Zero-plagiarism guarantee Each paper is composed from scratch, according to your instructions. Moreover, as we fircle down from produdts root, more and more internal nodes are absent. If this strategy always worked, then it would be easy to determine how to maximize profit: find the highest and lowest prices, and then work left from the highest price to find the lowest prior price, work right pproducts the lowest price to find the highest later price, and take the pair with the carter products circle cutting jig fail difference. This chapter gives several standard methods for simplifying the asymptotic anal- ysis of algorithms. In some cases, we assume that the inputs conform to a known probability carter products circle cutting jig fail, so that we are averaging the running time over all possible inputs. In Section 2.

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Although any two real numbers can be compared, not all functions are asymptot- ically comparable. That is, for two functions f. For example, we cannot compare the functions n and n1Csin n using asymptotic notation, since the value of the exponent in n1Csin n oscillates between 0 and 2, taking on all values in between. This section reviews some standard mathematical functions and notations and ex- plores the relationships among them. It also illustrates the use of the asymptotic notations.

For all real x,. The floor function f. For an asymptotically positive polynomial p. We say that a function f. When convenient, we shall assume 00 D 1.

We can relate the rates of growth of polynomials and exponentials by the fol- lowing fact. Using e to denote : : :, the base of the natural logarithm function, we have for all real x,. For all real x, we have the inequality. We have for all x, lim. An important notational convention we shall adopt is that logarithm functions will apply only to the next term in the formula, so that lg n C k will mean.

By equation 3. Computer scientists find 2 to be the most natural base for logarithms because so many algorithms and data structures involve splitting a problem into two parts. We can relate the growth of polynomials and polylogarithms by substituting lg n for n and 2a for a in equation 3. Thus, any positive polynomial function grows faster than any polylogarithmic function. As Exercise 3. We use the notation f. Formally, let f. For non- negative integers i , we recursively define.

Let lg. Because the log- arithm of a nonpositive number is undefined, lg. Be sure to distinguish lg. Then we define the iterated logarithm function as. Use the definitions of the asymptotic notations to prove the following properties. Rank the following functions by order of growth; that is, find an arrangement. Partition your list into equivalence classes such that functions f. Give an example of a single nonnegative function f. Prove or disprove each of the following conjectures.

We say that f. Show that for any two functions f. For each of the following functions f. Knuth [] traces the origin of the O-notation to a number-theory text by P. Bach- mann in The o-notation was invented by E. Landau in for his discussion of the distribution of prime numbers. Further dis- cussion of the history and development of asymptotic notations appears in works by Knuth [, ] and Brassard and Bratley [54].

Not all authors define the asymptotic notations in the same way, although the various definitions agree in most common situations. Some of the alternative def- initions encompass functions that are not asymptotically nonnegative, as long as their absolute values are appropriately bounded. Equation 3. Other properties of elementary math- ematical functions can be found in any good mathematical reference, such as Abramowitz and Stegun [1] or Zwillinger [], or in a calculus book, such as Apostol [18] or Thomas et al.

Knuth [] and Graham, Knuth, and Patash- nik [] contain a wealth of material on discrete mathematics as used in computer science. In Section 2.

Recall that in divide-and-conquer, we solve a problem recur- sively, applying three steps at each level of the recursion:. When the subproblems are large enough to solve recursively, we call that the recur- sive case. Sometimes, in addition to subproblems that are smaller instances of the same problem, we have to solve subproblems that are not quite the same as the original problem.

We consider solving such subproblems as part of the combine step. In this chapter, we shall see more algorithms based on divide-and-conquer. The first one solves the maximum-subarray problem: it takes as input an array of num- bers, and it determines the contiguous subarray whose values have the greatest sum. Then we shall see two divide-and-conquer algorithms for multiplying n n matri- ces.

Recurrences go hand in hand with the divide-and-conquer paradigm, because they give us a natural way to characterize the running times of divide-and-conquer algo- rithms. A recurrence is an equation or inequality that describes a function in terms. For example, in Section 2.

Recurrences can take many forms. For example, a recursive algorithm might. If the divide and combine steps take linear time, such an algorithm would give rise to the recurrence T. Subproblems are not necessarily constrained to being a constant fraction of the original problem size. For example, a recursive version of linear search see Exercise 2. Each recursive call would take con- stant time plus the time for the recursive calls it makes, yielding the recurrence T.

