Greatest Common Divisor
Jordi Cortadella
Department of Computer Science
Simplifying fractions
114 57 19
=
=
42
21
7
114 19 ∙ 6
=
42
7∙6
greatest
common
divisor
gcd(114, 42) = 6
gcd(19, 7) = 1
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The largest square tile
44
h=24
w = 60
What is the largest square tile that can exactly cover a wh rectangle?
90 tiles of side-length 4
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The largest square tile
66
h=24
w = 60
What is the largest square tile that can exactly cover a wh rectangle?
90 tiles of side-length 4
40 tiles of side-length 6
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The largest square tile
1212
h=24
w = 60
What is the largest square tile that can exactly cover a wh rectangle?
90 tiles of side-length 4
40 tiles of side-length 6
10 tiles of side-length 12
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The largest square tile
1212
h=24
w = 60
What is the largest square tile that can exactly cover a wh rectangle?
60  24 = (5  12)  (2  12) = (5  2)  (12  12)
gcd(60, 24)
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number
of tiles
largest
tile
6
Euclid (300 B.C.)
A fragment of Euclid’s Elements
http://www.math.ubc.ca/~cass/Euclid/papyrus/tha.jpg
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Euclidean algorithm for gcd
Observation:
if d divides 114 and 42, it also divides 114  42
114 42 72
−
=
𝑑
𝑑
𝑑
Therefore, all common divisors of 114 and 42
are also divisors of 72.
gcd 114, 42 = gcd(72, 42)
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Euclidean algorithm for gcd
147
21
1071
462
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Euclidean algorithm for gcd
• Properties:
gcd(a, a) = a; gcd(a, 0) = a
If a  b, then gcd(a, b) = gcd(a  b, b)
• Example
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a
114
72
b
42
42
30
30
18
6
6
42
12
12
12
6
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gcd (114, 42)
10
Euclidean algorithm for gcd
// Pre: a > 0, b > 0
// Returns the greatest common divisor of a and b
int gcd(int a, int
while (a != b)
if (a > b)
else b = b
}
return a;
}
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b) {
{
a = a – b;
– a;
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Euclidean algorithm for gcd
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iteration
a
b
0
10,000,001
154
1
9,999,847
154
2
9,999,693
154
3
9,999,539
154
…
…
154
64,934
165
154
64,935
11
154
64,936
11
143
64,937
11
132
64,938
11
121
64,939
11
110
…
11
…
64,947
11
22
64,948
11
11
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10000001 154
11 64935
12
Faster Euclidean algorithm for gcd
• Properties:
gcd(a, 0) = a
If b  0 then gcd(a, b) = gcd(a%b, b)
• Example
a
10,000,001
11
11
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b
154
154
0
13
Faster Euclidean algorithm for gcd
// Pre:
a  0, b  0
// Returns the greatest common divisor of a and b
int gcd(int a, int b) {
while (a != 0 and b != 0) {
if (a > b) a = a%b;
else b = b%a;
}
return a + b;
}
Termination: a == 0 or b == 0
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Faster Euclidean algorithm for gcd
// Pre:
a  0, b  0
// Returns the greatest common divisor of a and b
int gcd(int a, int b) {
while (b > 0) {
int r = a%b;
a = b;
b = r;
a
b
}
x
y
return a;
}
y > x%y
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a
42
114
42
30
12
b
114
42
30
12
6
6
0
15
Comparing algorithms for gcd
int gcd(int a, int b) {
while (a > 0 and b > 0) {
if (a > b) a = a%b;
else b = b%a;
}
return a + b;
}
Every iteration:
• 3 comparisons
• 1 mod operation
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int gcd(int a, int b) {
while (b > 0) {
int r = a%b;
a = b;
b = r;
}
return a;
}
Every iteration:
• 1 comparison
• 1 mod operation
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Efficiency of the Euclidean algorithm
How many iterations will Euclid’s algorithm need to
calculate gcd(a,b) in the worst case (assume a > b)?
– Subtraction version: a iterations
(consider gcd(1000000, 1))
– Modulo version:  5d(b) iterations,
where d(b) is the number of digits of b
(Gabriel Lamé, 1844)
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Binary Euclidean algorithm
Computers can perform 2 and /2 operations efficiently
(217)10 = (11011001)2
(2172)10 = (434)10 = (110110010)2
(217/2)10 = (108)10 = (1101100)2
(217%2)10 = 1 (least significant bit)
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Binary Euclidean algorithm
Assume a  b:
a
b
a
0
even even
even odd
2gcd(a/2, b/2)
gcd(a/2, b)
odd
odd
gcd(a,b/2)
gcd((a-b)/2, b)
even
odd
gcd(a,b)
a
a
132
66
33
6
3
3
3
3
b
84
42
21
21
21
9
3
0
2
2
3
12
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Binary Euclidean algorithm
// Pre:
a  0, b  0
// Returns the greatest common divisor of a and b
int gcd(int a, int b) {
int r = 1; // Accumulates common powers of two
while (a != 0 and b != 0) {
if (a%2 == 0 and b%2 == 0) {
a = a/2;
b = b/2;
r = r2;
}
else if (a%2 == 0) a = a/2;
else if (b%2 == 0) b = b/2;
else if (a > b) a = (a - b)/2;
else b = (b – a)/2;
}
return (a + b)r;
}
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Binary Euclidean algorithm
// Pre:
a  0, b  0
// Returns the greatest common divisor of a and b
int gcd(int a, int b) {
if (a == 0 or b == 0) return a + b;
int r
while
a
b
r
}
= 1; // Accumulates common powers of two
(a%2 == 0 and b%2 == 0) {
= a/2;
= b/2;
= r2;
while (a != b) {
if (a%2 == 0) a = a/2;
else if (b%2 == 0) b = b/2;
else if (a > b) a = (a - b)/2;
else b = (b – a)/2;
}
return ar;
}
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Summary
• Euclid’s algorithm is very simple.
• It is widely used in some applications related
to cryptography (e.g., electronic commerce).
• Euclid’s algorithm efficiently computes the
gcd of very large numbers.
• Question: how would you compute the least
common multiple of two numbers efficiently?
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