Re: [PHP] Math Question....
Supply a value and an optional maximum group size. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Math Question....
At 10:17 AM -0400 4/22/10, Dan Joseph wrote: On Thu, Apr 22, 2010 at 10:12 AM, Stephen wrote: 1,252,398 DIV 30 = 41,746 groups of 30. 1,252,398 MOD 30 = 18 items in last group Well, the only problem with going that route, is the one group is not equally sized to the others. 18 is ok for a group in this instance, but if it was a remainder of only 1 or 2, there would be an issue. Which is where I come to looking for a the right method to break it equally. -- -Dan Joseph _Dan: As I see it -- you are asking is "What would be the 'optimum' group size for 1,252,398?" "Optimum" here meaning: 1. A group size of 30 or under; 2. All groups being of equal size. Is that correct? There may not be an exact solution, but a first order attempt would be to divide the total number by 30 and check the remainder (i.e., MOD), the do the same for 29, 28, 27... and so on. The group size "solution" would be a number with a zero remainder OR with a remainder closest to your group size. That would be my first blush solution. Cheers, tedd -- --- http://sperling.com http://ancientstones.com http://earthstones.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Math Question....
On 23 April 2010 13:33, Dotan Cohen wrote: > What is wrong with 626,299 groups of 2 items each (done in my head, so > I might be off a little)? 2, 3, 6, 7, 14 and 21 are all valid. -- - Richard Quadling "Standing on the shoulders of some very clever giants!" EE : http://www.experts-exchange.com/M_248814.html EE4Free : http://www.experts-exchange.com/becomeAnExpert.jsp Zend Certified Engineer : http://zend.com/zce.php?c=ZEND002498&r=213474731 ZOPA : http://uk.zopa.com/member/RQuadling -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Math Question....
On 22 April 2010 17:07, Dan Joseph wrote: > Howdy, > > This is a math question, but I'm doing the code in PHP, and have expunged > all resources... hoping someone can guide me here. Â For some reason, I can't > figure this out. > > I want to take a group of items, and divide them into equal groups based on > a max per group. Â Example. > > 1,252,398 -- divide into equal groups with only 30 items per group max. > > Can anyone guide me towards an algorithm or formula name to solve this? Â PHP > code or Math stuff is fine. Â Either way... > > Thanks... > What is wrong with 626,299 groups of 2 items each (done in my head, so I might be off a little)? -- Dotan Cohen http://bido.com http://what-is-what.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Math Question....
On 22 April 2010 17:47, Developer Team wrote:
> Awesome source.
> Thanks
>
> On 4/22/10, Richard Quadling wrote:
>> On 22 April 2010 14:48, Dan Joseph wrote:
>>> On Thu, Apr 22, 2010 at 10:29 AM, Richard Quadling
>>> >>> wrote:
>>>
 >
> It sounds like you are looking for factors.
>
>
http://www.algebra.com/algebra/homework/divisibility/factor-any-number-1.solver
>
> Solution by Find factors of any number
>
> 1252398 is NOT a prime number: 1252398 = 2 * 3 * 7 * 29819
> Work Shown
>
> 1252398 is divisible by 2: 1252398 = 626199 * 2.
> 626199 is divisible by 3: 626199 = 208733 * 3.
> 208733 is divisible by 7: 208733 = 29819 * 7.
> 29819 is not divisible by anything.
>
> So 29819 by 42 (7*3*2)
>
> would be a route.
Aha. Missed the "30" bit.
So, having found the factors, you would need to process them to find
the largest combination under 30.
2*3
2*3*7
2*7
3*7
are the possibilities (ignoring any number over 30).
Of which 3*7 is the largest.
So, 1,252,398 divided by 21 = 59,638
Is that the sort of thing you are looking for?
>>>
>>> Yes, that looks exactly what like what I'm looking for. Â I'm going to try
>>> and wake up the algebra side of my brain that hasn't been used in years
>>> and
>>> see if I can digest all this.
>>>
>>> For the 2, 3, and 7, that is based solely on the last number being
>>> divisible
>>> by a prime number?
>>>
>>> Joao, Jason, thanks for the code.
>>>
>>> --
>>> -Dan Joseph
>>>
>>> www.canishosting.com - Unlimited Hosting Plans start @ $3.95/month. Â Promo
>>> Code "NEWTHINGS" for 10% off initial order
>>>
>>> http://www.facebook.com/canishosting
>>> http://www.facebook.com/originalpoetry
>>>
>>
>> This seems to be working ...
