I created this code because I was not able to find any functional that accomplishes my requirement.
If you can reduce it will be better.
Just enter de prefix lenght from 1 to 32 and you will get the decimal mask.
This code help me with my scripts for cisco.
import math
#Netmask octets
octet1 = [0,0,0,0,0,0,0,0]
octet2 = [0,0,0,0,0,0,0,0]
octet3 = [0,0,0,0,0,0,0,0]
octet4 = [0,0,0,0,0,0,0,0]
#POW list
pow_list = [7,6,5,4,3,2,1,0]
#Introduce prefix lenght
mask = int(input("Introduce the prefix lenght: "))
#According to the number of bits we will change the array elements from 0 to 1
while mask >= 25 and mask <= 32:
octet4[mask-25] = 1
mask -= 1
while mask >= 17 and mask <= 24:
octet3[mask-17] = 1
mask -= 1
while mask >= 9 and mask <= 16:
octet2[mask-9] = 1
mask -= 1
while mask >= 1 and mask <= 8:
octet1[mask-1] = 1
mask -= 1
#Obtain the number of ones
ones1 = octet1.count(1)
ones2 = octet2.count(1)
ones3 = octet3.count(1)
ones4 = octet4.count(1)
#Summary and reuslt of each octet.
sum1 = 0
for i in range(0,ones1):
sum1 = sum1 + math.pow(2,pow_list[i])
sum1 = int(sum1)
sum2 = 0
for i in range(0,ones2):
sum2 = sum2 + math.pow(2,pow_list[i])
sum2 = int(sum2)
sum3 = 0
for i in range(0,ones3):
sum3 = sum3 + math.pow(2,pow_list[i])
sum3 = int(sum3)
sum4 = 0
for i in range(0,ones4):
sum4 = sum4 + math.pow(2,pow_list[i])
sum4 = int(sum4)
#Join the results with a "."
decimal_netmask = str(sum1) + "." + str(sum2) + "." + str(sum3) + "." + str(sum4)
#Result
print("Decimal netmask is: "+ decimal_netmask)
Result:
Introduce the prefix lenght: 23
Decimal netmask is: 255.255.254.0
As you are probably doing more than just converting CIDR to netmask, I recommend checking out the built-in library ipaddress
from ipaddress import ip_network
cidr = input("Introduce the prefix length: ")
decimal_netmask = str(ip_network(f'0.0.0.0/{cidr}').netmask)
You can simplify your code by computing the overall mask value as an integer using the formula:
mask = 2**32 - 2**(32-prefix_length)
Then you can compute the 4 8-bit parts of the mask (by shifting and masking), appending the results to a list and then finally joining each element of the list with .:
def decimal_netmask(prefix_length):
mask = 2**32 - 2**(32-prefix_length)
octets = []
for _ in range(4):
octets.append(str(mask & 255))
mask >>= 8
return '.'.join(reversed(octets))
for pl in range(33):
print(f'{pl:3d}\t{decimal_netmask(pl)}')
Output:
0 0.0.0.0
1 128.0.0.0
2 192.0.0.0
3 224.0.0.0
4 240.0.0.0
5 248.0.0.0
6 252.0.0.0
7 254.0.0.0
8 255.0.0.0
9 255.128.0.0
10 255.192.0.0
11 255.224.0.0
12 255.240.0.0
13 255.248.0.0
14 255.252.0.0
15 255.254.0.0
16 255.255.0.0
17 255.255.128.0
18 255.255.192.0
19 255.255.224.0
20 255.255.240.0
21 255.255.248.0
22 255.255.252.0
23 255.255.254.0
24 255.255.255.0
25 255.255.255.128
26 255.255.255.192
27 255.255.255.224
28 255.255.255.240
29 255.255.255.248
30 255.255.255.252
31 255.255.255.254
32 255.255.255.255
This is what I've written:
def logic( arr , r , l ):
arr = []
i == 0
if (arr[i]) == (i +1) * i :
print('True')
i = i + 1
else :
print('False')
return(arr )
arr = list(map(int, input().split()))
r = int(input())
l = int(input())
print(logic(arr,r,l))
This is what we need:
You task is to calculate a boolean array b, where b[i] = true if there exists an integer x,
such that a[i] = (i + 1) * x and l ≤ x ≤ r. Otherwise, b[i] should be set to false.
Example
For a = [8, 5, 6, 16, 5], l = 1, and r = 3, the output should be logic(a, l, r) = [false, false, true, false, true].
For a[0] = 8, we need to find a value of x such that 1 * x = 8, but the only value that would work is x = 8 which doesn't satisfy the boundaries 1 ≤ x ≤ 3, so b[0] = false.
For a[1] = 5, we need to find a value of x such that 2 * x = 5, but there is no integer value that would satisfy this equation, so b[1] = false.
