CS代考 # Assessed Assignment 4 – cscodehelp代写

# Assessed Assignment 4

## Marking table

The exercises are defined so that it is hard to get a first-class mark.

1st class – 70 marks and above.
upper 2nd – 60-69 marks.
lower 2nd – 50-59 marks.
third class – 40-49 marks.
fail – 0-39 marks.

## Additional Required Libraries

The template for this assignment requires the use of the `random` and `mtl` libraries. To load these libraries when starting `ghci`, use the command

ghci -package mtl -package random Assessed4.hs

## Preparation
* Do __not__ modify the files `Types.hs` and `Assessed4-Template.hs`.
* Copy the file `Assessed4-Template.hs` to a new file called `Assessed4.hs` and write your solutions in `Assessed4.hs`.

__Don’t change the header of this file, including the module declaration, and, moreover, don’t change the
type signature of any of the given functions for you to complete.__

__If you do make changes, then we will not be able to mark your submission and hence it will receive zero marks!__
* Solve the exercises below in the file `Assessed4.hs`.

## Submissions should compile and run correctly on Jupyter Notebook

If your submission doesn’t compile or run correctly on Jupyter Notebook, it will get zero marks.

## Submission procedure

* Run the presubmit script to be provided to you on your submission **from Jupyter** by running `./presubmit.sh Assessed4` in the terminal (in the same folder as your submission).
* This will check that your submission is in the correct format.
* If it is, submit on Canvas.
* Otherwise fix and repeat the presubmission procedure.

## Plagiarism

Plagiarism will not be tolerated. Copying and contract cheating has led to full loss of marks, and even module or degree failure, in the past.

You will need to sign a declaration on Canvas, before submission, that you understand the [rules](/plagiarism) and are abiding by them, in order for your submission to qualify.

## Background material

– All of the questions on this assignment refer to the same background material.
– Read this material first, then implement the requested function.
– The corresponding type appears in the file `Assessed4-Template.hs` (to be copied by you).
– Replace the default function implementation of `undefined` with your own function.

## Question Difficulty

All questions have equal weight, but they are designed to increase in
difficulty. We have indicated an approximate difficulty level with
the �� symbols according to the following scheme:

– ð��¶ï¸� – Normal
– ð��¶ï¸�ð��¶ï¸� – Challenging
– ð��¶ï¸�ð��¶ï¸�ð��¶ï¸� – Very challenging

In particular, note that Exercise 4 is meant to be quite challenging, and that this has been done on purpose in view of the marking table described above.

## Background

### Playing cards

Some of the questions make reference to cards drawn from a [standard 52-card deck](https://en.wikipedia.org/wiki/Standard_52-card_deck).
A standard playing card has both a *rank* and a *suit*: the rank can be a number between 2 to 10 or a jack or queen or king or ace; the suit can be clubs or diamonds or hearts or spades.

We represent standard playing cards and a standard 52-card deck with the following definitions in Haskell:
“`haskell
data Rank = R2 | R3 | R4 | R5 | R6 | R7 | R8 | R9 | R10 | RJ | RQ | RK | RA
deriving (Eq,Ord,Enum)
data Suit = C | D | H | S
deriving (Eq,Ord,Enum)
data Card = Card { rank :: Rank, suit :: Suit }
deriving (Eq)
type Deck = [Card]

