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How do I count up using recursion in Java?

So, I'm auditing a Java course on EdX and have been doing really well up until this point. I've come to recursions and one of the tasks is to create a method that counts up to a specified number placing commas between each (i.e. System.out.println(writeNum(5)); should print "1, 2, 3, 4, 5, ... n" with no comma on the final number).
Additionally, there's supposed to be an IllegalArgumentException should a value less than 1 be passed.
I've been on it for 2 days wracking my brain and I cannot even comprehend how to start. I can do the factorial one :
public static int factorial (int n) {
if(n == 1){
return 1;
}
System.out.println(n);
return n*factorial(n-1);
}
No problems right? So, thinking about it I keep thinking, what is the base case? Is it 1? if so, then I'm just counting down and somehow need to count down, reorganize them, reprint them, then somehow figure it out that way. Or, I make my base case when n==n, which doesn't work either... I know I am probably overthinking this, but I have no idea how to even get it started... If anyone could go through it with me step by step to understand this, I would sincerely be grateful as I'm doing this because I genuinely want to learn and understand this stuff.

For recursion, you need to find when to return e.g. in the code given below, when n == 1, the method prints the value of n and returns. Apart from the terminating condition, another important aspect is where (i.e. whether before calling the method/function recursively or after it) you process (e.g. print) the parameter.
public class Main {
public static void main(String[] args) {
count(5);
}
static void count(int n) {
if (n == 1) {
System.out.print(n);
return;
}
count(n - 1);
System.out.print("," + n);
}
}
Output:
1,2,3,4,5
This is how it works:
count(5) -> calls count(4) with remaining thing to do is print ,5 once count(4) returns.
count(4) -> calls count(3) with remaining thing to do is print ,4 once count(3) returns.
count(3) -> calls count(2) with remaining thing to do is print ,3 once count(2) returns.
count(2) -> calls count(1) with remaining thing to do is print ,2 once count(1) returns.
count(1) -> prints 1 and returns.
The remaining thing of count(2) is done i.e. ,2 is printed.
The remaining thing of count(3) is done i.e. ,3 is printed.
The remaining thing of count(4) is done i.e. ,4 is printed.
The remaining thing of count(5) is done i.e. ,5 is printed.
Check this to learn more about recursion.

Recursive approach to solving the Towers of Hanoi puzzle

I'm trying to work out the "towers of hanoi" problem, which moves stacks of disks organized from smallest to largest from a start stack to a destination stack, by means of an auxiliary stack without ever placing a larger disk on top of a smaller disk.
I was given the algorithim:
If numberDisks == 1:
Display “Move the top disk from S to D”.
Else
Move(numberDisks -1, ‘S’, ‘A’, ‘D’);
Move(1, ‘S’, ‘D’, ‘A’);
Move(numberDisks -1, ‘A’, ‘D’, ‘S’);
However this seems to differ from most other examples which seem to work without using Move(1, ‘S’, ‘D’, ‘A’); in the recursive function.
As my code stands, I seem repeat the base case for every move, and im not sure how to structure my print statements to give the proper output which should look like:
Move disk 1 from S to D
Move disk 2 from S to A
Move disk 1 from D to A
Move disk 3 from S to D
Move disk 1 from A to S
Move disk 2 from A to D
Move disk 1 from S to D
When trying to move 3 disks.
// Recursively solve Towers of Hanoi puzzle
public static void main(String[] args) {
if (handleArguments(args)) {
System.out.println("numDisks is ok");
int numDisks = Integer.parseInt(args[0]);
Move(numDisks,'s', 'a', 'd' );
}
}
// recursive case
public static void Move(int disks, char start, char aux, char destination) {
// base case
if (disks == 1) {
System.out.println("Move disk 1 from S to D");
// if number of disks is 2 or greater
} else if(disks > 1) {
Move(disks - 1, start, aux, destination);
System.out.println("move disk " + disks + " from " + start + " to " + destination);
Move(1, start, destination, aux);
Move(disks - 1, aux, destination, start);
}
}

