Option info + IPT
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@ -11,97 +11,187 @@ class Question:
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def __exit__(self, exc_type, exc_value, exc_traceback):
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print((self.level*2)*' ' + f"<- {self.number}. : {self.name} -- Fin\n")
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if False:
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with Question(1):
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with Question(1,1):
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def dichotomie(f, n, a=0, b=2):
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iterations = 0
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while abs(b-a) >= 2*10**(-n):
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iterations += 1
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c = (a+b) / 2
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if f(c) == 0:
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return c
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elif f(a)*f(c) > 0:
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a = c
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else:
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b = c
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return c, iterations
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f = lambda x: x**3 - 3*x**2 + 1
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print(dichotomie(f,5,0,1.5))
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print(dichotomie(np.sin, 3, 3, 4))
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print(dichotomie(np.sin, 10, 3, 4))
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with Question(2,1):
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def newton(f, fp, x0, nbiter, epsilon=1e-10):
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for i in range(nbiter):
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x1 = x0 - f(x0) / fp(x0)
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if abs(x1 - x0) <= epsilon:
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break
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x0 = x1
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return x1
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print(newton(lambda x: x**3 - 3*x**2 + 1, lambda x: 3*x**2 - 6*x, 1.5, 200, epsilon=1e-100))
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print(newton(np.sin, np.cos, 3, 100))
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print(abs(newton(np.sin, np.cos, 3, 100) - dichotomie(np.sin, 10, 3, 4)[0]))
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with Question(2):
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from math import cos, exp, sin, log
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from scipy.integrate import quad
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from time import time
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with Question(1,2):
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def rectangles(f, a, b, n):
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res = 0
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for k in range(n):
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res += f(a + k*(b - a)/n)
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return res*(b - a)/n
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def f1(x):
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return x
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def f2(x):
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return x**10
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f3 = cos
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f4 = exp
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print(rectangles(f1, 0, 1, 1000), quad(f1, 0, 1)[0])
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print(rectangles(f2, 0, 1, 1000), quad(f2, 0, 1)[0])
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print(rectangles(f3, 0, np.pi/2, 1000), quad(f3, 0, np.pi/2)[0])
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print(rectangles(f4, -3, 3, 1000), quad(f4, -3, 3)[0])
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with Question(2,2):
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def trapezes(f, a, b, n):
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res = 0
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for k in range(n):
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res += f(a + k*(b - a)/n) + f(a + (k+1)*(b - a)/n)
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return res*(b - a)/(2*n)
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print(trapezes(f1, 0, 1, 1000), quad(f1, 0, 1)[0])
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print(trapezes(f2, 0, 1, 1000), quad(f2, 0, 1)[0])
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print(trapezes(f3, 0, np.pi/2, 1000), quad(f3, 0, np.pi/2)[0])
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print(trapezes(f4, -3, 3, 1000), quad(f4, -3, 3)[0])
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with Question(3, 2):
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def simpson(f, a, b, n):
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res = 0
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for k in range(n):
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res += f(a + k*(b - a)/n) + 4*f(((a+k*(b - a)/n) + (a + (k+1)*(b - a)/n))/2) + f(a + (k+1)*(b - a)/n)
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return res*(b - a)/(6*n)
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print(simpson(f1, 0, 1, 1000), quad(f1, 0, 1)[0])
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print(simpson(f2, 0, 1, 1000), quad(f2, 0, 1)[0])
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print(simpson(f3, 0, np.pi/2, 1000), quad(f3, 0, np.pi/2)[0])
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print(simpson(f4, -3, 3, 1000), quad(f4, -3, 3)[0])
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with Question(3):
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import matplotlib.pyplot as plt
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with Question(1, 2):
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def euler(F, t0, y0, t1, n):
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res = [y0]*(n+1)
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for i in range(1,n+1):
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res[i] = res[i-1] + (t1-t0)/n * F(t0 + i*(t1-t0)/n, res[i-1])
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return res
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def expo_euler(t, yt):
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return yt
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ne = 100
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n = 10
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plt.plot(np.linspace(0, 1, num=n+1), euler(expo_euler, 0, 1, 1, n), 'bo')
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plt.plot(np.linspace(0, 1, num=ne+1), np.exp(np.linspace(0, 1, num=ne+1)), 'r')
