Model solution of problem set 1

1. The algorithm:

Step 1: Initialize 5 accumulators as zero for five categories
	A1 : first year
	A2 : Sophomores
	A3 : Juniors
	A4 : Seniors
	A5 : Others

Step 2: For each student in the list 
	Do
	  If it is a freshman then increse A1 by 1
	  Else if it is a sophomore then increse A2 by 1
	  Else if it is a junior then increse A3 by 1
	  Else if it is a senior then increse A4 by 1
	  Else then increse A5 by 1
	End

Step 3: Output A1 to A5

Basic steps include creating accumulators for every group and
classifying student one by one. The input is the list of students
with their class years.

Total point is 10. One point would be taken off if more than one
pass was used in your algorithm. 


2. There are many possible ways to implement this. One possible
answer is as follows:

Step 1: Build a table with 5 columns of possible lab times and 15
	rows each of which for one lab assistant. For each row, 
	mark the time where earch lab assistant is available.

Step 2: Check if any column has only two or less assistants available.
	If so, this problem has no solution. Stop.

Step 3: Otherwise, start with the column with the fewest lab assistants,
	choose three of them. Cross their rows off.

Step 4: Repeat step 3 for other columns, in the order of increasing
	number of lab assistants available. If we can finish this 
	without running into the situation where a column doesn't
	have enough assistants, we are done.  Stop.

Step 5: Otherwise, backtrace assigned assistants(those who have
	been crossed off), and see if any swaps will solve the 
	problem. If so, go back to step 4.

Total point: 10.
a) Understand problem: 2
b) Clear description:  2
c) Sorting by some order: 2
d) Crossing off an assistant's name once he is assigned: 2
e) Backtracing once conflicts arise: 2


3.
http: hypertext transfer protocol.
frogs.animals.com: mnemonic name for the host that holds the file.
animals/moviestarts: directory on this host.
kermit.html: file name.

Total point is 10 and 2 for each answer.


4. The basic difference between the telephone network and Internet
is the format of message transmission. Telephone network is an end to
end connection. One private connection is created for each call, so
there is no other calls sharing the connection with a call.  In the
Internet, all information is broken up into "packets".  Packets are sent 
to routers and forwared
by routers. This is called a switching and forwarding system. Because
packets can be lost, the voice quality may be decreased in Internet.
Latency in a "crowded" Internet may be intolerate to telephone customers.
Security is also a problem in providing telephone service over the
Internet. 

Total point: 10.