CS 6601 Artificial Intelligence Fall Semester 2022 Midterm Examination Paper
Duration of Exam: 10 Oct 2022, 8:00 AM (EDT) – 17 Oct 2022, 8:00 AM (EDT)
Weight: 15% No. pages: 35 No. Questions: 6 Total marks: 103
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• Points breakdown is provided below.
Question No.
Problem 1: Game Playing [12 points]
For this question we will look at a variant of Isolation, which takes place on a 3×3 board. Player 1 controls piece ‘X’, while player 2 controls piece ‘O’. Squares which are no longer visited are grayed out. Both players have a single piece, and this piece can move to adjacent squares, but cannot move diagonally. For example, on player 1’s turn, with the board shown below in Figure 1, they have 4 valid moves:
Figure 1: Board Diagram
On the first turn, player 1 can place their piece in multiple locations. For this question, we will only look at the subtree where player 1 places their piece in the middle of the board. Because the rules for this variant of Isolation exhibit reflectional and rotational symmetry, we can greatly simplify the tree by removing duplicates:
Figure 2: The Game Tree
Minimax & Alpha-Beta Pruning:
Take a couple minutes to study this game tree in Figure 2 and get a feel for how it works. Use Figure 2 to answer the following two questions:
(a) (2 points) Use the Minimax algorithm to determine the value of this first move for Player 1. Does Player 1 have a guaranteed win?
(b) (4 points) Apply alpha-beta pruning to the tree (traversing branches from left to right). How many leaves are pruned?
Type Your Answer Here:
Evaluation Function:
Rather than searching the full tree, let’s limit our search to a fixed depth. In order to do this, we need a function to evaluate a non-terminal board state. Let’s use a simple function. Let A be the number of empty squares adjacent to player 1, and let B be the number of empty squares adjacent to player 2. Our evaluation function will simply be: A – B
For example, in the board shown below in Figure 3, player 1 has 2 adjacent empty squares, and player 2 has 1 empty adjacent square, so the heuristic function will evaluate to 2 – 1 = 1. Player 1 seeks to maximize this function, while player 2 seeks to minimize this function.
Figure 3: Example Board
We will also add one exception to this function – if it is player 1’s turn and player 1 has 0 adjacent squares, the function will evaluate to negative infinity, and if it is player 2’s turn and player 2 has 0 adjacent squares, it will evaluate to positive infinity. This ensures that win / loss conditions outweigh our adjacency heuristic.
Use this heuristic function to evaluate the board to a depth of 6, as shown below in Figure 4:
(c) (2 points) Use the minimax algorithm to evaluate the truncated tree in Figure 4. Apply alpha-beta
pruning to the tree (traversing branches from left to right). How many leaves are pruned? Type Your Answer Here:
Figure 4: Game Tree to a Depth of 6
For the next questions, you may use the trees shown in Figures 5, 6, 7, 8, 9 for computations.
Let’s say player 1 and player 2 use different algorithms to decide their moves. Player 1 uses the depth- limited (to depth 6) minimax search to pick their moves, while player 2 has the luxury of an infinite depth search (the search from the first part of this problem). Player 2 also knows that player 1 will only search the board to a depth of 6, and uses this knowledge in their minimax algorithm.
(d) (2 points) Given this first move, and assuming that both players strictly follow their algorithms, does player 2 win the game?
(e) (4 points) Apply alpha-beta pruning to player 2’s searches (traversing branches from left to right). How many leaves are pruned? (Hint: you will have to repeat player 1’s depth-limited searches as they proceed deeper into the game)
Type Your Answer Here:
Figure 5: Game Tree
Figure 6: Game Tree
Figure 7: Game Tree
Figure 8: Game Tree
Figure 9: Game Tree
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Problem 2: Search [20 points]
Part 1: Search Knowledge
You are given an admissible heuristic h1(n) for A∗, and that h2(n) = 2h1(n). Please answer the following questions:
(a) (1 point) Would h2(n) be always admissible? Yes No
(b) (2 points) A∗ graph-search with h2(n) would provide a guaranteed optimal solution. True False
(c) (2 points) An A∗ tree-search solution with h2(n) has a guaranteed cost ≤ 2(true cost of A∗ tree- search with h1(n))?