>>
>> > function findBestFactors($Value, $GroupSize, array &$Factors = null)
>> Â Â Â {
>> Â Â Â $Factors = array();
>> Â Â Â foreach(range(1, ceil(sqrt($Value))) as $Factor)
>> Â Â Â Â Â Â Â {
>> Â Â Â Â Â Â Â if (0 == ($Value % $Factor))
>> Â Â Â Â Â Â Â Â Â Â Â {
>> Â Â Â Â Â Â Â Â Â Â Â if ($Factor <= $GroupSize)
>> Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â {
>> Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â $Factors[] = $Factor;
>> Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â }
>> Â Â Â Â Â Â Â Â Â Â Â if ($Factor != ($OtherFactor = ($Value / $Factor)) &&
>> $OtherFactor
>> <= $GroupSize)
>> Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â {
>> Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â $Factors[] = $OtherFactor;
>> Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â }
>> Â Â Â Â Â Â Â Â Â Â Â }
>>
>> Â Â Â Â Â Â Â if ($Factor >= $GroupSize)
>> Â Â Â Â Â Â Â Â Â Â Â {
>> Â Â Â Â Â Â Â Â Â Â Â break;
>> Â Â Â Â Â Â Â Â Â Â Â }
>> Â Â Â Â Â Â Â }
>>
>> Â Â Â rsort($Factors);
>>
>> Â Â Â return reset($Factors);
>> Â Â Â }
>>
>> echo findBestFactors($argv[1], $argv[2], $Factors), PHP_EOL;
>> ?>
>>
>>
>> factors 1252398988 5000
>>
>> outputs  ...
>>
>> 4882
>>
>> and 21 for your value 1252398
>>
>> --
>> -
>> Richard Quadling
>> "Standing on the shoulders of some very clever giants!"
>> EE : http://www.experts-exchange.com/M_248814.html
>> EE4Free : http://www.experts-exchange.com/becomeAnExpert.jsp
>> Zend Certified Engineer : http://zend.com/zce.php?c=ZEND002498&r=213474731
>> ZOPA : http://uk.zopa.com/member/RQuadling
>>
>> --
>> PHP General Mailing List (http://www.php.net/)
>> To unsubscribe, visit: http://www.php.net/unsub.php
>>
>>
>
Thank you. It was a quick knock up, so could probably be optimized a
little more.
It will also not work beyond PHP_MAX_INT, unless the code is converted
to use the BCMath or GMP extension.
--
-
Richard Quadling
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Re: [PHP] Math Question....
On Thu, Apr 22, 2010 at 12:16 PM, Richard Quadling wrote:
> On 22 April 2010 14:48, Dan Joseph wrote:
> This seems to be working ...
>
> function findBestFactors($Value, $GroupSize, array &$Factors = null)
>{
>$Factors = array();
>foreach(range(1, ceil(sqrt($Value))) as $Factor)
>{
>if (0 == ($Value % $Factor))
>{
>if ($Factor <= $GroupSize)
>{
>$Factors[] = $Factor;
>}
>if ($Factor != ($OtherFactor = ($Value / $Factor))
> && $OtherFactor
> <= $GroupSize)
>{
>$Factors[] = $OtherFactor;
>}
>}
>
>if ($Factor >= $GroupSize)
>{
>break;
>}
>}
>
>rsort($Factors);
>
>return reset($Factors);
>}
>
> echo findBestFactors($argv[1], $argv[2], $Factors), PHP_EOL;
> ?>
>
>
> factors 1252398988 5000
>
> outputs ...
>
> 4882
>
> and 21 for your value 1252398
>
>
>
Wow! thanks... I just plopped it into phped and fired off some tests, and
I agree, seems to work fine. I appreciate your help today. I am still
looking over the algebra stuff, and am now comparing it to your code. This
will get me moving forward better in my project. Thank you!
--
-Dan Joseph
www.canishosting.com - Unlimited Hosting Plans start @ $3.95/month. Promo
Code "NEWTHINGS" for 10% off initial order
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Re: [PHP] Math Question....
On 22 April 2010 14:48, Dan Joseph wrote:
> On Thu, Apr 22, 2010 at 10:29 AM, Richard Quadling > wrote:
>
>> Â >
>> > It sounds like you are looking for factors.
>> >
>> >
>> http://www.algebra.com/algebra/homework/divisibility/factor-any-number-1.solver
>> >
>> > Solution by Find factors of any number
>> >
>> > 1252398 is NOT a prime number: 1252398 = 2 * 3 * 7 * 29819
>> > Work Shown
>> >
>> > 1252398 is divisible by 2: 1252398 = 626199 * 2.
>> > 626199 is divisible by 3: 626199 = 208733 * 3.
>> > 208733 is divisible by 7: 208733 = 29819 * 7.
