Transcript Lecture 11: Number Theory Algorithms

Number Theory Algorithms
Zeph Grunschlag
Copyright © Zeph Grunschlag,
2001-2002.
Agenda
Euclidean Algorithm for GCD
Number Systems


Decimal numbers (base-10)
Binary numbers (base-2)
 One’s complement
 Two’s complement

General base-b number systems
Arithmetic Algorithms



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Addition
Multiplication
Subtraction 1’s and 2’s complement
2
Euclidean Algorithm
m,n
Euclidean
Algorithm
gcd(m,n)
integer euclid(pos. integer m, pos. integer n)
x = m, y = n
while(y > 0)
r = x mod y
x=y
y=r
return x
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3
Euclidean Algorithm.
Example
gcd(33,77):
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Step
r = x mod y
x
y
0
-
33
77
4
Euclidean Algorithm.
Example
gcd(33,77):
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Step
r = x mod y
x
y
0
-
33
77
1
33 mod 77
= 33
77
33
5
Euclidean Algorithm.
Example
gcd(33,77):
Step
r = x mod y
x
y
0
-
33
77
77
33
33
11
1
2
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33 mod 77
= 33
77 mod 33
= 11
6
Euclidean Algorithm.
Example
gcd(33,77):
Step
r = x mod y
x
y
0
-
33
77
77
33
33
11
11
0
1
2
3
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33 mod 77
= 33
77 mod 33
= 11
33 mod 11
=0
7
Euclidean Algorithm.
Example
gcd(244,117):
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Step
r = x mod y
x
y
0
-
244
117
8
Euclidean Algorithm.
Example
gcd(244,117):
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Step
r = x mod y
x
y
0
-
244
117
1
244 mod 117 = 10
117
10
9
Euclidean Algorithm.
Example
gcd(244,117):
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Step
r = x mod y
x
y
0
-
244
117
1
244 mod 117 = 10
117
10
2
117 mod 10 = 7
10
7
10
Euclidean Algorithm.
Example
gcd(244,117):
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Step
r = x mod y
x
y
0
-
244
117
1
244 mod 117 = 10
117
10
2
3
117 mod 10 = 7
10 mod 7 = 3
10
7
7
3
11
Euclidean Algorithm.
Example
gcd(244,117):
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Step
r = x mod y
x
y
0
-
244
117
1
244 mod 117 = 10
117
10
2
3
4
117 mod 10 = 7
10 mod 7 = 3
7 mod 3 = 1
10
7
3
7
3
1
12
Euclidean Algorithm.
Example
gcd(244,117):
Step
r = x mod y
x
y
0
-
244
117
1
244 mod 117 = 10
117
10
2
3
4
117 mod 10 = 7
10 mod 7 = 3
7 mod 3 = 1
10
7
3
7
3
1
5
3 mod 1=0
1
0
By definition  244 and 117 are rel. prime.
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13
Euclidean Algorithm
Correctness
The reason that Euclidean algorithm
works is gcd(x,y ) is not changed from
line to line. If x’, y’ denote the next
values of x , y then:
gcd(x’,y’) = gcd(y, x mod y)
= gcd(y, x + qy)
(the useful fact)
= gcd(y, x )
(subtract y -multiple)
= gcd(x,y)
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14
Euclidean Algorithm
Running Time
EX: Compute the asymptotic running
time of the Euclidean algorithm in terms
of the number of mod operations:
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15
Euclidean Algorithm
Running Time
Assuming mod operation is O (1):
integer euclid(m, n)
x = m, y = n
while( y > 0)
r = x mod y
x=y
y=r
return x
O (1) +
?  ( O (1) +
O (1)
+ O (1)
+ O (1) )
+ O (1)
= ?  O(1)
Where “?” is the number of while-loop iterations.
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16
Euclidean Algorithm
Running Time
Facts: (x’ = next value of x, etc. )
1.
x can only be less than y at very
beginning of algorithm
–once x > y, x’ = y > y’ = x mod y
2. When x > y, two iterations of while loop
guarantee that new x is < ½ original x
–because x’’ = y’ = x mod y. Two cases:
I.
