How would the following pseudocode translate into Python?
function IntNoise(32-bit integer: x)
x = (x<<13) ^ x;
return ( 1.0 - ( (x * (x * x * 15731 + 789221) + 1376312589) & 7fffffff) / 1073741824.0);
end IntNoise function
I'm not sure about the following items: the 32-bit integer: x argument in the IntNoise call ; the <<, and the &7fffffff.
The function is a random number generator from this webpage: Perlin Noise.
Line by line, here are the changes:
function IntNoise(32-bit integer: x)
We don't need to declare the argument type, and prefer not to use CamelCase, so line one is:
def intnoise(x):
The only thing wrong with the next line is the semicolon. Removing it, we get:
x = (x << 13) ^ x
x will be shifted to the left by 13 bits and then the result will be bitwise exclusive-OR-ed with the starting value of x.
On the next line, once again no semicolon, and the 7ffffff needs to be prefixed with 0x, thus:
return ( 1.0 - ( (x * (x * x * 15731 + 789221) + 1376312589) & 0x7fffffff) / 1073741824.0)
Altogether, this makes:
def intnoise(x):
x = (x << 13) ^ x
return (1.0 - ((x * (x * x * 15731 + 789221) + 1376312589) & 0x7fffffff) / 1073741824.0)
The "32-bit integer" part doesn't, unless you use numpy.int32. Just mask the value to 32 bits when it makes sense.
The "<<" stands.
The "& 7fffffff" Needs to be converted. The "&" stands, but the hexadecimal literal needs a bit: 0x7fffffff.
As a learning exercise - great, but once you've understood it, just use os.urandom(4) for random numbers, or the functions in random for various pseudo-random generators.
From your code what i did understand is, the function IntNoise takes 32-bit integer as an input. the << is left shift operator from what i know. and it shifts the number's bit for 13 times to the left. ^ is exponential. And 7fffffff must be a number in hex format of representation. it is 8 digit number, every digit i.e. f and 7 takes 4 bits. f is value for 15 in hex.
Related
I have not really used maps before in my programming experience, so I am having trouble understanding the more complex versions of maps. So let's say that the problem is you are given an integer in minutes, in this case n = 808. What you are to do with this number is convert it to 24 hour time, so hh:mm. This would give you 13:28. Once this is done, add up the digits of that time to get the answer. So, the answer would be 14. I saw a really nice one-liner to this solution and I am trying to understand it because my solution took about 5 more lines of code.
This is the solution:
sum(map(int, str(n // 60 * 100 + n % 60)))
So I understand that maps apply the same function over an iteration of numbers, but what throws me off is the int,str(...) part. I am not sure what is going on behind the scenes.
There are two mathematical operators used here:
// represents floor division, i.e. extract the integer portion of the result after division.
% represents modulus, i.e. the remainder after division.
Therefore, for n = 808, the algorithm returns:
str(808 // 60 * 100 + 808 % 60) = str(13 * 100 + 28) = '1328'
map(int, '1328') then takes each character in the string '1328' and converts it into an integer, itself returning an iterable. map requires an iterable as its second (and subsequent) arguments. Strings can be iterated to extract each character one at a time.
Finally, sum takes each of the integers returned from map and adds them together.
An equivalent formulation of the logic is possible via sum with a generator expression:
sum(int(i) for i in str(n // 60 * 100 + n % 60))
map ,as you stated, applies a function over a iterable.
So, when you do
map(int, str(n // 60 * 100 + n % 60))
You are using the function int over the iterable str(n // 60 * 100 + n % 60). As you probably know, strings are iterables (because, of course, you can iterate over them) - that can be easily checked
for char in "abcd":
print(char)
a
b
c
d
The return from str(n // 60 * 100 + n % 60) is '1328'. When you apply int to each char, you transform each to an integer. You can easily see this by instead of taking the sum right away, getting a lsit
list(map(int, str(n // 60 * 100 + n % 60)))
[1, 3, 2, 8]
I guess now it is easy to see that the sum will get the sum of these numbers, which is what you wanted from the beginning :)
Both int() and str() are functions. In this particular example, when n=808 the argument to the str() function is calculated as 1328, which when converted to the string becomes '1328'. A string is iterable, so the map function is simply applying int to each character of the string, producing the sequence [1,3,2,8].
