# Investment strategies, lazy evaluation and memoization

This article will cover an interesting problem: given a set of possible investments, each with different tax rates, yearly rates and minimum time until withdrawal, what is the best investment strategy for the next 10, 20 or n years?

For instance, given the following investments:

• $i_1 = 9\%$ yearly rate, 25% taxes on profits upon withdrawal 1 year later
• $i_2 = 8\%$ yearly rate, 15% taxes on profits upon withdrawal 5 years later
• $i_3 = 7\%$ yearly rate, 0% taxes and withdrawal 3 years later

If we want to maximize earnings over 10 years, should we purchase $i_1$ ten times, $i_2$ twice, $i_3$ three times and one $i_1$, $i_3$ once and $i_2$ once and $i_1$ twice or some other combination?

Before we go into programming, let’s do some basic math/algorithms. This is all simple math, so don’t worry. You can also skip to programming if you prefer.

Theorem 1: The final value to withdraw with any investment i such as the ones exemplified can be written as product of the initial value and a factor defined by the investment i, $e(i) = 1 + (1 - taxes) \cdot ((1 + yearlyRate)^{time} - 1)$

Proof: Given some initial value v and composite yearly interest rates r, investment time t and taxes on profits taxes:
$profitsBeforeTaxes = v \cdot (1 + r)^t - v = v \cdot ((1 + r)^t - 1)$
$profitsAfterTaxes = profitsBeforeTaxes \cdot (1 - taxes)$
$profitsAfterTaxes = v \cdot ((1 + r)^t - 1) \cdot (1 - taxes)$

$finalValue = v + profitsAfterTaxes = v + v \cdot ((1 + r)^t - 1) \cdot (1 - taxes)$
$finalValue = v \cdot (1 + ((1 + r)^t - 1) \cdot (1 - taxes))$

The previous line proves the existence of the factor. Now, since $finalValue=v \cdot e(i)$:

$e(i) = finalValue / v = 1 + (1 - taxes) \cdot ((1 + r)^t - 1)$

Theorem 2: Given a set of possible investments and a deadline $n$, the best investment strategy $s_n=e(i_1) \cdot e(i_2) \cdot ... \cdot e(i_j)$, any sub-strategy contained in $s_n$ is the best strategy for the sum of the times of the investments contained in it.

Proof: Without loss of generality, let us consider $s_n=s_a \cdot s_b$, with $a+b=n$ and $0 and assume the contrary: $s_a$ is not the best strategy for time $a$, but $s_n$ is the best strategy for time $n$. So there must be $s'_a > s_a$, and that would mean $s'_n = s'_a \cdot s_b > s_n$, which contradicts $s_n$ being the best strategy. Therefore, there can be no $s'_a > s_a$ and $s_a$ is optimal.

### Programming

Maybe you skipped the last part, but don’t worry. I’ll just roll out the recursive solution to the problem. How do we describe the list of investments to be made that maximizes earnings after some time n?

$s_0 = 1$
$s_n = max \{s_1 \cdot s_{n-1}, s_2 \cdot s_{n-2}, ..., s_{n-1} \cdot s_1, i \}\text{, with } i =\text{ investment with largest }e(i)\text{ of all possible investments of } time=n$

This basically means we test every possible combination, which is not very smart, of course. The advantage of finding a recursive solution is that we can compute and store calculations for use later on. This is what we call memoization.

Also, we only need to check up to $s_{floor(n/2)} \cdot s_{n - floor(n/2)}$, since every other check is redundant.

How do we implement this? In a procedural language we could use an array of size n and fill it up with all solutions from 0 to n. In Haskell we generally don’t want to use mutable data structures nor do we want to specify the order of evaluation of things, so we must find another tool in the toolbox to do this; it turns out that lazy evaluation is just that!

Lazy evaluation is, roughly speaking, a mechanism by which a value is only computed when required by some function. This means that we can define a data structure in terms of itself, and that it can even be infinite. Take the following example:

repeat :: a -> [a]
repeat x = x : repeat x

What happens here is that the function repeat takes an object of some type a and returns a possibly infinite list of a. The list will grow in size as more elements of it are demanded by evaluation. Let us take this idea to implement our bestStrategyFunctionBad:

— We use the “investment” below to make sure the algorithm always returns some strategy, even if it means leaving you money in the bank
investmentLeaveInTheBank :: Investment
investmentLeaveInTheBank = Investment { name = “Leave it in the bank account”, rate = 0, taxes = 0.0, time = 1 }

withMax :: Ord a => (b -> a) -> [b] -> Maybe b
withMax f xs = snd maybeRes
where maybeRes = foldl’ (\acc el ->case acc of
Just (maxVal, maxEl) -> let cmp = f el in if cmp > maxVal then Just (cmp, el) else acc
Nothing -> Just (f el, el)) Nothing xs

withMax1 :: Ord a => (b -> a) -> b -> [b] -> b
withMax1 f firstEl xs = snd $foldl’ (\acc@(maxVal, _) el -> let cmp = f el in if cmp > maxVal then (cmp, el) else acc) (f firstEl, firstEl) xs bestStrategyBad :: Int -> [Investment] -> [Investment] bestStrategyBad timeInYears invs’ = go !! timeInYears where invs = investmentLeaveInTheBank : invs’ factorStrategyBad is = product$ fmap factorInvestment is
bestStrat desiredTime = withMax1 factorStrategyBad (maybeToList (bestInvestmentWithTime desiredTime)) (allCombinations desiredTime)
bestInvestmentWithTime desiredTime = withMax factorInvestment $filter (\i -> time i == desiredTime) invs — For desiredTime=7 “allCombinations” returns strategies e1 ++ e6, e2 ++ e5 and e3 ++ e4 allCombinations desiredTime = let halfTheTime = floor (fromIntegral desiredTime / 2) in fmap (\i -> go !! i ++ go !! (desiredTime – i)) [1..halfTheTime] go :: [[Investment]] go = [] : fmap bestStrat [1..] There is nothing magical about the code above. When demanding $\text{go !! 20}$, for instance, the function bestStrat will be called with the value 20, which will demand all possible strategy investments (as defined by our equations). Demanding all combinations will once again require $\text{go !! 19}$, $\text{go !! 18}$ and many others, which will repeat the process for a smaller n (the fact that they are smaller is crucial for our recursion to converge). What is different from recursion in imperative languages is that go is not a function: it is a list whose values are lazily calculated. As values are demanded from it, they are calculated only once, so you don’t have to worry about what order to evaluate things in. In C# this is sort of like a List<Lazy<Investment[]>>. This is nice! Still, there are two bad things about this solution: 1. go is a list, so accessing $\text{go !! n}$ is $O(n)$. If this were an array this would be better. We will not tackle this issue for now, but feel free to do so! 2. We are creating a large number of lists with the (++) function, not to mention that once we combine two strategies we have to go through every investment in the combined strategy to calculate its complete factor, when we could do better. So now let’s go and solve issue number 2. ### More lazy evaluation $\text{and a little abstraction}$ So, how can we solve issue number 2? Combining two strategies leads to a strategy with a factor that is the product of the factors of each strategy. There is no need to concatenate lists to discover the best strategy of some given size. To avoid needless work, we need more lazy evaluation. Let’s add some functions to our code and create the bestStrategyGood function: data StrategyCalc = StrategyCalc [Investment] Double factorStrategyGood (StrategyCalc _ x) = x combine :: StrategyCalc -> StrategyCalc -> StrategyCalc combine (StrategyCalc s1 f1) (StrategyCalc s2 f2) = StrategyCalc (s1 ++ s2) (f1 * f2) bestStrategyGood :: Int -> [Investment] -> [Investment] bestStrategyGood timeInYears invs’ = let StrategyCalc res _ = go !! timeInYears in res where invs = investmentLeaveInTheBank : invs’ bestStrat desiredTime = withMax1 factorStrategyGood (bestInvestmentWithTimeOr1 desiredTime) (allCombinations desiredTime) bestInvestmentWithTimeOr1 desiredTime = case withMax factorInvestment$ filter (\i -> time i == desiredTime) invs of
Nothing -> StrategyCalc [] 1
Just i -> StrategyCalc [i] (factorInvestment i)
— For desiredTime=7 “allCombinations” returns strategies e1 ++ e6, e2 ++ e5 and e3 ++ e4
allCombinations desiredTime = let halfTheTime = floor (fromIntegral desiredTime / 2) in fmap (\i -> combine (go !! i) (go !! (desiredTime – i))) [1..halfTheTime]
go :: [StrategyCalc]
go = StrategyCalc [] 1 : fmap bestStrat [1..]

Take your time to digest this: the list of investments in each StrategyCalc will only be evaluated when the caller needs it to be evaluated. However, the combine function will create a StrategyCalc whose factor is calculated in constant time when combining two strategies. In fact, you could even have the final factor of the optimal strategy without having ever constructed a non empty list. Nice!

### A little abstraction $\text{(skip to results if you prefer)}$

I thought a nice touch to finish this article would be to introduce an abstraction: the Monoid.

A Monoid is just a fancy name for a binary operation that is associative and a value that is an identity for this operation. The Int type, the sum function (+) and the value $0$ (zero) form an instance of Monoid, for instance, since any number plus zero equals itself and $(a+b)+c=a+(b+c)$ for any $a, b, c$ of type Int.

