Re: Project Euler
Andre Garzia-3 wrote: good question, have you given the revlet network permissions? Yes, , i selected Allow always. Double click one line in first card and stack go to next card with all fields empty. Just noticed that 141 kb takes too long to load. Does this works the same in every computer? Or only in my side? Same results using Firefox 3.5.8, Internet Explorer 8 and Google Chrome. -- View this message in context: http://n4.nabble.com/Project-Euler-tp1566143p1572042.html Sent from the Revolution - User mailing list archive at Nabble.com. ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler
Alejandro, For me the double-clicking doesn't work at all, but submitting a new solution is no problem. Poor Andre didn't know what he was getting himself into, but... Any chance Andre you could add a simple edit button on the front page instead of the double click? Maybe that would help. Andre Garzia-3 wrote: good question, have you given the revlet network permissions? Yes, , i selected Allow always. Double click one line in first card and stack go to next card with all fields empty. Just noticed that 141 kb takes too long to load. Does this works the same in every computer? Or only in my side? Same results using Firefox 3.5.8, Internet Explorer 8 and Google Chrome. ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler
Robert Brenstein wrote: revlet loading techniques? All I get is plugin not loaded. On my side of the web, double clicking one of the names of first card: Solutions Database shows all fields empty on second card: Test Solution. Why i am getting this error? Alejandro -- View this message in context: http://n4.nabble.com/Project-Euler-tp1566143p1571217.html Sent from the Revolution - User mailing list archive at Nabble.com. ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler
good question, have you given the revlet network permissions? On Fri, Feb 26, 2010 at 4:37 PM, Alejandro Tejada capellan2...@gmail.comwrote: Robert Brenstein wrote: revlet loading techniques? All I get is plugin not loaded. On my side of the web, double clicking one of the names of first card: Solutions Database shows all fields empty on second card: Test Solution. Why i am getting this error? Alejandro -- View this message in context: http://n4.nabble.com/Project-Euler-tp1566143p1571217.html Sent from the Revolution - User mailing list archive at Nabble.com. ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution -- http://www.andregarzia.com All We Do Is Code. ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler
Mike, 'scool, thanks. Brian, I agree, great project for math and coding. Sometimes it's an enlightening exercise to take an existing capability and code it oneself or to extend it. For instance, I have written routines to do infinite precision integer arithmetic (working on an infinite precision calculator), to do a sortof baseConvert, but which will accept decimals or fractions and bases larger than 36, intermediately large (10^12+) prime factorization and at one time in I wrote some routines to mimic trig functions. As far as sqrt, back in the old days when I was in high school, we learned to extract those by hand ... you, too? From: Michael Kann mikek...@yahoo.com Subject: Re: Project Euler To: How to use Revolution use-revolution@lists.runrev.com Message-ID: 327364.51145...@web56701.mail.re3.yahoo.com Content-Type: text/plain; charset=iso-8859-1 Mick, You beat me to it on the Fibonacci series. I also thought about using the golden ratio. Using a few other relationships allows one to compose a rather compact script. 1. the F numbers are always odd odd even, odd odd even is --- before four million 2. F1 + F2 + F3 . . . Fn = F(n+2)- 1 --- this allows you to do the summation at the end --- If you know Fn then you just go up a couple more --- numbers to get the sum of the first n numbers 3. If you stop the summation at F(even) then the sum of the even numbers will be 1/2 of the total sum -- 1,1,TWO,3,5,EIGHT,13,21,THIRTY-FOUR -- The two numbers sum to the EVENS (by definition) Here's the script -- 555 is just any large number on mouseUp put (1+ sqrt(5))/2 into g repeat with n = 0 to 555 step 3 if (g^n)/sqrt(5) 4E6 then exit repeat end repeat put (round((g^(n-1))/sqrt(5)) - 1) / 2 end mouseUp - The golden ratio technique is most valuable if you want to hit a very large F number. You don't have to count up to it. It is like random access memory. Here's another interesting property of a F series: 1 1 2 3 5 8 13 21 Take the square of 8 -- 64. It is off by one from the product 5*13. That holds for any 3 numbers in a row. Sometimes the square is one more than the products of its neighbors, sometimes one less: but always one off. ** Date: Thu, 25 Feb 2010 16:27:38 -0500 From: Brian Yennie bri...