This is my code currently:
from loguru import logger
def fibonacci(n, s="% s"):
"""
Using recursive method
"""
# logger.debug(f"Finding {n}th Fibonacci number")
logger.debug(s % ("fib(%d)" % (n)))
a = 0
b = 1
if n <= 0:
return a
elif n in (1, 2):
return b
else:
return fibonacci(n - 1, s % ("fib(%d) %%s" % (n - 1))) fibonacci(n - 2, s % ("fib(%d) %%s" % (n - 2)))
My goal is to show the recursive tree in the log such as for fibonacci(5)
:
fib(5)
fib(4) fib(3)
(fib(3) fib(2)) (fib(2) fib(1))
and so on...
Is this possible? Current code didn't produce the expected output.
Current output:
fib(5)
fib(4) fib(4)
fib(4) fib(3) fib(3)
fib(4) fib(3) fib(2) fib(2)
fib(4) fib(3) fib(1) fib(1)
fib(4) fib(2) fib(2)
fib(3) fib(3)
fib(3) fib(2) fib(2)
fib(3) fib(1) fib(1)
The idea:
CodePudding user response:
One way could be for each call in the recursion to "report" itself to its location in a record of the tree. Then we can iterate level by level. Something like:
def f(n):
t = ["None"] * (2**(n-1) 1)
def g(n, i):
l = "(" if i & 1 else ""
r = ")" if not (i & 1) else ""
t[i] = "%sfib(%s)%s" % (l, n, r)
if n > 1:
g(n-1, 2*i 1)
g(n-2, 2*i 2)
g(n, 0)
print(t[0][0:-1])
i = 1
while 2**i-1 < len(t):
print(" ".join(t[2**i-1:2**(i 1)-1]))
i = 1
Output:
f(5)
"""
fib(5)
(fib(4) fib(3))
(fib(3) fib(2)) (fib(2) fib(1))
(fib(2) fib(1)) (fib(1) fib(0)) (fib(1) fib(0)) None None
(fib(1) fib(0))
"""
CodePudding user response:
You could define a class to hold binary tree nodes and build the tree as the result of the recursive fibonacci function:
class BNode:
def __init__(self,value,left=None,right=None):
self.value = value
self.left = left
self.right = right
def print(self):
printBTree(self,nodeInfo=lambda n:(str(n.value),n.left,n.right))
from functools import lru_cache
@lru_cache() # optimize object count
def fiboTree(n): # (n is an index, not a count)
if n<2: return BNode(n)
a,b = fiboTree(n-2),fiboTree(n-1)
return BNode(a.value b.value,a,b)
Output:
fiboTree(7).print()
13
____________/ \____________
5 8
_____/ \____ _______/ \______
2 3 3 5
/ \ __/ \_ __/ \_ _____/ \____
1 1 1 2 1 2 2 3
/ \ / \ / \ / \ / \ / \ __/ \_
0 1 0 1 1 1 0 1 1 1 1 1 1 2
/ \ / \ / \ / \ / \
0 1 0 1 0 1 0 1 1 1
/ \
0 1
You can find the printBTree
function here
If you only need to illustrate the call hierarchy, you can use the printBTree function directly:
def fibo(n):
n=int(n) # linking with strings to let zero come out as a node
return (f"fibo({n})",[None,str(n-2)][n>1], [None,str(n-1)][n>1])
printBTree(5,fibo)
fibo(5)
____________/ \____________
fibo(3) fibo(4)
/ \ _____/ \____
fibo(1) fibo(2) fibo(2) fibo(3)
/ \ / \ / \
fibo(0) fibo(1) fibo(0) fibo(1) fibo(1) fibo(2)
/ \
fibo(0) fibo(1)
To print as you go, I would suggest using indentation to convey the call hierarchy otherwise the repeated additions will be hard to relate to their callers.
def fibo(n,indent=""):
if n<2: return n
print(indent[:-3] "|_ "*bool(indent)
f"fibo({n}) = fibo({n-2}) fibo({n-1})")
return fibo(n-2,indent "| ") fibo(n-1,indent " ")
fibo(7)
fibo(7) = fibo(5) fibo(6)
|_ fibo(5) = fibo(3) fibo(4)
| |_ fibo(3) = fibo(1) fibo(2)
| | |_ fibo(2) = fibo(0) fibo(1)
| |_ fibo(4) = fibo(2) fibo(3)
| |_ fibo(2) = fibo(0) fibo(1)
| |_ fibo(3) = fibo(1) fibo(2)
| |_ fibo(2) = fibo(0) fibo(1)
|_ fibo(6) = fibo(4) fibo(5)
|_ fibo(4) = fibo(2) fibo(3)
| |_ fibo(2) = fibo(0) fibo(1)
| |_ fibo(3) = fibo(1) fibo(2)
| |_ fibo(2) = fibo(0) fibo(1)
|_ fibo(5) = fibo(3) fibo(4)
|_ fibo(3) = fibo(1) fibo(2)
| |_ fibo(2) = fibo(0) fibo(1)
|_ fibo(4) = fibo(2) fibo(3)
|_ fibo(2) = fibo(0) fibo(1)
|_ fibo(3) = fibo(1) fibo(2)
|_ fibo(2) = fibo(0) fibo(1)