https://gist.github.com/ambuc/731f2d9b789a5e4e32bdafbd60bf7ff8

- The Puzzle
- Abstract Syntax Trees
- Evaluation
- Maybes and Monads
- Solving the Puzzle
- Putting it all together
- Runtime
- The Puzzles
- Return Volley

# The Puzzle

Josh Mermelstein and I have decided to begin challenging each other to a series of a math/programming puzzles. Here was his most recent puzzle:

Given

a)the list of numbers $[1, 2, 3, 4, 5]$,b)unlimited parentheses, andc)the operations $+$, $-$, $\times$, $/$, $\wedge$, and $!$, it is possible to construct many integers. For example,

- $1 = 1 + 2 - 3 - 4 + 5$, and
- $2 = 1 + (((2 + 3) * 4!) / 5!)$, etc.
What is the least positive integer that cannot be written this way?

- Numbers must appear in order, each exactly once.
- You may not take the factorial of a factorial (i.e. $(3!)! = 720$).
- Concatination is forbidden (i.e. $1 + 23 + 4 + 5$).
- Unary negative is forbidden (i.e. $13 = (-1) + 2 + 3 + 4 + 5$)

Let’s define this function a little more. I’d like to call the output the *least
nonconstructable positive integer*, or $\text{LNPI}$ of an ordered list. In this
case, the ordered list is $[1,2,3,4,5]$.

OK, how do we start?

# Abstract Syntax Trees

Let’s represent an expression like $(1!+2)-3$ as an abstract syntax tree; terminal nodes are integers, and all the other nodes are operators.

It makes sense to have two types of operators; functions of one argument, like the factorial, and combining functions of two arguments, like all the rest.

Haskell makes it easy to create abstract data types and start playing with them.

## AST Implementation

We can represent our values `Val`

as unrounded ratios of long ints, or
`Integer`

s. This means we never lose precision to rounding / near-zero numbers.

```
type Val = Ratio Integer
```

As we discovered before, there are two things we can do to `Val`

s:

- take a
*function*of one of them: $y = f(x)$ - perform an
*operation*on two of them: $y = f(x_1, x_2)$.

```
data Fn = Id | Fact deriving (Bounded, Enum)
data Op = Plus | Sub | Mult | Div | Exp deriving (Bounded, Enum)
```

Here `Id`

is the identity function $y(x) = x$. The rest should be
self-explanatory.

Finally we need an *expression* type `Expr`

, which can be either:

- a raw
`Val`

:`V (Ratio Integer)`

, - an expression
`E1`

of 1 argument, which takes a function`Fn`

and an expression to operate on, or - an expression
`E2`

of 2 arguments, which takes an operation`Op`

and a tuple of two expressions to operate on.

```
data Expr = V Val | E1 Fn Expr | E2 Op (Expr,Expr)
```

Let’s even derive custom `Show`

instances so we can pretty-print them, for
debugging purposes.

```
instance Show Op where
show Plus = "+"
show Sub = "-"
show Mult = "*"
show Div = "/"
show Exp = "^"
instance Show Expr where
show (V a) = show $ round a
show (E1 Id e) = show e
show (E1 Fact e) = show e ++ "!"
show (E2 o (e1,e2)) = "(" ++ show e1 ++ show o ++ show e2 ++ ")"
```

If we play with this in a `ghci`

shell, we see:

```
> E2 Sub ( E2 Plus ( E1 Fact ( V 1 ), V 2 ), V 3 )
((1!+2)-3)
```

This is good. Let’s figure out how to evaluate these expressions.

# Evaluation

The type signature of `eval`

should be `eval :: Expr -> Maybe Val`

, since not
all expressions return values. (Imagine taking the factorial of a non-integer,
or dividing by zero). To accomplish this in an elegant way we make use of monad
stuff like `=<<`

