# SICP Solutions

### Section - 2.1 - Introduction to Data Abstraction

#### Exercise 2.9

Lets consider the interval $I_1 = [l_1, u_1]$ and $I_2 = [l_2, u_2]$ with width $w_1 = \frac {u_1 - l_1} 2$ and $w_2 = \frac {u_2 - l_2} 2$.

$I_{add} = I_1 + I_2 = [l_1 + l_2, u_1 + u_2]$. Thus the width is:

$w_{add} = \frac {(u_1+u_2) - (l_1+l_2)} 2$
$= { \frac {u_1 - l_1} 2 } + { \frac {u_2 - l_2} 2 }$
$= w_1 + w_2$

Subtraction:

$I_{sub} = I_1 - I_2 = [l_1 - u_2, u_1 - l_2]$. Thus the width is:

$w_{sub} = \frac {(l_1-u_2) - (u_1-l_2)} 2$
$= { \frac {l_1 - u_1} 2 } + { \frac {l_2 - u_2} 2 }$
$= - (w_1 + w_2)$

Clearly, width of addition and subtraction of intervals is a function of the width of individual intervals.

If the width of the result of an operation(+, -, *, or /) between two intervals is a function of the widths of individual intervals, then applying same operation(+, -, *, or /) on two different pairs of intervals such that both pairs have same widths, will result in intervals with equals widths.

Lets take two different intervals $I_1 = [l_1, u_1]$ and $I_2 = [l_2, u_2]$ with equal widths say $w_p = \frac {u_1 - l_1} 2 = \frac {u_2 - l_2} 2$.

Now consider another interval $I_3 = [l_3, u_3]$ with width say $w_q = \frac {u_3 - l_3} 2$ such that $w_p \neq w_q$.

Lets say width of interval $(I_1 \text{ op } I_3)$ is $w_1$ and width of $(I_2 \text{ op } I_3)$ is $w_2$, where $\text{ op }$ denotes the arithmetic operation i.e. +, -, *, or /.

If width is a function of individual intervals, then it must be the case that $w_1 = w_2$ because width of pair $(I_1, I_3)$ and $(I_2, I_3)$ is same i.e. $(w_p, w_q)$. Thus $w_1 = w_2 = f(w_p, w_q)$ where $f$ is some function.

$I_1 = [0, 10], \; I_2 = [20, 30], \; I_3 = [100, 200]$. Thus we have $w_p = w_{I_1} = w_{I_2} = 5$ and $w_q = w_{I_3} = 50$.
$I_1 + I_3 = [100, 210]$. Thus width $w_1 = (210-100)/2 = 55 = w_p + w_q$. Also, $I_1 + I_3 = [120, 230]$. Thus width $w_2 = (230-120)/2 = 55 = w_p + w_q$.
Clearly $w_1 = w_2$ in case of addition operation.
$I_1 \times I_3 = [0, 2000]$. Thus width $w_1 = (2000-0)/2 = 1000$. Also, $I_2 \times I_3 = [2000, 6000]$. Thus width $w_2 = (6000-2000)/2 = 2000$.
Here $w_1 \neq w_2$. Hence width of the result of multiplication of two intervals is not a function of individual intervals.