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• Postgresql - one database for everyone, or one-database per customer

  - by user337876

  I'm working on a web-based business application where each customer will need to have their own data (think basecamphq.com type model) For scalability and ease-of-upgrades, I'd prefer to have a single database where each customer gets a filtered version of the data. The problem is how to guarantee that they stay sandboxed to their own data. Trying to enforce it in code seems like a disaster waiting to happen. I know Oracle has a way to append a where clause to every query based on a login id, but does Postgresql have anything similar? If not, is there a different design pattern I could use (like creating a view of each table for each customer that filters)? Worse case scenario, what is the performance/memory overhead of having 1000 100M databases vs having a single 1Tb database? I will need to provide backup/restore functionality on a per-customer basis which is dead-simple on a single database but quite a bit trickier if they are sharing the database with other customers.

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• Server database -> client database update based on version

  - by user296191

  Hi, What is the recommended method of collecting items in a server database, versioning the database then deploying only the version differences to a client ? Should it by a field in the table (ie. Version: 3.3.9876) against each record ? Should it be DateTime (server based) in each record ? And whats the best way to just deploy the changes to a client with an older version of the database ? Is it a DUMP to a file with a Bulk import of some description ? Open to comments.. Suggestions. Database can be anything (firebird, mysql, sqlserver, sqlite)... Any info greatly appreciated.

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• Transferring a flat file database to a MySQL database

  - by Jon

  I have a flat file database (yeah gross I know - the worst part is that it's 1.4GB), and I'm in the process of moving it to a MySQL database. The problem is that I'm not sure how to go about doing this - and I've checked through every related question on here but none relate to what I want to do, nor how my database is currently setup. My current flat file database is setup to where a normal MySQL row is its own file, and a MySQL table would be the directory. So for example if you have a user named Jon, there would be a file for the user in a directory named /members/. Within that file would be various information for the user including the users id, rank etc - all separated by tabs, all on separate lines (userid\t4). So here's an example user file: userid 4 notes staff notes: bla bla staff2 notes: bla bla bla username Example So how can I convert the above into their own rows and fields in MySQL? And if possible, could I do thousands of these files at once? Thanks.

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• I need to convert my Mysql database to sql server 2005 database

  - by Ayyappan.Anbalagan

  I need to convert my Mysql database to sql server 2005 database, Is there any tools available for this conversion.

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• Sun Oracle Database Machine a román Banca Transilvaniánál

  - by Fekete Zoltán

  Oracle sajtóhír: Banca Transilvania, first institution in Romania to use Sun Oracle Database Machine (English version) Sikersztori, ügyféltörténet pdf-ben. Az Database Machine V2 megjelenését 2009 szeptemberben jelentette az Oracle. A világon az elso bank, ahol már élesben muködik a Database Machine V2, a romániai Banca Transilvania! Olvassa el a sajtóhírt. A Banca Transilvania 1,5 milló ügyféllel rendelkezik. "This system, product of Oracle and Sun, is the fastest server in the world for data storage, online transactions processing and data warehousing applications." Robert C. Rekkers, Banca Transilvania CEO, ezt nyilatkozta:"Business information is accessed 30 times faster using the new system, leading to quicker decisions and a better data base segmentation", azaz a Database Machine segítségével az üzleti kérséseket 30-szor gyorsabban tudják megválaszolni, mint a korábbi rendszerrel. Leontin Toderici, Banca Transilvania COO mondta a következot: "The acquisition price was excellent, as the costs were below those of an ordinary system", azaz a rendszer ára kiváló volt, kisebb volt a kötsége, mint a hagyományos rendszereké. Sorin Mindrutescu, az Oracle Romania vezetoje büszke arra, hogy egy romániai cég is az innovatív rendszer felhasználói között lehet.: "Oracle Exadata V2 is the result of over 30 years of experience in hardware and software development of two leader companies. I am glad that a top Romanian company is amongst the first in the world to use this innovative product." Az Exadata termékcsalád és a Database Machine kiváló eszköz OLTP rendszerek, adattárházak, konszolidációs megoldások adatbázisainak futtatására. Egy csomagban a tartalmazza a szoftvert és az "okos" hardvert, az adatfeldoldozó, a tároló (storage) komponenseket, mindezt az extrém gyors Infiniband kapcsolatokkal összekötve. A Banca Transilvani az Oracle readingi (Nagy-Britannia) központjában tesztelte a Database Machine rendszert, s a korábbi rendszernél tízszer, néhol hetvenkettoször gyorsabb teljesítményt kaptak, 10-72-szeres teljesítménynövekedés!, említette Tudor Iliescu, Trend Import - Export CEO. A központi Oracle sajtóhír: Customers Select Oracle® Exadata for Extreme Performance of Data Warehouse and OLTP Applications

