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python help

post #1 of 20
Thread Starter 
Hello, I am just starting to learn python and I wrote this program to display the odd numbers after a user inputs a starting and ending value( this program also displays the starting and ending value if it is odd). I am just not sure if I am doing this in the best way possible and if there is a more practical way to accomplish this that I don't know, or if this is completely wrong? Thanks you to anyone who reads this and responds.

Code:
def main():
  x = input('Enter a starting number: ')
  y = input('Enter an ending number: ')
  for i in range(x, y , 2):
    if (x%2==0):
      i = i + 1
      print i
    elif (x%2!=0):
      print (i)
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post #2 of 20
I don't know python at all (don't think I will ever use it, just don't see the point in it). But I can read your code and understand it.

Why are you using the for loop adding 2 to value of X and then checking it again and again. Have the odd check first and then do the loop to print.

Also I tend to buffer my outputs in loops like that as printing slows things down. If you say buffer that in to a string it and print it out in the end it will will have the same effect to the user.

Also mod 2 isnt the most efficient way of testing if a value is odd.



Notice a patern.. all odd numbers have a 1 in the first position?
Code:
bin(x)¶
Convert an integer number to a binary string. The result is a valid Python expression. If x is not a Python int object, it has to define an __index__() method that returns an integer.

New in version 2.6.

http://docs.python.org/library/functions.html
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post #3 of 20
Division is computationally expensive (for example, in ARM chips add and subtract take 1 cycle. multiply takes 3 cycles. division takes 34). However, sometimes there are much faster ways to do things normally done with division. Let's take a look.

So, there are operations in computers called bitwise functions. They either apply logic gates to a number or reposition a number.

The ones in python are (I'll explain them in a minute)
a >> b which shifts a right by b amount
a << b which shifts a left by b amount
a & b which gives the logical AND of a and b (this is the one we'll actually use here though I briefly explain all but negation)
a | b which gives the logical OR of a and b
a ^ b which gives the logical XOR of a and b
~ a which gives the logical NOT of a
- a which gives the negative of a (negation is different than subtraction to a computer) I won't be explaining this concept here (maybe I'll go into 1's and 2's complements later).

Let's work with 6-bit binary numbers
Here's the numbers 0-15 and their 4-digit binary equivalents. The extra zeroes in the front are used to make the numbers the same amount of digits long.
00 0000
01 0001
02 0010
03 0011
04 0100
05 0101
06 0110
07 0111
08 1000
09 1001
10 1010
11 1011
12 1100
13 1101
14 1110
15 1111

a = 13
that's 001101 in binary

b = 55
that's 110111

c = 3
that's 000011

a >> c shifts the bits in 'a' right 3 times since c = 3 (or does all three at the same time, but we'll do it step by step)

001101 # no shift
000110 # that's the first shift. 0's are added to the left and any 1's at the end are lost. 'a' is now 6 (which is 13/2 as remainders are dropped)
000011 # that's shift 2. 'a' is now 3 (which is 6/2)
000001 # that's shift 3, so 'a' is now equal 1 (which is 3/2)

a << (c -1) shifts the bits in 'a' to the left 2 times
001101
011010 #first shift. a = 26 (2 * 13)
110100 #second shift. a = 52 (26 * 13)
If I had shifted a third time, the one on the left would have been dropped because my number would be too big (called an overflow error)

logic gates take two inputs and give one output

AND (only outputs 1 when all inputs are 1's)

a b | output
0 0 | 0
0 1 | 0
1 0 | 0
1 1 | 1

OR (outputs 1 except when all inputs are 0's)

a b | output
0 0 | 0
0 1 | 1
1 0 | 1
1 1 | 1

XOR (stands for exclusive OR. Only inputs with an odd number of 1's output a 1)

a b | output
0 0 | 0
0 1 | 1
1 0 | 1
1 1 | 0

NOT (while most gates can take any amount of numbers, NOT only takes one argument)

a | output
1 | 0
0 | 1


Now let's do a & b (a AND b)

001101
110111 # find the bitwise AND. Each column is it's own miniature AND function.
000101 #only two columns were all 1's, so only those output a 1. (another way to do this is recognize that any column with a 0 outputs 0)

for a | b (a OR b)

001101
110111 # find the bitwise OR. Each column is it's own miniature OR function.
111111 # since every column had a 1, the answer was all 1's (the easy way here is that any column with a 1 outputs a 1)

for a ^ b (XOR, don't confuse with the power symbol in many languages and TI calculators)

001101
110111 # find the bitwise XOR. Each column is it's own miniature XOR function.
111010 #only columns with an odd number of 1's output 1

~a

001101
110010 # just make all 1's into 0's and all 0's into 1's

Try the AND, OR, and XOR for a,b, and c together yourself


001101
110111
000011


the answers are
AND 000001
OR 111111
XOR 111001

Now let's take a look at your loop. Notice that all the odd numbers end in a 1 and all even numbers end in 0.

Let's say that your user inputs a number (which the computer converts to binary). No matter how many bits are used, the last bit will be 1 if the number is odd. We need a way to find out if that last digit is odd or even. The answer is to AND it to 00000001 (or however many zeroes you need to be long enough, but the computer will add them as needed). Let's look at the numbers 44, 35, 52 and 23 when they're ANDed with 1.