I want to play around with procedural content generation algorithms, and decided to start with noises (Perlin, value, etc)
For that, I want have a generic n-dimensional noise function. For that I wrote a function that returns a noise generation function of the given dimension:
small_primes = [1, 83, 97, 233, 61, 127]
def get_noise_function(dimension, random_seed=None):
primes_list = list(small_primes)
if dimension > len(primes_list):
primes_list = primes_list * (dimension / len(primes_list))
rand = random.Random()
if random_seed:
rand.seed(random_seed)
# random.shuffle(primes_list)
rand.shuffle(primes_list)
def noise_func(*args):
if len(args) < dimension:
# throw something
return None
n = [a*b for a, b in zip(args, primes_list)]
n = sum(n)
#n = (n << 13) ** n
n = (n << 13) ^ n
nn = (n * (n * n * 60493 + 19990303) + 1376312589) & 0x7fffffff
return 1.0 - (nn / 1073741824.0)
return noise_func
The, problem, I believe, is with the calculations. I based my code on these two articles:
Hugo Elias' value noise implementation (end of the page)
libnoise documentation
Example of one of my tests:
f1 = get_noise_function(1, 10)
print f1(1)
print f1(2)
print f1(3)
print f1(1)
It always returns -0.281790983863, even on higher dimensions and different seeds.
The problem, I believe, is that in C/C++ there is overflow is some of the calculations, and everything works. In python, it just calculates a gigantic number.
How can I correct this or, if possible, how can I generate a pseudo-random function that, after being seeded, for a certain input always returns the same value.
[EDIT] Fixed the code. Now it works.
Where the referenced code from Hugo Elias has:
x = (x<<13) ^ x
you have:
n = (n << 13) ** n
I believe Elias is doing bitwise xor, while you're effectively raising 8192*n to the power of n. That gives you a huge value. Then
nn = (n * (n * n * 60493 + 19990303) + 1376312589) & 0x7fffffff
takes that gigantic n and makes it even bigger, until you finally throw away everything but the last 31 bits. It doesn't make much sense ;-)
Try changing your code to:
n = (n << 13) ^ n
and see whether that helps.
I am trying to write a program using Python v. 2.7.5 that will compute the area under the curve y=sin(x) between x = 0 and x = pi. Perform this calculation varying the n divisions of the range of x between 1 and 10 inclusive and print the approximate value, the true value, and the percent error (in other words, increase the accuracy by increasing the number of trapezoids). Print all the values to three decimal places.
I am not sure what the code should look like. I was told that I should only have about 12 lines of code for these calculations to be done.
I am using Wing IDE.
This is what I have so far
# base_n = (b-a)/n
# h1 = a + ((n-1)/n)(b-a)
# h2 = a + (n/n)(b-a)
# Trap Area = (1/2)*base*(h1+h2)
# a = 0, b = pi
from math import pi, sin
def TrapArea(n):
for i in range(1, n):
deltax = (pi-0)/n
sum += (1.0/2.0)(((pi-0)/n)(sin((i-1)/n(pi-0))) + sin((i/n)(pi-0)))*deltax
return sum
for i in range(1, 11):
print TrapArea(i)
I am not sure if I am on the right track. I am getting an error that says "local variable 'sum' referenced before assignment. Any suggestions on how to improve my code?
Your original problem and problem with Shashank Gupta's answer was /n does integer division. You need to convert n to float first:
from math import pi, sin
def TrapArea(n):
sum = 0
for i in range(1, n):
deltax = (pi-0)/n
sum += (1.0/2.0)*(((pi-0)/float(n))*(sin((i-1)/float(n)*(pi-0))) + sin((i/float(n))*(pi-0)))*deltax
return sum
for i in range(1, 11):
print TrapArea(i)
Output:
0
0.785398163397
1.38175124526
1.47457409274
1.45836902046
1.42009115659
1.38070223089
1.34524797198
1.31450259385
1.28808354
Note that you can heavily simplify the sum += ... part.
First change all (pi-0) to pi:
sum += (1.0/2.0)*((pi/float(n))*(sin((i-1)/float(n)*pi)) + sin((i/float(n))*pi))*deltax
Then do pi/n wherever possible, which avoids needing to call float as pi is already a float:
sum += (1.0/2.0)*(pi/n * (sin((i-1) * pi/n)) + sin(i * pi/n))*deltax
Then change the (1.0/2.0) to 0.5 and remove some brackets:
sum += 0.5 * (pi/n * sin((i-1) * pi/n) + sin(i * pi/n)) * deltax
Much nicer, eh?