The same thing happens with investment strategies! So we can replace the combine function by the Monoidal append:

instance Monoid StrategyCalc where
mempty = StrategyCalc [] 1
mappend (StrategyCalc i1 f1) (StrategyCalc i2 f2) = StrategyCalc (i1 ++ i2) (f1 * f2)

Don’t forget that <> is an infix alias for mappend!

### Results

When taking the code for bestStrategyGood and the three investments from the beginning of the article, let us devise the best strategy to maximize gains over the next 11 years:

$ghci$ :l Investments.hs
ghci> let availableInvestments = [ Investment { name = “Investment 1”, rate = 0.09, taxes = 0.25, time = 1 } , Investment { name = “Investment 2”, rate = 0.08, taxes = 0.15, time = 5 } , Investment { name = “Investment 3”, rate = 0.07, taxes = 0, time = 3 } ]
ghci> fmap name \$ bestStrategyGood 11 availableInvestments
[“Investment 3″,”Investment 3″,”Investment 2”]

So it seems that buying investment 3, rebuying it and then buying investment 2 is the best strategy in this case.

That’s it! I hope you liked it, and if it helps, do know that this problem is still solvable with the same algorithm if the tax of each investment is a function of the amount of time since the investment title was purchased and if the time until withdrawal is either an exact time or a minimum time. It is also possible to include inflation-correcting investments if you pass around some estimated inflation; all of this with only minor modifications. Also, feel free to change the time unit to months and get something much more precise for your investments!

# Interfaces and typeclasses: Number APIs in C# and Haskell

In C# sometimes I sorely miss something like an INumber<T> interface with methods Add, Subtract, Multiply and others. The lack of this means it is cumbersome to write generic code on numbers. It means that instead of writing something like:

T Sum(IEnumerable<T> numbers) where T : INumber<T>;


We have to write all possible overloads:

double Sum(IEnumerable<double> numbers);
float Sum(IEnumerable<float> numbers);
decimal Sum(IEnumerable<decimal> numbers);


The implementation body of all these functions will be exactly the same, but we have to write it multiple times anyway. Some people work this out by creating generic methods for the operations they need while resorting to runtime type-checking:

T Add(T a, T b) {
if (a is double) {
return (double)a + (double)b;
}
else if (a is int) {
return (int)a + (int)b;
}
// .. and so on
}


This is not a terribly good solution, however, since we have no compile-time guarantees that the type T is a number at all, not to mention the performance costs of runtime type-checking. What’s more, this solution can’t be extended to new types; what if someone writes a Complex class to represent complex numbers? The implementation of Add would have to be open for modification, so this code could never be packaged in a library.

Sadly, it is hard to solve this conundrum without changing the Base Class Libraries themselves. The only thing we could hope for is for the people at Microsoft to design numeric interfaces such as INumber<T> (or others) and make our well-known primitive numeric types implement these interfaces.

Haskell is the programming language I’ve been playing with for the last year, and except for the steep learning curve, I have only good things to say. It can be extremely expressive and it is amusing to see that when my code builds it almost certainly works! It is also extremely terse, as you can easily see by this window manager’s less than 2000 lines of code, xmonad, and by the code on this post.

In Haskell, the problem shown above can be solved with typeclasses, which we can think of for now as something similar to interfaces, since they specify a contract that concrete types must obey. The big difference here is that when we create a typeclass, we can make types we don’t own implement (in Haskell: instantiate) it! This means we can design our numeric typeclasses and have Haskell’s standard numeric types, such as Data.Int and Data.Complex, instantiate them! What’s more: in Haskell we can create functions named “+”, “*”, “/”, “-” with infix application. No need to differentiate operators from regular functions: they are one and the same!

module Numeric where -- "Numeric" will be the namespace in which the definitions below will live

import qualified Prelude -- The prelude is a base set of types, typeclasses and functions that are used for common tasks

-- The "class" construct actually creates a typeclass (similar to an interface). Here we say that concrete types that instantiate this typeclass must implement functions called "+" and "*", both of them receiving two parameters of type "t" and returning an object of type "t" as well
class Number t where
(+) :: t -> t -> t
(*) :: t -> t -> t
zero :: t

-- Specifies the type "Int", which we don't own, as an instance of "Number". The Add and Multiply functions already exist in Haskell inside the Prelude. We'll use those.
instance Number Prelude.Int where
a + b = (Prelude.+) a b
a * b = (Prelude.*) a b
zero = 0

-- Now we can define a generic "sum" function with sums all numbers in a list. The following line says that type "t" must be an instance of the typeclass "Number", and that it receives a list of t and returns t. There are better ways to write this in Haskell, but that is not important right now
sum :: Number t => [t] -> t
sum [] = zero -- This is our base case: empty list sums to zero
sum (x:xs) = x + sum xs -- This separates the first element in the list, "x", from the remaining list, "xs"


Note to the reader: Haskell’s prelude already comes with a Num typeclass with more than just addition and multiplication, and existing numeric types already implement those.

