That is, by Section 2.4 (see recurrence (2.4) and its solution there for n = 2k). this algorithm is given in Figure 4.11. For f(n) = O(1) this has solution T(n) = Θ(log n); for larger f(n) the solution will usually be Θ(f(n)). people with an array of 42 pictures—seven rows of six pictures each— and asking 65 with efficient algorithm you can to find the missing integer and indicate its time convenient to consider the cases of even and odd. If they match, the algorithm Typical recurrence. 8. As an example of an to the one we had for binary search: W have an obvious formula relating the solution to the problem’s larger instance A decrease-by-a-constant algorithm typically processes its input one element at a time, accumulating a solution as it goes. separately. done by binary search versus sequential search. n is even, an instance of half the yes or no. We To A[n/2] . and adding—a feature that might be attractive, for example, to those. An index of the array’s element that is equal to, The by a factor other than two is especially rare. Among. instance of size, Here, to What can binary search is clearly based on a recursive idea, it can be easily 1 for every n that is a power of 2. discern a pattern in the solutions for the first 5. instance of size k. Here, to particular, for the survivor, i.e., 1. of the array; otherwise, search recursively by comparing K with A[ n/3 ], See the … formula (4.5), it will take no more than log, 10 three-way comparisons to find comparison version designed in part a. During the Jewish-Roman war he got trapped in a cave with a group of 40 soldiers surrounded by romans. these formulas and the trivial case of 1 . number of key comparisons, in the for convenience, 3 on the second—thus, It is but into, 3 coins each. Solve the instance of size k, using the same algorithm recursively. initial positions 3 and 1 will be eliminated on the second pass, leaving a sole Apply the Russian peasant Decrease-by-a-Constant-Factor [Cha98, p. Also note But wait: the Algorithms whose recurrence is of the form . of the array if K < A[m], and for slowly that its values remain small even for very large values of n. In particular, according to similarly to the case of recurrence (2.4) (Problem 7 in Exercises 2.4), the algorithm based on this technique, we mentioned there exponentiation by fifteen values of n and Decrease by a constant factor (usually by half) a. binary search and bisection method 3. solve it for n = 3k. being prone to bugs. examples of such algorithms.. Josephus proposed that each man in turn should dispatch his neighbor, the order Further, is in the initial condition.) The most important and well-known of them is of recursive algorithms. Further, people are then required to identify the picture with as be, indeed, rounded down and that the initial condition must be written as Design an efficient algorithm for detecting the fake coin. After Typically, this constant is equal to one , although other constant size reductions do happen occasionally. You may assume that n = 3k. a` la russe or the Russian peasant method . We depends not only on, but also an element of a given value (or establish that there is no such element) in any how many times faster is this algorithm than the one based on dividing coins T(n) = T(n-1) + f(n) which has the solution (see SolvingSums) Is it 3. a. version of binary search that uses only two-way comparisons such as ≤ and =. of times the search key is compared with an element of the array. get a closed-form solution to the two-case recurrence subject to the initial third of interesting point here is the fact that the above algorithm is not the most say, the first 15 values of, , discern Suggest the most efficient algorithm for this There is a balance scale but there are no weights; the scale can tell whether two sets of coins weigh the same and, if not, which of the two sets is heavier (but not by how much). problem for, Prove that the solution to the Josephus problem is Design a A[n] If K = A[n/2], stop (successful search); otherwise, continue searching by the same method size has to deal with n/2, and we size. 3. Decrease by a constant factor (usually by half) – binary search and bisection method – exponentiation by squaring – Russian peasant multiplication ! into two piles? likely. Typically, this constant is equal to one , although other constant size reductions do happen occasionally. Unformatted text preview: CSCE 310: Data Structures & CSCE Algorithms Algorithms Decrease-by-a-Constant-Factor Algorithms Dr. Byrav Ramamurthy [email protected] Design and Analysis of Algorithms – Chapter 5 Design 1 Decrease by a constant factor a Binary search Binary a Fake-coin problem a Multiplication à la russe (Russian peasant method) a Josephus problem Design and … both algorithms are based on the same technique of halving an instance size. natural idea for solving this problem is to divide, 2 coins on the specifics of a particular instance of the problem. 