Discuss and compare shared and switch Ethernet
1 answer:
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Answer:
Follows are the code to this question:
def FindPair(Values,SUM):#defining a method FindPair
found=False;#defining a boolean variable found
for i in Values:#defining loop for check Value
for j in Values:#defining loop for check Value
if (i+j ==SUM):#defining if block that check i+j=sum
found=True;#assign value True in boolean variable
x=i;#defining a variable x that holds i value
y=j;#defining a variable x that holds j value
break;#use break keyword
if(found==True):#defining if block that checks found equal to True
print("(",x,",",y,")");#print value
else:#defining else block
print("Sorry there is no such pair of values.");#print message
Values=[3,8,13,2,17,18,10];#defining a list and assign Values
SUM=20;#defining SUM variable
FindPair(Values,SUM);#calling a method FindPair
Output:
please find the attachment:
Explanation:
In the above python code a method, "FindPair" is defined, which accepts a "list and SUM" variable in its parameter, inside the method "found" a boolean variable is defined, that holds a value "false".
- Inside the method, two for loop is defined, that holds list element value, and in if block, it checks its added value is equal to the SUM. If the condition is true, it changes the boolean variable value and defines the "x,y" variable, that holds its value.
- In the next if the block, it checks the boolean variable value, if the condition is true, it will print the "x,y" value, otherwise, it will print a message.
Answer:
The answer is "64"
Explanation:
There has been 1 Private Key 1 Public Key for every person. The private key is used for other app messages. All else uses the Public Key to encrypt messages with that user. Those keys are numerically related. Even if you have 5 users, you will be given 5 Personal keys and 5 Public keys.
- All user will have a duplicate of other public keys, which indicates (n*n-1) copies of n public keys on different systems to make sure shared contact between all users, plus n private keys.
- The unique key count is 2n, with n^2 distributed private and public keys on different systems.
- 8 people, where 8 is a number, that is equal to n so,