L = {w | 2|w|a != 3|w|b + 2} ∪ {aaab, bbba}.
|w|a = number of a's, same for b.
How do I use the top of the stack when only the number of a's/b's counts and they can be in any order?
It is not completely clear what you mean by "use the top of the stack."
To construct a PDA may start with one for the language {w: |w|a = |w|b}.
When it reads an a it
puts an a on the stack if there is already an a or the stack is empty
removes a b from the stack
For the case of reading a b symmetrically. The PDA accepts if the stack is empty when the entire input has been read. So the stack indicates whether so far more a or more b have been read, because the majority symbol is the one that it contains.
With the factors 2 and 3 and the added 2 b it becomes a bit more complicated. I would not handle this in the stack but in the states. This means, we implement a counter for 0 or 1 a and for 0,1 or 2 b there. When we read a an input symbol x, we first try to increment the respective counter in the state. If this is possible, this is the only thing we do. If the counter is full, we set it to zero and take the action corresponding to this symbol in the PDA above for the stack.
For the +2, we count the first two b in the states before we actually start filling the counter.
Related
I have some activities with weights, and I would like to select non overlapping activities by maximizing the total weight. This is known problem and solution exists.
In my case, I am allowed to shift the start time of activities in some extend while duration remains same. This will give me some flexibility and I might increase my utilization.
Example scenario is something like the following where all activities are supposed to be in interval (0-200):
(start, end, profit)
a1: 10 12 120
a2: 10 13 100
a3: 14 18 150
a4: 14 20 100
a5: 120 125 100
a6: 120 140 150
a7: 126 130 100
Without shifting flexibility, I would choose (a1, a3, a6) and that is it. On the other hand I have shifting flexibility to the left/right by at most t units for any task where t is given. In that case I might come up with this schedule and all tasks can be selected except a7 since conflict cannot be avoided by shift .
t: 5
a1: 8 10 120 (shifted -2 to left)
a2: 10 13 100
a3: 14 18 150
a4: 18 24 100 (shifted +4 to right)
a5: 115 120 100 (shifted -5 to left)
a6: 120 140 150
In my problem, total time I have is very big with respect to activity duration. While activities are like 10sec on average, total time I have would even be 10000sec. However that does not mean all of activities can be selected since shifting flexibility would not be enough for some activities to non-overlap.
Also in my problem, there are clusters of activities which overlaps and very big empty space where no activities and there comes another cluster of overlapping activities i.e a1, a2, a3 and a4 are let say cluster1 and a5, a6 and a7 is cluster2. Each cluster can be expanded in time by shifting some of them to left and right. By doing that, I can select more activities than the original activity selection problem. However, I do not know how to decide which tasks to be shifted to left or right.
My expectation is to find an near-optimal solution where total profit would be somehow local optima. I do not need global optimum value. Also I do not have any criteria about cluster utilization., i.e I do not have a guarantee about a minimum number of activity per cluster etc. Actually, these clusters something I visually describe. There is not defined cluster. However, in time domain, activities are separated as clusters somehow.
Also activity start and end times are all integers since I can dismiss fractions. I would have around 50 activities whose duration would be 10 on average. And time window is like 10000.
Are there any feasible solution to this problem?
You mentioned that you can partition the activities into clusters that don't overlap even if activities within them are shifted to the extent. Each of these clusters can be considered independently, and the optimal results computed for each cluster simply summed up for the final answer. So the first step of the algorithm could be a trial run that extends all activities in both directions, finds which ones form clusters, and process each cluster independently. In the worst case, all of the activities might form a single cluster.
Depending on the maximum size of the remaining clusters, there are several approaches. If it's under 20 (or even 30, depending on whether you want your program to run in seconds or minutes), you could combine search over all subsets of activities in the given cluster with a greedy approach. In other words: if you are processing a subset of N elements, try every one of its 2^N possible subsets (okay, 2^N-1 if we forget the empty subset), check whether the activities in this specific subset can be scheduled in non-overlapping manner, and pick the subset that is eligible and has maximum sum.
How do we check that a given subset of activities can be scheduled in non-overlapping manner? Let's sort them in ascending order of their end and process them from left to right. For every activity, we try to schedule it as early as possible, making sure it does no intersect with activities we already considered. So, the first activity in the cluster is always started time t earlier than originally planned, the second one is started either when the first one ends, or t earlier than originally planned, whichever is larger, and so on. If at any point we can't schedule the next activity in a way that it does not overlap with previous one, then there is no way to schedule the activities in this subset in a non-overlapping manner. This algorithm takes O(NlogN) time, and overall each cluster is processed in O(2^N * NlogN). Once again, note that this function grows very quickly, so if you are dealing with large enough clusters, this approach goes out the window.