, `<$>`

, and `<*>`

, which is worth exploring in some detail in a
moment.

```
eval :: Expr -> Maybe Val
eval (V a) = Just a
eval (E1 f e) = calc1 f =<< eval e
eval (E2 o (e1, e2)) = calc2 o =<< (,) <$> eval e1 <*> eval e2
calc1 :: Fn -> Val -> Maybe Val
calc1 Id a = Just a
calc1 Fact a = guard (isInt a && a<100 && a>0) >> Just f
where f = facts !! (pred . fromIntegral . numerator) a
calc2 :: Op -> (Val, Val) -> Maybe Val
calc2 Plus (a,b) = Just $ a + b
calc2 Sub (a,b) = Just $ a - b
calc2 Mult (a,b) = Just $ a * b
calc2 Div (a,b) = guard (b/=0) >> Just (a/b)
calc2 Exp (a,b) = makeExp
where makeExp
| a == 0 = Just 0
| a == 1 = Just 1
| not (isInt b) = Nothing
| abs b > 1023 = Nothing
| otherwise = Just $ a ^^ numerator b
```

Obviously not all computation is valid. We refuse to:

- take the factorial of any negative number,
- take the factorial of any non-integer, or
- divide by zero.

One hurdle of this problem was the discovery that it is possible to quickly generate numbers too large for the computer to handle: for example,

`1+(2^(3^(4^5)))`

= $1 + 2^{3^{4^5}} \approx 10^{10^{488}}$ So we end up doing a fair bit of bounds checking. We also refuse to:

- compute the factorial of any integer greater than a hundred,
- take an exponent to the power of any non-integer,
- take an exponent to the power of any number outside $[-1023..1023]$,

Additionally, we also precompute the factorial of all numbers for which the factorial
is allowed, $[0..100]$, and store them in `facts`

; thus `calc1 Fact a`

is just a
lookup with guards.

Finally, we avoid some computation by noticing that $a^1 = a$ and $0^b = 0$.

OK, let’s try evaluating the above expression!

```
input output
> eval $ E2 Sub (E2 Plus (E1 Fact (V 1), V 2), V 3) Just (0 % 1)
> eval $ E2 Sub (E2 Plus (E1 Fact (V 1), V 2), V 4) Just ((-1) % 1)
> eval $ E2 Div (V 1, V 0) Nothing
```

Perfect: when an expression is valid, we compute the `Val`

as a ```
Just (Ratio
Integer)
```

; when an expression is invalid, we return Nothing.

# Maybes and Monads

Let’s take a moment and explore some of the above Haskell in detail. I’m going to try and explain how I accidentally invented the Monad while trying to solve a subproblem, as was predicted in You Could Have Invented Monads! (And Maybe You Already Have.), a wonderful article about this exact phenomenon.

Imagine a function like `calc1 :: Fn -> ? -> ?`

. How should we write this?

## First Try

Writing it like `calc1 :: Fn -> Maybe Val -> Maybe Val`

essentially requires us
to write two cases:

```
calc1 :: Fn -> Maybe Val -> Maybe Val
calc1 _ Nothing = Nothing
calc1 Id (Just a) = Just a
calc1 Fact (Just a) = if (canFactorial a) then Just (factorial a) else Nothing
```

I’m not sure that will carry over well to `calc2`

:

```
calc2 :: Fn -> (Maybe Val, Maybe Val) -> Maybe Val
calc2 _ (Nothing, _ ) = Nothing
calc2 _ (_, Nothing) = Nothing
calc2 Plus (Just a, Just b ) = Just $ a + b
calc2 Minus ...
```

This is OK, but let’s find a better way – maybe by assuming we won’t pass invalid expressions in the first place.

## Second Try

Writing it like `calc1 :: Fn -> Val -> Maybe Val`

assumes a valid expression, so
we can write:

```
calc1 :: Fn -> Val -> Maybe Val
calc1 Id a = Just a
calc1 Fact a = if (canFactorial a) then Just (factorial a) else Nothing
calc2 :: Fn -> (Val, Val) -> Maybe Val
calc2 Plus (a,b) = Just $ a + b
calc2 Minus ...
```

This is much better. So if our type signatures look like

```
calc1 :: Fn -> Val -> Maybe Val
calc2 :: Fn -> (Val, Val) -> Maybe Val
```

then what does `eval`

look like? It would need to do something like

```
eval :: Expr -> Maybe Val
eval (V a) = Just a
eval (E1 f e) = calc1 f (???)
```

And here’s where our trouble begins. Remember that we want to evaluate `e`

before passing it into `calc1`

. So we’d have to check if it was valid first.

