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Beginner Boolean compiling error

I am very new to Java (doing a beginners university module) so sorry for the probably silly question. I am trying to verify whether a ragged array is a 'tridiagonal matrix'.
It is valid if it is of length 3 at the first level and of length n − 1, n, and n − 1 at the second level. I intended to come up with a code to firstly verify the length is 3, then find the longest length array within it for n, then finally verify each length.
For whatever reason my code won't compile but I'm not seeing an error message, just a red exclamation mark on the class. I assume this means there are multiple errors. If anyone could point them out it would be a massive help.
static boolean isValidTridiagonal ( double [][] m)
{
if (double [][]=new double [3][])
{
int n = 0;
for(int i = 0; i < m.length; i++)
{
if(m[i].length > n)
{
n = m[i].length;
if( (m[0].length = n-1) && (m[1].length = n) &&(m[2].length=n-1))
{
return true
}
else
{
return false
}
}
else
{
return false
}
}
Thanks very much!

I agree with Foolish in the comments that it's helpful to use an IDE that can highlight syntax errors and other problems with the code, it really makes a huge difference. Apart from that, another general strategy is to always code in "baby steps": do only the minimal thing to test if the code works, compile and test often. And if you still have troubles, you can always comment out chunks of your code when searching for the offending bits.
Having said that, the errors that I see in your code are:
if (double [][]=new double[3][])
If you want to test the length of the input, you can do if (m.length == 3)
In
if( (m[0].length = n-1) && (m[1].length = n) &&(m[2].length=n-1))
you're not testing for equality, but rather trying to put the values n-1 etc into m[0].length, which is not going to work. What you probably meant was
if( (m[0].length == n-1) && (m[1].length == n) &&(m[2].length==n-1))
In
return true
you're missing a semicolon. The compiler is whiny about things like that and unless you use an IDE or learn to interpret the compiler error messages, it can be really painful to find such errors.
Finally, of course, the answer by vasste provides a much simpler solution to your actual task, so it's worth looking into that :).

Why do you need all that loops? If all arrays cannot be null, than
static boolean isValidTridiagonal(double[][] m) {
return m.length == 3 && m[0].length == m[1].length - 1 && m[2].length == m[0].length;
}

You're missing a few braces at the end but, judging from your indentation, you just forgot to copy them.
You're missing semicolons from the end of the return lines.
The condition within this if statement if (double [][] = new double [3][]) is not a valid expression. You simply want to evaluate the length, which you can do like if (m.length == 3). You did the same thing later on.
The line including (m[0].length = n-1) && (m[1].length = n) && (m[2].length=n-1) is not valid because you are performing assignment (=) in all three cases. An equality check is the double equals operator ==.
You do not return a value in every case. You can fix this by adding return false; after the closing brace of your first if statement, i.e. the last line of the function.
This is enough to get your code to compile. As mentioned in another answer though, your logic is confusing and without actually tracing it through I would speculate that it will not work as you would expect.
If I have understood your requirements correctly, you can rewrite the entire function as:
static boolean isValidTridiagonal ( double [][] m)
{
return m.length == 3 &&
m[0].length + 1 == m[1].length &&
m[2].length + 1 == m[1].length;
}
A proper IDE - Netbeans, Eclipse, etc. - will give you fairly descriptive error messages to show you where you've gone wrong.
This is basically completely stylistic but I wish someone had pointed this out to me earlier. If you ever find yourself writing code in this form:
if( (m[0].length == n-1) && (m[1].length == n) && (m[2].length == n-1))
{
return true;
}
else
{
return false;
}
know that you can save yourself so many lines without losing any readability by instead writing:
return (m[0].length == n-1) && (m[1].length == n) && (m[2].length == n-1);

Looking for help on creating more efficient way on doing a lot of checks

Note: Not a duplicate of How do I compare strings in java as I am taking about going through some checks to determine if something is inheriting something something else
Is their a better and more efficient way to do this:
As you can see I am inputting 2 strings then checking them of on a list, as if current = three then it returns true for checking for one, two and three
NOTE: these values(one,two,three) are just placeholders for the example in my use their is no relation between them except that they have a different priority.
public boolean checker(String current, String check) {
if (check.equals("one")) {
if (current.equals("one") || current.equals("two")
|| current.equals("three")) {
return true;
}
}
if (check.equals("two")) {
if (current.equals("two") || current.equals("three")) {
return true;
}
}
if (check.equals("three")) {
if (current.equals("three")) {
return true;
}
}
return false;
}

Here are a few pointers
As Frisch mentioned in comments, use .equals rather than == for String comparison.
Use switch/case
switch (check) {
case "one":
if (current.equals("one")) return true;
case "two":
if (current.equals("two")) return true;
case "three":
if (current.equals("three")) return true;
}
Apart from that, there doesn't seem to be much to do.