II.
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y > ½ x  x mod y = x – y < ½ x
y ≤ ½ x  x mod y < y ≤ ½ x
17
Euclidean Algorithm
Running Time
(1&2) After first iteration, size of x
decreases by factor > 2 every two
iterations.
I.e. after 2m+1 iterations,
x < original_x / 2m
Q: When –in terms of m– does this
process terminate?
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18
Euclidean Algorithm
Running Time
After 2m+1 steps, x < original_x / 2m
A: While-loop exits when y is 0, which is
right before “would have” gotten x =
0. Exiting while-loop happens when
2m > original_x, so definitely by:
m = log2 ( original_x )
Therefore running time of algorithm is:
O(2m+1) = O(m) = O (log2 (max(a,b)) )
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19
Euclidean Algorithm
Running Time
Measuring input size in terms of n = number of
digits of max(a,b):
n = (log10 (max(a,b)) ) = (log2 (max(a,b)) )
Therefore running time of algorithm is:
O(log2 (max(a,b)) ) = O(n)
(assumed naively that mod is an O(1)
operation, so estimate only holds for fixedsize integers such as int’s and long’s)
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20
Number Systems
Q: What does the string of symbols
2134
really mean as a number and why?
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21
Number Systems
A: 2 thousands 1 hundreds 3 tens and 4
= 2  103 + 1  102 + 3  101 + 4  100
But on the planet Ogg, the intelligent life
forms have only one arm with 5 fingers.
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22
Number Systems
So on Ogg, numbers are counted base 5.
I.e. on Ogg 2134 means:
2  53 + 1  52 + 3  51 + 4  50
To distinguish between these systems,
subscripts are used:
(2134)10 for Earth
(2134)5 for Ogg
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23
Number Systems
DEF: A base b number is a string of
symbols
u = ak ak-1 ak-2 … a2 a1 a0
With the ai in {0,1,2,3,…,b-2,b-1}.
The string u represents the number
(u )b = ak bk + ak-1 bk-1 + . . . + a1 b + a0
NOTE: When b > 10, run out of decimal
number symbols so after 7, 8, 9 use
capital letters, starting from A, B, C, …
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24
Number Systems
EG: base-2 (binary) 101, 00010
base-8 (octal ) 74, 0472
base-16 (hexadecimal ) 12F, ABCD
Q: Compute the base 10 version of these.
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25
Number Systems
A: base-2 (binary) 101, 00010
(101)2 = 1  22 + 0  21 + 1  20 = 5
(00010)2 = 0(24+23+22+20) + 121 = 2
base-8 (octal ) 74, 0472
(74)8 = 7  81 + 4  80 = 60
(0472)8 = 4  82 + 7  81 + 2  80 = 314
base-16 (hexadecimal ) 12F, ABCD
(12F)16 = 1162+2161+15160 = 303
(ABCD)16=10163+11162+12161+13160
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26
= 43981
Number Systems
Binary most natural system for bit-strings
and hexadecimal compactifies byte-strings
(1 byte = 2 hexadecimals)
EG in HTML:
<font color="ff00ff"> Nice Color </font>
Q: What color will this become?
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27
Number Systems
A: "ff00ff" represents the rgb –value:
The first byte is for redness, the second
byte is for green-ness, and the last for
blue-ness. The HTML above specifies that
1516 + 15 = 255 redness and blueness
values, but 016 + 0 = 0 green-ness. Red
and blue give purple, and 255 is the top
brightness so this is bright purple.
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28
Number Systems
Reverse Conversion
Convert arbitrary decimal numbers into
various bases, (calculator-functions
typically limited to base-2, 8, 16 and 10).
EG: Back at Ogg. Convert 646 to base-5.
Try to do all operations base-5.
