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by: necessitypro
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Interval partitioning problem different approach – arrange lectures in minimum number of classrooms

The problem of scheduling lectures in minimum number of classrooms is as follows: Find minimum number of classrooms to schedule all lecture so that no two occur at the same time in the same room.

The common algorithm that I find in books is:

Sort intervals by starting time so that s1 ≤ s2 ≤ ... ≤ sn. d ← 0 //number of classrooms for j = 1 to n {  if (lecture j is compatible with some classroom k)  schedule lecture j in classroom k  else  allocate a new classroom d + 1  schedule lecture j in classroom d + 1  d ← d + 1 } 

Now, I was thinking of an alternate approach where I sort my lectures by finishing times in ascending order and every time I check if lecture j is compatible with some classroom k and there are multiple classrooms that are compatible with that lecture, I schedule it in the classroom which the last jobs finish time in that classroom is closest to that jobs start time, i.e minimise the time a classroom is empty.

Sort intervals by starting time so that f1 ≤ f2 ≤ ... ≤ fn. d ← 0 //number of classrooms for j = 1 to n {  if (lecture j is compatible with some classroom k)  schedule lecture j in classroom k which was used last  else  allocate a new classroom d + 1  schedule lecture j in classroom d + 1  d ← d + 1 } 

I would like to know if this approach is right(not necessarily optimal). I have dry run it on a couple of cases, and looks to be okay. If yes, how can I prove its correctness? If not, how can what changes can I make the algorithm work.

Minimum time to assemble multipart object

I’ve been asked to do this task in an online assessment. I’ve passed, so my solution is supposedly correct, but I am unable to prove it. The task is:

Given a set of parts (array of integer part sizes), worker must put them all together. Parts are assembled in pairs. To put together two parts of sizes A and B worker needs A+B minutes. The resulting part’s size is also A+B. Write a program to determine the minimum time required to put together given set of parts.

My solution was:

MinHeap h for p in parts:   h.push(p) time = 0 while h.size() > 1:   v1 = h.pop()   v2 = h.pop()   time += v1 + v2   ### assembly time   h.push(v1 + v2)   ### adding new part return time 

This solution passed all tests.


Does this solution produce correct min time, and if yes, how to prove it?

Compare two plots by finding the minimum distance among points

I have a question about comparing the points within two plots. I would like to compare two plots and find the minimum distance among their points, in order to find the nearest/common points (i.e. those ones with minimum -or zero-distance) and plot it (overlapping). What I did is to extract the coordinates of their respectively points. But I do not know how to compare them and/or the two plots. I used the following line of code, but the result is completely different from what I am looking for.

Outer[EuclideanDistance, seq1, seq2, 1] // Flatten 

enter image description here

enter image description here

Could you please help me? Many thanks, Val

Algorithm to count the number of subsets of size k with sum of all its elements minimum possible

An array is given eg:-1 2 2 2 and we need to count the number of subsets for it of size k which has the sum of elements minimum possible

here the subsets of size k=3 are:- 122 122 122 222

we see that there are 3 subsets having the minimum sum

I did this in bruteforce approach..first I stored the subsets of size k in a vector and then found the sum for each of the subsets and stored them in another vector and then counted the frequency of the minimum element

how to optimize it?

Compare 2 images with minimum difference value

Sorry if this is the wrong place to ask this question, I’ll update/move my question if it is.

There is a linux command compare that I’m using to compare a difference between 2 images, this is the command:

compare 1.jpg 2.jpg diff.jpg

It’s working very well, the problem is: if both images has no difference, it still creates a diff file, and you open this file and it’s just a blank image.

Is there a way to tell this command to just create this diff file only if there is actually a difference between these 2 images?


Recommended minimum dimensions for touch areas for use with winter gloves

Are there any guidelines, recommondations or research about using touch screens with winter gloves? I’m aware of the normal recommended defaults (44pt according to Apple’s HIG, 48pt in Material Design), but those are minima.

Winter gloves come in different sizes and thicknesses, so I guess bigger touch areas are better, but I’m not sure of the thickness actually linearly correlates with precision. After all, it’s not about the area covered by the glove on press, but about the pixel position that’s converted to by the touch screen driver. Theoretically, a trained user with gloves could be as precise as one without any.

Maximum and Minimum distance from query point within bounding box

I’m reading an article regarding approximating sums using KD-trees (similar to FMM).

As part of the effort I’m trying to make sense of this article , which is cited.

I’m having trouble understanding this part:

Computing the maximum variation of weights over all points below node ND is easy. We know the location of xquery and we know the bounding hyperrectangle of the current node. A simple algorithm costing O(Number of tree dimensions) can compute the shortest and largest possible distances to any point in the node. From these two values, and the assumption that the weight function is non-increasing, we can compute the minimum and maximum possible weights wmin and wmax of any datapoint below node ND.

The only solution I can think of is to bound the variation (wmax-wmin) from above using the four corners of the bounding box, and then computing the distance from the query point to each of the four corners is indeed O(d). But this doesn’t seem to be the author’s intention. Could anybody point me to what I”m missing here?

Subwoofer Sound Settings revert back to default when sound level is set to minimum (mute/off)

I am having an annoying and destructive problem with my Subwoofer Sound Settings, it keep reverting back to default when the sound level is set to minimum (mute/off). This problem has blown my sub-woofer on my Macbook Pro. Before I replace the subwoofer, I would like to see if I could remedy this problem.

Background – At times I need to raise the volume for some shows or movies. I noticed my subwoofer about to explode so I adjusted the Subwoofer level to medium. This helped, BUT, at times I would lower the level and hit the minimum level (basically muting it). Well, Ubuntu eventually blew out my subwoofer. When I went back to these settings, I noticed the subwoofer level was back to the same level as the Output Volume. I also noticed the PROBLEM… if the output volume level hits minimum, this does the same with the subwoofer, and in turn, raising this level also raises the subwoofer in tandem, at times to maximum. I would like my subwoofer to NEVER exceed the level in which I set it too.

I assume I am not the only one whom has experienced this destructive hardware setting.

Cheers, – Ian