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Old 2011-08-31, 09:45   #1
JohnFullspeed
 
May 2011
France

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I work on the Goldback conjecture and the authors
of the verify explain that they compute the list of the prime

Why in their method they need this list?

N even number
repeat
for i:=1 to 100
M:=N-Primes(i)
If IsPrime(M) then N:=n+2;
until N > ?????
John
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Old 2011-09-01, 00:48   #2
Christenson
 
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That bit of programming is a simple verification of Goldbach's conjecture:

"Every even number can be expressed as the sum of two primes."

The program (if you add the correct braces) searches for small counterexamples. It needs some editing, since it needs to make output when it finds a counterexample, and a number of limits have been omitted.

They use the list of primes as a way to avoid repeatedly testing numbers for primality. Undoubtedly, the function IsPrime() also uses the list.

Last fiddled with by Christenson on 2011-09-01 at 00:49
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Old 2011-09-01, 07:26   #3
JohnFullspeed
 
May 2011
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To don't answer to the question

Many people use the list but why here?
What is the algo use?

(Sorry to use my bad english but i don't found traducer for your dialect)

John

N could have 1000 digits it is not a problem

Last fiddled with by JohnFullspeed on 2011-09-01 at 07:33
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Old 2011-09-01, 12:52   #4
Christenson
 
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Hint: I can read french, and your english translation makes things worse.
Hint: complete the program, and, for given N, tell me how many times I actually need to execute the inner loop, assuming a counterexample exists. Give me numbers with N=50.
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Old 2011-09-01, 19:03   #5
JohnFullspeed
 
May 2011
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Code:
Procedure Veri_GoldBack(L:Qword);
 Var I,J,W:integer;
     M,N,R: Qword;
     e2: double;
    Label LN;
 begin
    I:=0;
    HH:=0; // Le nombre  - le NP
    N:=Primes[200]+1;  /// for testing Not 0  1223
    a1 := GetTickCount;     //timer
    Offset:=0; // sortie
    repeat
     LN:
     N:=  N+2;
     If N>200000 then Break;// stop here
     j:=1;
     repeat
       HH:=N-Primes[j];   // test des divisions
       if not Is_Div(HH) then  Goto LN;
       J:=J+1;
     until J> 20;
     I:=I+1; // not found
   until  i>1000;
    a1 := GetTickCount-a1;

IsPrime:
Code:
Function Is_Div(A: QWord):boolean;
  var i,R,S: Cardinal;
     E5:extended;
   begin
     Result:=True;
     E5:=A;
     R:= Round(Sqrt(E5));
     i:=2;
     repeat
      If A mod Primes[i]=0 then exit;
      I:=I+1;
     until Primes[i] > r;
    Result:=False;
   end;
Like you can see ,I verify since
N=1223+1 even. The step is 2 (even) and I stop at 200 000
So I do 99 383 verifs


And now like Jasonp,axn,Five,silver,Bssquare... you stop to answer because your times are not in the same range that me, and you thinks that I most better that you You stop to answer.
I m not better than other but it's my job.


With the bad test of division I need 0,8 ... seconds
Perhaps I can learn to you if you want
The code is not magic

The time is on my RISC for a CISC the code will be other

PS can you give me the time for
for i:=1 to 100000 do
a=isprime(i);
They all say No.
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Old 2011-09-01, 21:03   #6
bsquared
 
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Quote:
Originally Posted by JohnFullspeed View Post
Code:
Procedure Veri_GoldBack(L:Qword);
 Var I,J,W:integer;
     M,N,R: Qword;
     e2: double;
    Label LN;
 begin
    I:=0;
    HH:=0; // Le nombre  - le NP
    N:=Primes[200]+1;  /// for testing Not 0  1223
    a1 := GetTickCount;     //timer
    Offset:=0; // sortie
    repeat
     LN:
     N:=  N+2;
     If N>200000 then Break;// stop here
     j:=1;
     repeat
       HH:=N-Primes[j];   // test des divisions
       if not Is_Div(HH) then  Goto LN;
       J:=J+1;
     until J> 20;
     I:=I+1; // not found
   until  i>1000;
    a1 := GetTickCount-a1;


Like you can see ,I verify since
N=1223+1 even. The step is 2 (even) and I stop at 200 000
So I do 99 383 verifs
I won't bother debunking the inner loop, but how can you claim to verify up to 200000 if you only run 1000 iterations of a loop that increments N by 2?
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Old 2011-09-01, 22:20   #7
Christenson
 
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John, slow down a bit...what language is this?
Since our hardware runs at very different speeds, depending on what else is going on on the computer, and the computer itself, we count operations, rather than clock time.

And re-read carefully what I asked...we need to begin with code which is correct, and do some small counting exercises before we start trying to time things. I can compute anything faster than you if it doesn't have to be right!!!

What you have there is called a mess...and a classic example of what I was talking about in terms of giving people the wrong idea about you.

In particular, suppose Goldbach's conjecture is false for N=50. How long will the program run, and how would we know if Goldbach's conjecture is false from looking at the output of the code?

I suggest you re-start from the original fragment you copied into this thread.
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Old 2011-09-02, 03:31   #8
LaurV
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Code:
 
begin
     N:=Primes[200]+3;  /// for testing Not 0  1223 {you better do postincrement of it}
     a1 := GetTickCount;     //timer
     repeat
        j:=0; {here preincrement works better, to exit the loop when is found}
        repeat
           J:=J+1;
        until (J>200) or not Is_Div(N-Primes[j]);  {not 20, as in your program}
        N:=N+2;                            
     until  N>20000; {not 1000!}
     a1 := GetTickCount-a1;
     {printing stuff}
end;
I think you just wanna start flame wars. Your program is bullshit, and I believe you just typed it in some "random" fashion, and you never ran it in a computer, first of all because "comments" (in all Pascal versions I know) are given between braces {} and not by // like in C-style. Beside of the fact that your program is full of shit variables you never use (did you claim you can actually.... optimize any program???), it only checked 1001 even numbers between 1226 and 3226, and for each of these numbers only tried if it can be written as a sum of two primes "a+b", with "a" being one of the first 20 primes. Please see my comment in red.

