CMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program Correctness

Recursive Definition & Algorithms,
and Program Correctness
Lecture 12, CMSC 56
Allyn Joy D. Calcaben

Recursion
A method where the solution to a problem depends on solutions
to smaller instances of the same problem.

We use 2 steps to define a function with the set of nonnegative
integers as its domain:
BASIS STEP: Specify the value of the function at zero

We use 2 steps to define a function with the set of nonnegative
integers as its domain:
BASIS STEP: Specify the value of the function at zero
RECURSIVE STEP: Give a rule for finding its value at an integer
from its values at smaller integers.

Example
Give the first four terms of the following sequence:
f(1) = 5
f(n + 1) = f(n) + 3
That is, we find f(1), f(2), f(3), and f(4).

Example
BASIS STEP: f(1) = 5
f(n + 1) = f(n) + 3

Example
RECURSIVE STEP: f(n + 1) = f(n) + 3

Solution
f(1) = 5

Solution
f(1) = 5
f(2) = f(1) + 3 = 5 + 3 = 8

Solution
f(1) = 5
f(2) = f(1) + 3 = 5 + 3 = 8
f(3) = f(2) + 3 = 8 + 3 = 11

Solution
f(1) = 5
f(2) = f(1) + 3 = 5 + 3 = 8
f(3) = f(2) + 3 = 8 + 3 = 11
f(4) = f(3) + 3 = 11 + 3 = 14

Example
Suppose that f is defined recursively by:
f(1) = 12
f(2) = 4
f(n + 1) = f(n) + 2 ⋅ f(n - 1)
Find f(1), f(2), f(3), and f(4).

Example
f(2) = 4
f(n + 1) = f(n) + 2 ⋅ f(n - 1)
Find f(1), f(2), f(3), and f(4).

Example
f(2) = 4
RECURSIVE STEP: f(n + 1) = f(n) + 2 ⋅ f(n - 1)
Find f(1), f(2), f(3), and f(4).

Solution
f(2) = 4
f(1) = 12

Solution
f(2) = 4
f(1) = 12
f(2) = 4

Solution
f(2) = 4
f(1) = 12
f(2) = 4
f(3) = f(2) + 2 ⋅ f(1) = 4 + 2 ⋅ 12 = 28

Solution
f(2) = 4
f(1) = 12
f(2) = 4
f(3) = f(2) + 2 ⋅ f(1) = 4 + 2 ⋅ 12 = 28
f(4) = f(3) + 2 ⋅ f(2) = 28 + 2 ⋅ 4 = 36

Example
f(0) = 3
f(n + 1) = 2 ⋅ f(n) + 3
Find f(1), f(2), f(3), and f(4).

Example
f(n + 1) = 2 ⋅ f(n) + 3
Find f(1), f(2), f(3), and f(4).

Example
RECURSIVE STEP: f(n + 1) = 2 ⋅ f(n) + 3
Find f(1), f(2), f(3), and f(4).

Solution
f(1) = 2 ⋅ f(0) + 3 = 2 ⋅ 3 + 3 = 9

Solution
f(1) = 2 ⋅ f(0) + 3 = 2 ⋅ 3 + 3 = 9
f(2) = 2 ⋅ f(1) + 3 = 2 ⋅ 9 + 3 = 21

Solution
f(1) = 2 ⋅ f(0) + 3 = 2 ⋅ 3 + 3 = 9
f(2) = 2 ⋅ f(1) + 3 = 2 ⋅ 9 + 3 = 21
f(3) = 2 ⋅ f(2) + 3 = 2 ⋅ 21 + 3 = 45

Solution
f(1) = 2 ⋅ f(0) + 3 = 2 ⋅ 3 + 3 = 9
f(2) = 2 ⋅ f(1) + 3 = 2 ⋅ 9 + 3 = 21
f(3) = 2 ⋅ f(2) + 3 = 2 ⋅ 21 + 3 = 45
f(4) = 2 ⋅ f(3) + 3 = 2 ⋅ 45 + 3 = 93

Example
Find a recursive definition for the following sequence with n as a
positive integer:
20, 17, 14, 11, 8, . . . .

Solution
positive integer:
20, 17, 14, 11, 8, . . . .

Solution
positive integer:
20, 17, 14, 11, 8, . . . .
RECURSIVE STEP: f(n + 1) = f(n) - 3

Example
Give a recursive definition of an, where a is a nonzero real number
and n is a nonnegative integer.

Solution

Solution
RECURSIVE STEP: f(n) = a ⋅ f(n - 1)

Recursively Defined Sets
Just as in the recursive definition of functions, recursive definitions
of sets have two parts:

BASIS STEP: An initial collection of elements is specified.

