## Project Euler

A place to discuss the implementation and style of computer programs.

Moderators: phlip, Moderators General, Prelates

Numquam
Posts: 162
Joined: Wed Nov 21, 2007 3:13 am UTC

### Re: Project Euler

Having trouble with number 7 which is:

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6^(th) prime is 13.

What is the 10001^(st) prime number?

I'm a very novice programmer and not very good at math anyways, so take caution - this code is not for the weak of heart. Shade your eyes if you can't take badly written code. (Its in Python)

Spoiler:

Code: Select all

n = 1x = 0check = 0while x <= 10001:    check = (2**(n-1))%n    if check == 1:        x = x + 1        print (x)    n = n + 1print (n-1)print ("done")

My "Solution"
Spoiler:
103919

What have I done wrong?
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Berengal
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### Re: Project Euler

Your check is wrong. What's 2^340 Mod 341? What's 11*31?
It is practically impossible to teach good programming to students who are motivated by money: As potential programmers they are mentally mutilated beyond hope of regeneration.

keeperofdakeys
Posts: 658
Joined: Wed Oct 01, 2008 6:04 am UTC

### Re: Project Euler

you may want to do some more research on prime numbers

Ended
Posts: 1459
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Location: The Tower of Flints. (Also known as: England.)

### Re: Project Euler

Specifically, your check (the Fermat test with a=2) is necessary for primality but not sufficient. That is, all prime numbers will pass it but some composite numbers will also pass it.
Generally I try to make myself do things I instinctively avoid, in case they are awesome.
-dubsola

Numquam
Posts: 162
Joined: Wed Nov 21, 2007 3:13 am UTC

### Re: Project Euler

Thanks for the help, got the right answer and learned a thing or two in the process.
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qui tot scriptorum taedia sustineas

fungfat
Posts: 5
Joined: Mon Jul 06, 2009 8:13 pm UTC

### Re: Project Euler

I was desperate about problem 3. Can someone help?

Problem 3 said that "A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99.
Find the largest palindrome made from the product of two 3-digit numbers."

I found the answer is 995 x 583 = 580085. But Project Euler said it is wrong. I tried another method and still came up with the same answer 580085. What have I done wrong.

Eddie

JBJ
Posts: 1263
Joined: Fri Dec 12, 2008 6:20 pm UTC
Location: a point or extent in space

### Re: Project Euler

fungfat wrote:I was desperate about problem 3. Can someone help?

Problem 3 said that "A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99.
Find the largest palindrome made from the product of two 3-digit numbers."

I found the answer is 995 x 583 = 580085. But Project Euler said it is wrong. I tried another method and still came up with the same answer 580085. What have I done wrong.

Eddie

There is a larger number. Of course the spirit of Euler prevents us (at least me) from just giving out the answer. Can you post your code?
So, you sacked the cocky khaki Kicky Sack sock plucker?
The second cocky khaki Kicky Sack sock plucker I've sacked since the sixth sitting sheet slitter got sick.

fungfat
Posts: 5
Joined: Mon Jul 06, 2009 8:13 pm UTC

### Re: Project Euler

I have done it in html, and then excel macro. I attached both codes here.

rem ------------------------------------------------------------------------
rem this is visual basic code
rem ------------------------------------------------------------------------
For i = 999 To 100 Step -1
For j = 999 To 100 Step -1
n = i * j
f3 = Mid(n, 1, 3)
b3 = Mid(n, 6, 1) & Mid(n, 5, 1) & Mid(n, 4, 1)
If f3 = b3 Then
MsgBox "answer is: " & i & " * " & j & " = " & n & ", and f3 = " & f3 & " and b3 = " & b3
i = 0
j = 0
End If
Next j
Next i

<html>

<!- ----------------- ->
<!- This is html code ->
<!- ----------------- ->
<Script language="JavaScript" >

function calsum()
{
mainform.bsum.value = "Calculating greatest numeral palindrome for " + mainform.maxnum.value + ". Please wait...";

var sd = new Date();
var stime = sd.getTime();

var nroote = parseInt(Math.sqrt(mainform.maxnum.value));
var newnum = mainform.maxnum.value;
var n1 = 999;
var n2 = 999;
var countl=0;
var result_text = "** cannot find **";

