guacamole LDAP creation
0 likes • Nov 18, 2022 • 0 views
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import anytree as atimport random as rm# Generate a tree with node_count many nodes. Each has a number key that shows when it was made and a randomly selected color, red or white.def random_tree(node_count):# Generates the list of nodesnodes = []for i in range(node_count):test = rm.randint(1,2)if test == 1:nodes.append(at.Node(str(i),color="white"))else:nodes.append(at.Node(str(i),color="red"))#Creates the various main branchesfor i in range(node_count):for j in range(i, len(nodes)):test = rm.randint(1,len(nodes))if test == 1 and nodes[j].parent == None and (not nodes[i] == nodes[j]):nodes[j].parent = nodes[i]#Collects all the main branches into a single tree with the first node being the rootfor i in range(1, node_count):if nodes[i].parent == None and (not nodes[i] == nodes[0]):nodes[i].parent = nodes[0]return nodes[0]
from itertools import productV='∀'E='∃'def tt(f,n) :xss=product((0,1),repeat=n)print('function:',f.__name__)for xs in xss : print(*xs,':',int(f(*xs)))print('')# p \/ (q /\ r) = (p \/ q) /\ (p \/ r)def prob1(p,q,r) :x=p or (q and r)y= (p or q) and (p or r)return x==ytt(prob1,3)# p/\(q\/r)=(p/\q)\/(p/\r)def prob2(p,q,r) :x=p and ( q or r )y=(p and q) or (p and r)return x==ytt(prob2,3)#~(p/\q)=(~p\/~q)def prob3(p,q) :x=not (p and q)y=(not p) or (not q)return x==ytt(prob3,2)#(~(p\/q))=((~p)/\~q)def prob4(p, q):x = not(p or q)y = not p and not qreturn x == ytt(prob4, 2)#(p/\(p=>q)=>q)def prob5(p,q):x= p and ( not p or q)return not x or qtt(prob5,2)# (p=>q)=((p\/q)=q)def prob6(p,q) :x = (not p or q)y=((p or q) == q)return x==ytt(prob6,2)#((p=>q)=(p\/q))=qdef prob7(p,q):if ((not p or q)==(p or q))==q:return 1tt(prob7,2)#(p=>q)=((p/\q)=p)def prob8(p,q):if (not p or q)==((p and q)==p):return 1tt(prob8,2)#((p=>q)=(p/\q))=pdef prob9(p,q):if ((not p or q)==(p and q))==p:return '1'tt(prob9,2)#(p=>q)/\(q=>r)=>(p=>r)def prob10(p,q,r) :x = not ((not p or q) and (not q or r)) or (not p or r)return xtt(prob10, 3)# (p = q) /\ (q => r) => (p => r)#answer 1def prob11(p,q,r) :x = not((p is q) and (not q or r)) or (not p or r)return xtt(prob11, 3)#(p=q)/\(q=>r)=>(p=>r)#answer 2def prob11(p,q,r):x=(p==q) and (not q or r)y=not p or rreturn not x or ytt(prob11,3)#((p=>q)/\(q=r))=>(p=>r)def prob12(p,q,r):x=(not p or q) and ( q==r )y=not p or rreturn not x or ytt(prob12,3)#(p=>q)=>((p/\r)=>(q/\r))def prob13(p,q,r):x=not p or qy=(not(p and r) or ( q and r))return not x or ytt(prob13,3)#Question#2----------------------------------------#(p=>q)=>r=p=>(q=>r)def prob14(p,q,r):x=(not(not p or q) or r)y=(not p or (not q or r))return x==ytt(prob14,3)def prob15(p, q):x = not(p and q)y = not p and not qreturn x == ytt(prob15, 2)def prob16(p, q):x = not(p or q)y = not p or not qreturn x == ytt(prob16, 2)def prob17(p):x = py = not preturn x == ytt(prob17, 1)
from collections import defaultdictdef collect_dictionary(obj):inv_obj = defaultdict(list)for key, value in obj.items():inv_obj[value].append(key)return dict(inv_obj)ages = {'Peter': 10,'Isabel': 10,'Anna': 9,}collect_dictionary(ages) # { 10: ['Peter', 'Isabel'], 9: ['Anna'] }
def max_n(lst, n = 1):return sorted(lst, reverse = True)[:n]max_n([1, 2, 3]) # [3]max_n([1, 2, 3], 2) # [3, 2]
# Deleting all even numbers from a lista = [1,2,3,4,5]del a[1::2]print(a)
import itertoolsdef compute_permutations(string):# Generate all permutations of the stringpermutations = itertools.permutations(string)# Convert each permutation tuple to a stringpermutations = [''.join(permutation) for permutation in permutations]return permutations# Prompt the user for a stringstring = input("Enter a string: ")# Compute permutationspermutations = compute_permutations(string)# Display the permutationsprint("Permutations:")for permutation in permutations:print(permutation)