Multiple exit points and cyclomatic complexity

I don't understand what is the rationale behind always using a single exit point in a function. I can only see the problem: it may increase cyclomatic complexity.

Compare these two abstract examples and their cyclomatic complexity:

Single exit point

  int f(boolean a, boolean b) {
    // the compiler forces us to have a default value
    // even if we know it will never be used
    // (there is always a default value whether it is
    //  explicit or implicit)
    int value = DEFAULT;
    if( a ) {
      if( b ) {
        value = A_AND_B;
      } else {
        value = A_AND_NOT_B;
      }
    } else {
      value = NOT_A;
    }
    return value;
  }

Multiple exit points

  int f(boolean a, boolean b) {
    if( !a ) {
      return NOT_A;
    }

    // here we know a is true (otherwise we wouldn't be here)
    // so we don't need to increase the cyclomatic complexity by
    // introducing another level of indentation
    // (without a return statement we could never be so confident):
    if( b ) {
      return A_AND_B;
    }

    // here we know both a and b, no need for DEFAULT:
    return A_AND_NOT_B;
  }