>> > 29819 is not divisible by anything.
>> >
>> > So 29819 by 42 (7*3*2)
>> >
>> > would be a route.
>>
>> Aha. Missed the "30" bit.
>>
>> So, having found the factors, you would need to process them to find
>> the largest combination under 30.
>>
>> 2*3
>> 2*3*7
>> 2*7
>> 3*7
>>
>> are the possibilities (ignoring any number over 30).
>>
>> Of which 3*7 is the largest.
>>
>> So, 1,252,398 divided by 21 = 59,638
>>
>>
>> Is that the sort of thing you are looking for?
>>
>>
>
> Yes, that looks exactly what like what I'm looking for. Â I'm going to try
> and wake up the algebra side of my brain that hasn't been used in years and
> see if I can digest all this.
>
> For the 2, 3, and 7, that is based solely on the last number being divisible
> by a prime number?
>
> Joao, Jason, thanks for the code.
>
> --
> -Dan Joseph
>
> www.canishosting.com - Unlimited Hosting Plans start @ $3.95/month. Â Promo
> Code "NEWTHINGS" for 10% off initial order
>
> http://www.facebook.com/canishosting
> http://www.facebook.com/originalpoetry
>
This seems to be working ...
= $GroupSize)
{
break;
}
}
rsort($Factors);
return reset($Factors);
}
echo findBestFactors($argv[1], $argv[2], $Factors), PHP_EOL;
?>
factors 1252398988 5000
outputs ...
4882
and 21 for your value 1252398
--
-
Richard Quadling
"Standing on the shoulders of some very clever giants!"
EE : http://www.experts-exchange.com/M_248814.html
EE4Free : http://www.experts-exchange.com/becomeAnExpert.jsp
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Re: [PHP] Math Question....
On Thu, 22 Apr 2010 10:49:11 -0400
Peter van der Does wrote:
>
> My take on it:
>
> $Items=1252398;
> $MaxInGroup=30;
> for ($x=$MaxInGroup; $x>1;$x--) {
> $remainder=$Items % $x;
> // Change 17 to the max amount allowed in the last group
> if ($remainder == 0 || $remainder >= 17) { //
> $groups = (int) ($Items /$x)+1;
> echo $groups."\n";
> echo $remainder;
> break;
> }
> }
>
Bugfixed LOL:
$Items=1252398;
$MaxInGroup=30;
for ($x=$MaxInGroup; $x>1;$x--) {
$remainder=$Items % $x;
// Change 17 to the max amount allowed in a group
if ($remainder == 0 || $remainder >= 17) {
$groups = (int) ($Items /$x);
if ($remainder > 0 ) {
$groups++;
}
echo $groups."\n";
echo $remainder;
break;
}
}
--
Peter van der Does
GPG key: E77E8E98
IRC: Ganseki on irc.freenode.net
Twitter: @petervanderdoes
WordPress Plugin Developer
Blog: http://blog.avirtualhome.com
Forums: http://forums.avirtualhome.com
Twitter: @avhsoftware
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Re: [PHP] Math Question....
On Thu, 22 Apr 2010 10:17:10 -0400
Dan Joseph wrote:
> On Thu, Apr 22, 2010 at 10:12 AM, Stephen
> wrote:
>
> > 1,252,398 DIV 30 = 41,746 groups of 30.
> >
> > 1,252,398 MOD 30 = 18 items in last group
> >
> Well, the only problem with going that route, is the one group is not
> equally sized to the others. 18 is ok for a group in this instance,
> but if it was a remainder of only 1 or 2, there would be an issue.
> Which is where I come to looking for a the right method to break it
> equally.
>
My take on it:
$Items=1252398;
$MaxInGroup=30;
for ($x=$MaxInGroup; $x>1;$x--) {
$remainder=$Items % $x;
// Change 17 to the max amount allowed in the last group
if ($remainder == 0 || $remainder >= 17) { //
$groups = (int) ($Items /$x)+1;
echo $groups."\n";
echo $remainder;
break;
}
}
--
Peter van der Does
GPG key: E77E8E98
IRC: Ganseki on irc.freenode.net
Twitter: @petervanderdoes
WordPress Plugin Developer
Blog: http://blog.avirtualhome.com
Forums: http://forums.avirtualhome.com
Twitter: @avhsoftware
--
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Re: [PHP] Math Question....