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29
Number Systems
Reverse Conversion
Back at Ogg. Convert 646 to base-5. Try to
do all operations as an Oggian (base-5):
(646)10 = (6)10(10)102 + (4)10(10)10+ (6)10
Each quantity easy to convert into base-5:
(6)10 =(11)5 since 6 = 5 + 1
(4)10 =(4)5 since 4 < 5
(10)10 =(20)5 since 10 = 25 + 0
So convert whole expression and do Oggian
arithmetic:
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30
Number Systems
Reverse Conversion
Back at Ogg. Convert 646 to base-5. Try to
do all operations as an Oggian (base-5):
(646)10 = (11)5(20)52 + (4)5(20)5+ (11)5
=
L11
31
Number Systems
Reverse Conversion
Back at Ogg. Convert 646 to base-5. Try to
do all operations as an Oggian (base-5):
(646)10 = (11)5(20)52 + (4)5(20)5+ (11)5
= (11)5(400)5 + (130)5+ (11)5
=
L11
32
Number Systems
Reverse Conversion
Back at Ogg. Convert 646 to base-5. Try to
do all operations as an Oggian (base-5):
(646)10 = (11)5(20)52 + (4)5(20)5+ (11)5
= (11)5(400)5 + (130)5+ (11)5
= (4400)5 + (141)5
=
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33
Number Systems
Reverse Conversion
Back at Ogg. Convert 646 to base-5. Try to
do all operations as an Oggian (base-5):
(646)10 = (11)5(20)52 + (4)5(20)5+ (11)5
= (11)5(400)5 + (130)5+ (11)5
= (4400)5 + (141)5
= (10041)5
Thinking like an Oggian hurts brain too
much…
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34
Number Systems
Reverse Conversion
Given an integer n and a base b find the string u
such that (u )b = n.
Pseudocode:
string represent(pos. integer n, pos. integer b)
q = n, i = 0
while( q > 0 )
ui = q mod b
q = q/b 
i = i +1
return ui ui-1 ui-2 … u2 u1 u0
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35
Number Systems
Reverse Conversion
EG: Convert 646 to Oggian (base-5):
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i
ui = q mod b
q = q/b 
-
-
646
36
Number Systems
Reverse Conversion
EG: Convert 646 to Oggian (base-5):
L11
i
ui = q mod b
q = q/b 
0
-
646
646/5=129
646 mod 5 = 1
37
Number Systems
Reverse Conversion
EG: Convert 646 to Oggian (base-5):
L11
i
ui = q mod b
q = q/b 
0
1
-
646
646/5=129
129/5=25
646 mod 5 = 1
129 mod 5 = 4
38
Number Systems
Reverse Conversion
EG: Convert 646 to Oggian (base-5):
L11
i
ui = q mod b
q = q/b 
0
1
2
-
646
646/5=129
129/5=25
25/5=5
646 mod 5 = 1
129 mod 5 = 4
25 mod 5 = 0
39
Number Systems
Reverse Conversion
EG: Convert 646 to Oggian (base-5):
L11
i
ui = q mod b
q = q/b 
0
1
2
3
-
646
646/5=129
129/5=25
25/5=5
5/5= 1
646 mod 5 = 1
129 mod 5 = 4
25 mod 5 = 0
5 mod 5 = 0
40
Number Systems
Reverse Conversion
EG: Convert 646 to Oggian (base-5):
i
ui = q mod b
q = q/b 
0
1
2
3
4
-
646
646/5=129
129/5=25
25/5=5
5/5= 1
1/5=0
646 mod 5 = 1
129 mod 5 = 4
25 mod 5 = 0
5 mod 5 = 0
1 mod 5 = 1
Reading last column in reverse: 10041
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41
Number Systems
In-Class Exercise
Some number-theory facts are base-dependent.
