# 0.6 Discrete structures recursion  (Page 4/8)

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Algorithm 1:   Even(positive integer k)

Input: k , a positive integer

Output: k-th even natural number (the first even being 0)

Algorithm:

if k = 1, then return 0;

else return Even(k-1) + 2 .

Here the computation of Even(k) is reduced to that of Even for a smaller input value, that is Even(k-1). Even(k) eventually becomes Even(1) which is 0 by the first line. For example, to compute Even(3), Algorithm Even(k) is called with k = 2. In the computation of Even(2), Algorithm Even(k) is called with k = 1. Since Even(1) = 0, 0 is returned for the computation of Even(2), and Even(2) = Even(1) + 2 = 2 is obtained. This value 2 for Even(2) is now returned to the computation of Even(3), and Even(3) = Even(2) + 2 = 4 is obtained.

As can be seen by comparing this algorithm with the recursive definition of the set of nonnegative even numbers, the first line of the algorithm corresponds to the basis clause of the definition, and the second line corresponds to the inductive clause.

By way of comparison, let us see how the same problem can be solved by an iterative algorithm.

Algorithm 1-a:   Even(positive integer k)

Input: k, a positive integer

Output: k-th even natural number (the first even being 0)

Algorithm:

int   i, even;

i := 1;

even := 0;

while( i<k ) {

even := even + 2;

i := i + 1;

}

return even .

Example 2: Algorithm for computing the k-th power of 2

Algorithm 2   Power_of_2(natural number k)

Input: k , a natural number

Output: k-th power of 2

Algorithm:

if k = 0, then return 1;

else return 2*Power_of_2(k - 1) .

By way of comparison, let us see how the same problem can be solved by an iterative algorithm.

Algorithm 2-a   Power_of_2(natural number k)

Input: k , a natural number

Output: k-th power of 2

Algorithm:

int   i, power;

i := 0;

power := 1;

while( i<k ) {

power := power * 2;

i := i + 1;

}

return power .

The next example does not have any corresponding recursive definition. It shows a recursive way of solving a problem.

Example 3: Recursive Algorithm for Sequential Search

Algorithm 3   SeqSearch(L, i, j, x)

Input: L is an array, i and j are positive integers, i ≤j, and x is the key to be searched for in L.

Output: If x is in L between indexes i and j, then output its index, else output 0.

Algorithm:

if i ≤j , then

{

if L(i) = x, then return i ;

else return SeqSearch(L, i+1, j, x)

}

else return 0.

Recursive algorithms can also be used to test objects for membership in a set.

Example 4: Algorithm for testing whether or not a number x is a natural number

Algorithm 4   Natural(a number x)

Input: A number x

Output: "Yes" if x is a natural number, else "No"

Algorithm:

if x<0,   then return "No"

else

if x = 0,   then return "Yes"

else return Natural( x - 1 )

Example 5: Algorithm for testing whether or not an expression w is a proposition (propositional form)

Algorithm 5   Proposition( a string w )

Input: A string w

Output: "Yes" if w is a proposition, else "No"

Algorithm:

if w is 1(true), 0(false), or a propositional variable, then return "Yes"

else if w = ~w1, then return Proposition(w1)

#### Questions & Answers

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
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Salomon
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Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
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