Transcript pptx

The power of logarithmic
computations
Jordi Cortadella
Department of Computer Science
Calculate 𝑥
𝑦
// Pre: 𝒙 ≠ 𝟎, 𝒚 ≥ 𝟎
// Returns 𝒙𝒚
int power(int x, int y) {
int p = 1;
for (int i = 0; i < y; ++i) p = px;
return p;
}
The algorithm performs 𝑦 multiplications.
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Can we reduce the number of multiplications?
Let us consider the calculation of 𝑥 32 :
For 𝑥 𝑦 we will need about log 2 𝑦 multiplications
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Recursive
𝑦
𝑥
• Basic case (𝑦 = 0): 𝑥 0 = 1
• Recursive cases (𝑦 > 0):
 𝑦 is even:
 𝑦 is odd:
𝑥 𝑦 = 𝑥2 𝑦 2
𝑥 𝑦 = 𝑥 ∙ 𝑥 𝑦−1 = 𝑥 ∙ 𝑥 2
(𝑦−1) 2
• For integer division:
𝑦−1 𝑦
=
2
2
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when 𝑦 is odd
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Recursive 𝑥
𝑦
// Pre: 𝒙 ≠ 𝟎, 𝒚 ≥ 𝟎
// Returns 𝒙𝒚
int power(int x, int y) {
if (y == 0) return 1;
if (y%2 == 0) return power(xx, y/2);
return xpower(xx, y/2);
}
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Example: 2
(exponents are powers of 2)
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Iterative
𝑦
𝑥
Key strategy: find a good invariant.
Let 𝑟 be the result of the computation.
int power(int x, int y) {
int z = 1;
// Invariant: 𝒓 = 𝒙𝒚 ∙ 𝒛
while (y != 0) {
if (y%2 == 0) {
x = xx; y = y/2;
} else {
z = zx; y = y – 1;
}
}
return z;
}
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𝑦
Iterative 𝑥 (improved version)
// Pre: 𝒙 ≠ 𝟎, 𝒚 ≥ 𝟎
// Returns 𝒙𝒚
int power(int x, int y) {
int z = 1;
// Invariant: 𝒓 = 𝒙𝒚 ∙ 𝒛
while (y != 0) {
if (y%2 == 1) z = zx;
x = xx;
y = y/2;
}
return z;
Example: 219
x
2
4
16
y
19
9
4
z
1
2
8
256
65536
4294967296
2
1
0
8
8
524288
}
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Fibonacci numbers
Leonardo Fibonacci
Pisa, 1170 - 1250
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Fibonacci numbers
• The Fibonacci sequence is defined as follows:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, …
• In mathematical terms, it is defined by the following
recurrence relation:
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Fibonacci numbers
Tiling with Fibonacci squares
https://en.wikipedia.org/wiki/Fibonacci_number
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Fibonacci numbers
The Fibonacci spiral
https://en.wikipedia.org/wiki/Fibonacci_number
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Fibonacci numbers
Number of petals in flowers
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Fibonacci numbers
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Fibonacci numbers
Shallow diagonals of Pascal’s Triangle
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Fibonacci numbers
// Pre: n  0
// Post: Returns the Fibonacci number of order n.
int fib(int n);
• Basic case:
n = 0 ⇒ return 0.
n = 1 ⇒ return 1.
• General case:
n > 1 ⇒ return fib(n - 1) + fib(n – 2)
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Fibonacci numbers: recursive version
// Pre: n  0
// Returns the Fibonacci number of order n.
int fib(int n) { // Recursive solution
if (n <= 1) return n;
else return fib(n - 1) + fib(n - 2);
}
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Fibonacci numbers
8
7
6
6
5
5
4
3
4
3
3
4
2
3
5
3
2 2 1
2 2 1 2 110 2 11010
2 11010 10
10
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4
3
4
3
3
2
2 2 1 2 110
2 11010 10
10
How many recursive calls?
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Fibonacci numbers
8
7
6
6
5
5
4
3
4
3
3
2
3
3
2 2 1
2 2 1 2 110 2 11010
2 11010 10
10
4
5
10
4
3
4
3
3
2
2 2 1 2 110
2 11010 10
10
For example, fib(5) is re-calculated 3 times.