===
Another approach is specific to the additional restrictions you provided. If the activities' starts and ends and parameter t are all measured in integer number of seconds, and t is about 2 minutes, then the problem for each cluster is set in a small discrete space. Even though you could position a task to start at a non-integer second value, there always is an optimal solution that uses only integers. (To prove it, consider an optimal solution that does not use integers - since t is integer, you can always shift tasks, starting from the leftmost, to the left a bit so that it starts at an integer value.)
Knowing that the start and end times are discrete, you can build a DP solution: process the activities in the ascending order of their end*, and memoize the maximum possible sum of weights you can obtain from the first 1, 2, ..., N activities for each x from activity_start - t to activity_start + t if a given activity ends at time x. If we denote this memoized function as f[activity][end_time], then the recurrence relation is f[a][e] = weight[a] + max(f[i][j] over all i < a, j <= e - (end[a] - start[a]), which roughly translates to "if activity a ended at time e, the previous activity must have ended at or before start of a - so let's pick the maximum total weight over previous activities and their ends, and add the current activity's weight".
*Again, we can prove that there is at least one optimal answer where this ordering is preserved, even though there might be other optimal answers which do not possess this property
We could go further and eliminate the iteration over previous activities, instead encoding this information in f. Its definition would then change to "f[a][e] is the maximum possible total weight of the first a activities if none of them ends after e", and recurrence relation would become f[a][e] = max(f[a-1][e], weight[a] + max(f[a-1][i] over i <= e - (end[a] - start[a])])), and its computational complexity would be O(X * N), where X is the total span of the discrete space where task starts/ends are placed.
I assume you need to compute not just the maximum possible weight, but also the activities you need to select to obtain it, and possibly even the exact time each of them needs to be started. Thankfully, we can derive all of this from the values of f, or compute it at the same time as we compute f. The latter is easier to reason about, so let's introduce a second function g[activity][end]. g[activity][end] returns a pair (last_activity, last_activity_end), essentially pointing us to the exact activity and its timing that the optimal weight in f[activity][end] uses.
Let's go through the example you provided to illustrate how this works:
(start, end, profit)
a1: 10 12 120
a2: 10 13 100
a3: 14 18 150
a4: 14 20 100
a5: 120 125 100
a6: 120 140 150
a7: 126 130 100
We order the activities by their end time, thereby swapping a7 and a6.
We initialize the values of f and g for the first activity:
f[1][7] = 120, f[1][8] = 120, ..., f[1][17] = 120, meaning that the first activity could end anywhere from 7 to 17, and costs 120. f[1][i] for all other i should be set to 0.
g[1][7] = (1, 7), g[1][8] = (1, 8), ..., g[1][17] = (1, 17), meaning that the last activity that was included in f[1][i] values was a1, and it ended at i. g[1][i] for all i outside [7, 17] is undefined/irrelevant.
That's where something interesting begins. For each i such that a2 cannot end at time i, let's assign f[2][i] = f[1][i], g[2][i] = g[1][i], which essentially means that we wouldn't be using activity a2 in those answers. For all other i, namely, in [8..18] interval, we have:
f[2][8] = max(f[1][8], 100 + max(f[1][0..5])) = f[1][8]
f[2][9] = max(f[1][9], 100 + max(f[1][0..6])) = f[1][9]
f[2][10] = max(f[1][10], 100 + max(f[1][0..7])). This is the first time when the second clause is not just plain 100, as f[1][7]>0. It is, in fact, 100+f[1][7]=220, meaning that we can take activity a2, shift it in a way that puts its end at time 10, and get a total weight of 220. We continue computing f[2][i] this way for all i <= 18.
The values of g are: g[2][8]=g[1][8]=(1, 8), g[2][9]=g[1][9]=(1, 9), g[2][10]=(2, 10), because it was optimal to take activity a2 and end it at time 10 in this case.
I hope the pattern of how this continues is visible - we compute all the values of f and g through the end, and then pick the maximum f[N][e] over all possible end times e of the last activity. Armed with the auxiliary function g, we can traverse the values backwards to figure out the exact activities and times. Namely, the last activity we use and its timing is in g[N][e]. Let's call them A and T. We know that A began at T-(end[A]-start[A]). Then, the previous activity must have ended at that point or before - so let's look at g[A-1][T-(end[A]-start[A]) for it, and so on.
