HOMEWORK FILE SOLUTIONS

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1 Data Structures Course (CSCI-UA 102) Professor Yap Spring 2012 HOMEWORK FILE February 13, 2012 SOLUTIONS 1 Homework 2: Due on Thu Feb 16 Q1. Consider the following function called crossproduct: int crossproduct(int[] A, int[] B){ int sum = 0; for (int i=0; i<a.length; i++) for (int j=0; j<b.length; j++) sum += A[i]*B[j]; return sum; } (a) Let the complexity of this function be T(n,m) where the lengths of A and B are n and m, respectively. Determine the complexity T(n, m) of this function exactly, not just asymptotically. For this purpose, make reasonable assumptions about the cost of each step. (b) Next, state the Θ-order of T(n,m) this means you write T(n,m) = Θ(f(n,m)) where f(n,m) is as simple as possible. (c) Finally prove your statement in (b). (d) Write a class called CrossProduct that has the above method, and also a main method to test the function. Your main function should just test the following two arrays A= [1,2,3,4], B= [2,4,6]. But sure to initialize the arrays to these values in the most effective way (look up if you do not know). Your Makefile must have a target called crossproduct (with short form cp) to compile AND run this class. (e) Can you improve the complexity of the method by computing the result in a different way? Indicate how to do this and give a Θ-order of the complexity if you can. Q2. The largest value of a Java int is 2,147,483,647. This value is Integer.MAX_VALUE in the wrapper class Integer. (a) First, use our rule of thumb (2 10 is 1000) to show why this is a roughly what you expect. (b) Next, explain why it is exactly this value, and not 2,147,483,646 or 2,147,483,648 (c) Now, we want to determine the largest value of n such that F n is less than MAX VALUE. You are told that for n > 10, F n is close to φ n where φ = (1+ 5)/ is the golden ratio. You may use a pocket calculator this part, but DO NOT use a program that computes F n. YOU MUST TELL US DETAILS ABOUT HOW YOU DETERMINE n. (d) Suppose we use the primitive type double. A Java double uses has a budget of 64 bits: 53 bits for mantissa, 1 bit for sign and 11 bits for exponent (that is a total of 65 not 64 bits the mysterious extra bit comes from the implicit leading bit in the mantissa). Using your rule of thumb, how many digits 1

2 do you expect the largest representable integer should be? (e) (THIS PART IS NOT REQUIRED) Confirm your answer for (d) experimentally, and tell us your experiment. Q3. We have given you code for two classes MyArrayList and MyLinkedList, both implementing the List ADT. Suppose we want to introduce a new method to the List ADT and to modify an original method as follows: The new method public void append( List<T> alist ) will modifies this list by appending the elements of alist to the end of itself, and alist becomes empty. E.g., if this list originally contains (1,3,5) and alist contains (a,b,c,d) then after calling append, this will contains (1,3,5,a,b,c,d). You need to implement this new method for MyLinkedList only, but rename your class to NewLinkedList. In MyArrayList, we want you to modify the behavior of remove() so that whenever the number of items in the list is less than half of the the length of the current array, we get a new array whose length is half the length of current array. Of course, it should not be less than the DEFAULT_CAPACITY Note that this choice half of the length of the current array is known to be a bad choice (why?), but it will serve our purpose. Rename your class to NewArrayList. You are to submit a Makefile that has targets called app and rem to test the above two modifications. 2 Homework 1: Due on Wed Feb 1 Q1. (4 Points for each part) Recall the big-oh and big-theta notations (the book s definition is in Section 2.1, which also has the big-omega notation). Let f(n) = 2n 100. Show that f(n) = Θ(n). In general, to show f(n) = Θ(g(n)), you need to explicitly tell us the 3 implicit positive constants C,c,n 0 in this notation. Show that f(n) = O(g(n)) and g(n) = O(h(n)) implies f(n) = O(h(n)). Note that showing f = O(g) only need 2 implicit constants, so it is a bit easier than showing f = Θ(g). Let f(n) = n 2. Show that it is NOT true that f(n) = O(n). Show that it is NOT true that 2 n = O(n k ), for any fixed value of k. Note that a function f(n) is called polynomial if f(n) = O(n k ) for some constant k > 0. So this shows 2 n is non-polynomial. 2