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plt.show()
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print("--- Calcul Matriciel ---\n\n")
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with Question(1):
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with Question(1,1):
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def dichotomie(f, n, a=0, b=2):
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iterations = 0
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while abs(b-a) >= 2*10**(-n):
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iterations += 1
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c = (a+b) / 2
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if f(c) == 0:
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return c
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elif f(a)*f(c) > 0:
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a = c
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with Question(2, 2):
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with Question(1,3):
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def matrice_nulle(n, m):
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return [[0 for _ in range(m)] for _ in range(n)]
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with Question(2,3):
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def dimension(A):
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if len(A):
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return len(A),len(A[0])
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else:
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b = c
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return c, iterations
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f = lambda x: x**3 - 3*x**2 + 1
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print(dichotomie(f,5,0,1.5))
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print(dichotomie(np.sin, 3, 3, 4))
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print(dichotomie(np.sin, 10, 3, 4))
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with Question(2,1):
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def newton(f, fp, x0, nbiter, epsilon=1e-10):
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for i in range(nbiter):
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x1 = x0 - f(x0) / fp(x0)
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if abs(x1 - x0) <= epsilon:
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break
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x0 = x1
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return x1
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print(newton(lambda x: x**3 - 3*x**2 + 1, lambda x: 3*x**2 - 6*x, 1.5, 200, epsilon=1e-100))
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print(newton(np.sin, np.cos, 3, 100))
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print(abs(newton(np.sin, np.cos, 3, 100) - dichotomie(np.sin, 10, 3, 4)[0]))
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with Question(2):
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from math import cos, exp, sin, log
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from scipy.integrate import quad
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from time import time
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with Question(1,2):
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def rectangles(f, a, b, n):
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res = 0
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for k in range(n):
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res += f(a + k*(b - a)/n)
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return res*(b - a)/n
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def f1(x):
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return x
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def f2(x):
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return x**10
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f3 = cos
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f4 = exp
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print(rectangles(f1, 0, 1, 1000), quad(f1, 0, 1)[0])
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print(rectangles(f2, 0, 1, 1000), quad(f2, 0, 1)[0])
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print(rectangles(f3, 0, np.pi/2, 1000), quad(f3, 0, np.pi/2)[0])
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print(rectangles(f4, -3, 3, 1000), quad(f4, -3, 3)[0])
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with Question(2,2):
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def trapezes(f, a, b, n):
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res = 0
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for k in range(n):
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res += f(a + k*(b - a)/n) + f(a + (k+1)*(b - a)/n)
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return res*(b - a)/(2*n)
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print(trapezes(f1, 0, 1, 1000), quad(f1, 0, 1)[0])
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print(trapezes(f2, 0, 1, 1000), quad(f2, 0, 1)[0])
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print(trapezes(f3, 0, np.pi/2, 1000), quad(f3, 0, np.pi/2)[0])
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print(trapezes(f4, -3, 3, 1000), quad(f4, -3, 3)[0])
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return 0
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with Question(3,3):
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def addition(A, B):
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n,m = dimension(A)
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return [[A[i][j] + B[i][j] for j in range(m)] for i in range(n)]
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A3=[[1,2,3],
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[0,1,0],
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[-1,2,-6]]
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B3=[[-1,5,0],
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[1,0,-2],
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[1,2,3]]
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print(addition(A3, B3), '\n', np.array(A3)+np.array(B3))
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print("Le résultat pour cette exemple est le même.")
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with Question(4,3):
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def transposee(A):
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n,m = dimension(A)
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return [[A[j][i]for j in range(n)] for i in range(m)]
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print(transposee([[1,2,3],
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[4,5,6]]))
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print(list(map(list,zip(*[[1,2,3],[4,5,6]])))) # Parce que c'est amusant de chercher à faire une version courte.