@qldlearning.com Subject: Re: Project Euler To: How to use Revolution use-revolution@lists.runrev.com Message-ID: a4db603e-27d8-4404-b01d-8d24b9b97...@qldlearning.com Content-Type: text/plain; charset=us-ascii Good stuff! I particularly like this project, because it allows for a combination of clever coding AND pure math. The best problems surely require both. It also depends what level of computation you force yourself to contain in you code. For example, Rev has a sqrt() function, but what if you had to write this code without that function built in? Would that lead to a whole different approach? Is it ok to use known formulas, or does your code have to derive them itself? Etc. With that said, it would be pretty darn cool if we got a near-complete set of solutions all written in Rev... what a nice reference for tight mathematical coding. ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler [SPOILER #3]
It's hurting my head trying to figure out why yours works. ;-) That said, it's very slow. The simple brute force method is roughly 100 times faster (since it doesn't iterate 10,000 times but only 100): put 0 into total put 0 into summer repeat with i = 1 to 100 subtract i^2 from total add i to summer end repeat add summer^2 to total _That_ said, I believe there is yet a faster way. I believe yours is O(n^5), mine is O(n^3), but O(n^2) is possible. (If someone understands big O notation better than I do feel free to jump in here) gc On Wed, Feb 24, 2010 at 1:47 PM, Brian Yennie bri...@qldlearning.com wrote: No, we're actually talking about #6 =). I put them in order of difficulty and this becomes the 3rd one... but it's actually id 6. Are we talking about the same #3? The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? On Tue, Feb 23, 2010 at 11:04 AM, Brian Yennie bri...@qldlearning.com wrote: I'm pretty proud of this one for #3... SPOILER ALERT SPOILER ALERT... scroll down if you want to see. Great site find, Andre!! put 0 into total repeat with i=1 to 100 repeat with j=1 to 100 if (i=j) then next repeat add i*j to total end repeat end repeat _ ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler [SPOILER #3]
Ah, now I understand how yours works -- that _is_ quite clever! gc On Thu, Feb 25, 2010 at 8:21 AM, Geoff Canyon Rev gcanyon+...@gmail.com wrote: It's hurting my head trying to figure out why yours works. ;-) That said, it's very slow. The simple brute force method is roughly 100 times faster (since it doesn't iterate 10,000 times but only 100): put 0 into total put 0 into summer repeat with i = 1 to 100 subtract i^2 from total add i to summer end repeat add summer^2 to total _That_ said, I believe there is yet a faster way. I believe yours is O(n^5), mine is O(n^3), but O(n^2) is possible. (If someone understands big O notation better than I do feel free to jump in here) gc On Wed, Feb 24, 2010 at 1:47 PM, Brian Yennie bri...@qldlearning.com wrote: No, we're actually talking about #6 =). I put them in order of difficulty and this becomes the 3rd one... but it's actually id 6. Are we talking about the same #3? The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? On Tue, Feb 23, 2010 at 11:04 AM, Brian Yennie bri...@qldlearning.com wrote: I'm pretty proud of this one for #3... SPOILER ALERT SPOILER ALERT... scroll down if you want to see. Great site find, Andre!! put 0 into total repeat with i=1 to 100 repeat with j=1 to 100 if (i=j) then next repeat add i*j to total end repeat end repeat _ ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler [SPOILER #3]
put 100 into n put n * (n + 1) * (2 * n - 2) / 6 into total ;) ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler [SPOILER #3]
Ian Macphail wrote: put 100 into n put n * (n + 1) * (2 * n - 2) / 6 into total ;) _ Oops! forgot it was the square of the sum. put 100 into n put n * (n + 1) / 2 into a put n * (n + 1) * (2 * n + 1) / 6 into b put a * a - b into total ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler [SPOILER #3]