```
eval (E1 f e) = if (isJust $ eval e)
then (calc1 f (fromJust $ eval e))
else Nothing
```

This is nuts. Luckily this type of implicit `Maybe`

checking is easy, since the
`Maybe`

type is a
monad,
and implements the Monad, Functor, and Applicative types.

### Functors, really quick

Quick functor recap:

`map`

lets us apply a function over a list:`map fn [x,y] = [fn x, fn y]`

`fmap`

lets us apply a function inside a container type:`fmap f (m a) = m (f a)`

.

Note also that `<$>`

is an infix synonym for `fmap`

. Confused? See the
following examples:

```
> map (+1) [1,2] [2,3]
> fmap (+1) (Just 1) Just 2
> (+1) <$> Just 1 Just 2
> (+1) <$> Nothing Nothing
```

One benefit of `fmap`

ping over the `Maybe`

monad is that if we pass it
`Nothing`

, it doesn’t need to unwrap and apply; it’ll just pass `Nothing`

through unscathed.

`fmap`

, really quick

So remember the above problem:

```
> calc1 Fact (4%1) Just (24%1)
> calc1 Fact Nothing <error>
```

OK, that’s expected. Let’s try using `<$>`

```
> calc1 Fact <$> Just (4%1) Just (Just (24%1))
> calc1 Fact <$> Nothing Nothing
```

So, good news and bad news. We can unwrap `Just a`

and apply the function to the
interior, but we pass it back re-wrapped, since that’s what `<$>`

does. Sadly,
that’s also what `calc1`

does. Here’s a clue:

```
> :type (<$>) (<$>) :: Functor f => (a -> b) -> f a -> f b
```

This makes sense: `<$>`

expects a function like `(a -> b)`

which doesn’t have an
opinion on how to re-wrap `b`

itself. So maybe `<$>`

isn’t the function we want?

### Monads, really quick

At this point I searched Hoogle for
`(a -> f b) -> f a -> f b`

and got, of
course…
the Monad.

```
> :type (=<<) (=<<) :: Monad m => (a -> m b) -> m a -> m b
> :type (>>=) (>>=) :: Monad m => m a -> (a -> m b) -> m b
```

There’s lots more to the monad, but for now we can use it as a way to take a value in context, apply a context-aware function, and return some output in context.

```
> calc1 Fact =<< Just (4%1) Just (24%1)
> calc1 Fact =<< Nothing Nothing
```

What about `calc2`

? It needs to take a tuple of `Expr`

s, and neither of them can
be `Nothing`

. Turns out we can use the fmap infix `<$>`

and their friend the
sequential application infix `<*>`

. Here’s a set of three trials:

```
(+) <$> Just 1 <*> Just 2 Just 3
(+) <$> Just 1 <*> Nothing Nothing
(+) <$> Nothing <*> Just 2 Nothing
```

So finally we can simply write `calc2 o =<< (,) <$> eval e1 <*> eval e2`

. Thus,

```
eval :: Expr -> Maybe Val
eval (V a) = Just a
eval (E1 f e) = calc1 f =<< eval e
eval (E2 o (e1, e2)) = calc2 o =<< (,) <$> eval e1 <*> eval e2
```

# Solving the Puzzle

Technically this is all we need to start solving the puzzle. Now that we have a way to compose and evaluate expressions, we just need to generate and evaluate all of them.

## Generating all expressions

Let’s talk about partitioning a strictly ordered list.

One interesting aspect of this problem is that because our numbers have to be
*in order*, the nodes containing the values (for example) $[1..3]$ will appear
in order, from left to right, as the terminal nodes of our *AST*. This means
that we can generate all possible ASTs by taking our root node and splitting the
numbers $[1..3]$ at some point, throwing the lesser half on the left, and the
greater half on the right.

It’s easy to see how we can apply this recursively to generate all tree shapes.
Then we can insert all possible iterations at each `op`

node, and all
possible functions *at* each function. There are only two functions, and one is
`Id`

which leaves the value untouched.

We implement the partition algorithm as:

```
mkPart :: [a] -> [([a],[a])]
mkPart xs = tail $ init $ zip (inits xs) (tails xs)
```

such that