True False
Now, you are given two different admissible A∗ heuristics, a(n) and b(n). Please answer the following questions:
(d) (2 points) Which one of the functions given below will combine the two heuristics into a single heuristic, which guarantees admissibility and results in A* expanding the fewest number of nodes?
1. F1 = min(a(n), b(n)) 2. F2 = max(a(n), b(n)) 3. F3 = avg(a(n), b(n)) 4. F4 = ln(a(n), b(n))
F1 F2 F3 F4
(e) (3 points) Let h∗(n) be the true cost function, what are the relationships between h∗(n) to a(n) and h∗(n) to b(n)? How can you use these relationships to arrive at a “Conclusion”, which is a relationship between a(n), b(n), and h∗(n) that guarantees admissibility and expansion of fewest nodes for your function above? E.g.(=, ≤, ≥, a + b)
Relationship between a(n) and h∗(n): Relationship between b(n) and h∗(n): Conclusion:
Part 2: Robotic Assembly
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.
Figure 10: Robot with 3DOF
The following (Figure 11) is a relaxed grid representation that shows the parts on the conveyor and the goal location. Assume that you have programmed the robot to lift, lower, grasp and release the parts. This allows you to write a program that can ignore the robot movement which will discretize the search space and avoid solving for continuous movement. Although you are concerned over producing efficient algorithms, and must optimize your steps, there must be a way to limit the complexity. You’ve decided to limit the ability of the robot to move according to a compass directions of N, NE, E, SE, S, SW, W, and NW. Based on that, answer the following question:
Figure 11: Grid of the Conveyor (f) (2 points) What is the branching factor for this space?
Type Your Answer Here:
That seems like a challenge, but you consider what you have learned in and decide you should be able to limit this to the expected branching factor you should consider.
(g) (2 points) What number of states would you expect to consider at depth k, given the rules below? Type Your Answer Here:
The rules of the assembly are as follows:
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
(a) 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.
Answer the following questions given the problem formulation:
(h) (1 point) For memory space optimality which type of search would you use?
Tree Search Graph Search
(i) (1 point) What is the state space assuming each part originates in its cell as shown in the grid above?
Type Your Answer Here:
(j) (1 point) What is the state space assuming each part originates randomly in the grid?
Type Your Answer Here:
(k) (1 point) What are the admissible heuristics for this problem? (multiple-option correct MCQ)
Manhattan Distance
((x2 − x2)2 + (y2 − y2)2)
Max(steps horizontally, vertically or diagonally)
Number of misplaced parts
|Part distance to current robot discrete position|+|Part distance to goal|
(l) (1 point) What would be the most effective strategy to transform the search to solve for optimal sequence and path concurrently? (multiple-option correct MCQ)
Use a priority queue that kept paths and costs from init to end-goals 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
(m) (1 point) Enter your choice for the best informed search algorithm to use for this problem that will return an optimal solution. It must minimize the cost of your movements individually and determine the optimal ordering of the parts moved. [Note: ID refers to Iterative-Deepening]
A∗ ID-DFS Backtracking DFS ID-A∗ Multi-goal A∗
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Problem 3: Optimization Algorithms [20 points]
We have introduced hill climbing in the lectures. Note that hill climbing is an analogy for optimizing a general function, that is, to find its minima and maxima. Let’s investigate a concrete algorithm, and its useful variants.
First we introduce our general problem formulation. For x ∈ X, where X can be some arbitrary set, define f (x) : X → R a function takes elements from X and produces a real number so that f (x) ∈ R. Our goal is to find a minimum solution f(x∗) over the entire set X. Assume the derivative of f at any point x exists and denoted f′(x). Recall that we learned hill climbing is essentially performing the following Gradient Descent iterations: xk+1 = xk − ηk f ′(xk), where xk is the current estimate of the optimal solu- tion, and xk+1 is the next estimate. ηk is the step size, which is a real number. Notice that in general, if X is n-dimensional, i.e., X ⊆ Rn, then we define the gradient of the function f at x to be:
∇f(x)=[∂f , ∂f ,…, ∂f ]T. ∂x1 ∂x2 ∂xn
As iterations progress, with proper step size, f(xk) would decrease. Hence the name Gradient Descent.
