================ Start Lecture #6 ================
Note from Nelya
The CAS Learning Center is offering tutoring services
for students needing help in 002, 004, 201, or 310.
You can announce this in your classes. The website for
the CAS Learning Center is http://www.nyu.edu/cas/clc/.
The schedule is listed on the site.
End of Note
The book is quite clear. I have little to add.
You might want to know
Assume you believe the running time t(n) of an algorithm is Θ(nd) for some specific d and you want to both verify your assumption and find the multiplicative constant.
Make a plot of (n, t(n)/nd). If you are right the points should tend toward a horizontal line and the height of this line is the multiplicative constant.
Homework: R-1.29
What if you believe it is polynomial but don't have a guess for d?
Ans: Use ...
Plot (n, t(n)) on log log paper. If t(n) is Θ(nd), say t(n) approaches bnd, then log(t(n)) approaches log(b)+d(log(n)).
So when you plot (log(n), log(t(n)) (i.e., when you use log log paper), you will see the points approach (for large n) a straight line whose slope is the exponent d and whose y intercept is the multiplicative constant b.
Homework: R-1.30
Stacks implement a LIFO (last in first out) policy. All the action occurs at the top of the stack, primarily with the push(e) and pop operation.
The stack ADT supports
There is a simple implementation using an array A and an integer s (the current size). A[s-1] contains the TOS.
Objection (your honor). The ADT says we can always push. A simple array implementation would need to signal an error if the stack is full.
Sustained! What do you propose instead?
An extendable array.
Good idea.
Homework: Assume a software system has 100 stacks and 100,000 elements that can be on any stack. You do not know how the elements are to be distributed on the stacks. However, once an element is put on one stack, it never moves. If you used a normal array based implementation for the stacks, how much memory will you need. What if you use an extendable array based implementation? Now answer the same question, but assume you have Θ(S) stacks and Θ(E) elements.
Stacks work great for implementing procedure calls since procedures have stack based semantics. That is, last called is first returned and local variables allocated with a procedure are deallocated when the procedure returns.
So have a stack of "activation records" in which you keep the return address and the local variables.
Support for recursive procedures comes for free. For languages with static memory allocations (e.g., fortran) one can store the local variables with the method. Fortran forbids recursion so that memory allocation can be static. Recursion adds considerably flexibility to a language as some cost in efficiency.
I am reviewing modulo since I believe it is no longer taught in high school.
The top diagram shows an almost ordinary analog clock. The major difference is that instead of 12 we have 0. The hands would be useful if this was a video, but I omitted them for the static picture. Positive numbers go clockwise (cw) and negative counter-clockwise (ccw). The numbers shown are the values mod 12. This example is good to show arithmetic. (2-5) mod 12 is obtained by starting at 2 and moving 5 hours ccw, which gives 9. (-7) mod 12 is (0-7) mod 12 is obtained by starting at 0 and going 7 hours ccw, which gives 5.
To get mod 8, divide the circle into 8 hours
instead of 12.
The bottom picture shows mod 5 in a linear fashion. In pink are the 5 values one can get when doing mod 5, namely 0, 1, 2, 3, and 4. I only illustrate numbers from -3 to 11 but that is just due to space limitations. Each blue bar is 5 units long so the numbers at its endpoints are equal mod 5 (since they differ by 5). So you just lay off the blue bar until you wind up in the pink.
My favorite high level language, ada, gets it right in the obvious way: Ada defines both mod and remainder (ada extends the math definition of mod to the case where the second argument is negative).
In the familiar case when x≥0 and y>0 mod and remainder are
equal. Unfortunately the book uses mod sometimes when x<0 and
consequently needs to occasionally add an extra y to get the true mod.
End of personal rant
Homework: Using the real mod (let's call it RealMod) evaluate
Unlike past years, I will use the real mod this year (and call it simply mod). The only disadvantage is that the real mod isn't a java primitive. Although I am confident it must be in some java library, I like programming more than searching library listings when the desired method is so simple. So I actually implemented two versions of mod in java. The first one is based on the definition of mod(a,b) as the number n such that 0≤n<b and |n-a| is a multiple of b (the latter condition is normally stated “n is congruent to a modulo b)”. This implementation uses only addition and subtraction. The second implementation, which is only one line long, uses the java %. However, it is quite ugly.