```
> mkPartitions [1,2,3] [([1],[2,3]),([1,2],[3])]
```

`valuesFrom`

in Theory

This is excellent. Let’s write the function `valuesFrom`

, which takes an ordered
list and returns all possible values which can be generated from combining it
as described above. Note:

`valuesFrom 1`

will just be`[1, 1!] = [1,1]`

`valuesFrom 3`

will just be`[3, 3!] = [3,6]`

```
valuesFrom :: [Val] -> [Val]
valuesFrom [x] = mapMaybe eval [ E1 f (V x) | f <- functions ]
valuesFrom xs = mapMaybe eval [ E1 f $ E2 o (V va, V vb)
| f <- functions
, o <- operations
, (as, bs) <- mkPartitions xs
, va <- valuesFrom as
, vb <- valuesFrom bs
]
```

We use `mapMaybe`

to map `eval`

over a list and discard the `Nothing`

s (and
`fromJust`

the `Just a`

s). This is great! It doesn’t check for duplicates or
anything, but it’s a good start.

## Caching

In practice, we would like to cache `valuesFrom`

.

Here’s why: in calling `valuesFrom [1,2,3,4,5]`

, we end up evaluating
`valuesFrom [2,3,4]`

twice: once as part of `[1,2,3,4]`

and once as part of
`[2,3,4,5]`

. See the above tree, which is an expansion of the earlier one for a
larger ordered list.

You can see we end up evaluating `f([2,3])`

twice, `f([1,2])`

twice, `f([3,4])`

twice, etc.

You can imagine how this inefficiency bares its teeth for higher-length ordered
lists; evaluating `f([1,2,3,4,5])`

will evaluate `[2,3]`

four times; as a part
of `[1,2,3]`

,`[1,2,3,4]`

,`[2,3,4]`

, and `[2,3,4,5]`

.

So let’s generate a big `Map`

. On that note, `Map`

is Haskell’s version of a
key/value pairing, what would be a dict in Python. Here’s what it will contain:

- the keys in this map should be consecutive sequences like
`[1], [1,2], [2,3,4], [4,5]`

, etc., and - the values should be sets of values
`S.Set Val`

which can be created from combining those numbers in the key.

To generate this recursively from the bottom up, we can first generate a map of all keys of length 1, and then use it to quickly generate a map of all keys of length 2, and so on and so on and so on.

Let’s do it.

`valuesFrom`

in Practice

Here’s how it all looks together: `valuesFrom range`

does a ```
M.findWithDefault
(S.empty) range
```

on a map, which is the union of two smaller maps:

where `prevMap`

is the `mkMap`

of a smaller subset, and `thisMap`

is the set of
the values of the expressions generated in `exprsFrom`

, which uses `valsFrom`

the partitioned right- and left-available values. Check it out:

```
valuesFrom :: [Val] -> S.Set Val
valuesFrom range = M.findWithDefault S.empty range
$ mkMap (length range)
where
mkMap :: Int -> M.Map [Val] (S.Set Val)
mkMap 0 = M.empty
mkMap n = M.union prevMap thisMap
where
prevMap = mkMap $ pred n
thisMap = foldr (\sq -> M.insert sq $ mkSet sq) M.empty $ subsequences n
mkSet = S.fromList . mapMaybe eval . exprsFrom prevMap
exprsFrom :: M.Map [Val] (S.Set Val) -> [Val] -> [Expr]
exprsFrom prevMap [x] = [ E1 f (V x) | f <- functions ]
exprsFrom prevMap xs = [ E1 f $ E2 o (V va, V vb)
| f <- functions
, o <- operations
, (as, bs) <- mkPartitions xs
, va <- valsFrom as
, vb <- valsFrom bs
]
where valsFrom xs = S.toList $ M.findWithDefault S.empty xs prevMap
subsequences n = filter ((== n) . length) $ map (take n) $ tails range
```

So, with caching, here’s what each `valuesFrom`

function call ends up looking
like:

# Putting it all together

We’re so close to the end here.

