Memoization with Vavr seems incosistent

Question

When the function is defined as following

static Function1<BigInteger, BigInteger> fibonacci = Function((BigInteger value) ->
            value.equals(BigInteger.ZERO) ? BigInteger.ZERO
                    : value.equals(BigInteger.ONE) ? BigInteger.ONE
                    : value.equals(BigInteger.valueOf(2)) ? BigInteger.ONE
                    : Program.fibonacci.apply(value.subtract(BigInteger.ONE)).add(Program.fibonacci.apply(value.subtract(BigInteger.valueOf(2))))
    ).memoized();

And called

System.out.println(fibonacci.apply(BigInteger.valueOf(1000)));

It calculated very fast. However if I move the memoized() to function variable as follows

static Function1<BigInteger, BigInteger> fibonacci = Function((BigInteger value) ->
            value.equals(BigInteger.ZERO) ? BigInteger.ZERO
                    : value.equals(BigInteger.ONE) ? BigInteger.ONE
                    : value.equals(BigInteger.valueOf(2)) ? BigInteger.ONE
                    : Program.fibonacci.apply(value.subtract(BigInteger.ONE)).add(Program.fibonacci.apply(value.subtract(BigInteger.valueOf(2))))
    ); // Removed memoized() from here

And called

fibonacci.memoized().apply(BigInteger.valueOf(1000));

It takes very long as if memoized() was not applied.

What might be the reason for that?

Because a) the recursion isn't called on the memoized form, b) the whole point of memoizing is that you need to save the memoization, not create a new memoization each time. — Louis Wasserman, Nov 02 '17 at 19:32
I guess I got it. My Second example is memoized for only 1000. The other values are called on raw **fibonacci()**. Can you answer the question so that I accep it? — ozgur, Nov 02 '17 at 19:38
Why you don't use Binet's formula for calculating fibonacci? http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html — Samir, Nov 02 '17 at 21:37
@SamirAghayarov I guess this is more about experimenting functional programming using Fibonacci as a toy example, rather than a genuine need for the function's values. — Nándor Előd Fekete, Nov 03 '17 at 03:37

score 2 · Accepted Answer · answered Nov 02 '17 at 20:02

Because a) the recursion isn't called on the memoized form, b) the whole point of memoizing is that you need to save the memoization, not create a new memoization each time.

Program.fibonacci is defined in terms of itself, so the recursion calls that version, not the memoized version.

Memoization with Vavr seems incosistent

1 Answers1