On Thu, Apr 22, 2010 at 10:29 AM, Richard Quadling wrote: > > > > It sounds like you are looking for factors. > > > > > http://www.algebra.com/algebra/homework/divisibility/factor-any-number-1.solver > > > > Solution by Find factors of any number > > > > 1252398 is NOT a prime number: 1252398 = 2 * 3 * 7 * 29819 > > Work Shown > > > > 1252398 is divisible by 2: 1252398 = 626199 * 2. > > 626199 is divisible by 3: 626199 = 208733 * 3. > > 208733 is divisible by 7: 208733 = 29819 * 7. > > 29819 is not divisible by anything. > > > > So 29819 by 42 (7*3*2) > > > > would be a route. > > Aha. Missed the "30" bit. > > So, having found the factors, you would need to process them to find > the largest combination under 30. > > 2*3 > 2*3*7 > 2*7 > 3*7 > > are the possibilities (ignoring any number over 30). > > Of which 3*7 is the largest. > > So, 1,252,398 divided by 21 = 59,638 > > > Is that the sort of thing you are looking for? > > Yes, that looks exactly what like what I'm looking for. I'm going to try and wake up the algebra side of my brain that hasn't been used in years and see if I can digest all this. For the 2, 3, and 7, that is based solely on the last number being divisible by a prime number? Joao, Jason, thanks for the code. -- -Dan Joseph www.canishosting.com - Unlimited Hosting Plans start @ $3.95/month. Promo Code "NEWTHINGS" for 10% off initial order http://www.facebook.com/canishosting http://www.facebook.com/originalpoetry
RE: [PHP] Math Question....
-Original Message- From: Ashley Sheridan [mailto:a...@ashleysheridan.co.uk] Sent: 22 April 2010 15:13 To: Dan Joseph Cc: PHP eMail List Subject: Re: [PHP] Math Question On Thu, 2010-04-22 at 10:17 -0400, Dan Joseph wrote: > On Thu, Apr 22, 2010 at 10:12 AM, Stephen wrote: > > > 1,252,398 DIV 30 = 41,746 groups of 30. > > > > 1,252,398 MOD 30 = 18 items in last group > > > Well, the only problem with going that route, is the one group is not > equally sized to the others. 18 is ok for a group in this instance, but if > it was a remainder of only 1 or 2, there would be an issue. Which is where > I come to looking for a the right method to break it equally. > > > >How do you mean break it equally? If the number doesn't fit, then you've >got a remainder, and no math is going to change that. How do you want >that remainder distributed? > >Thanks, >Ash >http://www.ashleysheridan.co.uk > > > Perhaps a round-robin approach is called for? HTH J -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Math Question....
On 22 April 2010 15:26, Richard Quadling wrote: > On 22 April 2010 15:13, Ashley Sheridan wrote: >> On Thu, 2010-04-22 at 10:17 -0400, Dan Joseph wrote: >> >>> On Thu, Apr 22, 2010 at 10:12 AM, Stephen wrote: >>> >>> > 1,252,398 DIV 30 = 41,746 groups of 30. >>> > >>> > 1,252,398 MOD 30 = 18 items in last group >>> > >>> Well, the only problem with going that route, is the one group is not >>> equally sized to the others. Â 18 is ok for a group in this instance, but if >>> it was a remainder of only 1 or 2, there would be an issue. Â Which is where >>> I come to looking for a the right method to break it equally. >>> >> >> >> How do you mean break it equally? If the number doesn't fit, then you've >> got a remainder, and no math is going to change that. How do you want >> that remainder distributed? >> >> Thanks, >> Ash >> http://www.ashleysheridan.co.uk >> >> >> > > It sounds like you are looking for factors. > > http://www.algebra.com/algebra/homework/divisibility/factor-any-number-1.solver > > Solution by Find factors of any number > > 1252398 is NOT a prime number: 1252398 = 2 * 3 * 7 * 29819 > Work Shown > > 1252398 is divisible by 2: 1252398 = 626199 * 2. > 626199 is divisible by 3: 626199 = 208733 * 3. > 208733 is divisible by 7: 208733 = 29819 * 7. > 29819 is not divisible by anything. > > So 29819 by 42 (7*3*2) > > would be a route. > > > Take note of > http://www.algebra.com/algebra/homework/divisibility/Prime_factorization_algorithm.wikipedia, > which has the comment ... > > "Many cryptographic protocols are based on the difficultly of > factoring large composite integers or a related problem, the RSA > problem. An algorithm which efficiently factors an arbitrary integer > would render RSA-based public-key cryptography insecure.". > > > > > -- > - > Richard Quadling > "Standing on the shoulders of some very clever giants!" > EE : http://www.experts-exchange.com/M_248814.html > EE4Free : http://www.experts-exchange.com/becomeAnExpert.jsp > Zend Certified Engineer : http://zend.com/zce.php?c=ZEND002498&r=213474731 > ZOPA : http://uk.zopa.com/member/RQuadling > Aha. Missed the "30" bit. So, having found the factors, you would need to process them to find the largest combination under 30. 2*3 2*3*7 2*7 3*7 are the possibilities (ignoring any number over 30). Of which 3*7 is the largest. So, 1,252,398 divided by 21 = 59,638 Is that the sort of thing you are looking for? -- - Richard Quadling "Standing on the shoulders of some very clever giants!" EE : http://www.experts-exchange.com/M_248814.html EE4Free : http://www.experts-exchange.com/becomeAnExpert.jsp Zend Certified Engineer : http://zend.com/zce.php?c=ZEND002498&r=213474731 ZOPA : http://uk.zopa.com/member/RQuadling -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Math Question....