Note that this approach works even if you don't partition anything into clusters, but with the partitioning, the size of the space in which tasks can be scheduled is reduced, and with it the runtime.
You might notice that neither of these solutions is polynomial in the size of input. I have a feeling that your problem doesn't have a general polynomial solution, but I was unable to prove it by reducing another NP-complete problem to it. Would be really curious to read a reduction / better general solution!
I have read this: https://www.topcoder.com/community/competitive-programming/tutorials/binary-search.
I can't understand some parts==>
What we can call the main theorem states that binary search can be
used if and only if for all x in S, p(x) implies p(y) for all y > x.
This property is what we use when we discard the second half of the
search space. It is equivalent to saying that ¬p(x) implies ¬p(y) for
all y < x (the symbol ¬ denotes the logical not operator), which is
what we use when we discard the first half of the search space.
But I think this condition does not hold when we want to find an element(checking for equality only) in an array and this condition only holds when we're trying to find Inequality for example when we're searching for an element greater or equal to our target value.
Example: We are finding 5 in this array.
indexes=0 1 2 3 4 5 6 7 8
1 3 4 4 5 6 7 8 9
we define p(x)=>
if(a[x]==5) return true else return false
step one=>middle index = 8+1/2 = 9/2 = 4 ==> a[4]=5
and p(x) is correct for this and from the main theory, the result is that
p(x+1) ........ p(n) is true but its not.
So what is the problem?
We CAN use that theorem when looking for an exact value, because we
only use it when discarding one half. If we are looking for say 5,
and we find say 6 in the middle, the we can discard the upper half,
because we now know (due to the theorem) that all items in there are > 5
Also notice, that if we have a sorted sequence, and want to find any element
that satisfies an inequality, looking at the end elements is enough.
I don't know whether the term "Lazy" Binary Search is valid, but I was going through some old materials and I just wanted to know if anyone can explain the algorithm of a Lazy Binary Search and compare it to a non-lazy Binary Search.
Let's say, we have this array of numbers:
2, 11, 13, 21, 44, 50, 69, 88
How to look for the number 11 using a Lazy Binary Search?
Justin was wrong.
The common binarySearch algorithm first checks whether the target is equal to current middle entry before proceeding to the left or right halves if required. Lazy binarySearch algorithm postpones the equality check until the very end.
algorithm lazyBinarySearch(array, n, target)
left<-0
right<-n-1
while (left<right) do
mid<-(left+right)/2
if(target>array[mid])then
left<-mid+1
else
right<-mid
endif
endwhile
if(target==array[left])then
display "found at position", left
else
display "not found"
endif
In your case, in an array,
2 11 13 21 44 50 69 88
and you want to search for 11
First we do a trace of common binary search,
index 0 1 2 3 4 5 6 7
2 11 13 21 44 50 69 88
left mid right
First while loop:
left <= right, we enter the first while loop. We calculated the mid index by (0+7)/2=3.5=3 by integer division, mid = 3. straight away we check if target 11 is equal to the mid index entry, 11 != 21, then we decide whether to go left or right, we finds out 11 < 21, should go left. left index remains unchanged, right index becomes mid index -1, right = 3 - 1 = 2. Done this step.
Second while loop:
left <= right, 0 <= 2, we enter the seond while loop. Mid index is recalcuated: (0+2)/2=1, mid = 1. At once we do the equality check, target 11 is the same as the index 1 entry, 11 == 11. We found this entry, leaving behind all the left right mid indexes (don't care) and breaks out the while loop, return index 1.
Now we trace this search by lazy binazySearch algorithm, initial array with left/right indexes set up the same as previous.
First while loop:
left < right, we enter the first while loop. Mid index is calculated as the same above = 3. Instead of doing an equality check in common binarySearch we do a comparison with the mid index entry this time. And the comparison only checks if our target 11 is greater than the mid index entry, leaving whether they equal or not to the very end outside the while loop. So we find out 11 < 21, right index is reset to the mid index, right = 3. Done this step.
Second while loop:
left < right, 0 < 3, we enter the second while loop. mid index is recalculated as mid = (0+3)/2 = 1 by integer division. Again we do a comparison with mid index entry 11, we realise it's not greater than mid index entry. We fall into the else part of the while loop and reset the right index to be mid index, right = 1. Done this step.
