Another Infinite Login Loop Problem - need help badly

Hello. I need help.
When I start up my iMac (24 inch last generation model) I can't log into the OS. When I enter my password I see the default OS wallpaper (nothing else) then it kicks me to a blue screen then right back to the login. Same thing over and over. I've already ran disk utility off of the install CD to repair permissions, etc... but no luck. This even happens when I enter into safe mode. I'm running the latest version of the OS.
Can anyone help me?

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Need help bad, screen problem

heres my problem. i setup arch on my desktop last night (64bit), i installed my nvidia drivers. i have a 21" crt sun monitor, not sure the model number. but i brought it to a friends this morning to update and install kdemod since they have highspeed. now i was hooked up to their monitor, they have a cheap 15" crt. so it worked fine there, now when i got it home, i get a black screen when it boots up on my monitor, i can get to a console however
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Last edited by ssl6 (2007-09-16 07:00:40)

nvm, i just deleted the xorg.conf and rebooted, its fine now
edit// nvm, its not using my nvidia driver.........i might still need help, im trying something right now, not sure what im doing though
ok, i tried to run nvidia-xconfig, and when i rebooted, im back to square one with my monitor flashin an invalid scan frequency
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# nvidia-xconfig: X configuration file generated by nvidia-xconfig
# nvidia-xconfig: version 1.0 (buildmeister@builder26) Wed Jun 13 16:54:14 PDT 2007
Section "ServerLayout"
Identifier "Layout0"
Screen 0 "Screen0" 0 0
InputDevice "Keyboard0" "CoreKeyboard"
InputDevice "Mouse0" "CorePointer"
EndSection
Section "Files"
RgbPath "/usr/lib/X11/rgb"
EndSection
Section "Module"
Load "dbe"
Load "extmod"
Load "type1"
Load "freetype"
Load "glx"
EndSection
Section "InputDevice"
# generated from default
Identifier "Mouse0"
Driver "mouse"
Option "Protocol" "auto"
Option "Device" "/dev/psaux"
Option "Emulate3Buttons" "no"
Option "ZAxisMapping" "4 5"
EndSection
Section "InputDevice"
# generated from default
Identifier "Keyboard0"
Driver "kbd"
EndSection
Section "Monitor"
Identifier "Monitor0"
VendorName "Unknown"
ModelName "Unknown"
HorizSync 30.0 - 110.0
VertRefresh 50.0 - 150.0
Option "DPMS"
EndSection
Section "Device"
Identifier "Device0"
Driver "nvidia"
VendorName "NVIDIA Corporation"
EndSection
Section "Screen"
Identifier "Screen0"
Device "Device0"
Monitor "Monitor0"
DefaultDepth 24
SubSection "Display"
Depth 24
Modes "1600x1200" "1280x1024" "1024x768" "800x600" "640x480"
EndSubSection
EndSection
here's that part of the xorg.conf from my other computer running 32bit arch with a geforce 3 connected to the same monitor via kvm switch. this machine is using a 96xx driver though
Section "Monitor"
# HorizSync is in kHz unless units are specified.
# HorizSync may be a comma separated list of discrete values, or a
# comma separated list of ranges of values.
# NOTE: THE VALUES HERE ARE EXAMPLES ONLY. REFER TO YOUR MONITOR'S
# USER MANUAL FOR THE CORRECT NUMBERS.
# HorizSync 30-64 # multisync
# HorizSync 31.5, 35.2 # multiple fixed sync frequencies
# HorizSync 15-25, 30-50 # multiple ranges of sync frequencies
# VertRefresh is in Hz unless units are specified.
# VertRefresh may be a comma separated list of discrete values, or a
# comma separated list of ranges of values.
# NOTE: THE VALUES HERE ARE EXAMPLES ONLY. REFER TO YOUR MONITOR'S
# USER MANUAL FOR THE CORRECT NUMBERS.
Identifier "sun"
HorizSync 31.5 - 82.0
VertRefresh 50.0 - 90.0
EndSection
Section "Device"
# The chipset line is optional in most cases. It can be used to override
# the driver's chipset detection, and should not normally be specified.
# Chipset "generic"
# The Driver line must be present. When using run-time loadable driver
# modules, this line instructs the server to load the specified driver
# module. Even when not using loadable driver modules, this line
# indicates which driver should interpret the information in this section.
# The BusID line is used to specify which of possibly multiple devices
# this section is intended for. When this line isn't present, a device
# section can only match up with the primary video device. For PCI
# devices a line like the following could be used. This line should not
# normally be included unless there is more than one video device
# intalled.
# BusID "PCI:0:10:0"
# VideoRam 256
# Clocks 25.2 28.3
Identifier "Standard VGA"
Driver "vga"
VendorName "Unknown"
BoardName "Unknown"
EndSection
Section "Device"
#VideoRam 65536
# Insert Clocks lines here if appropriate
Identifier "geforce"
Driver "nvidia"
EndSection
Section "Screen"
Identifier "Screen 1"
Device "geforce"
Monitor "sun"
DefaultDepth 24
Option "AddARGBGLXVisuals" "True"
SubSection "Display"
Viewport 0 0
Depth 8
Modes "1280x1024" "1024x768" "800x600" "640x480"
EndSubSection
SubSection "Display"
Viewport 0 0
Depth 16
Modes "1280x1024" "1024x768" "800x600" "640x480"
EndSubSection
SubSection "Display"
Viewport 0 0
Depth 24
Modes "1280x1024" "1024x768" "800x600" "640x480"
EndSubSection
EndSection
Section "Extensions"
Option "Composite" "Enable"
EndSection
any ideas here?
Last edited by ssl6 (2007-09-16 07:53:48)