standard52 :: Deck
standard52 = [Card {rank = r, suit = s} | r <- [R2 .. RA], s <- [C .. S]] Custom instances of the `Show` class are implemented in [Types.hs](Types.hs) in order to improve readability when printing strings representing playing cards. These instances have been tested on Jupyter, and should work in most modern terminals. ### Picking monads and probability distributions Some of the questions make reference to the `PickingMonad` class that was introduced in the [problem sheet](../../ProblemSheets/ProblemSheet-Week7.md) for weeks 7 and 8. Recall that this class builds on the `Monad` class by adding an operation `pick lo hi`, which is supposed to represent a choice of value ranging from the lower bound `lo` to the upper bound `hi` (inclusive in both bounds). ```haskell class Monad m => PickingMonad m where
pick :: Int -> Int -> m Int
In that problem sheet, we defined an instance of `PickingMonad` for the `IO` monad where `pick` was implemented by calling the system random number generator.
Here we’ll tweak the implementation a bit so that it also ensures that `lo <= hi` (raising an error otherwise). ```haskell instance PickingMonad IO where pick lo hi | lo <= hi = getStdRandom (randomR (lo, hi)) | otherwise = error ("pick lo hi: lo = " ++ show lo ++ " is greater than hi = " ++ show hi) For the purpose of testing your code in this assignment (testing by you and by the TAs), it will be helpful to introduce a couple other instances of `PickingMonad`. The [List monad](../../../LectureNotes/monads.md) supports an easy implementation of the `pick` operation where we just return the list of all possible values (after checking that `lo <= hi`): ```haskell instance PickingMonad [] where pick lo hi | lo <= hi = [lo..hi] | otherwise = error ("pick lo hi: lo = " ++ show lo ++ " is greater than hi = " ++ show hi) A more sophisticated implementation of `pick` is based on the monad of *finite probability distributions*. This monad can be seen as a refinement of the List monad supporting probabilistic computation: now we keep track not only of the list of possible values, but also their associated probabilities. In Haskell, the monad of finite probability distributions can be defined as follows: ```haskell newtype Dist a = Dist { dist :: [(a,Rational)] } deriving (Show) instance Monad Dist where return x = Dist [(x,1)] xm >>= f = Dist [(y,p*q) | (x,p) <- dist xm, (y,q) <- dist (f x)] together with the standard boilerplate: ```haskell instance Functor Dist where fmap f xm = xm >>= return . f

instance Applicative Dist where
pure = return
xm <*> ym = xm >>= x -> ym >>= return . x
And then the `pick` operation can be implemented as follows:
“`haskell
instance PickingMonad Dist where
pick lo hi | lo <= hi = Dist [(x,1 / fromIntegral (hi - lo + 1)) | x <- [lo..hi]] | otherwise = error ("pick lo hi: lo = " ++ show lo ++ " is greater than hi = " ++ show hi) This defines `pick lo hi` as a probability distribution taking any value between `lo` and `hi` with equal probability, in other words, as the [uniform distribution](https://en.wikipedia.org/wiki/Discrete_uniform_distribution) on the interval `[lo .. hi]`. To illustrate these different implementations of picking monads, consider the following simple program that picks a number between 0 and 3 and then uses it to index into the string "hello": ```haskell code :: PickingMonad m => m Char
i <- pick 0 3 return ("hello" !! i) Here is a sample transcript of running `code` using the `IO` monad by default (multiple runs can give different results): > code — this runs in the IO monad by default
Now running it using the List monad (always returns the same result):
> code :: [Char]
And running it using the monad of finite probability distributions (always returns the same result):
> code :: Dist Char
Dist {dist = [(‘h’,1 % 4),(‘e’,1 % 4),(‘l’,1 % 4),(‘l’,1 % 4)]}
In the last sample run, observe that our representation of finite probability distributions allows the same value to occur multiple times in the list of value/probability pairs.
The following function will compute the *total* probability of a value occurring in a given distribution, assuming that the values come from an `Eq` type:

“`haskell
prob :: Eq a => Dist a -> a -> Rational
prob xm x = sum [p | (y,p) <- dist xm, x == y] Similarly, the following function will "normalise" a distribution by first computing the list of values in its [support](https://en.wikipedia.org/wiki/Support_(mathematics)#In_probability_and_measure_theory), and then returning the probabilities of all those values: ```haskell normalise :: Eq a => Dist a -> Dist a
normalise xm = Dist [(x,prob xm x) | x <- support xm] support :: Eq a => Dist a -> [a]
support xm = nub [x | (x,p) <- dist xm, p > 0] — “nub” removes duplicates from a list

> prob code ‘l’
> normalise code
Dist {dist = [(‘h’,1 % 4),(‘e’,1 % 4),(‘l’,1 % 2)]}
(We didn’t need to put any type annotations above, since the type `code :: Dist Char` is automatically inferred from the calls to `prob` and `normalise`.)

