Haskell: Getting a single bit from a String of bits - haskell

So I'm using a Grey-code generator to generate all possible bit Strings of length 6. The generator is as follows:
gray :: Integer -> String
gray n
| n == 0 = [""]
| n > 0 = map (++"0") (gray (n-1)) ++
map (++"1") (reverse (gray (n-1)))
recipes = gray 6
Then, I'm attempting to get a specific bit from each of these Strings and convert that bit to an Integer. I'm doing this in the following way:
cost' :: String -> Cost
cost' r i = toInteger( ord ( r!!i ) )
Now, for some reason this isn't working. Regardless of what 'i' value I use, the function cost' will either result in 48 (if the bit in position 2 of the list is 0 --> ex. '100000') or 49 (if the bit in position 2 of the list is 1 --> ex. '101000').
It doesn't make any sense to me why this is.. It's my understanding that Strings are stored as lists in Haskell, and that to get a certain element 'i' from a list 'r' in Haskell, you execute 'r!!i'.

That's because ord returns the code point number of the character, and '0' is code point 48, '1' is code point 49. The function you want to use is digitToInt.

Related

List comprehension not ending in a square bracket, console freezing

Entering a list comprehension into GHCi does not generate a list, the final square brackets are missing, and the console freezes. This is what I have come up with:
[13*x + 3 | x <- [1..], rem (13*x + 3) 12 == 5, mod (13*x + 3) 11 == 0, 13*x + 3 <= 1000]
I believe the problem lies either with x <- [1..], or 13*x + 3 <= 1000. By 13*x + 3 <= 1000 I meant to determine the upper limit of the values x in x <- [1..] can take.
I'm given back a result [341, but it does the second square bracket is missing, and the console freezes.
Your program enters an infinite loop.
The first number is 341, but in order to produce the next number, your program keeps looking through all the subsequent values of x, evaluates all the guards for those values, and checks if all the guards are true. The very last guard, 13*x + 3 <= 1000 never becomes true again, so the program just keeps enumerating values of x forever. It's looking for the next such x for which all guards are true, and as soon as it finds one, it's going to print it. But such x never comes.
If you want the list to end once x*13 + 3 > 1000, you have to use takeWhile:
... | x <- takeWhile (\y -> y*13 + 3 <= 1000) [1..], ...
That way the list will actually stop when it reaches 1000. No more values of x would be produced.
You're giving the compiler way too much credit. It isn't going to carefully analyse your list comprehension in order to deduce that past a certain point there will be no more results, and it should call the list complete. It only does what you tell it to do.
In this case what you told it to do is:
[ 13*x + 3 -- produce numbers of the form 13*x + 3
| x <- [1..] -- by searching all x from [1..]
, rem (13*x + 3) 12 == 5 -- allowing only x that meet this condition
, mod (13*x + 3) 11 == 0 -- and this condition
, 13*x + 3 <= 1000 -- and this condition
]
So it prints [341 and "freezes" because it's still trying to compute the rest of that list. You don't see anything happening, but internally it's drawing ever bigger x from [1..] and diligently checking those conditions to realise that the number shouldn't be included. But it never hits the end of [1..] in order to stop, so it never gets up to printing the ] and waiting for more input.
With your code you are explicitly telling the compiler that you want to search every number in the infinite1 list [1..]. You are then expecting it to notice that 13*x + 3 <= 1000 can only be true for x drawn from a finite prefix of [1..] and thus actually not search the entire list [1..] as you instructed2.