//------------------------------------------------------
// processing loop...
//------------------------------------------------------
var nproduct = n1 * n2;
var sproduct = nproduct.toString();
var s1 = sproduct.substring(0,3);
var s2 = sproduct.substring(3,3);

var j=0;
var i=0;
var t="";
for (j=999; j>100; j--) {
n1 = j;
for (i=999; i>500; i--) {
countl++;
nproduct = n1 * i;
sproduct = nproduct.toString();
s1 = sproduct.substring(0,3);
s2 = sproduct.substring(5,6) + sproduct.substring(4,5) + sproduct.substring(3,4);
if ( s1 == s2 ) {
n2 = i;
i = -999;
result_text = " " + sproduct + " = " + n1 + " * " + n2;
}
} // inner for...

if ( i < -99 ) { j = -999; } // exit out loop if answer is found...
} // outer for...

//------------------------------------------------------
// End processing loop...
//------------------------------------------------------

var ed = new Date();
var etime = ed.getTime();
var dtime = etime-stime;
var showtime = dtime + " ms";
var sec = 0;

if ( dtime > 1000 ) {
sec = parseInt(dtime / 1000); // parseInt requires capital I...
dtime = dtime - sec * 1000;
showtime = sec + " sec " + dtime + " ms";
}

mainform.bsum.value = "\nLargest numeral palindrome for " + mainform.maxnum.value + " digit number is " + result_text + " and looped " + countl + " times and ran for " + showtime + ".";
mainform.bsum.value = mainform.bsum.value + ". \n\nTry another x value and click this button.\n\n";
mainform.maxnum.focus();
return false;
}
</script>
<title>Project Euler - Problem one</title>

<body>
<center>
<FORM NAME="mainform" onSubmit="return(calsum())">
<table>
<tr>
<td bgcolor=yellow>Problem 4: Padindrome Number</td>
</tr>
<tr>
<td> A palindromic number reads the same both ways. The largest palindrome made from <br>
the product of two 2-digit numbers is 9009 = 91 99. <br><br>
What is the largest palindrome number by multipying two x digits numbers (default to 3) ?
</td>
</tr>
<tr>
<td> <br>
My way:
<br><br>
Enter the x value to calculate the greatest numeral palindrome for production of two x digits numbers: &nbsp;&nbsp;&nbsp;
<INPUT TYPE="TEXT" SIZE=30 NAME="maxnum" value=3>
<br><br>
</td>
</tr>
<tr>
<td>
<INPUT type="button" name=bsum value="Click to find answer for the above x value: " OnClick=calsum() >
</td>
</tr>
</table>
</form>

<script>
mainform.maxnum.focus();
</script>
</body>
</html>

fungfat
Posts: 5
Joined: Mon Jul 06, 2009 8:13 pm UTC

### Re: Project Euler

JBJ wrote:
fungfat wrote:I was desperate about problem 3. Can someone help?

Problem 3 said that "A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99.
Find the largest palindrome made from the product of two 3-digit numbers."

I found the answer is 995 x 583 = 580085. But Project Euler said it is wrong. I tried another method and still came up with the same answer 580085. What have I done wrong.

Eddie

There is a larger number. Of course the spirit of Euler prevents us (at least me) from just giving out the answer. Can you post your code?

Actually, if you can tell me what the largest palindromic number is for problem 3, I can figure out what is wrong with my code.

Eddie

phlip
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Contact:

### Re: Project Euler

So, your code tries 999 * everything, and if none of them are palindromic it then tries 998 * everything, and so on... and you end up with 995 * 583.

But that only maximises one of the two multiplicands, not the product... imagine if, say, 991 * 990 was palindromic (it's not, but consider if it was)... it would be much bigger than 995 * 583, since both of the multiplicands are large numbers... but your algorithm wouldn't find it, because 991 < 995.

Code: Select all

enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
[he/him/his]

stephentyrone
Posts: 778
Joined: Mon Aug 11, 2008 10:58 pm UTC
Location: Palo Alto, CA

### Re: Project Euler

phlip wrote:So, your code tries 999 * everything, and if none of them are palindromic it then tries 998 * everything, and so on... and you end up with 995 * 583.