You have some indentation issues with your code but that could just be because of copy paste. Anyways adding a line sum = 0 at the beginning of your TrapArea function should solve your current error. But as #Blender pointed out in the comments, you have another issue, which is the lack of a multiplication operator (*) after your floating point division expression (1.0/2.0).
Remember that in Python expressions are not always evaluated as you would expect mathematically. Thus (a op b)(c) will not automatically multiply the result of a op b by c like you would expect with a mathematical expression. Instead this is the function call notation in Python.
Also remember that you must initialize all variables before using their values for assignment. Python has no default value for unnamed variables so when you reference the value of sum with sum += expr which is equivalent to sum = sum + expr you are trying to reference a name (sum) that is not binded to any object at all.
The following revision to your function should do the trick. Notice how I place multiplication operators (*) between every expression that you intend to multiply.
def TrapArea(n):
sum = 0
for i in range(1, n):
i = float(i)
deltax = (pi-0)/n
sum += (1.0/2.0)*(((pi-0)/n)*(sin((i-1)/n*(pi-0))) + sin((i/n)*(pi-0)))*deltax
return sum
EDIT: I also dealt with the float division issue by converting i to float(i) within every iteration of the loop. In Python 2.x, if you divide one integer type object with another integer type object, the expression evaluates to an integer regardless of the actual value.
A "nicer" way to do the trapezoid rule with equally-spaced points...
Let dx = pi/n be the width of the interval. Also, let f(i) be sin(i*dx) to shorten some expressions below. Then interval i (in range(1,n)) contributes:
dA = 0.5*dx*( f(i) + f(i-1) )
...to the sum (which is an area, so I'm using dA for "delta area"). Factoring out the 0.5*dx, makes the whole some look like:
A = 0.5*dx * ( (f(0) + f(1)) + (f(1) + f(2)) + .... + (f(n-1) + f(n)) )
Notice that there are two f(1) terms, two f(2) terms, on up to two f(n-1) terms. Combine those to get:
A = 0.5*dx * ( f(0) + 2*f(1) + 2*f(2) + ... + 2*f(n-1) + f(n) )
The 0.5 and 2 factors cancel except in the first and last terms:
A = 0.5*dx(f(0) + f(n)) + dx*(f(1) + f(2) + ... + f(n-1))
Finally, you can factor dx out entirely to do just one multiplication at the end. Converting back to sin() calls, then:
def TrapArea(n):
dx = pi/n
asum = 0.5*(sin(0) + sin(pi)) # this is 0 for this problem, but not others
for i in range(1, n-1):
asum += sin(i*dx)
return sum*dx
That changed "sum" to "asum", or maybe "area" would be better. That's mostly because sum() is a built-in function, which I'll use below the line.
Extra credit: The loop part of the sum can be done in one step with a generator expression and the sum builtin function:
def TrapArea2(n):
dx = pi/n
asum = 0.5*(sin(0) + sin(pi))
asum += sum(sin(i*dx) for i in range(1,n-1))
return asum*dx
Testing both of those:
>>> for n in [1, 10, 100, 1000, 10000]:
print n, TrapArea(n), TrapArea2(n)
1 1.92367069372e-16 1.92367069372e-16
10 1.88644298557 1.88644298557
100 1.99884870579 1.99884870579
1000 1.99998848548 1.99998848548
10000 1.99999988485 1.99999988485
That first line is a "numerical zero", since math.sin(math.pi) evaluates to about 1.2e-16 instead of exactly zero. Draw the single interval from 0 to pi and the endpoints are indeed both 0 (or nearly so.)
Just to be fully up front, this is regarding a homework, but this in itself is not the assignment. The assignment is using noise to create cool graphics. I have a bit of experience in Python but not enough to figure this simple thing out.
I'm having trouble generating a seeded-random [-1,1]. The pseudocode my teacher gave me is from Hugo Elias.
