# [Aaron McDaid](https://aaronmcdaid.github.io/) Statistics and C++, and other fun projects aaron.mcdaid@gmail.com [@aaronmcdaid](https://twitter.com/aaronmcdaid)
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***by [Aaron](https://aaronmcdaid.github.io/), 29th October 2017*** [(code for this document)](index.Rmd)
#   (A demo program of this, with all the code, is available as [countdown.hs](https://raw.githubusercontent.com/aaronmcdaid/aaronmcdaid.github.io/master/blog.posts/countdown.in.haskell/countdown.hs) .) I loved the British game show [*Countdown*](https://en.wikipedia.org/wiki/Countdown_(game_show)) as a child, which is based on [*Des chiffres et des lettres*](https://en.wikipedia.org/wiki/Des_chiffres_et_des_lettres). My first attempt at a non-trivial C program was, if I remember correctly, and attempt to solve the "Numbers Round" on Countdown. Given a target number, and six input numbers, the goal is to combine the inputs using only addition, subtraction, multiplication and division and reach the target value. For example, given target $562$ and inputs $9,8,2,10,4,4$ $$562 = 2+10*(4*4-9)*8$$ Let's assume, for now, that each input must be used once, and only once, in the expression. In reality, it is not necessary to use them all and I will describe that near the end of this post. I'll call this pair, the target and the inputs, a *game*. A given game may have many solutions, or even zero solutions. In this post, I'll describe a program written in [Haskell](https://en.wikipedia.org/wiki/Haskell_(programming_language)). I don't claim to be an expert in Haskell; this is just a bit of fun. I've recently become much more interested in it via the [Haskell Dublin Meetup](https://www.meetup.com/Haskell-Dublin-Meetup/). Any feedback appreciated! aaron.mcdaid@gmail.com or Twitter: [@aaronmcdaid](https://twitter.com/aaronmcdaid) ## Expressions At first, let's consider using just two inputs, instead of the six that are allowed in the real game show. That gives us four expressions, where $a$ and $b$ will take the place of the inputs $$a + b$$ $$a - b$$ $$a \times b$$ $$a \div b$$ Allowing three inputs, the number of expressions increases dramatically, here are some examples $$a + b \times c$$ $$(a + b) \times c$$ $$(c \div a) + b$$ The first step in Haskell is to represent the expressions. There are three kinds of expression: 1. Letter expressions. Just a Single letter in isolation $$a$$ $$b$$ 2. ExprT expressions. Term-expressions involving two terms with addition or subtraction $$a+c$$ $$d-b$$ 3. ExprF expressions. Factor-expressions involving two factors with multiplication and division $$a \times c$$ $$d \div b$$  data Expr = ExprT Terms -- term-expression , i.e. two or more terms separated by (+) or (-) | ExprF Factors -- factor-expression, i.e. two or more factors separated by (*) or (/) | Single Letter -- where 'Letter' (see below) represents 'a' through 'f'.  with a convenient Letter wrapper to store a single character, 'a','b','c','d','e','f':  data Letter = Letter Char -- the Char is between 'a' and 'f' inclusive, for the six allowed variables letter2offset :: Char -> Int -- 'a'=0, 'b'=1, 'c'=2, ... letter2offset l = (fromEnum l) - (fromEnum 'a')  ## Complex expressions More complex expressions are classified according to the 'top-level' operation, respecting the typical [Order of operations](https://en.wikipedia.org/wiki/Order_of_operations) in arithmetic. As a result, this is a term-expression: $$b \times c + a$$ because the addition is performed last. This is more clear with parentheses and spacing $$(b{\times}c) + a$$ or with this tree, where * represents multiplication:  + -- the top-level operation / \ * a / \ b c  On the other hand, the following is a factor-expression: $$b \times (c{+}a)$$  * -- the top-level operation / \ b + / \ c a  ## Recursive definition of Expr Focusing on ExprT, we need to represent the collection of two, or more, terms which are added or subtracted together. Each term is simply a subexpression  data Terms = Terms Expr AddOrSubtract Expr -- | -- ---- second sub-Expr-ession -- | -- \----------------------- first sub-Expr-ession  where AddOrSubtract is simply Add or Sub:  data AddOrSubtract = Add | Sub  You might notice that this only allows two sub-Expressions, when I said "two or more". To allow more than two terms, we allow that the first sub-Expr-ession might also be an ExprT expression. i.e. $$d+a+e$$ is represented as $$(d+a)+e$$  + / \ + + / \ \ d a e  A longer example is $a+b-c+d-e+f$, which we can represent as $((((a+b)-c)+d)-e)+f$. To avoid redundantly generating many expressions which are obviously identical to each other, we ban expression such as $$a*b+(c+d)$$ which are identical to $$(a*b+c)+d$$ To enforce this, we require that the second sub-expression inside a ExprT must **_not_** itself be a Term-expression. It may be Letter or ExprF. The second sub-expression in $a*b+(c+d)$ is $(c+d)$, which is a term-expression and therefore this expression is disallowed ## Constraints To recap, with more clarity about the restriction we've just explained. There are three kinds of Expr: 1. ExprT - a term-expression made up of exactly two sub-Expressions, combined by either addition or subtraction, where the second subexpression must not be a ExprT. 