November 15, 1995

                      cs330 - Discrete Structures
                      ===========================
                               Fall 1995

Hw #4 - Solutions
=================
In the following, bits have been sometimes grouped only for the sake of
readability.

1
=
a) 2^31 + 2^27 + 2^26 + 2^25 + 2^24 + 2^15 + 2^11 + 2^0 = 2,399,176,705

Or you can use hexadecimal for the conversion to get the same result.

   8*16^7 + 15*16^6 + 8*16^3 + 8*16^2

b) same result
c)
complement	0111 0000 1111 1111 0111 0111 1111 1110
add one                                               1
		---------------------------------------
		0111 0000 1111 1111 0111 0111 1111 1111

to get the absolute value of the number

   7*16^7 + 15*16^5 + 15*16^4 + 7*16^3 + 7*16^2 + 15*16^1 + 15*16^0 = 
= 1895790591

The string 1000 1111 0000 0000 1000 1000 0000 0001, in 2's complement
representation is the decimal -1895790591

d)
sign: 1 -> negative number
exp: 0001 1110 + 127 = 30 + 127 = -97 
significand: 1.000 0000 1000 1000 0000 0001
             ^
             |-- the hidden bit

The number is -1.00000001000100000000001*(2^-97) = 
            = -(1 + (2^-8) + (2^-12) + (2^-23))*(2^-97)
	    = -6.3370806*(10^-30) =
	    = -6.3370806e-30

2
=
a)
7*16^7 + 15*16^3 + 15*16^2 = 1879113472

b) 1879113472

c) because the MSB bit (the sign bit) is zero the number is positive
and there is no need to take the number's 2's complement.
  +1879113472

d)
sign: 0 -> positive number
exp: 1110 0000 - 127 = 97
significand: 1.000 0000 1111 1111 0000 0000 = 
	   = 1.000 0000 1111 1111

The number is +1.000000011111111*(2^97) = 
	    = +1000000011111111*(2^97)*(2*-15)  (moved the point to right)
	    = +33023*2^82 =
	    = 159689429364535596945288200192
	    = +1.5968942936*(10^29)	(approximative value)


3
=
a)
binary:	11 1111 1111		(10 bits)
octal: 1 111 111 111 -> 1777	(or 01777 to indicate it's octal)
hexa:  3ff	(or 0x3ff to indicate it's hexadecimal)

b) can't represent using only 8 bits
c) 
binary:	0000 0011 1111 1111
octal:	0 000 001 111 111 111 -> 001777
hexa:	0x03ff


4
=
a) can't represent negative numbers using standard binary
b) can't represent with 8 bits
c)
convert the absolute value (+1995) to standard binary
	1995 -> 11111001011
pad with zero to the left to get a 16 bit representation of the absolute
value
		0000 0111 1100 1011
take the 2's comp representation of the absolute value to get the
representation of -1995

complement:	1111 1000 0011 0100
add one:                          1
	        -------------------
		1111 1000 0011 0101	(-1995 in 2's comp representation)
octal: 1 111 100 000 110 101 -> 174065	(0174065)
hexa:  f835	(0xf835)

5
=
a) 10.25 -> 1010.01
b) 1010.01 -> 1.01001*2^3

6
=

1.0 -> 1.0      -> 1.0*2^0
 ^      ^
decimal binary
Positive number: sign=0 (bit31)
exp: 0 + 127 = 127 -> 0111 1111	(bits 30 to 23)
          ^
	  the bias

significant: 000 0000 0000 0000 0000 0000 (bits 22 to 0)
remember we don't represent the 1 left to the binary point.

So the representation is
	0 01111111 00000000000000000000000
	^    ^               ^
     sign   exp            significand

7
=
0.1 -> 0.0(0011) in binary. We must make sure we'll keep enough
fractional bits for the representation. The parantheses indicate the 
periodic fraction.
Normalize
0.0(0011) -> 1.(1001)*(2^-4)

sign: 0
exp: -4+127 = 123 -> 0111 1011
significand: 100 1100 1100 1100 1100 1100

So the representation is
	0 01111011 10011001100110011001100	(0x3dcccccc)

Note: In reality the representation is 0x3dcccccd (the LSB is 1) because
of the way the standard deals with the LSB. But this has not been 
discussed in class.

8
=
0x1001154f357428


9
=
Probably most SSNs can't be represented using 16 bits.