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• Oracle Database 11g R2 támogatott SAP alatt is

  - by Lajos Sárecz

  Húsvét óta már SAP alatt is használható az Oracle Database 11g R2. Köztudott, hogy az SAP csak a Release 2-re ad ki támogatást, így ez most egy igazán örömteli hír az SAP felhasználóknak, hiszen az alábbi 11g R2 újdonságokat tudják alkalmazni SAP környezetben: • Advanced Compression opció (táblára, RMAN mentésre, expdp-re, Data Guard hálózatra) • Real Application Testing • Oracle Database 11g Release 2 Database Vault • Oracle Database 11g Release 2 RAC • Advanced Encryption táblaterekre, RMAN mentésekre, expdp-re, Data Guard hálózatra • Direct NFS • Deferred Segments • Online Patching Azaz például tömöríthetové válik az SAP adatbázisa, vagy az abból készített mentések. Az eddigi tapasztalatok szerint a tömörítés aránya adatbázistól függoen 2-4-szeres. Az adatbázis upgrade és minden egyéb adatbázis infrastruktúrát érinto változatatás kockázata jelentosen csökkentheto lesz a Real Application Testing alkalmazásával. A rendszergazdai szerepkörök szeparaláhatóvá válnak a Database Vault felhasználásával. A Real Application Clusters 11g R2 újdonságai is elérheto lesznek. A Transparent Data Encryption révén a táblaterek és a mentések titkosíthatók úgy, hogy az alkalmazás számára mindez transzparens, azonban a médiához közvetlenül hozzáférve nem lesznek visszafejthetok az adatok. Támogatott lesz a Direct NFS kliens, ezzel NFS elérési sebesség jelentosen javul. A Deffered Segments révén pedig a tábla szegmensek csak akkor kerülnek lefoglalásra, amikor adat kerül a táblába. Ez azért hasznos, mert általában alkalmazások telepítésekor létrejön minden tábla, azonban sok táblába nem kerül adat. Ezáltal mind a telepítés ideje, mind az adatbázis mérete csökkentheto. Az Online Patching pedig lehetové teszi a leállításmentes patch telepítést. Hát azt gondolom ezek vonzó lehetoségek, érdemes betervezni a közeljövobe az SAP rendszerek alatti adatbázis frissítését, hiszen a 10g verzió Premier Support idén nyáron lejár. Az upgrade-hez pedig mindenképp javaslom a Real Application Testing használatát, amivel az éles terhelés mellett teszthelheto teszt környezetben az upgrade. A Sun Oracle Database Machine és az Exadata sajnos még nem támogatott SAP alatt, mivel az ASM certifikáció még nem zárult le. A hírek szerint 2011 elejére várható, hogy ez megtörténik.

