Answer:
No. This is not more secure.
Explanation:
this is not more secure than having all of Carol, bob and Alice having the same key.
alice would be able to know Carol's key for her to be able to verify the answer Carol gave to a challenge by her. she would also have to know bob's key to do the same.
any of these 3 would have to know someone else's secret key to verify answers to any challenge.
the challenge is to know the secret keys if the other two and if done, decryption can easily be done and they can impersonate.
Answer:
Code is in the attached screenshot.
Explanation:
Assumed it was written in Java based on your other question asked.
CCIE refers to Cisco Certified Internetwork Expert, a technical certification that demonstrates high proficiency of managing and establishing computer networks.
CSPM may refer to Certified Security Project Manager, which is a certification to demonstrate the individual’s capability in managing project in the field of cybersecurity.
MCITP refers to Microsoft Certified IT Professional, which demonstrates an individual’s ability to be a database or enterprise messaging administrator.
Oracle DBA refers to Oracle Database Administrator, and this certification demonstrates an individual’s ability to manage Oracle’s database from retrieving, maintaining, and editing them.
While PMP refers to Project Management Professional certification, one that you can use to prove your abilities as a project manager in various business contexts.
Thus, the best certification for him to use to get to a managerial career path is PMP.
Answer:
Check the explanation
Explanation:
#!usr/bin/python
#FileName: sieve_once_again.py
#Python Version: 2.6.2
#Author: Rahul Raj
#Sat May 15 11:41:21 2010 IST
fi=0 #flag index for scaling with big numbers..
n=input('Prime Number(>2) Upto:')
s=range(3,n,2)
def next_non_zero():
"To find the first non zero element of the list s"
global fi,s
while True:
if s[fi]:return s[fi]
fi+=1
def sieve():
primelist=[2]
limit=(s[-1]-3)/2
largest=s[-1]
while True:
m=next_non_zero()
fi=s.index(m)
if m**2>largest:
primelist+=[prime for prime in s if prime] #appending rest of the non zero numbers
break
ind=(m*(m-1)/2)+s.index(m)
primelist.append(m)
while ind<=limit:
s[ind]=0
ind+=m
s[s.index(m)]=0
#print primelist
print 'Number of Primes upto %d: %d'%(n,len(primelist))
if __name__=='__main__':
sieve()
