Symbolic computation 2: Linear recurrences

Transcription

1 Bachelor of Ecole Polytechique Computatioal Mathematics, year 2, semester Lecturer: Lucas Geri (sed mail) Symbolic computatio 2: Liear recurreces Table of cotets Warm-up Solvig liear recurreces with Sympy The fuctio rsolve Applicatios: the towers of Haoi Implemetatio of the towers of Haoi # execute this part to modify the css style from IPytho.core.display import HTML def css_stylig(): styles = ope("./style/custom2.css").read() retur HTML(styles) css_stylig() ## loadig pytho libraries # ecessary to display plots ilie: %matplotlib ilie # load the libraries import matplotlib.pyplot as plt # 2D plottig library import umpy as p # package for scietific computig from pylab import * from math import * import sympy as sympy from sympy import * # package for mathematics (pi, arcta, sqrt, # package for symbolic computatio Warm-up

2 Exercise. Solvig a recurrece (almost) by had Do it yourself. We cosider the sequece de ed by u 0. Use SymPy to d α, a, b, c such that for every 0 3 To solve a system of equatios with SymPy with ukows example =, { u = 2 u ( ). u = α 2 + a 2 + b + c. x, y, write for solve([x-y-2,3*y+x],[x,y]) 0 2. Prove with SymPy that your formula is true for every.

3 # Questio var('',iteger=true) var('alpha a b c') def u(,alpha,a,b,c): retur alpha*2** +a***2 +b* +c Solutios=solve( # Quatities which are to be zero: [u(0,alpha,a,b,c)-, # iitial coditio u(,alpha,a,b,c)-2*u(0,alpha,a,b,c)-3***2, u(2,alpha,a,b,c)-2*u(,alpha,a,b,c)-3*2**2, u(3,alpha,a,b,c)-2*u(2,alpha,a,b,c)-3*3**2, ], # Ukows: [alpha,a,b,c] ) prit(solutios) # Questio c_s=-8 alpha_s=9 b_s=-2 a_s=-3 prit('with these coefficiets the simplificatio of u()-2u(-)-3**2 gives prit('first values : '+str([u(i,alpha_s,a_s,b_s,c_s) for i i rage(0)])) # I order to check our results: def Recurrece(): if ==0: retur else: retur 2*Recurrece(-)+3***2 prit('we compare with : '+str([recurrece() for i rage(0)])) {c: -8, alpha: 9, b: -2, a: -3} With these coefficiets the simplificatio of u()-2u(-)-3**2 gives : 0 First values : [, 5, 22, 7, 90, 455, 08, 283, 4558, 9359] We compare with : [, 5, 22, 7, 90, 455, 08, 283, 4558, 935 9] Aswers.. With the rst values we obtai that if the formula is true the ecessarily 5 2. With the above script we proved that the formula is true: for every that u()-2u(-)-3**2=0. u = we have

4 Exercise 2. A double liear recurrece I Lab 2 we proved that if the a, b are itegers de ed by a + b 2 = ( + 2 ), a ( ) = 2 b ( ) ( ) Do it yourself.. Use SymPy to d a explicit formula for a. (I SymPy matrices are de ed by Matrix([[a,b],[c,d]]).) c, R 2. Use the formula obtaied at Questio to d real umbers such that + a c R. #Questio A=Matrix([[,2],[,]]) var('', iteger=true) Power=A** prit(power) a=latex(simplify(power[0,0]+power[0,])) prit(a) Matrix([[( + sqrt(2))**/2 + (-sqrt(2) + )**/2, sqrt(2)*( + sqrt(2))**/2 - sqrt(2)*(-sqrt(2) + )**/2], [sqrt(2)*( + sqrt (2))**/4 - sqrt(2)*(-sqrt(2) + )**/4, ( + sqrt(2))**/2 + (- sqrt(2) + )**/2]]) \frac{}{2} \left( + \sqrt{2}\right)^{} + \frac{\sqrt{2}}{2} \ left( + \sqrt{2}\right)^{} - \frac{\sqrt{2}}{2} \left(- \sqrt{ 2} + \right)^{} + \frac{}{2} \left(- \sqrt{2} + \right)^{} Aswers.. We export the result i LateX: a = ( + 2) 2 2 ( + 2) 2 2 ( 2 + ) 2 ( 2 + ) 2 + < ( + ( + 2 ). It follows that a = + + o() 2 ( + 2) 2 2 ( + 2) ( + 2) 2. As, we have that 2 ) is eglictible compared to

