I'm given the task to define a function to find the largest pyramidal number. For context, this is what pyramidal numbers are:
1 = 1^2
5 = 1^2 + 2^2
14 = 1^2 + 2^2 + 3^2
And so on.
The first part of the question requires me to find iteratively the largest pyramidal number within the range of argument n. To which, I successfully did:
def largest_square_pyramidal_num(n):
total = 0
i = 0
while total <= n:
total += i**2
i += 1
if total > n:
return total - (i-1)**2
else:
return total
So far, I can catch on.
The next part of the question then requires me to define the same function, but this time recursively. That's where I was instantly stunned. For the usual recursive functions that I have worked on before, I had always operated ON the argument, but had never come across a function where the argument was the condition instead. I struggled for quite a while and ended up with a function I knew clearly would not work. But I simply could not wrap my head around how to "recurse" such function. Here's my obviously-wrong code:
def largest_square_pyramidal_num_rec(n):
m = 0
pyr_number = 0
pyr_number += m**2
def pyr_num(m):
if pyr_number >= n:
return pyr_number
else:
return pyr_num(m+1)
return pyr_number
I know this is erroneous, and I can say why, but I don't know how to correct it. Does anyone have any advice?
Edit: At the kind request of a fellow programmer, here is my logic and what I know is wrong:
Here's my logic: The process that repeats itself is the addition of square numbers to give the pyr num. Hence this is the recursive process. But this isn't what the argument is about, hence I need to redefine the recursive argument. In this case, m, and build up to a pyr num of pyr_number, to which I will compare with the condition of n. I'm used to recursion in decrements, but it doesn't make sense to me (I mean, where to start?) so I attempted to recall the function upwards.
BUT this clearly isn't right. First of all, I'm sceptical of defining the element m and pyr_num outside of the pyr_num subfunction. Next, m isn't pre-defined. Which is wrong. Lastly and most importantly, the calling of pyr_num will always call back pyr_num = 0. But I cannot figure out another way to write this logic out
Here's a recursive function to calculate the pyramid number, based on how many terms you give it.
def pyramid(terms: int) -> int:
if terms <=1:
return 1
return terms * terms + pyramid(terms - 1)
pyramid(3) # 14
If you can understand what the function does and how it works, you should be able to figure out another function that gives you the greatest pyramid less than n.
def base(n):
return rec(n, 0, 0)
def rec(n, i, tot):
if tot > n:
return tot - (i-1)**2
else:
return rec(n, i+1, tot+i**2)
print(base(NUMBER))
this output the same thing of your not-recursive function.
def fac(n):
if (n < 1):
return 1
else:
n * fac(n-1)
print fac(4)
Why does return command cause the function to go back up and multiply the factorials? I have trouble understanding this.
1). You need to write a return in your code.
def fac(n):
if (n < 1):
return 1
else:
return n * fac(n-1)
print(fac(4))
2). I am uploading a picture which will help you to understand the concept of recursion. Follow the arrow from start to end in the picture.
You need to consider the call stack. The way this code needs to operate is for every fac(n) is equal to n * (n-1) * (n-2) ... (n - n + 1).
Your code was wrong. For a recursive function to work, you must be returning a value each time. You were getting None back because you never returned anything.
def fac(n):
if n == 1:
return 1
return n * fac(n - 1)
if __name__ == "__main__":
num = 2
print(fac(num))
Keep practicing recursion. It is extremely useful and powerful.
For more of a why it works than how it works perspective:
Calling a function recursively is no different than calling a different function: it just happens to execute the same set of instructions. You don't worry about Python getting confused between locals and parameters of different functions, or knowing where to return to; this is no different. In fact, if each recursive invocation had its own name instead of re-using the same name, you probably wouldn't be here.
Showing that it works properly is a bit different, because you need to make sure the recursion ends at some point. But this is no different than worrying about infinite loops.
I am trying to write a function which calculates multiple iteration hashes of a specific value (and output each iteration in the meantime).
