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That’s OK, I was already upset by someone else's comment, which is why I reacted the way I did.
RDSchaefer wrote: BTW, you really need to develop thicker skin.
Actually, I have a thick skin. Growing up with the surname "Darling" ensures that you develop one quickly
I am a little tired today, and when I received a comment from someone on here telling me that I was "NOT a scientist" when they don’t know me, and have never spoke to me, was what made me flip a little.
I can understand that it happens, and normally it would be water off a ducks back, however, on this occasion it caused me to respond.
Kind Regards,
Stephen
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you shouldn't add 5 or 10 but 5 or 0.
In the case the random generator provides 9999 then you would get
9999 * 10 + 10 = 99990 + 10 = 100000
six digits
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This solution is not correct - it exhibits a possible overflow error into 6 digits.
Consider: 4 Random digits all come out as 9 (multiply by 10 = 99990), then add 10 you get 100000 which does not meet the requirements...
The correct solution is:
Generate a four digit random number
Multiply it by 10 (to shift into correct position)
Then randomly choose to make the last digit a 5 or a 0 (adding 10 is the incorrect step)
The result is divisible by 5.
Regards
David.
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See my correction - you don't need to add 10 at all (you can add 5 without fear).
That was a silly on my part.
Regards
David R
---------------------------------------------------------------
"Every program eventually becomes rococo, and then rubble." - Alan Perlis
The only valid measurement of code quality: WTFs/minute.
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David1987 wrote: And make sure you reuse a single instance of Random ,
How exactly do I ensure that I am doing this?
I am using rand a number of times, and although I get different values, it does appear that they are very simular.
Regards,
Stephen
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Is this a joke question? Generate a four digit number and add an extra number to the end, being either 5 or 0. Then don't check it because it will be correct!!!
e.g. 4678; add 5 on the end to give 46785.
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Exactly my take on it
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Random random = new Random();
return (random.Next(17999) + 2000) * 5;
this gives a number divisable by 5 between 10000 and 99995
Paulo Gomes
Over and Out
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Generate a five digit random number, convert to string, replace the right most digit with 5 (or 0), parse the string back to an int.
No need to check as anything ending in 5 or 0 is divible by 5.
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This would work as an alternative but the math approach is better (more performant) unless you want the final value to be a string. And, even then the math approach would be faster.
(nnnn*10)=nnnn0
It's the same as appending a "0" to a four-digit number but it's much faster.
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I'm sorry but a lot of the answers made me laugh...
By definition any number that has been multiplied by 5 is divisible by 5. So the solution is:
x=random(low_limit, high_limit)*5; no need to check this!
The five digit condition is a little bit more complicated, because it is not clearly defined. is 00005 a five digit number by this definition? Low limit will then be 0, high limit 19999 (the highest 5 digit number fullfilling the condition is 99995).
To get a number that is divisable by 5, the ending needs to be 0 or 5. If you need checking, the modulo division is correct.
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// To generate as random numbers as possible, this variable must only
// be initialized once and then reused as much as possible.
Random rand = new Random();
/// <summary>
/// Generates a random 5 digit number that is always divisible by 5.
/// </summary>
/// <returns>Returns a random integer 5 digits long that is evenly divisible by 5.</returns>
private int GetDivBy5()
{
// rnd.Next(minValue, maxValue) returns a random between minValue
// and maxValue, inclusive. Use min/max values of 1000 and 9999 to
// yield a random 4 digit number between 1000 and 9999 then multiply
// by 10 to get a 5 digit number. This number will always be
// divisible by 5 but will always end in 0.
int divBy5 = rand.Next(1000, 9999) * 10;
// To get more random and double the possible result set, randomly
// add on 5. This will yield the total possible set of integers
// between 10000 and 99995 that are divisible by 5.
if (rand.NextDouble() < 0.5)
return divBy5;
else
return divBy5 + 5;
}
Stephen, you mention you are a coding noobie. So, a short word about random number generators - they're not really very random.
All they do is take a seed (beginning) value and run it through a mathmatical algorithm that generates a new number. If you use the same seed value every time, you will get the same "random" number every time. Most random number generators can either self-seed - generally using the current clock value - or you can pass them a seed value when you initialize them. This is why you only want to initialize your random number generator once and reuse it over and over.
Once you get past that, the rest is pretty easy. Just use the math several others have already provided. See the comments in the code...
<edit>
Um, just looked at the thread in the other forum where you defined your requirements a little better. It seems you would like to include 0-9999 in your result set too. You can do this easily enough by changing the min value passed to rand.Next(minValue, maxValue) to 0 instead of 1000. This last bit, is a bit tricky in that in programming terms "00005" is actually not a number but is really a string. To get the leading zeros, you need to convert your final value to a string and use a formatting mask to append the leading zeros. This also assumes "00000" is valid. If it's not, test for it with an "if" and call GetDivBy5 again.