101100 #44
000001 #now we AND
000000 #the even number outputs zero

100011 #35
000001 #now we AND
000001 #the odd number outputs one

110100 #52
000001 #now we AND
000000 #the even number outputs zero

010111 #23
000001 #now we AND
000001 #the odd number outputs one


Based on this, we can now say
Code:
x = int(input("start: ")) # don't forget to tell python to change the input from a string into an integer
y = int(input("end: "))
if 1 == (x & 1): # then the number is odd
    while (x <= y): # if our initial value of x is bigger than y, then we will output nothing
        print(x)
        x = x + 2 # odd no. plus even no. equals another odd no.
else:
    x = x + 1 # since x must be even, we add one to it to make it odd while staying within our parameters
    while (x <= y):
        print(x)
        x = x + 2


Edited by hajile - 9/12/12 at 7:06pm
post #4 of 20
Guys, I'm sure he'll appreciate your efforts wink.gif but I doubt he, as a beginner, is really asking for the information you are providing.

@nova4005, basically what is redundant in your code is the double modulus (remainder) test (to check if the number is odd). You can eliminate that and just use something like this:
Code:
def main():
...     x = input("Starting number: ")
...     y = input("Ending number: ")
...     for i in range(x,y):
...             if (i % 2 != 0):
...                     print i

This would simply test each number between the range provided (inclusive of starting and ending numbers) to see if it is odd. If it is odd, it will print the number. If not, it will move on to the next.
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post #5 of 20
Thread Starter 
Thank you everyone, I do appreciate all the answers. I still have a lot of reading and studying to do before all of this makes complete sense. Thank you again for your time and answers.
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post #6 of 20
Quote:
Originally Posted by andrewmchugh View Post

I don't know python at all (don't think I will ever use it, just don't see the point in it)...


That's a shame. Python is one of the funnest languages to play with, and is very powerful. Throw an awesome framework like Django at it and web development becomes a blast, too.

Your loss. thumb.gif
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post #7 of 20
Quote:
Originally Posted by {Unregistered} View Post

Guys, I'm sure he'll appreciate your efforts wink.gif but I doubt he, as a beginner, is really asking for the information you are providing.
@nova4005, basically what is redundant in your code is the double modulus (remainder) test (to check if the number is odd). You can eliminate that and just use something like this:



This would simply test each number between the range provided (inclusive of starting and ending numbers) to see if it is odd. If it is odd, it will print the number. If not, it will move on to the next.

True, he didn't specifically ask, but he'll be much better with both logic and programming if he reads and asks a few questions. Looking at good ways to solve the problem is important and in the case of the code I showed, the explanation was megaposted to introduce him to the simple principles of logic gates. Logic gates have no pre-requisite (except that the student be living or undead). In addition, alll the "bit twiddling" is explained step by step and I'm more than willing to explain any parts that aren't clear -- he has only to ask.

That said, taking out the range function (he should at least use rangex since he is using python 2.x which transforms range into a linked list of numbers rather than just providing an iterating integer, but using x for the iteration is still much faster) and taking out the division (a very important lesson for any programmer) will make the code many orders of magnitude faster (my adding 2 method is very straight forward while eliminating an expensive mod every loop). The only catchy part of my code is the use of the & which I go to huge lengths to explain.
post #8 of 20
Thread Starter 
Code:
x = int(input("start: ")) # don't forget to tell python to change the input from a string into an integer
y = int(input("end: "))
if 1 == (x & 1): # then the number is odd
    while (x <= y): # if our initial value of x is bigger than y, then we will output nothing
        print(x)
        x = x + 2 # odd no. plus even no. equals another odd no.
else:
    x = x + 1 # since x must be even, we add one to it to make it odd while staying within our parameters
    while (x <= y):
        print(x)
        x = x + 2


Hajile, I do have one question about the line "if 1 == (x & 1):" , if I understand correctly any number that is supplied to x that is odd will make this comparison true. So I could write "if 0 ==(x & 1):" and this would do the same for even number comparison, or am I still confused? I am also trying to think of some other small programs I could apply this to for practice. Do you have any suggestions that I could use to create some programs with this in mind for practice? Thank you again for your time as I'm sure you have better things to do than answer my simple questions.
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post #9 of 20
You don't need to have the two while. Its not going make the code any more efficient, however its more elegant.

pseudo answer.
Code:

x = int(input("start: ")) # don't forget to tell python to change the input from a string into an integer
y = int(input("end: "))
if 0 == (x & 1): # then the number is odd
x+=1;

while (x <= y): # if our initial value of x is bigger than y, then we will output nothing
{
    print(x)
    x = x + 2 # odd no. plus even no. equals another odd no.
}
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post #10 of 20
Thread Starter 
Thank you Andrew, i appreciate any help to help me learn better coding skills.
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Q9550 2.83Ghz @ 4.0Ghz 1.328v Gigabyte EP45T-DS3R Ati Radeon hd 6970 Sapphire HD 7970 
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Corsair XMS3 seagate Baracuda LG Blu-Ray RW rosewill air cooler 
OSMonitorKeyboardPower
Windows Pro 64-Bit none Micosoft Wireless OCZ 1000w 
CaseMouse
Antec 900 Microsoft Wireless 
CPUCPUMotherboardGraphics
Xeon E5-26XX V2 2.3-2.7Ghz 10core/20 thread Xeon E5-26XX V2 2.3-2.7Ghz 10core/20 thread SuperMicro x9DAi EVGA GTX 580 
RAMRAMHard DriveCooling
G.SKILL Ripjaws X Series 8GB (4 x 2GB) 240-Pin ... G.SKILL Ripjaws X Series 8GB (4 x 2GB) 240-Pin ... Patriot 16gb Flash drive Hyper 212+ evo 
CoolingOSPowerCase
Hyper 212+ evo Ubuntu 12.04 OCZ Z series 1000w gold Fractal design define xl 
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