That is, the 20th prime is 71. Your program was checking if an even number in the interval 1226..3226 (in Pascal notation) can be written as a sum of a prime smaller then 72 and another prime. That's all. You could check that much easier and faster in Pascal using "set" operators (Pascal can do a lot of stuff with sets of objects).

See how much time do you need if you give the right values for limits, like I modified above in your code, and anyhow this won't be enough correct to verify Goldbach, if you want to go to N=200000 then you have to store the primes to 100000, just in case you run in some number which is a sum of two primes of about the same size (try playing on some superior limit of a very large prime gap). But for the solely purpose of timing, you should use the right number of iterations....

Tip: If you just copy the codesnap from above, be careful how you set the compiler's behaviour of evaluating boolean expressions, otherwise you could get an "array out of bounds" when you evaluate Primes[201] for j=201 at the most inner loop.

Last fiddled with by LaurV on 2011-09-02 at 03:43
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Old 2011-09-02, 12:14   #9
JohnFullspeed
 
May 2011
France

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The original question is:

Why Zimmerman need to compute the primelist to verify Golback
In a Brute force method like thyis(I never say thatt I use it, that it is speederr orr anything like this

Christenson answer that the Brute force method is bad but don't give a method using prime list

To prove that the brute force method is false He say me try the sample withs N=50 maxi

I understand that he want to see how many time the brute force needs to verif that all even numbers <52 can be write like a sum
I write a little code to find
I do'nt say that I use It but I compute the time


For those that think that they are better than bullshiit I add a vector with the 25 sums. Before to give lessons on Pascal open a book


http://en.wikipedia.org/wiki/Turbo_Pascal
Historically, Pascal comments are indicated { like this }, or (* like this *), and these can span any number of lines. Later versions of Borland Pascal also supported C++-style comments // like this, which finish at the end of the line.

You forget to give the name of your Pascal compiler that you use to try the code

You see Christenson : no code publish, no times and more simple I never Use pascal
But when I ask how many time you need no answer: just bullshit

You are speeder than me : show me .
You give the math problem:


Choose an hard problem for the computer
or a classic suject: twin search,Aliquot,factorisation
You can use all the library you want the langage you want

To give me a handicap I make a version on windows 32.

or

You know that my job is to make YOUR code better for the computer
without change the math

So write a code
publish it
An every one can optimize it

I take the challenge to go 2 time speeder than you only because I know well the computer
I said 2 but I thinks 5...

Before to say that I m bullshit try: I wait your code

And don't forget it's the computer which make tiime not the coder


In this version N <50 but the chrono doesn't move
I set N to 10000 to have a small time 0,050 second
But it's easy to make many more faster without bullshit

Procedure Veri_GoldBack(L:Qword);
Var I,J,W:integer;
M,N,R: Qword;
Label LN;
begin
HH:=0;
N:=0; // Begin with N=0
a1 := GetTickCount; //timer
repeat
LN:
N:= N+2;
If N> 50 then Break;// stop when N > 50
j:=1;
repeat
HH:=N-Primes[j]; // test des divisions
if odd(hh) then // speeder for mod 2 and mod 5
if not Is_Div(HH) then
begin
Q[N-1]:= N;// the sum
Q[N]:= hh;// a prime
Goto LN;
end;
J:=J+1;// No limit to J
until Primes[j] > N; // overflow N<0
I:=I+1; // not found
NotFound[i] :=N; //if the Goldbak is false you arrive here and can stop
until false; // infinity loop
a1 := GetTickCount-a1;
end;
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Old 2011-09-02, 14:15   #10
Christenson
 
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Quote:
Originally Posted by Christenson View Post
Hint: I can read french, and your english translation makes things worse.
Hint: complete the program, and, for given N, tell me how many times I actually need to execute the inner loop, assuming a counterexample exists. Give me numbers with N=50.
Notes:
@Laurv: The first program could work, if the braces in the original (or the begin/end pairs) had been preserved, and the code that would execute if Goldbach's conjecture were to prove false were included.
John has completely lost it with the second program...it's hopeless and not even worth debugging.

@John
1) I haven't said brute-force is good or bad....it's just one more tool.....we all work in terms of properties of the programs. (If you think I think brute force is bad, you have to ask why I have a high rank at TF, which is a brute-force method...)
2) I wanted you to give the original program...I promise you that what you posted wasn't it. This way we can CHECK your analysis.
3) I wanted you to analyse the original program, in terms of what would happen if, say, 50 was a counterexample to Goldbach's conjecture. How many loops would be required?
4) Talk about the program, not the people. Improve the program....get it right FIRST AND FOREMOST.
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Old 2011-09-02, 17:28   #11
JohnFullspeed
 
May 2011
France

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Where is your code ? I don't search juge but idea
But you are not an 'intlectual '
You need concret (French expression not insult)

I verify the conjecture with
0 < N < 1 223 193 544 (( 10^ 9) I need 1' 32
(I can give all the sums: no need of verify)
It's slow? You are far? How many works before joint the WC of the forum

PS Don't say it to Christensen : the code use only 6 lines
In the version I use prime list: I found a use( not the use of Zimmer)

I wait code...
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