First thing: Understand the algorithm for moving n discs (from S to D, with A)
If there is only one disc to move, just move it then stop*.
Move n - 1 discs from S to A, with D.
Move nth disc from S to D
Move n - 1 discs from A to D, with S.
*Equally: If there are 0 discs, just stop. (Some would argue that this is the superior termination condition because, in your code, it will prevent the print statement when there is one from being a special case, it will be handled naturally by the print statement you cover step 3 with. When, for example, you decide to change this method to return a list of the steps instead of printing it, this change will only have to be applied in one place)
You mention that "I seem repeat the base case for every move." If you look at your code, and compare to my statements above. You will see that you call Move(1, start, destination, aux); between my steps 3 and 4. This is why you are repeating your base case, because you are explicitly calling, repeating your base case, but it doesn't make any sense to.
The other main problem I can see:
System.out.println("Move disk 1 from S to D"); will always print 'S' and 'D', in that order, when often you will need to specify a different move, ensure to use the arguments for this statement.
I don't think that there is anything else, but try it and see.
In response to the example you were given, in the start of your post. It produces subtly different output than your version.
Yours specifies (or attempts to) which size disc should be moved at each step, the example only specifies which stack to move a disc from and to, regardless of its size.
The recursive call with 1 as its argument in the middle there is to print the move instruction for moving the final disc in the stack (my step 3).

Recursive Pascal's Triangle Row big O cost

I'm studying for CS interviews and I decided to try making up my own problem and solving it recursively.
The question I am trying to solve is this: I want to be able to write a recursive function that finds the nth row of pascal's triangle.
Ex pascals(1) -> 1
pascals(2) -> 1,1
pascals(3) -> 1,2,1
I believe I have solved this function. It takes a helper function to get it started with the base case.
function int[] nthPascal(int[] a, int n){
// in the case that the row has been reached return the completed array
if(n==1){
return a;
}
// each new row of pascal's triangle is 1 element larger. Ex 1, 11, 121,1331 etc.
int[] newA = new int[a.length()+1];
//set the ends. Technically this will always be 1.
// I thought it looked cleaner to do it this way.
newA[0]=a[0];
newA[newA.length-1]=a[a.length-1];
//so I loop through the new array and add the elements to find the value.
//ex 1,1 -> -,2,- The ends of the array are already filled above
for(int i=1; i<a.length; i++){
// ex 1+1 = 2 for 1,1 -> 1,2,1
newA[i]=a[i-1]+a[i]
}
//call the recursive function if we are not at the desired level
return nthPascal(newA,n-1);
}
/**
*The helper function
*/
public int[] helperPascal(int n){
return nthPascal(int[] {1},n);
}
My question is how do I find the bigO cost?
I am familiar with common algorithm costs and how to find them. The recursion in this makes it confusing for me.
I know it's clearly not constant, n, nlogn, etc. My thought was that it was 3^n?
I searched for an example and found:
Pascal's Triangle Row Sequence
and
What is the runtime of this algorithm? (Recursive Pascal's triangle).
But they are trying to find a specific element in a given location I believe. I couldn't find anyone implementing pascal's triangle this way and talking about bigO cost.
Is this because there is a better way to write a recursive row finding pascal's function? If so please share :)

Each time you call nthPascal, it calls itself recursively just once. Therefore, you can get the time complexity by adding the time complexities of each invocation of the function (excluding the recursive call). (If you had a function that traverses a binary tree, it would typically call itself recursively twice, which makes the computation more complicated. Here, though, since it calls itself only once, the computation is pretty simple.)
Each invocation of the function has a loop that executes a.length times, and the body of the loop executes in constant time. There aren't any other loops or any other statements that execute in anything but constant time, except for when you allocate the array intA because Java will intialize every element of the array. The result, though, is that when you call nthPascal with an array of length M, the execution time will be O(M) not counting the recursive call.
So assuming that the execution time is roughly M*k for some constant k, that means the total execution time will be 1*k + 2*k + 3*k + ... + (n-1)*k. And 1 + 2 + 3 + ... + n - 1 is (n * (n - 1)) / 2, which is O(n2). So O(n2) is the answer you're looking for.

For each recursive call, you are doing a for loop of size k, where k is the depth of the recursive call (at depth k, you have an array of size k).
To get the complete row at depth n, you call nthPascal at depth 1, 2, ..., n.
Therefore, your total complexity will be 1+2+...+n=n(n+1)/2 = O(n²).

Java Homework Help - Recursion with array

I've got a set of recursion problems that I need to do. I've completed 3 out of the 4 of them we were given, but I'm having a hard time wrapping my head around this last one. I don't necessarily want the actual answer, but maybe just point me in the right direction, because I'm not even seeing what my stop condition should be on this one. And note, it has to be recursive, no loops, etc.
Thanks in advance for any help provided!
Write recursive method arrayRange that returns the maximum integer minus the minimum integer in the filled array of ints. Use recursion; do not use a loop. The following assertions must pass (note the shortcut way to pass a reference to a new array--it saves your writing a bit of code (this passes an array built as a parameter):
assertEquals(2, rf.arrayRange(new int[] { 1, 2, 3 } ));
assertEquals(2, rf.arrayRange(new int[] { 3, 2, 1 } ));
assertEquals(0, rf.arrayRange(new int[] { 3 } ));
assertEquals(3, rf.arrayRange(new int[] { -3, -2, -5, -4 } ));
// Precondition: a.length > 0
public int arrayRange(int[] a)

The stop condition is when there are only two items left: the maximum and minimum. Then just return the difference. (Also handle the case of 1 or 0 items, consider input such as in the test cases.)
Now .. how to reduce the list each pass? :) I would consider inspecting the first three values at a time (of the three, only two should remain in the recursive step).
Happy homeworking.