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with Question(5,3):
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def multiple(A, coef):
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n,m = dimension(A)
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return [[coef * A[i][j] for j in range(m)] for i in range(n)]
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print(multiple([[1,2],
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[3,4]], 2))
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with Question(6,3):
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def multiplication(A, B):
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n,p = dimension(A)
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_,m = dimension(B)
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return [[sum(A[i][k] * B[k][j] for k in range(p))for j in range(m)] for i in range(n)]
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A6 = [[1,2,3],
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[4,5,6]]
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B6 = [[3,6],
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[2,5],
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[1,4]]
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print(multiplication(A6,B6))
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with Question(7,3):
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def puissance(A, n):
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_,m = dimension(A)
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res = [[1 if j==i else 0 for j in range(m)] for i in range(m)]
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for _ in range(0, n):
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print(n)
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res = multiplication(res, A)
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return res
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print(puissance([[0,1],
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[0,0]], 2))
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with Question(3, 2):
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def simpson(f, a, b, n):
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res = 0
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for k in range(n):
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res += f(a + k*(b - a)/n) + 4*f(((a+k*(b - a)/n) + (a + (k+1)*(b - a)/n))/2) + f(a + (k+1)*(b - a)/n)
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return res*(b - a)/(6*n)
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print(simpson(f1, 0, 1, 1000), quad(f1, 0, 1)[0])
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print(simpson(f2, 0, 1, 1000), quad(f2, 0, 1)[0])
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print(simpson(f3, 0, np.pi/2, 1000), quad(f3, 0, np.pi/2)[0])
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print(simpson(f4, -3, 3, 1000), quad(f4, -3, 3)[0])
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with Question(3):
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import matplotlib.pyplot as plt
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with Question(1, 2):
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def euler(F, t0, y0, t1, n):
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res = [y0]*(n+1)
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for i in range(1,n+1):
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res[i] = res[i-1] + (t1-t0)/n * F(t0 + i*(t1-t0)/n, res[i-1])
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return res
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def expo_euler(t, yt):
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return yt
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n = 100
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plt.plot(np.linspace(0, 1, num=n+1), euler(expo_euler, 0, 1, 1, n))
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plt.plot(np.linspace(0, 1, num=n+1), np.exp(np.linspace(0, 1, num=n+1)))
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plt.show()
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with Question(2,3):
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with Question(1,4):
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def echange_ligne(A, i, j):
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A[i], A[j] = A[j], A[i]
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A1 = [[1,2,3],
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[4,5,6]]
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echange_ligne(A1, 0, 1)
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print(A1)
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with Question(2,4):
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def transvection(A, i, j, mu):
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A[i] = addition([A[i]], multiple([A[j]], mu))[0]
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transvection(A1, 0, 1, -1)
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print(A1)
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with Question(3, 4):
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def pivot_partiel(A, j0):
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n, m = dimension(A)
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if n<j0 or m<j0:
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raise IndexError("j0 trop grand")
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maxi = A[j0][j0]
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for i in range(j0, n):
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maxi = max(A[i][j0], maxi)
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return maxi
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print(pivot_partiel([[1,2,3],
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[4,5,6],
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[7,3,9]], 1))
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with Question(4,4):
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def devient_triangle(A):
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pass
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28
SPE/OPT/Logique.ml
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28
SPE/OPT/Logique.ml
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@ -0,0 +1,28 @@