5 Solvig recurreces with SymPy: rsolve The fuctio rsolve The aim of this lab sessio is to use Sympy to obtai explicit formulas for some sequeces de ed by liear recurreces. More precisely, we will see how to obtai a explicit formula for u i two cases:. Liear recurrece of order oe: this is a sequece (u) 0 is de ed by u 0 = a, { u = α u + f (), ( ), where a, α are some give costats ad f is a arbitrary fuctio. 2. Liear recurrece of order two: this is a sequece (u ) 0 is de ed by u 0 = a, u = b, u = α u + β u 2 + f (), ( 2), where are some give costats ad is a arbitrary fuctio. a,, b, α, β f Some kow examples t i this settigs: u 0 = a, u = ru. Geometric sequeces:. 2. Arithmetic sequeces: u 0 = a, u = u + r. 3. The Fiboacci sequece:. F =, F 2 =, F = F + F 2 The followig script shows how to solve the liear recurrece of a geometric sequece, usig fuctio rsolve :

6 # First example: a geometric sequece u = Fuctio('u') # declares the ame of the sequece = symbols('',iteger=true) # declares the variable f = u()-3*u(-) # the expressio which is to be zero ExplicitFormula=rsolve(f,u(),{u(0):5}) prit('the formula for u() is '+str(explicitformula)+'') # Secod example with a o-liear term prit(' ') v = Fuctio('v') # declares the ame of the sequece = symbols('',iteger=true) # declares the variable f = 2*v()-v(-)-2** # the expressio which is to be zero ExplicitFormula=rsolve(f,v(),{v(0):}) prit('the formula for v() is '+str(explicitformula)+'') The formula for u() is 5*3** The formula for v() is 2*2**/3 + 2**(-)/3 Exercise 3. A rst example with rsolve Do it yourself.. Use SymPy to solve the recurrece of the Fiboacci sequece ad d a explicit formula for F. (To set up two iitial coditios you must write {u():,u(2):}.) 2. Use the formula obtaied at Questio to prove that F lim = φ =. F 2 u = Fuctio('u') # declares the ame of the sequece = symbols('',iteger=true) # declares the variable f = u()-u(-)-u(-2) # the expressio which is to be zero ExplicitFormula=simplify(rsolve(f,u(),{u():,u(2):})) Formula2=latex(simplify(ExplicitFormula)) prit('the formula for u() is '+str(formula2)+'') The formula for u() is \frac{2^{- }}{5} \sqrt{5} \left(\left( + \sqrt{5}\right)^{} - \left(- \sqrt{5} + \right)^{}\right)

7 Aswers.. We put the above aswer i LateX: 2 F 5 2. We obtai (by settig ): F 5 φ ψ 5 =, F 5 φ 5 ψ φ ψ (ψ/φ) 5 5 sice. = ( ). 5 ( + 5) ( 5 + ) ψ = ( 5)/2 =, (we divide everythig by φ.) (ψ/φ) 5 φ, ψ/φ < 5 Remark. The output of rsolve is a expressio which depeds o the symbolic variable. If we wat to evaluate this expressio (for istace for = 0) we must write: Value=ExplicitFormula.subs(,0) prit('0th Fiboacci umber = '+str(value)) prit('after simplificatio : '+str(simplify(value))) 0th Fiboacci umber = sqrt(5)*(-(-sqrt(5) + )**0 + ( + sqrt (5))**0)/520 After simplificatio : 55

8 Applicatio: the Towers of Haoi (This remaider of this sessio is devoted to the study of the Towers of Haoi, which you have studied i "Discrete Mathematics" (Bachelor ).) Exercise 4. The puzzle with three rods Let us recall this mathematical puzzle. We are give a stack of disks arraged from largest o the bottom to smallest o top placed o a rod a, together with two empty rods b, c. The Tower of Haoi puzzle asks for the miimum umber of moves required to move the etire stack, oe disk at a time, from rod a to aother ( b or c). A move is allowed oly if it moves a smaller disk o top of a larger oe. Here is a example which shows that the Tower of Haoi with solvable i moves: 7 = 3 disks is More geerally we use the followig recursive strategy to solve the Tower of Haoi with disks: Assume that you have a strategy for disks; Move the smallest disks from rod to rod with your strategy Move disk from to a b a c b c Move the smallest disks from rod to rod with your strategy As there is a obvious strategy for Haoi of Tower for ay. Let H be the sequece de ed by = Of course H, the above gure gives H 3 = 7. this recursive algorithm allows to solve the H = Number of moves for solvig Haoi with disks, with the recursive strateg =

9 Do it yourself. Fid a recursive formula for the sequece H : write H as a fuctio of H. Aswers. The three steps take respectively H, ad H moves. Therefore H = 2 H +. Do it yourself.. Write a recursive fuctio Haoi() which returs the value of H. 5 H 2. Write a small script which returs the rst values of. Ca you guess a formula? def Haoi(): # iput: positive iteger # output: umber of moves eeded to solve Haoi with disks if ==: retur else: retur 2*Haoi(-)+ # First few values of Haoi() prit('first values of H: '+str([haoi() for i rage(,5)])) First values of H: [, 3, 7, 5, 3, 63, 27, 255, 5, 023, 2 047, 4095, 89, 6383] Aswers. The guess is H = 2.