1. ExprF - a factor-expression made up of exactly two sub-Expressions, combined by either multiplication or division, where the second subexpression must not be a ExprF. 1. Letter - a single letter from the six allowed: $a,b,c,d,e,f$ The other constraint, of course, is that a given Letter may not appear more than once, i.e. we don't allow this: $$a*b + a$$  + / \ * a -- 'a' is (Letter 0) / \ a b -- 'a' is (Letter 0), 'b' is (Letter 1)  $a$ (i.e. Letter 0) appears twice in that expression, therefore that expression is invalid. # Recap the data representation To recap the data representation:  data Expr = ExprT Terms -- additive expression, i.e. two or more terms separated by +- | ExprF Factors -- multiplicative expression, i.e. two or more factors separated by +- | Single Letter -- where 'Letter' (see below) represents 'a' through 'f', the first six letters data Terms = Terms Expr AddOrSubtract Expr -- second Expr must *not* be ExprT data Factors = Factors Expr MultiplyOrDivide Expr -- second Expr must *not* be ExprF data AddOrSubtract = Add | Sub data MultiplyOrDivide = Mul | Div data Letter = Letter Char -- the Char is between 'a' and 'f' inclusive, for the six allowed variables letter2offset :: Char -> Int -- 'a'=0, 'b'=1, 'c'=2, ... letter2offset l = (fromEnum l) - (fromEnum 'a')  ## Generating all expressions, subject to those constraints Given three letters $a,b,c$, represented as a list in Haskell with [Letter 0, Letter 1, Letter 2], we want to generate all the valid Expr-essions. We first need to define a useful utility function, allPartitions. Given a list, it will return a list of pairs of lists, with every possible way to split the list into two lists. If there are $n$ items in the original list, there will be $2^n$ items in the list of all partitions.  allPartitions :: [a] -> [([a],[a])] allPartitions [] = [([],[])] allPartitions [x] = [([x],[]),([],[x])] allPartitions (x:xs) = [ (x:l,r ) | (l,r) <- allPartitions xs ] ++ [ (l ,x:r) | (l,r) <- allPartitions xs ]  {haskell} let { allPartitions [] = [([],[])] ; allPartitions [x] = [([x],[]),([],[x])] ; allPartitions (x:xs) = [ (x:l,r ) | (l,r) <- allPartitions xs ] ++ [ (l ,x:r) | (l,r) <- allPartitions xs ] } allPartitions ['a'..'c']  ## allExprs So, allExprs is a function from a list of Letters to a list of Expressions:  allExprs :: [Letter] -> [Expr]  If the list of letters has zero or one elements, the list of expressions is relatively simple  allExprs [] = [] -- no expressions possible allExprs [l] = [Single l] -- just one expression is possible, made up of a Single Letter  But if there are more than two letters in the lists, we have to build a list of all the possible term-expressions and all the possible factor-expressions. Here is the full function before I discuss it a little further:  allExprs :: [Letter] -> [Expr] allExprs [] = [] allExprs [l] = [Single l] allExprs ls = [ ExprT (Terms e1 addOrSubtract e2) | (l,r) <- allPartitions ls , not . null $l , not . null$ r , e2 <- allExprs r , case e2 of -- to enforce that the second Expr in an ExprT is *not* an ExprT (ExprT _) -> False _ -> True , e1 <- allExprs l , addOrSubtract <- [Add, Sub] ] ++ [ ExprF (Factors e1 multiplyOrDivide e2) | (l,r) <- allPartitions ls , not . null $l , not . null$ r , e2 <- allExprs r , case e2 of -- to enforce that the second Expr in an ExprF is *not* an ExprF (ExprF _) -> False _ -> True , e1 <- allExprs l , multiplyOrDivide <- [Mul, Div] ]  A *list-comprehension* in Haskell is [ e | ... ], where the ... are made up of several loop variables and condition expressions separated by commas. For example, to make a list of $x^2$ for each *odd* integer in the range [0..10], you would do  squares_of_some_odd_numbers = [ x^2 | x <- [0..10] , x mod 2 == 1]  Detailed explanation of the above: 1. (l,r) <- allPartitions ls - for each possible partition of ls, store the two sides into l and r. 1. , not . null $l - condition ignoring an empty left-hand-side. 1. , not . null$ r - condition ignoring an empty right-hand-side. 1. , e2 <- allExprs r - recursively call allExprs with the right-hand-side. 1. , case e2 of { (ExprT _) -> False; _ -> True } - skip (via False) when the second sub-expression would be a term-expression. 