Path: csiph.com!usenet.pasdenom.info!weretis.net!feeder4.news.weretis.net!feeds.phibee-telecom.net!newsfeed.xs4all.nl!newsfeed3.news.xs4all.nl!xs4all!newsgate.cistron.nl!newsgate.news.xs4all.nl!post.news.xs4all.nl!not-for-mail Return-Path: X-Original-To: python-list@python.org Delivered-To: python-list@mail.python.org X-Spam-Status: OK 0.071 X-Spam-Evidence: '*H*': 0.86; '*S*': 0.00; 'root': 0.05; 'see:': 0.07; 'calculating': 0.09; 'item.': 0.09; 'name?': 0.09; 'subject:number': 0.09; 'def': 0.12; '(3,': 0.16; '(9,': 0.16; '3)]': 0.16; '999': 0.16; 'division,': 0.16; 'function?': 0.16; 'integer.': 0.16; 'inverse': 0.16; 'output?': 0.16; 'reversing': 0.16; 'subject:skip:m 10': 0.16; 'superfluous': 0.16; 'tuple': 0.16; 'wrote:': 0.18; 'obviously': 0.18; 'written': 0.21; 'input': 0.22; 'header:User-Agent:1': 0.23; 'stopping': 0.24; '---': 0.24; 'header:In-Reply-To:1': 0.27; 'function': 0.29; 'url:wiki': 0.31; 'decimal': 0.31; 'exclude': 0.31; 'there.': 0.32; 'subject:all': 0.32; "i'd": 0.34; 'could': 0.34; 'something': 0.35; 'but': 0.35; 'there': 0.35; 'done': 0.36; 'next': 0.36; 'possible': 0.36; 'url:org': 0.36; 'should': 0.36; 'list': 0.37; 'thank': 0.38; 'to:addr:python-list': 0.38; 'pm,': 0.38; 'rather': 0.38; 'does': 0.39; 'to:addr:python.org': 0.39; 'how': 0.40; 'dave': 0.60; 'most': 0.60; 'break': 0.61; 'numbers': 0.61; "you'll": 0.62; "you've": 0.63; 'real': 0.63; 'soon': 0.63; 'more': 0.64; 'charset:windows-1252': 0.65; 'received:74.208': 0.68; 'prime': 0.74; 'repeat': 0.74; 'square': 0.74; 'gain': 0.79; 'divide': 0.84; 'end.': 0.84; 'factors,': 0.84; 'improvement': 0.84; 'multiplying': 0.84; 'received:74.208.4.194': 0.84; 'subject:find': 0.84; 'angel': 0.91; 'items,': 0.91; 'factors': 0.97 Date: Fri, 10 Apr 2015 21:16:57 -0400 From: Dave Angel User-Agent: Mozilla/5.0 (X11; Linux x86_64; rv:31.0) Gecko/20100101 Thunderbird/31.5.0 MIME-Version: 1.0 To: python-list@python.org Subject: Re: find all multiplicands and multipliers for a number References: <890bd388-f50a-4dec-9ef5-27715427472a@googlegroups.com> <55287387.4070501@davea.name> In-Reply-To: <55287387.4070501@davea.name> Content-Type: text/plain; charset=windows-1252; format=flowed Content-Transfer-Encoding: 7bit X-Provags-ID: V03:K0:/kzdUfqu8XKMaxg2MAgcYzMc1APh7bzp/ITTqaRfXXJM7D2d8om DjMJHpYL0Pqc3N8JvcND5b3ECUq3A3hom9Sk+0wBG6KHJ8o2pWo2PX7ScxsgAdD9ccNGaC2 VEWfoFBS1QbYCYwmr/KmOmlFkoY20DkBm6Wzrx3fmXVIW0sv3xzFtYa9Au8Rk2khp+Tp3DK Fml0vGVyz5EGNDQ2omV4Q== X-UI-Out-Filterresults: notjunk:1; X-BeenThere: python-list@python.org X-Mailman-Version: 2.1.20 Precedence: list List-Id: General discussion list for the Python programming language List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , Newsgroups: comp.lang.python Message-ID: Lines: 65 NNTP-Posting-Host: 2001:888:2000:d::a6 X-Trace: 1428715027 news.xs4all.nl 2847 [2001:888:2000:d::a6]:42926 X-Complaints-To: abuse@xs4all.nl Xref: csiph.com comp.lang.python:88789 On 04/10/2015 09:06 PM, Dave Angel wrote: > On 04/10/2015 07:37 PM, ravas wrote: >> def m_and_m(dividend): >> rlist = [] >> dm = divmod >> end = (dividend // 2) + 1 >> for divisor in range(1, end): >> q, r = dm(dividend, divisor) >> if r is 0: >> rlist.append((divisor, q)) >> return rlist >> >> print(m_and_m(999)) >> --- >> output: [(1, 999), (3, 333), (9, 111), (27, 37), (37, 27), (111, 9), >> (333, 3)] >> --- >> >> How do we describe this function? >> Does it have an established name? >> What would you call it? >> Does 'Rosetta Code' have it or something that uses it? >> Can it be written to be more efficient? >> What is the most efficient way to exclude the superfluous inverse tuples? >> Can it be written for decimal numbers as input and/or output? >> >> Thank you! >> > > I'd call those factors of the original number. For completeness, I'd > include (999,1) at the end. > > If it were my problem, I'd be looking for only prime factors. Then if > someone wanted all the factors, they could derive them from the primes, > by multiplying all possible combinations. > > The program can be sped up most obviously by stopping as soon as you get > a tuple where divisor > q. At that point, you can just repeat all the > items, reversing divisor and q for each item. Of course, now I notice > you want to eliminate them. So just break out of the loop when divisor > > q. > > You can gain some more speed by calculating the square root of the > dividend, and stopping when you get there. > > But the real place to get improvement is to only divide by primes, > rather than every possible integer. And once you've done the division, > let q be the next value for dividend. So you'll get a list like > > [3, 3, 3, 37] > > for the value 999 > > See: > http://rosettacode.org/wiki/Factors_of_an_integer#Python > > And http://rosettacode.org/wiki/Prime_decomposition#Python There the function that you should grok is decompose() -- DaveA