However, I can't get my head over how to perform, for instance, md5 hash function on itself multiple times. For instance:
a = hashlib.md5('fun').hexdigest()
b = hashlib.md5(a).hexdigest()
c = hashlib.md5(b).hexdigest()
d = hashlib.md5(c).hexdigest()
.......
I think the recursion is the solution, but I just can't seem to implement it properly. This is the general factorial recursion example, but how do I adapt it to hashes:
def factorial(n):
if n == 0:
return 1
else:
return n * factorial(n - 1)
This is a classic application of generators. Python allows a maximum of 500 recursions due to its unusually heavy stack. For anything which might be executed anywhere near that many times, iteration will often be faster. Using a generator allows you to break after any desired number of executions and allows flat usage of the desired logic in your code. The following example prints the output of 10 such executions.
from itertools import islice
def hashes(n):
while True:
n = hashlib.md5(n).hexdigest()
yield n
for h in islice(hashes('fun'), 10):
print(h)
In general, you are looking for a loop like
while True:
x = f(x)
where you repeatedly replace the input with the result of the most recent application.
For your specific example,
def iterated_hash(x):
while True:
x = hashlib.md5(x).hexdigest()
return x
However, since you don't really want to do this an infinite number of times, you need to supply a count:
def iterated_hash(x, n):
while True:
if n == 0:
return x
x = hashlib.md5(x).hexdigest()
or with a for loop,
def iterated_hash(x, n):
for _ in range(n):
x = hashlib.md5(x).hexdigest()
return x
(Practically speaking, you want to use the for loop, but it's nice to see how the for loop is just a finite special case of the more general infinite loop.)
Just iterate as many times as needed:
def make_hash(text, iterations):
a = hashlib.md5(text).hexdigest()
for _ in range(iterations):
a = hashlib.md5(a).hexdigest()
return a
a = make_hash('fun', 5) # 5 iterations
My code is as follows.
I tried coding out for each case first, so given n = 4, my code looks like this:
a = overlay_frac(0,blank_bb,scale(1/4,rune))
b = overlay_frac(1/4,blank_bb,scale(1/2,rune))
c = overlay_frac(1/2,blank_bb,scale(3/4,rune))
d = overlay_frac(3/4,blank_bb,scale(1,rune))
show (overlay(a,(overlay(b,(overlay(c,d))))))
My understanding is that the recursion pattern is:
a = overlay_frac((1/n)-(1/n),blank_bb,scale(1/n,rune))
b = overlay_frac((2/n)-(1/n),blank_bb,scale(2/n,rune))
c = overlay_frac((3/n)-(1/n),blank_bb,scale(3/n,rune))
d = overlay_frac((4/n)-(1/n),blank_bb,sale(4/n,rune))
Hence, the recursion pattern that I came up with is:
def tree(n,rune):
if n==1:
return rune
else:
for i in range(n+1):
return overlay(overlay_frac(1-(1/n),blank_bb,scale(i/n,rune)),tree(n-1,rune))
When I hardcode this, everything turns out just fine, but I suspect I'm not doing the recursion properly. Where have I gone wrong?
You are in fact trying to do an iteration within a recursive call. In stead of using loop, you can use an inner function to memorize your status. The coefficient you defined is actually changed with both n and i, but for a given n it changed with i only. The status you need to memorize with inner function is then i, which is the same as you looping through i.
You can still achieve your goal by doing so
def f(i, n):
return overlay_frac((i/n)-(1/n),blank_bb,scale(i/n,rune))
# for each iteration, you check if i is equal to n
# if yes, return the result (base case)
# otherwise, you apply next coefficient to the previous result
# you start with i = 0 and increase by one every iteration until i reach to n (base case)
# notice how similar this recursive call looks like a loop
# the only difference is the status are updated within the function call itself
# therefore you will not have the problem of earlier return
def recursion(n):
def iteration(i, out):
if i == n:
return out
else:
return iteration(i+1, overlay(f(n-1, n), out))
return iteration(0, f(n, n))
Here, n is assumed to be the times of overlay you want to apply. When n = 0, no function applied on the last coefficient f(n, n). When n = 1, the output would be overlay applied once on coefficient with i = n - 1 and coefficient with i = n.
