+- An A* tree search solution with $h_2(n)$ has a guaranteed cost $\le 2C^*$
+
+> You are given two different admissible A∗ heuristics, $a(n)$ and $b(n)$.
+
+- You can combine them using $max(a(n), b(n))$. Doing so guarantees admissibility, and results in the fewest node expansions.
+	- $min$ - guarantees admissibility, but is less efficient
+	- $avg$ - guarantees admissibility, but is less efficient
+	- $ln(a(n), b(n))$ - `TypeError: ln() takes 1 positional argument but 2 were given`
+
+> Let $h^∗(n)$ be the true cost function, what are the relationships between $h^∗(n)$ to $a(n)$ and $h^∗(n)$ to $b(n)$?
+
+- $a(n) \le h^*(n)$
+- $b(n) \le h^*(n)$
+- $max(a(n), b(n)) \le h^*(n)$
+
+### Part 2: Robotic Assembly
+![[Pasted image 20240225160754.png]]
+
+> As a lead engineer at Ploetz-Tharner, you are selected to design the new robotic assembly program. A 5-part sequencing will be performed by a robotics system with 3DOF (3 degrees of freedom). The robot can move in the x,y,z space, and the parts are arranged in a grid on a conveyor. The robot will optimally move the parts to the target positions.
+> 
+> \[...\]
+
+- The robot can move in one of 8 directions at a time. This means the branching factor is 8.
+- The number of states at depth $k$ is bounded by $8^k$ as an upper limit.
+
+> The parts must be moved one at a time toward their prescribed goal position aligning the parts for assembly. Only one part can occupy a cell, and your robot can sense what cell a part is in. Your robot can perform the following actions:
+> 
+> 1. Ready (positions the robot at i0) no cost (Only available for step 0, result returns i0)
+> 2. Move (direction, # of cells) (Result returns current cell)  
+	(a) Moving N, S, W, E has a cost of 1 per cell moved.  
+	(b) Diagonal movement (i.e. NW, NE, SW, SE) has a cost of 1 per cell moved.
+
+- The state space of the parts is $20 \space nCr \space 5$, which is $20*19*18*17*16=1860480$
+- The state space of the robot is $20$
+- The full state space is $1860480*20=37209600$
+- Doesn't matter where the initial positions of the parts are, the answer is the same.
+- I think the sample exam is missing a 0.
+- Admissible heuristics (they did not do a good job of describing this problem)
+	- Manhattan is not an admissible heuristic. A single diagonal move costs 1. Manhattan distance for a diagonal move would cost 2.
+	- Euclidean distance is not an admissible heuristic. A single diagonal move costs 1. ED for a diagonal move is $\sqrt{2}$ which is $\gt 1$.
+	- $max(VertDist, HorizDist, DiagDist)$ would be admissible according to the answer key, but they did not describe what $DiagDist$ means.
+	- Number of misplaced parts would be admissible, just not very efficient. I don't know why it's not marked as such in the answer key. This whole thing needs a fucking redo too.
+	- $|dist(part, robot)| + |dist(part, goal)|$ is not admissible. In the event that all of the parts are on their goal and the robot is far away from some of the parts, the heuristic would return $>0$, when $h^*(n)=0$
+
+> What would be the most effective strategy to transform the search to solve for optimal sequence and path concurrently? (multiple-option correct MCQ)
+
+The study guide says we can
+- Add a final goal to the end of the search
+- Set all of the g-values to infinity
+- Add linking nodes between the end-goals and initial locations
+
+Interestingly the guide does _not_ include using a PQ that keeps paths from init to end goals.
+
+The study guide also says ID-A* is the best algorithm of the bunch, even better than regular A*.
+
+## Problem 3: Optimization Algorithms
+### Part 1: Gradient Descent
+- $x_{k+1}=x_k - \eta_{k} f'(x_k)$
+- $f(x)=2x^2 - 4x + 5$
+- $f'(x) = 4x - 4$
+- $x_k = 3$
+- $\eta_k=0.25$
+- $\forall k = 1, 2, 3, ...$
+- $f(3)$
+	- $=2(3^2)-(4)(3)+5$
+	- $=2(9)-12+5$
+	- $=18-12+5$
+	- $=6+5$
+	- $=11$
+- $x_{k+1}$
+	- $= 3 - (0.25)(4(3)-4)$
+	- $= 3 - (0.25)(12 - 4)$
+	- $= 3 - (0.25)(8)$
+	- $= 3 - 2$ 
+	- $= 1$
+- $x_{k+2}$
+	- $=1-(0.25)(4(1)-4)$
+	- $=1-(0.25)(0)$
+	- $=1-0$
+	- $=1$
+
+- Assuming an arbitrary function which does not go to negative infinity
+	- You can always find a local minimum through one round of gradient descent.
+	- It's not guaranteed that the local minimum you find is the global minimum.
+
+### Part 2: Line Search
+- $f(x)=2x^2-4x+5$
+- $x_1=3$
+- $f(x_k-t'\triangledown f(x_k))=min_{t\in R}f(x_k-t\triangledown f(x_k))$
+
+> What is the optimal step size for $k=2$?
+
+The question fundamentally here seems to be "What value of $\eta_k$ causes GD to immediately arrive at the global minimum in one step, starting at $x_k$?"
+
+$min_{t\in R}f(x_k-t\triangledown f(x_k))$, essentially we're trying to find a value $x_{k+1}$ which minimizes the function $f$.
+
+From https://personal.math.ubc.ca/~CLP/CLP1/clp_1_dc/ssec_find_maxmin.html
+
+> If $f(x)$ has a global maximum or global minimum, for $a≤x≤b$, at $x=c$, then there are 3 possibilities. Either $f'(c)=0$, $f'(c)$ does not exist, $c=a$, or $c=b$
+
+- $min_{x \in R} \space f(x)=min_{x \in R} \space 2x^2-4x+5$
+- $= x : \space 4x-4=0$
+- $min_{x \in R} \space f(x)=1$
+
+- $x_{k+1}=x_k-(\eta_k)(4)(x_k-1)$
+- $1=3-(\eta_k)(4)(3-1)$
+- $-2=-(\eta_k)(4)(2)$
+- $2=\eta_k(8)$
+- $\eta_k=0.25$
+
+### Part 3: Acceleration Scheme