We can get a set of all possible numbers from a range `xs`

with `valuesFrom xs`

.
This returns a set we can filter for just positive integers; then we want to
turn that into a list and find its first gap; where

```
-- > firstGap [1,2,3,5,6] = 4
firstGap :: (Num a, Eq a) => [a] -> a
firstGap ls = 1 + ls !! (head . findIndices (/=1) . gaps) ls
where gaps :: Num t => [t] -> [t]
gaps [x] = []
gaps (x:xs) = (head xs - x) : gaps xs
```

Finally, we can define

```
lnpi :: [Val] -> Integer
lnpi = numerator . firstGap
. S.toList . S.filter (>0)
. S.filter isInt. valuesFrom
```

So, let’s try it!

```
> lnpi [1..5] 159
```

and that’s our final answer; 159 is the smallest integer we can’t construct from the numbers $[1..5]$.

For the complete code, please see this gist.

# Runtime

Finally, the runtime…

```
j@mes $ ghc -O2 lnpi.hs && time ./lnpi
[1 of 1] Compiling Main ( lnpi.hs, lnpi.o )
Linking lnpi ...
159
real 0m0.332s
user 0m0.327s
sys 0m0.003s
```

This is phenomenal. Early drafts of this took anywhere from one to ten minutes.

The secret to this low runtime is evaluating the values early for re-use in
other expressions, and nubbing down lists to sets as early as possible, for
fewer iterations.

# The Puzzles

So Josh actually posed me a set of puzzles. Once we discussed how the performance was and how easy it was to play around with different ideas in this space, we settled on a set of bonus problems; provided here are their questions, answers, runtimes, and some comments on performance.

## Puzzle One

**Find the least nonconstructable integer from the ordered list [1,2,3,4,5]**:

This was the original problem. The solution `01.hs`

was trivial:

```
```

$ ghc -O2 operationsLibrary.hs 01.hs && time ./01 159 0m0.333s

```
## Puzzle Two
**Find the least nonconstructable integer from any length five, strictly
increasing sublist of `[0..9]`**:
```

main = print $ first (map numerator)
$ minimumBy (compare `on`

snd) $ map (id &&& lnpi) $ sets [0..9] 5 ```

```
$ ghc -O2 operationsLibrary.hs 02.hs && time ./02
([0,2,6,8,9],2)
0m39.127s
```

So I actually played a lot with trying to parallelize this one. A first
attempt to compute one large cached map of all sublists of `[0..9]`

took
a very large amount of time, and was too large to fit entirely in memory; the
process of swapping bits of it in and out was expensive. It turned out to be
faster to compute `lnpi([])`

from scratch each time, at anywhere from `0.03s`

to `3s`

each, depending on how many possible products were exponentially
large and could be dropped immediately.

I did play with the
Control.Parallel
library for a bit, and ended up using `pseq`

and `par`

to parallelize this
independent computation of `lnpi([])`

across four cores, for a slightly
better runtime:

```
main = print
$ first (map numerator)
$ a `par` b `par` c `par` d `pseq` reduceFunc [a,b,c,d]
where [a,b,c,d] = map (reduceFunc . map mapFunc)
$ segmentInto (sets [0..9] 5) 4
reduceFunc = minimumBy (compare `on` snd)
mapFunc = id &&& lnpi
segmentInto xs n = map (\x -> take y $ drop (y*x) xs) [0..pred n]
where y = succ $ div (length xs) n ```
```

$ ghc -j -O2 -threaded -rtsopts operationsLibrary.hs 02b.hs –make -fforce-recomp && time ./02b +RTS -N4 ([0,2,6,8,9],2) 0m23.407s

```
## Puzzle Three (a,b)
**What lists can't make one?**
Here we end up using `valuesFrom` directly; we don't care about the `lnpi`
at all. We can use `map (f &&& g)` which takes an input `x` and returns a
tuple `(f x, g x)`. We use list function `elem` to see that `1` isn't a
possible value, and print the (empty) list.
```

main = print $ filter (not.snd) $ map ( id &&& elem 1 . valuesFrom ) $ sets [0..9] 5 ```

```
$ ghc -O2 operationsLibrary.hs 03.hs && time ./03
[]
0m36.4s
```

This performance was a little lacking, but we can reoptimize by fetching the
list of `expressionsFrom`

and evaluating one-by-one to see if they are ```
Just
(1%1)
```

; we skip the process of mapping `id`

s and the cost of holding all the
values in memory instead of printing an empty list lazily.