3 SOLUTION To show that f(n) = 2n 100 = Θ(n), we split this into two parts: (i) First show that f(n) = O(n). Note that f(n) = 2n 100 2n for n 0. Hence we can choose C = 2 and n 0 = 0 in the definition of f(n) = O(n). (ii) Next we show that f(n) = Ω(n). Observe f(n) = n+(n 100) n for n 100. So c = 1 and n 0 = 100 will do. Show that f(n) = O(g(n)) and g(n) = O(h(n)) implies f(n) = O(h(n)). By definition, there is some constant C > 0 and n 0 such that f(n) C g(n) for all n n 0. Similarly, there is some constant C > 0 and n 1 such that g(n) C h(n) for all n n 1. Let n 2 = max{n 0,n 1 }. Then for all n n 2, we have f(n) C g(n) C C h(n). If we choose C = C C, then f(n) C h(n) for all n n 2. This proves f = O(h). Let f(n) = n 2. Show that it is NOT true that f(n) = O(n). If f(n) = O(n), this means there is some C > 0 such that f(n) C n for n large enough. That is, n 2 C n for n large enough. But if we choose n > C, then n 2 = n n > C n, contradiction. To show 2 n O(n k ) for any k, we show the stronger claim that 2 n = Ω(n k ). It is a bit tricky to show if you do not use Calculus tools (and we do want to avoid them). We will bootstrap ourselves, each time by induction. Initially assume that n and k are integers. First we show that 2 n n for all n 2. This is clear for n = 1. Suppose n 2. Then 2 n = 2(2 n 1 ) 2(n 1) (by induction hypothesis). But 2(n 1) = n+(n 2) n (since n 2). Fix any C > 1. Next we show that 2 n Cn (eventually). First, choose n 0 such that 2 n0 2C. Then 2 2n0 = 2 n0 2 n0 (2C)(n 0 ) (by previous part). Hence 2 2n0 C(2n 0 ). This proves our claim for n = 2n 0. Then for n > 2n 0, 2 n = 2(2 n 1 ) 2(C(n 1)) (by induction hypothesis). Thus 2 n C(2n 2) Cn (ev.) Now we show that 2 n Cn k (ev.) for any k 2 and C > 1. We may choose C > 2 k. By induction hypothesis, we know that 2 n Cn k 1 (ev.) for any C > 1. First choose n 0 such that 2 n0 Cn0 k 1 (induction hypothesis). Then 2 2n0 = 2 n0 2 n0 (Cn0 k 1 )(Cn0 k 1 C 2 n k 0 C(2n 0 ) k since C 2 k. So we have show our basis case when n = 2n 0. For n > 2n 0, we have 2 n = 2(2 n 1 ) 2(C(n 1) k ) = 2C(n k kn k 1 + ) obtained by expanding (n 1) k. We may write (n 1) k = n k p(n) where p(n) is a polynomial of degree k 1. Then 2 n 2C(n k p(n)) = Cn k +C(n k 2p(n)). But n k 2p(n) 0 (ev). Therefore 2 n Cn k (ev). We have now shown that 2 n Cn k (ev) for integer values of n and k, and for any C > 1. If these are not necessarily integers, we use the fact that 2 n 2 n C n k (ev.) (as proved above) C(n/2) k = C n k (C = C2 k ) I have since discovered a much simpler proof: Use the fact that 2 n = n ( n ) i=0 i (why is this true? use the interpretation that 2 n is the number of subsets of a set of size n). Then 2 n > ( n k) (n/2) k /k! (if n > 2k). But (n/2) k /k! = C n k if we choose C = 1/(2 k k!). Q2. (9 Points) In class, we described a Java class Fib that has a method fib(n) to compute F n, the nth 3