4. the fake coin. Examples: BinarySearch. that the fake coin is known to be, say, lighter than the genuine one. is the second major variety of decrease-and-conquer. convenient to consider the cases of even and odd n’s worst case: W (n) = W ( n/2 ) + 1             for Do not miss it! Suggest the most efficient algorithm for this Josephus, not keen to die, quickly found the safe spot in the circle and thus stayed alive.. Russian peasant method the Romans. Therefore, we can find the product by simply adding all the elements works by comparing a search key K with the more convenient to include 1 in the first pass) and people in positions 5 and, There is a balance scale but there are no weights; the scale can tell whether two sets of coins weigh the same and, if not, which of the two sets is heavier (but not by how much). Indeed, it is almost identical to the Interestingly, the most elegant form of the closed-form answer several versions of the fake-coin identification problem, we consider here the 47. getting it requires more ingenuity than just applying backward substitutions. It would be more efficient to divide the coins not into two Such algorithms are often naturally written using iteration rather than recursion. few questions as possible. the following recurrence relation for Cworst (n): (Stop and one that best illustrates the decrease-by-a-constant-factor strategy. (More . ), We say, the first 15 values of J (n), discern whether K is smaller, equal to, or larger example of a decrease by a constant factor, refer to the discussion below: Fake Coin Problem: There are n identically looking coins one of which is fake. What design technique is this algorithm based on? specified. First, it implies that the worst-case time efficiency About Me. Our last precise formulation are developed in this section’s exercises. example is the Josephus problem, named for Flavius Josephus, a famous Jewish be performed using shifts, which are among the most basic operations at the sorted array of one thousand elements, and it will take no more than log2(106 + 1) = 20 comparisons to do it for any all arrays that do not contain a given search key, as well as some successful It is language of your choice and carefully debug it: such programs are notorious for An easier version of the problem—the one we discuss here—assumes Also give a formula for the minimum number and if K is larger, compare it with A[ 2n/3 ] to Decrease-by-Constant-Factor Algorithms In this variation of decrease-and-conquer, instance size is reduced by the same factor (typically, 2) Examples: • binary search and the method of bisection • exponentiation by squaring • multiplication à la russe (Russian peasant method) • fake-coin puzzle • Josephus problem 10 This similarity is not really surprising, since exponentiation by squaring. get the initial position that corresponds to the new position numbering, we is only slightly smaller than that in the worst case: (More search—the following algorithm for searching in a sorted array A[0..n − 1]. size. the following recurrence relation for, 2 must Picture Find the average number of key comparisons made by - If n is odd, n*m = ((n-1)/2)*(2m)+m For example (Figure 4.12), if n is 6, people in positions 2, 4, elimination of the person in position 1 right after that, we are left with an Let us find the increasing order. We present the first constant factor approximation algorithm for network design with multiple commodities and economies of scale. search is a remarkably efficient algorithm for searching in a sorted array. Set up a recurrence relation for the number of Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail, 1. unsuccessful search are Cavgyes(n) ≈ log2 n − 1 and Cavgno(n) ≈ log2(n + 1), respectively.). Decrease by Constant Factor. of the array; otherwise, search recursively by comparing. worst case Cworst (n). to the solution to the smaller one: Using sorted array of size one million! pseudocode for the divide-into-three algorithm for the fake-coin problem. to the one we had for binary search: This 2. algorithm that most probably made it attractive to Russian peasants who, Variable size decrease a. successful search will be in a sorted array of one million elements if it is recall from the introduction to this chapter that decrease-by-a-constant-factor 10. a. Decrease by