```
main = print
$ map (map numerator)
$ filter (isNothing . find1)
$ sets [0..9] 5
where find1 xs = if null ls then Nothing else Just (head ls)
where ls = filter (\e -> eval e == Just (1%1)) $ expressionsFrom xs
```

```
$ ghc -O2 operationsLibrary.hs 03b.hs && time ./03b
[]
0m7.087s
```

### What’s `expressionsFrom`

?

It’s a lot like `valuesFrom`

, except we don’t evaluate the expressions at every
possible step. To write both `valuesFrom`

and `expressionsFrom`

I was able to
extract a lot of shared abstractions. Here is the code for them both, all in
one place. This looks different from the form of `valuesFrom`

above, but it has
the same performance and interface.

```
makeMap :: Ord a => [a] -> ((M.Map [a] b, [a]) -> b) -> M.Map [a] b
makeMap range insertFn = mkMap (length range)
where mkMap 0 = M.empty
mkMap n = M.union prev this
where prev = mkMap (n-1)
this = foldr f z ls
where f sq = M.insert sq $ insertFn (prev, sq)
z = M.empty
ls = filter ((==n).length) $ map (take n) $ tails range
valuesFrom :: [Val] -> [Val]
valuesFrom range = S.toList $ M.findWithDefault S.empty range $ makeMap range insFn
where insFn = S.fromList . mapMaybe eval . exprsFrom
exprsFrom :: (M.Map [Val] (S.Set Val), [Val]) -> [Expr]
exprsFrom (prevMap, [x] ) = [ E1 f (V x) | f <- functions ]
exprsFrom (prevMap, range) = [ E1 f $ E2 o (V va, V vb)
| f <- functions
, o <- operations
, (as, bs) <- mkPart range
, va <- valsFrom as
, vb <- valsFrom bs
]
where valsFrom range = S.toList $ M.findWithDefault S.empty range prevMap
expressionsFrom :: [Val] -> [Expr]
expressionsFrom range = M.findWithDefault [] range $ makeMap range exprsFrom
where exprsFrom :: (M.Map [Val] [Expr], [Val]) -> [Expr]
exprsFrom (prevMap, [x] ) = [ E1 f (V x) | f <- functions ]
exprsFrom (prevMap, range) = [ E1 f $ E2 o (ea, eb)
| f <- functions
, o <- operations
, (as, bs) <- mkPart range
, ea <- exprsFrom (prevMap, as)
, eb <- exprsFrom (prevMap, bs)
]
```

## Puzzle Three (c)

**Print how each list can make one.**

By using `expressionsFrom`

we can find the first expression that evaluates
to $1$, and avoid evaluating the entire possible tree.

```
main = mapM_ print $ map (map numerator &&& find1) $ sets [0..9] 5
where find1 xs = if null ls then Nothing else Just (head ls)
where ls = filter (\e -> eval e == Just (1%1)) $ expressionsFrom xs
```

```
$ ghc -O2 operationsLibrary.hs 03c.hs && time ./03c
([0,1,2,3,4],Just (0+(1+((2-3!)+4))))
([0,1,2,3,5],Just (0+(1+(2+(3-5)))))
([0,1,2,4,5],Just (0+((1*2)+(4-5))))
([0,1,3,4,5],Just (0+(((1+3)!/4)-5)))
([0,2,3,4,5],Just (0+(2+((3-4)^5))))
([1,2,3,4,5],Just (1+((2-(3+4))+5)))
([0,1,2,3,6],Just (0+(1+((2*3)-6))))
([0,1,2,4,6],Just (0+(1+(2+(4-6)))))
([0,1,3,4,6],Just (0+((1*3)+(4-6))))
([0,2,3,4,6],Just (0+(2+(3-(4!/6)))))
...
0m7.085s
```

# Return Volley

Josh, here is my puzzle:

I saw a Sol Lewitt exhibition at a museum a few years ago which consisted of a series of variations of incomplete cubes. There were 122 of them, and each was a contiguous subset of lines which traced the twelve edges of the skeleton of a cube. The trick was that these subsets were unique across rotation but not reflection.

Create a script / renderer to generate the table of Varations of Incomplete Open Cubes, and then apply it to the skeleton of each platonic solid.

Some light reading: Analysis of Variations of Incomplete Open Cubes by Sol Lewitt, a paper by Michael Allan Reb (2011).