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 2011-08-31, 09:45 #1 JohnFullspeed   May 2011 France 7·23 Posts New Code 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
 2011-09-01, 00:48 #2 Christenson     Dec 2010 Monticello 179510 Posts 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
 2011-09-01, 07:26 #3 JohnFullspeed   May 2011 France 2418 Posts Thanks 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
 2011-09-01, 12:52 #4 Christenson     Dec 2010 Monticello 5·359 Posts 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.
 2011-09-01, 19:03 #5 JohnFullspeed   May 2011 France 16110 Posts Goldback 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.
2011-09-01, 21:03   #6
bsquared

"Ben"
Feb 2007

D2116 Posts

Quote:
 Originally Posted by JohnFullspeed 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?

 2011-09-01, 22:20 #7 Christenson     Dec 2010 Monticello 5·359 Posts 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.
 2011-09-02, 03:31 #8 LaurV Romulan Interpreter     Jun 2011 Thailand 24·571 Posts 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
2011-09-02, 14:15   #10
Christenson

Dec 2010
Monticello

5·359 Posts

Quote:
 Originally Posted by Christenson 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.

 2011-09-02, 17:28 #11 JohnFullspeed   May 2011 France 16110 Posts Word... 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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