## ð��¶ï¸� Implementation Task 1 – Some Useful Picking Functions

1. Write a function which chooses an arbitrary element from a list.
“`haskell
choose :: PickingMonad m => [a] -> m a
choose = undefined
More precise requirements:
* `choose xs` should run without error for any non-empty list `xs :: [a]` (for the empty list it can do anything)
* in the case of the monad `m = IO`, `choose xs` should run in time proportional to the length of `xs`
* in the case of the monad `m = Dist`, `choose xs :: Dist a` should compute a (not necessarily normalised) distribution where each value in `xs` is assigned a probability proportional to the number of times it occurs in `xs`. That is, `prob xs x` should be equal to `k / n`, where `k = length [y | y <- xs, x == y]` and `n = length xs`. > choose standard52 — multiple runs in the IO monad can return different results
> choose standard52
> choose [True,False] :: Dist Bool
Dist {dist = [(True,1 % 2),(False,1 % 2)]}
> prob (choose “hello”) ‘l’

2. Write a function which takes a monadic computation of a boolean (an “experiment”) and runs it repeatedly, returning how many times it evaluates to `True`.
“`haskell
simulate :: Monad m => m Bool -> Integer -> m Integer
simulate = undefined
More precise requirements:
* `simulate bm n` should run without error for any non-negative integer `n` (for `n < 0` it can do anything) * in the case of the monad `m = IO`, `simulate bm n` should run in time proportional to `n` * in the case of the monad `m = Dist`, `prob (simulate bm n) k` should give the probability that if the experiment `bm` is repeated `n` times, it will return `True` exactly `k` times. > simulate (choose [True,False]) 100000
> normalise (simulate (choose [True,False]) 3)
Dist {dist = [(3,1 % 8),(2,3 % 8),(1,3 % 8),(0,1 % 8)]}

## ð��¶ï¸�ð��¶ï¸� Implementation Task 2 – [Shuffling](https://en.wikipedia.org/wiki/Shuffling) our deck

1. Write a function which returns an arbitrary splitting of a list into two contiguous pieces.
“`haskell
cut :: PickingMonad m => [a] -> m ([a],[a])
cut = undefined
More precise requirements:
* `cut xs` should run without error for any list `xs :: [a]` (including the empty list)
* in the case of the monad `m = IO`, `cut xs` should return a pair of lists `(ys,zs)` whose concatenation `ys ++ zs` is `xs`, in time proportional to the length of `xs`
* in the case of the monad `m = [ ]`, `cut xs :: [([a],[a])]` should compute the list of all possible pairs `(ys,zs)` whose concatenation `ys ++ zs` is `xs` (without duplicates)
* in the case of the monad `m = Dist`, `cut xs :: Dist ([a],[a])` should compute the uniform distribution on all possible pairs of lists `(ys,zs)` whose concatenation `ys ++ zs` is `xs`.

> cut standard52
([2�,2�,2�,2�,3�,3�,3�,3�,4�,4�,4�,4�,5�,5�,5�],[5�,6�,6�,6�,6�,7�,7�,7�,7�,8�,8�,8�,8�,9�,9�,9�,9�,10�,10�,10�,10�,J�,J�,J�,J�,Q�,Q�,Q�,Q�,K�,K�,K�,K�,A�,A�,A�,A�])
> cut [1..5] :: [([Int],[Int])]
[([],[1,2,3,4,5]),([1],[2,3,4,5]),([1,2],[3,4,5]),([1,2,3],[4,5]),([1,2,3,4],[5]),([1,2,3,4,5],[])]
> cut [1..5] :: Dist ([Int],[Int])
Dist {dist = [(([],[1,2,3,4,5]),1 % 6),(([1],[2,3,4,5]),1 % 6),(([1,2],[3,4,5]),1 % 6),(([1,2,3],[4,5]),1 % 6),(([1,2,3,4],[5]),1 % 6),(([1,2,3,4,5],[]),1 % 6)]}

2. Write a function which returns an arbitrary shuffle of a pair of lists.
“`haskell
shuffle :: PickingMonad m => ([a],[a]) -> m [a]
shuffle = undefined
More precise requirements:
* `shuffle (ys,zs)` should run without error for any pair of lists `ys, zs :: [a]` (including the empty lists)
* in the case of the monad `m = IO`, `shuffle (ys,zs)` should return a list that is a possible interleaving of `ys` with `zs`, in time proportional to the sum of the lengths of `ys` and `zs`
* in the case of the monad `m = Dist`, `shuffle (ys,zs) :: Dist [a]` should give the uniform distribution on all possible interleavings of `ys` with `zs`.