On 22 April 2010 15:13, Ashley Sheridan wrote: > On Thu, 2010-04-22 at 10:17 -0400, Dan Joseph wrote: > >> On Thu, Apr 22, 2010 at 10:12 AM, Stephen wrote: >> >> > 1,252,398 DIV 30 = 41,746 groups of 30. >> > >> > 1,252,398 MOD 30 = 18 items in last group >> > >> Well, the only problem with going that route, is the one group is not >> equally sized to the others. 18 is ok for a group in this instance, but if >> it was a remainder of only 1 or 2, there would be an issue. Which is where >> I come to looking for a the right method to break it equally. >> > > > How do you mean break it equally? If the number doesn't fit, then you've > got a remainder, and no math is going to change that. How do you want > that remainder distributed? > > Thanks, > Ash > http://www.ashleysheridan.co.uk > > > It sounds like you are looking for factors. http://www.algebra.com/algebra/homework/divisibility/factor-any-number-1.solver Solution by Find factors of any number 1252398 is NOT a prime number: 1252398 = 2 * 3 * 7 * 29819 Work Shown 1252398 is divisible by 2: 1252398 = 626199 * 2. 626199 is divisible by 3: 626199 = 208733 * 3. 208733 is divisible by 7: 208733 = 29819 * 7. 29819 is not divisible by anything. So 29819 by 42 (7*3*2) would be a route. Take note of http://www.algebra.com/algebra/homework/divisibility/Prime_factorization_algorithm.wikipedia, which has the comment ... "Many cryptographic protocols are based on the difficultly of factoring large composite integers or a related problem, the RSA problem. An algorithm which efficiently factors an arbitrary integer would render RSA-based public-key cryptography insecure.". -- - Richard Quadling "Standing on the shoulders of some very clever giants!" EE : http://www.experts-exchange.com/M_248814.html EE4Free : http://www.experts-exchange.com/becomeAnExpert.jsp Zend Certified Engineer : http://zend.com/zce.php?c=ZEND002498&r=213474731 ZOPA : http://uk.zopa.com/member/RQuadling -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Math Question....
On Thu, 2010-04-22 at 10:17 -0400, Dan Joseph wrote: > On Thu, Apr 22, 2010 at 10:12 AM, Stephen wrote: > > > 1,252,398 DIV 30 = 41,746 groups of 30. > > > > 1,252,398 MOD 30 = 18 items in last group > > > Well, the only problem with going that route, is the one group is not > equally sized to the others. 18 is ok for a group in this instance, but if > it was a remainder of only 1 or 2, there would be an issue. Which is where > I come to looking for a the right method to break it equally. > How do you mean break it equally? If the number doesn't fit, then you've got a remainder, and no math is going to change that. How do you want that remainder distributed? Thanks, Ash http://www.ashleysheridan.co.uk
Re: [PHP] Math Question....
On Thu, Apr 22, 2010 at 10:12 AM, Stephen wrote: > 1,252,398 DIV 30 = 41,746 groups of 30. > > 1,252,398 MOD 30 = 18 items in last group > Well, the only problem with going that route, is the one group is not equally sized to the others. 18 is ok for a group in this instance, but if it was a remainder of only 1 or 2, there would be an issue. Which is where I come to looking for a the right method to break it equally. -- -Dan Joseph www.canishosting.com - Unlimited Hosting Plans start @ $3.95/month. Promo Code "NEWTHINGS" for 10% off initial order http://www.facebook.com/canishosting http://www.facebook.com/originalpoetry
Re: [PHP] Math Question....