That is a perfectly reasonable thing to want, and I can imagine a system capable of pulling that off (at least with simple conditions like this). So testing it out like this to see if it works is a good idea! However unless someone actually told you that figuring out enumeration upper bounds from conditions in list comprehensions is a feature that GHC can provide, it shouldn't be surprising that it never completes when you tell it to search an infinite list.
For this style of list comprehension (getting all numbers in a range meeting certain conditions) you normally shouldn't use [1..] and then try to impose a stopping condition. Just figure out that the last number that will pass 13*x + 3 <= 1000 and use [1..76] as your generator instead. You can even have Haskell figure it out for you with [1 .. (1000 - 3) div 13].
You use a generator like [1..] when you want to get all numbers of the right form. Then you can use functions like take or takeWhile to get a finite section at the point where you want to use it for something. e.g.
Prelude> let xs = [13*x + 3 | x <- [1..], rem (13*x + 3) 12 == 5, mod (13*x + 3) 11 == 0]
Prelude> takeWhile (<= 1000) xs
[341]
Prelude> take 5 xs
[341,2057,3773,5489,7205]
In fact the simplest and most direct way to express what you want in a single expression is this:
takeWhile (<= 1000) [13*x + 3 | x <- [1..], rem (13*x + 3) 12 == 5, mod (13*x + 3) 11 == 0]
Everything in a list comprehension (except the generator expression) is only talking about a single element at a time. There's just no way to express concepts that are talking about the returned list as a whole, like "stop searching once the returned numbers go out of this range". But that concept is trivial to express outside of list comprehension as a normal function (takeWhile (<= 1000)). Don't feel like you have to shoehorn your entire computation into a single list comprehension.
1 Strictly speaking it's infinite if you're using a type like Integer (which is the type Haskell will pick without any other code using the result to impose other constraints on the type). If you're using Int then it's technically finite, and your list comprehension will eventually end when it "runs out of numbers". [1..] as a list of Int is still impractically vast for an exhaustive search, however.
But if you use a smaller type, like Word16 (needs to be imported from Data.Word) then you can in fact finish your original list comprehension in a practical amount of time. (Though I had to tweak it a little to make sure the 13*x stuff was computed in a larger type so it doesn't overflow)
Prelude> import Data.Word
Prelude Data.Word> [13*x + 3 | x <- [1 :: Word16 ..], let x' = fromIntegral x, rem (13*x' + 3) 12 == 5, mod (13*x' + 3) 11 == 0, 13*x' + 3 <= 1000]
[341]
2 While I'm being pedantic in the footnotes, if your original list comprehension is being evaluated as a list of Int it wouldn't even be valid to just stop after x grows high enough that 13*x + 3 <= 1000 fails for the first time. Try this:
Prelude Data.Word> let x = 768614336404564650 :: Int
Prelude Data.Word> 13*x + 3 <= 1000
True
This happens because Int does in fact have an upper bound, so a large enough Int will overflow back to negative when you multiply it by 13. So when searching [1..] as [Int] the compiler is in fact right to keep looking past x = 77; there are almost certainly more numbers in your original list comprehension if it's [Int], they just take a long time to reach.
Again a good way to demonstrate is to use a smaller finite type, like Word16. If I use your original list comprehension as [Word16] without modifying it to avoid overflow in the conditions, you get this:
Prelude Data.Word> [13*x + 3 | x <- [1..], rem (13*x + 3) 12 == 5, mod (13*x + 3) 11 == 0, 13*x + 3 <= 1000] :: [Word16]
[341,605,209,869,473,77,737]
Even if the compiler was smart enough to know the regions of [1..] that could possibly pass 13*x + 3 <= 1000 condition, it's never going to be able to read your mind and know whether the overflow-produced numbers are solutions you intended or are the result of a bug in your code. It just does what you tell it to do.