Pseudocode:
function Noise1(integer x, integer y)
n = x + y * 57
n = (n<<13) ^ n;
return ( 1.0 - ( (n * (n * n * 15731 + 789221) + 1376312589) & 7fffffff) / 1073741824.0);
end function
My attempt in Python:
def noise(x, y):
n = x + y * 57
n = (n<<5) ^ n;
return ( 1.0 - ( (n * (n * n * 15731 + 789221) + 1376312589) & 7fffffff) / 1073741824.0)
The problem is the & 7fffffff bit in return statement. First, I'm not sure what that operation is. Maybe a bit-shift? Second, I'm not sure how to do that operation in Python. I had just removed that part, but I am getting huge negative numbers, nowhere near a [-1,1]
The & symbol stands for bitwise AND
The ^ symbol stands for bitwise XOR
and the 7FFFFFFF is a HEX number.
In programming you can represent a hex number using 0x where 7FFFFFFF is 0x7FFFFFFF
Some further reading on hex numbers.
To do a binary AND in python you just use & 0x7FFFFFFF
see more here about bitwise operations in python
I replaced 7FFFFFFF with 0x7FFFFFFF in your existing code and tried plugging in some random values and all the answers which I got were in [-1, 1].
The problem is the & 7fffffff bit in return statement. First, I'm not sure what that operation is. Maybe a bit-shift?
A bit mask. << is used for shifting. & is bitwise-AND. 7fffffff, interpreted as a hexadecimal number, is a quantity that, if re-written in binary, has 31 bits, all of which are 1. The effect is to select the low-order 31 bits of the value.
To tell Python that 7fffffff should be interpreted as a hexadecimal number, you must prefix it with 0x, hence 0x7fffffff.
Second, I'm not sure how to do that operation in Python.
The same way as in the pseudocode (i.e. with &).
In Python 3, I am checking whether a given value is triangular, that is, it can be represented as n * (n + 1) / 2 for some positive integer n.
Can I just write:
import math
def is_triangular1(x):
num = (1 / 2) * (math.sqrt(8 * x + 1) - 1)
return int(num) == num
Or do I need to do check within a tolerance instead?
epsilon = 0.000000000001
def is_triangular2(x):
num = (1 / 2) * (math.sqrt(8 * x + 1) - 1)
return abs(int(num) - num) < epsilon
I checked that both of the functions return same results for x up to 1,000,000. But I am not sure if generally speaking int(x) == x will always correctly determine whether a number is integer, because of the cases when for example 5 is represented as 4.99999999999997 etc.
As far as I know, the second way is the correct one if I do it in C, but I am not sure about Python 3.
There is is_integer function in python float type:
>>> float(1.0).is_integer()
True
>>> float(1.001).is_integer()
False
>>>
Both your implementations have problems. It actually can happen that you end up with something like 4.999999999999997, so using int() is not an option.
I'd go for a completely different approach: First assume that your number is triangular, and compute what n would be in that case. In that first step, you can round generously, since it's only necessary to get the result right if the number actually is triangular. Next, compute n * (n + 1) / 2 for this n, and compare the result to x. Now, you are comparing two integers, so there are no inaccuracies left.
The computation of n can be simplified by expanding
(1/2) * (math.sqrt(8*x+1)-1) = math.sqrt(2 * x + 0.25) - 0.5
and utilizing that
round(y - 0.5) = int(y)
for positive y.
def is_triangular(x):
n = int(math.sqrt(2 * x))
return x == n * (n + 1) / 2
You'll want to do the latter. In Programming in Python 3 the following example is given as the most accurate way to compare
def equal_float(a, b):
#return abs(a - b) <= sys.float_info.epsilon
return abs(a - b) <= chosen_value #see edit below for more info
Also, since epsilon is the "smallest difference the machine can distinguish between two floating-point numbers", you'll want to use <= in your function.
Edit: After reading the comments below I have looked back at the book and it specifically says "Here is a simple function for comparing floats for equality to the limit of the machines accuracy". I believe this was just an example for comparing floats to extreme precision but the fact that error is introduced with many float calculations this should rarely if ever be used. I characterized it as the "most accurate" way to compare in my answer, which in some sense is true, but rarely what is intended when comparing floats or integers to floats. Choosing a value (ex: 0.00000000001) based on the "problem domain" of the function instead of using sys.float_info.epsilon is the correct approach.
Thanks to S.Lott and Sven Marnach for their corrections, and I apologize if I led anyone down the wrong path.
Python does have a Decimal class (in the decimal module), which you could use to avoid the imprecision of floats.
floats can exactly represent all integers in their range - floating-point equality is only tricky if you care about the bit after the point. So, as long as all of the calculations in your formula return whole numbers for the cases you're interested in, int(num) == num is perfectly safe.