For a[2] = 6, we can choose x = 2 because 3 * 2 = 6 and 1 ≤ 2 ≤ 3, so b[2] = true.
Below is the error:
UnboundLocalError: local variable 'i' referenced before assignment
You are using i == 0. This is an equals to. This is not asigning the variable. I think you are looking for i = 0. == sign is used in If statements etc.
Also may I ask why you are using the parameters arr , r & l in your logic function. But assign them inside the function?
EDIT: I have not programmed in python for a while and missed the indent
I am trying to understand the recursion function. I would like to know how that answer is coming with steps
def tri_recursion(k):
if(k>0):
result = k+tri_recursion(k-1)
print(result)
else:
result = 0
return result
print("\n\nRecursion Example Results")
tri_recursion(6)
results are this just want to know how its coming
1
3
6
10
15
21
The function computes the sum of all numbers between 0 and n, and prints intermediate results. The first 1 is 0+1, the 3 = 0+1+2, 6 = 0+1+2+3, 10 = 0+1+2+3+4, ...
To understand a recursive function, you need 2 points : how is the recursive call done, and when does the recursion stop.
The recursive call is given by result = k+tri_recursion(k-1)and the recursion stops when k <= 0and returns 0. So if we assume only positive numbers, we could describe tri_recursion so:
tri_recursion(k) = k + tri_recursion(k-1) if k > 0
tri_recursion(0) = 0
So tri_recursion(k) = k + tri_recursion(k-1) = k + (k-1) + tri_recursion(k-2) = k + (k-1) + (k-2) + tri_recursion(k-3) ... = k + (k-1) + (k-2) + ... + 0
So tri_recursion(k) is the sum of all numbers between 0 and k.
Note that the sum on all numbers between 0 and k equals k*(k+1) / 2 so tri_recursion(6) = 6 * 7 / 2 = 21
I was playing with the recursive Ackermanns function. For certain values my prompt whould not show every calculated output 'cause Python whould exceed its recursive limit so fast that whould freeze the prompt before the "easy" parts whould catch up with it.
So I thought I could add a recursive counter and a quick pause after a full execution of the function. I was getting the anticipated outputs until it reached the values (1,0). After that I got a TypeError: can only concatenate tuple (not "int") to tuple.
My code is as follows:
import time
import sys
sys.setrecursionlimit(3000)
def ackermann(i,j,rec):
output = None
if i==0:
output = j+1
elif j==0:
output = ackermann(i-1,1,rec)
rec=rec+1
else:
output = ackermann(i-1,ackermann(i,j-1,rec),rec)
rec=rec+1
return output,rec
rec=0
for i in range(5):
for j in range(5):
print("(",i,",",j,")= ",ackermann(i,j,rec))
time.sleep(2)
Notice that removing all instances of rec (my recurence counter), the program runs fine. (You can see all outputs for values i,j = 3)
Can someone point out how to correct my code or propose a different method of finding how many times the Ackermann function has calls itself ?
Also, I've noticed that putting a limit of 5000 whould crash my python kernel very fast. Is there an upper limit ?
I use the latest Anaconda.
EDIT
I tried to implement the same function using a list as a parameter with the following data [i,j,output,#recursion]
import time
import sys
sys.setrecursionlimit(3000)
def ackermann(*rec):
rec=list(rec)
print(rec) # see the data as they initialize the function
if rec[0][0]==0:
rec[0][1]=rec[0][1]+1
rec[0][2] = rec[0][1]+1
elif rec[0][1]==0:
rec[0][0]=rec[0][0]-1
rec[0][1]=1
rec = ackermann()
rec[0][3]=rec[0][3]+1
else:
rec[0][0]=rec[0][0]-1
rec[0][1] = ackermann()
rec = ackermann()
rec[0][3]=rec[0][3]+1
return rec
for i in range(5):
for j in range(5):
rec=[i,j,0,0]
print(ackermann(rec))
time.sleep(1)
But this time I get a IndexError: list index out of rangebecause for some unknown reason my list gets emptied
OUTPUT:
[[0, 0, 0, 0]]
[[0, 1, 2, 0]]
[[0, 1, 0, 0]]
[[0, 2, 3, 0]]
[[0, 2, 0, 0]]
[[0, 3, 4, 0]]
[[0, 3, 0, 0]]
[[0, 4, 5, 0]]
[[0, 4, 0, 0]]
[[0, 5, 6, 0]]
[[1, 0, 0, 0]]
[]
The problem with the original implementation is that
return output, rec
