The for loop

1.5 The for loop

20 Definition The for loop has either of the following syntaxes:1 for indexvariable ← lowervalue to uppervalue

do statements or

for indexvariable ← uppervalue downto lowervalue do statements

Here lower value and upper value must be non-negative integers with uppervalue ≥ lowervalue.

21 Example (Factorial Integers) Recall that for a non-negative integer n the quantity n! (read “ n factorial”) is defined as follows. 0! = 1 and if n > 0 then n! is the product of all the integers from 1 to n inclusive:

n! = 1 ·2 ···n.

For example 5! = 1 ·2 ·3 ·4 ·5 = 120. Write an algorithm that given an arbitrary non-negative integer n outputs n!.

22 Example (Positive Integral Powers 1) Write an algorithm that will compute xn, where x is a given real number and n is a given positive integer.

Solution: We can approach this problem as we did the factorial function in example 21. Thus a possible answer would be

In example 34 we shall examine a different approach.

23 Example (Reversing an Array) An array (X [0], . . . X [n −1]) is given. Without introducing another array, put its entries in reverse order.

Solution: Observe that we exchange

X [0] ↔ X [n −1],

X [1] ↔ X [n −2],

1 The syntax in C, C++, and Java is slightly different and makes the for loop much more powerful than the one we are presenting here.

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This holds as long as i < n − i − 1, that is 2i < n − 1, which happens if and only if 2i ≤ n − 2, which happens if and only if i ≤ b(n −2)/2c. We now use a swapping algorithm, say the one of example 13.Thus a possible answer is

24 Definition The command break stops the present control statement and jumps to the next control statement. The command output(. . . ) prints whatever is enclosed in the parentheses.

Many a programmer considers using the break command an ugly practice. We will use it here and will abandon it once we study the while loop.

25 Example What will the following algorithm print?

Solution: We have, in sequence,

The programme ends up printing 3456.

26 Example (Maximum of n Numbers) Write an algorithm that determines the maximum element of a 1-dimensional array of n elements.

Solution: We declare the first value of the array (the 0-th ent ry) to be the maximum (a sentinel value). Then we successively compare it to other n −1 entries. If an entry is found to be larger than it, that entry is declared the maximum.

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Recall that a positive integer p > 1 is a prime if its only positive factors of p are either 1 or p. An integer greater than 1 which is not prime is said to be composite.2 To determine whether an integer is prime we rely on the following result.

√

27 Theorem Let n > 1 be a positive integer. Either n is prime or n has a prime factor ≤ n.

Proof: If n is prime there is nothing to prove. Assume then than n is composite. Then n can be written as the

√

28 Example To determine whether 103 is prime we proceed as follows. Observe that b 103c = 10.3 We now divide 103 by every prime ≤ 10. If one of these primes divides 103 then 103 is not a prime. Otherwise, 103 is a prime. A quick division finds

103 mod 2 = 1,

103 mod 3 = 1,

103 mod 5 = 3,

103 mod 7 = 5,

whence 103 is prime since none of these remainders is 0.

29 Definition (Boolean Variable) A boolean variable is a variable that only accepts one of two possible values: true or false.

The not unary operator changes the status of a boolean variable from true to false and viceversa.

30 Example (Eratosthenes' Primality Testing) Given a positive integer n write an algorithm to determine whether it is prime.

Solution: Here is a possible approach. The special cases n = 1, n = 2, n = 3 are necessary because in our version of the for loop we need the lower index to be at most the upper index.

2Thus 1 is neither prime nor composite.

3Here bxc denotes the floor of x, that is, the integer just to the left of x if x is not an integer and x otherwise.

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From a stylistic point of view, this algorithm is unsatisfactory, as it uses the break statement. We will see in example 35 how to avoid it.

31 Example (The Locker-room Problem) A locker room contains n lockers, numbered 1 through n. Initially all doors are open. Person number 1 enters and closes all the doors. Person number 2 enters and opens all the doors whose numbers are multiples of 2. Person number 3 enters and if a door whose number is a multiple of 3 is open then he closes it; otherwise he opens it. Person number 4 enters and changes the status (from open to closed and viceversa) of all doors whose numbers are multiples of 4, and so forth till person number n enters and changes the status of door number n. Write an algorithm to determine which lockers are closed.

Solution: Here is one possible approach. We use an array Locker of size n + 1 to denote the lockers (we will ignore Locker[0]). The value true will denote an open locker and the value false will denote a closed locker.4

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34 Example (Positive Integral Powers 2) Write an algorithm that will compute an, where a is a given real number and n is a given positive integer.

Solution: We have already examined this problem in example 22. From the point of view of computing time, that solution is unsatisfactory, as it would incur into n multiplications, which could tax the computer memory if n is very large. A more efficient approach is the following. Basically it consists o f writing n in binary. We successively square x getting a sequence

x → x2 → x4 → x8 → ··· → x2k ,

and we stop when 2k ≤ n < 2k+1. For example, if n = 11 we compute x → x2 → x4 → x8. We now write 11 = 8 + 2 + 1 and so x11 = x8x2x.

The§ while loop can be used to replace the for loop, and in fact, it is more efficient than it. For, the code for i ← k to n do something

is equivalent to

i ← k

while i <= n

do

i ← i + 1 something

But more can be achieved from the while loop. For instance, instead of jumping the index one-step-at-a-time, we could jump t steps at a time by declaring i ← i + t. Also, we do not need to use the break command if we incorporate the conditions for breaking in the test of the loop.

35 Example Here is the ISPRIME1 programme from example 30 with while loops replacing the for loops. If n > 3, then n is divided successively by odd integers, as it is not necessary to divide it by even integers.

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