|
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
|
|
|
|
|
Now you investigate to see what gives the best solution
- no sort of cubes
- sort from biggest to smallest
- sort from smallest to biggest
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
|
|
|
|
|
Thanks
Yes; Best way is here:
LinkBinsOrItems.sortItemList();
LinkBinsOrItems.sortBinList();
Best Sort Mode:
private void btnBestMode_Click(object sender, EventArgs e)
{
if (LinkBinsOrItems.listOfBins == null)
{
if (LinkBinsOrItems.listOfItems == null)
{
MessageBox.Show("Item List And Also Bin List Are Empty!!"
+"You Must Enter Bins And Items..."
, "Bin List And Item list Empty Error", MessageBoxButtons.OK
, MessageBoxIcon.Information);
return;
}
MessageBox.Show("Bin List Is Empty!! You Must Enter Bins...", "Bin List Empty Error"
, MessageBoxButtons.OK, MessageBoxIcon.Information);
return;
}
if (LinkBinsOrItems.listOfItems == null)
{
MessageBox.Show("Item List is Empty!! You Must Enter Items..."
, "Item List Empty Error", MessageBoxButtons.OK, MessageBoxIcon.Information);
return;
}
DialogResult result = MessageBox.Show("Are you sure you want to do sorting "
+"for creat more free space?(This can change items places"
, "Sort For Best Mode",MessageBoxButtons.YesNo,MessageBoxIcon.Question);
if (result == DialogResult.No)
return;
LinkedListNode<Item> itemNode;
LinkedListNode<Bin> binNode;
itemNode = LinkBinsOrItems.listOfItems.First;
for (int i = 1; i <= LinkBinsOrItems.listOfItems.Count; i++)
{
itemNode.Value.BinNumber = -1;
}
binNode = LinkBinsOrItems.listOfBins.First;
for (int i = 1; i <= LinkBinsOrItems.listOfBins.Count; i++)
{
binNode.Value.clearRelations();
}
LinkBinsOrItems.sortItemList();
LinkBinsOrItems.sortBinList();
listBox1.Items.Add("");
itemNode = LinkBinsOrItems.listOfItems.First;
for (int i = 1; i <= LinkBinsOrItems.listOfItems.Count; i++)
{
listBox1.Items.Add(itemNode.Value);
listBox2.Items.Add(LinkBinsOrItems.placeItemInBin(itemNode.Value));
itemNode = itemNode.Next;
}
itemReportInitaling();
BinReportInitaling();
}
|
|
|
|
|
Cenator wrote: Thanks [Rose] Thank you.
CP have a standard way to reward contributors: click on the green arrow on left of the useful answer/advice. the author is rewarded with reputation points and you also gain some organizer points.
--------------------------------------------
Depending on reality of the problem size of the warehouse and number of cubes to fit, you may want to optimize the answer even further.
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
|
|
|
|
|
I wasn't sure if this forum was the best one for this question, so I apologize in advance if it isn't.
I'm currently learned Gene Expression Programming and I'm a little unclear as to how the mutation rate is applied to a population. I've looked at source code from different implementations and I've seen it used different ways and would like to know if there is actually a proper way to implement it or if it can be implement at the programmers discretion.
As an example, say there are 10 chromosomes, each with 2 genes, and each gene has 10 nodes and the mutation rate (mr) is 0.1. I've seen code that mutates nodes implemented in these way:
- Loop through each chromosome and generate a random number (rn) between 0 - 1 on each loop. If the rn <= mr then select the gene for mutation. Then loop through every node of the chromosome, ignoring gene boundaries (so in this example there are 20 nodes = 10 nodes/gene * 2 genes) and generate another rn and if the rn <= mr, mutate the node.
- Loop through each chromosome and generate a random number (rn) between 0 - 1 on each loop. If the rn <= mr then select the gene for mutation. Then loop through each gene, again generating a rn on each loop, and if the rn <= mr select the gene for mutation, otherwise continue the loop. If the gene is selected for mutation, loop through each node of the selected gene (10 nodes since there are 10 nodes/gene), generating a rn on each loop and mutate the selected node if rn <= mr.
- Randomly select 10% of the nodes in the entire populate (in this case 20 nodes since there are 200 nodes = 10 chromosomes * 2 genes/chromosome * 10 nodes/gene).
- Randomly select 10% of the chromosomes (in this case 1 chromosome) and then select the nodes for mutation following the same procedure as Item 1 or Item 2 above, starting from the second loop.
I believe that either Item 1 or Item 2 would be the correct way to apply the mutation operator, but I haven't been able to find a definitive answer either through searches on Google/CP or in the book I have been reading.