Two things.
Don't check strings using equality. Use the .equals() method. You can call it off the string literal. So something like this. Calling it off the string literal is safe even with nulls, which is generally a good thing.
if ("one".equals(check))
You can take advantage of Java's short circuit operators && and ||
if ("one".equals(check)) {
if ("one".equals(current) || "two".equals(current) || "three".equals(current)) {
return true;
}
}
Can become
if ("one".equals(check) && ("one".equals(current) || "two".equals(current) || "three".equals(current))) {
return true;
}
Which will be evaluated from left to right. Since the "one".equals(check) is on the far most left, and is &&'ed with the rest of the statement, Java will bail out of the condition checking if "one".equals(check) is not true, and will not evaluate the rest of the statement.
Since you're just returning true or false, you can also take this a step further and reduce all of your if statements into a single one using De Morgan's laws (http://en.wikipedia.org/wiki/De_Morgan's_laws). Generally you state your boolean if statement in the way that is most natural to you, and then you start simplifying it by applying transformations that keep the logical if statement the same.
A good example of this is, stolen from the below link.
In the context of the main method's program body, suppose the following data is defined:
int score, min = 0, max = 20;
boolean bad, good;
Further suppose that a value is assigned to score, perhaps from a keyboard entry, and I would like to test, with a Boolean expression whether the score is a valid number or not. A good score is in the closed range [0 .. 20], which includes 0 and 20.
good = (score >= min && score <= max);
I would like to get the score from the keyboard in a do while loop, so that I can validate the entry. The logic in my control structure is to demand another entry for the score while the entry is bad. I have a definition of a good entry, and I will use definitions of operators and De Morgan's Law to help me write an expression that represents a bad entry.
good = (score >= min && score <= max); // original definition of good from the logic of my task
good = !(score < min) && !(score > max); // by definition, >= means ! < , <= means ! >
good = !(score < min || score > max); // by De Morgan's' Law
bad = !good ; // bad is not good
bad = !!(score < min || score > max); // substituting for good
bad = score < min || score > max; // double negation is dropped
http://fcmail.aisd.net/~JABEL/1DeMorgansLaw.htm

I would like to update you some thing.
1. We can apply switch cases only on primitive data types but not on objects. As string is object we can't use strings in case/switch statement.
I would like to suggest you to enums/maps in this case.
Please find the below sample programm how i implemented using maps.
public static void main(String[] args) {
Map<String,Integer> map = new HashMap<String, Integer>();
map.put("one", 1);
map.put("two", 2);
map.put("three", 3);
String current = "one";
String check = "one";
if(map.get(check)<=map.get(current)){
System.out.println("Our condition is success");
}
}
Instead of multiple comparison this is better.
---Santhosh

|| and && in a single if statement

I'm getting some peculiar output from a program in which I have a few if statements like this:
if((m.getLeft(position).state == position.state.wall || m.getLeft(position).state == position.state.border)
&& (m.getBelow(position).state == position.state.open || m.getBelow(position).state == position.state.visited)){
check = true;
}
where I have both && and || used in the same if condition. I want the boolean check to be made true if the cell at getLeft() is either a wall or a border AND the cell at getBelow() is either open or visited.
Does this code, the way I have it written now, perform this check properly? Perhaps more importantly, is it poor form to write a statement like this? Would it be better to break it up into two statements?
I'm not sure whether the peculiarities I'm seeing are resulting from these statements or something else, my question is more about general best practices.
EDIT: Thanks for all the input. I suspected as much (that it was too complicated), which was why I framed my question the way I did.
EDIT (a year later, looking back) (to restate the above more strenuously) for the love of god, don't write anything like the if statement above. If you find yourself with a similar line of code, remember occam's razor is perhaps nowhere more applicable than in programming.

Your complicated if statement can be refactored as:
if((m.getLeft(position).state == position.state.wall || m.getLeft(position).state == position.state.border)) {
if((m.getBelow(position).state == position.state.open || m.getBelow(position).state == position.state.visited)){
check = true;
}
}
Thanks to #Makoto:
check = ((m.getLeft(position).state == position.state.wall || m.getLeft(position).state == position.state.border)) && ((m.getBelow(position).state == position.state.open || m.getBelow(position).state == position.state.visited));
And your code, as well as this code "works".
But also, remember, stick to a naming convention that is mentioned in Java's Style Guidelines. I have no idea what m in your code is. This kind of naming an object reference must be avoided. Also, state seems to be a public field in your class (assuming). Such public access fields should also be avoided. Instead use getters() and setters().