will happily create a tuple when output and rec are both numbers, which is true whenever i=0. But once you get to i=1, j=0 the function calls Ackerman on (0,1,rec), which returns a tuple, to which it then cannot add the integer rec, hence the error message. I believe I have worked with that idea, though, almost unchanged, except rather than trying to pass and return rec, I made it global (ugly, I know). I also reformatted the output so I could read it better. Thus:
import time
import sys
sys.setrecursionlimit(3000)
def ackermann(i,j):
global rec
output = None
if i==0:
output = j+1
elif j==0:
output = ackermann(i-1,1)
rec=rec+1
else:
output = ackermann(i-1,ackermann(i,j-1))
rec=rec+1
return output
for i in range(5):
for j in range(5):
rec = 0
print
print("ack("+str(i)+","+str(j)+") = "+str(ackermann(i,j)))
print("rec = "+str(rec))
print
time.sleep(1)
and the output, before erroring out, is,
ack(0,0) = 1
rec = 0
ack(0,1) = 2
rec = 0
ack(0,2) = 3
rec = 0
ack(0,3) = 4
rec = 0
ack(0,4) = 5
rec = 0
ack(1,0) = 2
rec = 1
ack(1,1) = 3
rec = 2
ack(1,2) = 4
rec = 3
ack(1,3) = 5
rec = 4
ack(1,4) = 6
rec = 5
ack(2,0) = 3
rec = 3
ack(2,1) = 5
rec = 8
ack(2,2) = 7
rec = 15
ack(2,3) = 9
rec = 24
ack(2,4) = 11
rec = 35
ack(3,0) = 5
rec = 9
ack(3,1) = 13
rec = 58
ack(3,2) = 29
rec = 283
ack(3,3) = 61
rec = 1244
ack(3,4) = 125
rec = 5213
ack(4,0) = 13
rec = 59
It seems to me there are only one or two other values (it will choke on 4,2 I believe, no matter what, so you would need to get 5, 0 first) you could hope to get out this way, no matter how much you tinker.
I am a little troubled that rec appears to exceed the recursion limit, but I think Python must be interpreting along the way somehow, so that it gets deeper than one might think, or that I don't fully understand sys.recursionlimit (I looked at rec a few times, and at the very least I followed your lead on calculating it; also, as a sanity check I switched the order of incrementing it and the function call and got the same results).
EDIT: I added another parameter to track how deeply any particular call ever recurses. That turns out to be typically less than (and at most one more than) "rec." rec represents (actually 1 less than) how many times the function is called to make the particular calculation, but not all of these need be on the Python interpreter stack simultaneously.
Revised code:
import time
import sys
sys.setrecursionlimit(3000)
def ackermann(i,j,d):
global rec
global maxDepth
if ( d > maxDepth ) : maxDepth = d
output = None
if i==0:
output = j+1
elif j==0:
rec=rec+1
output = ackermann(i-1,1, d+1)
else:
rec=rec+1
output = ackermann(i-1,ackermann(i,j-1, d+1),d+1)
return output
for i in range(5):
for j in range(5):
rec = 0
maxDepth=0
print
print("ack("+str(i)+","+str(j)+") = "+str(ackermann(i,j,1)))
print("rec = "+str(rec))
print("maxDepth = "+str(maxDepth))
print
time.sleep(1)
revised output (before it gives up)
ack(0,0) = 1
rec = 0
maxDepth = 1
ack(0,1) = 2
rec = 0
maxDepth = 1
ack(0,2) = 3
rec = 0
maxDepth = 1
ack(0,3) = 4
rec = 0
maxDepth = 1
ack(0,4) = 5
rec = 0
maxDepth = 1
ack(1,0) = 2
rec = 1
maxDepth = 2
ack(1,1) = 3
rec = 2
maxDepth = 3
ack(1,2) = 4
rec = 3
maxDepth = 4
ack(1,3) = 5
rec = 4
maxDepth = 5
ack(1,4) = 6
rec = 5
maxDepth = 6
ack(2,0) = 3
rec = 3
maxDepth = 4
ack(2,1) = 5
rec = 8
maxDepth = 6
ack(2,2) = 7
rec = 15
maxDepth = 8
ack(2,3) = 9
rec = 24
maxDepth = 10
ack(2,4) = 11
rec = 35
maxDepth = 12
ack(3,0) = 5
rec = 9
maxDepth = 7
ack(3,1) = 13
rec = 58
maxDepth = 15
ack(3,2) = 29
rec = 283
maxDepth = 31
ack(3,3) = 61
rec = 1244
maxDepth = 63
ack(3,4) = 125
rec = 5213
maxDepth = 127
ack(4,0) = 13
rec = 59
maxDepth = 16
In your edited version of the code, you used a *arg in your def for ackerman and made it explicitly a list, and you get eleven output lists containing a four-element list in each until on the twelfth recursion you get an empty list. So, did the first eleven lists contain the expected elements according to the ackermann constraints? Also, on the twelfth recursion, you say the list was "emptied." I wonder for analytical purposes if it might make sense to say instead it wasn't filled in the first place. That is, not that something emptied it but that something didn't fill it as expected on the twelfth time through.