Problem description
Given vertices V which can be seen as named "propositions".
Given weights:
data W
= Requires -- ^ Denotes that a "proposition" depends on another.
| Invalidates -- ^ Denotes that a "proposition" invalidates another.
In a linear ordering, if A requires B, then B must come before A, conversely, if A invalidates B, then B must come after A.
Given a weighted directed multigraph (multidigraph) with at most 2 parallel edges... Where a vertex can only require the inclusion of another vertex once, and only invalidates another vertex once...
G = (V, E)
E = (V, V, W)
Or alternatively represented as a directed cyclic graph with no self-loops and where the only cycles form directly between one vertex and another. With weights changed to:
data W
= Requires -- ^ Denotes that a "proposition" depends on another.
| InvalidatedBy -- ^ Denotes that a "proposition" is invalidated by another.
Given that vertices may occur more than once in the ordering...
How can a linear ordering be constructed from such a graph?
Additionally, if the tail of the linear ordering ends with a vertex V which was included due to being InvalidatedBy another vertex, then it may be omitted if the head of the ordering starts with V.
Some desired properties are:
Minimality - there should be as little duplication of vertices as possible
Stability - the ordering should be as similar as possible to the order between vertices on the same "level" in which the graph was constructed
Run-time complexity - The number of vertices are not that high, but still... the run-time complexity should be as low as possible.
If various algorithms fulfill these to varying degrees, I'd love to see all of them with their trade offs.
Algorithms written in any language, or pseudocode, are welcome.
Example graphs:
Example graph 1:
B `requires` A
C `requires` A
D `requires` A
E `invalidates` A
F `invalidates` A
G `invalidates` A
With minimal linear ordering: [A, B, C, D, E, F, G]
Example graph 2:
C `requires` A
C `invalidates` A
B `requires` A
With minimal linear ordering: [A, B, C]
Example graph 3:
B `requires` A
B `invalidates` A
C `requires` A
C `invalidates` A
With minimal linear ordering: [A, B, A, C]
Naive implementation
A naive implementation constructs a linear ordering by starting with all nodes with no incoming edges and for all of those nodes:
fetches all outgoing edges
partitions those by requires/invalidates
constructs the linear ordering of "requires" and puts that first
adds the current node
constructs the linear ordering of "invalidates" and adds that.
Here's a Haskell implementation of this description:
import Data.List (partition)
import Data.Maybe (fromJust)
import Control.Arrow ((***))
import Data.Graph.Inductive.Graph
fboth :: Functor f => (a -> b) -> (f a, f a) -> (f b, f b)
fboth f = fmap f *** fmap f
outs :: Graph gr => gr a b -> Node -> (Adj b, a)
outs gr n = let (_, _, l, o) = fromJust $ fst $ match n gr in (o, l)
starts :: Graph gr => gr a b -> [(Adj b, a)]
starts gr = filter (not . null . fst) $ outs gr <$> nodes gr
partW :: Adj W -> (Adj W, Adj W)
partW = partition ((Requires ==) . fst)
linearize :: Graph gr => gr a W -> [a]
linearize gr = concat $ linearize' gr <$> starts gr
linearize' :: Graph gr => gr a W -> (Adj W, a) -> [a]
linearize' gr (o, a) = concat req ++ [a] ++ concat inv
where (req, inv) = fboth (linearize' gr . outs gr . snd) $ partW o
The ordering can then be optimized by removing equal consecutive like so:
-- | Remove consecutive elements which are equal to a previous element.
-- Runtime complexity: O(n), space: O(1)
removeConsequtiveEq :: Eq a => [a] -> [a]
removeConsequtiveEq = \case
[] -> []
[x] -> [x]
(h:t) -> h : ug h t
where
ug e = \case
[] -> []
(x:xs) | e == x -> ug x xs
(x:xs) | otherwise -> x : ug x xs
Edit: Using DCG, SCC, and topsort
With the algorithm described by #Cirdec :
Given a directed cyclic graph (DCG) where edges of form: (f, t) denote that f must come before t in the ordering.