One other thing that is mentioned in the book I am reading, but isn't explained in detail is the following: there is a population of 10 chromosomes, each with 1 gene, and a total length of 14 nodes. The book states that the mutation rate should be equivalent to two one-point mutations per chromosome, which in this case (because the chromosome's length is 14) should be 0.143. How was a mr of 0.143 derived from the chromosome length of 14? This isn't explained anywhere so far, even though this equivalence has been mentioned multiple times.
Any information, or links to answers, would be greatly appreciated. Thank you in advance for any help.
A black hole is where God tried to divide by zero.
There are 10 kinds of people in the world; those who understand binary and those who don't.
|
|
|
|
|
IMHO the type of mutation doesn't matter that much. Its purpose is to "jump" to nearby areas in the problem space. Any variation will accomplish that. The best mutation rate can't be stated as a hard and fast rule, as in that book; it will depend on the specific problem you're trying to solve.
(BTW, there's no "singularity" at the center of a black hole; the "infinite density" is an artifact of using continuous equations to describe a discrete system. The physicists have mistaken their equations for the reality they're trying to model.)
|
|
|
|
|
Hi all,
I have three arrays: X[i], Y[i] and Value[i], which are the position (x, y) of the pixel "i" and its value (pixel actived as 1, otherwise 0). Pleae have a look at this graph linked:
[See here, please: "graph"]
Data format is also shown at the right side of the graph. I would like to extract the sub-array which represents connected-components (or spots) in the image since the data is sorted in order of ascending x rather than in order of spots.
It is easy to achieve my purpose by programing in Matlab which already has some advance functions. But for the reason that it is huge data and low processing speed in matlab), I have to do carry out it by using C/C++.
Does anyone have some good idea to extract sub-array for each spot? Thanks in advanced.
With best regards
feifeihanyu
modified 10-Jul-15 10:31am.
|
|
|
|
|
You need some extra information to connect the pixels that make a spot. As it stands there is no way to be certain that, for example, spot(5, 8) connects to the other two. What would happen if the blue spot was touching the red one, how would you tell where the separation occurs?
|
|
|
|
|
Hi, thanks for reply!
All connected-pixels make a spot! If the blue spot touchs the red one, then they are actually one spot!
There is an algorithm to label connected-components in binary image. see here Blob detection
The difference from that example is that I already have arrays representing non-zero pixel positions with pixel values instead of a full binary image! The problem is that (x, y) of spots/blobs is sorted in order. so the positions (x, y) of different spots and blobs may mix together if they local at the same raw or column! for example, the positions of the red and the purple,please see my graph and right side, I labeled the positions from the same spot by lines.
Actually, I can transform the arrays into a full binary image, then perform a similar algorithm like Blob detection . But it will be a problem if I have many images which have few blobs but with big dimension, for example, 4096x4096. then it not wise to do that becasue many time will be wasted in loop search. I need an algorithm dealing with the arrays directly in stead of a full binary image. This is why I post a question here for asking for help.
regards
feifeihanyu
modified 10-Jul-15 12:21pm.
|
|
|
|
|
can somebody tell me who i can talk to regarding Mathematics and Algorithms
|
|
|
|
|
Member 11849553 wrote: who i can talk to regarding Mathematics and Algorithms You can talk to anybody about it. However, if you have a proper question, then feel free to post it in the appropriate forum.
|
|
|
|
|
If u take a tour of MATLAB groups you`ll most probably find some'C' export tool that converts your m-files to c-files. whether or not the generated code has the optimum performance is source of discussion.
for this special example, you might also look 4 3rd party graphic processing/rendering libraries(boost::gli,Qt,mapserver,gdal...).
|
|
|
|
|
on second thought, you can set up an output hash table that collects all connected points under the same hash. then use the flood-fill algorithm to paint the spots; ie. gather points under hash values in the table and remove painted points from input container. this way beginning of the input container always contains the next first unpainted point.
|
|
|
|
|
My maths is quite poor and I need some help writing an algorithm/pseudo for a simple cost estimator:
rules
I have three inputs x, y and z. The inputs each have a separate max value that if breached return an error. These max values, X = 20, Y = 15, Z = 30.
Now I need a function that compares the in-putted values and if possible swaps the values to keep them under the constraints. Also,
Z should always contain the smallest value as long as the swapped value still meets the max value constraints.
So for a few examples if the user entered X=15,Y=5,Z=10 the values for y and Z would be swapped to make z the lowest value. With X=5 Y=18 Z=2 X and Y would be swapped with x to fit under max values. With X=10 Y=8 Z=2 nothing would be swapped. X=31 Y=18 Z=10 nothing would be swapped and you would get an error.
