@Dave: Can you throw some light on random() function?? How it is generating
numbers between 0.0 and 1.0, and how many digits are there after the
point.......because if there is only one digit then we will not be able to
print words after 10th place because 10*0.1(lowest number generated by
random())=1...since lowest number generated by the random() function is not
able to convert 10th word into printable situation,so we can't print words
above 10th place (inclusive) ............ so it's not solving the original
problem???
On Sun, Aug 26, 2012 at 3:55 AM, Dave <dave_and_da...@juno.com> wrote:
> @Kunal: Yes. You are missing that you don't know the number of words in
> the file in advance. So, to use your method, you would have to read the
> file once to find n, and then read through it again to select the lucky_pos
> th word. The method I proposed only requires reading the file once.
>
> Furthermore, assuming that rand() produces a random non-negative integer,
> rand()%n is not equiprobable for all values of n. Consider n = 3. Then
> since rand() takes on 2^31 values, rand()%3 cannot take on the values 0, 1,
> and 2 with equal probability since 2^31 is not divisible by 3.
>
> Dave
>
> On Saturday, August 25, 2012 1:44:03 PM UTC-5, Kunal Patil wrote:
>
>> How about using rand()%n ??
>> Like, calculate lucky_pos = rand()%n
>> Then print word at lucky_pos th position...
>> Am I missing anything? All words are still equiprobable to get printed
>> right?
>> On Aug 20, 2012 11:45 AM, "Dave" <dave_an...@juno.com> wrote:
>>
>>> @Navin: Okay. Here is a paraphrase. Assume double function random()
>>> returns a uniformly distributed random number >= 0.0 and < 1.0.
>>>
>>> read first word from file into string save;
>>> int i = 1
>>> while not EOF
>>> {
>>> read next word from file into string temp;
>>> i++;
>>> if( i * random() < 1.0 )
>>> copy temp to save;
>>> }
>>> print save;
>>>
>>> Dave
>>>
>>> On Monday, August 20, 2012 12:02:54 AM UTC-5, Navin Kumar wrote:
>>>
>>>> @Dave sir, I didn't get your logic. Can you please elaborate it?
>>>>
>>>> On Sun, Aug 19, 2012 at 4:08 AM, Dave <dave_an...@juno.com> wrote:
>>>>
>>>>> @Navin: Here is the algorithm:
>>>>>
>>>>> Save the first word.
>>>>> For i = 2, 3, ..., n = number of words in the file
>>>>> replace the saved word with the i-th word with probability 1/i.
>>>>> When EOF is reached, every word in the file will have probability 1/n
>>>>> of being the saved word. Print it.
>>>>>
>>>>> Dave
>>>>>
>>>>> On Saturday, August 18, 2012 1:28:56 AM UTC-5, Navin Kumar wrote:
>>>>>
>>>>>> Print a *Random word* from a file. Input is "path to a file",
>>>>>>
>>>>>> constraints- No extra memory like hashing etc. All the words in the
>>>>>> file should have equal probability.
>>>>>
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‘Kailash Bagaria’
B-tech 4th year
Computer Science & Engineering
Indian Institute of Technology, Roorkee
Roorkee, India (247667)
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