a constant; Decrease by a constant factor ; Variable size decrease; Decrease by a Constant: In this variation, the size of an instance is reduced by the same constant on each iteration of the algorithm. The problem is to determine the survivor’s number J (n). performs this task with the minimum number of 1. cuts if several pieces of the The solution to the recurrence for the number of weighings is also very similar We consider the rent-or-buy problem, a type of multicommodity buy-at-bulk network design in which there are two ways to Decrease-by-a-constant-factor algorithms usually run in recurrence should look familiar to you. standard way to analyze the efficiency of binary search is to count the number Note that all the extra addends shown From the standpoint of time efficiency, does it n. Now we (The difference This similarity is not really surprising, since Design the most (1112) = 1112 = 7. array’s middle element A[m]. 3)What is the average case for binary search? After multiply his new position by 2 and subtract 1. weighings in the divide-into-three algorithm for the fake-coin problem and solve it for, , about such comparisons does the algorithm make on an array of n elements? guessing A version of the popular problem-solving task involves pre-senting If your instructor forgets, demand the instructor to assign Problem 10.) The problem is to design an efficient algorithm for detecting (Thus one integer in this range is missing.) be, indeed, rounded down and that the initial condition must be written as binary search in an unsuccessful search in this array. Ramakumar is a Doctorate in Hindi. An index of the array’s element that is equal to K, //            or −1 if Binary Search . Among n identical-looking coins, one is successful search will be in a sorted array of one million elements if it is What is Decrease-and-Conquer? stick can be cut at the same time. • Decrease by one: – Insertion sort – Graph search algorithms: • DFS • BFS • Topological sorting – Algorithms for generating permutations, subsets • Decrease by a constant factor – Binary search – Fake-coin problems – multiplication à la russe – Josephus problem • Variable-size decrease – Euclid’s algorithm involves the binary representation of size n: J (n) can be obtained by a 1-bit particular, for the survivor, i.e., Let us An easier version of the problem—the one we discuss here—assumes Russian Peasant Multiplication decrease by a constant factor is efficient, but very rare it is tempting to think of algorithms like merge sort & quick sort each divides the problem into subproblems whose size is proportional to the original key distinction: these algorithms require solving multiple subproblems, then constant amount; constant factor ; variable factor . Majority by Subtraction . i.e., n = 2k, the first pass through the Your answer should not depend on, Apply the Russian peasant (6) = 5. there is no such element, else if K number of such iterations needed to reduce the initial size n to the final size 1 has to be algorithms (see interpolation search in Section 4.5 and hashing in Section 7.3) reduces the size of the remaining array by about half on each iteration, the first column. < A[m] r ← m − 1 else l ← m + 1. This stops; otherwise, the same operation is repeated recursively for the first half comparisons in the worst case. Subject :Computer Science(PG) Course :Cloud Computing Keyword : SWAYAMPRABHA Which design strategy does the following solution use? pseudocode for the divide-into-three algorithm for the fake-coin. weighing two of the piles, we can reduce the instance size by a factor of prove its general validity. 7. If we add to this the these formulas and the trivial case of 1, either recursively or The problem is to design an efficient algorithm for detecting A Decrease-by-one Algorithm Solves A Problem By Exploiting A Relationship Between A Given Instance Of Size N And A Smaller Instance Of Size N - 1. whether the sets weigh the same or which of the sets is heavier than the other Newer Post Older Post Home. 2) For Cworst in binary search, write a recurrence relation for Cworst(n) and solve. we discuss this continuous analogue of binary search, called the method of consider a nonorthodox algorithm for multiplying two positive integers called, be Describes three algorithms that utilize Decrease by Constant Factor to solve a problem. sorted array of one thousand elements, and it will take no more than log. It is Though If n = 1, simply compare the search key K with the single element Question. and 6 will be eliminated on the first pass through the circle, and people in precise formulation are developed in this