udev 0.1 release
A message that is not in the archive (sent off-list or not yet received): [email protected]
Yes, 1998 3 into 1 splitters would be needed. The break down is as follows:
Three of the original power connectors would need 400 splitters, while the remaining two would just need 399 splitters. This would result in 4001 power connectors, or 4001 disks. Because it's a Friday, here is a simple program that verified my findings.
int main(int argc,char **argv)
{
int con_a,con_b,con_c,con_d,con_e;
int pow_a,pow_b,pow_c,pow_d,pow_e;
con_a=con_b=con_c=con_d=con_e=0;
pow_a=pow_b=pow_c=pow_d=pow_e=1;
while(1) {
con_a++;
pow_a=con_a*3-(con_a==1?0:con_a-1);
if(pow_a+pow_b+pow_c+pow_d+pow_e>=4000)
break;
con_b++;
pow_b=con_b*3-(con_b==1?0:con_b-1);
if(pow_a+pow_b+pow_c+pow_d+pow_e>=4000)
break;
con_c++;
pow_c=con_c*3-(con_c==1?0:con_c-1);
if(pow_a+pow_b+pow_c+pow_d+pow_e>=4000)
break;
con_d++;
pow_d=con_d*3-(con_d==1?0:con_d-1);
if(pow_a+pow_b+pow_c+pow_d+pow_e>=4000)
break;
con_e++;
pow_e=con_e*3-(con_e==1?0:con_e-1);
if(pow_a+pow_b+pow_c+pow_d+pow_e>=4000)
break;
}
fprintf(stderr,"con_a = %d, pow_a = %d\n",con_a,pow_a);
fprintf(stderr,"con_b = %d, pow_b = %d\n",con_b,pow_b);
fprintf(stderr,"con_c = %d, pow_c = %d\n",con_c,pow_c);
fprintf(stderr,"con_d = %d, pow_d = %d\n",con_d,pow_d);
fprintf(stderr,"con_e = %d, pow_e = %d\n",con_e,pow_e);
fprintf(stderr,"pow_a+pow_b+pow_c+pow_d+pow_e =
%d\n",pow_a+pow_b+pow_c+pow_d+pow_e);
return 0;
}-----Original Message-----
From: [email protected] [mailto:[email protected]]
Sent: Friday, April 11, 2003 1:18 PM
To: [email protected]
Cc: Steve Lee; [email protected]
Subject: Re: [ANNOUNCE] udev 0.1 release
Quoted 4 lines · Oliver Neukum
Am Freitag, 11. April 2003 20:03 schrieb Michael Buesch:
Quoted 3 lines · John Bradford
Quoted 2 lines
[Puzzle]
Say the power supply had five 5.25" drive power connecters, how
many 1
Quoted 1 line
Quoted 1 line
Quoted 1 line
into 3 power cable splitters would you need to connect all 4000
disks?
Quoted 2 lines
You are hereby hit. All drivers need to be connected to a splitter. Thus any answer under 4000/3 must be wrong.
Yea, but: "Say the power supply had five 5.25" drive power connecters" :)
Signature · 2 lines
My homepage: http://www.8ung.at/tuxsoft fighting for peace is like fu**ing for virginity
Quoted 1 line · Steve Lee
Yes, 1998 3 into 1 splitters would be needed.
Yes, I think this is correct. :)
My previously posted 400 ment "400 to each powersupply cable" (but I wasn't correct with it, because as you said to some are only 399 connected.
PS: We should do something more productive than this. :)
Regards
Michael Buesch.
Signature · 2 lines
My homepage: http://www.8ung.at/tuxsoft fighting for peace is like fu**ing for virginity