Math Question....
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From: "Jo�o C�ndido de Souza Neto" on 22 Apr 2010 10:26 there�s an error in the forget the former: for ($g = $maxpergroup; $g > 0; $g--) { if ($items mod $g <> 0) continue; $Groups = $items div $g; } ""Jo�o C�ndido de Souza Neto"" <joao(a)consultorweb.cnt.br> escreveu na mensagem news:87.84.21218.63C50DB4(a)pb1.pair.com... > for ($g = $maxpergroup; $g > 0; $g++) { > if ($items mod $g <> 0) continue; > $Groups = $items div $g; > } > > Maybe it can helps you. > > "Dan Joseph" <dmjoseph(a)gmail.com> escreveu na mensagem > news:q2oa20394491004220707x980cef5ej2b310c97d123056b(a)mail.gmail.com... >> 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. 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. 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? >> PHP >> code or Math stuff is fine. Either way... >> >> Thanks... >> >> -- >> -Dan Joseph >> >> www.canishosting.com - Unlimited Hosting Plans start @ $3.95/month. >> Promo >> Code "NEWTHINGS" for 10% off initial order >> >> http://www.facebook.com/canishosting >> http://www.facebook.com/originalpoetry >> > >
From: "Jo�o C�ndido de Souza Neto" on 22 Apr 2010 10:29 I think today is not may day to write. hehehe sorry. ""Jo�o C�ndido de Souza Neto"" <joao(a)consultorweb.cnt.br> escreveu na mensagem news:12.B5.21218.F0D50DB4(a)pb1.pair.com... > there�s an error in the forget the former: > > for ($g = $maxpergroup; $g > 0; $g--) { > if ($items mod $g <> 0) continue; > $Groups = $items div $g; > } > > > ""Jo�o C�ndido de Souza Neto"" <joao(a)consultorweb.cnt.br> escreveu na > mensagem news:87.84.21218.63C50DB4(a)pb1.pair.com... >> for ($g = $maxpergroup; $g > 0; $g++) { >> if ($items mod $g <> 0) continue; >> $Groups = $items div $g; >> } >> >> Maybe it can helps you. >> >> "Dan Joseph" <dmjoseph(a)gmail.com> escreveu na mensagem >> news:q2oa20394491004220707x980cef5ej2b310c97d123056b(a)mail.gmail.com... >>> 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. 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. 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? >>> PHP >>> code or Math stuff is fine. Either way... >>> >>> Thanks... >>> >>> -- >>> -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 >>> >> >> > >
From: Richard Quadling on 22 Apr 2010 10:26 On 22 April 2010 15:13, Ashley Sheridan <ash(a)ashleysheridan.co.uk> wrote: > On Thu, 2010-04-22 at 10:17 -0400, Dan Joseph wrote: > >> On Thu, Apr 22, 2010 at 10:12 AM, Stephen <stephen-d(a)rogers.com> 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
From: Richard Quadling on 22 Apr 2010 10:29 On 22 April 2010 15:26, Richard Quadling <rquadling(a)googlemail.com> wrote: > On 22 April 2010 15:13, Ashley Sheridan <ash(a)ashleysheridan.co.uk> wrote: >> On Thu, 2010-04-22 at 10:17 -0400, Dan Joseph wrote: >> >>> On Thu, Apr 22, 2010 at 10:12 AM, Stephen <stephen-d(a)rogers.com> 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
From: "Jason" on 22 Apr 2010 10:37 -----Original Message----- From: Ashley Sheridan [mailto:ash(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 <stephen-d(a)rogers.com> 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? <? $items=1252398; $groupsize=30; for ($i=0;$i<$items;$i++) $grouparray[$i % $groupsize][]=$i; print_r($grouparray); ?> HTH J
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