For example First-Grade Teacher’s Rule:
A base-10 number is divisible by 3 iff the sum of
its digits are. Formally, let
n = (uk uk-1 uk-2 … u2 u1 u0)10. Then:


n mod 3    ui  mod 3
 i 0 
k
EG: 3|12135 because 3|(1+2+1+3+5 = 12)
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42
Arithmetical Algorithms
Addition
Numbers are added from least significant
digit to most, while carrying any
overflow resulting from adding a
column:
base-10
base-16
Carry:
x
+y
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+
7
4
6
3
2
9
0
9
+
A
4
F
0
C
B
0
9
44
Arithmetical Algorithms
Addition
Numbers are added from least significant
digit to most, while carrying any
overflow resulting from adding a
column:
base-10
Carry:
1
x
+y
L11
base-16
+
7
4
6
3
2
9
0
9
2
+
A
4
F
0
C
B
0
9
9
45
Arithmetical Algorithms
Addition
Numbers are added from least significant
digit to most, while carrying any
overflow resulting from adding a
column:
base-10
Carry:
1
x
+y
L11
base-16
+
7
4
6
3
2
9
0
9
7
2
+
A
4
F
0
C
B
0
9
F
9
46
Arithmetical Algorithms
Addition
Numbers are added from least significant
digit to most, while carrying any
overflow resulting from adding a
column:
base-10
Carry:
1
x
+y
L11
+
base-16
1
7
4
6
3
2
9
0
9
3
7
2
+
A
4
F
0
C
B
0
9
F
F
9
47
Arithmetical Algorithms
Addition
Numbers are added from least significant
digit to most, while carrying any
overflow resulting from adding a
column:
base-10
Carry:
1
x
+y
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+
1
base-16
1
1
7
4
6
3
2
9
0
9
0
3
7
2
+
A
4
F
0
C
B
0
9
6
F
F
9
48
Arithmetical Algorithms
Addition
Numbers are added from least significant
digit to most, while carrying any
overflow resulting from adding a
column:
base-10
Carry:
1
L11
1
1
7
4
6
3
+
2
9
0
9
1
0
3
7
2
x
+y
1
base-16
A
4
F
0
+
C
B
0
9
1
6
F
F
9
49
Arithmetical Algorithms
Addition of Positive Numbers
string add(strings xk xk-1…x1x0, yk yk-1…y1y0 , int base)
carry = 0, xk+1 = yk+1 = 0
for(i = 0 to k+1)
digitSum = carry + xi + yi
zi = digitSum mod base
carry = digitSum /base 
return zk+1zk zk-1…z1z0
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50
1’s Complement
2’s Complement
The binary number system makes some
operations especially simple and efficient
under certain representations.
Two such representations are


1’s complement
2’s complement
Each makes subtraction much simpler.
Each has disadvantage that number length is
pre-determined.
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51
1’s Complement
Fix k bits. (EG, k = 8 for bytes)
Represent numbers with |x | < 2k-1
Left-most bit tells the sign


0 –positive (so positive no.’s as usual)
1 –negative (but other bits change too!)
Positive numbers the same as standard binary
expansion
Negative numbers gotten by taking the
boolean complement, hence nomenclature
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52
1’s Complement
Examples
k = 8:
00010010 represents 18
11101101 represents -18
Notice: when add these representations as
usual get 11111111, i.e. negative 00000000
or -0 = 0.
