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Number Representations and
Computer Arithmetic
“Computing” with Computers

Back in the “old days”
(WW II time), the
word “computer”
meant this
• “Computers” were used
to compute things such
as trajectory tables,
etc.
– redundant “computers”
for reliability and speed
• Quite effective!
CS 21a
9/23/02
– new computers were tested
against human computers
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 2
“Computing” with Computers
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Of course, we don’t do things
that way anymore!
But HOW do (electronic)
computers compute?
The trick is to use BINARY
numbers
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CS 21a
9/23/02
use 1’s and 0’s
corresponds on current/voltage
being ON or OFF
easy to implement
resilient against noise
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 3
Java types: int and double
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int range:
-2,147,483,648 to 2,147,483,647
double range:
4.94e-324 to 1.80e+308
These ranges depend on:
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CS 21a
9/23/02
Storage size
Internal data representation
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 4
Review: Decimal Numbers
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Integer Representation

number is sum of DIGIT * “place value”
d7 d6 d5 d4 d3 d2 d1 d0
Range
0 to 10n - 1

107 106 105 104 103 102 101 100
379210 = 3  103 + 7  102 + 9 
= 3000 + 700 + 90 + 2
101
+ 2 
100
Adding two decimal numbers

add by “place value”, one digit at a time
3792
+ 531
???
CS 21a
9/23/02
0 0 0 0 3 7 9 2
1
3 7 9 2
+ 0 5 3 1
0 4 3 2 3
“carry 1”
because
9+3 = 12
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 5
Binary Numbers
• Humans can naturally count up to 10 values,
• But computers can count only up to 2 values
(OFF and ON, or 0 and 1)

(Unsigned) Binary Integer Representation

“base” of place values is 2, not 10
b7 b6 b5 b4 b3 b2 b1 b0
Range
0 to 2n - 1
CS 21a
9/23/02
0 1 1 0 0 1 0 0
27 26 25 24 23 22 21 20
011001002 = 26 + 25 +
22
= 64 + 32 + 4
aka “0b01100100”
= 10010
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 6
Converting from Binary to Decimal
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VERY USEFUL trick for a CS/MIS person …
Memorize powers of 2 from 20 up to 210
Lets you approximate any power of 2
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Example 1: what is 216?
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“1 KB” is actually 1,024 bytes, not 1000 bytes
“1 MB” is 1K*1KB = 220 bytes
= 1,048,576 bytes
= approximately 1 million bytes
“1 GB” is 1K * 1MB = 230 bytes = (approx 1 billion)
26 * 210 = 64 * 1024 = 64 K = 65,536
Example 2:
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CS 21a
9/23/02
20
21
22
23
24
25
26
27
28
29
210
=
=
=
=
=
=
=
=
=
=
=
1
2
4
8
16
32
64
128
256
512
1,024
or “1K”
The Pentium processor does integer math
with 32-bit numbers. What’s the highest
unsigned number it can handle (approximately)?
range = 2n-1 = 232 - 1
232 = 22 * 230
= 4 * 1 G = approximately 4 billion
= actually 4*1024*1024*1024 = 4,294,967,296 (minus 1)
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 7
Converting from Binary to Decimal
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More practice
0b00000010
0b00001110
0b00110001
0b10110001
CS 21a
9/23/02
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
20
21
22
23
24
25
26
27
28
29
210
=
=
=
=
=
=
=
=
=
=
=
1
2
4
8
16
32
64
128
256
512
1,024
or “1K”
Odds and Ends
Slide 8
From Decimal to Binary
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General rule:
Divide and write remainder
from right to left
Why this works
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remainder gives bit 0
odd numbers have a 1 in bit 0
Note that this rule works in
converting decimal to any base

e.g., HEX numbers (more later)
3792 / 2 = 1896 rem 0
1896 / 2 = 948 rem 0
948 / 2 = 474 rem 0
474 / 2 = 237 rem 0
237 / 2 = 118 rem 1
118 / 2 =
59 rem 0
59 / 2 =
29 rem 1
29 / 2 =
14 rem 1
14 / 2 =
7 rem 0
7 / 2 =
3 rem 1
3 / 2 =
1 rem 1
1 / 2 =
0 rem 1
0b111011010000
= 211+210+29+27+26+24
= 3792
CS 21a
9/23/02
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 9
Binary Arithmetic

Adding two decimal numbers
3792
+ 531
???