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type variables = char;;
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type formule =
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| Var of variables
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| Non of formule
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| Implique of formule*formule
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| Equivaut of formule*formule
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| Et of formule list
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| Ou of formule list;;
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let f = Implique (Et [Var 'p';Non (Var 'q')], Ou [Var 'p';Non (Var 'r')]);;
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let exemple_de_contexte v = match v with
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|'p' -> true
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|'q' -> false
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|'r' -> false
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|_ -> true;;
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let rec interpretation contexte formule = match formule with
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| Var v-> contexte v
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| Non f -> not (interpretation contexte f)
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| Implique (a,b) -> not (interpretation contexte a) || (interpretation contexte b)
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| Equivaut (a,b) -> not (interpretation contexte a) = (interpretation contexte b)
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| Et [] -> true
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| Et (h::t) -> (interpretation contexte h) && (interpretation contexte (Et t))
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| Ou [] -> false
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| Ou (h::t) -> (interpretation contexte h) || (interpretation contexte (Ou t));;
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interpretation (exemple_de_contexte) (Et [Var 'q'; Non (Var 'r')]);;
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12
SPE/OPT/TD 1/main.ml
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12
SPE/OPT/TD 1/main.ml
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@ -0,0 +1,12 @@
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(*1 *)
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let rec est_un_systeme c = match c with
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| [] -> false
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| [h] -> h=1
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| h1::h2::t -> h1 > h2 && est_un_systeme (h2::t);;
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est_un_systeme [3;4;3];;
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(*2*)
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(* Mq $x \geq M(x)$ par l'absurde
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Supposons qu'on a M(x) > x
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alors il existe $k \in \mathbb{N}^m$ tel que
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41
SPE/OPT/TD 1/main.tex
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41
SPE/OPT/TD 1/main.tex
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\documentclass[a4paper,10pt]{article}
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%\documentclass[a4paper,10pt]{scrartcl}
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\usepackage[utf8]{inputenc}
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\usepackage{mathtools}
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\usepackage{amssymb}
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\usepackage{amsmath}
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\title{TD Option Info}
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\author{}
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\date{}
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\pdfinfo{%
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/Title (TD option info)
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/Author ()
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/Creator ()
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/Producer ()
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/Subject ()
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/Keywords ()
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}
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\begin{document}
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\maketitle
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\section{Systèmes de pièces}
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\subsection{}
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\section{}
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Montrons qu'on a $ x \geq M(x) $\\
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$M(x)$ est défini, alors il existe $k \in \mathbb{N}^m$ tel que $x = \sum_{i=1}^m k_i c_i$ avec $\sum_{i=1}^m k_i = M(x)$\\
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On sait que $\forall\text{ } 1\leq i\leq m, c_i >= 1$
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Donc $x = \sum_{i=0}^m k_i c_i \geq \sum_{i=0}^m k_i = M(x)$\\\\
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Montrons qu'on a $\lceil\frac{x}{c_1} \rceil \leq M(x)$\\
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Par définition $\lceil\frac{x}{c_1} \rceil \leq \frac{x}{c_1} + 1$\\
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Montrons qu'on a $\frac{x}{c_1} + 1 \leq M(x)$\\
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$M(x)$ est défini, alors il existe $k \in \mathbb{N}^m$ tel que $x = \sum_{i=1}^m k_i c_i$ avec $\sum_{i=1}^m k_i = M(x)$\\
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Ainsi, $\frac{x}{c_1} + 1 = \left( \sum_{i=1}^m \frac{k_i c_i}{c_1}\right) + 1 = \left( \sum_{i=2}^m \frac{k_i c_i}{c_1}\right) + k_1 + 1 $\\
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or on a $c_1 > c_2 > ... > c_m$ donc
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$\left( \sum_{i=2}^m \frac{k_i c_i}{c_1}\right) + k_1 + 1 \leq \left( \sum_{i=2}^m k_i\right) + k_1 + 1 = M(x) + 1 $\\
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CQFD.
