Gcd Which Again Is the Second Number Read in From the File Each Time

Recursion

Generally speaking, recursion is the concept of well-defined self-reference. It is the decision of a succession of elements past operating on one or more preceding elements according to a rule or a formula involving a finite number of steps.

In calculator science, recursion is a programming technique using function or algorithm that calls itself one or more than times until a specified condition is met at which time the rest of each repetition is processed from the final one called to the first.

For example, let's wait at a recursive definition of a person's ancestors:

• One's parents are one'southward ancestors
• The parents of whatever ancestor are also ancestors of the person under consideration

We can write pseudocode to determine whether somebody is someone'due south ancestor.

Function isAncestor(Person ten, Person y):
IF x is y's parent, THEN:
render true
ELSE
return isAncestor(x, y's mom) OR isAncestor(x, y'south dad)
}

This is a recursive function that calls itself. Notice that in that location'due south a example when the office does non call itself recursively, otherwise, the office will go on calling itself and volition never stop to return a value. Thus, a recursive role ordinarily has a sure structure: (1) a base case, which does non telephone call the function itself; and (2) a recursive step, which calls the function itself and moves closer to the base case. Fifty-fifty with the correct structure, we however need to be conscientious to infiniteness. You may have noticed that the higher up part isAncestor still has some problem. What if 10 is not an ancestor of y? So the program will keep asking about if 10 is y's parents' ancestor, and and then on. Information technology will never reach the base case, and the list of parents will go much farther back. The program will never stop. The problem hither is the base case here is not complete. We should add together a new base case:

Role isAncestor(Person x, Person y):
IF x is y's parent, THEN:
return true
ELSE IF x was not built-in earlier y was born, And then:
render imitation
ELSE
return isAncestor(x, y'due south mom) OR isAncestor(x, y'south dad)
}

Important: Every recursion must have at least one base instance, at which the recursion does non recur (i.e., does not refer to itself).

More than examples of recursion:

• Russian Matryoshka dolls. Each doll is made of solid wood or is hollow and contains another Matryoshka doll inside it.
• Mod OS defines file system directories in a recursive way. A file system consists of a top-level directory, and the contents of this directory consists of files and other directories.
• Much of the syntax in modern programming languages is defined in a recursive way. For example, an argument listing consists of either (1) an argument or (2) an statement list followed by a comma and an argument.

Practice: requite a recursive definition of the following data structures:

• A linked listing
• Number of nodes in a linked list

Defining Problems in Means That Facilitate Recursion

To blueprint a recursive algorithm for a given problem, information technology is useful to call up of the different ways we can subdivide this problem to define problems that have the same general structure equally the original trouble. This procedure sometimes ways we need to redefine the original problem to facilitate similar-looking subproblems.

Some observations: (ane) Avoid if possible recursive functions that brand multiple overlapping calls to themselves, which leads to an exponential complexity; and (2) repetition in code tin be achieved through recursion.

In the following examples, you should ever ask yourself what are the base of operations example and the recursive step, note the naturalness of the implementation, empathise how the loop replacement feature of recursion is involved, and maybe think most what is the running time and space usage.

Instance one: Factorial Calculation

We know that the factorial of n (north >= 0) is calculated by north! = n * (n-1) * (northward-2) * ... * 2 * one.  Notation that the production of (n-1) * (n-2) * ... * two * 1 is exactly (north-one)!. Thus we tin can write the expression as n! = n * (n-1)!, which is the recursive expression of the factorial adding.

What is the base case? What is the recursive step?

public class RecursiveFactorial {
public static void main (String[] args) {
for (int i = 1; i < 10; i++)
Arrangement.out.println(i + "\t" + factorial(i));
}
static int factorial (int n) {
if (due north < ii) return ane;                     // base instance
else return north * factorial(n-one);    // recursive case
}
}

The above recursion is chosen a linear recursion since it makes i recursive call at a fourth dimension. The loop equivalent:

public static int factorial(int due north) {
int result = one;
for (int i = 2; i <= north; i++)
result *= i;
return issue;
}

Recursion and Stacks

Let's take a shut look at the mechanism with which a recursive program is actually implemented past the compiler. In the previous example, we have seen how a recursion executes its forward and backing-out phases. The gild in which the recursive process backs out is the opposite of the order in which it goes forward. Thus some activity may be performed that involves recalling something that has been stored in the forrad process. The compiler uses a stack to implement recursion.