So, we need to prove that for any triangular number, every piece of maths you do can be done with integer arithmetic (and anything coming out as a non-integer must imply that x is not triangular):
To start with, we can assume that x must be an integer - this is required in the definition of 'triangular number'.
This being the case, 8*x + 1 will also be an integer, since the integers are closed under + and * .
math.sqrt() returns float; but if x is triangular, then the square root will be a whole number - ie, again exactly represented.
So, for all x that should return true in your functions, int(num) == num will be true, and so your istriangular1 will always work. The only sticking point, as mentioned in the comments to the question, is that Python 2 by default does integer division in the same way as C - int/int => int, truncating if the result can't be represented exactly as an int. So, 1/2 == 0. This is fixed in Python 3, or by having the line
from __future__ import division
near the top of your code.
I think the module decimal is what you need
You can round your number to e.g. 14 decimal places or less:
>>> round(4.999999999999997, 14)
5.0
PS: double precision is about 15 decimal places
It is hard to argue with standards.
In C99 and POSIX, the standard for rounding a float to an int is defined by nearbyint() The important concept is the direction of rounding and the locale specific rounding convention.
Assuming the convention is common rounding, this is the same as the C99 convention in Python:
#!/usr/bin/python
import math
infinity = math.ldexp(1.0, 1023) * 2
def nearbyint(x):
"""returns the nearest int as the C99 standard would"""
# handle NaN
if x!=x:
return x
if x >= infinity:
return infinity
if x <= -infinity:
return -infinity
if x==0.0:
return x
return math.floor(x + 0.5)
If you want more control over rounding, consider using the Decimal module and choose the rounding convention you wish to employ. You may want to use Banker's Rounding for example.
Once you have decided on the convention, round to an int and compare to the other int.
Consider using NumPy, they take care of everything under the hood.
import numpy as np
result_bool = np.isclose(float1, float2)
Python has unlimited integer precision, but only 53 bits of float precision. When you square a number, you double the number of bits it requires. This means that the ULP of the original number is (approximately) twice the ULP of the square root.
You start running into issues with numbers around 50 bits or so, because the difference between the fractional representation of an irrational root and the nearest integer can be smaller than the ULP. Even in this case, checking if you are within tolerance will do more harm than good (by increasing the number of false positives).
For example:
>>> x = (1 << 26) - 1
>>> (math.sqrt(x**2)).is_integer()
True
>>> (math.sqrt(x**2 + 1)).is_integer()
False
>>> (math.sqrt(x**2 - 1)).is_integer()
False
>>> y = (1 << 27) - 1
>>> (math.sqrt(y**2)).is_integer()
True
>>> (math.sqrt(y**2 + 1)).is_integer()
True
>>> (math.sqrt(y**2 - 1)).is_integer()
True
>>> (math.sqrt(y**2 + 2)).is_integer()
False
>>> (math.sqrt(y**2 - 2)).is_integer()
True
>>> (math.sqrt(y**2 - 3)).is_integer()
False
You can therefore rework the formulation of your problem slightly. If an integer x is a triangular number, there exists an integer n such that x = n * (n + 1) // 2. The resulting quadratic is n**2 + n - 2 * x = 0. All you need to know is if the discriminant 1 + 8 * x is a perfect square. You can compute the integer square root of an integer using math.isqrt starting with python 3.8. Prior to that, you could use one of the algorithms from Wikipedia, implemented on SO here.
You can therefore stay entirely in python's infinite-precision integer domain with the following one-liner:
def is_triangular(x):
return math.isqrt(k := 8 * x + 1)**2 == k
Now you can do something like this:
>>> x = 58686775177009424410876674976531835606028390913650409380075
>>> math.isqrt(k := 8 * x + 1)**2 == k
True
>>> math.isqrt(k := 8 * (x + 1) + 1)**2 == k
False
>>> math.sqrt(k := 8 * x + 1)**2 == k
False
The first result is correct: x in this example is a triangular number computed with n = 342598234604352345342958762349.
Python still uses the same floating point representation and operations C does, so the second one is the correct way.
Under the hood, Python's float type is a C double.
The most robust way would be to get the nearest integer to num, then test if that integers satisfies the property you're after:
import math
def is_triangular1(x):
num = (1/2) * (math.sqrt(8*x+1)-1 )
inum = int(round(num))
return inum*(inum+1) == 2*x # This line uses only integer arithmetic