We use techniques for bounding summations to solve the recurrence. Such recurrences arise frequently. A recurrence of the form in equation 4. To use the master method, you will need to memorize three cases, but once you do that, you will easily be able to determine asymptotic bounds for many simple recurrences. We will use the master method to determine the running times of the divide-and-conquer algorithms for the maximum-subarray problem and for matrix multiplication, as well as for other algorithms based on divide- and-conquer elsewhere in this book.

Occasionally, we shall see recurrences that are not equalities but rather inequal- ities, such as T. Because such a recurrence states only an upper bound on T. Similarly, if the inequality were reversed to T. In practice, we neglect certain technical details when we state and solve recur- rences.

Boundary conditions represent another class of details that we typically ignore. Since the running time of an algorithm on a constant-sized input is a constant, the recurrences that arise from the running times of algorithms generally have T.

Consequently, for convenience, we shall generally omit statements of the boundary conditions of recurrences and assume that T. For example, we normally state recurrence 4. The reason is that although changing the value of T. When we state and solve recurrences, we often omit floors, ceilings, and bound- ary conditions.

We forge ahead without these details and later determine whether or not they matter. They usually do not, but you should know when they do. Ex- perience helps, and so do some theorems stating that these details do not affect the asymptotic bounds of many recurrences characterizing divide-and-conquer algo- rithms see Theorem 4.

In this chapter, however, we shall address some of these details and illustrate the fine points of recurrence solution methods. Suppose that you been offered the opportunity to invest in the Volatile Chemical Corporation.

Like the chemicals the company produces, the stock price of the Volatile Chemical Corporation is rather volatile. You are allowed to buy one unit of stock only one time and then sell it at a later date, buying and selling after the close of trading for the day. To compensate for this restriction, you are allowed to learn what the price of the stock will be in the future. Your goal is to maximize your profit.

Figure 4. Unfortunately, you might not be able to buy at the lowest price and then sell at the highest price within a given period. In Figure 4. You might think that you can always maximize profit by either buying at the lowest price or selling at the highest price. For example, in Figure 4. If this strategy always worked, then it would be easy to determine how to maximize profit: find the highest and lowest prices, and then work left from the highest price to find the lowest prior price, work right from the lowest price to find the highest later price, and take the pair with the greater difference.

The horizontal axis of the chart indicates the day, and the vertical axis shows the price. The bottom row of the table gives the change in price from the previous day. Again, the horizontal axis indicates the day, and the vertical axis shows the price.

We can easily devise a brute-force solution to this problem: just try every possible pair of buy and sell dates in which the buy date precedes the sell date. A period of n days has. Can we do better? In order to design an algorithm with an o. We want to find a sequence of days over which the net change from the first day to the last is maximum. The table in Figure 4. If we treat this row as an array A, shown in Figure 4. We call this contiguous subarray the maximum subarray. For example, in the array of Figure 4.

So let us seek a more efficient solution to the maximum-subarray problem. The maximum-subarray problem is interesting only when the array contains some negative numbers.

If all the array entries were nonnegative, then the maximum-subarray problem would present no challenge, since the entire array would give the greatest sum. Divide-and-conquer suggests that we divide the subarray into two subarrays of as equal size as possible. As Figure 4. Thus, all that is left to do is find a. This problem is not a smaller instance of our original problem, because it has the added restriction that the subarray it chooses must cross the midpoint.

This procedure works as follows. Line 1 tests for the base case, where the subarray has just one element. A subar- ray with just one element has only one subarray—itself—and so line 2 returns a tuple with the starting and ending indices of just the one element, along with its value.

Lines 3—11 handle the recursive case. Line 3 does the divide part, comput- ing the index mid of the midpoint. Lines 4 and 5 conquer by recur- sively finding maximum subarrays within the left and right subarrays, respectively. Lines 6—11 form the combine part. Line 6 finds a maximum subarray that crosses the midpoint. Recall that because line 6 solves a subproblem that is not a smaller instance of the original problem, we consider it to be in the combine part.

Line 7 tests whether the left subarray contains a subarray with the maximum sum, and line 8 returns that maximum subarray. Otherwise, line 9 tests whether the right subarray contains a subarray with the maximum sum, and line 10 returns that max- imum subarray.