How can I use Haskell exception for index negative

I am trying to solve one of my Haskell question. The question asks me to that extracts a slice of a list of integers. Function should take a list and two indexes new list number contains between two indexes.
For this function;
First index should smaller than second index
First index cannot be negative number
we cannot use any built-in functions
example:
makeSlice [1,2,3,4,5] 2 3
[3,4]
makeSlice [1,2,3,4,5] (-1) 3
*** Exception: First index cannot be negative
I tried a few option but below function if I give positive number I am getting "First index cannot be negative" exception
makeSlice :: [a] -> Int -> Int -> [a]
makeSlice [] _ _ =[]
makeSlice (h:t) i k
|k < 0 = []
| i>k = error "First index cannot be greater than second index (i > k)"
| i< 0 = error "First index cannot be negative (i < 0)!"
| i>0 = makeSlice t (i - 1) (k - 1)
| otherwise = h:makeSlice t (i -1 ) (k - 1)
Can you help me to find where I am making wrong?
Add terminating condition for your recursion. On each call you subtract one from i and when it reaches below 0 you just throw error.
Probably the easiest approach to this uses the Prelude take and drop functions to process the list. Then you just need to do the bounds checking:
slice :: Int -> Int -> [a] -> Either String [a]
slice from to lst
| from < 0 = Left "First index cannot be negative (i < 0)!"
| to < from = Left "First index cannot be greater than second index (i > k)"
| otherwise = Right $ take (to - from) $ drop from $ lst
Here I'm using Either to report either success or failure. On the one hand that disagrees with the problem as stated; on the other, it gives callers a chance to handle the error without immediately terminating the program, which is more polite.

How can i use conditionals in list comprehension?

I am trying to build a list of 0's using list comprehension. But i also want to make an index 1 where i choose in the list. For example myList 5 2 = [0,1,0,0,0] where 5 is the number of elements and 2 is the index.
myList el index = [0 | n <- [1..el], if n == index then 1 else 0]
but this results in an error.
The smallest change that fixes that is
myList el index = [if n == index then 1 else 0 | n <- [1..el]]
Note that what's at the left of | is what generates the list elements. A list comprehension of the form [ 0 | ...] will only generate zeros, and the ... part only decides how long is the resulting list.
Further, in your code the compiler complains because at the right of | we allow only generators (e.g. n <- someList), conditions (e.g. x > 23), or new definitions (let y = ...). In your code the if ... is interpreted to be a condition, and for that it should evaluate to a boolean, but then 1 makes the result a number, triggering a type error.
Another solution could be
myList el index = replicate (index-1) 0 ++ [1] ++ replicate (el-index) 0
where replicate m 0 generates a list with m zeros, and ++ concatenates.
Finally, note that your index is 1-based. In many programming languages, that's unconventional, since 0-based indexing is more frequently used.

How to minimize a string's length by iteratively removing all occurrences of some specified words from the string