Part 1: Gradient Descent (GD)
Now generally attributed its invention to Cauchy and Hadamard in the 1800s, gradient descent has been one of the most influential and useful methods in modern mathematical computation. In the era of machine learning, gradient descent has been one of the driving forces behind computational models, including but not restricted to deep neural networks, support vector machines, and maximum likelihood estimation. In this subsection, we first look at two examples of gradient descent, followed by some conceptual questions.
(a) (1 point) Let f(x) = 2×2 − 4x + 5, f : R → R. Suppose xk = 3, and we set constant step size ηk = 14,∀k = 1,2,3,… What is the value of xk+1?
(b) (1 point) True or False: for an arbitrary function that does not go to negative infinity, you can always find a global optimal solution by applying gradient descent one time with proper step sizes.
True False
(c) (1 point) True or False: for an arbitrary function that does not go to negative infinity, you can always find a local optimal solution by applying gradient descent with proper step sizes (without restart).
True False
It is known that if Gradient Descent converges, it would converge in sub-linear speed for arbitrary func-
tions, that is: f (xk) − f (x∗) ≤ Ck , ∀k = 1, 2, 3, … where C is a problem dependent constant. It takes more
effort to improve precision in later iterations (notice that the ratio k/(k + 1) approaches 1 as k grows). We wish to improve this dreadful result, which follows two variants of Gradient Descent.
Part 2: Line Search
Let X be a subset of Rn. f : X → R is a real-valued function. Consider the GD iteration update rule defined as: xk+1 = xk − ηk∇ f (xk). Now we wish to know: what is the optimal step size for this particular iteration? The answer can be formulated as another optimization problem: find t′ ∈ R such that
f(xk − t′∇f(xk)) = min f(xk − t∇f(xk)) t∈R
The optimal step size should decrease the function value the furthest along the direction f′(xk). Hence this method is termed line search.
Let f(x) = 2×2 − 4x + 5, f : R → R. Suppose that x1 = 3. Answer the following questions.
(d) (2 points) What is the value for the optimal step size for k = 2? Type Your Answer Here:
(e) (2 points) Using the step size you found above, what is value of x2? Type Your Answer Here:
Part 3: Acceleration Scheme
First proposed in the 60s by Polyak, the Heavy Ball method is inspired by physics. Imagine you are rolling a ball down a hill. The ball would accelerate due to gravity, so that towards the bottom of the hill, velocity would gradually increase. Reflected in the algorithm, the current estimate of the solution would get to the optimal point faster in later iterations. Now consider the following update rule for minimizing a real-valued function f defined over subsets of Rn:
xk+1 = xk − αk∇f(xk) + βk(xk − xk−1)
where αk , βk are real numbers, usually αk , βk ∈ (0, 1). The additional parameter takes inspiration from
physics. If we image our current iterate xk as an object with mass, then adding this additional term
causes acceleration toward the bottom of the hill.
Now, consider the following problem in R2 and answer the following questions.
min f (x) = xT Ax − bT x x∈R2
where the matrix A = [a1 a2] ∈ R2x2, having the first column a1 = [1 3]T , and the second column a2 =[2 2]T.Letb=[3 5]T.Supposeαk =1/4,βk =1/2,andx1 =[4 4]T,x0 =[0 1]T.
(f) (1 point) Compute f (x1) Type Your Answer Here:
(g) (2 points) What is the value of x2? Say x2 = [a b]T , then fill up the following blanks: a:
(h) (2 points) Compute f (x2)
Type Your Answer Here:
(i) (3 points) Let f (x) = 2×2 − 4x + 5, f : R → R. Suppose that x1 = 3, xk−1 = 4 and we set constant step size αk = 1/4,∀k = 1,2,3,… and βk = 1/3. What is the value of xk+1?
