Help with func

[flatfish]

I’m taking an online course in Java programming and one of the problems on my open book test is to find the values of func1

Private int func1(int m, int n) {
If (m==n || n==1)
Return 1;
Else
Return func1(m-1,n-1) + n*func1(m-1,n);
}

The thing is I can’t find this anywhere in my text book, and I haven’t had any algebra in 10 years. (if it is indeed algebra and I'm going off on a tangent) If you walk me thru it, don't use the values of 5, 3 because I don't want the answer.
Thank you

[r035198x]

I’m taking an online course in Java programming and one of the problems on my open book test is to find the values of func1

Private int func1(int m, int n) {
If (m==n || n==1)
Return 1;
Else
Return func1(m-1,n-1) + n*func1(m-1,n);
}

The thing is I can’t find this anywhere in my text book, and I haven’t had any algebra in 10 years. (if it is indeed algebra and I'm going off on a tangent) If you walk me thru it, don't use the values of 5, 3 because I don't want the answer.
Thank you

Why don't you walk us through what you think first and then people can correct you if you are wrong.

[flatfish]

OK, the first part is easy if both values equal each other or the second value equals 1 than "1" is returned until some other values are inputed until the "else" is run.
So lets say the values are 4 and 2
(m-1,n-1) + n*func1(m-1,n)
(3,1) + 2*func1(3,2)
since there's no space between the comas which isn't normally done in Java, I assume the're multiplied.
(3) + 2*func1(6)
(3) + 2*4, 2(6)
(3) + 8 , 12
11 ,12
23 ?

[r035198x]

OK, the first part is easy if both values equal each other or the second value equals 1 than "1" is returned until some other values are inputed until the "else" is run.
So lets say the values are 4 and 2
(m-1,n-1) + n*func1(m-1,n)
(3,1) + 2*func1(3,2)
since there's no space between the comas which isn't normally done in Java, I assume the're multiplied.
(3) + 2*func1(6)
(3) + 2*4, 2(6)
(3) + 8 , 12
11 ,12
23 ?

You need to look at a recursion tutorial.
In func1(3,1) + 2*func1(3,2), 3 and 1 are not multiplied together. func1(3, 1) means call the function func1 with values 3 and 1.

func1(3, 1) + 2*(func1(3, 2))

now for func1(3, 1), n == 1
so func1(3, 1) = 1

You now have

1 + 2*(func1(2, 1) + 2*(func1(2, 2))

Which you should now see is

1 + 2*(1 + 2*(func1(1, 1) + 2*(func1(1, 2))))

= 1 + 2*(1 + 2 + 2)

[flatfish]

Thanks for the help. That makes sense. I figured I was going down the wrong path. My text book touches on recursion briefly, but never had any examples like the one depicted.