This question appeared in a programming contest and we still have no idea how to solve it.
Question:
Given a string S and a list of strings L, we want to keep removing all occurences of substrings that may be in L. And we have to minimize the length of the final string formed. Also note that removal of a string may initiate more removals.
For example,
S=ccdedefcde, L={cde}
then answer = 1. Because we can reduce S by ccdedefcde -> cdefcde -> fcde -> f.
S=aabaab, L={aa, bb} then answer = 0 as reduction can be carried out by aabaab -> aabb -> aa -> ‘Empty String’
S=acmmcacamapapc, L={mca, pa} then answer=6 as reduction can be carried out by acmmcacamapapc-> acmcamapapc -> acmapapc -> acmapc.
The maximum length of S can be 50 and the maximum length of list L can be 50.
My approach is a basic recursive traversal in which I return the minimum length that I can get by removing different sub-strings. Unfortunately this recursive approach will time out in the worst case input as we have 50 options at each step and the recursion depth is 50.
Please suggest an efficient algorithm that may solve this problem.
Here's a polynomial-time algorithm that yields optimal results. Since it's convenient for me, I'm going to use the polynomial-time CYK algorithm as a subroutine, specifically the extension that computes a minimum-weight parse of a string according to a context-free grammar with weighted productions.
Now we just have to formalize this problem with a context-free grammar. The start symbol is A (usually S, but that's taken already), with the following productions.
A -> N (weight 0)
A -> A C N (weight 0)
I'll explain N shortly. If N and C were terminals, then A would accept the regular language N (C N)*. The nonterminal C matches a single terminal (character).
C -> a (weight 1)
C -> b (weight 1)
C -> c (weight 1)
...
The nonterminal N matches strings that are nullable, that is, strings that can be reduced to the empty string by deleting strings in L. The base case is obvious.
N -> (weight 0)
We also have a production for each element of L. When L = {mca, pa}, for example, we have the following productions.
N -> N m N c N a N (weight 0)
N -> N p N a N (weight 0)
I hope that it's clear how to construct the one-to-one correspondence between iterative removals and parses, where the parse weight is equal to the length of the residual string.
Note: this is not an optimal solution, since it doesn't work for the example S=ABAABABAA, L={ABA}
Algorithm
RECURSIVE_FUNCTION ( STRING STR, STRING PATTERN) :
1. STRING LEFT = STR.SUBSTR (0, STR.FIND(PATTERN))
2. STRING RIGHT = STR.SUBSTR(STR.FIND(PATTERN), STR.LENGTH)
3. IF (RIGHT is empty) THEN RETURN LEFT
4. STRING FIN = RECUR(LEFT) + RECUR(RIGHT)
5. RETURN RECUR(FIN)
function SUBSTR(A,B) will return substring of the string, from index A inclusive to index B exclusive
Operation A + B is concatenation of string A and B
function RECUR(A) call the same function, aka recurrence
Example: ccdedefcde
First it will branch down with RECUR(LEFT) + RECUR(RIGHT):
c[cde]defcde
/ \
c def[cde]
/
def
Then it will RECUR(FIN) on merge:
cdef*
/ \
c def
/
def
* will RECUR to do the following before that MERGE completes:
[cde]f
\
f
and finally the ROOT call returns f

Haskell not in scope list comprehension

all_nat x = [ls| sum ls == x]
I'd like to write a function that given an integer x it returns all the lists that the result of their elements when summed is the integer x but I always get the error "not in scope: 'ls' " for both times it apperas. I'm new to haskell. What's the syntax error here?
The problem is that you need to define all used variables somewhere, but ls is undefined. Moreover, it can't be defined automatically, because the compiler doesn't know about the task — how the list should be generated? Ho long can it be? Are terms positive or not, integral or not? Unfortunately your code definition of the problem is quite vague for modern non-AI languages.
Let's help the compiler. To solve such problems, it's often useful to involve some math and infer the algorithm inductively. For example, let's write an algorithm with ordered lists (where [2,1] and [1,2] are different solutions):
Start with a basis, where you know the output for some given input. For example, for 0 there is only an empty list of terms (if 0 could be a term, any number could be decomposed as a sum in infinitely many ways). So, let's define that:
allNats 0 = [[]] --One empty list
An inductive step. Assuming we can decompose a number n, we can decompose any number n+k for any positive k, by adding k as a term to all decompositions of n. In other words: for numbers greater than 0, we can take any number k from 1 to n, and make it the first term of all decompositions of (n­-k):
allNats n = [ k:rest --Add k as a head to the rest, where
| k <- [1 .. n] --k is taken from 1 to n, and
, rest <- allNats (n - k)] --rest is taken from solutions for (n—k)
That's all! Let's test it:
ghci> allNat 4
[[1,1,1,1],[1,1,2],[1,2,1],[1,3],[2,1,1],[2,2],[3,1],[4]]
Let's break this up into two parts. If I've understood your question correctly, the first step is to generate all possible (sub)lists from a list. There's a function to do this, called subsequences.
The second step is to evaluate the sum of each subsequence, and keep the subsequences with the sum you want. So your list comprehension looks like this:
all_nat x = [ls| ls <- subsequences [1..x], sum ls == x]
What about
getAllSums x = [(l,r)| l <- partial_nat, r <- partial_nat, l + r == x ]
where partial_nat = [1..x]

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