Type Your Answer Here:
In his seminal work, Nestrov proposed the Accelerated Gradient Descent (AGD) method, which uses the following update:
yk = xk + βk(xk − xk−1)
xk+1 =yk −αk∇f(yk)
In general, both Polayk’s Heavy Ball method and Nestrov’s Accelerated Gradient Descent method will be faster than vanilla gradient descent, i.e., they require fewer iterations to reach the same level of pre- cision to the optimal solution. Such results would be reflected on convergence speed, often pushing the convergence rate to linear, that is, one requires as many iterations to reach the next digit precision as in pushing previous digit precisions. For example, if we need 3 iterations to drive f(xk − fx∗) ≤ 10−1 to 10−2, then we only need roughly three iterations to push from 10−6 to 10−7 when using an optimization method with linear convergence speed. This is a significant improvement from a sub-linear rate (hence the name “sub”-linear).
Using the Nestrov’s AGD method defined above, solve the following question. Consider the same problem in R2
min f (x) = xT Ax − bT x x∈R2
where the matrix A = [a1 a2] ∈ R2x2, having the first column a1 = [1 3]T , and the second column a2 =[2 2]T.Letb=[3 5]T.Supposeαk =1/5,βk =1/4,andxk =[2 3]T,xk−1 =[1 0]T.
(j) (1 point) Compute f (xk) Type Your Answer Here:
(k) (2 points) What is the value of xk+1? Say xk+1 = [a b]T , then fill up the following blanks: a:
(l) (2 points) Compute f (xk+1)
Type Your Answer Here:
Hint: To compute the gradient of xT Ax − bT x, notice that this is a quadratic function plus a linear function. Results follow similar differentiation rules in R, for example, ∇(x2 + tx) = 2x + t.
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Problem 4: Constraint Satisfaction Problem [10 points]
The Ploetzobots and Starnticons are two factions in the world of the Transformers who fight for power of Sourishetron, their home planet. The Ploetzobots are led by Opthomas Prime and the Starnticons are led by Thadotron.
Figure 12: Ploetzobots vs. Starnticons
In one of the final battles, Opthomas Prime and Thadotron have engaged in a final battle, whereas Opthomas Prime’s team was in charge of defeating giant Trypticon. His team decided to use the Ploet- zobots spaceship, the Ark, to fight Trypticon.
Figure 13: Trypticon Figure 14: The Ark
However, in order to operate the Ark, his team needs to figure out which part of the Ark they should operate. There are seven parts of the ship to operate:
1. The cannons
2. The laser guns
3. To fix the broken parts of the ship
4. The shield
5. The energon battery cells, which ensure power to all parts of the ship 6. The navigation, essentially being the pilot of the ship
7. The thrusters of the ship
There are a few constraints of the ship to consider.
1. Whoever works in the cannons and in the laser guns need to be great shooters
2. Whoever works in the navigation needs to have very good reflexes in order to avoid all of the attacks
3. Whoever is in charge of fixing the ship needs to know how to properly operate the Teletraan II, which is the computer which will help fix different parts of the ship.
4. The following pairing constraints need to be addressed for easy communication, since they sit next to each other.
(a) The bots who take care of the cannons and the laser guns need to be good buddies
(b) The bots who take care of the shields and the energon battery cells need to be good buddies (c) The bots who take care of the navigation and the thrusters need to be good buddies
Additionally, the Ploetzobots needed to attend the bot academy before becoming soldiers. For this, they needed to study well or improve their skills well. Next we will cover the Ploetzobots during their time in their academy:
1. Bumblebee has great reflexes, but it was never good at shooting nor was it good at understanding the technical side of their central computer, the Teletraan II. It was great friends with
Jazz, Buckhead, and Arcee.
2. Wheeljack didn’t have great reflexes, but he was a great shooter and pretty much invented the Teletraan II. But it didn’t really interact with many of his peers, so it was pretty much a great partner with Ratchet and Arcee.