It looks to me that it performs properly, however it is rather difficult to read.
If you're going to be using an && in an if statement, you may as well nest a new if statement instead, it is essentially the same thing in most cases.

Breaking down your boolean statement, it reads something like this:
(a == x_1 || a == x_2) && (b == x_3 || b == x_4)
Unfortunately, that's about as simple as that particular boolean statement will get. There are options to make the pain a lot easier:
Refactor your code to not need such complex statements, by breaking it up into two if-blocks (as shown in my refactor below), or
Extract that as a method and assign the return value to check. Honestly, it's either going to be true or false.
Here's an example of the refactor method. I don't know what m is precisely:
public boolean isCheck(M m, Position p) {
boolean retVal = false;
if(m.getLeft(p).state == p.state.wall || m.getLeft(p).state == p.state.border)) {
if((m.getBelow(p).state == p.state.open || m.getBelow(p).state == p.state.visited))
retVal = true;
}
}
return retVal;
}
// call it as such
check = isCheck(m, position);

Use another IF insted of using &&:
if(m.getLeft(position).state == position.state.wall ||
m.getLeft(position).state == position.state.border){
if(m.getBelow(position).state == position.state.open ||
m.getBelow(position).state == position.state.visited){
check = true;
}
}

If-else block in nested for loops - Compiler claims I need a return statement

How do I make this work? It says that I need to add a return statement but I have one.
public boolean clockFactCheck(int a, int b){
for (int i = 0; i <= 276; i++){
for (int h = 0; h <= 55; h++){
if (a == i + 186 && b == h + 133){
return true;
} else {
return false;
}
}
}
}

The code provided may not reach one of the returns for any input a,b and that's what the compiler is complaining about.
Actually in your case the if-else will be reached with the very first iteration - unfortunately something which the compiler cannot deduce. Therefore, it goes the save way and issues this error.
Comment: Therefore, in your loop seems not to make much sense since it will not iterate at all but stop within the first iteration i==0 and h==0. Did you meant to code something like that?

You don't have a return statement after the for loop but even then h++ is dead code becaue it will never get past the first iteration.

Place the return statement outside the loop. (declare a boolean in your function, modify it and return it at the end)

My guess is that the compiler is not clever enough to figure out that there is no codepath around the loops.
If this is real code, simplify it to
return (a == 186 && b == 133);
If this is not the real code, there is probably another path that does not return (if you made an error pasting), or there really is a compiler bug or limitation. At one point, the halting problem bites you, and the code is too complex for the compiler to figure out.
In the latter case I would place a
throw new RuntimeException(
String.format("should never get here (a = %d, b = %d) !",a,b));
at the last statement.
That makes both the compiler happy and does not introduce an "undefined" return value for a case that should either never happen, or if it does, has not been thought of.

Oh yeah my bad, I'm sorry this was a stupid question it was 5am when I posted this. I figured out the answer to my problem. If you made the same stupid mistake as me here is the fix
public boolean clockFactCheck(int a, int b){
for (int i = 0; i <= 276; i++){
for (int h = 0; h <= 55; h++){
if (a == i + 186 && b == h + 133){
return true;
}
}
}
return false;
}

Java requires every path return a value. The compiler could not judge if the circulation will return a path. If your code is like this:
public boolean add(){
for(int i=0;i<100;i++)
if(i==5000)
return true;}
the function add will not return a value. Instead, an error will occur.

Why switch is faster than if

Lots of Java books describe the switch statement as being faster than the if else statement. But I did not find out anywhere why switch is faster than if.
Example
I have a situation I have to choose any one item out of two. I can use either use
switch (item) {
case BREAD:
//eat Bread
break;
default:
//leave the restaurant
}
or
if (item == BREAD) {
//eat Bread
} else {
//leave the restaurant
}
considering item and BREAD is a constant int value.
In the above example which is faster in action and why?

Because there are special bytecodes that allow efficient switch statement evaluation when there are a lot of cases.
If implemented with IF-statements you would have a check, a jump to the next clause, a check, a jump to the next clause and so on. With switch the JVM loads the value to compare and iterates through the value table to find a match, which is faster in most cases.

A switch statement is not always faster than an if statement. It scales better than a long list of if-else statements as switch can perform a lookup based on all the values. However, for a short condition it won't be any faster and could be slower.

