I was trying to get this recursion faster but when I use numbers 50 and 44.4 it takes too long my desired outcome for those numbers is -800555.6302016332
z = int(input())
x = float(input())
def rec(n):
global x
l = {}
if n == 0:
return -1
elif n == 1:
return x
elif n == 2:
return -(x+1)/3
else:
if n in l:
return l[n]
value = float((n/x)*rec(n-1) + ((-1)**n)*((n+1)/(n-1)) * rec(n-2) + ((n-1)/(2*x))*rec(n-3))
l[n] = value
return value
print(rec(z))
You are reinitializing your dictionary l = {} each time you recurse. Making l a global var should fix your problem:
l = {}
def rec(n):
global x
global l
if n == 0:
return -1
elif n == 1:
return x
elif n == 2:
return -(x+1)/3
else:
if n in l:
return l[n]
value = float((n/x)*rec(n-1) + ((-1)**n)*((n+1)/(n-1)) * rec(n-2) + ((n-1)/(2*x))*rec(n-3))
l[n] = value
return value
You could also use functools.lru_cache which does memoization for you:
import functools
#functools.lru_cache
def rec(n):
global x
if n == 0:
return -1
elif n == 1:
return x
elif n == 2:
return -(x+1)/3
else:
return float((n/x)*rec(n-1) + ((-1)**n)*((n+1)/(n-1)) * rec(n-2) + ((n-1)/(2*x))*rec(n-3))
I would also suggest avoiding the use of global variables:
import functools
def rec(n, x):
#functools.lru_cache
def recurse(n):
if n == 0:
return -1
elif n == 1:
return x
elif n == 2:
return -(x+1)/3
else:
return float((n/x)*recurse(n-1) + ((-1)**n)*((n+1)/(n-1)) * recurse(n-2) + ((n-1)/(2*x))*recurse(n-3))
return recurse(n)
def main():
n = int(input())
x = float(input())
print(rec(n, x))
if __name__ == "__main__":
main()
Related
I try to make a binary heap in python but I get trouble printing it. I make sure that the logic in the program is right but when I want to try printing it I get the wrong result. This is what I want for program output:
Input:
6
1 2
1 3
1 7
2
3
2
Output:
7
3
command 1 to insert number
command 2 to display the highest number
command 3 to delete the highest number
This is my program:
class BinaryHeap:
def __init__(self):
self.items = []
def size(self):
return len(self.items)
def parent(self, i):
return (i - 1)//2
def left(self, i):
return 2*i + 1
def right(self, i):
return 2*i + 2
def get(self, i):
return self.items[i]
def get_max(self):
if self.size() == 0:
return None
return self.items[0]
def extract_max(self):
if self.size() == 0:
return None
largest = self.get_max()
self.items[0] = self.items[-1]
del self.items[-1]
self.max_heapify(0)
return largest
def max_heapify(self, i):
l = self.left(i)
r = self.right(i)
if (l <= self.size() - 1 and self.get(l) > self.get(i)):
largest = l
else:
largest = i
if (r <= self.size() - 1 and self.get(r) > self.get(largest)):
largest = r
if (largest != i):
self.swap(largest, i)
self.max_heapify(largest)
def swap(self, i, j):
self.items[i], self.items[j] = self.items[j], self.items[i]
def insert(self, key):
index = self.size()
self.items.append(key)
while (index != 0):
p = self.parent(index)
if self.get(p) < self.get(index):
self.swap(p, index)
index = p
bheap = BinaryHeap()
n = int(input())
for i in range (n):
operation= input('What would you like to do? ').split()
if operation == 1:
data = int(operation[1])
bheap.insert(data)
elif operation == 2:
print('Maximum value: {}'.format(bheap.get_max()))
elif operation == 3:
print('Maximum value removed: {}'.format(bheap.extract_max()))
elif operation == 4:
break
I need your opinion to fixed it.
operation is a list (you called split), but you compare it as an int in your if statements. Also, you should compare it against "1", "2", ... not 1, 2, ...