That's not quite it, because 999*91 = 90909. fungfat, you really should just post your code so that we don't need to engage in speculation.

As long as we're speculating though, my guess is what phlip said together with the check for palindromes only working for 6-digit numbers.

Edit: oh, his code is already there, I just didn't recognize that as code when skimming through the thread.
Last edited by stephentyrone on Tue Jul 07, 2009 1:32 am UTC, edited 1 time in total.
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phlip
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Contact:

### Re: Project Euler

stephentyrone wrote:That's not quite it, because 999*91 = 90909.

Yeah, but his code only counts down to 100... so it doesn't try 91.

stephentyrone wrote:fungfat, you really should just post your code so that we don't need to engage in speculation.

He... already has. It's right there.

stephentyrone wrote:As long as we're speculating though, my guess is what phlip said together with the check for palindromes only working for 6-digit numbers.

The 6-digit thing is reasonable though... since we know there's at least one suitable 6-digit palindrome, so the answer must be bigger than that, and the number must be smaller than 999*999 which is also 6-digits.

So while it would be better form to have a length-agnostic "is this a palindrome" function, it's still correct as-is.

Code: Select all

enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵​╰(ಠ_ಠ ⚠) {exit((int)⚠);}
[he/him/his]

Agent_Irons
Posts: 213
Joined: Wed Sep 10, 2008 3:54 am UTC

### Re: Project Euler

I'm going to go with the early termination problem. Just find all palindromes, and if you find one larger than the current largest palindrome replace it. And so on. When you've tested all the combinations all the way down to 100x100 (definitely not the answer) the remaining largest palindrome must be the answer.

python code snippet:
Spoiler:

Code: Select all

if list(str(x*y)) == list(str(x*y))[::-1]:    listpalindromes.append(x*y)#loops and initializations left as an exercise for the reader.

I believe you might also mean problem 4. Problem three is the one about factoring a very large number.

JBJ
Posts: 1263
Joined: Fri Dec 12, 2008 6:20 pm UTC
Location: a point or extent in space

### Re: Project Euler

First, I went back through my Euler folder, and this isn't problem #3, it's #4.
Project Euler #4 wrote:A palindromic number reads the same both ways.
The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.

Find the largest palindrome made from the product of two 3-digit numbers.

A brute force method works okay in this instance, there are 899 3 digit numbers (100-999) which would be just over 800,000 iterations if you for-next from 100-999. That should only take a couple of seconds.

And as Philip pointed out, you are only testing the first multiplicand, not the product.

Your test for a six digit number is okay, but if you're using VB, why not use the StrReverse function?
n = i*j
If CStr(n) = StrReverse(CStr(n)) Then '// proof of palindrome

Then, you need to test if n is larger than any previous palindromic n's found.
Use another variable, say x, and initially assign it a very small value like 101.
Test if n is greater than x, and if so, reassign x as n.
When the loops finish, x will be your answer.

Note: The solution in VB with brute force and no optimization takes about 2-3 seconds for 810,000 iterations. Optimized, it runs in less than 1/10th of a second and goes through less than 10,000 iterations. That should give you something to shoot for if you are using Euler to brush up your optimization skills.

EDIT - (ninja'd on the problem number)
Agent_Irons wrote:I'm going to go with the early termination problem. Just find all palindromes, and if you find one larger than the current largest palindrome replace it. And so on. When you've tested all the combinations all the way down to 100x100 (definitely not the answer) the remaining largest palindrome must be the answer.

The problem there is that not all palindromes are the product of two 3-digit numbers. 999,999 is palindromic, but is 999 * 1001, 998,899 is also palindromic, but is 781 * 1279
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Qoppa
Posts: 694
Joined: Sat Nov 24, 2007 9:32 pm UTC
Location: Yes.