1. , e1 <- allExprs l - recursively call allExprs with the left-hand-side. 1. , addOrSubtract <- [Add, Sub] - to try both possibilities Each of those items in the list above is either something to loop over (e2,e1, or addOrSubtract) or a condition to skip over some of those. For every combination that is acceptable, we evaluate ExprT (Terms e1 addOrSubtract e2) (or ExprF (Factors e1 multiplyOrDivide e2) when computing all the factor-expressions) and return the entire list. # Evaluating an expression with a given set of inputs With an expression $a+b \times c$ and inputs [10,30,100] we "fill" the expression to get a "filled-expression" $$10+30 \times 100$$ with value $3010$. Bear in mind that we only allow integer division in this game. And therefore, we cannot fill $a \div b$ with inputs [5,2] as $\frac52$ is not an integer and we cannot allow division by zero either. Therefore, our eval function doesn't return Int for the value; instead it returns Maybe Int (known as Optional Int, or similar, in other programming languages).  eval :: [Int] -> Expr -> Maybe Int -- | | | -- | | ------ returned Int, or 'Nothing' if there was a division problem -- | ----------------- expression to evaluate -- ------------------------- a list of integers to take the place of 'a' to 'f' in the expression  We must deal with each of the three kinds of expression. For those not familiar with Haskell's do notation, I will not attempt to explain its many varied and wonderful uses! But, in the context of a function returning Maybe Int, it allows us to 'see through' the Maybe and just directly perform operations on the contained Ints:  eval is (Single (Letter l)) = do return (is !! (letter2offset l))  Next the Term-expression, simply add or subtract. But, remember that eval is e1 and eval is e2 may be Nothing (if there a division problem in the sub-expression), in which case the do will automatically *short-circuit* as the <- will fail to extract and Int from the Maybe Int, and therefore we simply return Nothing without attemping to do any addition or subtraction:  eval is (ExprT (Terms e1 addOrSubtract e2)) = do t1 <- eval is e1 t2 <- eval is e2 case addOrSubtract of Add -> return (t1+t2) Sub -> return (t1-t2)  Multiplication is straightforward:  eval is (ExprF (Factors e1 Mul e2)) = do f1 <- eval is e1 f2 <- eval is e2 return (f1*f2)  However, division must be treated specially. If the second expression (the number to divide by) is zero, or it does not divide evenly into the numerator, then the guard will force the *short-circuit* to occur and return Nothing:  eval is (ExprF (Factors e1 Div e2)) = do f1 <- eval is e1 f2 <- eval is e2 guard $f2 /= 0 guard ( f1 mod f2 == 0 ) return (f1 div f2)  ## Bringing it all together We can now bring everything together, attemping to "fill in" every possible expression with the inputs and testing if the eval-uated result equals our target. Remember I said earlier that we don't actually need to use all six inputs exactly one each? We are allowed to use a subset instead if we wish. With a target of$4$and inputs$(1,2,3)$we are entitled to use$4=1+3$and ignore the$2$. Therefore, we do another partitioning, simply dividing the list into those that we will use and those we will not use:  allPartialExprs :: [Letter] -> [Expr] allPartialExprs ls = [ es | (used, ignored) <- allPartitions ls , not . null$ used , es <- allExprs used ]  which is then called by findSolutions:  findSolutions target inputs = do let range_of_chars = ['a' .. (toEnum . (subtract 1) . (+ (length inputs)) . fromEnum $'a')] -- if (length inputs == 6), then range_of_chars will be ['a'..'f'] e <- allPartialExprs$ map Letter range_of_chars guard . (== (Just target)) \$ (eval inputs e) return (e, showWithInputs inputs e)  For each successful solution, findSolutions returns the raw Expr object as well as a nice string represenation of the expression "filled-in" with the inputs. Finally, a nice demo where it's all printed nicely:  findSolutionsAndPrintThemNicely target inputs = findSolutions target inputs & mapM (\(e,filledIn) -> do let showe = show e let padding = replicate (24 - length showe) ' ' putStrLn (padding ++ showe ++ " " ++ filledIn) ) main = do findSolutionsAndPrintThemNicely 562 [9,8,2,4,10,4]  This gives hundreds of solutions, here are a few:  (a*b*c*d-e-f) (9*8*2*4-10-4) (a*b*d*c-f-e) (9*8*4*2-4-10) (a*c*f*b-d-e) (9*2*4*8-4-10) (b*c*f*a-e-d) (8*2*4*9-10-4) ((e+f)*(a-d)*b+c) ((10+4)*(9-4)*8+2) (c+(d*f-a)*b*e) (2+(4*4-9)*8*10) (c+(a-f)*(d+e)*b) (2+(9-4)*(4+10)*8) (c-(e+f)*b*(d-a)) (2-(10+4)*8*(4-9)) 
***by [Aaron](https://aaronmcdaid.github.io/), 29th October 2017***