The current JVM has two kinds of switch byte codes: LookupSwitch and TableSwitch.
Each case in a switch statement has an integer offset, if these offsets are contiguous (or mostly contiguous with no large gaps) (case 0: case 1: case 2, etc.), then TableSwitch is used.
If the offsets are spread out with large gaps (case 0: case 400: case 93748:, etc.), then LookupSwitch is used.
The difference, in short, is that TableSwitch is done in constant time because each value within the range of possible values is given a specific byte-code offset. Thus, when you give the statement an offset of 3, it knows to jump ahead 3 to find the correct branch.
Lookup switch uses a binary search to find the correct code branch. This runs in O(log n) time, which is still good, but not the best.
For more information on this, see here: Difference between JVM's LookupSwitch and TableSwitch?
So as far as which one is fastest, use this approach:
If you have 3 or more cases whose values are consecutive or nearly consecutive, always use a switch.
If you have 2 cases, use an if statement.
For any other situation, switch is most likely faster, but it's not guaranteed, since the binary-search in LookupSwitch could hit a bad scenario.
Also, keep in mind that the JVM will run JIT optimizations on if statements that will try to place the hottest branch first in the code. This is called "Branch Prediction". For more information on this, see here: https://dzone.com/articles/branch-prediction-in-java
Your experiences may vary. I don't know that the JVM doesn't run a similar optimization on LookupSwitch, but I've learned to trust the JIT optimizations and not try to outsmart the compiler.

So if you plan to have loads of packets memory isn't really a large cost these days and arrays are pretty fast. You also cannot rely on a switch statement to auto generate a jump table and as such it's easier to generate the jump table scenario yourself. As you can see in below example we assume a maximum of 255
packets.
To get the below result your need abstraction.. i'm not going to explain how that works so hopefully you have an understanding of that.
I updated this to set the packet size to 255 if you need more then that you'll have to do a bounds check for (id < 0) || (id > length).
Packets[] packets = new Packets[255];
static {
packets[0] = new Login(6);
packets[2] = new Logout(8);
packets[4] = new GetMessage(1);
packets[8] = new AddFriend(0);
packets[11] = new JoinGroupChat(7); // etc... not going to finish.
}
public void handlePacket(IncomingData data)
{
int id = data.readByte() & 0xFF; //Secure value to 0-255.
if (packet[id] == null)
return; //Leave if packet is unhandled.
packets[id].execute(data);
}
Edit since I use a Jump Table in C++ a lot now i'll show an example of a function pointer jump table. This is a very generic example, but I did run it and it works correctly. Keep in mind you must set the pointer to NULL, C++ will not do this automatically like in Java.
#include <iostream>
struct Packet
{
void(*execute)() = NULL;
};
Packet incoming_packet[255];
uint8_t test_value = 0;
void A()
{
std::cout << "I'm the 1st test.\n";
}
void B()
{
std::cout << "I'm the 2nd test.\n";
}
void Empty()
{
}
void Update()
{
if (incoming_packet[test_value].execute == NULL)
return;
incoming_packet[test_value].execute();
}
void InitializePackets()
{
incoming_packet[0].execute = A;
incoming_packet[2].execute = B;
incoming_packet[6].execute = A;
incoming_packet[9].execute = Empty;
}
int main()
{
InitializePackets();
for (int i = 0; i < 512; ++i)
{
Update();
++test_value;
}
system("pause");
return 0;
}
Also another point i'd like to bring up is the famous Divide and Conquer. So my above 255 array idea could be reduced to no more then 8 if statements as a worst case scenario.
I.e. but keep in mind it get's messy and hard to manage fast and my other approach is generally better, but this is utilize in cases where arrays just won't cut it. You have to figure out your use case and when each situation works best. Just as you wouldn't want to use either of these approaches if you only have a few checks.
If (Value >= 128)
{
if (Value >= 192)
{
if (Value >= 224)
{
if (Value >= 240)
{
if (Value >= 248)
{
if (Value >= 252)
{
if (Value >= 254)
{
if (value == 255)
{
} else {
}
}
}
}
}
}
}
}

At the bytecode level, subject variable is loaded only once into processor register from a memory address in the structured .class file loaded by Runtime,and this is in a switch statement; whereas in an if-statement, a different jvm instruction is produced by your code-compiling DE, and this requires that each variable be loaded in to registers although same variable is used as in next preceeding if-statement. If you know of coding in assembly language then this would be commonplace; although java compiled coxes are not bytecode, or direct machine code, the conditional concept hereof is still consistent.
Well, I tried to avoid deeper technicality upon explaining. I hope I had made the concept clear and demystified. Thank you.