Algorithm for finding subsets. This program outputs empty lists, since it most likely takes the global variable 'subset'. How to take a local variable 'subset' and add it to the list?
subsets = []
subset = []
n = int(input())
k = int(input())
def search(k):
if k == n+1:
subsets.append(subset)
else:
subset.append(k)
search(k+1)
subset.pop()
search(k+1)
search(k)
print(subsets)
You need to pass subset as a parameter, or else the global variable will be used.
def search(k, subset):
if k == n+1:
subsets.append(subset)
else:
subset.append(k)
search(k+1, subset)
subset.pop()
search(k+1, subset)
Related
This is a standart permutation function. Im tring to return the list of the lists of the permutations)
Could you help me with storaging the result of the recursion iterations? for example this code returns nonsense. It would be perfect if there was no global variable and rezulting list was inside the func
Thanks!
'''
z=[]
def func(N,M=-1,pref=None):
global z
if M == -1:
M = N
pref = pref or []
if M==0:
z.append(pref)
print(pref)
for i in range(N):
if i not in pref:
pref.append(i)
func(N,M-1,pref)
pref.pop()
func(3)
print(z)
'''
You are passing a list (pref variable in for loop) reference to your function and you are removing a single item from that and that's why you are ending with an empty list z.
Create a new list or copy the list before passing it to the function to avoid this situation.
z = []
def func(N, M=-1, pref=None):
global z
if M == -1:
M = N
pref = pref or []
if M == 0:
z.append(pref)
print(pref)
for i in range(N):
if i not in pref:
pref.append(i)
func(N, M - 1, pref[:])
pref.pop()
func(3)
print(z)
For better understand please read this one. List changes unexpectedly after assignment. How do I clone or copy it to prevent this?
If you want to have some kind of accumulator you must pass it to the recursion function, beware it could be a little nightmare.
I have an excercise about prime numbers that requires me to write a function which takes a list of elements and a number p and marks elements False which are in the range 2p, 3p...N
First I create a list of True and False:
true_value = [False, False] + [True for x in range(n-1)] #Let assumme that n=16
And then I write the function that find the even number in this list (with p = 2)
def mark_false(bool_list, p):
range_new = [x for x in range(len(bool_list))]
for i in range(2, len(range_new)):
for j in range(p, len(range_new), p):
if (i*p == range_new[j]) & (i*p <= len(range_new)):
bool_list[j] = False
return bool_list
This function help me to find the location of the even number (>2) and return to False
Example: a = list_true(16)
a = [False,False,True,True,True,True,True,True,True,True,True,True,True,True,True,True,True]
b = mark_false(a, 2)
b = [False,False,True,True,False,True,False,True,False,True,False,True,False,True,False,True]
This function mark_false does work but the problem is everytime I run it I have to create a list range_new which takes a lot of time to calculate. So how do I rewrite this function so it can run faster without creating new lists?
You seem to be doing things the long way around, searching for the j value that matches the multiple of p you want to set to False. But since you already know that value already, there's no need to search for it, just set it directly.
I'd do:
def mark_false(bool_list, p):
for i in range(p, len(bool_list), p): # p, 2*p, 3*p, ...
bool_list[i] = False # do the assignment unconditionally
You probably shouldn't need a return statement, since you're modifying the list you are passed in-place. Returning the list could make the API misleading, as it might suggest that the returned list is a new one (e.g. a modified copy).
If you did want to return a new list, you could create one with a list comprehension, rather than modifying the existing list:
def mark_false_copy(bool_list, p):
return [x if i % p else False for i, x in enumerate(bool_list)]
So, I was given this question where have to enter two numbers and find the binary representation of all the numbers between them and we have to check if there is consecutive 1's in the binary form, if present then skip otherwise print it.