• In the forwarding phase, the values of local variables and parameters, and the return address are pushed on the stack for every level of the recursion
• In the backing-out phase, the stacked accost is popped and used to return to executing the residuum of the code in the calling level, and the stacked local variables and parameters are popped and used to restore the land of that call

Practise

• Exponentiation. Calculate 10ⁿ using both iteration and recursion. (Assume x > 0 and north >= 0)

Case 2: Reversing an Assortment

Let us consider the trouble of reversing the n elements of an array, A, so that the offset element becomes the final, the 2d element becomes the second to the concluding, and then on. We can solve this problem using the linear recursion, by observing that the reversal of an array can be achieved by swapping the kickoff and last elements and then recursively reversing the remaining elements in the assortment.

Algorithm ReverseArray(A, i, j):
Input: An array A and nonnegative integer indices i and j
Output: The reversal of the elements in A starting at alphabetize i and catastrophe at j
if i < j then
Bandy A[i] and A[j]
ReverseArray(A, i+one, j-1)
render

Exercises

• Summing the elements of an assortment recursively
• Finding the maximum element in an array A of n elements using recursion

Example 3: Towers of Hanoi

This is a standard trouble where the recursive implementation is piddling but the non-recursive implementation is almost impossible.

In the Towers of Hanoi puzzle, we are given a platform with iii pegs, a, b, and c, sticking out of it. On peg a is a stack of due north disks, each larger than the next, so that the smallest is on the summit and the largest is on the bottom. The puzzle is to move all the disks from peg a to c, moving one disk at a time, so that nosotros never place a larger deejay on superlative of a smaller one. The following figures requite an example of the starting position and the catastrophe position of the disks with north = 4. Let's look at an example of moving 4 disks.

a                       b                 c                                                a                 b                   c
(source)              (spare)         (dest)                                        (source)       (spare)          (dest)

Recall about what is the base of operations example? What is the recursive stride?

At the top level, we want to move 4 disks from peg a to c, with a spare peg b. We tin can suspension the problem of moving iv disks into three steps:

1. Motion disk three and smaller from peg a to b, using c as a spare peg. This can exist done past recursively calling the same procedure but with 3 disks instead. After this procedure, we will have 3 smaller disks on peg b.
2. Move disk four from peg a to peg c. Afterwards this procedure, nosotros will have three smaller disks on peg b, disk four on peg c, and peg a empty.
3. Motion disk 3 and smaller from peg b to c, using a as spare peg. Again, this can exist washed by recursively calling the same procedure on iii disks with unlike source and destination. Later this procedure, we will take all the disks on peg c without breaking the rules.

The pseudocode looks like the following. Nosotros call this function to move 4 disks past MoveDisk(iv, a, c, b).

Algorithm MoveDisk(disk, source, dest, spare) {
if (disk = = 1) then
movement disk from source to dest
else
MoveDisk(disk-1, source, spare, dest)      // pace 1 above
move disk from source to dest                    // footstep 2 above
MoveDisk(disk-ane, spare, dest, source)       // footstep 3 higher up
}

Let's trace our solution. To visualize the recursive calling process, we generate a telephone call tree. This is a call tree for moving 3 disks from peg a to c.

Notice that, each MoveDisk phone call will branch into two functions calls unless information technology'southward the base case. If we want to motion n disks, how many movements practice nosotros need with this recursive function?

Assume M(i) represents the number of movement for the disks, let's calculate how long does it accept to move n disks.

• Thou(1) = 1
• Chiliad(2) = 2M(one) + 1 = 3
• Chiliad(iii) = 2M(2) + i = 7
• M(4) = 2M(3) + 1 = 15
• ...
• We tin can guess 1000(n) = 2 ⁿ - 1

This can be verified by plugging it into our role.

• One thousand(1) = 2¹ - ane
• Thousand(northward) = 2M(n-i) + 1 = 2[2M(north-ii) + 1] + 1 = ... = two ^k K(n-k) + 2 ^k-1 + ii ^k-2 + ... + 2 + 1
• Chiliad(n) = 2 ⁿ - ane when k = n-one (stopping at the base instance)

A 64-disk version of the puzzle lies in a Hanoi monastery, where monks continuously toward solving the puzzle. When they consummate the puzzle, the world will come to an finish. Now, yous know the respond. How long will the world last? roughly 585.442 billion years. The universe is currently about 13.7 billion years former.

Instance 4: Fibonacci Sequence

Fibonacci sequence is the sequence of numbers ane, 1, 2, 3, 5, eight, 13, 21, 34, 55, .... The beginning two numbers of the sequence are both one, while each succeeding number is the sum of the 2 numbers before it. We tin can ascertain a function F(northward) that calculates the nth Fibonacci number.

First, the base cases are: F(0) = ane and F(1) = 1.

Now, the recursive case: F(north) = F(n-1) + F(n-two).

