Introduction to Artificial Intelligence CAP 4601 Summer 2013 Midterm Exam

Transcription

1 Itroductio to Artificial Itelligece CAP 601 Summer 013 Midterm Exam 1. Termiology (7 Poits). Give the followig task eviromets, eter their properties/characteristics. The properties/characteristics of the Checkers task eviromet is provided as a example. Task Eviromet: Checkers Fully Observable or Partially Observable: Fully Observable Sigle Aget or Multiaget: Multiaget Determiistic or Stochastic: Determiistic Episodic or Sequetial: Sequetial Static or Dyamic: Static Discrete or Cotiuous: Discrete Kow or Ukow: Kow Task Eviromet: Robotic Car (Machie Drivig) Fully Observable or Partially Observable: Partially Observable Sigle Aget or Multiaget: Multiaget Determiistic or Stochastic: Stochastic Episodic or Sequetial: Sequetial Static or Dyamic: Dyamic Discrete or Cotiuous: Cotiuous Kow or Ukow: Kow

2 . Liear Regressio (1 Poits). Give the followig poits: { ( 0, 1 ), (, 1 ), (, -7 ), (, -7) } We have the followig plot: Calculate the Sample Mea for the x's ( a umber ): x i1 0 8 x i

3 Calculate the Sample Mea for the y's ( a umber ): -3 y i Calculate the Biased Sample Variace for the x's ( a umber ): y i var x,biased 1 i x i x 0 Calculate the Biased Sample Variace for the y's ( a umber ): 16 var y,biased 1 i y i y

4 Calculate the Biased Sample Covariace betwee the x's ad the y's ( a umber ): - cov 1 x,,biased i x xy yi y i Calculate the w0 from y = w0 + w1*x ( i.e. calculate the Itercept, a umber ): 1 cov w1 var xy,,biased x,biased w w x y Calculate the w1 from y = w0 + w1*x ( i.e. calculate the Slope, a umber ): - cov w1 var xy,,biased x,biased

5 3. Classificatio (1 Poits). Give the followig two classes each with three poits, fid the mea of each class ad the fid the equatio of the plae that separates those classes halfway betwee their respective meas. Esure that the separatig plae produces positive values for Class Poits (Blue) ad egative values for Class 1 Poits (Red). I other words, fid the separatig plae that could be used for classificatio with the maximum margi betwee the classes ad where the separatig plae z is z > 0 for Class Poits (Blue) ad z < 0 for Class 1 Poits (Red). Class 1 Poits (Red): { ( -, ), ( 0, ), (, ) } Class Poits (Blue): { ( -, 0 ), ( 0, 0 ), (, 0 ) } The followig is the plot of these two classes: We wat the followig classificatio:

6 x x, y, Class1 i i1 i , 3 3 0, y i Class1, 3 3 xy, midpoit x x, y, Class x, y x, y 0, 0,0 i i1 i , 3 3 0,0 Class1 Class 0 0, 0 0, 0, 0,1 y i Class, 3 3 Calculate the x value of the Midpoit betwee the Sample Mea of the Class 1 Poits ad the Sample Mea of the Class Poits ( i.e. the value of x i the midpoit ( x, y ) ): 0 Calculate the y value of the Midpoit betwee the Sample Mea of the Class 1 Poits ad the Sample Mea of the Class Poits ( i.e. the value of y i the midpoit ( x, y ) ): 1