Compute the condensation of the DCG in 1.
Turn each SSC in the condensation in 2. into a palindrome.
Compute the topsort of the graph in 3.
Concatenate the computed ordering.
In Haskell:
{-# LANGUAGE LambdaCase #-}
import Data.List (nub)
import Data.Maybe (fromJust)
import Data.Graph.Inductive.Graph
import Data.Graph.Inductive.PatriciaTree
import Data.Graph.Inductive.NodeMap
import Data.Graph.Inductive.Query.DFS
data MkEdge = MkEdge Bool Int Int
req = MkEdge True
inv = MkEdge False
toGraph :: [MkEdge] -> [(Int, Int, Bool)] -> Gr Int Bool
toGraph edges es = run_ empty nm
where ns = nub $ edges >>= \(MkEdge _ f t) -> [f, t]
nm = insMapNodesM ns >> insMapEdgesM es
-- | Make graph into a directed cyclic graph (DCG).
-- "Requires" denotes a forward edge.
-- "Invalidates" denotes a backward edge.
toDCG :: [MkEdge] -> Gr Int Bool
toDCG edges = toGraph edges $
(\(MkEdge w f t) -> if w then (t, f, w) else (f, t, w)) <$> edges
-- | Make a palindrome of the given list by computing: [1 .. n] ++ [n - 1 .. 1].
-- Runtime complexity: O(n).
palindrome :: [a] -> [a]
palindrome = \case
[] -> []
xs -> xs ++ tail (reverse xs)
linearize :: Gr Int a -> [Int]
linearize dcg = concat $ topsort' scc2
where scc = nmap (fmap (fromJust . lab dcg)) $ condensation dcg
scc2 = nmap palindrome scc
For the graph g2:
g2 = [ 2 `req` 1
, 2 `inv` 1
, 3 `req` 1
, 3 `inv` 1
, 4 `req` 1
, 5 `inv` 1
]
> prettyPrint $ toDCG g2
1:2->[(False,2)]
2:1->[(True,1),(True,3),(True,4)]
3:3->[(False,2)]
4:4->[]
5:5->[(False,2)]
> prettyPrint $ condensation $ toDCG g2
1:[5]->[((),2)]
2:[1,2,3]->[((),3)]
3:[4]->[]
> linearize $ toDCG g2
[5,2,1,3,1,2,4]
This ordering is neither minimal nor valid since the ordering violates the dependencies. 5 invalidates 1, which 2 depends on. 2 invalidates 1 which 4 depends on.
A valid and minimal ordering is: [1,4,2,1,3,5]. By shifting the list to the right, we get [5,1,4,2,1,3] which is also a valid ordering.
If the direction of the graph is flipped, the ordering becomes: [4,2,1,3,1,2,5]. This is not a valid ordering either... At the boundaries, 5 can happen, and then 4, but 5 invalidates 1 which 4 depends on.
I believe the following algorithm will find a minimal string of vertices in linear time:
Decompose the graph into its strongly connected components. Existing algorithms do this in linear time.
In each strongly connected component each node needs to be listed both before and after every other node. List the nodes [1..n] of each strongly connected component in the following order [1..n] ++ [n-1..1]
Concatenate the strongly connected components together in order by a topological sort. Existing algorithms topologically sort directed acylic graphs like this in linear time.
Click here for the answer. Turing Machine
The question is to construct a Turing Machine which accepts the regular expression,
L = {a^n b^n | n>= 1}.
I am not sure if my answer is correct or wrong. Thank you in advance for your reply.
You cannot "accept the regular expression", only the language it describes. And what you provide is not a regular expression, but a set description. In fact, the language is not regular and therefore cannot be described by standard regular expressions.
The machine from your answer accepts the language described by the regular expression a^+ b^+.
A TM could mark the first a (e.g. by converting it to A) then delete the first b. And for each n one loop. If you and up with a string only of A, then accept.