### Re: Project Euler

Arrggh... My efficient algorithm for problem 18/67 is so close. It gives me the correct answer for the 4 row triangle example, and it can get me an answer for both 18 and 67 in a reasonable amount of time (instantly for 18, 30 seconds for 67), but it's not quite working and I can't figure out why. The answer it gives me for 18 is off by 10, and I don't know how off my answer for 67 is... I've been staring at my code for at least an hour now, and I still can't see anything obviously wrong with my algorithm. Baah!

Code: Select all

_=0,w=-1,(*t)(int,int);a()??<char*p="[gd\~/d~/\\b\x7F\177l*~/~djal{x}h!\005h";(++w<033)?(putchar((*t)(w??(p:>,w?_:0XD)),a()):0;%>O(x,l)??<_='['/7;{return!(x%(_-11))?x??'l:x^(1+ ++l);}??>main(){t=&O;w=a();}

xulaus
Posts: 136
Joined: Thu Jul 03, 2008 11:09 am UTC

### Re: Project Euler

Qoppa wrote:Arrggh... My efficient algorithm for problem 18/67 is so close...

Oh man, I remember that one. I had the hardest time just thinking of an efficent algorithm. I was so pleased with myself when I got it. Do you need some help, or was that just a rant?
Meaux_Pas wrote:I don't even know who the fuck this guy is

jaap
Posts: 2094
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Contact:

### Re: Project Euler

Qoppa wrote:Arrggh... My efficient algorithm [...] can get me an answer for both 18 and 67 in a reasonable amount of time (instantly for 18, 30 seconds for 67)

My program answers problem 67 instantly (less than a millisecond). I'm not saying this to brag, but just to point out that you are probably not doing this in the best or easiest way. It is only half a dozen lines of code (excluding the initialisation of the triangular array).
Minor hint:
Spoiler:
Try working through the triangle from the bottom upwards. Although you can do it in either direction, thinking about doing it bottom-up might make things click.

Qoppa
Posts: 694
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Location: Yes.

### Re: Project Euler

Mostly just a rant. The problem I have has to be a stupid small thing, because the algorithm I've coded up should work provided I actually typed everything out correctly.
Last edited by Qoppa on Tue Jul 07, 2009 5:26 pm UTC, edited 1 time in total.

Code: Select all

_=0,w=-1,(*t)(int,int);a()??<char*p="[gd\~/d~/\\b\x7F\177l*~/~djal{x}h!\005h";(++w<033)?(putchar((*t)(w??(p:>,w?_:0XD)),a()):0;%>O(x,l)??<_='['/7;{return!(x%(_-11))?x??'l:x^(1+ ++l);}??>main(){t=&O;w=a();}

fungfat
Posts: 5
Joined: Mon Jul 06, 2009 8:13 pm UTC

### Re: Project Euler

phlip wrote:So, your code tries 999 * everything, and if none of them are palindromic it then tries 998 * everything, and so on... and you end up with 995 * 583.

But that only maximises one of the two multiplicands, not the product... imagine if, say, 991 * 990 was palindromic (it's not, but consider if it was)... it would be much bigger than 995 * 583, since both of the multiplicands are large numbers... but your algorithm wouldn't find it, because 991 < 995.

Thanks Phlip, it works. I found the answer
Spoiler:
906609 = 993 * 913
in 405450 iteration using about 8 seconds running in Javascript. I will play with some optimisation such as concentrate the iteration in the higher number operands.
Last edited by phlip on Wed Jul 08, 2009 12:08 am UTC, edited 1 time in total.
Reason: Put a spoiler around answer

Agent_Irons
Posts: 213
Joined: Wed Sep 10, 2008 3:54 am UTC

### Re: Project Euler

JBJ wrote:
Agent_Irons wrote:I'm going to go with the early termination problem. Just find all palindromes, and if you find one larger than the current largest palindrome replace it. And so on. When you've tested all the combinations all the way down to 100x100 (definitely not the answer) the remaining largest palindrome must be the answer.

The problem there is that not all palindromes are the product of two 3-digit numbers. 999,999 is palindromic, but is 999 * 1001, 998,899 is also palindromic, but is 781 * 1279

I meant all palindromes that are products of two three digit numbers. I don't really see how to optimize this, other than trying for the largest products first, and trimming out the duplicates. 999x998 = 998x999, after all.