Still stuck with the same old producer consumer weight problem need help

Hello All,
This is the problem I am stuck with right now.
I have two array lists one producer array list and one consumer array list denoted by a and b
P1 P2 P3 P4 P5
5 6 7 8 9
C1 C2 C3 C4 C5
2 3 4 5 6
Now we find all those producer consumer pairs which satisfy the criteria Pi>=Ci
We have the following sets
(5,2)(6,2)(7,2),(8,2),(9,2)
(5,3)(6,3)(7,3),(8,3),(9,3)
(5,4)(6,4)(7,4),(8,4),(9,4)
(5,5)(6,5)(7,5),(8,5),(9,5)
(6,6)(7,6)(8,6),(9,6)
Let us done each of them with Si
so we have S1,S2,S3,S4,S5
we assign a third parameter called weight to each element in Si which has satisfied the condition Pi>=Ci;
so we we will have
(5,2,ai),(6,2,bi),(7,2,ci)....etc for S1
similarly for S2 and so on.
We need to find in each set Si the the pair which has the smallest weight.
if we have (5,2,3) and (6,2,4) then 5,2,3 should be chosen.We should make sure that there is only one pair in every set which is finally chosen on the basis of weight.
Suppose we get a pair (5,2,3) in S1 and (5,2,3) in S2 we should see that (5,2,3) is not used to compare to compare with any other elements in the same set S2,
Finally we should arrive at the best element pair in each set.They should be non repeating in other sets.
Given a problem
P0 P1 P2 P3 P4
9 5 2 2 8
6 5 4 5 3
C0 C1 C2 C3 C4
we have So as (P0,C0) and (P4,C0)
assuming that the one with the smaller index has lesser weight PO is selected.In the program I have used random weights.from set S1 we select the pair PO,CO
S1 =(P0,C1),(P1,C1) and (P4,C1)
since P0 and P4 are already used in previous set we dont use them for checking in S1 so we have (P1,C1) as best.
S2=(P0,C2),(P1,C2) and (P4,C2) so we dont use P0,C2 and P1 and C2 because PO and P1 are already used in S1.
So we choose P4,C2
in S3 and S4 ae have (P0,C3),(P1,C3),(P4,C3) so we dont choose anything
and same in S4 also.
So answer is
(P0,C0),(P1,C1) and (P4,C2).
My program is trying to assign weights and I am trying to print the weights along with the sets.It doesnt work fine.I need help to write this program to do this.
Thanks.
Regards.
NP
What I have tried till now.
I have one more question could you help me with this.
I have an array list of this form.
package mypackage1;
import java.util.*;
public class DD
private int P;
private int C;
private int weight;
public void set_p(int P1)
P=P1;
public void set_c(int C1)
C=C1;
public void set_weight(int W1)
weight=W1;
public int get_p()
return P;
public int get_c()
return C;
public int get_x()
return weight;
public static void main(String args[])
ArrayList a=new ArrayList();
ArrayList min_weights_int=new ArrayList();
ArrayList rows=new ArrayList();
ArrayList temp=new ArrayList();
Hashtable h=new Hashtable();
String v;
int o=0;
DD[] d=new DD[5];
for(int i=0;i<4;i++)
d=new DD();
for(int i=0;i<4;i++)
d[i].set_p(((int)(StrictMath.random()*10 + 1)));
d[i].set_c((int)(StrictMath.random()*10 + 1));
d[i].set_weight(0);
System.out.println("Producers");
for(int i=0;i<4;i++)
System.out.println(d[i].get_p());
