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!
So I'm struggling to wrap my head around something I'm trying to write. Nest for-next loops are clearly the only way to go (as far as I can tell) but I just can't get any sort of pseudo code worked out. My problem is this, given a fixed number (lets say 100 for simplicity) I want to iterate through all combinations of sets of numbers upto 5 that totals 100. Lets say in steps of 5. So to be clear I would want to run the following first few example:
100
95 - 5
90 - 10
---
10 - 90
5 - 95
90 - 5 - 5
85 - 5 - 10
80 - 5 - 15
---
5 - 5 - 85
85 - 10 - 5
80 - 10 - 10
75 - 10 - 15
---
---
80 - 5 - 5 - 5 - 5
75 - 5 - 5 - 5 - 10
---
Hopefully that give you the idea of my my target is. My problem is that I just can't work out an effective way to program. I'm pretty competent in actually writing code (usually) but every time I sit down and do this I end up with 10's of nested for-next loops that just don't work!
To remove that nesting problem there's a simple approach : use a queue or a stack.
Here's some pseudo-code:
// add 1 item to start with in the queue
while(queue.Count > 0)
{
// 1. dequeue item from queue
// 2. do your work on it
// 3. if there's another combination emerging from step 2, enqueue it in the queue
}
// 4. this point will be reached once finished
Using a custom type to store whatever information you need for that job and by having a single loop instead of many should help you tackle the problem relatively quickly :D
I'm executing some code and then waiting somewhere between 1 second and 1 minute. I'm currently using random 0:01:00 /seed but what I really need is to be able to set a floor so that its waiting between 30 seconds and 1 minute.
If you want 0:0:30 to be the minimum and 0:1:0 to be the maximum, try the formula:
0:0:29 + random 0:0:31
This formula yields a "discretely distributed (pseudo)random value". If you want a "continuously distributed (pseudo) random value", you can use (just in R3) the formula:
0:0:30 + random 30.0
R2 does not have a native support for "continuously distributed (pseudo)random values".
Not my area of expertise, but:
00:00:30 + to time! (random 100% * (to integer! 00:00:30))
...appears to work, I think.
>>random/seed now/precise
>> t1: now wait 30 + random 30 difference now t1
== 0:00:39
How about the following:
0:00:30 + random 0:00:30
You could generate a whole number from 1 to 30 and subtract that number in seconds from 1 minute and 1 second.
(and about seeding, use that, but not constantly)
I have forked many child processes and assigned priority and core to each of them. Porcess A executes at period of 3 sec and process B at a period of 6 sec. I want them to execute in such a way that the higher priority processes should preempt lower priority ones only at predefined points and tried to acheive it with semaphores. I have used this same code snippets within the 2 processes with different array values in both.
'bubblesort_desc()' sorts the array in descending order and prints it. 'bubblesort_asc()' sorts in ascending order and prints.
while(a<3)
{
printf("time in sort1.c: %d %ld\n", (int)request.tv_sec, (long int)request.tv_nsec);
int array[SIZE] = {5, 1, 6 ,7 ,9};
semaphore_wait(global_sem);
bubblesort_desc(array, SIZE);
semaphore_post(global_sem);
semaphore_wait(global_sem);
bubblesort_asc(array, SIZE);
semaphore_post(global_sem);
semaphore_wait(global_sem);
a++;
request.tv_sec = request.tv_sec + 6;
request.tv_nsec = request.tv_nsec; //if i add 1ms here like an offset to the lower priority one, it works.
semaphore_post(global_sem);
semaphore_close(global_sem); //close the semaphore file
//sleep for the rest of the time after the process finishes execution until the period of 6
clk = clock_nanosleep(CLOCK_MONOTONIC, TIMER_ABSTIME, &request, NULL);
if (clk != 0 && clk != EINTR)
printf("ERROR: clock_nanosleep\n");
}
I get the output like this whenever two processes get activated at the same time. For example at time units of 6, 12,..
time in sort1.c: 10207 316296689
time now in sort.c: 10207 316296689
9
99
7
100
131
200
256
6
256
200
5
131
100
99
1
1
5
6
7
9
One process is not supposed to preempt the other while one set of sorted list is printing. But it's working as if there are no semaphores. I defined semaphores as per this link: http://linux.die.net/man/3/pthread_mutexattr_init
Can anyone tell me what can be the reason for that? Is there a better alternative than semaphores?
Its printf that's causing the ambiguous output. If the results are printed without '\n' then we get a more accurate result. But its always better to avoid printf statements for real time applications. I used trace-cmd and kernelshark to visualize the behaviour of the processes.
I've been melting my brains over a peculiar request: execute every two minutes a certain query and if it returns rows, send an e-mail with these. This was already done and delivered, so far so good. The result set of query is like this:
+----+---------------------+
| ID | last_update |
+----+---------------------|
| 21 | 2011-07-20 13:03:21 |
| 32 | 2011-07-20 13:04:31 |
| 43 | 2011-07-20 13:05:27 |
| 54 | 2011-07-20 13:06:41 |
+----+---------------------|
The trouble starts when the user asks me to modify it so the solution so that, e.g., the first time that ID 21 is caught being more than 5 minutes old, the e-mail is sent to a particular set of recipients; the second time, when ID 21 is between 5 and 10 minutes old another set of recipients is chosen. So far it's ok. The gotcha for me is from the third time onwards: the e-mails are now sent each half-hour, instead of every five minutes.
How should I keep track of the status of Mr. ID = 43 ? How would I know if he has already received an e-mail, two or three? And how to ensure that from the third e-mail onwards, the mails are sent each half-hour, instead of the usual 5 minutes?
I get the impression that you think this can be solved with a simple mathematical formula. And it probably can be, as long as your system is reliable.
Every thirty minutes can be seen as 360 degrees, or 2 pi radians, on a harmonic function graph. That's 12 degrees = 1 minute. Let's take cosin for instance:
f(x) = cos(x)
f(x) = cos(elapsedMinutes * 12 degrees)
Where elapsed minutes is the time since the first 30 minute update was due to go out. This should be a constant number of minutes added to the value of last_update.
Since you have a two minute window of error, it will be time to transmit the 30 minute update if the the value of f(x) (above) is between the value you would get at less than one minute before or after the scheduled update. Which would be = cos(1* 12 degrees) = 0.9781476007338056379285667478696.
Bringing it all together, it's time to send a thirty minute update if this SQL expression is true:
COS(RADIANS( 12 * DATEDIFF(minutes,
DATEADD(minutes, constantNumberOfMinutesBetweenSecondAndThirdUpdate, last_update),
CURRENT_TIMESTAMP))) > 0.9781476007338056379285667478696
If you need a wider window than exactly two minutes, just lower this number slightly.