Guess: adding numbers with mixed sign
works the same as adding positive numbers
Trade-off: 0 not unique
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53
1’s Complement
Addition
Addition is the same as usual binary addition except:
if the final carry is 1, cycle the carry to the least
significant digit:
00010010 represents 18, 11110011 represents -12
Sum 00000110 represents 6:
Carry:
x
+y
pre-sum
overflow
answer
L11
0
1
0
1
0
1
1
1
0
0
0
0
1
1
0
1
54
1’s Complement
Addition
Addition is the same as usual binary addition except:
if the final carry is 1, cycle the carry to the least
significant digit:
00010010 represents 18, 11110011 represents -12
Sum 00000110 represents 6:
Carry:
x
+y
pre-sum
overflow
answer
L11
0
1
0
1
0
1
1
1
0
0
0
0
1
1
0
1
1
55
1’s Complement
Addition
Addition is the same as usual binary addition except:
if the final carry is 1, cycle the carry to the least
significant digit:
00010010 represents 18, 11110011 represents -12
Sum 00000110 represents 6:
Carry:
x
+y
pre-sum
overflow
answer
L11
1
0
1
0
1
0
1
1
1
0
0
0
0
1
1
0
0
1
1
56
1’s Complement
Addition
Addition is the same as usual binary addition except:
if the final carry is 1, cycle the carry to the least
significant digit:
00010010 represents 18, 11110011 represents -12
Sum 00000110 represents 6:
Carry:
x
+y
pre-sum
overflow
answer
L11
1
0
1
0
1
0
1
1
1
0
0
0
0
1
1
1
0
0
1
1
57
1’s Complement
Addition
Addition is the same as usual binary addition except:
if the final carry is 1, cycle the carry to the least
significant digit:
00010010 represents 18, 11110011 represents -12
Sum 00000110 represents 6:
Carry:
x
+y
pre-sum
overflow
answer
L11
1
0
1
0
1
0
1
1
1
0
0
0
0
0
1
1
1
0
0
1
1
58
1’s Complement
Addition
Addition is the same as usual binary addition except:
if the final carry is 1, cycle the carry to the least
significant digit:
00010010 represents 18, 11110011 represents -12
Sum 00000110 represents 6:
Carry:
x
+y
pre-sum
overflow
answer
L11
1
0
1
0
1
0
1
1
1
1
0
0
0
0
0
0
1
1
1
0
0
1
1
59
1’s Complement
Addition
Addition is the same as usual binary addition except:
if the final carry is 1, cycle the carry to the least
significant digit:
00010010 represents 18, 11110011 represents -12
Sum 00000110 represents 6:
Carry:
x
+y
pre-sum
overflow
answer
L11
1
0
1
0
1
1
0
1
0
1
1
1
0
0
0
0
0
0
1
1
1
0
0
1
1
60
1’s Complement
Addition
Addition is the same as usual binary addition except:
if the final carry is 1, cycle the carry to the least
significant digit:
00010010 represents 18, 11110011 represents -12
Sum 00000110 represents 6:
Carry:
x
+y
pre-sum
overflow
answer
L11
1
0
1
1
0
1
0
1
0
1
0
1
1
1
0
0
0
0
0
0
1
1
1
0
0
1
1
61
1’s Complement
Addition
Addition is the same as usual binary addition except:
if the final carry is 1, cycle the carry to the least
significant digit:
00010010 represents 18, 11110011 represents -12
Sum 00000110 represents 6:
Carry:
x
+y
pre-sum
overflow
answer
L11
1
0
1
0
1
0
1
0
1
0
1
0
1
1
1
0
0
0
0
0
0
1
1
1
0
0
1
1
62
1’s Complement
Addition
Addition is the same as usual binary addition except:
if the final carry is 1, cycle the carry to the least
significant digit:
00010010 represents 18, 11110011 represents -12
Sum 00000110 represents 6:
Carry:
x
+y
pre-sum
overflow
answer
L11
1
0
1
0
1
0
1
0
1
0
1
0
1
1
1
0
0
0
0
0
0
1
1
1
0
0
1
1
1
63
1’s Complement
Addition
Addition is the same as usual binary addition except:
if the final carry is 1, cycle the carry to the least
significant digit:
00010010 represents 18, 11110011 represents -12
Sum 00000110 represents 6:
Carry:
x
+y
pre-sum
overflow
answer
L11
1
0
1
0
1
0
1
0
1
0
1
0
1
1
1
0
0
0
0
0
0
1
1
1
0
0
1
1
1
1
0
64
1’s Complement
Addition
Addition is the same as usual binary addition except:
if the final carry is 1, cycle the carry to the least
significant digit:
00010010 represents 18, 11110011 represents -12
Sum 00000110 represents 6:
Carry:
x
+y
pre-sum
overflow
answer
L11
1
0
1
0
1
0
1
0
1
0
1
0
1
1
1
0
0
0
0
0
0
1
1
1
0
0
1
1
1
1
1
0
65
1’s Complement
Addition
Addition is the same as usual binary addition except:
if the final carry is 1, cycle the carry to the least
significant digit:
00010010 represents 18, 11110011 represents -12
Sum 00000110 represents 6:
Carry:
x
+y
pre-sum
overflow
answer
L11
1
0
1
0
1
0
1
0
1
0
1
0
1
1
1
0
0
0
0
0
0
1
1
1
0
0
1
1
1
1
1
1
0
66
1’s Complement
Addition
Addition is the same as usual binary addition except:
if the final carry is 1, cycle the carry to the least
significant digit:
00010010 represents 18, 11110011 represents -12
Sum 00000110 represents 6:
Carry:
x
+y
pre-sum
overflow
answer
L11
1
0
1
0
1
0
1
0
1
0
1
0
1
1
1
0
0
0
0
0
0
1
1
1
0
0
1
1
1
0
0
0
0
0
1
1
1
0
67
2’s Complement
Fixes the non-uniqueness of zero problem
Adding mixed signs still easy
No cycle overflow (pre-computed)
Java’s approach (under the hood)
Same fixed length k , sign convention, and
definition of positive numbers as with 1’s
complement
Represent numbers with -2(k-1)  x < 2(k-1)

L11
EG. Java’s byte ranges from -128 to +127
68
2’s Complement
Negatives (slightly harder than 1’s comp.):
1. Compute 1’s complement
2. Add 1
Summarize: -x = ¬x + 1.