Dan Joseph wrote: I want to take a group of items, and divide them into equal groups based on a max per group. Example. 1,252,398 -- divide into equal groups with only 30 items per group max. 1,252,398 DIV 30 = 41,746 groups of 30. 1,252,398 MOD 30 = 18 items in last group Stephen -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Math Question
Jeremy Schroeder wrote: Hey group Is there a function that when you divide 2 numbers you drop the remainder and are left with the whole number. A whole section of the manual dedicated to mathematical functions! http://us4.php.net/manual/en/ref.math.php floor() ceil() round() -- By-Tor.com It's all about the Rush http://www.by-tor.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Math Question
Thanks for all the help, floor() was the correct choice for this problem . -Blake Vincent Jansen wrote: Hi Richard I agree But you always want to round down ;) Blake> Is there a function that when you divide 2 numbers you drop the Blake> remainder and are left with the whole number. Still seems floor() to me Vincent -Original Message- From: Richard Davey [mailto:[EMAIL PROTECTED] Sent: woensdag 11 februari 2004 15:23 To: Vincent Jansen Cc: [EMAIL PROTECTED] Subject: Re[2]: [PHP] Math Question Hello Vincent, Wednesday, February 11, 2004, 2:15:15 PM, you wrote: VJ> Seems to me you need VJ> floor($number1 / $number2) Only if you always want to round *down* the equation. For example flooring a result of 4.99 will give you an integer of 4 whereas round() (or ceil()) will give you 5. It depends on the situation as to which is most useful. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Math Question
Seems to me you need floor($number1 / $number2) Vincent -Original Message- From: Richard Davey [mailto:[EMAIL PROTECTED] Sent: woensdag 11 februari 2004 15:07 To: Jeremy Schroeder Cc: [EMAIL PROTECTED] Subject: Re: [PHP] Math Question Hello Jeremy, Wednesday, February 11, 2004, 2:00:32 PM, you wrote: JS> Is there a function that when you divide 2 numbers you drop the JS> remainder and are left with the whole number. round($number1 / $number2) See the manual for the precision value if you need it. -- Best regards, Richardmailto:[EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Math Question
Hello Jeremy, Wednesday, February 11, 2004, 2:00:32 PM, you wrote: JS> Is there a function that when you divide 2 numbers you drop the JS> remainder and are left with the whole number. round($number1 / $number2) See the manual for the precision value if you need it. -- Best regards, Richardmailto:[EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: PHP Math Question
"Website Managers.Net" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
I like Eric's idea of showing the values and added a form to it so you can
select the number of items to show before running the script.
"
Maximum Number to Show:
";
}
else {
for($i = 1, $j = 1; $i <= $_POST["i"]; $i++, $j++){
if($j == 5){
$j = 1;
print "\n";
}
print $i." is to ".$j."\n";
} // end for
?>
Jim
- Original Message -
From: "Bronislav Kluèka" <[EMAIL PROTECTED]>
To: "Eric Bolikowski" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Thursday, December 11, 2003 5:22 PM
Subject: RE: [PHP] Re: PHP Math Question
| I do not know if I understand well, but what about
|
| $group=$number % 4;
| if ($group==0) $group=4;
|
| Brona
|
| > -Original Message-
| > From: Eric Bolikowski [mailto:[EMAIL PROTECTED]
| > Sent: Thursday, December 11, 2003 10:53 PM
| > To: [EMAIL PROTECTED]
| > Subject: [PHP] Re: PHP Math Question
| >
| >
| >
| > "Mike D" <[EMAIL PROTECTED]> wrote in message
| > news:[EMAIL PROTECTED]
| > > I'm am completely stumped on a seemingly simple math formula
| > >
| > > I need to find a way to map a set of numbers up to 4 (e.g.
| > 1,2,3,4 or 1,2)
| > > to any number in a range of up to 10,000 (ideally, unlimited).
| > Such that,
| > >
| > > (e.g. 1,2,3,4)
| > >
| > > 1 is to 1
| > > 2 is to 2
| > > 3 is to 3
| > > 4 is to 4
| > >
| > > 5 is to 1
| > > 6 is to 2
| > > 7 is to 3
| > > 8 is to 4
| > >
| > > 9 is to 1
| > > 10 is to 2
| > > 11 is to 3
| > > 12 is to 4
| > >
| > > 13 is to 1
| > > 14 is to 2
| > > 15 is to 3
| > > 16 is to 4
| > >
| > > And so on...
| > >
| > > Is anyone good at math, that can throw me a bone?
| > >
| > > Thanks y'all,
| > > Mike D
| > >
| > >
| > >
| > > Mike Dunlop
| > > AWN, Inc.
