online compiler and debugger for c/c++

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#[TEX]\text{Cas(m,n)=}\prod_{j=1}^{\lfloor \frac{m}{2} \rfloor} \prod_{k=1}^{\lfloor {n} \rfloor}2 sqrt{\left(\cos^2\left(\frac{\pi \cdot j}{m+1}\right) + \cos^2\left(\frac{\pi \cdot k}{n+1}\right)\right)}[/TEX] # source : http://math.uchicago.edu/~may/REU2015/REUPapers/Borys.pdf import math from itertools import combinations # Définir la fonction def fonction(m, n, pi, k, j): return 2*math.sqrt(math.cos((pi * j) / (m + 1))**2 + math.cos((pi * k) / (n + 1))**2) # Définir les valeurs de m et n m = 8 # pair n = 7 # pair ou impair pi_value = math.pi # Calculer le produit def cas(m,n): resultat = 1 for j_value in range(1, m // 2 + 1): for k_value in range(1, n + 1): resultat *= fonction(m, n, pi_value, k_value, j_value) return resultat # Afficher le résultat C0 = cas(8,7) print("\nPavages possibles = ", C0) C1 = cas(8,6) print("\nnbr de cas route en L1 = ", C1) C2 = cas(8,5) # corr : + 1 était faux print( "nbr de cas route en L2 = ", C2) C3 = (cas(8,2) + cas(8,4)) print( "nbr de cas route en L3 = ", C3) C4 = 2 * cas(8,3) print( "nbr de cas route en L4 = ", C4) #C5=C3, C6=C2, C7=C1 P1 = C1 / C0 print("\nP1 route en ligne_1 = ", P1) P2 = C2 / C0 print( "P2 route en ligne_2 = ", P2) P3 = C3 / C0 print( "P3 route en ligne_3 = ", P3) P4 = C4 / C0 print( "P4 route en ligne_4 = ", P4) #P5=P3, P6=P2, P7=P1 P = P4 + 2 *(P1 + P2 + P3) print("\nP au moins 1 route H = ", P) # Cas Liet Cas(8,n = 2 to 7) for i in [2,3,4,5,6,7]: print (f"cas(8C, {i}L) = ", cas(8,i)) # Methode S_i , C_i S_1 = 0 S_2 = cas(m, 2) - 1 S_3 = cas(m, 3) - 2 * S_2 - 1 S_4 = cas(m, 4) - 3 * S_3 - 2 * S_2 - 1 S_5 = cas(m, 5) - 2 * S_4 - 2 * S_3 - 2 * S_2 - S_2**2 - 1 S_6 = cas(m, 6) - 2 * S_5 - 2 * S_4 - 2 * S_3 - 2 * S_2 - S_2**2 - 2 * S_2 * S_3 - 1 S_7 = cas(m, 7) - 2 * S_6 - 2 * S_5 - 2 * S_4 - 2 * S_3 - 2 * S_2 - S_3**2 - S_2**2 - 2 * S_2 * S_4 - 2 * S_2 * S_3 - 1 print ("\nC_7 = ",cas(m, 7)) print ( "S_7 = ",S_7) print ( "Cas Favorables C_7 - S_7 = ", cas(m, 7) - S_7) print ( "Proba = (C_7 - S_7)/C_7 = ", (cas(m, 7) - S_7)/cas(m, 7)) # Poincaré Crible traduit en Python par GPT def Pav(n): nb = 1 for j in range(1, 5): nb *= math.prod(4 * math.cos(j * math.pi / 9) ** 2 + 4 * math.cos(k * math.pi / (n + 1)) ** 2 for k in range(1, math.ceil(n / 2) + 1)) return round(nb) def PLH(L, n, Pav_list): nb = Pav_list[L[0]] for i in range(len(L) - 1): nb *= Pav_list[L[i + 1] - L[i] - 1] nb *= Pav_list[n - 1 - L[-1]] return nb def H(n): Pav_list = [Pav(n) for n in range(8)] print("Pav_list:", Pav_list) H_val = 0 for k in range(n): C = combinations(range(n), k + 1) H_val += (-1) ** k * sum(PLH(L, n, Pav_list) for L in C) return H_val print ("\n\nCrible de Poincaré") result = H(7) print (result)

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