If neither the left nor right subarrays contain a subarray achieving the maximum sum, then a maximum subarray must cross the midpoint, and line 11 returns it. As we did when we analyzed merge sort in Section 2. We denote by T. For starters, line 1 takes constant time. The base case, when n D 1, is easy: line 2 takes constant time, and so.

Lines 1 and 3 take constant time. Because we have to solve two subproblems—for the left subarray and for the right subarray—the contribution to the running time from lines 4 and 5 comes to 2T. As we have. For the recursive case, therefore, we have. Combining equations 4. This recurrence is the same as recurrence 4. As we shall see from the master method in Section 4. You might also revisit the recursion tree in Figure 2.

Thus, we see that the divide-and-conquer method yields an algorithm that is asymptotically faster than the brute-force method. With merge sort and now the maximum-subarray problem, we begin to get an idea of how powerful the divide- and-conquer method can be. Sometimes it will yield the asymptotically fastest algorithm for a problem, and other times we can do even better. As Exercise 4. What problem size n0 gives the crossover point at which the recursive algorithm beats the brute-force algorithm?

Then, change the base case of the recursive algorithm to use the brute-force algorithm whenever the problem size is less than n0. Does that change the crossover point? How would you change any of the algorithms that do not allow empty subarrays to permit an empty subarray to be the result?

Start at the left end of the array, and progress toward the right, keeping track of the maximum subarray seen so far. If you have seen matrices before, then you probably know how to multiply them.

Otherwise, you should read Section D. We must compute n2 matrix entries, and each is the sum of n values. The following procedure takes n n matrices A and B and multiplies them, returning their n n product C. We assume that each matrix has an attribute rows, giving the number of rows in the matrix. The for loop of lines 3—7 computes the entries of each row i , and within a given row i , the. Line 5 initializes cij to 0 as we start computing the sum given in equation 4.

You would be incorrect, however: we have a way to multiply matrices in o. We can use these equations to create a straightforward, recursive, divide-and-conquer algorithm:. This pseudocode glosses over one subtle but important implementation detail.

How do we partition the matrices in line 5? In fact, we can partition the matrices without copying entries. The trick is to use index calculations. We identify a submatrix by a range of row indices and a range of column indices of the original matrix. We end up representing a submatrix a little differently from how we represent the original matrix, which is the subtlety we are glossing over.

Let T. In the base case, when n D 1, we perform just the one scalar multiplication in line 4, and so. As discussed, partitioning the matrices in. We also must account for the four matrix additions in lines 6—9.

Since the number of matrix additions is a constant, the total time spent adding ma-. The total time for the recursive case, therefore, is the sum of the partitioning time, the time for all the recursive calls, and the time to add the matrices resulting from the recursive calls:.

When we account for the eight recursive calls, however, we cannot just sub- sume the constant factor of 8. In other words, we must say that together they take 8T. You can get a feel for why by looking back at the recursion tree in Figure 2. The factor of 2 determined how many children each tree node had, which in turn determined how many terms contributed to the sum at each level of the tree.

If we were to ignore. Bear in mind, therefore, that although asymptotic notation subsumes constant multiplicative factors, recursive notation such as T. This might be the biggest understate- ment in this book. It has four steps:. Using the submatrices created in step 1 and the 10 matrices created in step 2, recursively compute seven matrix products P1; P2; : : : ; P7.

Compute the desired submatrices C11; C12; C21; C22 of the result matrix C by adding and subtracting various combinations of the Pi matrices. Hence, we obtain the following recurrence for the running time T. We have traded off one matrix multiplication for a constant number of matrix ad- ditions.

Once we understand recurrences and their solutions, we shall see that this tradeoff actually leads to a lower asymptotic running time. By the master method in Section 4. The right-hand column just shows what these products equal in terms of the original submatrices created in step 1. Expanding out the right-hand side, with the expansion of each Pi on its own line and vertically aligning terms that cancel out, we see that C11 equals. Since we shall see in Section 4.

Note: Although Exercises 4. What would the running time of this algorithm be? Pan has discovered a way of multiplying 68 68 matrices using , mul- tiplications, a way of multiplying 70 70 matrices using , multiplications, and a way of multiplying 72 72 matrices using , multiplications. Which method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication algorithm? Answer the same question with the order of the input matrices reversed.