Third while loop:
left < right, 0 < 1, this time we have to re-enter the while loop again since it still satisfies the while condition. Mid index becomes (0+1)/2=0, mid = 0. After comparing target 11 with mid index entry 2 we found out it's greater than it (11 > 2), left index is reset to mid + 1, left = 0 + 1 = 1. Done this step.
With left index = 1 and right index = 1, left = right, while loop condition is no longer satisfied, so there's no need to re-enter. We fall into the if part down below. Target 11 equals left index entry 11, we found the target and returns left index 1.
As you can see, lazy binarySearch does one more while loop step to finally realise the index is actually 1. And this is how the word "postpones the equality check" means in the definition I mentioned in the very beginning. Clearly the lazy binarySearch algorithm does more things than common binarySearch before reaching the program termination. And the term "lazy" is reflected in the time of when to check the target equals the mid index entry.
However lazy binarySearch is more preferable to use under some other circumstances, but it's not in the context of this case.
(The reasoning part of the algorithm is done by me, anyone wishes to copy please do credit)
source: "Data Structures Outside In With Java" by Sesh Venugopal, Prentice Hall
As far as I am aware "Lazy binary search" is just another name for "Binary search".
I read in a book on non-deterministic mapping there is mapping from Q*∑ to 2Q for M=(Q,∑,trans,q0,F)
where Q is a set of states.
But I am not able to understand how it's 2Q;
if there are 3 states a, b, c, how does it map to 8 states?
I always found that the easiest way to think about these (since the set of states is finite) is as having each of those subsets be an encoding of a base-2 number that ranges from 0 (all bits zero) to 2|Q|-1 (all bits one), where there are as many bits in the number as there are members in the state set, Q. Then, you can just take one of these numbers and map it into a subset by using whether a particular bit in the number is set. Easy!
Here's a worked example where Q = {a,b,c}. In this case, |Q| is 3 (there are three elements) and so 23 is 8. That means we get this if we say that the leading bit is for element a, the next bit is for b, and the trailing bit for c:
0 = 000 = {}
1 = 001 = {c}
2 = 010 = {b}
3 = 011 = {b,c}
4 = 100 = {a}
5 = 101 = {a,c}
6 = 110 = {a,b}
7 = 111 = {a,b,c}
See? That initial three states has been transformed into 8, and we have a natural numbering of them that we could use to create the labels of those states if we chose.
Now, to the interpretations of this within a non-deterministic context. Basically, the non-determinism means that we're uncertain about what state we're in. We represent this by using a pseudo-state that is the set of “real” states that we might be in; if we have total non-determinism then we are in the pseudo-state where all real-states are possible (i.e., {a,b,c}) whereas the pseudo-state where no real-states are possible (i.e., {}) is the converse (and really ought to be impossible to reach in the transition system). In a real system, you're usually not dealing with either of those extremes.
The logic of how you convert the deterministic transition system into a non-deterministic one is rather more complex than I want to go into here. (I had to read a substantial PhD thesis to learn it so it's definitely more than an SO answer's worth!)
2Q means the set of all subsets of Q. For each state q and each letter x from sigma, there is a subset of Q states to which you can go from q with letter x. So yeah, if there are three states abc the set 2Q consists of 8 elements {{}, {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}}. It doesn't map to 8 states, it maps to one of these 8 sets. HTH
I want to combine multiple lists of items into a single list, retaining the overall order requirements. i.e.:
1: A C E
2: D E
3: B A D
result: B A C D E
above, starting with list 1, we have ACE, we then know that D must come before E, and from list 3, we know that B must come before A, and D must come after B and A.
If there are conflicting orderings, the first ordering should be used. i.e.
1: A C E
2: B D E
3: F D B
result: A C F B D E
3 conflicts with 2 (B D vs D B), therefore requirements for 2 will be used.
If ordering requirements mean an item must come before or after another, it doesn't matter if it comes immediately before or after, or at the start or end of the list, as long as overall ordering is maintained.
This is being developed using VB.Net, so a LINQy solution (or any .Net solution) would be nice - otherwise pointers for an approach would be good.
Edit: Edited to make example 2 make sense (a last minute change had made it invalid)
The keyword you are probably interested in is "Topological sorting". The solution based on that would look as follows:
Create an empty directed graph.
Process sequences in order, for each two consecutive elements X,Y in a sequence add an edge X->Y to the graph, unless this would form a cycle.
Perform a topological sort on the vertices of the graph. The resulting sequence should satisfy your requirements.