[QUOTE=CRGreathouse;265082]Yes, I had to interrupt the code to see what it was doing as it had fallen into an infinite loop. If your third version works, good for you.[/QUOTE]

no my first version worked better than my last one the loop isn't infinite it just has to recalculate a[#a] twice on the fly. that's basically the major difference I see.

[QUOTE=science_man_88;265083]no my first version worked better than my last one the loop isn't infinite it just has to recalculate a[#a] twice on the fly. that's basically the major difference I see.[/QUOTE]

The version that you posted in post #2252 is an infinite loop, since each prime it adds to a is 3. For example, if I add
[code]print1(b[z]" ");[/code]
just before
[code]if(b[z]!=0,a=concat(a,b[z]);break)[/code]
I get
[code]3 3 3 3 3 3 3 3 [3000 more 3s skipped] 3 3 3 3 3 3 3 ^C
*** at top-level: Erastosthenes1(100)
*** ^-------------------
*** in function Erastosthenes1: ...int1(b[z]" ");if(b[z]!=0,
*** a=concat(a,b[z]);bre
*** ^--------------------
*** user interrupt after 330 ms.

*** Break loop: type <Return> to continue; 'break' to go back to GP[/code]

My browser's having a hard time rendering that uber-long string of 3's. Is it really necessary?

[QUOTE=CRGreathouse;265085]The version that you posted in post #2252 is an infinite loop, since each prime it adds to a is 3. For example, if I add
[code]print1(b[z]" ");[/code]
just before
[code]if(b[z]!=0,a=concat(a,b[z]);break)[/code]
I get
[code]3 3 3 3 3 3 3 3 3 (skipped 3's ) 3 3 3 ^C
*** at top-level: Erastosthenes1(100)
*** ^-------------------
*** in function Erastosthenes1: ...int1(b[z]" ");if(b[z]!=0,
*** a=concat(a,b[z]);bre
*** ^--------------------
*** user interrupt after 330 ms.

*** Break loop: type <Return> to continue; 'break' to go back to GP[/code][/QUOTE]

I don't get any of that on my end except me stopping it.

What do you get if you add the print statement I suggested?

[QUOTE=bsquared;265086]My browser's having a hard time rendering that uber-long string of 3's. Is it really necessary?[/QUOTE]

After my computer lovingly crafted each and every one of those 3s? I'm shocked by your audacity.

[QUOTE=CRGreathouse;265089]After my computer lovingly crafted each and every one of those 3s? I'm shocked by your audacity.[/QUOTE]

They were a beautiful thing... I'm almost sorry to see them go.

:max:But now they're back thanks to sm88's quote. :max:

[QUOTE=CRGreathouse;265088]What do you get if you add the print statement I suggested?[/QUOTE]

I get a whole bunch of 3's but I don't see how it's possible outside of one thing:

for the start of the forstep you need a if statement:

[CODE](x)->a=[2];b=vector(x-2,n,n+2);until(a[#a]>sqrt(x),forstep(y=if(a[#a]==2,a[#a],a[#a]-2),#b,a[#a],b[y]=0);for(z=1,#b,if(b[z]!=0,a=concat(a,b[z]);break())));a[/CODE]

this happened because I made a have 2 in it to start I think. and because trying to eliminate multiples of three with a n+2 setup in the array made 3 in position 1 not 3 so it never gets eliminated.

[QUOTE=science_man_88;265091]I get a whole bunch of 3's but I don't see how it's possible outside of one thing:

for the start of the forstep you need a if statement:

[CODE](x)->a=[2];b=vector(x-2,n,n+2);until(a[#a]>sqrt(x),forstep(y=if(a[#a]==2,a[#a],a[#a]-2),#b,a[#a],b[y]=0);for(z=1,#b,if(b[z]!=0,a=concat(a,b[z]);break())));a[/CODE]

this happened because I made a have 2 in it to start I think. and because trying to eliminate multiples of three with a n+2 setup in the array made 3 in position 1 not 3 so it never gets eliminated.[/QUOTE]

now it's missing the part to add in the values greater than sqrt(x).

Let's do this again. Use only one array and do something like this:

[code]b=vector(x,unused,1);
b[1]=0;
ptr=2;
while(b[ptr] <= sqrt(x),
\\ cross off numbers divisible by b[ptr]

ptr++;
while(b[ptr] == 0, ptr++);
);

\\ Now b[n] is 1 if n is prime and 0 otherwise

\\ Output the primes
numberOfPrimes=sum(i=1,#b,b[i]);
n=0;
vector(numberOfPrimes,unused,
while(b[n]==0,n++);
n
)[/code]

[QUOTE=CRGreathouse;265096]Let's do this again. Use only one array and do something like this:

[code]b=vector(x,unused,1);
b[1]=0;
ptr=2;
while(b[ptr] <= sqrt(x),
\\ cross off numbers divisible by b[ptr]

ptr++;
while(b[ptr] == 0, ptr++);
);

\\ Now b[n] is 1 if n is prime and 0 otherwise

\\ Output the primes
numberOfPrimes=sum(i=1,#b,b[i]);
n=0;
vector(numberOfPrimes,unused,
while(b[n]==0,n++);
n
)[/code][/QUOTE]

[CODE], ptr++););numberOfPrimes=sum(i=1,#b,b[i]);n=0;vector
(x)->b=vector(x,unused,1);b[1]=0;ptr=2;while(b[ptr]<=
>eratosene4(100)
hile: array index (101) out of allowed range [1-100].[/CODE]