Note the last requirement is a bit subtle.
One way to get a uniform distribution is via the [Gilbert-Shannon-Reeds model](https://en.wikipedia.org/wiki/Gilbert%E2%80%93Shannon%E2%80%93Reeds_model) of shuffling, where the probability of picking the head of the shuffle from `ys` (respectively, from `zs`) is `m/(m+n)` (respectively, `n/(m+n)`), where `m = length ys` and `n = length zs`.

> shuffle (“hello”, “world!”)
“hewolrllod!”
> normalise (shuffle (“Aa”,”Bb”))
Dist {dist = [(“AaBb”,1 % 6),(“ABab”,1 % 6),(“ABba”,1 % 6),(“BAab”,1 % 6),(“BAba”,1 % 6),(“BbAa”,1 % 6)]}

3. A higher-order function which performs n iterations of a “[riffle shuffle](https://en.wikipedia.org/wiki/Riffle_shuffle_permutation)” by repeatedly cutting a list in two and then shuffling the two halves back together, where the specific cut and shuffle functions are passed as extra arguments.
“`haskell
riffles :: PickingMonad m => ([a] -> m ([a],[a])) -> (([a],[a]) -> m [a]) -> Int -> [a] -> m [a]
riffles = undefined
More precise requirements:
* `riffles cf sf n xs` should run without error for any list `xs` and integer `n >= 0`, assuming the functions `cf` and `sf` run without error for all inputs
* in the case of the monad `m = IO`, `riffles cf sf n xs` should return a list that is a possible `n`-fold riffle shuffle of `xs` using the cut function `cf` and shuffle function `sf`, making `n` total calls each to the functions `cf` and `sf`
* in the case of the monad `m = Dist`, `riffle cf sf n xs :: Dist [a]` should compute the correct distribution over all possible `n`-fold riffle shuffles of `xs`, according to the cut function `cf` and shuffle function `sf`.

> riffles cut shuffle 7 standard52
[10�,10�,9�,K�,10�,A�,2�,4�,2�,6�,8�,2�,A�,K�,6�,6�,K�,2�,3�,8�,Q�,Q�,4�,8�,10�,A�,J�,7�,J�,6�,J�,3�,3�,5�,9�,5�,9�,A�,3�,7�,5�,4�,7�,7�,4�,K�,Q�,J�,9�,5�,Q�,8�]
> riffles cut ((xs,ys) -> return (xs++ys)) 7 standard52 — cf. https://en.wikipedia.org/wiki/Zarrow_shuffle
[2�,2�,2�,2�,3�,3�,3�,3�,4�,4�,4�,4�,5�,5�,5�,5�,6�,6�,6�,6�,7�,7�,7�,7�,8�,8�,8�,8�,9�,9�,9�,9�,10�,10�,10�,10�,J�,J�,J�,J�,Q�,Q�,Q�,Q�,K�,K�,K�,K�,A�,A�,A�,A�]
> normalise (riffles cut shuffle 5 [1..3])
Dist {dist = [([1,2,3],1889 % 7776),([2,1,3],31 % 192),([2,3,1],31 % 192),([1,3,2],31 % 192),([3,1,2],31 % 192),([3,2,1],865 % 7776)]}

## ð��¶ï¸�ð��¶ï¸� Exercise 3 – Random Permutations

Write function which returns a random permutation of a list.
“`haskell
permute :: PickingMonad m => [a] -> m [a]
permute = undefined
More precise requirements:
* `permute xs` should run without error for any list `xs :: [a]` (including the empty list)
* in the case of the monad `m = IO`, `permute xs` should return a permutation of `xs`, in time at worst quadratic in the length of `xs`
* in the case of the monad `m = []`, `permute xs :: [[a]]` should compute the list of all possible permutations of `xs` (without duplicates)
* in the case of the monad `m = Dist`, `permute xs :: Dist [a]` should give the uniform distribution on all possible permutations of `xs`.

