The numbers in computer memory are typically represented as floating point numbers
A floating point number is represented as
$$\textrm{number} = \textrm{significand} \times \textrm{base}^{\textrm{exponent}},$$where $\textrm{significand}$ is integer, $\textrm{base}$ is positive integer and $\textrm{exponent}$ is integer (can be negative), i.e.
$$ 1.2 = 12 \cdot 10^{-1}.$$A: In most cases, they work just fine.
However, fixed point represents numbers within specified range and controls absolute accuracy.
Floating point represent numbers with relative accuracy, and is suitable for the case when numbers in the computations have varying scale
(i.e., $10^{-1}$ and $10^{5}$).
In practice, if speed is of no concern, use float32 or float64.
In modern computers, the floating point representation is controlled by IEEE 754 standard which was published in 1985 and before that point different computers behaved differently with floating point numbers.
IEEE 754 has:
$ 0 \leq c \leq b^p - 1, \quad 1 - emax \leq q + p - 1 \leq emax$
The relative accuracy of single precision is $10^{-7}-10^{-8}$,
while for double precision is $10^{-14}-10^{-16}$.
Crucial note 1: A float32 takes 4 bytes, float64, or double precision, takes 8 bytes.
Crucial note 2: These are the only two floating point-types supported in hardware.
Crucial note 3: You should use double precision in CSE and float on GPU/Data Science.
Some demo (for division accuracy)
In [6]:
import numpy as np
import random
#c = random.random()
#print(c)
c = np.float32(0.925924589693)
a = np.float32(8.9)
b = np.float32(c / a)
print('{0:10.16f}'.format(b))
print a * b - c
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#a = np.array(1.585858585887575775757575e-5, dtype=np.float)
a = np.array(5.0, dtype=np.float32)
b = np.sqrt(a)
print('{0:10.16f}'.format(b ** 2 - a))
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a = np.array(2.28827272710, dtype=np.float32)
b = np.exp(a)
print np.log(b) - a
Many operations lead to the loss of digits [loss of significance] (https://en.wikipedia.org/wiki/Loss_of_significance)
For example, it is a bad idea to subtract two big numbers that are close, the difference will have fewer correct digits.
This is related to algorithms and their properties (forward/backward stability), which we will discuss later.
However, the rounding errors can depend on the algorithm.
Consider the simplest problem: given $n$ numbers floating point numbers $x_1, \ldots, x_n$
compute their sum
$$S = \sum_{i=1}^n x_i = x_1 + \ldots + x_n.$$The simplest algorithm is to add one-by-one.
What is the actual error for such algorithm?
Naive algorithm adds numbers one-by-one,
$$y_1 = x_1, \quad y_2 = y_1 + x_2, \quad y_3 = y_2 + x_3, \ldots.$$The worst-case error is then proportional to $\mathcal{O}(n)$, while mean-squared error is $\mathcal{O}(\sqrt{n})$.
The Kahan algorithm gives the worst-case error bound $\mathcal{O}(1)$ (i.e., independent of $n$).
Can you find the $\mathcal{O}(\log n)$ algorithm?
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n = 10 ** 8
#x = #np.random.randn(n)
#x = (-1) ** np.arange(n) + 1e-3 * np.random.randn(n)
sm = 1e-10
x = np.ones(n, dtype=np.float32) * sm
x[0] = 1.0
#x16 = np.array(x, dtype=np.float32)
#x = np.array(x16, dtype=np.float64)
true_sum = 1.0 + (n - 1)*sm
approx_sum = np.sum(x)
from numba import jit
@jit
def dumb_sum2(x):
s = np.float32(0.0)
for i in range(len(x)):
s = s + x[i]
return s
@jit
def kahan_sum(x):
s = np.float32(0.0)
c = np.float32(0.0)
for i in range(len(x)):
y = x[i] - c
t = s + y
c = (t - s) - y
s = t
return s
k_sum = kahan_sum(x)
d_sum = dumb_sum2(x)
print('Error in sum: {0:3.1e}, kahan: {1:3.1e}, dumb_sum: {2:3.1e} '.format(approx_sum - true_sum, k_sum - true_sum, d_sum - true_sum))
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import math
print math.fsum([1, 1e20, 1, -1e20] * 10000), np.sum([1, 1e20, 1, -1e20] * 10000)
Vectors typically provide an (approximate) description of a physical (or some other) object.