The java code is here. Please note that almost all of it is scaffolding to generate tests and print results. The implementations of mod are at the end. A run of the tests is here.
Queues implement a FIFO (first in first out) policy. Elements are inserted at the rear and removed from the front using the enqueue and dequeue operations respectively.
The queue ADT supports
For queues we need a
front and rear "pointers" f and r. Since we are using arrays f and r
are actually indexes not pointers. Calling the array Q, Q[f] is the
front element of the queue, i.e., the element that would be returned
by dequeue(). Similarly, Q[r] is the element into which enqueue(e)
would place e. There is one exception: if f=r, the queue is empty so
Q[f] is not the front element.
Without writing the code, we see that f will be increased by each dequeue and r will be increased by every enqueue.
Assume Q has n slots Q[0]…Q[N-1] and the queue is initially
empty with f=r=0. Now consider enqueue(1); dequeue(); enqueue(2);
dequeue(); enqueue(3); dequeue(); …. There is never more than
one element in the queue, but f and r keep growing so after N
enqueue(e);dequeue() pairs, we cannot issue another operation.
The solution to this problem is to treat the array as circular,
i.e., right after Q[N-1] we find Q[0]. The way to implement this is
to arrange that when either f or r is N-1, adding 1 gives 0 not N.
Similarly for r. So the increment statements become
Note: Recall that we had some grief due to our starting arrays and loops at 0. For example, the fifth slot of A is A[4] and the fifth iteration of "for i←0 to 30" occurs when i=4. The updates of f and r directly above show one of the advantages of starting at 0; they are less pretty if the array starts at 1.
The size() of the queue seems to be r-f, but this is not always
correct since the array is circular.
For example let N=10 and consider an initially empty queue with f=r=0 that has
enqueue(10)enqueue(20);dequeue();enqueue(30);dequeue();enqueue(40);dequeue()
applied. The queue has one element, r=4, and f=3, so r-f=1 as expected.
But now apply 6 more enqueue(e) operations
enqueue(50);enqueue(60);enqueue(70);enqueue(80);enqueue(90);enqueue(100)
At this point the array has 7 elements, r=0, and f=3.
Clearly the size() of the queue is not r-f=-3.
It is instead 7, the number of elements in the queue.
The problem is that r in some sense is 10 not 0 since there were 10
enqueue(e) operations. In fact if we kept 2 values for f and 2 for r,
namely the value before the mod and after, then size() would be
rBeforeMod-fBeforeMod. Instead we, use the following formula.
size() = (r-f) mod N
Remark: If we used java's definition of mod so
that -3 mod 10 gave -3, we would need the inelegant formula
size() = (r-f+N) mod N.
Since isEmpty() is simply an abbreviation for the test size()=0, it is just testing if r=f.
Algorithm front():
if isEmpty() then
signal an error // throw QueueEmptyException
return Q[f]
Algorithm dequeue():
if isEmpty() then
signal an error // throw QueueEmptyException
temp←Q[f]
Q[f]←NULL // for security or debugging
f←(f+1) mod N
return temp
Algorithm enqueue(e):
if size() = N-1 then
signal an error // throw QueueFullException
Q[r]←e
r←(r+1) mod N
Round Robin processor scheduling is queue based as is fifo disk arm scheduling.
More general processor or disk arm scheduling policies often use priority queues (with various definitions of priority). We will learn how to implement priority queues later this chapter (section 2.4).
Homework (unofficial): (You may refer to your 202 notes if you wish; mine are on-line based on my home page). How can you interpret Round Robin processor scheduling and fifo disk scheduling as priority queues. That is what is the priority? Same question for SJF (shortest job first) and SSTF (shortest seek time first). If you have not taken an OS course (202 or equivalent at some other school), you are exempt from this question. Just write on you homework paper that you have not taken an OS course.
Problem Set #1, Problem 2:
Please read this question carefully; it is similar to but different from the one in class. You are given an Abstract Data Type (ADT) for some kind of container that includes an insert(x) method and a clean() method. Insert(x) adds an item to the container; clean() removes all items. The complexity of insert(x) is k, the number of items in the container after the insert has completed. The complexity of clean() is k3, where k is the number of items in the container before the clean is executed. Assume the container is initially empty and that you are to perform a total of N operations, each an insert(x) or a clean() (all the x's are distinct).