Edit: @v I like that solution very much. It goes against my algorithm sense, because I have operations with primes filed under "computationally expensive" in my head. A lot of these project Euler problems boil down to factoring things. No huge surprises there I guess.
Last edited by Agent_Irons on Tue Jul 07, 2009 6:26 pm UTC, edited 1 time in total.

Berengal
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### Re: Project Euler

Agent_Irons wrote:I'm going to go with the early termination problem. Just find all palindromes, and if you find one larger than the current largest palindrome replace it. And so on. When you've tested all the combinations all the way down to 100x100 (definitely not the answer) the remaining largest palindrome must be the answer.
Actually, if you count down, you only need to find the first, since all other palindromes will be smaller.

Code: Select all

def p4():    for n in range(999,99,-1):        n = int(str(n)+str(n)[::-1])        t = [d for d in divisors(n) if len(str(d)) == 3]        for (x,y) in ((x, y) for x, y in product(t, t) if x*y == n):            return x*y

'product' found in itertools. Implementation of 'divisors' is left as an excercise for the reader. This code is faster on my machine than going the other way around (looping through possible factors and filtering the non-palindrome-producing ones), probably because of early-termination. My code's outer loop is run at most 899 times, my code's inner loop returns once it's run. There are som hidden loops in 'product' and 'divisor', but they're asymptotically constant.
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JBJ
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Location: a point or extent in space

### Re: Project Euler

Agent_Irons wrote:
JBJ wrote:
Agent_Irons wrote:I'm going to go with the early termination problem. Just find all palindromes, and if you find one larger than the current largest palindrome replace it. And so on. When you've tested all the combinations all the way down to 100x100 (definitely not the answer) the remaining largest palindrome must be the answer.

The problem there is that not all palindromes are the product of two 3-digit numbers. 999,999 is palindromic, but is 999 * 1001, 998,899 is also palindromic, but is 781 * 1279

I meant all palindromes that are products of two three digit numbers. I don't really see how to optimize this, other than trying for the largest products first, and trimming out the duplicates. 999x998 = 998x999, after all.

I see now. I tried that, but still got better performance the other way.

VB Script - runs in .03 seconds on my laptop, and loops 6,124 times
Spoiler:

Code: Select all

answer = 100*100for a = 999 to 100 step -1for b = a to 100 step -1   prod = a * b   if prod < answer Then      Exit For   End If   If CStr(prod) = StrReverse(Cstr(prod)) Then      if prod > answer then         answer = prod      End If   End IfnextnextWScript.echo answer

I tried your approach of going top down finding palindromes along the way, checking if they are the product of two 3-digit numbers, and terminating early on discovery. It took .25 seconds on my laptop and loops 52,869 times.
Spoiler:

Code: Select all

for a = 999999 to 100000 step -1   If CStr(a) = StrReverse(CStr(a)) Then   for b = 999 to 100 step - 1      If a mod b = 0 Then         If Len(a/b) = 3 Then            WScript.echo a            WScript.quit         End If         Exit For      End If   Next   End IfNext
Only major optimization is exiting the inner loop on discovery of the first 3 digit integer factor. I can't think of a way to make it faster than that at the moment.

Both are considerably faster than a true brute force, and both could be enhanced a little more by refining the bounds (i.e. we know it will likely be two numbers in the 900's, so we could just check from 900-999) but the performance gained is almost trivial at that point.

*Edit - fixed small bug that Berengal pointed out.
Last edited by JBJ on Tue Jul 07, 2009 7:44 pm UTC, edited 1 time in total.
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eric.lifka
Posts: 10
Joined: Thu May 21, 2009 1:58 am UTC

### Re: Project Euler

inspired by the previous posts on problem 4, I worked on getting it's time down as much as I could. I think I've got it as low as it will go, about .015 seconds per run.

Can anyone see anything I could improve or am I knocking on the lower limits?