One of the main question is how accurate the approximation is (1%, 10%).
What is an acceptable representation, of course, depends on the particular applications. For example:
Norm is a qualitative measure of smallness of a vector and is typically denoted as $\Vert x \Vert$.
The norm should satisfy certain properties:
The distance between two vectors is then defined as $$ d(x, y) = \Vert x - y \Vert. $$
Euclidean norm, or $2$-norm, is a subclass of an important class of $p$-norms: $$ \Vert x \Vert_p = \Big(\sum_{i=1}^n |x_i|^p\Big)^{1/p}. $$ There are two very important special cases:
We will give examples where Manhattan is very important: it all relates to the compressed sensing methods that emerged in the mid-00s as one of the most popular research topics.
All norms are equivalent in the sense that
$$
C_1 \Vert x \Vert_* \leq \Vert x \Vert_{**} \leq C_2 \Vert x \Vert_*
$$
for some constants $C_1(n), C_2(n)$, $x \in \mathbb{R}^n$ for any pairs of norms $\Vert \cdot \Vert_*$ and $\Vert \cdot \Vert_{**}$. The equivalence of the norms basically means that if the vector is small in one norm, it is small in another norm. However, the constants can be large.
In [55]:
import numpy as np
n = 100
a = np.ones(n)
b = a + 1e-3 * np.random.randn(n)
print 'Relative error:', np.linalg.norm(a - b, np.inf) / np.linalg.norm(b, np.inf)
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%matplotlib inline
import seaborn as sns
import matplotlib.pyplot as plt
import numpy as np
p = 0.5 #Which norm do we use
M = 40000 #Number of sampling points
a = np.random.randn(M, 2)
b = []
for i in xrange(M):
if np.linalg.norm(a[i, :], p) <= 1:
b.append(a[i, :])
b = np.array(b)
plt.fill(b[:, 0], b[:, 1])
plt.axis('equal')
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$L_1$ norm, as it was discovered quite recently, plays an important role in compressed sensing.
The simplest formulation is as follows:
The question: can we find the solution?
The solution is obviously non-unique, so a natural approach is to find the solution that is minimal in the certain sense:
$$ \Vert x \Vert \rightarrow \min, \quad \mbox{subject to } Ax = f$$Typical choice of $\Vert x \Vert = \Vert x \Vert_2$ leads to the linear least squares problem (and has been used for ages).
The choice $\Vert x \Vert = \Vert x \Vert_1$ leads to the [compressed sensing]
(https://en.wikipedia.org/wiki/Compressed_sensing) and what happens, it typically yields the sparsest solution.
And we finalize the lecture by the concept of stability.
Let $x$ be an object (for example, a vector). Let $f(x)$ be the function (functional) you want to evaluate.
You also have a numerical algorithm alg(x) that actually computes approximation to $f(x)$.
The algorithm is called forward stable, if $$\Vert alg(x) - f(x) \Vert \leq \varepsilon $$
The algorithm is called backward stable, if for any $x$ there is a close vector $x + \delta x$ such that
$$alg(x) = f(x + \delta x)$$and $\Vert \delta x \Vert$ is small.
A classical example is the solution of linear systems of equations using LU-factorizations
We consider the Hilbert matrix with the elements
$$a_{ij} = 1/(i + j + 1), \quad i = 0, \ldots, n-1, \quad j = 0, \ldots n-1.$$And consider a linear system
$$Ax = f.$$(We will look into matrices in more details in the next lecture, and for linear systems in the upcoming weeks, but now you actually see the linear system)
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import numpy as np
n = 500
a = [[1.0/(i + j + 1) for i in range(n)] for j in range(n)] #Hil
a = np.array(a)
rhs = np.random.randn(n)
sol = np.linalg.solve(a, rhs)
print np.linalg.norm(a.dot(sol) - rhs)/np.linalg.norm(rhs) #Ax - y
#print sol
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plt.plot(sol)
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from IPython.core.display import HTML
def css_styling():
styles = open("./styles/custom.css", "r").read()
return HTML(styles)
css_styling()
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