| > > // www.awn.com
| > > [ e ] [EMAIL PROTECTED]
| > > [ p ] 323.606.4237
| >
| > Here is some simple code to solve that problem(if i have
| > understood right):
| >
| >
| > for($i = 1, $j = 1; $i <= 1; $i++, $j++){
| >
| > if($j == 5){
| > $j = 1;
| > print "\n";
| > }
| >
| > print "$i is to $j\n";
| >
| > }
| >
| > ?>
| >
| > Eric
| >
| > --
| > PHP General Mailing List (http://www.php.net/)
| > To unsubscribe, visit: http://www.php.net/unsub.php
| >
|
| --
| PHP General Mailing List (http://www.php.net/)
| To unsubscribe, visit: http://www.php.net/unsub.php
|
|
Mike sent me an email, and it seemed that he was satisfied with the script
that I made:
";
}
print "$i is to $j\n";
}
?>
Good idea to rather make an interface for this, Jim!
Eric
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Re: [PHP] Re: PHP Math Question
"Mike D" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
I'm am completely stumped on a seemingly simple math formula
I need to find a way to map a set of numbers up to 4 (e.g. 1,2,3,4 or 1,2)
to any number in a range of up to 10,000 (ideally, unlimited). Such that,
(e.g. 1,2,3,4)
1 is to 1
2 is to 2
3 is to 3
4 is to 4
5 is to 1
6 is to 2
7 is to 3
8 is to 4
9 is to 1
10 is to 2
11 is to 3
12 is to 4
13 is to 1
14 is to 2
15 is to 3
16 is to 4
etc.
Here is an (untested) function that may work:
function map_four( $num )
{
$map_val = $num % 4;
if ($map_val) {
return $map_val;
} else {
return 4;
}
}
Basically, it looks like a modulus with 4, except that you want
multiples of four to return four instead of zero.
Unless, of course, I completely misunderstood the question. Which is not
uncommon as of late.
--
Rob
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Re: [PHP] Re: PHP Math Question
I like Eric's idea of showing the values and added a form to it so you can
select the number of items to show before running the script.
"
Maximum Number to Show:
";
}
else {
for($i = 1, $j = 1; $i <= $_POST["i"]; $i++, $j++){
if($j == 5){
$j = 1;
print "\n";
}
print $i." is to ".$j."\n";
} // end for
?>
Jim
- Original Message -
From: "Bronislav Klučka" <[EMAIL PROTECTED]>
To: "Eric Bolikowski" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Thursday, December 11, 2003 5:22 PM
Subject: RE: [PHP] Re: PHP Math Question
| I do not know if I understand well, but what about
|
| $group=$number % 4;
| if ($group==0) $group=4;
|
| Brona
|
| > -Original Message-
| > From: Eric Bolikowski [mailto:[EMAIL PROTECTED]
| > Sent: Thursday, December 11, 2003 10:53 PM
| > To: [EMAIL PROTECTED]
| > Subject: [PHP] Re: PHP Math Question
| >
| >
| >
| > "Mike D" <[EMAIL PROTECTED]> wrote in message
| > news:[EMAIL PROTECTED]
| > > I'm am completely stumped on a seemingly simple math formula
| > >
| > > I need to find a way to map a set of numbers up to 4 (e.g.
| > 1,2,3,4 or 1,2)
| > > to any number in a range of up to 10,000 (ideally, unlimited).
| > Such that,
| > >
| > > (e.g. 1,2,3,4)
| > >
| > > 1 is to 1
| > > 2 is to 2
| > > 3 is to 3
| > > 4 is to 4
| > >
| > > 5 is to 1
| > > 6 is to 2
| > > 7 is to 3
| > > 8 is to 4
| > >
| > > 9 is to 1
| > > 10 is to 2
| > > 11 is to 3
| > > 12 is to 4
| > >
| > > 13 is to 1
| > > 14 is to 2
| > > 15 is to 3
| > > 16 is to 4
| > >
| > > And so on...
| > >
| > > Is anyone good at math, that can throw me a bone?
| > >
| > > Thanks y'all,
| > > Mike D
| > >
| > >
| > >
| > > Mike Dunlop
| > > AWN, Inc.