Now that we have seen how recurrences characterize the running times of divide- and-conquer algorithms, we will learn how to solve recurrences. We can use the substitution method to establish either upper or lower bounds on a recurrence.

As an example, let us determine an upper bound on the recurrence. We guess that the solution is T. The substitution method requires us to prove that T. Substituting into the recurrence yields T. Mathematical induction now requires us to show that our solution holds for the. Typically, we do so by showing that the boundary condi- tions are suitable as base cases for the inductive proof. For the recurrence 4. This requirement can sometimes lead to problems.

Let us assume, for the sake of argument, that T. Then for n D 1, the bound T. Consequently, the base case of our inductive proof fails to hold. We can overcome this obstacle in proving an inductive hypothesis for a spe- cific boundary condition with only a little more effort. In the recurrence 4. We keep the troublesome boundary condition T. Thus, we can replace T. Note that we make a distinction between the base case of the recurrence n D 1 and the base cases of the inductive proof n D 2 and n D 3.

With T. Now we can complete the inductive proof that T. For most of the recurrences we shall examine, it is straightforward to extend boundary conditions to make the inductive assumption work for small n, and we shall not always explicitly work out the details. Unfortunately, there is no general way to guess the correct solutions to recurrences. Guessing a solution takes experience and, occasionally, creativity.

Fortunately, though, you can use some heuristics to help you become a good guesser. You can also use recursion trees, which we shall see in Section 4. If a recurrence is similar to one you have seen before, then guessing a similar solution is reasonable.

As an example, consider the recurrence. Intuitively, however, this additional term cannot substantially affect the.

Consequently, we make the guess that T. Another way to make a good guess is to prove loose upper and lower bounds on the recurrence and then reduce the range of uncertainty. For example, we might start with a lower bound of T. Then, we can gradually lower the upper bound and raise the lower bound until we converge on the correct, asymptotically tight solution of T. Sometimes you might correctly guess an asymptotic bound on the solution of a recurrence, but somehow the math fails to work out in the induction.

The problem frequently turns out to be that the inductive assumption is not strong enough to prove the detailed bound. If you revise the guess by subtracting a lower-order term when you hit such a snag, the math often goes through. Substituting our guess in the recurrence, we obtain. We might be tempted to try a larger guess, say T. Although we can make this larger guess work, our original guess of T. In order to show that it is correct, however, we must make a stronger inductive hypothesis. Intuitively, our guess is nearly right: we are off only by the constant 1, a lower-order term.

Nevertheless, mathematical induction does not work unless we prove the exact form of the inductive hypothesis.

We overcome our difficulty by subtracting a lower-order term from our previous guess. Our new guess is T. We now have T. As before, we must choose the constant c large enough to handle the boundary conditions. You might find the idea of subtracting a lower-order term counterintuitive. Af- ter all, if the math does not work out, we should increase our guess, right? Not necessarily!

When proving an upper bound by induction, it may actually be more difficult to prove that a weaker upper bound holds, because in order to prove the weaker bound, we must use the same weaker bound inductively in the proof.

In our current example, when the recurrence has more than one recursive term, we get to subtract out the lower-order term of the proposed bound once per recursive term. In the above example, we subtracted out the constant d twice, once for the T.

We ended up with the inequality T. It is easy to err in the use of asymptotic notation. For example, in the recur- rence 4.

The error is that we have not proved the exact form of the inductive hypothesis, that is, that T. We therefore will explicitly prove that T. Sometimes, a little algebraic manipulation can make an unknown recurrence simi- lar to one you have seen before. We can simplify this recurrence, though, with a change of variables. For convenience, we shall not worry about rounding off values, such as p. Renaming m D lg n yields T. Indeed, this new recurrence has the same solution: S.

Changing back from S. Show that a substitution proof with the assumption T. Then show how to subtract off a lower-order term to make a substitution proof work. Your solution should be asymptotically tight. Do not worry about whether values are integral. Although you can use the substitution method to provide a succinct proof that a solution to a recurrence is correct, you might have trouble coming up with a good guess. Drawing out a recursion tree, as we did in our analysis of the merge sort recurrence in Section 2.

In a recursion tree, each node represents the cost of a single subproblem somewhere in the set of recursive function invocations. We sum the costs within each level of the tree to obtain a set of per-level costs, and then we sum all the per-level costs to determine the total cost of all levels of the recursion. A recursion tree is best used to generate a good guess, which you can then verify by the substitution method. If you are very careful when drawing out a recursion tree and summing the costs, however, you can use a recursion tree as a direct proof of a solution to a recurrence.