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• Project Euler 10: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 10.  As always, any feedback is welcome. # Euler 10 # http://projecteuler.net/index.php?section=problems&id=10 # The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. # Find the sum of all the primes below two million. import time start = time.time() def primes_to_max(max): primes, number = [2], 3 while number < max: isPrime = True for prime in primes: if number % prime == 0: isPrime = False break if (prime * prime > number): break if isPrime: primes.append(number) number += 2 return primes primes = primes_to_max(2000000) print sum(primes) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• New P6 Reporting Database R2

  - by mark.kromer

  Along with our announced GA release of P6 Analytics R1 recently, you may have noticed that when you purchase P6 Analytics, we provide a restricted use license for P6 Reporting Database R2. This represent an updated version of the previous P6 Reporting Database 6.2 and can be purchased individually on a per-CPU basis. Typically, you will want just the reporting database if you would like the P6 data warehouse components such as the ETL, data models, ODS and star schemas in order to report on that data with another reporting tool other than Oracle. The P6 Analytics solution will only work on Oracle BI (OBI). But I pasted below some examples of a simplistic matrix report that I built from the P6 Reporting Database using Microsoft SQL Server Reporting Services. This is the Report Builder tool which is very similar to other similar tools to build reports on the market today such as Crystal Reports or Oracle BI Publisher. This is an example of what you can do (in a very simple format) by using the P6 Reporting Database without P6 Analytics: Here is a quick run-down of some of the key new features in P6 Reporting Database R2 that were added as enhancements to the 6.2 version: • 4 new star schemas (improved projects star, project history, resource utilization and resource allocation) • Improved ETL performance and reliability • P6 security is inherited at the star schema level • Custom P6 project, activity & resource codes are now available as customizable dimensions in the star schemas • Time-phase data down to the data is now available from the star schemas • An updated Operational Data Store (ODS) for operational reporting that includes the WBS hierarchy • The ODS now includes daily spreads for activity and resource assignments

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• SQL SERVER – Installing AdventureWorks Sample Database – SQL in Sixty Seconds #010 – Video

  - by pinaldave

  SQL Server has so many enhancements and features that quite often I feel like playing with various features and try out new things. I often come across situation where I want to try something new but I do not have sample data to experiment with. Also just like any sane developer I do not try any of my new experiments on production server. Additionally, when it is about new version of the SQL Server, there are cases when there is no relevant sample data even available on development server. In this kind of scenario sample database can be very much handy. Additionally, in many SQL Books and online blogs and articles there are scripts written by using AdventureWork database. The often receive request that where people can get sample database as well how to restore sample database. In this sixty seconds video we have discussed the same. You can get various resources used in this video from http://bit.ly/adw2012. More on Errors: SQL SERVER – Install Samples Database Adventure Works for SQL Server 2012 SQL SERVER – 2012 – All Download Links in Single Page – SQL Server 2012 SQLAuthority News – SQL Server 2012 – Microsoft Learning Training and Certification SQLAuthority News – Download Microsoft SQL Server 2012 RTM Now I encourage you to submit your ideas for SQL in Sixty Seconds. We will try to accommodate as many as we can. If we like your idea we promise to share with you educational material. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: Database, Pinal Dave, PostADay, SQL, SQL Authority, SQL in Sixty Seconds, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, T SQL, Technology, Video

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• Project Euler 15: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 15.  As always, any feedback is welcome. # Euler 15 # http://projecteuler.net/index.php?section=problems&id=15 # Starting in the top left corner of a 2x2 grid, there # are 6 routes (without backtracking) to the bottom right # corner. How many routes are their in a 20x20 grid? import time start = time.time() def factorial(n): if n == 0: return 1 else: return n * factorial(n-1) rows, cols = 20, 20 print factorial(rows+cols) / (factorial(rows) * factorial(cols)) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 9: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 9.  As always, any feedback is welcome. # Euler 9 # http://projecteuler.net/index.php?section=problems&id=9 # A Pythagorean triplet is a set of three natural numbers, # a b c, for which, # a2 + b2 = c2 # For example, 32 + 42 = 9 + 16 = 25 = 52. # There exists exactly one Pythagorean triplet for which # a + b + c = 1000. Find the product abc. import time start = time.time() product = 0 def pythagorean_triplet(): for a in range(1,501): for b in xrange(a+1,501): c = 1000 - a - b if (a*a + b*b == c*c): return a*b*c print pythagorean_triplet() print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 5: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 5.  As always, any feedback is welcome. # Euler 5 # http://projecteuler.net/index.php?section=problems&id=5 # 2520 is the smallest number that can be divided by each # of the numbers from 1 to 10 without any remainder. # What is the smallest positive number that is evenly # divisible by all of the numbers from 1 to 20? import time start = time.time() def gcd(a, b): while b: a, b = b, a % b return a def lcm(a, b): return a * b // gcd(a, b) print reduce(lcm, range(1, 20)) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Large invoice database structure and rendering