Spoiler:
I started with 999, working down, finding the largest palindrome of each number, so for 999 the first time I find a palindrome (again working down from 999) I take that as the max palindrome for 999, moving on to find the max for 998. But the other thing I do, as I'm working down the inner list, trying to find the largest palindrome for say 997, I check to see if I've fallen below the potential for a new max. ie, if the old max was 9009, then if the number I'm testing and the current number in the test list (999...99) cannot surpass that max, then I won't find a new maximum anyway so I take 0 for that number. Maybe I should just let the code talk, I don't feel like I'm explaining myself very well. Anyway, does anyone see any improvements?

Code: Select all

#! pythondef findBigestPalindrome():   largest = 0   for i in range(999, 99, -1):      if i * 999 < largest:         return largest      tmp = getPalindrome(i, largest)      if tmp > largest:         largest = tmp   return largestdef getPalindrome(num, max):   for i in range(999, 99, -1):      test = i * num      if test < max:         return 0      if str(test) == str(test)[::-1]:         return test   return 0if __name__ == "__main__":   from time import time   tstart = time()   result = findBigestPalindrome()   tstop = time()   print ("result: " + str(result) + "\nfound in " + str(tstop - tstart) + " seconds.\n")

Thanks!

EDIT: I realized I didn't really take bounds into account, so just to see pushed the lower bound up to 900 instead of 99, but the time didn't change at all, staying right around .015.

Berengal
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### Re: Project Euler

JBJ wrote:Both are considerably faster than a true brute force
The last piece of code is a true brute force. It checks every number if it's palindrome, then checks every possible three-digit factor to see if it's a factor. The only way it could be even more brute force would be if you brute-forced the second factor as well, but I'm not sure how to do that in a plausible way.

A good implementation loops at most 10^n - 10^(n-1) - 1 times for the highest n-digit. That is an upper bound almost seven times as lower than your actual loop count, and in the case of n = 3, it has an actual loop count of 94, almost 65 times lower than your actual.

Also, your first code piece has a bug: it doesn't check the cases where both three-digit factors are equal (which is the case for eight six-digit palindromes).

eric.lifka wrote:Can anyone see anything I could improve or am I knocking on the lower limits?

Code: Select all

def p4(digits = 3):    for n in range((10**digits)-1, (10**(digits-1))-1, -1):        n = int(str(n)+str(n)[::-1])        t = [d for d in divisors(n) if len(str(d)) == digits]        for (x,y) in ((x, y) for x, y in product(t, t) if x*y == n):            yield x, y, x*ydef p4alt(digits = 3):    return ((x, y, x*y) for x in range(10**digits - 1,10**(digits-1) - 1,-1) for y in range(x, 10**(digits-1) - 1, -1) if str(x*y) == str(x*y)[::-1])def findBigestPalindrome():   largest = 0   for i in range(999, 99, -1):      if i * 999 < largest:         return largest      tmp = getPalindrome(i, largest)      if tmp > largest:         largest = tmp   return largest########>>> timeit(lambda:findBigestPalindrome(), number=10)0.083815686>>> timeit(lambda:next(p4()), number = 10)0.05191168200000007>>> timeit(lambda:max(p4alt()), number = 10)3.8289639840000014

Mine is p4 (needs next called on it), the "for x in [999..100]; for y in [x .. 100];" is p4alt (needs max). I haven't looked at optimizing beyond the foundational algorithm.
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fungfat
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Joined: Mon Jul 06, 2009 8:13 pm UTC

### Re: Project Euler

Agent_Irons wrote:
JBJ wrote:
Agent_Irons wrote:I'm going to go with the early termination problem. Just find all palindromes, and if you find one larger than the current largest palindrome replace it. And so on. When you've tested all the combinations all the way down to 100x100 (definitely not the answer) the remaining largest palindrome must be the answer.

The problem there is that not all palindromes are the product of two 3-digit numbers. 999,999 is palindromic, but is 999 * 1001, 998,899 is also palindromic, but is 781 * 1279

I meant all palindromes that are products of two three digit numbers. I don't really see how to optimize this, other than trying for the largest products first, and trimming out the duplicates. 999x998 = 998x999, after all.

Edit: @v I like that solution very much. It goes against my algorithm sense, because I have operations with primes filed under "computationally expensive" in my head. A lot of these project Euler problems boil down to factoring things. No huge surprises there I guess.