In this section, we will use recursion trees to generate good guesses, and in Section 4. For example, let us see how a recursion tree would provide a good guess for the recurrence T. We start by focusing on finding an upper bound for the solution. For convenience, we assume that n is an exact power of 4 another example of tolerable sloppiness so that all subproblem sizes are integers.

Part c shows this process carried one step further by expanding each node with cost T. The cost for each of the three children of the root is c. We continue expanding each node in the tree by breaking it into its constituent parts as determined by the recurrence.

The fully expanded tree in part d has height log4 n it has log4 nC 1 levels. Because subproblem sizes decrease by a factor of 4 each time we go down one level, we eventually must reach a boundary condition. How far from the root do we reach one? Thus, the tree has log4 nC 1 levels at depths 0; 1; 2; : : : ; log4 n. Next we determine the cost at each level of the tree.

Each level has three times more nodes than the level above, and so the number of nodes at depth i is 3i. The bottom level, at depth log4 n, has 3. This last formula looks somewhat messy until we realize that we can again take advantage of small amounts of sloppiness and use an infinite decreasing geometric series as an upper bound.

Backing up one step and applying equation A. Thus, we have derived a guess of T. In this example, the coefficients of cn2 form a decreasing geometric series and, by equation A. In other words, the cost of the root dominates the total cost of the tree.

In fact, if O. Now we can use the substitution method to verify that our guess was cor- rect, that is, T. We want to show that T. As before, we let c represent the constant factor in the O. When we add the values across the levels of the recursion tree shown in the figure, we get a value of cn for every level. The longest simple path from the root to a leaf is n!

Intuitively, we expect the solution to the recurrence to be at most the number of levels times the cost of each level, or O. Consider the cost of the leaves. Moreover, as we go down from the root, more and more internal nodes are absent.

Consequently, levels toward the bottom of the recursion tree contribute less than cn to the total cost. We could work out an accu- rate accounting of all costs, but remember that we are just trying to come up with a guess to use in the substitution method.

Let us tolerate the sloppiness and attempt to show that a guess of O. Indeed, we can use the substitution method to verify that O. We show that T. We have. Thus, we did not need to perform a more accurate accounting of costs in the recursion tree. Use the substitution method to verify your answer. Verify your bound by the substi- tution method. To use the master method, you will need to memorize three cases, but then you will be able to solve many recurrences quite easily, often without pencil and paper.

The recurrence 4. The a subproblems are solved recursively, each in time T. The function f. Replacing each of the a terms T. We will prove this assertion in the next section. We normally find it convenient, therefore, to omit the floor and ceiling functions when writing divide-and-conquer recurrences of this form.

Theorem 4. Then T. In each of the three cases, we compare the function f. Intuitively, the larger of the two functions determines the solution to the recurrence.

If, as in case 1, the function nlogb a is the larger, then the solution is T. If, as in case 3, the function f. If, as in case 2, the two func- tions are the same size, we multiply by a logarithmic factor, and the solution is T. Beyond this intuition, you need to be aware of some technicalities. In the first case, not only must f. In the third case, not only must f. This condition is satisfied by most of the polynomially bounded functions that we shall encounter.

Note that the three cases do not cover all the possibilities for f. There is a gap between cases 1 and 2 when f. Similarly, there is a gap between cases 2 and 3 when f. If the function f. To use the master method, we simply determine which case if any of the master theorem applies and write down the answer.

Since f. Case 2 applies, since f. For sufficiently large n, we have that af. Consequently, by case 3, the solution to the recurrence is T. You might mistakenly think that case 3 should apply, since. The problem is that it is not polynomially larger. The ratio f. Consequently, the recurrence falls into the gap between case 2 and case 3. See Exercise 4. Recurrence 4. As is our practice, we omit stating the base case in the recurrence. Since n3 is polynomially larger than f.

Again, case 1 applies, and we have the solution T. If his algorithm creates a subproblems, then the recurrence for the running time T. Basically, more complex assignments will cost more than simpler ones. The level of expertise is also a major determinant of the price of your assignment. Delivering a high-quality product at a reasonable price is not enough anymore. This describes us perfectly.

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