7 Sice we wat the separatig plae to produces positive values for Class Poits (Blue), the we wat the ormalized ormal vector to poit toward Class Poits (Blue). Sice we wat to fid the separatig plae used for classificatio with the maximum margi betwee the classes, the we wat to start poit from the midpoit we just calculated; hece, the agle we eed to use for this ormalized ormal vector is 90. Therefore, we have: ormalized cos,si cos 90,si 90 0, 1 Calculate the x value of the Normalized Normal Vector that poits towards the Class Poits (Blue) (i.e. the value of the x i the ormalized ormal vector ormalized ): 0 Calculate the y value of the Normalized Normal Vector that poits towards the Class Poits (Blue) (i.e. the value of the y i the ormalized ormal vector ormalized ): -1 Sice we have the followig: f x, y cos,si x, y x, y classificatio 0 0 0, 1 xy, 0,1 0, 1 x 0, y1 0 x 0 1 y1 0 x 00 1 y x 1 y 1 0 x 1 y 1 0 x 1 y Calculate the w0 ( a umber ) i the equatio of the separatig plae z = w0 + w1*x + w*y that creates the separatig lie i the level set z = 0: 1 Calculate the w1 ( a umber ) i the equatio of the separatig plae z = w0 + w1*x + w*y that creates the separatig lie i the level set z = 0: 0 Calculate the w ( a umber ) i the equatio of the separatig plae z = w0 + w1*x + w*y that creates the separatig lie i the level set z = 0: -1

8 . Naive Bayes Network ( Poits). Usig a small subset of data from a experimet at the Ft. Beig Battle Lab, costruct a Naïve Bayes Network to predict a subject's first actio (move or fire) i close rage egagemets uder various circumstaces. "Urba" meas the subject was i a city, vice beig i a forest. "Day" meas the ecouter happeed i daylight, vice at ight usig ight-visio equipmet. "Withi5m" meas that the target was first observed at a rage less tha 5 meters, vice up to 50 meters. "MoveFirst" meas that the subject's first actio o detectig the threat was to move to cover, vice firig immediately. HINT: Draw the graph of this Naïve Bayes Network. Rememberig that a Naïve Bayes Network assumes coditioal idepedece. QUESTION: Usig the data table above, do the followig (Do ot write ay decimal approximatios): 1) Write the exact fractio for P( MoveFirst = true ). 3/10

9 ) Write the exact fractio for P( MoveFirst = false ). 7/10 3) Write the exact fractio for P( Urba = false MoveFirst = false ). /7 ) Write the exact fractio for P( Urba = false MoveFirst = true ). 1/3 5) Write the exact fractio for P( Urba = true MoveFirst = false ). 3/7

10 6) Write the exact fractio for P( Urba = true MoveFirst = true ). /3 7) Write the exact fractio for P( Day = false MoveFirst = false ). 3/7 8) Write the exact fractio for P( Day = false MoveFirst = true ). /3 9) Write the exact fractio for P( Day = true MoveFirst = false ). /7

11 10) Write the exact fractio for P( Day = true MoveFirst = true ). 1/3 11) Write the exact fractio for P( Withi5m = false MoveFirst = false ). 3/7 1) Write the exact fractio for P( Withi5m = false MoveFirst = true ). /3 13) Write the exact fractio for P( Withi5m = true MoveFirst = false ). /7

12 1) Write the exact fractio for P( Withi5m = true MoveFirst = true ). 1/3 15) Write the equatio to predict the probability that a subject's first actio is to shoot whe he ecouters a threat withi 5 meters i a city at ight. I order words, write the equatio that could calculate: P( MoveFirst = false Urba = true, Day = false, Withi5m = true ) i terms of oly the probabilities above. Just write the probabilities (usig the rules of Probability that you leared about i Chapter 13). Do NOT write the exact fractios or the decimal approximatio. P( MoveFirst = false Urba = true, Day = false, Withi5m = true ) = ( P( Urba = true MoveFirst = false ) * P( Day = false MoveFirst = false ) * P( Withi5m = true MoveFirst = false ) * P( MoveFirst = false ) ) / ( P( Urba = true MoveFirst = false ) * P( Day = false MoveFirst = false ) * P( Withi5m = true MoveFirst = false ) * P( MoveFirst = false ) + P( Urba = true MoveFirst = true ) * P( Day = false MoveFirst = true ) * P( Withi5m = true MoveFirst = true ) * P( MoveFirst = true ) ) Or writte out so that we ca actually have a hope of beig able to read it:,, 5 5 PWithi5m true MoveFirst false P MoveFirst false P MoveFirst false Urba true Day false Withi m true P Urba true MoveFirst false P Day false MoveFirst false P Withi m true MoveFirst false P MoveFirst false P Urba true MoveFirst false P Day false MoveFirst false P Urba true MoveFirst true P Day false MoveFirst true P Withi5m true MoveFirst true P MoveFirst true