To meet these requirements, you might consider the following simple recursive algorithm, which generates a uniformly random permutation of a list in quadratic time:

0. if the list is empty, return the empty list
1. otherwise, generate a random permutation of the tail of the list
2. insert the head of the list into the resulting permutation at a
uniformly random position

> permute standard52
[10�,8�,8�,7�,6�,4�,J�,2�,Q�,4�,2�,3�,2�,7�,9�,9�,3�,5�,A�,9�,3�,K�,K�,7�,10�,J�,J�,4�,3�,6�,6�,5�,5�,8�,9�,A�,A�,K�,Q�,4�,K�,10�,5�,10�,8�,Q�,A�,J�,7�,2�,6�,Q�]
> permute [1..3] :: [[Int]]
[[1,2,3],[2,1,3],[2,3,1],[1,3,2],[3,1,2],[3,2,1]]
> normalise (permute [1..3])
Dist {dist = [([1,2,3],1 % 6),([2,1,3],1 % 6),([2,3,1],1 % 6),([1,3,2],1 % 6),([3,1,2],1 % 6),([3,2,1],1 % 6)]}

## ð��¶ï¸�ð��¶ï¸�ð��¶ï¸� Exercise 4 – Random Binary Trees

Write a function which returns a random binary tree with a given list of leaves.
“`haskell
data Bin a = L a | B (Bin a) (Bin a) deriving (Show,Eq)

genTree :: PickingMonad m => [a] -> m (Bin a)
genTree = undefined
More precise requirements:
* `genTree xs` should run without error for any non-empty list `xs :: [a]` (for the empty list it can do anything)
* in the case of the monad `m = IO`, `genTree xs` should return a binary tree whose canopy is a permutation of `xs`, in time at worst quadratic in the length of `xs` (The _canopy_ of a tree is the list of leaves of the tree as they appear in an in-order traversal.)
* in the case of the monad `m = []`, `genTree xs :: [Bin a]` should compute the list of all possible binary trees whose canopy is a permutation of `xs` (without duplicates)
* in the case of the monad `m = Dist`, `genTree xs :: Dist (Bin a)` should give the uniform distribution on all possible binary trees whose canopy is a permutation of `xs`.

To meet these requirements, you might consider the following simple recursive algorithm (known as *Rémy’s algorithm*), which generates a uniformly random binary tree with a given list of leaves in quadratic time (or even linear time with a clever implementation, see p.16 of [Knuth, volume 4a, pre-fascicle 4a](http://web.archive.org/web/20190713015719/http://www.cs.utsa.edu/~wagner/knuth/fasc4a.pdf)):

0. If the list is `[x]`, return a leaf labelled by `x`.
1. Otherwise, generate a random binary tree whose canopy is a permutation of the tail
2. Uniform randomly pick a subtree of the resulting tree, and replace it by a binary node which has the old subtree as one child, and a leaf labelled by the head of the list as the other child (flip a coin to decide whether the new leaf goes to the left or right).

> genTree [1,2] :: [Bin Int]
[B (L 1) (L 2),B (L 2) (L 1)]
> prob (genTree [1..4]) (B (L 3) (B (B (L 1) (L 4)) (L 2)))
> prob (genTree [1..4]) (B (B (L 4) (L 1)) (B (L 2) (L 3)))
> genTree standard52
B (L 10�) (B (L 4�) (B (B (B (B (B (L J�) (B (B (L 10�) (B (L 6�) (L J�))) (L 2�))) (L 6�)) (B (B (B (B (B (B (L 6�) (B (B (B (L K�) (L 8�)) (B (L 5�) (L Q�))) (L 3�))) (B (L 7�) (L 3�))) (L 6�)) (B (B (B (B (B (B (B (L 9�) (B (L A�) (L 4�))) (L 2�)) (L 9�)) (L K�)) (B (L 8�) (L 7�))) (B (B (B (B (L 5�) (L Q�)) (B (L 7�) (B (L 5�) (B (L 2�) (L K�))))) (B (L Q�) (L 9�))) (B (L 8�) (L 8�)))) (L J�))) (L A�)) (B (B (B (L J�) (B (L 3�) (L K�))) (L 9�)) (L 10�)))) (B (L A�) (L 4�))) (B (B (B (L 5�) (B (L 3�) (B (B (L 7�) (B (L A�) (L Q�))) (L 2�)))) (L 4�)) (L 10�))))