10 Do it yourself.. Use rsolve to d a explicit formula for the sequece H. 2. Compare with your ow fuctio Haoi() ad with your guess. The output should be like Sympy says the formula for H() is... Sympy says the first values for H() are [, 3, 7,...] Haoi() says the first values for H() are [, 3, 7,...] H = Fuctio('H') # declares the ame of the sequece = symbols('',iteger=true) # declares the variable f = H()-2*H(-)- # the expressio which is to be zero ExplicitFormula=rsolve(f,H(),{H():}) Formula=simplify(ExplicitFormula) prit('sympy says the formula for H() is '+str(formula)+'') FirstValuesofH= [Formula.subs(,i) for i i rage(,5)] prit('sympy says the first values for H() are '+str(firstvaluesofh)) prit('haoi() says the first values for H() are '+str([haoi() for i r Sympy says the formula for H() is 2** - Sympy says the first values for H() are [, 3, 7, 5, 3, 63, 27, 255, 5, 023, 2047, 4095, 89, 6383] Haoi() says the first values for H() are [, 3, 7, 5, 3, 63, 27, 255, 5, 023, 2047, 4095, 89, 6383] Exercise 5. The tower of Haoi with four rods We ow cosider the towers of Haoi with four rods a, b, c, d. I that case, formulas are more di cult to guess ad we will really eed rsolve to obtai exact formulas ad asymptotics.

11 The recursive algorithm for fours rods The recursive strategy is as follows. If or, use the algorithm of Exercise 4 to move the disk(s) to peg (it takes move for, moves for ). If = = 2 d 3 = 3 = 2, use the followig recursio: a, b, c, d 2 a b Use the fours rods to move the smallest disks from. a, c, d a d Use rods to move the two largest disks. a, b, c, d 2 b d Use the fours rods to move the smallest disks from. Let J be the sequece de ed by = Number of moves for solvig Haoi with disks ad 4 rods. J = As before, J, J 2 = 3. Do it yourself. Fid a recursive formula for the sequece J. Aswers. The three steps take respectively J 2, 3 ad J 2 moves. Therefore J = 2 J Do it yourself.. Write a recursive fuctio HaoiFour() which returs the value of J. 2. Write a small script which returs the te rst values of J. Ca you guess a formula? (You ca ask (

12 def HaoiFour(): if ==: retur elif ==2: retur 3 else: retur 2*HaoiFour(-2)+3 # First few values Haoi() prit('first values of J: '+str([haoifour() for i rage(,5)])) First values of J: [, 3, 5, 9, 3, 2, 29, 45, 6, 93, 25, 8 9, 253, 38] Aswers. There is o obvious guess! Do it yourself.. Use rsolve to d a explicit formula for the sequece J. 2. Compare with your ow fuctio HaoiFour(). J = Fuctio('J') # declares the ame of the sequece = symbols('',iteger=true) # declares the variable f = J()-2*J(-2)-3 # the expressio which is to be zero ExplicitFormula=rsolve(f,J(),{J():,J(2):3}) Formula=simplify(ExplicitFormula) prit('sympy says the formula for J() is '+str(formula)+'') FirstValuesofJ= [simplify(formula.subs(,i)) for i i rage(,5)] prit('sympy says the first values for J() are '+str(firstvaluesofj)) prit('haoifour() says the first values for J() are '+str([haoifour() for Sympy says the formula for J() is 2**(/2 - )*(2*sqrt(2) + 3) + (-sqrt(2))***(-sqrt(2) + 3/2) - 3 Sympy says the first values for J() are [, 3, 5, 9, 3, 2, 29, 45, 6, 93, 25, 89, 253, 38] HaoiFour() says the first values for J() are [, 3, 5, 9, 3, 2, 29, 45, 6, 93, 25, 89, 253, 38] Aswers. The explicit formula for J is 2 ( ) + ( 2) ( ) 3. 2

13 Do it yourself. (Theory). Accordig to the formula foud by SymPy, what happes whe J for the sequece? 2 J 0 2. Usig the formula foud by SymPy, prove that. H + Aswers.. We rewrite the above formula as J 2 2 J Therefore oscillates betwee ad. 2 = ( + 3/2 + ( ) ( 2 + 3/2)) We have that J 4 for large eough. Therefore J H 2 2