I actually can optimise and get the answer in 8100 iterations in 0.2 seconds (compared to 405450 in 8 seconds). I started with 9xx times 9xx...1xx, and do not do 8xx times 8xx..1xx if a palindomic number is obtained. I also use only the first operand if they are multiple of 11. This idea is obtained from the Project Euler official solution, which is only available if you entered the correct answer for the problem.

Agent_Irons
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### Re: Project Euler

As far as time attacks go I'm sure we could get the speed down farther. Anyone want to implement this in Assembly?

The Black Hand
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### Re: Project Euler

jaap wrote:My program answers problem 67 instantly (less than a millisecond). I'm not saying this to brag, but just to point out that you are probably not doing this in the best or easiest way. It is only half a dozen lines of code (excluding the initialisation of the triangular array).

Your hint got to a solution for this, in just four lines of code, excluding setup and output, with times as fast as you could care for.
It could probably be twerked for speed, but I usually prefer* pythonic, rather than fast, code.
Spoiler:

Code: Select all

#input is a list of lists containing the triangle data.output = input[:]for i, row in reversed(list(enumerate(input[:-1]))):    for j, val in enumerate(row):        output[i][j] = input[i][j] + max(output[i+1][j], output[i+1][j+1])print "Answer is", output[0][0]

What language is your <1ms solution in?

*See my prime iterator in coding: hacks and snippets for an example of the alternative.
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jaap
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### Re: Project Euler

jaap wrote:My program answers problem 67 instantly
Your hint got to a solution for this, in just four lines of code, excluding setup and output, with times as fast as you could care for.
[...]
What language is your <1ms solution in?
I did it in Java, and you found the exact same algorithm as me.

Josephine
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### Re: Project Euler

I uh... think I did about the most inefficient way of solving problem 10 ever. I basically took an existing primality tester and modified it with total = total + i. Any suggestions?

Code: Select all

import mathprint "2,3,"state = 1total = 0for i in range(4, 2000000):   upper = math.sqrt(i)   upper = int(upper)   for thing in range(2, upper):      state = 1      if (i % thing == 0):         state = 0         break   if (state == 1):       total = total + i      print total,exit
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jaap
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### Re: Project Euler

nbonaparte1 wrote:I uh... think I did about the most inefficient way of solving problem 10 ever. I basically took an existing primality tester and modified it with total = total + i. Any suggestions?

http://en.wikipedia.org/wiki/Eratosthenes_Sieve

Josephine
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### Re: Project Euler

ah, right. That. Forgot about that. I'm still going to leave Python running overnight, see if it gets the answer in any reasonable timeframe.
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The Black Hand
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### Re: Project Euler

Remove the print statement from the loop. It'll speed up considerably.
73, de KE8BSL loc EN26.

Qoppa
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### Re: Project Euler

Well I got number 67 finally, and interestingly enough, what I ended up doing to fix it brought the execution time down to 16 milliseconds for the 100 line triange. Nice! I also used a different algorithm than the one posted here (mine started off as a variation on Dijkstra's algorithm, but it's now quite different).

Code: Select all

_=0,w=-1,(*t)(int,int);a()??<char*p="[gd\~/d~/\\b\x7F\177l*~/~djal{x}h!\005h";(++w<033)?(putchar((*t)(w??(p:>,w?_:0XD)),a()):0;%>O(x,l)??<_='['/7;{return!(x%(_-11))?x??'l:x^(1+ ++l);}??>main(){t=&O;w=a();}

xulaus
Posts: 136
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### Re: Project Euler

Qoppa wrote:(mine started off as a variation on Dijkstra's algorithm, but it's now quite different).

That's a novel approach. Kudos. How well did Dijkstra's algorithm perform, or did you modify it before testing?
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Qoppa
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### Re: Project Euler

Well Dijkstra's didn't work (I was 10 off for whatever reason on the smaller triangle), and it was what took 30 seconds for the larger triangle, though my implementation was admittedly very unoptimized. I'm still unsure why it didn't work though, since I can't see any flaw in what I was doing. I tried both just modifying it to find the longest path rather than the shortest, and also taking 1/v and finding the shortest path, but neither worked for whatever reason.
Spoiler:
I eventually got it to work by just processing the vertices in order from top to bottom rather than taking the vertex with the current longest/shortest path.