00010010 represents 18
11101101 + 1 = 11101110 represents -18.
Add together without over-flow: 00000000
Q: What are the ranges of Java’s 32-bit int and
64-bit long? (All of Java’s integer types use
2’s complement)
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2’s Complement
A:
1) 32-bit int’s: Largest int =
011111….1 = 231-1 = 2,147,483,647
Smallest int =
100000….0 = -231 = -2,147,483,648
2) 64-bit long’s: Largest long =
011111….1 = 263-1 =
9,223,372,036,854,775,807
Smallest int =
100000….0 = -263 =
-9,223,372,036,854,775,808
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70
2’s Complement
Addition
Addition is the same as usual binary addition
no exceptions!!
11101110 = (-18)10, 11110100 = (-12)10
Sum together (11100010) = (-30)10:
Carry:
x
+y
L11
1
1
1
1
1
1
0
1
1
0
1
1
1
0
0
0
71
2’s Complement
Addition
Addition is the same as usual binary addition
no exceptions!!
11101110 = (-18)10, 11110100 = (-12)10
Sum together (11100010) = (-30)10:
Carry:
x
+y
L11
1
1
1
1
1
1
0
1
1
0
1
1
1
0
0
0
0
72
2’s Complement
Addition
Addition is the same as usual binary addition
no exceptions!!
11101110 = (-18)10, 11110100 = (-12)10
Sum together (11100010) = (-30)10:
Carry:
x
+y
L11
1
1
1
1
1
1
0
1
1
0
1
1
1
0
1
0
0
0
73
2’s Complement
Addition
Addition is the same as usual binary addition
no exceptions!!
11101110 = (-18)10, 11110100 = (-12)10
Sum together (11100010) = (-30)10:
Carry:
x
+y
L11
1
1
1
1
1
1
1
0
1
1
0
1
1
0
1
0
1
0
0
0
74
2’s Complement
Addition
Addition is the same as usual binary addition
no exceptions!!
11101110 = (-18)10, 11110100 = (-12)10
Sum together (11100010) = (-30)10:
Carry:
x
+y
L11
1
1
1
1
1
1
1
0
1
1
1
0
0
1
1
0
1
0
1
0
0
0
75
2’s Complement
Addition
Addition is the same as usual binary addition
no exceptions!!
11101110 = (-18)10, 11110100 = (-12)10
Sum together (11100010) = (-30)10:
Carry:
x
+y
L11
1
1
1
1
1
1
1
1
0
1
0
1
1
0
0
1
1
0
1
0
1
0
0
0
76
2’s Complement
Addition
Addition is the same as usual binary addition
no exceptions!!
11101110 = (-18)10, 11110100 = (-12)10
Sum together (11100010) = (-30)10:
Carry:
x
+y
L11
1
1
1
1
1
1
1
1
1
1
0
1
0
1
1
0
0
1
1
0
1
0
1
0
0
0
77
2’s Complement
Addition
Addition is the same as usual binary addition
no exceptions!!