System.out.println("Consumers");
for(int i=0;i<4;i++)
System.out.println(d[i].get_c());
System.out.println("Weights");
for(int i=0;i<4;i++)
System.out.println(d[i].get_x());
for(int i=0;i<4;i++ )
int bi =d[i].get_c();
ArrayList row=new ArrayList();
for(int j=0;j<4;j++)
if( d[j].get_p() >=bi)
d[j].set_weight((int)(StrictMath.random()*10 + 1));
row.add("(" + bi + "," + d[j].get_p() + "," +d[j].get_x() + ")");
else
d[j].set_weight(0);
row.add("null");
rows.add(row);
System.out.println(rows);
int f=0;
for(Iterator p=rows.iterator();p.hasNext();)
temp=(ArrayList)p.next();
String S="S" +f;
h.put(S,temp);
String tt=new String();
for(int j=0;j<4;j++)
if(temp.get(j).toString() !="null")
// System.out.println("In if loop");
//System.out.println(temp.get(j).toString());
String l=temp.get(j).toString();
System.out.println(l);
//System.out.println("Comma matches" + l.lastIndexOf(","));
//System.out.println(min_weights);
f++;
for(Enumeration e=h.keys();e.hasMoreElements();)
//System.out.println("I am here");
int ii=0;
int smallest=0;
String key=(String)e.nextElement();
System.out.println("key=" + key);
temp=(ArrayList)h.get(key);
System.out.println("Array List" + temp);
for( int j=0;j<4;j++)
String l=(temp.get(j).toString());
if(l!="null")
System.out.println("l=" +l);
[\code]

In your example you selected the pair with the greatest
distance from the first set, and the pair with the least
distance from the second. I don't see how the distance
function was used.
Also it's not clear to me that there is always a solution,
and, if there is, whether consistently choosing the
furthest or the closest pairs will always work.
The most obvious approach is to systematically try
all possibilities until the answer is reached, or there
are no possibilities left. This means backtracking whenever
a point is reached where you cannot continue. In this case
backtrack one step and try another possibility at this
step. After all possible choices of the previous step,
backtrack one more step and so on.
This seems rather involved, and it probably is.
Interestingly, if you know Prolog, it is ridiculously
easy because Prolog does all the backtracking for you.
In Java, you can implement the algorithm in much the same
way as Prolog implementations do it--keep a list of all the
choice points and work through them until success or there
are none left.
If you do know Prolog, you could generate lots of random
problems and see if there is always a solution.

Another Infinite Birthday Loop Issue

I have been a loyal Photoshop user since version 2.5.
I just purchased the Photoshop Photography Program and I am also unable to get beyond the verify my date of birth.
No problem when wanting to charge my card but a problem when trying to access the download center.
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I just wasted over 30 on the phone with customer support with any help.
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Another Infinite Login Loop Problem - need help badly

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