600851475143 = 71* 839 * 1471 * 6857 It's factors are 1, 71, 839, 1471, 6857, 59569, 104441, 486847, 1234169, 5753023, 10086647, 87625999, 408464633, 716151937, 8462696833, and 600851475143 I wrote a stack that will do this for any number whose largest prime factor is around 100,000,000 or which is a prime up to about the square of that. This one took more time to display the results than to calculate them. - Mick Date: Wed, 24 Feb 2010 12:01:17 -0600 From: Geoff Canyon Rev gcanyon+...@gmail.com Subject: Re: Project Euler [SPOILER #3] To: How to use Revolution use-revolution@lists.runrev.com Message-ID: 8ea148a01002241001h68210441i388f66f67ba71...@mail.gmail.com Content-Type: text/plain; charset=ISO-8859-1 Are we talking about the same #3? The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler
First, thanks, Andre for starting the thread and turning us / me onto the Euler Project. I enjoy implementing algorithms, but math is fun, too! Since Euler was a mathematician, how about a more mathematical approach. I haven't seen these approaches yet in my quick scan of the posts, so I will put them out there: For the first problem *** mathematical note The number of natural numbers below X that are divisible by 3 or 5 is: the number divisible by 3, plus the number divisible by 5, minus the number divisible by 15 (since these number have already been counted twice). For combining in a more general fashion, i.e. other than 3 and 5, rather than using the multiple you would use the Least Common Multiple (not implemented here). For instance, if you wanted the number divisible by 6 or 4, you would use numdiv(4) + numdiv(6) - numdiv(12), since if you used numdiv(24) you would miss subtracting off many numbers divisible by both 4 and 6 but not by 24 (12*1, 12*3, 12*5 ...) Here is a testing routine, followed by a generalization of the problem (no error correction). For the testing routine, as soon as you are satisfied that it works as advertised, hold the shift key to jump out of the loop. Then it will display the results of divisibility by 3 or 5. on doTest repeat until the shiftkey is down ask X with 1000 put it into X ask N with 3 put it into N answer = or plain with Eq or NoEq put it into EorNo put numOfNumsBelowXandDivisByN(X,N,EorNo) into theResult if EorNo is NoEq then put Number [[X]] and divisible by [[N]] is [[theResult]]. into mergeStr else put Number = [[X]] and divisible by [[N]] is [[theResult]]. into mergeStr end if answer merge(mergeStr) end repeat put 1000 into X put NoEq into EorNo put 3 into N put numOfNumsBelowXandDivisByN(X,N,EorNo) into theResult3 put 5 into N put numOfNumsBelowXandDivisByN(X,N,EorNo) into theResult5 put 15 into N put numOfNumsBelowXandDivisByN(X,N,EorNo) into theResult15 put Number [[X]] and divisible by 3 or 5 is \ [[theResult3 + theResult5 - theResult15]]. into mergeStr answer merge(mergeStr) end doTest function numOfNumsBelowXandDivisByN X,N, EqOrNoEq put X into X_ put X_ mod N into XmodN -- X div N will be the number returned EXCEPT ... if EqOrNoEq = NoEq and XmodN = 0 then -- X divisible by N and it matters subtract 1 from X_ end if return X_ div N end numOfNumsBelowXandDivisByN For the second problem ** The even Fibonacci numbers form a Fibonacci-like sequence E(1) = 0, E(2) = 2, for n2: E(n) = 4E(n-1) + E(2) 0, 2, 8, 34, 144, 610, 2584, etc. Also, the sums of the even Fibs form a slightly less Fibonacci-like sequence S(1) = 0, S(2)=2, S(3) = 10, for n3: S(n) = 5*S(n-1) - 3*S(n-2) - S(n-3) 0, 2, 10, 44, 188, 798, 3382, etc. There may be an even more direct approach to one or both of these sequences (I don't know if there is), such as with the plain vanilla (no offense intended) Fibonacci sequence. The i-th member of the sequence is: (g^i-(-1/g)^i) / rt5 where rt5 is the square root of 5 and g is the golden ratio: ( (1+rt5)/2 ). So that sequence can be calculated explicitly (rather than recursively). In the testing routine, I start by displaying the explicit calculation of the first 8 members of the Fibonacci sequence Then the recursive method of the even Fibs, keeping track of the sums. Then the recursive method of the sums, only using the even Fibs, by calculation to test whether to stop. on doTest put sqrt(5) into rt5 put (1+rt5)/2 into g repeat with i = 1 to 8 put (g^i-(-1/g)^i) / rt5 into theF answer theF end repeat -- even Fibs sums by first method answer First Method put 2 into F1 put 8 into F2 put 10 into theSum repeat while F2 = 400 put 4*F2 + F1 into newFib add newFib to theSum put F2 into F1 put newFib into F2 end repeat answer Fib F1 , Sum theSum answer Second Method put 0 into S1 put 2 into S2 put 10 into S3 repeat while S3 - S2 = 400 put 5*S3 - 3*S2 - S1 into newFib put S2 into S1 put S3 into S2 put newFib into S3 end repeat answerSum S3 end doTest I'd welcome any comments on these. Date: Tue, 23 Feb 2010 13:28:37 -0300 From: Andre Garzia an...