Question 50

The procedure is:

void find_and_replace (char *A, char *oldc, char *newc) {
    for (int i=0; i<5; i++)
        for (int j=0; j<3; j++)
            if (A[i] == oldc[j]) A[i] = newc[j];
}

The intended specification is to replace each character in $A$ matching oldc[j] with newc[j] exactly once.
However, the inner loop over $j$ continues even after a match occurs and $A[i]$ is modified. If the newly written character newc[j] matches a subsequent oldc[k] with $k > j$, the character $A[i]$ will be erroneously replaced a second time during the same pass!

Therefore, a test case can expose the flaw if and only if there exist indices $j < k$ such that:

Let us check each of the four test cases:

1. Test case 1: oldc = "abc", newc = "dab"

   • $j = 0$: oldc[0] = 'a', newc[0] = 'd'. 'd' is not in oldc[1..2] ('b', 'c').
   • $j = 1$: oldc[1] = 'b', newc[1] = 'a'. 'a' is not in oldc[2] ('c').
   • No cascading match exists for any input string. Flaw is NOT captured.

2. Test case 2: oldc = "cde", newc = "bcd"

   • $j = 0$: oldc[0] = 'c', newc[0] = 'b'. 'b' is not in oldc[1..2] ('d', 'e').
   • $j = 1$: oldc[1] = 'd', newc[1] = 'c'. 'c' is not in oldc[2] ('e').
   • No cascading match exists. Flaw is NOT captured.

3. Test case 3: oldc = "bca", newc = "cda"

   • $j = 0$: oldc[0] = 'b', newc[0] = 'c'.
   • Notice that for $k = 1$, oldc[1] = 'c'! Here $j < k$ and newc[0] == oldc[1].
   • If $A[i] = \text{'b'}$, at $j = 0$ it is replaced by 'c', and then at $j = 1$ it matches oldc[1] and is replaced again by 'd'. The flaw is exposed!

4. Test case 4: oldc = "abc", newc = "bac"

   • $j = 0$: oldc[0] = 'a', newc[0] = 'b'.
   • Notice that for $k = 1$, oldc[1] = 'b'! Here $j < k$ and newc[0] == oldc[1].
   • If $A[i] = \text{'a'}$, at $j = 0$ it becomes 'b', and then at $j = 1$ it matches oldc[1] and becomes 'a' again. The flaw is exposed!

Thus, exactly two test cases (Test cases 3 and 4) will be able to capture the flaw. Option (B) is correct.