| > > // www.awn.com
| > > [ e ] [EMAIL PROTECTED]
| > > [ p ] 323.606.4237
| >
| > Here is some simple code to solve that problem(if i have
| > understood right):
| >
| >
| > for($i = 1, $j = 1; $i <= 1; $i++, $j++){
| >
| > if($j == 5){
| > $j = 1;
| > print "\n";
| > }
| >
| > print "$i is to $j\n";
| >
| > }
| >
| > ?>
| >
| > Eric
| >
| > --
| > PHP General Mailing List (http://www.php.net/)
| > To unsubscribe, visit: http://www.php.net/unsub.php
| >
|
| --
| PHP General Mailing List (http://www.php.net/)
| To unsubscribe, visit: http://www.php.net/unsub.php
|
|
--
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To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Re: PHP Math Question
I do not know if I understand well, but what about
$group=$number % 4;
if ($group==0) $group=4;
Brona
> -Original Message-
> From: Eric Bolikowski [mailto:[EMAIL PROTECTED]
> Sent: Thursday, December 11, 2003 10:53 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP] Re: PHP Math Question
>
>
>
> "Mike D" <[EMAIL PROTECTED]> wrote in message
> news:[EMAIL PROTECTED]
> > I'm am completely stumped on a seemingly simple math formula
> >
> > I need to find a way to map a set of numbers up to 4 (e.g.
> 1,2,3,4 or 1,2)
> > to any number in a range of up to 10,000 (ideally, unlimited).
> Such that,
> >
> > (e.g. 1,2,3,4)
> >
> > 1 is to 1
> > 2 is to 2
> > 3 is to 3
> > 4 is to 4
> >
> > 5 is to 1
> > 6 is to 2
> > 7 is to 3
> > 8 is to 4
> >
> > 9 is to 1
> > 10 is to 2
> > 11 is to 3
> > 12 is to 4
> >
> > 13 is to 1
> > 14 is to 2
> > 15 is to 3
> > 16 is to 4
> >
> > And so on...
> >
> > Is anyone good at math, that can throw me a bone?
> >
> > Thanks y'all,
> > Mike D
> >
> >
> >
> > Mike Dunlop
> > AWN, Inc.
> > // www.awn.com
> > [ e ] [EMAIL PROTECTED]
> > [ p ] 323.606.4237
>
> Here is some simple code to solve that problem(if i have
> understood right):
>
>
> for($i = 1, $j = 1; $i <= 1; $i++, $j++){
>
> if($j == 5){
> $j = 1;
> print "\n";
> }
>
> print "$i is to $j\n";
>
> }
>
> ?>
>
> Eric
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
--
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To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: PHP Math Question
"Mike D" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > I'm am completely stumped on a seemingly simple math formula > > I need to find a way to map a set of numbers up to 4 (e.g. 1,2,3,4 or 1,2) > to any number in a range of up to 10,000 (ideally, unlimited). Such that, > > (e.g. 1,2,3,4) > > 1 is to 1 > 2 is to 2 > 3 is to 3 > 4 is to 4 > > 5 is to 1 > 6 is to 2 > 7 is to 3 > 8 is to 4 > > 9 is to 1 > 10 is to 2 > 11 is to 3 > 12 is to 4 > > 13 is to 1 > 14 is to 2 > 15 is to 3 > 16 is to 4 > > And so on... > > Is anyone good at math, that can throw me a bone? > > Thanks y'all, > Mike D > > > > Mike Dunlop > AWN, Inc. > // www.awn.com > [ e ] [EMAIL PROTECTED] > [ p ] 323.606.4237 Here is some simple code to solve that problem(if i have understood right): "; } print "$i is to $j\n"; } ?> Eric -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] math question
If u don't want to use the pi() function. Try M_PI. This is a pi constant its value is 3.14159265358979323846 $theta=15*M_PI/360; - Original Message - From: Anon Y Mous <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Saturday, June 23, 2001 2:21 PM Subject: Re: [PHP] math question > pi is a function. Try this: > > $theta = 15*pi()/360; > > It should return 0.13089969389957 > > > -Evan Nemerson > > > > $theta = 15*pi/360; > > gives a value of 0 for $theta... it should be something like 0.131... > > > Julia > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] > To contact the list administrators, e-mail: [EMAIL PROTECTED] _ Do You Yahoo!? Get your free @yahoo.com address at http://mail.yahoo.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] math question
Have you defined the 'pi' constant? If you are trying to use PHP's built-in PI constant, it is named 'M_PI' Perhaps consider increasing your error reporting level so that PHP reports problems like undefined constants - see error_reporting() for more details. --zak - Original Message - From: "Julia A. Case" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Saturday, June 23, 2001 12:07 AM Subject: [PHP] math question > $theta = 15*pi/360; > gives a value of 0 for $theta... it should be something like 0.131... -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] math question
pi is a function. Try this: $theta = 15*pi()/360; It should return 0.13089969389957 -Evan Nemerson > $theta = 15*pi/360; > gives a value of 0 for $theta... it should be something like 0.131... > Julia -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] math question
""Julia A. Case"" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > $theta = 15*pi/360; > gives a value of 0 for $theta... it should be something like 0.131... > > Julia I get echo 15 * M_PI / 360; returns 0.13089969389957 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