I have tried this much I have created the required list of binary representation but now how to iterate over every number and check for that 1's possibility??
n = int(input())
m = int(input())
str=[]
for i in range(n+1,m):
x = bin(i)
str.append(x)
print(str)
Here is the output
3
7
['0b100', '0b101', '0b110']
So something like this:
def solve(a, b):
result = []
for x in range(a, b):
binary_str = bin(x)
if '11' not in binary_str:
result.append(binary_str)
return result
a = int(input())
b = int(input())
r = solve(a, b)
print(r) # ['0b100', '0b101']
Note that bin does not return binary but a binary representation (i.e. a string).
A slightly less efficient (because bin() is called twice) solution but shorter:
def solve(a, b):
return [bin(x) for x in range(a, b) if '11' not in bin(x)]
print(*solve(int(input()), int(input()))) # 0b100 0b101
POwer in Python. How to write code to display a ^ n using funсtion?
why doesn't this code working?
a = int(input())
n = int(input())
def power(a, n):
for i in range (n):
a=1
a *= n
print(power (a, n))
Few errors:
Changing a will lose your power parameter, use result (or something else).
Move setting result = 1 outside your loop to do so once.
Multiply by a not by n.
Use return to return a value from the function
def power(a, n):
result = 1 # 1 + 2
for _ in range (n):
result *= a # 3
return result # 4
Style notes:
Setting/mutating a parameter is considered bad practice (unless explicitly needed), even if it is immutable as is here.
If you're not going to use the loop variable, you can let the reader know by using the conventional _ to indicate it (_ is a legal variable name, but it is conventional to use it when not needing the variable).
Tip: you can simple use a**n
It doesn't work because your function doesn't return the end value. Add return a to the end of the function.
ALSO:
That is not how a to the power of n is is calculated.
A proper solution:
def power(a,n):
pow_a = a
if n is 0:
return 1
for _ in range(n-1): # Substracting 1 from the input variable n
pow_a *= a # because n==2 means a*a already.
return pow_a
and if you want to be really cool, recursion is the way:
def power_recursive(a,n):
if n is 0:
return 1
elif n is 1:
return a
else:
a *= power_recursive(a,n-1)
return a
from math import ceil
def merge(all_lst):
sorted_lst = []
while all_lst:
min_value,index = all_lst[0][0],0
for lst in all_lst:
if lst[0]<min_value:
min_value = lst[0]
index = all_lst.index(lst)
sorted_lst.append(min_value)
all_lst[index].pop(0)
if not all_lst[index]:
all_lst.remove(all_lst[index])
return sorted_lst
def merge_sort(lst, k):
def split(lst):
split_lst = []
j = ceil(len(lst)/k) if len(lst)>=k else 1
for i in range(0,len(lst),j):
split_lst.append(lst[i:i+j])
return split_lst
lst=split(lst)
if len(lst[0])==1:
return lst
else:
for i in range(len(lst)):
lst[i]=merge(merge_sort(lst[i],k))
return merge(lst)
Above is my code for k-way merge sort. Basically what it does is split the list into k smaller list by calling the split function until each sublist in the list is a single element. Then the list containing sublists will be merged into one single list.
My code works fine when splitting is done twice. (eg.[3,6,8,5,2,1,4,7] -> [3,6,8],[5,2,1],[4,7] -> [3],[6],[8],[5],[2],[1],[4],[7]). But when the splitting is done more than twice, (eg,[3,6,8,5,2,1,4,7] -> [3,6,8,5],[2,1,4,7] -> [3,6],[8,5],[2,1],[4,7] -> [3],[6],[8],[5],[2],[1],[4],[7]), the code will fail. Can anyone help find me find out what goes wrong in my code? Thanks in advance.
I believe the problem you're having is that merge_sort sometimes returns a flattened list and other times returns a list of lists. You should probably return a flat list in all cases. There's some other cruft: You don't need split to be its own function, since you only call it the one time.
Here's a greatly simplified version of your code:
def merge_sort(lst, k):
if len(lst) == 1: # simpler base case
return lst
j = ceil(len(lst)/k) # no need to check for k < len(lst) (ceil handles it)
#split and recursively sort in one step
lst = [merge_sort(lst[i:i+j], k) for i in range(0, len(lst), j)]
return merge(lst) # always return a merged list (never a list of lists)