@Kailash: I assumed a function random() which generates uniformly
distributed double precision random numbers between 0 and 1. Usually such a
generator would produce numbers with 53 bit precision, so I guess that is
between 15 and 16 digits after the decimal point. Just google "random
number generator" for a variety of methods to do this.
Dave
On Sunday, August 26, 2012 9:03:21 AM UTC-5, kailash wrote:
> @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_an...@juno.com
> <javascript:>>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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