0 4 3 2 3
“carry 1”
because
9+3 = 12
Adding two binary numbers

CS 21a
9/23/02
1
3 7 9 2
+ 0 5 3 1
same, but “1 + 1 = 10”
14
+ 7
1 1 1
1 1 1 0
+ 0 1 1 1
21
1 0 1 0 1
“carry 1”
because
1+1 = 10
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 10
Binary Arithmetic
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In general:
Add up to 3 bits at a time
per place value
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1 1 1 0 0
1 1 1 0
+ 0 1 1 1
sum bits
carry-out bits
0 1 0 1
1 1 1 1 0
Output 2 bits at a time
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CS 21a
9/23/02
A and B
“carry in”
carry-in bits
A bits
B bits
sum bit for that place value
“carry out” bit
(becomes carry-in
of next bit)
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 11
So what about negative values?
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In math, it is easy to represent negative values
Just put a negative sign (-) prefix
In computers, we extend the binary notation in
order to support signed values
We can use the following 3 methods:
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CS 21a
9/23/02
Sign and magnitude (using the Most Significant Bit
or MSB)
1’s complement
2’s complement
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 12
Signed Integers
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Sign-and-Magnitude
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210
-210
0 010
1 010
010
-010
0 000
1 000
flip bits
easy to do subtraction
STILL has 2 “zeros”
210
-210
0 010
1 101
010
-010
0 000
1 111
210
-210
0 010
1 110
010
-010
0 000
0 000
1’s complement
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MSB represents sign bit (0 for
positive, 1 for negative)
has 2 “zeroes”
2’s complement
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flip bits, then add 1
easy subtraction
(just negate, then add)
has only 1 zero
Another interpretation:

CS 21a
9/23/02
add place values as before,
except that MSB is negative
n-1)John Paul Vergara,
(i.e., MSB©isLuis
place
is 2and
F. G.value
Sarmenta
Ateneo de Manila University
Odds and Ends
Slide 13
Binary Subtraction

1’s complement subtraction: flip the negative
5 - 3 = 2
0 1 0 1
0101
0011 flip
1100
-3 in 1’s complement form
3 - 5 = -2
0011
0101 flip
1010
+1 1 0 0
1 0 0 0 1
1
0 0 1 0
Add the
carry
overflow
2
0 0 1 1
+1 0 1 0
1 1 0 1
-2
-5 in 1’s complement form
CS 21a
9/23/02
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 14
Binary Subtraction

2’s complement subtraction: flip then add 1
5 - 3 = 2
0101
0011 flip
1100 +1
1101
-3 in 2’s complement form
3 - 5 = -2
0011
0101 flip
1010 +1
1011
-5 in 2’s complement form
CS 21a
9/23/02
0 1 0 1
+1 1 0 1
1 0 0 1 0
2
ignore
overflow
0 0 1 1
+1 0 1 1
1 1 1 0
0001 flip
0010 +1
-2
(flip+1 also gives positive of negative number)
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 15
Range of binary numbers
Java (and most computer platforms
today) use 2’s comp.
 Hence the range for the different int
types

byte (8 bits): ?
 short (16 bits): ?
 int (32 bits): -2,147,483,648 to
2,147,483,647
 long (64 bits): ?

CS 21a
9/23/02
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 16
Hexadecimal (Hex) Numbers
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aka “0x64”
Base 16
Each hex digit goes
6416 = 6*1610 + 4 =
from 0-9, then A-F
Hex is a convenient
shortform for binary
b7 b6 b5 b4 b3 b2 b1
4 bits = 1 hex digit
0 1 1 0 0 1 0
(aka nibble)
27 2 6 25 24 23 22 21
1 byte = 8 bits = 2 hex digits
Another useful trick:
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
b0
0
20
memorize binary of 0 to F
Addition and conversion
to/from decimal is similar

CS 21a
9/23/02
10010
except use base 16 instead of 2
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 17
Some Exercises
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What’s the range of an n-bit sign-mag integer?
How about 1’s comp? 2’s comp?
Convert the ff signed 8-bit values to decimal
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Convert the ff to 16-bit binary and hex numbers:
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CS 21a
9/23/02
413, 39, 1045, -3, -124, -134
Add the following pairs
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0b01010101, 0b1110111, 0x14, 0x41
0b00001011 + 0b00100100, 0x3F + 0x2F
If 0x7F and 0x32 are signed 8-bit integers, what’s
wrong with adding them and storing the result in a
byte?
Puzzle: How can I use my 10 fingers to count up to
1000?
© Luis F. G. Sarmenta and John Paul Vergara,
Ateneo de Manila University
Odds and Ends
Slide 18