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\subsection{}
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\end{document}
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46
SPE/OPT/TP 1/main.ml
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46
SPE/OPT/TP 1/main.ml
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(*Q1*)
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let h_exemple s = (Char.code (Char.uppercase_ascii s.[0])) - 65;;
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h_exemple "z";;
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type ('a, 'b) dict = {
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hache: 'a -> int;
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table_hachage: ('a *'b) list array;
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largeur: int};;
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(*Q2*)
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let creer_dict_vide n hache = {
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hache=hache;
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table_hachage=(Array.make n []);
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largeur=n};;
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let d_exemple = creer_dict_vide 25 h_exemple;;
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(*Q3*)
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let contient cle d =
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let h = d.hache cle in
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List.mem cle (List.map fst d.table_hachage.(h));;
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contient "blablab" d_exemple;;
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let inserer cle enregistrement d =
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if contient cle d then failwith "Déjà dans le dictionnaire" else ();
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let h = d.hache cle in
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d.table_hachage.(h) <- ((cle, enregistrement)::d.table_hachage.(h));;
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||||
let valeur cle d =
|
||||
let rec aux acc = match acc with
|
||||
|[] -> raise Not_found
|
||||
|(c, enregistrement)::t -> if c=cle then enregistrement else aux t in
|
||||
aux d.table_hachage.(d.hache cle);;
|
||||
|
||||
let supprimer cle d =
|
||||
let h = d.hache cle in
|
||||
let rec aux acc li = match li with
|
||||
| [] -> raise Not_found
|
||||
| (c, e)::t -> if c=cle then acc@t else aux ((c,e)::acc) t in
|
||||
d.table_hachage.(h) <- aux [] d.table_hachage.(h);;
|
||||
let modifier cle enregistrement d =
|
||||
supprimer cle d;
|
||||
inserer cle enregistrement d;;
|
||||
|
||||
inserer "rire" "voilà" d_exemple;;
|
||||
valeur "mais" d_exemple;;
|
||||
supprimer "rigolo" d_exemple;;
|
||||
modifier "blablab" "truc" d_exemple;;
|
||||
d_exemple;;
|
||||
42
SPE/OPT/TP 2/main.ml
Normal file
42
SPE/OPT/TP 2/main.ml
Normal file
@ -0,0 +1,42 @@
|
||||
type formule =
|
||||
| Var of int
|
||||
| Not of formule
|
||||
| Imply of formule*formule
|
||||
| Equiv of formule*formule
|
||||
| And of formule list
|
||||
| Or of formule list;;
|
||||
|
||||
let f = And [Or [Var 0; Var 1; Var 2]; Not (Var 0)];;
|
||||
let g = Equiv (Var 0, And [Var 1; Var 2]);;
|
||||
let rec evaluer_tab t form = match form with
|
||||
| Var i -> t.(i)
|
||||
| Not f -> not (evaluer_tab t f)
|
||||
| Imply (f1,f2) -> (not (evaluer_tab t f1)) || (evaluer_tab t f2)
|
||||
| Equiv (f1,f2) -> (evaluer_tab t f1) = (evaluer_tab t f2)
|
||||
| And [] -> true
|
||||
| And (h::tail) -> (evaluer_tab t h) && (evaluer_tab t (And tail))
|
||||
| Or [] -> false
|
||||
| Or (h::tail) -> (evaluer_tab t h) || (evaluer_tab t (Or tail));;
|
||||
5 lsr 1;;
|
||||
|
||||
let evaluer n c f =
|
||||
let contexte = Array.make n false in
|
||||
let j = ref c in
|
||||
for i = 0 to n-1 do
|
||||
contexte.(i) <- !j mod 2 = 0;
|
||||
j := !j lsr 1
|
||||
done;
|
||||
evaluer_tab contexte f;;
|
||||
evaluer 3 0 f;;
|
||||
|
||||
let table_de_verite n f =
|
||||
let rec puis n a = match n with
|
||||
| 0 -> 1
|
||||
| _ -> a * (puis (n-1) a) in
|
||||
let res = Array.make (puis n 2) false in
|
||||
for c=0 to ((Array.length res) -1) do
|
||||
res.(c) <- evaluer n c f
|
||||
done;
|
||||
res;;
|
||||
table_de_verite 3 g;;
|
||||
|
||||
Loading…
Reference in New Issue
Block a user