Lecture 2: Ints

Question 1

What are the objectives of this lecture? (3)

Answer 1

(1) working with and around resource limitations, (2) employing integer algorithms - binary addition, (3) identifying , describing, and effectively using integers as signed modular arithmetic and as fixed length bit vectors in c0

Question 2

Easy method to convert from binary to decimal

Answer 2

1 = 1
1 * 2 + 0 = 2
2* 2 + 0 = 4
4 * 2 + 1 = 9
9 * 2 + 0 = 18
18 * 2 + 1 = 37
37 * 2 + 1 = 75
75 * 2 + 0 = 150

Question 3

Why is overloading not a good idea for dealing with limited storage space?

Answer 3

(1) somewhat expensive because overflow condition must be explicitly detected, (2) the statement (n+n)-n = n+(n-n) would no longer be equal, as the first might give you an overload error while the second would give you n

Question 4

Why is modular arithmetic a good solution to dealing with limited storage space for ints?

Answer 4

Many of the usual laws of arithmetic continue to hold. Table of properties of the abstract algebra class of rings shared between ordinary integers and integers modulo a fixed number n
List of properties:
(1) commutativity of add
(2) associativity of add
(3) additive identity/unit
(4) additive inverse
(5) cancellation: -(-x)=x
(6) commutativity of mult
(7) associativity of mult
(8) multiplicative identity/unit
(9) distributivity
(10) annihilation: x * 0 = 0

Question 5

How do we define negative numbers?

Answer 5

as additive inverses: -x is the number y s.t. x + y = 0

Question 6

How does modular arithmetic lend itself to additive inverses?

Answer 6

The mod of a number x, is the additive inverse of -x.

x = y mod(16) implies x+y=0

Question 7

How can we write out the equivalence classes of numbers modulo 16 together with binary representation?

Answer 7

(-16, 0, 16, 0000)
(-15, 1, 17, 0001)
(-14, 2, 18, 0010)
(-13, 3, 19, 0011)
(-12, 4, 20, 0100)
(-11, 5, 21, 0101)
(-10, 6, 22, 0110)
(-9, 7, 23, 0111)
(-8, 8, 24, 1000)
(-7, 9, 25, 1001)
(-6, 10, 26, 1010)
(-5, 11, 27, 1011)
(-4, 12, 28, 1100)
(-3, 13, 29, 1101)
(-2, 14, 30, 1110)
(-1, 15, 31, 1111)

Question 8

How do you figure out the decimal value of a binary number that has a leading one? (ex. 1101, and 1001)

Answer 8

Get the mod number (2^4=9).
For 1101: 8+2+1=11 and 11 mod 16 is equal to 11 and -5. We use -5 because we want an even number of positive and negative numbers

Question 9

What is the binary procedure to convert from x to -x? How did we derive this procedure?

Answer 9

fir complement all the bits, then add 1; derived the procedure from y= -x and x + y = 10000 (base 2) = 0 (mod 16).
If we said -x was just the complement of x, then x + -x = 1111 (base 2) mod 16 = 15 mod (16) = -1 so we’re still one short

Question 10

What is the maximum number in 4-bit numbers? What is the minimum?

Answer 10

maximum: 0111 = 7; minimum: 1000 = 8, and 8 mod 16 = -8

Question 11

What is the generalized formula for the ranges of positive and negative numbers in a representation with p bits?

Answer 11

positives range from 0 to 2^p-1 and the negatives range from -2^p to -1

Question 12

What is the expression that relates integer division and integer modulus?

Answer 12

[need to be continued]