@andregarzia.com Subject: Project Euler To: How to use Revolution use-revolution@lists.runrev.com Message-ID: 7c87a2a11002230828ia2f0ff1s399ab3ae1b2d0...@mail.gmail.com Content-Type: text/plain; charset=ISO-8859-1 Hi There Folks, I don't know if you guys are familiar with http://www.projecteuler.net which is officially a very fun project according to my own concepts of fun. It is a repository of mathematical problems which require more than simple math to solve, it usually require some programming. You can go
Re: Project Euler
Mick, You beat me to it on the Fibonacci series. I also thought about using the golden ratio. Using a few other relationships allows one to compose a rather compact script. 1. the F numbers are always odd odd even, odd odd even --- this allows you to use step 3 in the loop --- you will only check even numbers and stop at the last one --- before four million 2. F1 + F2 + F3 . . . Fn = F(n+2)- 1 --- this allows you to do the summation at the end --- If you know Fn then you just go up a couple more --- numbers to get the sum of the first n numbers 3. If you stop the summation at F(even) then the sum of the even numbers will be 1/2 of the total sum -- 1,1,TWO,3,5,EIGHT,13,21,THIRTY-FOUR -- The two numbers sum to the EVENS (by definition) Here's the script -- 555 is just any large number on mouseUp put (1+ sqrt(5))/2 into g repeat with n = 0 to 555 step 3 if (g^n)/sqrt(5) 4E6 then exit repeat end repeat put (round((g^(n-1))/sqrt(5)) - 1) / 2 end mouseUp - The golden ratio technique is most valuable if you want to hit a very large F number. You don't have to count up to it. It is like random access memory. Here's another interesting property of a F series: 1 1 2 3 5 8 13 21 Take the square of 8 -- 64. It is off by one from the product 5*13. That holds for any 3 numbers in a row. Sometimes the square is one more than the products of its neighbors, sometimes one less: but always one off. --- On Thu, 2/25/10, Mick Collins mickc...@mac.com wrote: From: Mick Collins mickc...@mac.com Subject: Re: Project Euler To: use-revolution@lists.runrev.com Date: Thursday, February 25, 2010, 2:13 PM First, thanks, Andre for starting the thread and turning us / me onto the Euler Project. I enjoy implementing algorithms, but math is fun, too! Since Euler was a mathematician, how about a more mathematical approach. I haven't seen these approaches yet in my quick scan of the posts, so I will put them out there: For the first problem *** mathematical note The number of natural numbers below X that are divisible by 3 or 5 is: the number divisible by 3, plus the number divisible by 5, minus the number divisible by 15 (since these number have already been counted twice). For combining in a more general fashion, i.e. other than 3 and 5, rather than using the multiple you would use the Least Common Multiple (not implemented here). For instance, if you wanted the number divisible by 6 or 4, you would use numdiv(4) + numdiv(6) - numdiv(12), since if you used numdiv(24) you would miss subtracting off many numbers divisible by both 4 and 6 but not by 24 (12*1, 12*3, 12*5 ...) Here is a testing routine, followed by a generalization of the problem (no error correction). For the testing routine, as soon as you are satisfied that it works as advertised, hold the shift key to jump out of the loop. Then it will display the results of divisibility by 3 or 5. on doTest repeat until the shiftkey is down ask X with 1000 put it into X ask N with 3 put it into N answer = or plain with Eq or NoEq put it into EorNo put numOfNumsBelowXandDivisByN(X,N,EorNo) into theResult if EorNo is NoEq then put Number [[X]] and divisible by [[N]] is [[theResult]]. into mergeStr else put Number = [[X]] and divisible by [[N]] is [[theResult]]. into mergeStr end if answer merge(mergeStr) end repeat put 1000 into X put NoEq into EorNo put 3 into N put numOfNumsBelowXandDivisByN(X,N,EorNo) into theResult3 put 5 into N put numOfNumsBelowXandDivisByN(X,N,EorNo) into theResult5 put 15 into N put numOfNumsBelowXandDivisByN(X,N,EorNo) into