I have written this codepen so far:
http://codepen.io/FredHair/pen/gpvoPR?editors=101[^]
It finds the minVal, medVal and maxVal. I just need some help with the next logic steps for swapping the values.
The values themself represent a bounding box, orientating an inputted box that would lead to a cost estimate.
I have already had a go at writing this into a program but I made numerous logical errors.
If anyone can offer any insight into the maths and best way to write this I can then write the code.
Any help is greatly appriceted, I hope I have made myslef clear, feel free to look at codepen and ask any questions.
Regards,
F
|
|
|
|
|
Here's a piece of the solution:
void SwapValues (ref int a, ref int b)
{
int temp = a;
a = b;
b = temp;
}
Now you have to write the "if" statements to determine the calls to SwapValues to get your variables in numeric order.
modified 1-Aug-15 15:30pm.
|
|
|
|
|
This code sort X, Y and Z so that X > Y > Z
code between 'then' and 'endif' is swaping values
if x < y then t= x; x= y; y= t; endif
if x < z then t= x; x= z; z= t; endif
if y < z then t= y; y= z; z= t; endif
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
|
|
|
|
|
I would like to know why my previous message was downvoted ?
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
modified 1-Aug-15 22:58pm.
|
|
|
|
|
It's a high level member too - my counter-vote only brought the message back up to a 3.
Possibly the same person reported in B&S[^] who's been randomly abuse-voting Lounge messages.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
|
|
|
|
|
Thank you very much.
You probably right, and I have been a prey of choice since hi downvoted my answers 5 times in a row.
Nota: I am only Silver member, so my upvote will not be as good as yours
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
|
|
|
|
|
Hi All
I had to solve a specific problem for a client, this problem take some time to come with a original solution, that is working.
What i would like to know is if my solution is a original solution, if it fits as an algorithm or more like a recipe (since is very specific to this problem), and also if it is an original algorithm where should I publish it.
Thanks in avance for your answers.
Now the problem:
I Work with business process management, a common issue in BPM is who to define who does which tasks and how to implement a rule based on the responsabilities and role of the user.
In this case an aproval depend on the CEO (the one that must do the task). However he can delegate the task (everytime or in vacation) to the CFO and Manager A.
Also the Manager A delegates to both the Assistant Manager 1 (AM1) and Assistant Manager 2 (AM2), and the CFO delegates to Finantial Manager A (FMA) and Finantial Manager B (FMB).
So a process may be aproved by any of these groups:
- CEO
- CFO and Manager A
- CFO and AM1 and AM2
- FMA and AM1 and AM2
- FMB and AM1 and AM2
- FMA and Manager A
- FMB and Manager B
Second part:
(Assume CEO - Level 1, CFO - Level 2, Manager A - Level 2, The others - Level 3)
We seen that someone in level 2 can aprove with someone in level 3 of another branch.
However it gets tricky since AM1 (level 3) can delegate its on tasks on CFO (level 2 of the other branch)
So this line
- CFO and AM1 and AM2
turns into 2 lines:
- CFO and AM1 and AM2
- CFO and CFO and AM2 (because CFO only aproves once so it become: - CFO and AM2)
This seems hard, but belive i am not give all the requirements, the real problem as some issues more, but the main question is the one I explained above.
The Solution
I will give the solution, maybe write an article where. However first i want to know if it already exists, or something similar. But google for task delegation returns results obviously nothing to do with this problem.
|
|
|
|
|
Hi,
Seems related to oriented graph
Numerous mathematical tools exist for graphs.
So there is little chance that your solution is original, or at least efficient tools already exist for this family of problems.
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
|
|
|
|
|
Tanks ppolymorphe
I was always looking at trees and not graphs.
After your tip i did some research and i think my solution fits more or less in Direct Acyclic Graph more than oriented graph. The image (of the Hasse diagram) in this site is prety similiar with my solution https://en.wikipedia.org/wiki/Directed_acyclic_graph[^]
I am going to read some more about this.
|
|
|
|
|
Rui__Silva wrote: Tanks ppolymorphe You're welcome.
Rui__Silva wrote: After your tip i did some research and i think my solution fits more or less in Direct Acyclic Graph more than oriented graph. Oriented is a direct translation from French, the right translation is Directed
Glad you found what you were looking for.
- If you consider my answer useful, you can up vote it (green arrow when the mouse is on left at beginning of answer).
- If you think the problem solved, change the tittle of your question and add Answered or Solved
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
|
|
|
|
|
Anonymous down-vote countered.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
|
|
|
|
|
Thanks you ! Thanks you ! Thanks you all up voters
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
modified 1-Jul-15 14:36pm.
|
|
|
|
|