So, to get a string with leading zeros use the following;
private void button1_Click(object sender, EventArgs e)
{
this.textBox1.Text = this.GetDivBy5().ToString("00000");
}
</edit>
modified on Tuesday, August 23, 2011 11:03 AM
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Thank you very much.
Sorry I missed this post, and to be honest I am a little p***** off with the negative comments I am receiving, that is why I stopped looking at the responses as it upset me a little that people on here can be so damn rude and insulting, oh, and judgmental, for no apparent reason?
Anyway, thank you very much.
Kind Regards,
Stephen
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Everyone's a Legend in Their Own Lunchtime. You've learned a lesson - whether you wished to or not - about being precise in framing your question, so I hope that at least that, and the sensible answers you did get, helped.
As to the others who felt it was easier to ridicule than to stop for a second and realise there was more to it, I apologise.
cheers,
Chris Maunder
The Code Project | Co-founder
Microsoft C++ MVP
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Chris Maunder wrote: Everyone's a Legend in Their Own Lunchtime.
Agreed!
Chris Maunder wrote: You've learned a lesson - whether you wished to or not
Actually I learned two. Yes; one was indeed to think allot more before I pose my question!
Chris Maunder wrote: As to the others who felt it was easier to ridicule than to stop for a second and realise there was more to it, I apologise.
You have nothing to apologise for, and it was really just the one statement that I was not a scientist that caused me to react.
Never mind, I have moved on
Kind Regards,
Stephen
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You're welcome, Stephen.
Please keep in mind that mad geniuses as many programmers may be, few people tend to lump us programmers into the "high social skills" cliche. It's been a while since I worked with scientists but I seem to recall from my university days they can be somewhat socially lacking too. I remember this one physicist... Ahem... Uh, anyway, its probably best to just take the "water off a ducks back" approach - no one really means to insult - especially if you want any of us social deviants to actually help you.
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Michael A. Cochran wrote: It's been a while since I worked with scientists but I seem to recall from my university days they can be somewhat socially lacking too.
Could not agree more! At the end of the day we are all human.
Michael A. Cochran wrote: it’s probably best to just take the "water off a ducks back" approach
I normally do, however, being told by someone that I had never met or spoke to that I was not a scientist was what got to me. It was not just an insult and ignorant statement, but it was not even to do with programming and in my own opinion was a personal attack that was not called for.
Anyway, I have taken what I need from this post, got my code to work and moved on
Kind Regards,
Stephen
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x = generate random number between 10000 and 99995
remainder to be added = 5 - x MOD 5
Therefore, the resultant random number should be x + 5 - x MOD 5
Here's a sample code: http://ideone.com/LxgrC[^]
a.k.a. Vite Phoenix and Vite Zeus...
Proud member and co-founder of OlympianZ
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Why didn't I think of that? Too much business programming, I guess. We don't use modulus in business programming much - at least I haven't had the need.
Correct, of course. Much more elegant.
If Stephen is still listening, in C# it becomes;
Random rand = new Random();
private int GetDivBy5()
{
int divBy5 = rand.Next(0, 99999);
return divBy5 - (divBy5 % 5);
}
private void button1_Click(object sender, EventArgs e)
{
this.textBox1.Text = this.GetDivBy5().ToString("00000");
}
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nice, the shortest and the best so far
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You are getting all sorts of advice. Multiplying by 10 can result in overflow. The easiest is: num -= num%5;
IE subtract the remainder if you divided by five from the result.
Another way: num = (num/5) * 5;
That will NEVER cause an overflow in most languages. If you are using vb.net, besides the syntax replace "/" with "\".
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I guess a simple expression like
int random= (rand()%18000)*5 + 10000
should do the trick where rand() is a function that returns any positive random number for C# it can be replaced with
int random= (new Random().Next()%18000)*5 + 10000
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The solution I found most elegant had already been posted by David1984 but is deleted. So I'm going to post it now:
You want a number X that can be divided by 5.
Any number Y can be multiplied by 5. The result X is divisible by definition (and by 5).
You want X to have 5 digits.
5-digit-numbers range from 10000 to 99999. Divide both by 5 and throw away the result's decimal places to get the limits for Y (2000 to 19999).
Supposed you already have an instance of Random obeying the tips on its use, you get X by using this code (X changed to a meaningful name)
int randomDivisibleByFive = random.Next(2000, 19999) * 5;
Ciao,
luker
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