Code: Select all

_=0,w=-1,(*t)(int,int);a()??<char*p="[gd\~/d~/\\b\x7F\177l*~/~djal{x}h!\005h";(++w<033)?(putchar((*t)(w??(p:>,w?_:0XD)),a()):0;%>O(x,l)??<_='['/7;{return!(x%(_-11))?x??'l:x^(1+ ++l);}??>main(){t=&O;w=a();}

xulaus
Posts: 136
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### Re: Project Euler

Qoppa wrote:Well Dijkstra's didn't work (I was 10 off for whatever reason on the smaller triangle), and it was what took 30 seconds for the larger triangle, though my implementation was admittedly very unoptimized. I'm still unsure why it didn't work though, since I can't see any flaw in what I was doing. I tried both just modifying it to find the longest path rather than the shortest, and also taking 1/v and finding the shortest path, but neither worked for whatever reason.
Spoiler:
I eventually got it to work by just processing the vertices in order from top to bottom rather than taking the vertex with the current longest/shortest path.

It's a shame that didn't work, it would have been a neat little program if it had.
Meaux_Pas wrote:I don't even know who the fuck this guy is

Dongorath
Posts: 93
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### Re: Project Euler

I didn't want to work, so I decided to try PE#87. Quite easy if you don't do stupid errors in your code, but the worst part ? Checking for duplicate results ! Checking at insertion ? Slow as hell !!! Eventually, I remebered that :
Spoiler:
Sort() is your friend ! Insert every single result in a List<T> and then Sort() ! Then, just count every element which is different from the previous one !

JBJ
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### Re: Project Euler

I'm having a bugger of a time with #87
This is the approach I took, tell me if I'm way of base or not.

First I find the largest prime, which when squared, is still less than 50 million.
That's 7069, and 70692 = 49,970,761
I square and put all primes below that into an array.
There are 908 squared primes in that array (22,32,52,72...,70692) -> (4,9,25,49,...,49970761)

I then do the same for cubed primes, and primes to the fourth power, and they are in their respective arrays.
There are 73 primes when cubed remain less than 50 million. Top one being 367, and 3673 = 49,430,863
There are 23 primes when taken to the 4th power are less than 50 million. Top one is 83, 834 = 47,458,321

So I have 73*23 = 1679 possible combinations of p3 and p4 sums. Several of those sums will exceed 50 million, so I eliminate those and I end up with a total of 1,552 p3 + p4 combinations.

So now I have 1,552 * 908 = 1,409,216 possible combinations of p2 and (p3+p4), and of course some of those will exceed 50 million, so I only count the ones that don't. I end up with 1,139,575 which Euler tells me is wrong.

FYI, the largest number I get under 50 million is 49,999,927 which is 49672 + 1733 + 674
As best I can tell, I'm not getting any duplicates. I know for certain that all 1,552 p3 and p4 combinations are unique.

Am I even close?
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Berengal
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### Re: Project Euler

All the combinations are unique, but are all the results?

My approach was similar to your, but my reasoning was somewhat more simplistic.

Define the sets [imath]S[/imath], [imath]C[/imath] and [imath]F[/imath] to be the all squared, cubed and fourth-powered primes below 50,000,000. Define the set [imath]A[/imath]: $\forall a. a \in A \to a \lt 50000000 \wedge a = s + c + f \wedge (s, c, f) \in (S * C * F)$
The answer is the size of the set [imath]A[/imath]
It is practically impossible to teach good programming to students who are motivated by money: As potential programmers they are mentally mutilated beyond hope of regeneration.

Dongorath
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Joined: Tue Oct 16, 2007 1:17 pm UTC

### Re: Project Euler

@JBJ : as Berengal and I said, some numbers can be written as different sums, thus you have to check for duplicate results. I proposed a solution in my spoiler, Berengal proposes to use a set.

@Berengal : this one liner seems so simple... I wish it was this easy to do that in C#...