  - by user132624

  Our client has a MS SQL database that has 1 million customer invoice records in it. Using the database, our client wants its customers to be able to log into a frontend web site and then be able to view, modify and download their company’s invoices. Given the size of the database and the large number of customers who may log into the web site at any time, we are concerned about data base engine performance and web page invoice rendering performance. The 1 million invoice database is for just 90 days sales, so we will remove invoices over 90 days old from the database. Most of the invoices have multiple line items. We can easily convert our invoices into various data formats so for example it is easy for us to convert to and from SQL to XML with related schema and XSLT. Any data conversion would be done on another server so as not to burden the web interface server. We have tentatively decided to run the web site on a .NET Framework IIS web server using MS SQL on MS Azure. How would you suggest we structure our database for best performance? For example, should we put all the invoices of all customers located within the same 5 digit or 6 digit zip codes into the same table? Or could we set up a separate home directory for each customer on IIS and place each customer’s invoices in each customer’s home directory in XML format? And secondly what would you suggest would be the best method to render customer invoices on a web page and allow customers to modify for best performance? The ADO.net XML Data Set looks intriguing to us as a method, but we have never used it.

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• Project Management - Asana / activeCollab / basecamp / alternative / none

  - by rickyduck

  I don't know whether this should be on programmers - I've been looking at the above three apps over the past few weeks just for myself and I'm in two minds. All three look good, are easy to use, and I came to this conclusion; Asana is the easiest to use ActiveCollab is the feature rich and easiest flow BaseCamp is the best UX / design But I didn't really find my workflow was any more quicker / efficient, in fact it was a bit slower and organized. Is there a realistic place for them in workflow - should programmers use them for themselves, or only when a project manager can take control of it?

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• Project Euler 8: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 8.  As always, any feedback is welcome. # Euler 8 # http://projecteuler.net/index.php?section=problems&id=8 # Find the greatest product of five consecutive digits # in the following 1000-digit number import time start = time.time() number = '\ 73167176531330624919225119674426574742355349194934\ 96983520312774506326239578318016984801869478851843\ 85861560789112949495459501737958331952853208805511\ 12540698747158523863050715693290963295227443043557\ 66896648950445244523161731856403098711121722383113\ 62229893423380308135336276614282806444486645238749\ 30358907296290491560440772390713810515859307960866\ 70172427121883998797908792274921901699720888093776\ 65727333001053367881220235421809751254540594752243\ 52584907711670556013604839586446706324415722155397\ 53697817977846174064955149290862569321978468622482\ 83972241375657056057490261407972968652414535100474\ 82166370484403199890008895243450658541227588666881\ 16427171479924442928230863465674813919123162824586\ 17866458359124566529476545682848912883142607690042\ 24219022671055626321111109370544217506941658960408\ 07198403850962455444362981230987879927244284909188\ 84580156166097919133875499200524063689912560717606\ 05886116467109405077541002256983155200055935729725\ 71636269561882670428252483600823257530420752963450' max = 0 for i in xrange(0, len(number) - 5): nums = [int(x) for x in number[i:i+5]] val = reduce(lambda agg, x: agg*x, nums) if val > max: max = val print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Best approach for a clinic database