RECURSIVE STEP: Rules for forming new elements in the set from
those already known to be in the set are provided.

RECURSIVE STEP: Rules for forming new elements in the set from
those already known to be in the set are provided.
(Optional)
EXCLUSION RULE: Specifies that a recursively defined set contains
nothing other than those elements specified in the
basis step or generated by applications of the
recursive step.

Example
Consider the subset S of the set of integers recursively defined by:
BASIS STEP: 3 ϵ S
RECURSIVE STEP: If x ϵ S and y ϵ S, then x + y ϵ S
What is set S?

Solution
Consider the subset S of the set of integers recursively defined by:
BASIS STEP: 3 ϵ S
RECURSIVE STEP: If x ϵ S and y ϵ S, then x + y ϵ S
What is set S?
S is the set of all positive multiples of 3.

Example
Give a recursive definition of the set T = {2n – 2 | n is a natural
number greater than 0}

Solution
Elements: T = { 0, 2, 6, 14, 30, . . . . }

Solution
Elements: T = { 0, 2, 6, 14, 30, . . . . }
BASIS STEP: 0 ϵ T

Solution
Elements: T = { 0, 2, 6, 14, 30, . . . . }
BASIS STEP: 0 ϵ T
RECURSIVE STEP: if x ϵ T, then 2x + 2 ϵ T

Find f(2), f(3), f(4), and f(5) if f is defined recursively by:
f(0) = − 1, f(1) = 2, and for n = 1, 2, …
1. f (n + 1) = f (n) + 3f (n − 1)
2. f (n + 1) = f (n − 1)/f (n)
Challenge (1/4 Sheet Paper)

Recursive Algorithm
Definition 1
An algorithm is called recursive
if it solves a problem by reducing it to
an instance of the same problem with
smaller input.
1 #include <stdio.h>
2 void recurse() {
3 . . .
4 recurse();
5 . . .
6 }
7 int main() {
8 . . .
9 recurse();
10 . . .
11 }

Example
Give a recursive algorithm for computing n!, where n is a
nonnegative integer.

Solution
Give a recursive algorithm for computing n!, where n is a
nonnegative integer.

Example
Give a recursive algorithm for computing an, where a is a nonzero
real number and n is a nonnegative integer.

Solution
Give a recursive algorithm for computing an, where a is a nonzero
real number and n is a nonnegative integer.

Recursion and Iteration
2 void recurse() {
3 . . .
4 recurse();
5 . . .
6 }
7 int main() {
8 . . .
9 recurse();
10 . . .
11 }
2 int main() {
3 . . .
4 for ( initialization ; condition ; action) {
5 . . .
6 . . .
7 }
8 }

CMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program Correctness

Program Verification
A program is said to be correct if it produces the correct output for
every possible input.

A program is said to be correct if it produces the correct output for
every possible input.
Consists of 2 parts:
1. shows that the correct answer is obtained if the program
terminates. This establishes the partial correctness.
2. Shows that the program always terminates.

To specify what it means for a program to produce the correct output, two
propositions are used:
1. Initial Assertion - gives the properties that the input values must
have.
2. Final Assertion - gives the properties that the output of the
program should have, if the program did what was
intended.

Definition 1

Example
Show that the program segment
y ≔ 2
z ≔ x + y
is correct with respect to the initial assertion p : x = 1 and the final assertion
q : z = 3.

Solution
Show that the program segment
y ≔ 2
z ≔ x + y
is correct with respect to the initial assertion p : x = 1 and the final assertion
q : z = 3.
Suppose that p is true, so that x = 1 as the program begins. Then y is
assigned the value 2, and z is assigned the sum of the values of x and y,
which is 3. Hence, S is correct with respect to the initial assertion p and the
ﬁnal assertion q. Thus, p{S}q is true.

Example
Verify that the program segment
if x > y then
y ≔ x
is correct with respect to the initial assertion T and the final assertion y >= x.

Solution
Verify that the program segment
if x > y then
y ≔ x
is correct with respect to the initial assertion T and the final assertion y >= x.
When the initial assertion is true and x > y, the assignment y := x is carried
out. Hence, the final assertion, which asserts that y >= x, is true in this case.
Moreover, when the initial assertion is true and x > y is false, so that x <= y,
the final assertion is again true. Hence, using the rule of inference for
program segments of this type, this program is correct with respect to the
given initial and final assertions.

Announcement
2ND LONG EXAMINATION
OCTOBER 25, 2018
5:00PM – 7:00PM, CL3 & B6
COVERS LECTURE 7 – 12
“STUDY HARD!”

CMSC 56 | Lecture 12: Recursive Definition & Algorithms, and Program Correctness

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