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    <p>Here's a straightforward translation of the formula to Haskell: </p> <pre><code>fib n = round $ (phi^n - (1 - phi)^n) / sqrt 5 where phi = (1 + sqrt 5) / 2 </code></pre> <p>This gives correct values only up to <code>n = 75</code>, because it uses <code>Double</code> precision floating-point arithmetic.</p> <p>However, we can avoid floating-point arithmetic by working with numbers of the form <code>a + b * sqrt 5</code>! Let's make a data type for them:</p> <pre><code>data Ext = Ext !Integer !Integer deriving (Eq, Show) instance Num Ext where fromInteger a = Ext a 0 negate (Ext a b) = Ext (-a) (-b) (Ext a b) + (Ext c d) = Ext (a+c) (b+d) (Ext a b) * (Ext c d) = Ext (a*c + 5*b*d) (a*d + b*c) -- easy to work out on paper -- remaining instance methods are not needed </code></pre> <p>We get exponentiation for free since it is implemented in terms of the <code>Num</code> methods. Now, we have to rearrange the formula slightly to use this.</p> <pre><code>fib n = divide $ twoPhi^n - (2-twoPhi)^n where twoPhi = Ext 1 1 divide (Ext 0 b) = b `div` 2^n -- effectively divides by 2^n * sqrt 5 </code></pre> <p>This gives an exact answer.</p> <hr> <p>Daniel Fischer points out that we can use the formula <code>phi^n = fib(n-1) + fib(n)*phi</code> and work with numbers of the form <code>a + b * phi</code> (i.e. ℤ[φ]). This avoids the clumsy division step, and uses only one exponentiation. This gives a much nicer implementation:</p> <pre><code>data ZPhi = ZPhi !Integer !Integer deriving (Eq, Show) instance Num ZPhi where fromInteger n = ZPhi n 0 negate (ZPhi a b) = ZPhi (-a) (-b) (ZPhi a b) + (ZPhi c d) = ZPhi (a+c) (b+d) (ZPhi a b) * (ZPhi c d) = ZPhi (a*c+b*d) (a*d+b*c+b*d) fib n = let ZPhi _ x = phi^n in x where phi = ZPhi 0 1 </code></pre>
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    1. POFibonacci's Closed-form expression in Haskell
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    1. CO+1 for the algebraic number field idea
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    2. COYes, hurray for ℤ[√5] :)
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    3. COIs this the same trick used [here](http://stackoverflow.com/questions/4327846/calculating-fibonacci/4328194#4328194) ?
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