I had an issue where CrashPlan kept on crashing on Mac (OSX 10.8.4). The CrashPlan launch menu bar would also fail to show, and even when I started it manually, it would only stay active for no more than about 60 seconds before crashing again.
In some cases a large file selection (>1TiB or 1 million files) can cause CrashPlan to crash. This can be noticed by continuous stopping and starting. You may also see the CrashPlan application and System Tray icon disappearing in the middle of a backup. Or the main CrashPlan program will run for about 30 seconds then close down with no error message.
…
The CrashPlan engine by default is limited to 512MB of main memory. The CrashPlan engine won’t use more memory than that, even if the CrashPlan engine needs more working memory and the computer has memory available. When the CrashPlan engine is running out of memory it crashes.
The issue was that as a heavy user, I backup more than 1TB of data. However, CrashPlan only allocates 512MB of memory in Java, which is insufficient for my large backup size.
To copy homework with no understanding of how to do it. Typically accomplished in the 10 minutes before said homework is due.
Originates from the Colorado School of Mines, where cold spiking is so revered that every year they award a prize for the best spiker.
Hey Jeremy! Let me cold spike your thermo or I’ll beat you bloody.
by Spjorkster Mar 10, 2005
A couple thoughts: I don’t know who “Spjorkster” is. I don’t cold spike; it’s stupid and I rather not turn in the homework. The term has been around since at least 2005, that’s amazing.
So now the project. The object reads:
The objective of this take-home project is to design and implement an electronic combination lock. The combination lock is to have a start-up combination code of 1-2-3-4 that MUST be changed immediately upon first-time activation. Furthermore, if the new combination code has three numbers all the same then an error message is to be sent and a different code has to be entered as the combination. Each number in the four-number combination code is to be an eight-bit vector.
The requirements are thus:
The combination lock design MUST be implemented as a state machine (using VHDL), with a minimum of three states: open, lock, set_combo. More that three states may be used (may or may not require greater than 3 states). The outputs should include a lock signal, indicating that the lock is currently locked, an open signal indicating that the lock is currently unlocked. The data inputs are to include the combination code (for when setting a new unlock code.) Additional there is to be a mechanism for resetting the entire circuit. Plus your design is to handle the succession of three wrong guesses at the combination by going into a security mode in which the initial code (1-2-3-4) followed by the entry of the correct code is required to unlock the lock. In your simulation waveforms, you MUST display the state variable (this will display the states traversed in setting, locking and unlocking the combination) along with the inputs and outputs.
There must also be at least four cases in which the combination is incorrect and at least three cases in which the combination is correct. And there is to be a string of three consecutive wrong attempts to unlock putting the circuit in the the security mode in which the initial code (1-2-3-4) immediately followed by the entry of the correct code is required to unlock the lock Remember to include in your report all schematics and/or VHDL code, simulation waveforms and state diagram/table.
Doing this all in VHDL isn’t as painful as I thought, but it was still very painful. I think the only thing I hated programing more so far was threaded tasks in Java. The idea of using states is somewhat similar, but not the same functionally.
Here’s the plan in a nutshell:
Yes, I did that state diagram. Isn’t it pretty?
Here’s the code:
LIBRARY ieee;
USE ieee.std_logic_1164.all;
USE IEEE.STD_LOGIC_ARITH.all;
USE IEEE.STD_LOGIC_UNSIGNED.ALL;
ENTITY combinationLock IS
PORT( Clock, Reset : IN STD_LOGIC; -- use positive logic for the reset
w : IN STD_LOGIC_VECTOR (1 DOWNTO 0); -- user action: 00 is no action; 01 is action; 11 is lock
a : IN STD_LOGIC_VECTOR (7 DOWNTO 0); -- digit a input
b : IN STD_LOGIC_VECTOR (7 DOWNTO 0); -- digit b input
c : IN STD_LOGIC_VECTOR (7 DOWNTO 0); -- digit c input
d : IN STD_LOGIC_VECTOR (7 DOWNTO 0); -- digit d input
aCode : BUFFER STD_LOGIC_VECTOR (7 DOWNTO 0); -- code of the first digit
bCode : BUFFER STD_LOGIC_VECTOR (7 DOWNTO 0); -- code of the second digit
cCode : BUFFER STD_LOGIC_VECTOR (7 DOWNTO 0); -- code of the third digit
dCode : BUFFER STD_LOGIC_VECTOR (7 DOWNTO 0); -- code of the fourth digit
unlockAttempt : BUFFER STD_LOGIC_VECTOR (1 DOWNTO 0); -- to keep track of how many attmpts have been made to unlock
isLock : OUT STD_LOGIC; -- is the safe locked
isError : OUT STD_LOGIC); -- is there error -- both lights on indicate in combo setting mode!