3. Ratchet was the medic, so it spent a lot of time understanding how to repair its peers using the Teletraan II. Since it was the medic, it took an oath to not engage in battle, so it refuses to use any of the shooters (i.e. cannons or laser guns) and it didn’t have very good reflexes. It was really nice friends with Arcee, Wheeljack, and Jazz, but found Bumblebee’s immaturity hard to deal with, and Blurr spoke too fast for him to understand.
4. Blurr is a speedster, so he has amazing reflexes. Because of that, it was also a great shooter, but it didn’t really enjoy studying the Teletraan II, so it pretty much doesn’t know much about it. Because it speaks very quickly, not many people could understand him, so pretty much only Arcee learned to work well with him.
5. Arcee is friendly bot. It worked well with everyone; literally anyone and everyone. Above that, it was top of its class in shooting, although it reflexes weren’t great, and its knowledge in the Teletraan II wasn’t great either.
6. Jazz was a slick fellow. It was great at shooting, with very good reflexes, but it wasn’t great at operating the Teletraan II. It was a great friend of Bumblebee, Arcee, and Ratchet.
7. Finally, Buckhead was a very big and friendly fellow, but rather timid. It was great friends with Bumblebee and Arcee. Although he didn’t have the capabilities of having good reflexes or knowl- edge of the Teletraan II, he was an amazing fighter and great shooter.
The information about friendships and skills can also be found in the tables shown in Figure 15.
Figure 15: Friendship and Skill Tables
(a) (0.5 points) Given the listed constraints and the characteristics of each of the Ploetzobots, is it possible to find a solution that satisfies all of the constraints?
(b) (3.5 points) If yes, what are the possible bots that can seat in each of the listed roles, listed in alphabetical order and separated by a comma and a space (e.g. Arcee, Bumblebee, Jazz)? If not, leave the fields blank
Cannon: Laser Gun: Fixing: Shield: Energon:
程序代写 CS代考 加QQ: 749389476
Navigation:
Thrusters:
After a lot of fighting, the shields were heavily damaged. Due to that, greater communication was needed between the fixer and the shield user. In other words, they now need to be able to work well together. All of the previous constraints are to remain.
(c) (0.5 points) Given the listed constraints and the characteristics of each of the Ploetzobots, is it possible to find a solution that satisfies all of the constraints?
(d) (3.5 points) If yes, what are the possible bots that can seat in each of the listed roles, listed in alphabetical order and separated by a comma and a space (e.g. Arcee, Bumblebee, Jazz)? If not, leave the fields blank
Cannon: Laser Gun: Fixing: Shield: Energon: Navigation: Thrusters:
In the battle, the laser gun is hit, and now, it needs to be fixed. But it also needs to be closely monitored by the fixer, so the fixer and the laser gun user need to be able to work well together.
(e) (0.5 points) Given the listed constraints and the characteristics of each of the Ploetzobots, is it possible to find a solution that satisfies all of the constraints?
(f) (3.5 points) If yes, what are the possible bots that can seat in each of the listed roles, listed in alphabetical order and separated by a comma and a space (e.g. Arcee, Bumblebee, Jazz)? If not, leave the fields blank
Computer Science Tutoring
Laser Gun: Fixing: Shield: Energon: Navigation: Thrusters:
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Problem 5: Probability: Some memories are best forgotten! [18 points]
Leonard has a rare condition, which prevents him from forming new memories. Because of this, he tries to remind himself of what to do by writing himself notes. Additionally, as if his memory impairment weren’t challenging enough, he’s also trying to solve a murder mystery!
Each day after sleuthing, he arranges a plan for the following day: he either asks an informant to meet him at the City Grill restaurant, or he invites the informant to visit him in the hotel parking lot at the Discount Inn, where he stays. In either case, the meeting time is at noon each day. The problem is, once he wakes up the following day, he can’t remember the plan on his own; he can only remember it if he left himself a note the night before. But he doesn’t always remember to write the notes!
If Leonard wakes up without a note for himself, he goes to the City Grill 80% of the time, because he doesn’t like being alone in the parking lot. For the other times he wakes up without a note, he stays in the hotel parking lot because he has a hunch that is where he is supposed to be. Unfortunately, his hunches are independent of whether he actually asked that day’s informant to meet him at the parking lot or not.