theResult15 put Number [[X]] and divisible by 3 or 5 is \ [[theResult3 + theResult5 - theResult15]]. into mergeStr answer merge(mergeStr) end doTest function numOfNumsBelowXandDivisByN X,N, EqOrNoEq put X into X_ put X_ mod N into XmodN -- X div N will be the number returned EXCEPT ... if EqOrNoEq = NoEq and XmodN = 0 then -- X divisible by N and it matters subtract 1 from X_ end if return X_ div N end numOfNumsBelowXandDivisByN For the second problem ** The even Fibonacci numbers form a Fibonacci-like sequence E(1) = 0, E(2) = 2, for n2: E(n) = 4E(n-1) + E(2) 0, 2, 8, 34, 144, 610, 2584, etc. Also, the sums of the even Fibs form a slightly less Fibonacci-like sequence S(1) = 0, S(2)=2, S(3) = 10, for n3: S(n) = 5*S(n-1) - 3*S(n-2) - S(n-3) 0, 2, 10, 44, 188, 798, 3382, etc. There may be an even more direct approach to one or both of these sequences (I don't know if there is), such as with the plain vanilla (no offense intended) Fibonacci sequence. The i-th member of the sequence is: (g^i-(-1/g)^i) / rt5 where rt5 is the square root
Re: Project Euler
Good stuff! I particularly like this project, because it allows for a combination of clever coding AND pure math. The best problems surely require both. It also depends what level of computation you force yourself to contain in you code. For example, Rev has a sqrt() function, but what if you had to write this code without that function built in? Would that lead to a whole different approach? Is it ok to use known formulas, or does your code have to derive them itself? Etc. With that said, it would be pretty darn cool if we got a near-complete set of solutions all written in Rev... what a nice reference for tight mathematical coding. Mick, You beat me to it on the Fibonacci series. I also thought about using the golden ratio. Using a few other relationships allows one to compose a rather compact script. 1. the F numbers are always odd odd even, odd odd even --- this allows you to use step 3 in the loop --- you will only check even numbers and stop at the last one --- before four million 2. F1 + F2 + F3 . . . Fn = F(n+2)- 1 --- this allows you to do the summation at the end --- If you know Fn then you just go up a couple more --- numbers to get the sum of the first n numbers 3. If you stop the summation at F(even) then the sum of the even numbers will be 1/2 of the total sum -- 1,1,TWO,3,5,EIGHT,13,21,THIRTY-FOUR -- The two numbers sum to the EVENS (by definition) Here's the script -- 555 is just any large number on mouseUp put (1+ sqrt(5))/2 into g repeat with n = 0 to 555 step 3 if (g^n)/sqrt(5) 4E6 then exit repeat end repeat put (round((g^(n-1))/sqrt(5)) - 1) / 2 end mouseUp - The golden ratio technique is most valuable if you want to hit a very large F number. You don't have to count up to it. It is like random access memory. Here's another interesting property of a F series: 1 1 2 3 5 8 13 21 Take the square of 8 -- 64. It is off by one from the product 5*13. That holds for any 3 numbers in a row. Sometimes the square is one more than the products of its neighbors, sometimes one less: but always one off. ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler
I did about eighty project euler problems in rev about three years ago. In the process I wrote my own bignum library and many other utility routines. I don't know that any of it was release-worthy, or if I have the stacks anymore. Then I switched to coding in J (which has built-in unlimited precision integers). One thing to consider is that project euler specifically discourages sharing code before you've come up with your own answer. At the site you can't get help, or even enter the forum for a problem, until you've solved it yourself. gc On Tue, Feb 23, 2010 at 4:04 PM, Robert Brenstein r...@robelko.com wrote: On 23.02.10 at 15:54 -0300 Andre Garzia apparently wrote: Done! checkout http://wecode.org/euler you need RevWeb plugin for that. Cheers andre PS: check the source code to see some uberpretty revlet loading techniques! revlet loading techniques? All I get is plugin not loaded. Robert ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler [SPOILER #3]
Are we talking about the same #3? The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? On Tue, Feb 23, 2010 at 11:04 AM, Brian Yennie bri...@qldlearning.com wrote: I'm pretty proud of this one for #3... SPOILER ALERT SPOILER ALERT... scroll down if you want to see. Great site find, Andre!! put 0 into total repeat with i=1 to 100 repeat with j=1 to 100 if (i=j) then next repeat add i*j to total end repeat end repeat ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler [SPOILER #3]