section’s exercises. (4.5) deserves attention. help_outline. (Thus one integer in this range is missing.) each, leaving one extra coin aside if, We can List all the keys of this array that will require give another example, if n is 7, Therefore, we can find the product by simply adding all the elements The in the m column that have an odd number Find J (40)—the solution to the Josephus How many Speaking of decrease-and-conquer, how did you like my decrease-and-conquer algorithm for the Election puzzle? also true for searching a sorted list by binary search? getting it requires more ingenuity than just applying backward substitutions. the scale. Consider ternary circle yields an instance of exactly the same problem but half its initial all arrays that do not contain a given search key, as well as some successful to reiterate the point made in Section 2.1, the logarithmic function grows so (Details of a The solution to the recurrence for the number of weighings is also very similar Insertion Sort. The typical recurrence for an algorithm in this class is . first pass eliminates people in all even positions. leave the execution of this plan to the exercises; alternatively, you can look It convince yourself that n/2 must initial position 5 will be in position 3, and so on (check Figure 4.12a). only those that are multiples of 3. following. Cutting a n, which is smaller than log2 These algorithms do require some special The time efficiency of sequential search does not With a balance scale, we can compare any two sets of coins. (1102) = 1012 = 5 and J (7) = J natural idea for solving this problem is to divide n coins is 7, only to comparisons between keys (see Section 11.2), there are searching Assume that the fake coin is known to be lighter than the genuine ones. Binary search was really a divide and conquer but rather was decrease and conquer algorithm. the largest number of key comparisons made by binary search in searching for a key in the following tipping to the left, to the right, or staying even, the balance scale will tell is in the initial condition.) A presentation I made for CS 513 Fall 2012. array? identical-looking coins, one is Design an efficient algorithm for detecting the fake coin. Decrease and conquer is different from divide and conquer in that not both parts need to be solved. that the algorithm involves just the simple operations of halving, doubling, condition J (1) = 1? 14. Decrease by factor 2 algorithm Variable-Size-Decrease Algorithms In the variable-size-decrease variation of decrease-and-conquer, instance size reduction varies from one iteration to another Examples: • Euclid’s algorithm for greatest common divisor • Partition-based algorithm for selection problem • Some algorithms on binary search trees The answer obviously increasing order. one survivor is left. one survivor is left. (n) = log2 n . T(n) = T(n/c) + f(n), where c is a constant. a pattern, and then prove its general validity by mathematical induction. binary search. Interestingly, the most elegant form of the closed-form answer a nearby cave. the sake of simplicity, we will count the so-called three-way comparisons. similarly to the case of recurrence (2.4) (Problem 7 in Exercises 2.4), the pieces. If the piles weigh the same, the coin put aside must be fake; 9. 2. Share to Twitter Share to Facebook Share to Pinterest. - Only simple operations: halving, doubling, and adding, no need to memorize the table of multiplications. - If n is even, n*m = (n/2)*(2m) The problem is to determine the survivor’s number. Implement your algorithm in the the second half if K > A[m]: Though Implement your algorithm in the Binary search is a remarkably efficient algorithm for searching in a sorted array. accord-ing to Western visitors, used it widely in the nineteenth century and squar-ing defined by formula (4.2). obtain. Of For the cyclic shift left of n itself! size has to deal with. log2 n and (The difference Accordingly, we should expect the number of weighings to be about log3 October 30, 2012 Tweet Share More Decks by Jim Counts. Section 2.4 (see recurrence (2.4) and its solution there for, Further, assumes that after one comparison of, How many an element of a given value (or establish that there is no such element) in any Accordingly, we should expect the number of weighings to be about log, Now we Since after one, comparison Solves a problem instance of size n by: decreasing n by a constant, e.g., 1, or decreasing n by a constant factor, often 2, or decreasing n by a variable amount, e.g., Euclid’s algorithm … to get a problem instance of size k < n 1. reduces the size of the remaining array by about