This way avoids the earlier return inside your loop.
In fact you can omit the inner function by adding additional argument to your outer function. Then you need to assign the default initial i. The inner function is not really necessary here. The key is to use the function argument to memorize the status (variable i in this case).
def f(i, n):
return overlay_frac((i/n)-(1/n),blank_bb,scale(i/n,rune))
def recursion(n, i=0):
if i == n:
return f(n, n)
else:
return overlay(f(n-1, n), recursion(n, i+1))
Your first two code blocks don't correspond to the same operations. This would be equivalent to your first block (in Python 3).
def overlayer(n, rune):
def layer(k):
# Scale decreases linearly with k
return overlay_frac((1 - (k+1)/n), blank_bb, scale(1-k/n, rune))
result = layer(0)
for i in range(1, n):
# Overlay on top of previous layers
result = overlay(layer(i), result)
return result
show(overlayer(4, rune))
Let's look at your equations again:
a = overlay_frac(0,blank_bb,scale(1/4,rune))
b = overlay_frac(1/4,blank_bb,scale(1/2,rune))
c = overlay_frac(1/2,blank_bb,scale(3/4,rune))
d = overlay_frac(3/4,blank_bb,scale(1,rune))
show (overlay(a,(overlay(b,(overlay(c,d))))))
What you wrote as "recursion" is not a recursion formula. If you compare your formulas for the recursion with the ones you gave us, you can infer n=4 which makes no sense. For a recursion pattern you need to describe your inner variables as a manifestation of the same expression with only a different parameter. That is, you should replace:
f_n = overlay_frac((1/4)*(n-1),blank_bb,sale(n/4,rune))
such that f_1=a, f_2=b etc...
Then your recursion fomula that you want to calculate translates to:
show (overlay(f_1,(overlay(f_2,(overlay(f_3,f_4))))))
You can write the function f_n as f(n) (and maybe other paramters) in your code and then do
def recurse(n):
if n == 4:
return f(4)
else:
return overlay(f(n),recurse(n+1))
then call:
show( recurse (1))
You need to assert that n<5and integer, otherwise you'll end up in an infinity loop.
There may still be some mistake, but it should be along those lines. Once you've actually written it like this however, it (maybe) doesn't really make sense to do a recursion anyways. If you only want to do it for n_max=4, that is. Just call the function in one line by replacing a,b,c,d with f_1,f_2,f_3,f_4
I have a line code like this -
while someMethod(n) < length and List[someMethod(n)] == 0:
# do something
n += 1
where someMethod(arg) does some computation on the number n. The problem with this code is that I'm doing the same computation twice, which is something I need to avoid.
One option is to do this -
x = someMethod(n)
while x < length and List[x] == 0:
# do something
x = someMethod(n + 1)
I am storing the value of someMethod(n) in a variable x and then using it later. However, the problem with this approach is that the code is inside a recursive method which is called multiple times. As a result, a lot of excess instances of variables x are being created which slows the code down.
Here's the snipped of the code -
def recursion(x, n, i):
while someMethod(n) < length and List[someMethod(n)] == 0:
# do something
n += 1
# some condition
recursion(x - 1, n, someList(i + 1))
and this recursion method is called many times throughout the code and the recursion is quite deep.
Is there some alternative available to deal with a problem like this?
Please try to be language independent if possible.
You can use memoization with decorators technique:
def memoize(f):
memo = dict()
def wrapper(x):
if x not in memo:
memo[x] = f(x)
return memo[x]
return wrapper
#memoize
def someMethod(x):
return <your computations with x>
As i understand your code correctly you are looking for some sort of memorization.
https://en.wikipedia.org/wiki/Memoization
it means that on every recursive call you have to save as mush as possible past calculations to use it in current calculation.