Is there a more elegant syntax for these boolean expressions? - c++

Question

I was just writing an improved linear version of a recursive Fibonacci algorithm, and realized that my boolean expressions look really bad and unreadable. Is there a cleaner way to do what I'm trying to do?
int fibonacci(int num) {
if (num <= 1)
return num;
// starts looking ugly here
int a = intExists(num-1);
int b = intExists(num-2);
bool aAndB = (a != -1 && b != -1);
bool justA = (a != -1 && b == -1);
bool justB = (a == -1 && b != -1);
int number = 0;
if (aAndB)
number = (a + b);
else if (justA)
number = (a + fibonacci(num - 2));
else if (justB)
number = (fibonacci(num-1) + b);
else
number = (fibonacci(num - 1) + fibonacci(num - 2));
map.push_back(Pair(num, number));
return number;
}
Thanks

Answer 1

If you're talking about:
bool aAndB = (a != -1 && b != -1);
then I would say, "no."
This code looks perfectly expressive to me. aAndB is initialized at the moment it comes in to being, and the conditions are very clear. This might look a bit odd when you're first starting out in C++, but before you know it it will be second nature and other constructs will seem silly.
One thing I would suggest is to make aAndB const if you don't intend to change it:
const bool aAndB = (a != -1 && b != -1);
This is even more expressive.
It also might give the compiler an additional opportunity to optimize your code.
Remember -- write code for humans to understand. Not for computers to understand.

Answer 2

Why don't you make a and b as bools and assign those as true if a == -1 and false otherwise. Then, the expressions will become easier to handle.

Answer 3

Could do a switch statement to clean up the if else statements a little. Other than that just add comments

Answer 4

You could rewrite it to use conditional branching, like this:
int fibonacci(int num) {
if (num <= 1)
return num;
int a = intExists(num-1);
int b = intExists(num-2);
const bool isA = (a != -1); // change in the definition
const bool isB = (b != -1); // change in the definition
int number = 0;
if (isA && isB)
number = (a + b);
else if (isA) // conditionnal branching
number = (a + fibonacci(num - 2));
else if (isB) // conditionnal branching
number = (fibonacci(num-1) + b);
else
number = (fibonacci(num - 1) + fibonacci(num - 2));
map.push_back(Pair(num, number));
return number;
}

Answer 5

I'm assuming that intExists(n) looks up map and if finds n in there, returns fibonacci(n) else it returns -1. Then you could do this:
int fibonacci(int num) {
if (num <= 1)
return num;
int a = intExists(num-1);
int b = intExists(num-2);
if (a == -1) // if a wasn't found, then compute it
a = fibonacci(num-1);
if (b == -1) // if b wasn't found, then compute it
b = fibonacci(num-2);
int number = a + b;
map.push_back(std::make_pair(num, number));
return number;
}
Bonus:
Here is another completely different implementation of fibonnacci() based on Binet's formula:
#include <cmath>
int fibonacci(int n) {
static const double e1 = 1.6180339887498948482045868343656; // = (1 + sqrt(5)) / 2
static const double e2 = -0.61803398874989484820458683436564; // = (1 - sqrt(5)) / 2
static const double c = 0.44721359549995793928183473374626; // = 1 / sqrt(5);
double f = c * (std::pow(e1, n) - std::pow(e2, n));
return static_cast<int>(f + 0.5);
}
int main() {
for (int n = 1; n < 15; ++n)
std::cout << fibonacci(n) << ' ';
}
It outputs:
1 1 2 3 5 8 13 21 34 55 89 144 233 377

Answer 6

Plain C++ code is clean enough:
bool a = intExists(num-1);
bool b = intExists(num-2);
if (a && b) {
//
} else if (a) {
//
} else if (b) {
//
} else {
//
}

Answer 7

int a = intExists(num-1);
int b = intExists(num-2);
bool aAndB = (a != -1 && b != -1);
bool justA = (a != -1 && b == -1);
bool justB = (a == -1 && b != -1);
Quick look into the approach you took. Under what circumstances can justB be true? (Hint: never)
That should help you simplify your approach, although there are better approaches than memoization.

Answer 8

Changing intExists to return boolean values, you can do a switch-case statements like that:
bool a = intExists(num-1);
bool b = intExists(num-2);
switch ((a << 1) + b) {
default: // none exists
case 1: // only b exist
case 2: // only a exist
case 3: // both exists
}
The rationale is to transform those booleans in a binary number

Answer 9

A slightly drastic rewrite is to let an external function handle the lookup table.
That way you don't need to care about more than one value at a time.
This one uses map so I didn't have to write so much in order to test it, but it should be easy enough to adapt:
std::map<int, int> table;
int fibonacci(int num);
int value(int num)
{
int result = table[num];
if (!result)
{
result = fibonacci(num);
table[num] = result;
}
return result;
}
int fibonacci(int num)
{
if (num <= 2)
return 1;
return value(num - 1) + value(num - 2);
}