So:
operation = input('What would you like to do? ').split()
if operation[0] == "1":
data = int(operation[1])
bheap.insert(data)
elif operation[0] == "2":
print('Maximum value: {}'.format(bheap.get_max()))
elif operation[0] == "3":
print('Maximum value removed: {}'.format(bheap.extract_max()))
elif operation[0] == "4":
break
If you just want 7 and 3 in the output, and only after the input has been completely processed, then you should remove those verbose print statement (where you output phrases), and instead collect the output -- for instance in a list:
output = []
n = int(input())
for i in range(n):
operation = input('What would you like to do? ').split()
if operation[0] == "1":
data = int(operation[1])
bheap.insert(data)
elif operation[0] == "2":
output.append(bheap.get_max())
elif operation[0] == "3":
bheap.extract_max()
elif operation[0] == "4":
break
print("\n".join(map(str, output)))
I just want to know if there is a simple way to convert an int(negative number) to a list
Example
a = -2021
result : [-2,0,2,1]
def numberToList(n):
# base case
if n == 0:
return []
# recurse
return numberToList(n // 10) + [ n % 10 ]
def numberInput(n):
if n < 0:
n = numberToList(abs(n))
n[0] = n[0] * -1
return n
else:
return numberToList(n)
Just use a function like this:
from typing import List
def num_list(num: int) -> List[int]:
if num == 0:
return []
ns = [int(i) for i in str(num).lstrip('-')]
if num < 0:
ns[0] *= -1
return ns
I've made a hexadecimal converter to practice recursion/recursive thinking. I, however, The recurssion doesn't appear to be happening as the functions seems to just output the result of 9 as of current.The code is as follows:
import math
curr=0
def convert(x):
L=len(x)
L-=1
sol=0
if L == 0:
return 0
else:
if x[curr]==["A","a"]:
v=10
elif x[curr]==["B","b"]:
v=11
elif x[curr]==["C","c"]:
v=12
elif x[curr]==["D","d"]:
v=13
elif x[curr]==["E","e"]:
v=14
elif x[curr]==["F","f"]:
v=15
else:
v=int(x[curr])
sol+=((v)*(16**(L-1)))
return sol + convert(x[curr+1])
def main():
print(convert('98A'))
main()
You were setting L = len(x) everytime you call the function. Here is one solution:
import math
def convert(x, L):
c = len(x) - 1
sol=0
if L > c:
return 0
else:
if (x[L]=="A" or x[L]=="a"):
v=10
elif (x[L]=="B" or x[L]=="b"):
v=11
elif (x[L]=="C" or x[L]=="c"):
v=12
elif (x[L]=="D" or x[L]=="d"):
v=13
elif (x[L]=="E" or x[L]=="e"):
v=14
elif (x[L]=="F" or x[L]=="f"):
v=15
else:
v=int(x[L])
sol+=((v)*(16**(c - L)))
print(sol)
return sol + convert(x, L + 1)
def main():
print(convert('98A', 0))
main()
You can use something like this:
class HexMap:
# mapping char to int
d = { hex(n)[2:]:n for n in range(16)}
def convert(x):
s = 0
# use reverse string and sum up - no need for recursion
for i,c in enumerate(x.lower()[::-1]):
s += HexMap.d[c]*16**i
return s
def main():
print(convert('98A'))
main()
Output:
2442
Recursive version:
# class HexMap: see above
def convert(x):
def convert(x,fak):
if not x:
return 0
else:
return HexMap.d[x[-1]]*16**fak + convert(x[:-1],fak+1)
return convert(x.lower(),0)
def main():
print(convert('98A'))
main()
Same output.
I'm trying to write a function that would recursively hash a key for n times, alternating between sha224 and sha256. Each iteration would be hash_256(hash_224)--a hash256 for the hash224 of the key--so that it would yield n * (hash_256(hash_224)). However, I'm new to coding and can't figure out how to write a recursive function with these parameters.
import hashlib
def shasum(key, n):
key = str(key).encode('utf-8')
hash_a = hashlib.sha224(key).hexdigest().encode('utf-8'))
hash_b = hashlib.sha256(hash_a).hexdigest()
if n == 0 or 1:
return hash_b #one iteration of 256(224)
else:
return n-1
return hash_b #stuck here
Edited: now it behaves like a number generator. What's wrong?
import hashlib
n = 0
def sha480(seed):
hashed_224 = str(hashlib.sha224(seed)).encode('utf-8')
hashed_256 = hashlib.sha256(hashed_224).hexdigest()
hashed_480 = str(hashed_256)
print("hash: " + hashed_480)
def repeater(key, n):
if n == 0:
return key
seed = str(key).encode('utf-8')
while n > 0:
return sha480(repeater(seed, n-1))
repeater('what', 2)
You have no recursive calls at all. You could change it to:
def hash_a(key):
return hashlib.sha224(key).hexdigest().encode('utf-8')
def hash_b(key):
return hashlib.sha256(key).hexdigest()
def shasum(key, n):
if n == 0: # base case: 0 iterations -> return key itself
return key
key = str(key).encode('utf-8')
return hash_b(hash_a(shasum(key, n - 1))) # recursve call
A side note: n == 0 or 1 is equivalent to (n == 0) or 1 which is always true. For that pattern, use n == 0 or n == 1 or shorter n in (0, 1)
Your code is nearly correct. just some minor issues fixed as below
import hashlib
def shasum(key, n):
print ("n: " + str(n))
key = str(key).encode('utf-8')
hash_a = hashlib.sha224(key).hexdigest().encode('utf-8')
print ("hash_a: " + str(hash_a))
hash_b = hashlib.sha256(hash_a).hexdigest()
print ("hash_b: " + str(hash_b))
if n == 0:
return hash_b #one iteration of 256(224)
else:
return shasum(hash_b, n-1)
I tried making this code and run it, but the result showed "None".
I've edited by adding another else in the loop function, then it actually produces a int result of '57' which is not exactly the answer.
def double(n):
return n*2
def halve(n):
return n//2
def mult(m,n):
def loop(m,n):
if n>1:
if n%2 != 0:
return m + loop(double(m),halve(n))
else:
return m #added another else
else:
return m
if n>0:
return loop(m,n)
else:
return 0
print(mult(57,86))
by simple recursion
def double(n):
return n * 2
def halve(n):
return n // 2
def mult(m, n, a = 0):
if n % 2 != 0:
a = a + m
m = double(m)
n = halve(n)
if n % 2 == 0:
m = double(m)
n = halve(n)
if n != 0:
return mult(m, n, a)
return a
print(mult(57, 86))
by nested function
def double(n):
return n * 2
def halve(n):
return n // 2
def mult(m, n):
def loop(m, n, a = 0):
if n % 2 != 0:
a = a + m
m = double(m)
n = halve(n)
if n % 2 == 0:
m = double(m)
n = halve(n)
if n != 0:
return loop(m, n, a)
return a
return loop(m, n)
print(mult(57, 86))