13 5. k-nearest Neighbor ( Poits). If k 1, the our closest eighbors would be: ad majority rule would classify our query poit as. k, the our closest eighbors would be:, If If 3 k, the our closest eighbors would be:,, k, the our closest eighbors would be:,,, k, the our closest eighbors would be:,,,, k, the our closest eighbors would be:,,,,, k, the our closest eighbors would be:,,,,,, If If 5 If 6 If 7 7 ad majority rule would classify our query poit as. ad majority rule would classify our query poit as. ad majority rule would classify our query poit as. ad majority rule would classify our query poit as. ad our query poit would be arbitrarily classified. ad majority rule would classify our query poit as. 6. Liear Regressio vs. Logistic Regressio (6 Poits). Describe the differece betwee Liear Regressio ad Logistic Regressio. Detail what is beig miimized i both methods (i.e. what is beig regressed). Liear Regressio is a Supervised Learig method used for Regressio, ot Classificatio. Liear Regressio miimizes the distace of the y value of each poit with the y value of the lie of regressio. Logistic Regressio is a Supervised Learig method used for Classificatio, ot Regressio. Logistic Regressio miimizes the distace betwee the class of a poit (expressed as either a 0 or a 1) ad the value of that poit o the Logistic Fuctio.

14 7. Parametric vs. Noparametric Methods (8 Poits). For the methods below, idicate whether the method is parametric or oparametric usig a "P" for parametric ad a "N" for oparametric: Liear Regressio: P k-nearest Neighbor: N Bayesia Network: P Multivariate Liear Regressio: P Logistic Regressio: P Perceptro: P Multilayer Feed-Forward Neural Network: P Locally Weighted Regressio: N Support Vector Machies: P Kerel Desity Estimatio: N 8. Parametric vs. Noparametric ( Poits). Describe the differece betwee parametric methods ad oparametric methods. Parametric methods summarize data ito a fixed umber of parameters whose cout does ot icrease as the umber of data poits icrease. Noparametric methods do ot summarize data ito a fixed umber of parameters. The umber of parameters (if they're used at all) icreases as the umber of data poits icrease. 9. The Curse of Dimesioality (6 Poits). Describe the Curse of Dimesioality (esp. as described by Pedro Domigos i "A Few Useful Thigs to Kow about Machie Learig"). The curse of dimesioality, coied by Bellma i 1961, refers to the fact that may algorithms that work fie i low dimesios become itractable whe the iput is high-dimesioal. But i machie learig it refers to much more. Geeralizig correctly becomes expoetially harder as the dimesioality (umber of features) of the examples grows, because a fixed-size traiig set covers a dwidlig fractio of the iput space. Moreover, the similarity-based reasoig that machie learig algorithms deped o (explicitly or implicitly) breaks dow i high dimesios.

15 10. Supervised vs. Usupervised Learig ( Poits). Describe the differece betwee Supervised ad Usupervised Learig. Supervised Learig requires data that are labeled with their class durig learig. Oly class membership (classificatio) is determied by the supervised learig method. Usupervised Learig does ot require data that are labeled with their class durig learig. Both classes ad class membership are determied by the usupervised learig method. 11. A* Search (0 Poits). 1) The ode where h=15: 1 ) The ode where h=11: 0 3) The ode where h=8: ) The ode where h=7: 3 5) The ode where h=6: 6) The ode where h=10: 5 7) The ode where h=: 0 8) The ode where h=3: 0 9) The ode where h=9: 0

16 10) The ode where h=5: 0 11) The ode where h=0: 0 1) The ode where h=0 GOAL: 6 13) Is this heuristic admissible? If yes, why? If o, why? No, the heuristic is ot admissible because h=0 is a overestimate sice sigle steps oly cost 10 ad the ode where h=0 is oly oe step away from the goal. Here's the exhaustive aswer: First, we augmet the graph above with edge weights: Next, we add the top ode to the frotier: ad explored is empty: f 15 frotier g 0 h 15 explored =.