As stated before, language L = {a^nb^n; n >= 1} cannot be described by regular expressions, it doesn't belong into the category of regular grammars. This language in particular is an example of context-free grammar, and thus it can be described by context-free grammar and recognized by pushdown automaton (an automaton with LIFO memory, a stack).
Grammar for this language would look something like this:
G = (V, S, R, P)
Where:
V is finite set of non-terminal characters, V = { S }
S is finite set of terminal characters, S = { a, b }
R is relation that describes "rewrites" from non-terminal characters to non-terminals and terminals, in this case R = { S -> aSb, S -> ab }
P is starting non-terminal character, P = S
A pushdown automata recognizing this language would be more complex, as it is a 7-tuple M = (Q, S, G, D, q0, Z, F)
Q is set of states
S is input alphabet
G is stack alphabet
D is the transition relation
q0 is start state
Z is initial stack symbol
F is set of accepting states
For our case, it would be:
Q = { q0, q1, qF }
S = { a, b }
G = { z0, X }
D will take a form of relation (current state, input character, top of stack) -> (output state, top of stack) (meaning you can move to a different state and rewrite top of stack (erase it, rewrite it or let it be)
(q0, a, z0) -> (q0, Xz0) - reading the first a
(q0, a, X) -> (q0, XX) - reading consecutive a's
(q0, b, X) -> (q1, e) - reading first b
(q1, b, X) -> (q1, e) - reading consecutive b's
(q1, e, z0) -> (qF, e) - reading last b
where e is empty word (sometimes called epsilon)
q0 = q0
Z = z0
F = { qF }
The language L = {a^n b^n | n≥1} represents a kind of language where we use only 2 character, i.e., a, b. In the beginning language has some number of a’s followed by equal number of b’s . Any such string which falls in this category will be accepted by this language. The beginning and end of string is marked by $ sign.
Step-1:
Replace a by X and move right, Go to state Q1.
Step-2:
Replace a by a and move right, Remain on same state
Replace Y by Y and move right, Remain on same state
Replace b by Y and move right, go to state Q2.
Step-3:
Replace b by b and move left, Remain on same state
Replace a by a and move left, Remain on same state
Replace Y by Y and move left, Remain on same state
Replace X by X and move right, go to state Q0.
Step-5:
If symbol is Y replace it by Y and move right and Go to state Q4
Else go to step 1
Step-6:
Replace Y by Y and move right, Remain on same state
If symbol is $ replace it by $ and move left, STRING IS ACCEPTED, GO TO FINAL STATE Q4
I have a combinatorial problem that can be solved inefficiently using the cartesian
product of multiple sets. Concretely, I have multiple items and multiple elements that
satisfy each item. The problem consists of finding all possible combinations of elements
that satisfy all items. For example:
items -> elements
------------------------
1 -> {a,b} // a and b cover the item 1
2 -> {a,b} // a and b cover the item 2
3 -> {a,b,c} // a, b and c cover the item 3
4 -> {a,b,e,f} // a, b, e, f cover the item 4
Alternative representation:
element -> items covered
------------------------
a -> {1,2,3,4}
b -> {1,2,3,4}
c -> {3}
e -> {4}
f -> {4}
The goal is to find all combinations that cover items 1,2,3,4.
Valid solutions are:
{a},{a,b},{a,c},{a,e},{a,f},{a,b,c},{a,b,e},{a,b,f},{a,b,c,e},{a,b,c,f}
{b},{b,c},{b,e},{b,f},{b,c,e},{b,c,f}
Note that the order is not important, so {a,b} = {b,a} ({a,b} x {c,d} = {c,d} x {a,b}).
Also, note that {a,a,a,a}, {a,a,a,b}... are redundant combinations.
As you can see, this problem is similar to the set cover problem, where the universe
of elements for this example are the items U={1,2,3,4} and the set of subsets from U is S={ab={1,2,3,4},c={3},ef{4}}, where set {1,2,3,4} is the set of items covered by the element a and b, {3} is the set of elements covered by c, and {4} is the set of elements covered by elements e and f. However, the goal here is not finding the
minimal combination of sets from S that covers all elements from U, but finding all combinations of elements {a,b,c,e,f} that cover all items {1,2,3,4}.