  - by user18013

  As a practical assignment for the database course I'm taking I've been instructed to create a database for a local clinic, I've meet with the doctors a couple of times and discussed the information that needs to be stored in the database from personal to medical. Now I'm facing a tough decision because I've been given two choices: either to implement the database as a "local website" which only operates inside the clinic via WiFi, or to implement the front-end as a regular desktop application connecting to a shared database. Note: I've a 40 days deadline to deliver the first prototype and meet with my client. My questions are: 1- which approach should I go with given that I've more experience with desktop applications programming than web? 2- if I go with desktop front-ends what would be the best way to synchronize the database between all clients?? I've no experience and having searched for an answer a lot but came up with nothing detailed on this matter. 3- if I go with the web solution which choice would be best PHP & MySQL or ASP.NET & SQL Server or a different combination?? (given that my knowledge in both PHP & ASP.NET are nearly the same).

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• Project Euler 52: Ruby

  - by Ben Griswold

  In my attempt to learn Ruby out in the open, here’s my solution for Project Euler Problem 52.  Compared to Problem 51, this problem was a snap. Brute force and pretty quick… As always, any feedback is welcome. # Euler 52 # http://projecteuler.net/index.php?section=problems&id=52 # It can be seen that the number, 125874, and its double, # 251748, contain exactly the same digits, but in a # different order. # # Find the smallest positive integer, x, such that 2x, 3x, # 4x, 5x, and 6x, contain the same digits. timer_start = Time.now def contains_same_digits?(n) value = (n*2).to_s.split(//).uniq.sort.join 3.upto(6) do |i| return false if (n*i).to_s.split(//).uniq.sort.join != value end true end i = 100_000 answer = 0 while answer == 0 answer = i if contains_same_digits?(i) i+=1 end puts answer puts "Elapsed Time: #{(Time.now - timer_start)*1000} milliseconds"

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• Project Kapros: A Custom-Built Workstation Featuring an In-Desk Computer

  - by Jason Fitzpatrick

  While we’ve seen our fair share of case mods, it’s infrequent we see one as polished and built-in as this custom built work station. What started as an IKEA Galant desk, ended as a stunningly executed desk-as-computer build. High gloss paint, sand-blasted plexiglass windows, custom lighting, and some quality hardware all come together in this build to yield a gorgeous setup with plenty of power and style to go around. Hit up the link below for a massive photo album build guide detailing the process from start to finish. Project Kapros: IKEA Galant PC Desk Mod [via Kotaku] How to Stress Test the Hard Drives in Your PC or Server How To Customize Your Android Lock Screen with WidgetLocker The Best Free Portable Apps for Your Flash Drive Toolkit

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• Need database selection advise

  - by jacknad

  I know this is considered a bad question since there is no correct answer, but I need to decide on a database for embedded linux (DaVinci 368 based) hardware and I've never had to produce a design with a database before. Each record will probably contain less than 1000 images with associated alpha-numeric data and the mass storage will be some kind of flash drive. Only one user needs access to the data at a time. MySQL claims to be "The world's most popular open source database" but SQLite claims to be "the most widely deployed SQL database engine in the world." Perhaps there is another that is also the best in the world? Which is easiest to use for a database newbie? Should I just flip a coin? Does it really matter which one I pick? Do I even need to use a database software package or should I roll my own? I won't need bells and whistles like sorting, but I'll probably need to delete the oldest records to make room for new ones if the storage fills up.