How does the recursion here work?

Code 1:
public static int fibonacci (int n){
if (n == 0 || n == 1) {
return 1;
} else {
return fibonacci (n-1) + fibonacci (n-2);
}
}
How can you use fibonacci if you haven't gotten done explaining what it is yet? I've been able to understand using recursion in other cases like this:
Code 2:
class two
{
public static void two (int n)
{
if (n>0)
{
System.out.println (n) ;
two (n-1) ;
}
else
{
return ;
}
}
public static void main (String[] arg)
{
two (12) ;
}
}
In the case of code 2, though, n will eventually reach a point at which it doesn't satisfy n>0 and the method will stop calling itself recursively. In the case of code 2, though, I don't see how it would be able to get itself from 1 if n=1 was the starting point to 2 and 3 and 5 and so on. Also, I don't see how the line return fibonacci (n-1) + fibonacci (n-2) would work since fibonacci (n-2) has to contain in some sense fibonacci (n-1) in order to work, but it isn't there yet.
The book I'm looking at says it will work. How does it work?

Well, putting aside what a compiler actually does to your code (it's horrible, yet beautiful) and what how a CPU actually interprets your code (likewise), there's a fairly simple solution.
Consider these text instructions:
To sort numbered blocks:
pick a random block.
if it is the only block, stop.
move the blocks
with lower numbers to the left side,
higher numbers to the right.
sort the lower-numbered blocks.
sort the higher-numbered blocks.
When you get to instructions 4 and 5, you are being asked to start the whole process over again. However, this isn't a problem, because you still know how to start the process, and when it all works out in the end, you've got a bunch of sorted blocks. You could cover the instructions with slips of paper and they wouldn't be any harder to follow.

In the case of code 2 though n will eventualy reach a point at which it doesnt satisfy n>0 and the method will stop calling itself recursivly
to make it look similar you can replace condition if (n == 0 || n == 1) with if (n < 2)
Also i don't see how the line `return fibonacci (n-1) + fibonacci (n-2) would work since fibbonacci n-2 has to contain in some sense fibonacci n-1 in order to wrok but it isn't there yet.
I suspect you wanted to write: "since fibbonacci n-1 has to contain in some sense fibonacci n-2"
If I'm right, then you will see from the example below, that actually fibonacci (n-2) will be called twice for every recursion level (fibonacci(1) in the example):
1. when executing fibonacci (n-2) on the current step
2. when executing fibonacci ((n-1)-1) on the next step
(Also take a closer look at the Spike's comment)
Suppose you call fibonacci(3), then call stack for fibonacci will be like this:
(Veer provided more detailed explanation)
n=3. fibonacci(3)
n=3. fibonacci(2) // call to fibonacci(n-1)
n=2. fibonacci(1) // call to fibonacci(n-1)
n=1. returns 1
n=2. fibonacci(0) // call to fibonacci(n-2)
n=0. returns 1
n=2. add up, returns 2
n=3. fibonacci(1) //call to fibonacci(n-2)
n=1. returns 1
n=3. add up, returns 2 + 1
Note, that adding up in fibonacci(n) takes place only after all functions for smaller args return (i.e. fibonacci(n-1), fibonacci(n-2)... fibonacci(2), fibonacci(1), fibonacci(0))
To see what is going on with call stack for bigger numbers you could run this code.
public static String doIndent( int tabCount ){
String one_tab = new String(" ");
String result = new String("");
for( int i=0; i < tabCount; ++i )
result += one_tab;
return result;
}
public static int fibonacci( int n, int recursion_level )
{
String prefix = doIndent(recursion_level) + "n=" + n + ". ";
if (n == 0 || n == 1){
System.out.println( prefix + "bottommost level, returning 1" );
return 1;
}
else{
System.out.println( prefix + "left fibonacci(" + (n-1) + ")" );
int n_1 = fibonacci( n-1, recursion_level + 1 );
System.out.println( prefix + "right fibonacci(" + (n-2) + ")" );
int n_2 = fibonacci( n-2, recursion_level + 1 );
System.out.println( prefix + "returning " + (n_1 + n_2) );
return n_1 + n_2;
}
}
public static void main( String[] args )
{
fibonacci(5, 0);
}

The trick is that the first call to fibonacci() doesn't return until its calls to fibonacci() have returned.
You end up with call after call to fibonacci() on the stack, none of which return, until you get to the base case of n == 0 || n == 1. At this point the (potentially huge) stack of fibonacci() calls starts to unwind back towards the first call.
Once you get your mind around it, it's kind of beautiful, until your stack overflows.