17 Eterig the loop, we see that frotier is ot empty, so we pop frotier to ode: f 15 ode g 0 h 15 The, we test if ode is the goal. Sice it is t, we add ode to explored: f 15 explored = g 0. h 15 For each child of the ode, we test if it is i explored ad frotier. Sice they are t, we add each child to the frotier givig priority to the smallest f : f 16 f 17 f 18 f 0 f 1 frotier g 10, g 10, g 10, g 10, g 10 h 6 h 7 h 8 h 10 h 11 O the ext iteratio, we agai see that frotier is ot empty, so we pop frotier to ode: f 16 ode g 10 h 6

18 Sice this ode is t the goal, we add the ode to explored: f 15 f 16 explored = g 0, g 10 h 15 h 6 ad the we look at the childre of that ode. For each child, we check to see if the child is i the frotier or i explored. Sice each child is t, we add these childre to frotier: f 17 f 18 f 0 f 1 f 5 f 0 frotier g 10, g 10, g 10, g 10, g 0, g 0 h 7 h 8 h 10 h 11 h 5 h 0 O the ext iteratio, we see that frotier is ot empty, so we pop frotier to ode: f 17 ode g 10 h 7 Sice ode is t the goal, we add ode to explored: f 15 f 16 f 17 explored = g 0, g 10, g 10 h 15 h 6 h 7 Sice ode does t have ay childre, we go o to the ext iteratio.

19 O this ext iteratio, we see that frotier is still ot empty, so we pop frotier to ode: f 18 ode g 10 h 8 Sice ode is t the goal, we add ode to explored: f 15 f 16 f 17 f 18 explored = g 0, g 10, g 10, g 10 h 15 h 6 h 7 h 8 For each child of ode, we check to see if the child is i the frotier or i explored. Sice each child is t, we add these childre to frotier: f 0 f 1 f 3 f 5 f 9 f 0 frotier g 10, g 10, g 0, g 0, g 0, g 0 h 10 h 11 h 3 h 5 h 9 h 0 O the ext iteratio, we see that frotier is ot empty, so we pop frotier to ode: f 0 ode g 10 h 10 Sice ode is t the goal, we add ode to explored: f 15 f 16 f 17 f 18 f 0 explored = g 0, g 10, g 10, g 10, g 10 h 15 h 6 h 7 h 8 h 10

20 ad the we look at the childre of that ode. For each child, we check to see if the child is i the frotier or i explored. Sice each child is t, we add these childre to frotier: f 0 f 1 f 3 f 5 f 9 f 0 frotier g 0, g 10, g 0, g 0, g 0, g 0 h 0 h 11 h 3 h 5 h 9 h 0 O the ext iteratio, we see that frotier is ot empty, so we pop frotier to ode: f 0 ode g 0 h 0 Sice ode is the goal, the we have our solutio ad we re doe ad we put zeros i the odes that will ever be expaded:

21 Therefore, we have the followig: a. The ode where h=15: 1 b. The ode where h=11: 0 c. The ode where h=8: d. The ode where h=7: 3 e. The ode where h=6: f. The ode where h=10: 5 g. The ode where h=: 0 h. The ode where h=3: 0 i. The ode where h=9: 0 j. The ode where h=5: 0 k. The ode where h=0: 0 j. The ode where h=0 GOAL: 6 l. Is this heuristic admissible? If yes, why? If o, why? Highlightig the followig: We see that our heuristic has a higher value tha the actual edge cost to the goal because edge cost 10 0 heuristic ; hece, our heuristic overestimates the cost to reach the goal. Therefore, sice our heuristic overestimates the cost to reach the goal, the this heuristic is ot admissible. I other words: No, our heuristic is ot admissible because it overestimates the cost to reach the goal.