4 Fibonacci number. Recall that F 0 = 0,F 1 = 1 and F n+1 = F n + F n 1 for n 1. This definition is slightly different than the book s (p. 6), but the difference is unimportant. Please download the file Fib.java from our pickup site to study this class. Let T(n) denote the number of steps used by fib(n). You can interpret steps quite liberally, counting any constant time (i.e., O(1)) operation as one step. Observe (by studying the program f ib(n)) that { 1 if n 1 T(n) T(n 1)+T(n 2) else. (a) Show that T(n) 2 n/2. (b) Conclude that T(n) is non-polynomial. HINT: you may assume that T(n) T(m) whenever n m. SOLUTION (a) Note that the claim T(n) 2 n/2 is false when n = 1. But we claim that it is true eventually. Indeed, we only need n 2. For n = 2, it is true because T(2) = 2 2 2/2. For n = 3, it is true because T(3) = 3 2 3/2. Inductively, we use the observation that T(n) T(n 2) + T(n 2) = 2 T(n 2). Repeating this, T(n) 2 (2 T(n 4)) = 2 2 T(n 4). So T(n) 2 k T(n 2k). How large can k be? Well, about n/2. More precisely, we need n 2k = 3 or n 2k = 4. Thus, k = (n 3)/2 or k = (n 4)/2. So T(n) 2 (n i)/2 T(i) where i = 3 or i = 4. But we know that T(i) 2 i/2. Thus T(n) 2 (n i)/2)2i =2 n/2. (b) To say that T(n) is polynomial means T(n) = O(n k ) for some k > 0. But we already know from the previous question that 2 n C n k (ev.) for any k > 0. So (ev.) where C = C 2 k. T(n) 2 n/2 C (n/2) k C n k Q3. (30 Points) In class, we discussed two problems with the current implementation of f ib(n): (1) The use of int means that the output would be wrong for n > 60. The solution is to use the Java class Integer. (2) The fact that the recursive fib calls itself twice meant that this is an exponential time algorithm (see previous question). We can empirically see this by trying to compute F 45. We also sketched a version that only requires one recursive call (this one would be polynomial time). Please implement such an improved version that incorporates (1) and (2). The name ofyourjavaclassshouldbe calledfibonacci,andit should have(at least) twomethods: Integer fib(integer n) that computes F n, and a main method that reads an optional number n from the command line. If n is not given, then it defaults to 20. It should print the value fib(n). Your main method should also accept a second optional number m, and this means you compute a consecutive sequence of m Fibonacci numbers, starting from fib(n). The default value of m is 1. E.g., if we call java Fibonacci 4 2 then you should print F 4 = 3 and F 5 = 5. Download a Makefile from the pickup site, and put it in the same directory as Fibonacci.java. Modify it for your current assignment so that the following 12 behavior is achieved: > make compile --- this should call "javac Fibonacci.java" > make c --- this should be the same as "make compile" > make p=fib c --- this should be the same as "make c" but using Fib.java > make run --- this should call "java Fibonacci" > make r --- this should call "make run" > make --- same as "make c; make r" > make p=fib --- same as "make" but using Fib.java > make arg= same as "java Fibonacci 12" 4

5 > make arg=12 arg2=3 --- same as "java Fibonacci 12 3" > make test --- this should call "java Fibonacci 20 10" > make test1 --- this should call "java Fibonacci 30 10" > make test2 --- this should call "java Fibonacci 80" Note that we must use three variables in the Makefile, p, arg, arg2. To do this, you should read up our tutorial on make, and in particular how to use variables. Also, our sample Makefile already illustrates this. Here is one way to creat the TAR file to send us (either in Cygwin or in MacOS Windows or in Linux): > tar cvf hw1-chee-yap.tar README Q-1.pdf Q-2.pdf Makefile Fibonacci.java where > is just the command prompt (so do not type it!). SOLUTION The solution files Fibonacci.java and Makefile-hw1 are both found in our pickup site. NOTE that although the make file s name is non-standard, you can still use it by telling the make program which make file to use, as follows: 5

Data Structures Course (CSCI-UA 102) Professor Yap Spring 2012 HOMEWORK FILE. May 1, 2012

Data Structures Course (CSCI-UA 102) Professor Yap Spring 2012 HOMEWORK FILE May 1, 2012 Homework Instructions (H10 is new!) H1. Please read carefully. There is penalty for non-compliance. H2. This homework