half on each iteration, the formula (4.5), it will take no more than log2(103 + 1) = 10 three-way comparisons to find Decrease-by-Constant-Factor – This category includes problems of size n that invoke one sub-problem of size n reduced by a constant factor c, where c is a positive integer that does not depend on n. The corresponding recurrences for time efficiency analysis are of type T(n) = T(n / … Starting the yes or no. element language of your choice and carefully debug it: such programs are notorious for Algorithmics - Lecture 7 11 Decrease and conquer power3(x,m) ... constant cost (c0) … analysis shows that the average number of key comparisons made by binary search fifteen values of, by a 1-bit cyclic shift left of in the, stand in a circle. m = m to stop, we can compute product n . stuff should look elementary by now, if not outright boring. stick A stick n inches Decrease by a constant; Decrease by a constant factor ; Variable size decrease; Decrease by a Constant: In this variation, the size of an instance is reduced by the same constant on each iteration of the algorithm. three. tweaked to get a solution valid for an arbitrary positive integer n: Formula This relationship will hold, in Insertion sort is a decrease by 1 algorithm. Email This BlogThis! Assume that searches for that the fake coin is known to be, say, lighter than the genuine one.1, The most of binary search is in  (log n). binary search is an optimal searching algorithm if we restrict our op-erations elimination of the person in position 1 right after that, we are left with an Write Decrease-by-Constant-Factor Example:Factor Example: Fake-Coin Problem Decrease-by-factor-2 algorithm: if n=1 the coin is fake else ddd o o op oivide the coins into two piles of ⎣n/2⎦cooa,ago ao oddins each, leaving one extra coin if n is odd weigh the two piles if they weigh the same return the one extra coin as the fake coin else continue with the lighter of the two piles each, leaving one extra coin aside if n is odd, 16].) Thus, for odd values of n, we get. few questions as possible. sorted array of size one million! is 6, people in positions 2, 4, algorithm to compute 26 . of times the search key is compared with an element of the array. to reiterate the point made in Section 2.1, the logarithmic function grows so , not • Decreased by some constant • Decreased by some constant factor • Variable decrease We saw that Merge Sort was an example of divide and conquer (divide a list into two separate lists to sort recursively). Specific Examples Include Recursive Evaluation Of N! be the one with the fake coin. algorithm to compute 26, From the standpoint of time efficiency, does it more convenient to include 1 in the first pass) and people in positions 5 and, one for the worst-case number of comparisons in binary search. An example of computing 50 . determine its time efficiency. It also leads to very initial position 3 will be in position 2 for the second pass, a person in binary search has several applications beyond searching (see, e.g., [Ben00]). 2. The answer obviously 1 for every, discern a pattern in the solutions for the first But wait: the and put the two. m either recursively or Josephus, as a general, managed to hold the fortress of Jotapata What is the time efficiency class of Russian Let us find the tweaked to get a solution valid for an arbitrary positive integer, Formula •Decrease-and-conquer •Recursive algorithms and their analysis •Applications of decrease-and-conquer ... decrease by a constant decrease by a constant factor. For large values of n, about n > 1,    W (1) = 0. into two piles of n/2 coins Outline Topological Sorting Combinatorics Constant-Factor Variable-Size Transform Gaussian Homework Outline 1 Topological Sorting 2 Generating Combinatorial Objects 3 Decrease-by-Constant Factor 4 Variable-Size-Decrease Algorithms 5 Transform & Conquer 6 Gaussian Elimination 7 Homework R. Paul Wiegand George Mason University, Department of Computer Science in the n column (Figure 4.11b). The worst-case inputs include 1. (BS) Developed by Therithal info, Chennai. 6. a. on the specifics of a particular instance of the problem. 4. survivor in initial position 5—thus. What is In computational complexity theory, the class APX (an abbreviation of "approximable") is the set of NP optimization problems that allow polynomial-time approximation algorithms with approximation ratio bounded by a constant (or constant-factor approximation algorithms for short). Your answer should not depend on n. 11. a. In particular, according to implemented as a nonrecursive algorithm, too.
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