No, we're actually talking about #6 =). I put them in order of difficulty and this becomes the 3rd one... but it's actually id 6. Are we talking about the same #3? The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ? On Tue, Feb 23, 2010 at 11:04 AM, Brian Yennie bri...@qldlearning.com wrote: I'm pretty proud of this one for #3... SPOILER ALERT SPOILER ALERT... scroll down if you want to see. Great site find, Andre!! put 0 into total repeat with i=1 to 100 repeat with j=1 to 100 if (i=j) then next repeat add i*j to total end repeat end repeat _ ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler
Andre, on mouseUp put 0,1 into mySequence put 0 into mySum repeat put the last item of mySequence into myCurrentNumber if myCurrentNumber 4E6 then if myCurrentNumber mod 2 is 0 then add myCurrentNumber to mySum put myCurrentNumber comma (item 1 of mySequence + myCurrentNumber) into mySequence else exit repeat end repeat put mySum end mouseUp -- Best regards, Mark Schonewille Economy-x-Talk Consulting and Software Engineering Homepage: http://economy-x-talk.com Twitter: http://twitter.com/xtalkprogrammer Economy-x-Talk is always looking for new software development projects. Feel free to contact me for a quote. Op 23 feb 2010, om 17:28 heeft Andre Garzia het volgende geschreven: Hi There Folks, I don't know if you guys are familiar with http:// www.projecteuler.net which is officially a very fun project according to my own concepts of fun. It is a repository of mathematical problems which require more than simple math to solve, it usually require some programming. You can go solving your problems and getting scores. I just solved the first one in Rev. Add all the natural numbers below one thousand that are multiples of 3 or 5. Now, the second one is giving me trouble... rsrsrsrs Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed four million. I think I am going to reach the upper limits of rev math with it. :D ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler
Mark, you're the man! I was doing something so complex that I can't believe myself! That is very elegant, I like the way you replace the sequence, I was keeping them all doing lots of loops. Thanks for this solution! wonderful! :D Cheers andre On Tue, Feb 23, 2010 at 1:49 PM, Mark Schonewille m.schonewi...@economy-x-talk.com wrote: Andre, on mouseUp put 0,1 into mySequence put 0 into mySum repeat put the last item of mySequence into myCurrentNumber if myCurrentNumber 4E6 then if myCurrentNumber mod 2 is 0 then add myCurrentNumber to mySum put myCurrentNumber comma (item 1 of mySequence + myCurrentNumber) into mySequence else exit repeat end repeat put mySum end mouseUp -- Best regards, Mark Schonewille Economy-x-Talk Consulting and Software Engineering Homepage: http://economy-x-talk.com Twitter: http://twitter.com/xtalkprogrammer Economy-x-Talk is always looking for new software development projects. Feel free to contact me for a quote. Op 23 feb 2010, om 17:28 heeft Andre Garzia het volgende geschreven: Hi There Folks, I don't know if you guys are familiar with http://www.projecteuler.netwhich is officially a very fun project according to my own concepts of fun. It is a repository of mathematical problems which require more than simple math to solve, it usually require some programming. You can go solving your problems and getting scores. I just solved the first one in Rev. Add all the natural numbers below one thousand that are multiples of 3 or 5. Now, the second one is giving me trouble... rsrsrsrs Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed four million. I think I am going to reach the upper limits of rev math with it. :D ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution -- http://www.andregarzia.com All We Do Is Code. ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler [SPOILER #3]
I'm pretty proud of this one for #3... SPOILER ALERT SPOILER ALERT... scroll down if you want to see. Great site find, Andre!! put 0 into total repeat with i=1 to 100 repeat with j=1 to 100 if (i=j) then next repeat add i*j to total end repeat end repeat ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler
I was just about to say the same thing! ;-) Mark, you should post your solution, and quick! Otherwise Andre is going to do it! (j/k) Bob On Feb 23, 2010, at 8:49 AM, Mark Schonewille wrote: Andre, on mouseUp put 0,1 into mySequence put 0 into mySum repeat put the last item of mySequence into myCurrentNumber if myCurrentNumber 4E6 then if myCurrentNumber mod 2 is 0 then add myCurrentNumber to mySum put myCurrentNumber comma (item 1 of mySequence + myCurrentNumber) into mySequence else exit repeat end repeat put mySum end mouseUp -- Best regards, Mark Schonewille ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler
Oooh, oooh! In all seriousness, it would be fun if someone compiled a set of different correct solutions to these problems. What an incredible way of showing people's different coding styles in Rev. (my solution to #2 after the break) put 0 into tot put 1 into pre put 2 into fib repeat if (fib = 400) then exit repeat if (fib mod 2 = 0) then add fib to tot put fib into tmp add pre to fib put tmp into pre end repeat___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler
Ho Ho Ho, I am building a little stack with the solutions as buttons. There are problems 1, 2, 6 (with comments showing it was from Mark), I am working on 5 now... :D will post it to RevOnLine :D On Tue, Feb 23, 2010 at 2:05 PM, Bob Sneidar b...@twft.com wrote: I was just about to say the same thing! ;-) Mark, you should post your solution, and quick! Otherwise Andre is going to do it! (j/k) Bob On Feb 23, 2010, at 8:49 AM, Mark Schonewille wrote: Andre, on mouseUp put 0,1 into mySequence put 0 into mySum repeat put the last item of mySequence into myCurrentNumber if myCurrentNumber 4E6 then if myCurrentNumber mod 2 is 0 then add myCurrentNumber to mySum put myCurrentNumber comma (item 1 of mySequence + myCurrentNumber) into mySequence else exit repeat end repeat put mySum end mouseUp -- Best regards, Mark Schonewille ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution -- http://www.andregarzia.com All We Do Is Code. ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler
Brian, this is quite a good idea... I think we could do a RevLet, this way we could post execute the stuff. :D Let me cook something. :D On Tue, Feb 23, 2010 at 2:09 PM, Brian Yennie bri...@qldlearning.comwrote: Oooh, oooh! In all seriousness, it would be fun if someone compiled a set of different correct solutions to these problems. What an incredible way of showing people's different coding styles in Rev. (my solution to #2 after the break) put 0 into tot put 1 into pre put 2 into fib repeat if (fib = 400) then exit repeat if (fib mod 2 = 0) then add fib to tot put fib into tmp add pre to fib put tmp into pre end repeat___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution -- http://www.andregarzia.com All We Do Is Code. ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler
Done! checkout http://wecode.org/euler you need RevWeb plugin for that. Cheers andre PS: check the source code to see some uberpretty revlet loading techniques! On Tue, Feb 23, 2010 at 2:12 PM, Andre Garzia an...@andregarzia.com wrote: Brian, this is quite a good idea... I think we could do a RevLet, this way we could post execute the stuff. :D Let me cook something. :D On Tue, Feb 23, 2010 at 2:09 PM, Brian Yennie bri...@qldlearning.comwrote: Oooh, oooh! In all seriousness, it would be fun if someone compiled a set of different correct solutions to these problems. What an incredible way of showing people's different coding styles in Rev. (my solution to #2 after the break) put 0 into tot put 1 into pre put 2 into fib repeat if (fib = 400) then exit repeat if (fib mod 2 = 0) then add fib to tot put fib into tmp add pre to fib put tmp into pre end repeat___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution -- http://www.andregarzia.com All We Do Is Code. -- http://www.andregarzia.com All We Do Is Code. ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
Re: Project Euler
On 23.02.10 at 15:54 -0300 Andre Garzia apparently wrote: Done! checkout http://wecode.org/euler you need RevWeb plugin for that. Cheers andre PS: check the source code to see some uberpretty revlet loading techniques! revlet loading techniques? All I get is plugin not loaded. Robert ___ use-revolution mailing list use-revolution@lists.runrev.com Please visit this url to subscribe, unsubscribe and manage your subscription preferences: http://lists.runrev.com/mailman/listinfo/use-revolution
