C++ Recursion I need help?

Not one but two non-negative integers times and ful are equivalent if possibly:
Each are 0, or
x-1 in addition to y-1 usually are equal
Write the bool -function termed equals which recursively determines whether it has the two int parameters are matched and income true should they are as well as false if not.

bool compatible (int times, int y)

in the event (x==0 y==0)
returning true;
in addition equals(x-1, y-1);
returning false;

First, it says if *BOTH* usually are zero.You test if either are zero.

Next, you call up equals(x-1, y-1) and then return bogus.That’s not necessarily what the description affirms, it says they are equal if x-1 in addition to y-1 are equal, not that you need to test in the event x-1 as well as y-1 usually are equal and regardless of x along with y are not equal.

Also, you must test if they’re non-negative.Normally, if a person test if 1 as well as 2 will be equal, you can recurse always.Think about this:
1) ONE PARTICULAR, 2 -> possibly not both absolutely nothing, so look at 0, 1
2) 0, ONE PARTICULAR -> possibly not both absolutely nothing, so look at -1, 0
3) -1, 0 -> not both zero, so look at -2, -1
In addition to on permanently!

So putting everthing together:
bool compatible (int times, int y)

if( (x<0) (y<0) ) returning false;
if( (x==0) && y==0) ) returning true;
returning equals(x-1, y-1);

Change the actual function like this:

bool compatible (int times, int y)

in the event ((x==0 ALONG WITH y! =0) (x! =0 IN ADDITION TO y==0) )
returning false;
in the event (x==0 ALONG WITH y==0)
returning true;
in addition equals(x-1, y-1);
//return false;

This will likely recurse the actual function.Say the 2 main numbers usually are 3 in addition to 2.
Rec1:x:2, y:1
Rec2:x:1, y:0 returning FALSE

Say the 2 main numbers usually are 2 in addition to 2.
Rec1:x:1, y:1
Rec2:x:0, y:0 returning TRUE

That’s your silly recursion.

y-1 along with x-1 are equal if and if only y along with x are generally equal.

Are you sure in which you’ve stated the matter correctly

Moreover, if you absolutely must have both to become equal then you certainly want your && (and) certainly not the (or) rational operator.

If you insist on creating a recursion then this is what you would like:

bool equals( int x, int y )
…if( x==0 && y==0 ) returning true;
…in addition return equals(x-1, y-1);

All this will likely do will be keep decrementing both equally variables only there’re both matched.Like MY PARTNER AND I said, this will probably only end up being true iff times = y to begin with.