END combinationLock;
ARCHITECTURE Behavior OF combinationLock IS
TYPE State_type IS(SET_COMBO, LOCK_OPEN, LOCK_SECURE, SECURITY_MODE);
SIGNAL y : State_type;
BEGIN
PROCESS( Reset, Clock)
BEGIN
IF Reset = '1' THEN
aCode <= CONV_STD_LOGIC_VECTOR(1,8);
bCode <= CONV_STD_LOGIC_VECTOR(2,8);
cCode <= CONV_STD_LOGIC_VECTOR(3,8);
dCode <= CONV_STD_LOGIC_VECTOR(4,8);
y <= SET_COMBO;
ELSIF (Clock'EVENT AND Clock = '1') THEN
CASE y IS
WHEN SET_COMBO =>
IF (w(1) = '0' AND w(0) = '0') THEN
y <= SET_COMBO;
ELSIF (w(1) = '0' AND w(0) = '1') THEN
IF ( ( (a = b) AND (b = c) ) OR ( (a = c) AND (c = d) ) OR ( (a = b) AND (b = d) ) OR ( (b = c) AND (c = d) ) )THEN -- three numbers are repeated -- -----
isError <= '1';
y <= SET_COMBO;
ELSE -- program new code
aCode <= a;
bCode <= b;
cCode <= c;
dCode <= d;
isError <= '0';
y <= LOCK_OPEN;
END IF;
END IF;
WHEN LOCK_OPEN =>
isLock <= '0';
unlockAttempt <= CONV_STD_LOGIC_VECTOR(0,2); -- reset the security lockout
IF (w(1) = '0' AND w(0) = '0') THEN
y <= LOCK_OPEN;
ELSIF (w(1) = '0' AND w(0) = '1') THEN
y <= SET_COMBO;
ELSIF (w(1) = '1' AND w(0) = '1') THEN
y <= LOCK_SECURE;
END IF;
WHEN LOCK_SECURE =>
isLock <= '1'; -- set the lock indicator
isError <= '0'; -- clear all errors because it's locked now
IF (w(1) = '0' AND w(0) = '0') THEN
y <= LOCK_SECURE;
ELSIF ( w(1) = '0' AND w(0) = '1') THEN
IF (unlockAttempt = CONV_STD_LOGIC_VECTOR(0,2) ) THEN
IF (a = aCode AND b = bCode AND c = cCode AND d = dCode) THEN
y <= LOCK_OPEN;
ELSE
unlockAttempt <= CONV_STD_LOGIC_VECTOR(1,2);
y <= LOCK_SECURE;
END IF;
ELSIF (unlockAttempt = CONV_STD_LOGIC_VECTOR(1,2) ) THEN
IF (a = aCode AND b = bCode AND c = cCode AND d = dCode) THEN
y <= LOCK_OPEN;
ELSE
unlockAttempt <= CONV_STD_LOGIC_VECTOR(2,2);
y <= LOCK_SECURE;
END IF;
ELSIF (unlockAttempt = CONV_STD_LOGIC_VECTOR(2,2) ) THEN
IF (a = aCode AND b = bCode AND c = cCode AND d = dCode) THEN
y <= LOCK_OPEN;
ELSE
unlockAttempt <= CONV_STD_LOGIC_VECTOR(3,2);
y <= SECURITY_MODE;
END IF;
ELSIF (unlockAttempt = CONV_STD_LOGIC_VECTOR(3,2)) THEN
y <= SECURITY_MODE;
END IF;
END IF;
WHEN SECURITY_MODE =>
IF (w(1) = '0' AND w(0) = '0') THEN
y <= SECURITY_MODE;
ELSIF (w(1) = '0' AND w(0) = '1') THEN
IF(a = CONV_STD_LOGIC_VECTOR(1,8) AND b = CONV_STD_LOGIC_VECTOR(2,8) AND c = CONV_STD_LOGIC_VECTOR(3,8) AND d = CONV_STD_LOGIC_VECTOR(4,8)) THEN
unlockAttempt <= CONV_STD_LOGIC_VECTOR(2,2);
y <= LOCK_SECURE;
END IF;
END IF;
END CASE;
END IF;
END PROCESS;
END Behavior;
In all states (except Reset, which isn’t defined as typical state), ’00’ is used to stay within the state. ’01’ is used as an action key and ’11’ is used to lock the device (as noted with comments within the VHDL code). isError is ‘1’ when a user has entered a new invalid combination (any combination with where 3 digits are the same). Security Mode is entered when three unsuccessful attempts have been made to unlock the device. When this happens, the Security Mode-state is entered and the user must enter ‘1 2 3 4’. When this happens, the user is sent back to the Lock Secure-state and is granted one try before being kicked back to the Security Mode-state. However, if the attempt is correct, the user is sent to the Lock Open-state where they can change the combination or lock the device again.
Here’s the schematic:
…and the waveforms, which shows that it does indeed work:
Note: realACode, realBCode, realCCode, realDCode, realUnlock[1] and realUnlock[0] are debugging variables used to ensure that the device is working properly during testing. real*Code is used to display the stored combination code and realUnlock[*] is used to count how many attempts have been tried to unlock the device.
All in all, pretty cool. Minus the nine hours or so it took to do it.
Domain Name Servers (DNS) might be considered the single most important aspect of the Internet, yet many people don’t know what they are and those that do probably don’t know how they work. In the end, this ends up being a good thing. But over the last week or so, they’ve unknowingly been the bane of my internship.
Domain Name Servers really serve a single purpose: they take a domain name (such as AndrewFerguson.net) and return an IP address (such as 70.96.188.33). In reality, there are many DNS which talk with each other and share information and (usually) ensure that you get to the website you’re trying to reach.
But what happens if a DNS goes down? Or actually: What happens if the person responsible for add the names of a half dozen DNS to your TCP/IP settings fails to do this? Short answer is: You can’t do any work. Long answer is: you spend days trying to figure out why you don’t have permissions to access a network share despite the fact that permissions aren’t the issue.
So yea, all fixed now. Today was probably one of my most productive days yet. I started playing around with some code to get a proof-of-concept design for DeIcer and made some great headway on that.