14 Exercise 6. The recursive algorithm for fours rods: a better variat? We de e a variat of the previous strategy: If or or, use the previous algorithm with rods to move the disk(s) to peg (it takes move for, moves for, moves for If = = 2 = 3 4 = 3 ). 3 d = 3 = 2 5, use the followig recursio: a, b, c, d 3 a b Use the fours rods to move the smallest disks from. a, c, d a d Use rods to move the three largest disks. a, b, c, d 3 b d Use the fours rods to move the smallest disks from. Let K be the sequece de ed by K = Number of moves for solvig Haoi with disks ad 4 rods with that strate = We have, K, K 2 = 3, K 3 = 5. Do it yourself.. Fid a recursive formula for the sequece K. (Do ot try to solve the recurrece with SymPy, it does ot seem to work.) 2. Write a fuctio HaoiFourBis() which returs the value of K. 3. Plot o the same gure the rst values of K ad J. Which algorithm looks faster? Aswers. The three steps take respectively, ad moves. Therefore, for. K 3 7 K 3 4 = K K 3

15 def HaoiFourBis(): if ==: retur elif ==2: retur 3 elif ==3: retur 5 else: retur 2*HaoiFourBis(-3)+7 FirstJ=[HaoiFour() for i rage(,25)] FirstK=[HaoiFourBis() for i rage(,25)] prit('the first values for K() are '+str(firstk)+'') prit('the first values for J() are '+str(firstj)+'') plt.plot(firstk) plt.plot(firstj) plt.show() The first values for K() are [, 3, 5, 9, 3, 7, 25, 33, 4, 5 7, 73, 89, 2, 53, 85, 249, 33, 377, 505, 633, 76, 07, 2 73, 529] The first values for J() are [, 3, 5, 9, 3, 2, 29, 45, 6, 9 3, 25, 89, 253, 38, 509, 765, 02, 533, 2045, 3069, 4093, 6 4, 889, 2285] Do it yourself. We admit that there exist reals + K α R c > 0, α R, R > c. ( ) such that. Fid a sequece which coverges to (ad prove the covergece, assumig that ( ) holds). 2. Write a script which returs a umerical approximatio of R. (As the covergece might be very slow, you should write a o recursive fuctio to compute large values of K.) R

16 Aswers.. If K c the = R. Ideed, K / K lim K / = exp( log( )) = exp( log(c α R + o(c α R ))) = exp( log(c α R ( + o())) = exp( (log(c α R ) + log( + o()))) = exp( (log(c) + α log() + log(r) + o())), (sice log( + o( R. K + / K (Oe could also cosider the sequece.) ( ) Remark: I fact oe ca prove that does ot hold but we have that K = c α R where is a bouded (ad oscillatig) sequece, ad 2 /3 c R = = # Let us compute i a o-recursive way the values of K: HaoiFourBis_orecursive=[,3,5] N=200 for i i rage(4,n): m=haoifourbis_orecursive[-3] HaoiFourBis_orecursive.apped(2*m+7) FirstK_ormalized=[(HaoiFourBis_orecursive[])**(/(+0.0)) for i rage prit('st approximatio of R = limit K_^(/): '+str(firstk_ormalized[-]) prit('2d approximatio of R = limit (K_{+}/K_): '+str(haoifourbis_orec plt.plot(firstk_ormalized) plt.show() st approximatio of R = limit K_^(/): d approximatio of R = limit (K_{+}/K_):.25

17 For fu (ot graded) Exercise 7. Implemetatio for three rods Do it yourself. Write a fuctio SolvigHaoi(,x,y) which prits the x y sequece of displacemets eeded to move disks from rod to rod. For example: SolvigHaoi(3,'a','c') a -> c a -> b c -> b a -> c b -> a b -> c a -> c (Hit: You may begi with a fuctio ThirdRod(x,y) which returs the x, y {a, b, c} complemet rod of i.)

18 def ThirdRod(x,y): # iput: strigs x,y i the set {'a','b','c'} # output: the complemet i {'a','b','c'} if set([x,y])==set(['a','b']): retur 'c' elif set([x,y])==set(['a','c']): retur 'b' else: retur 'a' def SolvigHaoi(,x,y): # iput: iteger (umber of disks) strigs x,y i the set {'a','b','c'} # output: sequece of moves if ==: prit(str(x)+' -> '+str(y)) else: SolvigHaoi(-,x,ThirdRod(x,y)) prit(str(x)+' -> '+str(y)) SolvigHaoi(-,ThirdRod(x,y),y) SolvigHaoi(3,'a','c') a -> c a -> b c -> b a -> c b -> a b -> c a -> c