And that’s it! The syntax up there really is that short, and it really is type-checked! Also, it only scratches the surface of Haskell is capable of. Believe me, just a tiny scratch.

It is important to notice here that in C# it is entirely possible to make new types that can be added to existing types by defining a public static T operator +(T a, T2 b) in the new type T. What we can’t do is specify generic type constraints that allow us to work with numeric types. In reality, this is not just about numeric APIs: it is just a consequence of the fact that we can’t make types we don’t own implement interfaces, combined to the fact that parametric polymorphism only allows restrictions based on subclassing or interface implementation (with the exception of the new(), struct and class constraints).

It is not hard to think of how useful typeclasses can be. Why doesn’t IList and ICollection implement IReadOnlyCollection anyways? Maybe we want both StringBuilder and System.String to implement IString, allowing for generic code that doesn’t need to convert between one and another. There are many possibilities out there.

Let’s take this a little further, because  it can get pretty interesting: how about subtraction? In C# we can subtract a TimeSpan from a DateTime and get another DateTime. Subtracting an int from an int, however, yields another int. Can we encode this information in Haskell in a way that is checked by the compiler itself, allowing us to write generic code that is able to subtract one object from another? The answer is yes.

More: can we develop a set of typeclasses that makes sure that arithmetic operations will NEVER overflow? This would really help us write banking software, for example, allowing us to add and subtract enormous values without worrying about it. With two language extensions called MultiParamTypeClasses and TypeFamilies we can!

{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE TypeFamilies #-}
module Numeric where

import qualified Prelude
import Prelude (Int, Integer, toInteger, negate) -- These types and functions will be available without the need to be prefixed by "Prelude."
import Data.Time

class Subtractive t1 t2 where
type Difference t1 t2 :: * -- This is just a fancy way of saying that the combination of "Difference" and two types is meant to represent another type
(-) :: t1 -> t2 -> Difference t1 t2

-- In Haskell the type "Integer" represents arbitrarily large integers. They are like "BigInteger" in .NET or Java. Our implementation of (-) is exactly the same as Prelude's in this case.
instance Subtractive Integer Integer where
type Difference Integer Integer = Integer
(-) = (Prelude.-)

instance Subtractive Int Int where
type Difference Int Int = Integer;-- Here we say that the Difference between two Ints is an Integer, because no matter how small two Ints are, their difference is always representable as an Integer
a - b = (toInteger a) - (toInteger b)

-- Let's just enjoy ourselves a little and put in some date and time types in the mix
instance Subtractive UTCTime NominalDiffTime where
type Difference UTCTime NominalDiffTime = UTCTime
a - b = addUTCTime (negate b) a

-- The function below works for any two types t1 and t2 which allow for (t1 - t2). It takes in a list of tuples and returns a list of the differences between the two elements in each tuple.
someGenericDifferenceFunction :: Subtractive t1 t2 => [(t1, t2)] -> [Difference t1 t2]
someGenericDifferenceFunction [] = []
someGenericDifferenceFunction ((a, b) : xs) = (a - b) : someGenericDifferenceFunction xs -- Quick reminder: ":" is a function that takes an element and a list and prepends the element into the list


The example above is still incomplete: we need instances of Subtractive Integer Int and Subtractive Int Integer for this API to become more practical. This is left to the reader, however. Meanwhile, let’s try this out in ghci:

 terminal> ghci ghci> :l Numeric.hs *Numeric> let list = [(1, 2), (5, 5), (10, 3)] :: [(Int, Int)] -- We need to specify the type of "list" because literals could be any type that implements "Num", including Int and Integer. *Numeric> let difs = someGenericDifferenceFunction list *Numeric> difs [-1,0,7] *Numeric> :t difs difs :: [Integer] 

C# is great and a lot of what we achieved with Haskell could be achieved through an ISubtractive<T, T2, TResult>, if only we could make existing types implement it. We could also create structs that simply wrap existing types and write implicit coercion rules from (and to) them, making these new types implement our custom interfaces, and make a lot of things possible with that, but we wouldn’t be able to pass an instance of IEnumerable<WrapperType> as a replacement for a IEnumerable<WrappedType> without explicit casting, for example, and we’d also have to watch out and stay away from built-in arithmetic operators, since they might no longer obey the relations between types specified through the interfaces (we might want int + int = BigInteger).

So that’s it. I hope you’ve enjoyed your reading, but most of all I hope any C# or Java (or any other mainstream language) developer that reads this gives Haskell a shot. It really is an amazing language.

Any comments and corrections are very welcome!