He has a tattoo of the meeting time and two possible locations in case he forgets to make notes. So, if he guesses the location correctly on days without notes, he will meet them there at the correct time. If he goes to the wrong location, he does not meet his informant that day.
When he does leave a note for himself the night before, he always successfully meets the informant at the arranged location.
It turns out there is a pattern to when he leaves a note for himself. When he goes to the City Grill, there’s an 80% chance he’ll remember to leave a note for himself that night about the next day’s plans! But if he goes to the hotel parking lot during the day, there’s only a 30% chance he remembers to write the note. These probabilities hold whether or not he met his informant successfully earlier in the day.
Because there are suspects who don’t like Leonard snooping around, Leonard wants to avoid being too predictable. When he invites an informant to a location, he flips a fair coin (a fair coin has 0.5 probability of landing heads, and 0.5 probability landing tails). If it lands heads, he invites the informant to the restaurant for the following day; otherwise, to the hotel parking lot. The informants always go to the correct location that they were told to go to.
Earlier today (a Monday), he successfully met an informant at the City Grill after leaving himself a note Sunday evening.
For each answer, use only the information provided in the scenario above and follow the rounding rules for the exam.
(a) (2 points) Consider the table shown in Figure 16. Fill in the values for conditional probabilities of Leonard’s location each day:
Type the value of A here: Type the value of B here: Type the value of C here: Type the value of D here:
Github
Type the value of E here: Type the value of F here: Type the value of G here: Type the value of H here:
Figure 16: Table for Leonard’s Location
(b) (1 point) What is the probability of Leonard meeting the informant on any given day in which Leonard wakes up without a note from the previous night?
Type Your Answer here:
(c) (1.5 points) What is the probability that Leonard goes to the City Grill on Tuesday?
Type Your Answer here:
(d) (1.5 points) What is the probability Leonard goes to a different location than his informant on Tuesday?
Type Your Answer here:
(e) (1.5 points) What is the probability Leonard goes to the hotel parking lot and meets his informant on Tuesday?
Type Your Answer here:
(f) (2.5 points) What is the probability Leonard remembers to write a note for himself Tuesday night for Wednesday’s plan?
Type Your Answer here:
(g) (3 points) What is the probability Leonard goes to the City Grill restaurant on Wednesday? Type Your Answer here:
(h) (3 points) What is the probability Leonard successfully meets his informant on Wednesday? Type Your Answer here:
Now, Natalie, whom Leonard previously met, has some important evidence on the murder suspect she needs to share with Leonard! Leonard will not try to coordinate a meeting with her this week, but she knows that Leonard flips a coin to decide where to direct each informant and that the meeting time is noon each day. So, she also plans to flip a fair coin each day to decide which meeting spot to try to catch him. She intends to do this for up to two days, on Tuesday and Wednesday of the same week as described earlier in the problem.
(i) (1 point) What is the probability Natalie will successfully find Leonard at least once on Tuesday or Wednesday?
Type Your Answer here:
(j) (1point) Givenalltheinformation,whichofthefollowingstrategieswouldgiveNataliethehighest probability of finding Leonard at one of his two meeting spots by the end of Wednesday? (Single- Option Correct MCQ)
1. Strategy A: The stated strategy (flipping a fair coin each day to choose where to go) 2. Strategy B: Always going to the Discount Inn parking lot
3. Strategy C: Always going to the City Grill
Strategy A Strategy B Strategy C
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Problem 6: Bayes Nets: There are no accidents! [18 points]
As foretold by master Oogway, Shifu meets his destiny on the road he takes to avoid it. As a result of his own actions, Tai Lung has escaped the Chorh-Gom prison, and he is on his way toward the Jade palace to steal the dragon scroll. However, Shifu can barely concentrate now as he has to deal with a lot of things!
He has just witnessed the death of master Oogway, who requested him to trust Po as the Dragon Warrior. Po on the other hand is clueless about Kung Fu and Shifu needs to train him s