11101110 = (-18)10, 11110100 = (-12)10
Sum together (11100010) = (-30)10:
Carry:
x
+y
L11
1
1
1
1
1
1
1
1
1
1
1
1
0
1
0
1
1
0
0
1
1
0
1
0
1
0
0
0
78
2’s Complement
Addition
Addition is the same as usual binary addition
no exceptions!!
11101110 = (-18)10, 11110100 = (-12)10
Sum together (11100010) = (-30)10:
Carry:
x
+y
L11
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
0
1
1
0
0
1
1
0
1
0
1
0
0
0
79
2’s Complement
Addition
Addition is the same as usual binary addition no
exceptions!!
11101110 = (-18)10, 11110100 = (-12)10
Sum together (11100010) = (-30)10:
Carry:
x
+y
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
0
1
1
0
0
1
1
0
1
0
1
0
0
0
As a final check take the negative to see if get 30:
(¬11100010+1) = (00011101+1) = 00011110. YES!
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80
Arithmetical Algorithms
Positive Binary Multiplication
Long multiplication simplifies in binary because
multiplying by 2k amounts to left-shifting k-places
(<<k), and each time multiply either by 0·2k or 1·2k.
EG:
x
y
1·(x<<0)
0·(x<<1)
0·(x<<2)
1·(x<<3)
1
1
0
0
1
0
1
1
Add rows:
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81
Arithmetical Algorithms
Positive Binary Multiplication
Long multiplication simplifies in binary because
multiplying by 2k amounts to left-shifting k-places
(<<k), and each time multiply either by 0·2k or 1·2k.
EG:
x
y
1·(x<<0)
0·(x<<1)
0·(x<<2)
1·(x<<3)
1
1
1
0
0
0
1
0
1
1
1
1
Add rows:
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82
Arithmetical Algorithms
Positive Binary Multiplication
Long multiplication simplifies in binary because
multiplying by 2k amounts to left-shifting k-places
(<<k), and each time multiply either by 0·2k or 1·2k.
EG:
x
y
1·(x<<0)
0·(x<<1)
0·(x<<2)
1·(x<<3)
0
1
1
1
0
0
0
0
0
1
0
1
0
1
1
1
Add rows:
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83
Arithmetical Algorithms
Positive Binary Multiplication
Long multiplication simplifies in binary because
multiplying by 2k amounts to left-shifting k-places
(<<k), and each time multiply either by 0·2k or 1·2k.
EG:
x
y
1·(x<<0)
0·(x<<1)
0·(x<<2)
1·(x<<3)
0
0
0
1
1
1
0
0
0
0
0
0
0
1
0
1
0
1
1
1
Add rows:
L11
84
Arithmetical Algorithms
Positive Binary Multiplication
Long multiplication simplifies in binary because
multiplying by 2k amounts to left-shifting k-places
(<<k), and each time multiply either by 0·2k or 1·2k.
EG:
x
y
1·(x<<0)
0·(x<<1)
0·(x<<2)
1·(x<<3)
1
0
0
0
0
1
1
1
1
0
0
1
0
0
0
0
0
1
0
1
0
1
1
1
Add rows:
L11
85
Arithmetical Algorithms
Positive Binary Multiplication
Long multiplication simplifies in binary because
multiplying by 2k amounts to left-shifting k-places
(<<k), and each time multiply either by 0·2k or 1·2k.
EG:
x
y
1·(x<<0)
0·(x<<1)
0·(x<<2)
1·(x<<3)
Add rows:
L11
1
1
0
0
1
0
0
1
0
1
1
1
0
0
1
0
0
0
0
0
0
1
0
1
0
1
1
1
0
1
1
86
Arithmetical Algorithms
Binary Multiplication
bitstring multiply(bitstrings xk xk-1…x1x0, yk yk-1…y1y0)
x = xk xk-1…x1x0
p = 0 // the partial product
for(i = 0 to k+1)
if(yi == 1)
p = add(p , x << i ) // prev. algorithm
return p
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