So here's a puzzle for you.
You start with an empty string and clipboard. You can perform four operations:
Each operation takes a different amount of time:
CTRL-A-C)CTRL-P)The question is "What's the shortest string length that requires a delete to create most efficiently?".
If you wanted to figure this out for youself, well, you came to the wrong place.
-- we'll need these later
import Data.MemoCombinators (memo2, integral)
import Data.Monoid (mempty)
So let's encode what we know so far
data Operation = Append | Delete | Copy | Paste
deriving (Eq,Show)
type Length = Int
type Ticks = Int
type Chain = [ Operation ]
opCost :: Operation -> Int
opCost Append = 1
opCost Delete = 1
opCost Copy = 3
opCost Paste = 2
chainCost :: Chain -> Int
chainCost = sum . map opCost
Let's simplify our problem space. As the problem is stated, we're considering all chains of operations:
AACAAPAAADADCDDP
PDPPCA
AAAAACDPNote a couple things:
PA and AP increase the size of the string by one plus the clipboard length. PD and DP increase the size of the string by one minus the clipboard length.This lets us restrict us to chains of a certain form. We can rule out all the Pastes before a Copy since we're searching for the most efficient solution, and this only adds ticks. We can restrict pastes to only be after a copy. This gives these meta-operations:
Now all our chains look like this:
AAACPPPACPPD
AADCPADDD
CPPPACPDDThere are futher optimizations that could be made, but this turns out to be sufficient for our purposes.
If we work backwards from a string of length n , it could have been made by
n-1 in 1 tick,n+1 in t-1 tick,d and q such that q * d = n, by Copying a string of length d and Pasting it q -1 times in a total of 2*q + 1 ticks.That last bit deserves an example. A string of 12 could have been by
So at this point a helper function to discover all the divisors of a number seems useful:
-- find all the pairs that multiply to the given value
divisors :: Int -> [ (Int, Int) ]
divisors = integral $ \n -> [ (d,q) | d <- [1..n], let (q,r) = n `quotRem` d, r == 0 ]
That integral above is my first use of the
Data.MemoCombinators
library. It memoizes the function, so that the divisors for each input are only computed once (the first time they're requested) and then cached to be reused for future calls.
This saves me some time. For example, if I'm used divisors inside an m-fold loop, the loop would be O(m * n) without memoization, but only O(m + n) with memoization.
Consider again working backwards from a string of length n. Though the last meta-operation may have operated upon a shorter (Append, Copy/Paste) or longer (Delete) string, if were looking at the chain operations that leads
most efficiently to our string of length n, than the string that the meta-operation acted upon must be, by definition, more efficient, since we had to spend further ticks after reaching it. So if we know how to efficiently construct strings in less than t ticks, we can use that to efficiently construct strings in exactly t ticks.
This opens us up to Dynamic Programming, which is the real reason I broke out Data.MemoCombinators. Below we use memo2 to memoize a two argument function that calculates the chains required to generate a string of length n in exactly t ticks using the exact working backward method discussed above.
Note that chainsToIn calls itself at least four times, so if we used straight recursion, this would lead to a combinatorial explosion of work. Memoization saves that, letting us calculate a single chainsToIn n t in O(t2) time and all chainsToIn n t for 0 <= n < N and 0 <= t < T in O(NT + T2) time.
-- figure out how to get a string of the given length
-- in exactly the given amount of ticks
chainsToIn :: Length -> Ticks -> [ Chain ]
chainsToIn = memo2 integral integral $ \n t -> case (n,t) of
(0,0) -> return []
(_,t) | t <= 0 -> mempty
(c,_) | c <= 0 -> mempty
_ -> -- helper function to extend chains that generate
-- the given length by the given suffix
-- to create chains to the current location
let moveFrom :: Length -> Chain -> [ Chain ]
moveFrom n' c = map (++c) . chainsToIn n' $ t - chainCost c
-- find all the ways to get to the current location
in concat [ moveFrom (n-1) [Append]
, moveFrom (n+1) [Delete]
, do
(d,q) <- divisors n
moveFrom d $ Copy : replicate (q-1) Paste
]
Now that we can find all the ways (if any) of generating a string of length n in t ticks, we can determine the most efficient ways to generate a string of length n by simply iterating from t=0 up until we find a value of t that gives us a non-empty set of ways to find a string of that length.
-- find the optimal chain to e
optimalChainsTo :: Length -> [ Chain ]
optimalChainsTo = integral $ \n ->
head . dropWhile null $ map (chainsToIn n) [ 0.. ]
Our solution, then, is shortest length where all the chains of operations that generate that length in optimal time include a Delete.
-- our solution only has chains with a Delete in it
solution :: Length
solution = head $ filter (all (elem Delete) . optimalChainsTo) [ 0.. ]
Let's see what it is!
solution
And what was so special about the chain of operations that generate it so efficiently?
optimalChainsTo solution
And how many ticks did it require?
map chainCost it
Out of curiousity, lets see what the optimal chains look like for strings up to length 64.
-- need a few more libraries
import Text.Printf (printf)
import Control.Monad (forM_)
putStrLn "num ticks chains"
forM_ [ 0 .. 64 ] $ \i ->
let cs@(c:_) = optimalChainsTo i
in printf " %2d %2d %s\n" i (chainCost c) (unwords $ map (map $ head . show) cs)
This post was written using IHaskell. Feel free to download and play with the IHaskell notebook version of this post, or join the discussion on Reddit.