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Note: I cannot use any of the built in functions to solve this.
This is the question:
Given a function that returns a random integer number between 1 and 5, create a function that creates a random integer between 1 and 7.
As it has been answered in Adam Rosenfield's solution a while ago you may try to use an algoritm similar to this one below:
int i;
do
{
i = 5 * (rand5() - 1) + rand5(); // i is now uniformly random between 1 and 25
} while(i > 21);
// i is now uniformly random between 1 and 21
return i % 7 + 1; // result is now uniformly random between 1 and 7
EDIT: Translation to python:
def rand7():
i = 0;
while True:
i = 5 * (rand5() -1) + rand5()
if i > 21:
break
return i % 7 + 1
Apparently the following doesn't work. See the linked answer and the other responses as well.
Well, here is an approach I am thinking of. It could be wrong. I am going with the assumption that the result should be randomly distributed within the larger range.
Generate 7 random numbers using the given random functions (that generates numbers [1,5]) and calculate their sum.
This will result in a value between 7 (1 * 7) and 35 (5 * 7).
Because this value is evenly divisible by the target range and randomly distributed, then it seems like it would be valid to collapse the intermediate range [7,35] back to [1,7] without losing uniformity.
Or maybe that's not quite it - but it seems like exploiting a common multiple between the numbers is key.
import random
def winningFunc():
return originalFuncReturning1to5() + random.randint(0,2)
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I already know how does abs function works.I mean it only shows you how far the number is from zero
The only thing i can't just get is with this example:
print(abs(3 + 4j)) # prints 5 !! Why ?
Because for complex numbers abs(number) will return magnitude of Complex Numbers.
Magnitude value will counted as :
√x2+y2 = √(3² + 4²) = √(9 + 16) = √(25) = 5.0
So abs with complex number will return magnitude of complex numbers.
for further reference you can use https://www.geeksforgeeks.org/abs-in-python/.
As the rest of the answers stated above, 3+4j is a complex number and the formula of calculating the absolute value of a complex number x+yi is sqrt( (x^2) + (y^2) ). In your case it's:
sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5
Because in python, 3+4j is complex number and python calculates the absolute value or magnitude of the complex number when you do abs() on it. Magnitude of 3+4j is 5. Try this :
type(3+4j)
It should give <class 'complex'>.
Note : Magnitude of a complex number a+bj is ((a**2+b**2)**0.5)
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I have 2 questions I need to ask for help with. One question I don't entirely understand so if someone could help me to that would be great.
Question One (the one I don't entirely understand):
One definition of e is
Formula for e
This can be calculated as my_e = 1/math.factorial(0) + 1/math.factorial(1) + 1/math.factorial(2) + 1/math.factorial(3) + …
Let n be the input number of the math.factorial() function. n successively takes on 0, 1, 2, 3 and so on. Find the smallest n such that the absolute value of (my_e – math.e) is less than or equal to 10-10. That is, abs(my_e - math.e) <= (10 ** -10).
I just don't entirely understand what I am being asked to do. Clarification would be great. Thanks!
Question 2:
Ask the user to type in a series of integers. Sum them up and print out the sum and the number of integers the user has entered.
My code
So what should happen is after I enter the numbers I want to enter and hit the enter key, it should calculate and print out "sum = 25 count = 3". The screenshot shows what error message I am getting.
Any help you have is welcomed and greatly appreciated.
As far as you first question goes:
>>> import math
>>> math.e
2.718281828459045
>>> sum(1.0/math.factorial(i) for i in range(5))
2.708333333333333
>>> abs(sum(1.0/math.factorial(i) for i in range(5)) - math.e) < 10**-10
False
>>> abs(sum(1.0/math.factorial(i) for i in range(30)) - math.e) < 10**-10
True
So, somewhere between n == 5 and n == 30 you get 10 decimal places for e. Create the sum term by term in a while-loop (instead of by using the sum function as I have, since you are unlikely to have seen that syntax yet). At each pass through the loop, compare the sum with math.e. Stop when you get the target accuracy. Return the final n.
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ok, I am studying python in my computing course and i have been challenged with designing a code which validates a GTIN-8 code, I have came to a hurdle that i can not jump. I have looked on how to find the equal or higher multiple of a 10 and i have had no success so far, hope you guys can help me!
