A problem can involve a lot of details. Several of these details are irrelevant for solving the problem.

**Abstraction**

To ride a bicycle, it is
sufficient to understand the functioning of the pedal, handlebar, brakes and
bell. As a rider, we model a bicycle by these four features. A bicycle has a
lot more details, which the rider can ignore. Those details are irrelevant for
the purpose of riding a bicycles.

A problem can involve a lot of
details. Several of these details are irrelevant for solving the problem. Only
a few details are essential. Abstraction is the process of ignoring or hiding
irrelevant details and modeling a problem only by its essential features. In
our everyday life, we use abstractions unconsciously to handle complexity.
Abstraction is the most effective mental tool used for managing complexity. If
we do not abstract a problem adequately, we may deal with unnecessary details
and complicate the solution.

**Example 6.9. **A map is an abstraction of** **the things we find on the ground.
We do not represent every detail on the ground. The map-maker picks out the
details that we need to know. Different maps are drawn for different purposes
and so use different abstractions, i.e., they hide or represent different
features. A road map is designed for drivers. They do not usually worry about
hills so most hills are ignored on a road map. A walker's map is not interested
in whether a road is a one-way street, so such details are ignored.

**Example 6.10. **In medicine, different** **specialists work with different
abstractions of human body. An orthopaedician works with the abstraction of
skeletal system, while a gastroenterologist works with digestive system. A
physiotherapist abstracts the human body by its muscular system.

We use abstraction in a variety of
ways while constructing algorithms — in the specification of problems,
representing state by variables, and decomposing an algorithm to functions. An
algorithm designer has to be trained to recognize which features are essential
to solve the problem, and which details are unnecessary. If we include
unnecessary details, it makes the problem and its solution over-complicated.

Specification abstracts a problem
by the properties of the inputs and the desired input-output relation. We
recognize the properties essential for solving the problem, and ignore the
unnecessary details.

**State: **In
algorithms, the state of a** **computation is abstracted by a set
of variables.

**Functions: **When
an algorithm is very** **complex, we can decompose it into
functions and abstract each function by its specification.

State is a basic and important
abstraction. Computational processes have state. A computational process starts
with an initial state. As actions are performed, its state changes. It ends
with a final state. State of a process is abstracted by a set of variables in
the algorithm. The state at any point of execution is simply the values of the
variables at that point.

**Example 6.11. Chocolate Bars: **A** **rectangular chocolate bar is
divided into squares by horizontal and vertical grooves. We wish to break the
bar into individual squares.

To start with, we have the whole
of the bar as a single piece. A cut is made by choosing a piece and breaking it
along one of its grooves. Thus a cut divides a piece into two pieces. How many
cuts are needed to break the bar into its individual squares?

In this example, we will abstract
the essential variables of the problem. We solve the problem in Example 8.6.

**Essential variables: **The number of pieces** **and the number of cuts are the
essential variables of the problem. We will represent them by two variables, p
and c, respectively. Thus, the state of the process is abstracted by two
variables p and c.

1. The problem could be cutting a
chocolate bar into individual pieces or cutting a sheet of postage stamps into
individual stamps. It is irrelevant. The problem is simply cutting a grid of
squares into individual squares.

2. The sequence of cuts that have
been made and the shapes and sizes of the resulting pieces are irrelevant too.
From p and c, we cannot reconstruct the sizes of the individual pieces. But,
that is irrelevant to solving the problem

**Example 6.12. **Consider Example 1.2, Goat,** **grass and wolf problem. In this
example,

we will write a specification of
the problem. We will solve it in Example 2.1. The problem involves four
individuals, and each is at one of the two sides of the river. This means that
we can represent the state by four variables, and each of them has one of the
two values. Let us name the variables farmer, goat, grass and wolf, and their
possible values L and R. A value of L means "at the left side". A
value of R means "at the right side". Since the boat is always with
the farmer, it is important to not introduce a variable to represent its
position.

In the initial state, all four
variables farmer, goat, grass, wolf have the value L.

**farmer, goat, grass, wolf = L, L, L, L**

In the final state, all four
variables should have the value R.

**farmer, goat, grass, wolf = R, R, R, R**

The specification of the problem
is

cross_river

**--inputs: farmer, goat, grass, wolf = L, L, L, L**

**--outputs: farmer, goat, grass, wolf = R, R, R, R**

**subject to the two constraints that**

1. the goat cannot be left alone
with the grass:

**if goat = grass then farmer = goat**

2. the goat cannot be left alone
with the wolf:

**if goat = wolf then farmer = goat**

Variables are named boxes to store
values. Assignment statement is used to store a value in a variable. It is
written with the variable on the left side of the assignment operator and a
value on the right side.

**variable := value**

When this assignment is executed,
the value on the right side is stored in the variable on the left side. The
assignment

**m := 2**

stores value 2 in variable m.

If the variable already has a
value stored in it, assignment changes its value to the value on the right
side. The old value of the variable is lost.

The right side of an assignment
can be an expression.

**variable := expression**

In this case, the expression is
evaluated and the value of the expression is stored in the variable. If the
variable occurs in the expression, the current value of the variable is used in
evaluating the expression, and then the variable is updated. For example, the
assignment

**m := m + 3**

evaluates the expression m + 3
using the current value of m.

**m + 3**

**= 2 + 3**

**= 5**

and stores the value 5 in the
variable m.

The two sides of an assignment
statement are separated by the symbol :=, known as assignment operator, and
read as "becomes" or "is assigned". The assignment
statement

**v := e**

is read as v "becomes"
e. Note that assignment operator is not equality operator^{1}. The
meanings of v := e and v = e are different. Assignment does not state a
mathematical equality of a variable, but changes the value of a variable. The
assignment m := m + 3 does not state that m is equal to m + 3. Rather, it
changes the value of the variable m to the value of the expression m + 3.

An assignment statement can change
the values of multiple variables simultaneously. In that case, the number of
variables on the left side should match the number of expressions on the right
side. For example, if we wish to assign to three variables v1, v2 and v3, we
need 3 expressions, say, el, e2, e3.

**vl, v2, v3 := el, e2 , e3**

The left side is a comma-separated
list of variables. The right side is a comma-separated list of expressions. To
execute an assignment statement, first evaluate all the expressions on the
right side using the current values of the variables, and then store them in
the corresponding variables on the left side.

**Example 6.13. **What are the values of** **variables m and n after the
assignments in

line (1) and line (3)?

**1.
****m, n := 2 , 5**

**2.
****m, n = ? , ?**

**3.
****m,n:=m+3,n-1**

**4.
****m, n = ? , ?**

The assignment in line (1) stores
2 in variable m, and 5 in variable n.

The assignment in line (3)
evaluates the expressions m + 3 and n - 1 using the current values of m and n
as

**m + 3 , n - 1**

**=2+3,5-1**

**= 5,4**

and stores the values 5 and 4 in
the variables m and n, respectively.

**1. m, n := 2,5**

**2. -- m, n = 2 , 5**

**3. m, n := m + 3, n - 1**

**4. -- m, n = 2 + 3, 5-1 = 5, 4**

Values of the variables after the
two assignments are shown in in line (2) and line(4).

**Example 6.14**.
In Example 6.11, we** **abstracted the state of the
process by two variables p and c. The next step is to model the process of
cutting the chocolate bar. When we make a single cut of a piece, the number of
pieces (p) and the number of cuts both
increase by 1. We can model it by an assignment statement.

**p, c := p + 1, c+1**

which is read as p and c
"become" p + 1 and c + 1, respectively.

Tags : Computer Science , 11th Computer Science : Chapter 6 : Specification and Abstraction

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