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Fibonacci numbers
• When fib(8) is calculated:
–
–
–
–
–
–
–
–
fib(7) is called once
fib(6) is called twice
fib(5) is called 3 times
fib(4) is called 5 times
fib(3) is called 8 times
fib(2) is called 13 times
fib(1) is called 21 times
fib(0) is called 13 times
• When fib(n) is calculated, how many times will fib(1) and
fib(0) be called?
• Example: fib(50) calls fib(1) and fib(0) about 2.4·1010 times
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Fibonacci numbers: iterative version
// Pre: n  0
// Returns the Fibonacci number of order n.
int fib(int n) { // iterative solution
int f_i = 0;
int f_i1 = 1;
// Inv: f_i is the Fibonacci number of order i.
//
f_i1 is the Fibonacci number of order i+1.
for (int i = 0; i < n; ++i) {
int f = f_i + f_i1;
f_i = f_i1;
f_i1 = f;
}
return f_i;
Complexity: O(n)
}
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Fibonacci numbers
Algebraic solution: find matrix A such that
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Fibonacci numbers
Complexity: O(log n)
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Fibonacci numbers
typedef vector< vector<int> > M2x2;
// Pre: A and B are 2x2 integer matrices
// Returns AB
M2x2 MatrixMul(const M2x2& A, const M2x2& B) {
M2x2 C(2, vector<int>(2));
C[0][0] = A[0][0]B[0][0] + A[0][1]B[1][0];
C[0][1] = A[0][0]B[0][1] + A[0][1]B[1][1];
C[1][0] = A[1][0]B[0][0] + A[1][1]B[1][0];;
C[1][1] = A[1][0]B[0][1] + A[1][1]B[1][1];
return C;
}
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Fibonacci numbers
// Pre: A is a 2x2 integer matrix
// Returns An
M2x2 power(const M2x2& A, int n) {
if (n == 0) return Identity(); // returns I
if (n%2 == 0) return power(MatrixMul(A, A), n/2);
return MatrixMul(A, power(MatrixMul(A, A), n/2));
}
Complexity: O(log n)
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Fibonacci numbers
// Pre: n  0
// Returns the Fibonacci number of order n.
int fib(int n) {
if (n <= 1) return n;
// Creates the Fibonacci matrix [[1,1],[1,0]]
M2x2 A(2, vector<int>(2, 1));
A[1][1] = 0;
M2x2 Fn = power(A, n - 1); // Complexity O(log n)
return Fn[0][0];
}
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Fibonacci numbers and golden ratio
• Two quantities 𝑎 > 𝑏 > 0 are in the golden ratio if
𝑎+𝑏 𝑎
= =𝜑
𝑎
𝑏
• The golden ratio is:
1+ 5
𝜑=
= 1.6180339887 …
2
• The ratio of two consecutives Fibonacci numbers
converges to the golden ratio:
𝐹𝑛+1
lim
=𝜑
𝑛→∞ 𝐹𝑛
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⇒
𝐹𝑛+1 ≈ 𝜑 ∙ 𝐹𝑛
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Fibonacci numbers and golden ratio
• Let 𝐴 be the Fibonacci matrix
1 1
𝐴=
1 0
• The eigenvalues of 𝐴 are (Ax=x):
1 = 𝜑,
2 = 1 − 𝜑
• Example (𝐹15 = 610, 𝐹16 = 987, 𝐹17 = 1597):
𝜑 ∙ 𝐹15 = 987.00073313 … ≈ 𝐹16
𝜑 ∙ 𝐹16 = 1596.99954689 … ≈ 𝐹17
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Fibonacci numbers and golden ratio
2
1.9
Ratio Fn/Fn-1
1.8
1.7
1.6
1.5
1.4
1
2
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5
8
13
21
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89 144 233 377 610 987
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Conclusions
• Many naïve algorithms perform repeated
computations, often hidden behind the
natural computations.
• Identify repeated computations and re-design
algorithms accordingly. A deep knowledge of
the problem is required.
• Doubly-recursive functions usually generate
an explosion of computations (see Fibonacci).
Try to avoid them whenever possible.
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