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• How to determine number of resources to be allocated in a software project

  - by aditi

  Last day I have been interviewed and the interviwer asked me as given the outline of a project, how can we determine the number of resources to be needed for the same? I donot know to do do so? Is there any standard way of doing so? or is it based on the experience? or how.... I am pretty new in this activity and my knowledge is zero at present .... so any clear explanation with some example(simple) will help me(and people like me) to understand this. Thanks

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• Project Euler 2: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 2.  As always, any feedback is welcome. # Euler 2 # http://projecteuler.net/index.php?section=problems&id=2 # Find the sum of all the even-valued terms in the # Fibonacci sequence which do not exceed four million. # Each new term in the Fibonacci sequence is generated # by adding the previous two terms. By starting with 1 # and 2, the first 10 terms will be: # 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... # Find the sum of all the even-valued terms in the # sequence which do not exceed four million. import time start = time.time() total = 0 previous = 0 i = 1 while i <= 4000000: if i % 2 == 0: total +=i # variable swapping removes the need for a temp variable i, previous = previous, previous + i print total print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 16: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 16.  As always, any feedback is welcome. # Euler 16 # http://projecteuler.net/index.php?section=problems&id=16 # 2^15 = 32768 and the sum of its digits is # 3 + 2 + 7 + 6 + 8 = 26. # What is the sum of the digits of the number 2^1000? import time start = time.time() print sum([int(i) for i in str(2**1000)]) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 13: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 13.  As always, any feedback is welcome. # Euler 13 # http://projecteuler.net/index.php?section=problems&id=13 # Work out the first ten digits of the sum of the # following one-hundred 50-digit numbers. import time start = time.time() number_string = '\ 37107287533902102798797998220837590246510135740250\ 46376937677490009712648124896970078050417018260538\ 74324986199524741059474233309513058123726617309629\ 91942213363574161572522430563301811072406154908250\ 23067588207539346171171980310421047513778063246676\ 89261670696623633820136378418383684178734361726757\ 28112879812849979408065481931592621691275889832738\ 44274228917432520321923589422876796487670272189318\ 47451445736001306439091167216856844588711603153276\ 70386486105843025439939619828917593665686757934951\ 62176457141856560629502157223196586755079324193331\ 64906352462741904929101432445813822663347944758178\ 92575867718337217661963751590579239728245598838407\ 58203565325359399008402633568948830189458628227828\ 80181199384826282014278194139940567587151170094390\ 35398664372827112653829987240784473053190104293586\ 86515506006295864861532075273371959191420517255829\ 71693888707715466499115593487603532921714970056938\ 54370070576826684624621495650076471787294438377604\ 53282654108756828443191190634694037855217779295145\ 36123272525000296071075082563815656710885258350721\ 45876576172410976447339110607218265236877223636045\ 17423706905851860660448207621209813287860733969412\ 81142660418086830619328460811191061556940512689692\ 51934325451728388641918047049293215058642563049483\ 62467221648435076201727918039944693004732956340691\ 15732444386908125794514089057706229429197107928209\ 55037687525678773091862540744969844508330393682126\ 18336384825330154686196124348767681297534375946515\ 80386287592878490201521685554828717201219257766954\ 78182833757993103614740356856449095527097864797581\ 16726320100436897842553539920931837441497806860984\ 48403098129077791799088218795327364475675590848030\ 87086987551392711854517078544161852424320693150332\ 59959406895756536782107074926966537676326235447210\ 69793950679652694742597709739166693763042633987085\ 41052684708299085211399427365734116182760315001271\ 65378607361501080857009149939512557028198746004375\ 35829035317434717326932123578154982629742552737307\ 94953759765105305946966067683156574377167401875275\ 88902802571733229619176668713819931811048770190271\ 25267680276078003013678680992525463401061632866526\ 36270218540497705585629946580636237993140746255962\ 24074486908231174977792365466257246923322810917141\ 91430288197103288597806669760892938638285025333403\ 