A näive implementation could be done by performing a cartesian product between
sets for 1,2,3 and 4, and then filtering the combinations that are redundant. However,
this approach is very inefficient. Suppose I have this situation:
1 -> {a,b,c,d,e,f,g,h}
2 -> {a,b,c,d,e,f,g,h}
3 -> {a,b,c,d,e,f,g,h}
4 -> {a,b,c,d,e,f,g,h}
5 -> {a,b,c,d,e,f,g,h}
6 -> {a,b,c,d,e,f,g,h,i}
A cartesian product between the six sets will result in a 8^5*9=294912 combinations,
when there are actually many fewer combinations, which are: {a,b,c,d,e,f,g} U {a,b,c,d,e,f,g} x {i}.
Another way to solve this problem is to enumerate all elements, skipping
the combinations that are equivalent to other previously generated, and also
skipping repeated elements. This is kinda easy to compute and can be implemented
as an iterator that returns a combination at a time, but I don't know if there is
a better way to solve this problem, or if this problem was studied before.
How would you solve this problem?
First, realize that if a set of elements does not satisfy all items, neither does any of its subsets.
Second, realize that if a set satisfies all items, so do all its supersets.
Now, all you have to do is:
Let S be the set of all elements.
Let R be the empty set.
Define a function find( s, r ) which does:
If r includes s, return r.
If s does not satisfy all items, return r.
Otherwise add s to r.
For every item I in s,
let s' be s-I
let s be f(s', r)
return s.
Just call find(S,R) and you have your answer.
This method performs some duplicate tests, but always kills a branch whenever it is identified as such. This leads to a lot of pruning on a large set of elements.
Both lookup of whether r includes a particular set of elements and the check if s satisfies all items can be made very fast at the expense of extra memory.
What if you did this:
1 -> {a,b}
2 -> {b,c}
3 -> {a,b,c}
4 -> {a,e,f}
=>
a -> [1,3,4]
b -> [1,2,3]
c -> [2,3]
e -> [4]
f -> [4]
Then enumerate the combinations of the left side that provide (at least) [1,2,3,4]
For each item in the set of all-satisfying sets, enumerate combinations
with other items.
All-Satisfying-Sets: {{a,b},{b,e},{b,f}}
Combinations within All-Satisfiying-Sets: {{a,b,e},{a,b,f},{b,e,f},{a,b,e,f}}
Others: {c}
Combinations with Others: {{a,b,c},{b,e,c},{b,f,c}
,{a,b,e,c},{a,b,f,c},{b,e,f,c},{a,b,e,f,c}}
Or you could do this in Haskell:
import Data.List (union, subsequences, sort)
example1 = [(["a"],[1,2,3,4])
,(["b"],[1,2,3,4])
,(["c"],[3])
,(["e"],[4])
,(["f"],[4])]
example2 = [(["a"],[1,2,3,4,5,6])
,(["b"],[1,2,3,4,5,6])
,(["c"],[1,2,3,4,5,6])
,(["e"],[1,2,3,4,5,6])
,(["f"],[1,2,3,4,5,6])
,(["g"],[1,2,3,4,5,6])
,(["h"],[1,2,3,4,5,6])
,(["i"],[6])]
combs items list =
let unify (a,b) (a',b') = (sort (a ++ a'), sort (union b b'))
in map fst
. filter ((==items) . snd)
. map (foldr unify ([],[]))
. subsequences
$ list
OUTPUT:
*Main> combs [1..4] example1
[["a"],["b"],["a","b"],["a","c"],["b","c"],["a","b","c"],["a","e"],["b","e"],
["a","b","e"],["a","c","e"],["b","c","e"],["a","b","c","e"],["a","f"],["b","f"],
["a","b","f"],["a","c","f"],["b","c","f"],["a","b","c","f"],["a","e","f"],
["b","e","f"],["a","b","e","f"],["a","c","e","f"],["b","c","e","f"],
["a","b","c","e","f"]]