For non-negative integers m and n a function is defined as follows: f(m,n)={n+1,ifm=0andf(m−1,1)ifm≠0,n=0andf(m−1,f(m,n−1))ifm≠0,n≠0}Then the value of f(1,1) is:

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to evaluate the function \( f(1, 1) \) based on the definition provided. The function is defined as follows:

\[ f(m, n) = \begin{cases} n + 1 & \text{if } m = 0 \\ f(m - 1, 1) & \text{if } m \neq 0 \text{ and } n = 0 \\ f(m - 1, f(m, n - 1)) & \text{if } m \neq 0 \text{ and } n \neq 0 \end{cases} \]

### Step 1: Determine the case for \( f(1, 1) \)

Since \( m = 1 \) and \( n = 1 \), we fall into the third case of the function definition because both \( m \) and \( n \) are not zero:

\[ f(1, 1) = f(1 - 1, f(1, 1 - 1)) = f(0, f(1, 0)) \]

### Step 2: Evaluate \( f(1, 0) \)

Next, we need to evaluate \( f(1, 0) \). Here, \( m = 1 \) and \( n = 0 \), so we use the second case of the function:

\[ f(1, 0) = f(1 - 1, 1) = f(0, 1) \]

### Step 3: Evaluate \( f(0, 1) \)

Now we evaluate \( f(0, 1) \). Here, \( m = 0 \) and \( n = 1 \), so we use the first case of the function:

\[ f(0, 1) = 1 + 1 = 2 \]

### Step 4: Substitute back to find \( f(1, 1) \)

Now we substitute \( f(1, 0) \) back into our expression for \( f(1, 1) \):

\[ f(1, 1) = f(0, f(1, 0)) = f(0, 2) \]

### Step 5: Evaluate \( f(0, 2) \)

Finally, we evaluate \( f(0, 2) \). Here, \( m = 0 \) and \( n = 2 \):

\[ f(0, 2) = 2 + 1 = 3 \]

### Conclusion

Thus, we find that:

\[ f(1, 1) = 3 \]

### Final Answer

The value of \( f(1, 1) \) is **3**.

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