34413065578016127815921815005561868836468420090470\ 23053081172816430487623791969842487255036638784583\ 11487696932154902810424020138335124462181441773470\ 63783299490636259666498587618221225225512486764533\ 67720186971698544312419572409913959008952310058822\ 95548255300263520781532296796249481641953868218774\ 76085327132285723110424803456124867697064507995236\ 37774242535411291684276865538926205024910326572967\ 23701913275725675285653248258265463092207058596522\ 29798860272258331913126375147341994889534765745501\ 18495701454879288984856827726077713721403798879715\ 38298203783031473527721580348144513491373226651381\ 34829543829199918180278916522431027392251122869539\ 40957953066405232632538044100059654939159879593635\ 29746152185502371307642255121183693803580388584903\ 41698116222072977186158236678424689157993532961922\ 62467957194401269043877107275048102390895523597457\ 23189706772547915061505504953922979530901129967519\ 86188088225875314529584099251203829009407770775672\ 11306739708304724483816533873502340845647058077308\ 82959174767140363198008187129011875491310547126581\ 97623331044818386269515456334926366572897563400500\ 42846280183517070527831839425882145521227251250327\ 55121603546981200581762165212827652751691296897789\ 32238195734329339946437501907836945765883352399886\ 75506164965184775180738168837861091527357929701337\ 62177842752192623401942399639168044983993173312731\ 32924185707147349566916674687634660915035914677504\ 99518671430235219628894890102423325116913619626622\ 73267460800591547471830798392868535206946944540724\ 76841822524674417161514036427982273348055556214818\ 97142617910342598647204516893989422179826088076852\ 87783646182799346313767754307809363333018982642090\ 10848802521674670883215120185883543223812876952786\ 71329612474782464538636993009049310363619763878039\ 62184073572399794223406235393808339651327408011116\ 66627891981488087797941876876144230030984490851411\ 60661826293682836764744779239180335110989069790714\ 85786944089552990653640447425576083659976645795096\ 66024396409905389607120198219976047599490197230297\ 64913982680032973156037120041377903785566085089252\ 16730939319872750275468906903707539413042652315011\ 94809377245048795150954100921645863754710598436791\ 78639167021187492431995700641917969777599028300699\ 15368713711936614952811305876380278410754449733078\ 40789923115535562561142322423255033685442488917353\ 44889911501440648020369068063960672322193204149535\ 41503128880339536053299340368006977710650566631954\ 81234880673210146739058568557934581403627822703280\ 82616570773948327592232845941706525094512325230608\ 22918802058777319719839450180888072429661980811197\ 77158542502016545090413245809786882778948721859617\ 72107838435069186155435662884062257473692284509516\ 20849603980134001723930671666823555245252804609722\ 53503534226472524250874054075591789781264330331690' total = 0 for i in xrange(0, 100 * 50 - 1, 50): total += int(number_string[i:i+49]) print str(total)[:10] print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 7: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 7.  As always, any feedback is welcome. # Euler 7 # http://projecteuler.net/index.php?section=problems&id=7 # By listing the first six prime numbers: 2, 3, 5, 7, # 11, and 13, we can see that the 6th prime is 13. What # is the 10001st prime number? import time start = time.time() def nthPrime(nth): primes = [2] number = 3 while len(primes) < nth: isPrime = True for prime in primes: if number % prime == 0: isPrime = False break if (prime * prime > number): break if isPrime: primes.append(number) number += 2 return primes[nth - 1] print nthPrime(10001) print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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• Project Euler 4: (Iron)Python

  - by Ben Griswold

  In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 4.  As always, any feedback is welcome. # Euler 4 # http://projecteuler.net/index.php?section=problems&id=4 # Find the largest palindrome made from the product of # two 3-digit numbers. A palindromic number reads the # same both ways. The largest palindrome made from the # product of two 2-digit numbers is 9009 = 91 x 99. # Find the largest palindrome made from the product of # two 3-digit numbers. import time start = time.time() def isPalindrome(s): return s == s[::-1] max = 0 for i in xrange(100, 999): for j in xrange(i, 999): n = i * j; if (isPalindrome(str(n))): if (n > max): max = n print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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