Write the recursive part and the call tree for F(5).

Algorithm Fib(n) {
if (n < 2) render ane
else return Fib(n-1) + Fib(n-2)
}

The above recursion is called binary recursion since it makes two recursive calls instead of one. How many number of calls are needed to compute the kth Fibonacci number? Let northward_thousand denote the number of calls performed in the execution.

• n₀ = 1
• n₁ = 1
• n₂ = northward₁ + n₀ + 1  = iii > 2¹
• n₃ = n_ii + n₁   + 1 = 5 > ii²
• due north_iv = n₃ + due north₂ + 1 = 9 > 2^three
• due north₅ = n₄ + n₃ + 1 = fifteen > 2ⁱⁱⁱ
• ...
• northward_k   > 2 ^m/2

This means that the Fibonacci recursion makes a number of calls that are exponential in k. In other words, using binary recursion to compute Fibonacci numbers is very inefficient.  Compare this problem with binary search, which is very efficient in searching items, why is this binary recursion inefficient? The chief problem with the approach above, is that there are multiple overlapping recursive calls.

We can compute F(n) much more efficiently using linear recursion. I style to accomplish this conversion is to define a recursive function that computes a pair of consecutive Fibonacci numbers F(north) and F(n-ane) using the convention F(-1) = 0.

Algorithm LinearFib(northward) {
Input: A nonnegative integer north
Output: Pair of Fibonacci numbers (F _{due north} , F _n-one)
if (northward <= i) then
render (n, 0)
else
(i, j) <-- LinearFib(n-1)
render (i + j, i)
}

Since each recursive call to LinearFib decreases the argument n by 1, the original telephone call results in a series of due north-1 boosted calls. This functioning is significantly faster than the exponential fourth dimension needed past the binary recursion. Therefore, when using binary recursion, we should starting time endeavour to fully sectionalisation the problem in two or, nosotros should be sure that overlapping recursive calls are actually necessary. Usually, we can eliminate overlapping recursive calls by using more memory to keep track of previous values. In fact, this approach is a central part of a technique called dynamic programming. Let'south use iteration to generate the Fibonacci numbers. What'southward the complexity of this algorithm?

public static int IterationFib(int n) {
if (n < 2) return n;
int f0 = 0, fane = one, f2 = ane;
for (int i = 2; i < north; i++) {
f0 = f1;
f1 = fii;
f2 = f0 + f1;
}
render f2;
}

Example 5: Binary Search

What's the base case? What'due south the recursive case?

public grade TestBinarySearch {
TestBinarySearch() {
int[] arr = {22, 33, 44, 55, 66, 77, 88, 99};
System.out.println("search(" + 55 + "): " + BinarySearch(arr, 0, arr.length-one, 55));
System.out.println("search(" + 50 + "): " + BinarySearch(arr, 0, arr.length-1, 50));
}

    public static void main(Cord[] args) {
new TestBinarySearch();
}

    int BinarySearch(int[] arr, int starting time, int finish, int ten) {
int mid = (start + finish) / 2;
if (arr[mid] = = x) return mid;
if (offset > finish) return -ane;
if (arr[mid] < x) return BinarySearch(arr, mid+i, terminate, 10);
else return BinarySearch(arr, first, mid-1, 10);
}
}

Exercise

• Summing the elements in an array using the binary recursion

Drawbacks of Recursion

Recursion consumes stack space. Every recursive method telephone call produces a new instance of the method, one with a new set of local variables. The full stack space used depends upon the level of nesting of the recursion process, and the number of local variables and parameters.

Recursion may perform redundant computations. Consider the recursive computation of the Fibonacci sequence.

In sum, one has to weigh the simplicity of the code delivered by recursion confronting its drawbacks as described above. When a relatively elementary iterative solution is possible, it is definitely a better culling.

Tail Recursion

Nosotros can convert a recursive algorithm into a not-recursive algorithm and in that location are some instances when nosotros can do this conversion more than easily and efficiently. Specifically, we can hands convert algorithms that use tail recursion. An algorithm uses tail recursion if it uses linear recursion and the algorithm makes a recursive call as its very last operation. The recursive call must be admittedly the last thing the method does. For example, the examples 1, 2 and five are all tail recursion, and tin be easily implemented using iteration.

Exercises

Either write the pseudo-lawmaking or the Java lawmaking for the post-obit problems. Draw the recursion trace of a simple instance. What is the running time and space requirement?.

1. Recursively searching a linked list
2. Frontward printing a linked list
3. Opposite printing a linked list

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Source: https://www.cpp.edu/~ftang/courses/CS240/lectures/recursion.htm

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