"How can you use Fibonacci if you haven't gotten done explaining what it is yet?"
This is an interesting way to question recursion. Here's part of an answer: While you're defining Fibonacci, it hasn't been defined yet, but it has been declared. The compiler knows that there is a thing called Fibonacci, and that it will be a function of type int -> int and that it will be defined whenever the program runs.
In fact, this is how all identifiers in C programs work, not just recursive ones. The compiler determines what things have been declared, and then goes through the program pointing uses of those things to where the things actually are (gross oversimplification).

Let me walkthrough the execution considering n=3. Hope it helps.
When n=3 => if condition fails and else executes
return fibonacci (2) + fibonacci (1);
Split the statement:
Find the value of fibonacci(2)
Find the value of fibonacci(1)
// Note that it is not fib(n-2) and it is not going to require fib(n-1) for its execution. It is independent. This applies to step 1 also.
Add both values
return the summed up value
The way it gets executed(Expanding the above four steps):
Find the value of fibonacci(2)
if fails, else executes
fibonacci(1)
if executes
value '1' is returned to step 1.2. and the control goes to step 1.3.
fibonacci(0)
if executes
value '1' is returned to step 1.3. and the control goes to step 1.4.
Add both
sum=1+1=2 //from steps 1.2.2. and 1.3.2.
return sum // value '2' is returned to step 1. and the control goes to step 2
Find the value of fibonacci(1)
if executes
value '1' is returned
Add both values
sum=2+1 //from steps 1.5. and 2.2.
return the summed up value //sum=3

Try to draw an illustration yourself, you will eventually see how it works. Just be clear that when a function call is made, it will fetch its return value first. Simple.

Try debugging and use watches to know the state of the variable

Understanding recursion requires also knowing how the call stack works i.e. how functions call each other.
If the function didn't have the condition to stop if n==0 or n==1, then the function would call itself recursively forever.
It works because eventually, the function is going to petter out and return 1. at that point, the return fibonacci (n-1) + fibonacci (n-2) will also return with a value, and the call stack gets cleaned up really quickly.

I'll explain what your PC is doing when executing that piece of code with an example:
Imagine you're standing in a very big room. In the room next to this room you have massive amounts of paper, pens and tables. Now we're going to calculate fibonacci(3):
We take a table and put it somewhere in the room. On the table we place a paper and we write "n=3" on it. We then ask ourselves "hmm, is 3 equal to 0 or 1?". The answer is no, so we will do "return fibonacci (n-1) + fibonacci (n-2);".
There's a problem however, we have no idea what "fibonacci (n-1)" and "fibonacci (n-2)" actually do. Hence, we take two more tables and place them to the left and right of our original table with a paper on both of them, saying "n=2" and "n=1".
We start with the left table, and wonder "is 2 equal to 0 or 1?". Of course, the answer is no, so we will once again place two tables next to this table, with "n=1" and "n=0" on them.
Still following? This is what the room looks like:
n=1
n=2 n=3 n=1
n=0
We start with the table with "n=1", and hey, 1 is equal to 1, so we can actually return something useful! We write "1" on another paper and go back to the table with "n=2" on it. We place the paper on the table and go to the other table, because we still don't know what we're going to do with that other table.
"n=0" of course returns 1 as well, so we write that on a paper, go back to the n=2 table and put the paper there. At this point, there are two papers on this table with the return values of the tables with "n=1" and "n=0" on them, so we can compute that the result of this method call is actually 2, so we write it on a paper and put it on the table with "n=3" on it.
We then go to the table with "n=1" on it all the way to the right, and we can immediately write 1 on a paper and put it back on the table with "n=3" on it. After that, we finally have enough information to say that fibonacci(3) returns 3.
It's important to know that the code you are writing is nothing more than a recipe. All the compiler does is transform that recipe in another recipe your PC can understand. If the code is completely bogus, like this:
public static int NotUseful()
{
return NotUseful();
}
will simply loop endlessly, or as in my example, you'll keep on placing more and more tables without actually getting anywhere useful. Your compiler doesn't care what fibonacci(n-1) or fibonacci(n-2) actually do.