Here is a small piece of code, i need to find the equal or higher multiple of 10;
NewNumber = (NewGtin_1 + Gtin_2 + NewGtin_3 + Gtin_4 + NewGtin_5 + Gtin_6 + NewGtin_7)
print (NewNumber)
The easiest way, involving no functions and modules, could be using floor division operator //.
def neareast_higher_multiple_10(number):
return ((number // 10) + 1) * 10
Examples of usage:
>>> neareast_higher_multiple_10(15)
20
>>> neareast_higher_multiple_10(21)
30
>>> neareast_higher_multiple_10(20.1)
30.0
>>> neareast_higher_multiple_10(20)
30
We can also make a generalized version:
def neareast_higher_multiple(number, mult_of):
return ((number // mult_of) + 1) * mult_of
If you need nearest lower multiple, just remove + 1:
def neareast_lower_multiple(number, mult_of):
return (number // mult_of) * mult_of
To find the nearest multiple, you can call both these functions and use that one with a lower difference from the original number.
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n=iterations
for some reason this code will need a lot more iterations for more accurate result from other codes, Can anyone explain why this is happening? thanks.
n,s,x=1000,1,0
for i in range(0,n,2):
x+=s*(1/(1+i))*4
s=-s
print(x)
As I mentioned in a comment, the only way to speed this is to transform the sequence. Here's a very simple way, related to the Euler transformation (see roippi's link): for the sum of an alternating sequence, create a new sequence consisting of the average of each pair of successive partial sums. For example, given the alternating sequence
a0 -a1 +a2 -a3 +a4 ...
where all the as are positive, the sequences of partial sums is:
s0=a0 s1=a0-a1 s2=a0-a1+a2 s3=a0-a1+a2-a3 s4=a0-a1+a2-a3+a4 ...
and then the new derived sequence is:
(s0+s1)/2 (s1+s2)/2 (s2+s3)/2 (s3+s4)/2 ...
That can often converge faster - and the same idea can applied to this sequence. That is, create yet another new sequence averaging the terms of that sequence. This can be carried on indefinitely. Here I'll take it one more level:
from math import pi
def leibniz():
from itertools import count
s, x = 1.0, 0.0
for i in count(1, 2):
x += 4.0*s/i
s = -s
yield x
def avg(seq):
a = next(seq)
while True:
b = next(seq)
yield (a + b) / 2.0
a = b
base = leibniz()
d1 = avg(base)
d2 = avg(d1)
d3 = avg(d2)
for i in range(20):
x = next(d3)
print("{:.6f} {:8.4%}".format(x, (x - pi)/pi))
Output:
3.161905 0.6466%
3.136508 -0.1619%
3.143434 0.0586%
3.140770 -0.0262%
3.142014 0.0134%
3.141355 -0.0076%
3.141736 0.0046%
3.141501 -0.0029%
3.141654 0.0020%
3.141550 -0.0014%
3.141623 0.0010%
3.141570 -0.0007%
3.141610 0.0005%
3.141580 -0.0004%
3.141603 0.0003%
3.141585 -0.0003%
3.141599 0.0002%
3.141587 -0.0002%
3.141597 0.0001%
3.141589 -0.0001%
So after just 20 terms, we've already got pi to about 6 significant digits. The base Leibniz sequence is still at about 2 digits correct:
>>> next(base)
3.099944032373808
That's an enormous improvement. A key point here is that the partial sums of the base Leibniz sequence give approximations that alternate between "too big" and "too small". That's why averaging them gets closer to the truth. The same (alternating between "too big" and "too small") is also true of the derived sequences, so averaging their terms also helps.
That's all hand-wavy, of course. Rigorous justification probably isn't something you're interested in ;-)
That is because you are using the Leibniz series and it is known to converge very (very) slowly.
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Basically, the user will enter a probability of getting success p = .34, so how do I determine the number of trials it takes to get success?
I know that I could get the number of trial by making a counter like n+= 1 but not sure how to use this probability value. Help will be greatly appreciated.
You can use random() from the random module to generate a floating point value in the range [0,1]. Then you can compare that value to the user-inputted probability. If the random value is less than the user's given probability, you have a success.
Once you understand that, all you need is a simple while loop that will keep generating numbers until you get a success.
Here is an example of a function that might do what you want:
import random
# The function you wanted
def run_trials(p):
'''only accepts int or float in range [0.0, 1.0]'''
# Count trials
num_trials = 1
while True:
r = random.random()
if r <= p:
print "This took", num_trials, "trial(s) to get a success."
break
else:
num_trials += 1
#Input
while True:
p = input("Please enter a probability: ")
if isinstance(p, float) or isinstance(p, int):
if 0 <= p <= 1:
break
run_trials(p) # pass it to the function