Exercise 4: Markov Processes, Cellular Automata and Fuzzy Logic

Transcription

1 Exercie 4: Marko rocee, Cellular Automata and Fuzzy Logic Formal Method II, Fall Semeter 203 Solution Sheet Marko rocee Theoretical Exercie. (a) ( point) tanding 0.25 lying eating 0.4 (b) (2 point) The tranition matrix in the order (tanding, lying, eating), uing the notation from the cript i a follow: = The tationary ditribution ector x mut fulfill the following requirement: x = x In other term, the ditribution hall not change from tate X n to tate X n. Uing i x i = a contraint (the um of all probabilitie mut equal ), we obtain:

2 x x x x = x = x 2 = x 2 = x 2 x 3 x x x x 2 Thi equation can alo be formulated in term of the following linear equation ytem: x = 0.2x x ( x x 2 ) x 2 = 0.3x + 0.7x ( x x 2 ) x 3 = 0.5x x ( x x 2 ) which can eaily be oled uing tandard technique. Let hae a look at an alternatie way of oling the problem: If you compare the definition of the tationary ditribution x = x with the eigenalue problem = λ you will notice ome imilaritie. In particular, if you ubtitute x for and et λ =, you will realize that the tationary ditribution i in fact an eigenector of with a correponding eigenalue of λ =. Scaling the eigenector to = ( i > 0), lead to the tationary ditribution. The following octae code how how thi could be done: =[0.2,0.25,0.2; 0.3,0.7,0.4; 0.5,0.05,0.4] 2 [Q,L] = eig () 3 x = Q (:,)/ um (Q (:,)) Uing either one of the method, you hould arrie at a tationary ditribution of: % x = 53.9% % Hidden Marko Model ractical exercie 2. (5 point) The key to oling thi quetion i to realize that the hidden tate are the letter that are meant to be typed and the emiion token are the letter that hae been typed. Therefore, it i only neceary to introduce emiion token with enible emiion probabilitie for the letter. For example, could be ued for the correct letter and 0.3 for the urrounding letter. The algorithm itelf doe not require any change. An open ource clone of MATLAB. The aboe code alo run in MATLAB. 2

3 Cellular Automata 3. Firt, the decimal rule number 20 0 ha to be conerted to binary, which lead to (a) ( point) Thi binary number i then filled into the rule table, maintaining the proper order (the leat ignificant bit of the binary number alway correpond to (0, 0, 0)): a t n a t n a t n+ a t+ n (b) (2 point) The grid can then be drawn by checking the rule table. boundary condition. Note the (c) (2 bonu point) The inerted pattern i produced uing rule 80. The rule number can intuitiely be found by inerting the rule table found in part (a), and thu then reading the rule number in the reere direction (note that thi now i the correct order): 20 0 = inert = 80 0 reere 3

4 4. (4 point, 0.5 for each correct anwer, no penalty) Each cell of a CA can be looked upon a a Finite State Machine. In D CA, the complexity of the emerging pattern depend motly on the neighborhood radiu r. Gien an arbitrary elementary CA (with rule table) and the cell tate at iteration i, the initial configuration can alway be computed. Gien the cell tate of a D CA at arbitrary iteration i and i + one can in ome cae reproduce the rule table. In 2D CA, the Moore neighborhood conit of a central cell and 4 neighbor wherea the on Neumann neighborhood conit of the central cell and 8 neighbor. The rule table of a D CA with fixed boundary condition contain fewer rule than the rule table of a D CA with cyclic boundary condition. CA can be ued to imulate procee from e.g. phyic and biology. A majority of CA rule are capable of unieral computation. t f 4

5 Fuzzy Logic 5. (a) ( point) Memberhip function for input ariable θ and θ a well a the output ariable : m θ m N N θ V 0V 0V m θ N θ There i no unique olution to thi tak. The particular function are the deigner choice, that mot probably hae to be fine tuned by experimenting with the controller either in imulation or in the actual experimental etup. Note that the direction of the axi for the oltage i omewhat trange. Thi i to comply with the nomenclature in the additional lecture note on fuzzy logic where the direction of the cart i counter- intuitie too. (b) (2 point) One poible rule table: θ\θ N N N N N There might be other acceptable olution depending on how they are jutified. (c) (2 point) In each cycle/period, the kinetic energy i maximal when the pendulum i paing the poition at the bottom (table equilibrium point), i.e. θ = 80. We aume θ = 36 for implicity, although that alue i uppoed to increae during the wing-up proce. The only et to conider are θ and θ becaue all other memberhip function are 0, o the rule (, ; ) ha to be actiated. At thi point, the geometrical output function i contructed a follow: 5

6 m m m m m m V 0 V The lat tep i to compute the centroid 0V 0 V of the polygon haded in orange. One can eaily ee that thi polygon can be plit up into a triangle at the left and athe rectangle lat tep The ati lat the totep right. compute i tothe compute the area centroid the of the centroid of rectangle the of the polygon i 3 and haded the -coordinate orange. One of it cancentroid eailycan ee ieaily that 7. ee The thi that triangle polygon thi polygon ha canan be canare plit be aplit of up up 0.5 into and aa the triangle -coordinate the left left ofand it a rectangle at the right. The area of the rectangle i 3 and the -coordinate of it centroid a rectangle i roughly at the right. at TheBy area geometric of the decompoition, the -coordinate of centroid i 7. The triangle ha an area of 0.5 and the -coordinate of it centroid the it centroid polygon i roughly i therefore 7. at The triangle approximately By geometric ha andecompoition, are a of V. 0.5 the and-coordinate the -coordinate of the polygon of it On centroid the other i iroughly therefore hand, approximately at the potential By 6.56 geometric energy V. i decompoition, maximal in eachthe cycle -coordinate the point of at thewhich polygon the On i the elocity therefore other hand, change approximately the potential ign, that energy i i maximal V. in each cycle at the point at which the elocity change ign, that i On the other hand, the potential energy maximal θ = 0 = 0.. Let u aume = 90. Let u aume θ = 90 in each at. (Again, (Again, thi alue i uppoed to increae the point thi alue i uppoed to increae from cycle at which the elocity change ign, that i from cycle to cycle. A in the to cycle. = 0 A in the preiou example, preiou example, the only rule the that only i torule be conidered that i to i (, be ; conidered ):. Let i u(, aume ; ): = 90. (Again, thi alue i uppoed to increae from cycle to cycle. A in the m preiou example, the only rule that i to be conidered i (, ; ): m m m m m V 0 V 0 The centroid of 80the polygon haded in 0 blue76ha a -coordinate 0V of roughly 6. 0 V The centroid of the polygon haded in blue ha a -coordinate of roughly -6. The centroid of the polygon haded in blue ha a -coordinate of roughly -6. 6