
Here is the question requirement:
There is a map shown in below
WWWWWWWWWWDW
WWWW S W
WWWW WWWW W
WWWWSWWWW S
WWWW WWWW W
B ES E W
WWWWWWWWWWWW
W is the wall that the player cannot move to
S is the score can the player option
E is the energy can gain
B is beginning point
D is the ending point
First, player has 0 energy and they need to go to E to gain energy first. Then when they gain energy, they can go to S to gain score. However, every E and S contains different energies, energies required and scores. Here is the clear description of the map:
<5 0> is B (Begining point)
<5 4> is E (energy point, contain 20 energy)
<3 4> is S (Score, require 6 energy to gain and contain 30 scores)
<5 5> is S (Score, require 2 energy to gain and contain 230 scores)
<5 8> is E (energy, contain 30 energy)
<3 10> is S (Score, require 16 energy to gain and contain 30 scores)
<1 8> is S (Score, require 8 energy to gain and contain 10 scores)
The final score is in this formula:
energy + scores  steps*2
The best path is B > <5 4> > <3 4> > <5 5> > <5 8> > <3 10> > D
The key point is that how to let the program or algorithm detect the optimal path (With the minimum steps and know which score should get as in the above example, the <1 8> Score is ignored.)
I am now figuring how can let the computer know going the <3 4> and go back the <5 4> and go to <5 5> is the best path instead of going to <5 5> > <5 4> > <3 4> > <5 5>. I know the later one is a stupid path. But what I can think of is that to do the best path, we may need to search all path with factorial times loop which is not accepted in term of running time.





This requirement is incomplete, you only fave a vague description of what is the best path, how you compare 2 path op see which 1 is better.
and the map look weird.
We can't workout something from this.
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein





Here is the refine question:
There is a map shown in below
<pre>WWWWWWWWWWDW
WWWW S W
WWWW WWWW W
WWWWSWWWW S
WWWW WWWW W
B ES E W
WWWWWWWWWWWW
W is the wall that the player cannot move to
S is the score can the player option
E is the energy can gain
B is beginning point
D is the ending point
First, player has 0 energy and they need to go to E to gain energy first. Then when they gain energy, they can go to S to gain score. However, every E and S contains different energies, energies required and scores. Here is the clear description of the map:
<5 0> is B (Begining point)
<5 4> is E (energy point, contain 20 energy)
<3 4> is S (Score, require 6 energy to gain and contain 30 scores)
<5 5> is S (Score, require 2 energy to gain and contain 230 scores)
<5 8> is E (energy, contain 30 energy)
<3 10> is S (Score, require 16 energy to gain and contain 30 scores)
<1 8> is S (Score, require 8 energy to gain and contain 10 scores)
The final score is in this formula:
energy + scores  steps*2
The best path is B > <5 4> > <3 4> > <5 5> > <5 8> > <3 10> > D
The key point is that how to let the program or algorithm detect the optimal path (With the minimum steps and know which score should get as in the above example, the <1 8> Score is ignored.)
I am now figuring how can let the computer know going the <3 4> and go back the <5 4> and go to <5 5> is the best path instead of going to <5 5> > <5 4> > <3 4> > <5 5>. I know the later one is a stupid path. But what I can think of is that to do the best path, we may need to search all path with factorial times loop which is not accepted in term of running time.





I recently messed up this question and I can't quite figure out how to solve it. With divide and conquer we can easily tell if the substrings are palindromes (and counting them alone would be easy). But we're not supposed to count them. The long string below should yield "3" as is has 3 levels of palindromes. Not wrapping my head around how to keep track of the rank when unwinding.
"yx xy yx xy yx xy yx xy" //  not all are palindromes, so nothing here
"yxxy yxxy yxxy yxxy" // all 4 are palindromes  level 3
"yxxyyxxy yxxyyxxy" // both are palindromes  level 2
"yxxyyxxyyxxyyxxy" // whole string is palindrome  level 1
bool isPalindrome( const std::string& str ){
}
int palindromeCount( const std::string& str ) {
if ( !isPalindrome( str ) ) return 0;
int l = palindromeCount( firstHalfofStr );
int r = palindromeCount( secondHalfofStr );
int x{0};
if (l && r ) {
}
return 1 + x;
}





Maybe you should post the "original question" because your explanation isn't making much sense; particularly the "level" part. Or who said you need recursion? And what does "not counting" to get a count mean?
It was only in wine that he laid down no limit for himself, but he did not allow himself to be confused by it.
― Confucian Analects: Rules of Confucius about his food





Sorry for the confusion. I posted as best as I could remember. Actually was something quite silly on my part that was pointed out. WIll post the udpate soon.





When automatically checking coding exercises, we often run the programs in
question against test cases, which are made up of one input and one output file. As expected,
the program is fed the input file and then its output is compared with the output file. If these
match for all test cases, we deem the program correct.
Coming up with these test cases is tricky. Specifically, we want to create test cases for the
following problem: ‘Given a complete weighted graph G, compute its minimum spanning tree
T’. We have already created the desired output files containing the different minimum spanning
trees, we now want to find the corresponding input files.
It is known that a graph can have many different MST’s. To make testing easier, we want
to ensure that the minimum spanning tree in every output file is unique for the graph described
in the respective input file. The test cases also need to be able to tell apart algorithms that
are wrong, but happen to find the correct MST by chance. For example, if all edges not in the
MST have very large weights, a naive algorithm could find the MST by including only the light
edges. Of course this algorithm would be incorrect in general. To avoid this, we want the input
files to contain complete graphs G whose sum of edge weights is minimum. You need to design
a program that can generate these input files.
Goal: Given a weighted tree T with n nodes, find the complete graph G of minimum weight
such that T ⊆ G and T is the unique minimum spanning tree of G. Assume all edge weights
are integer.
1. Find an algorithm whose run time is polynomial in n.
2. Improve the complexity to O(n · log n).
Hint: One way to do this is by thinking of Kruskal’s algorithm and the cut property of minimum
spanning trees.
can you help me to find a solution ?





As I said to your other copy of this:
Quote: We are more than willing to help those that are stuck: but that doesn't mean that we are here to do it all for you! We can't do all the work, you are either getting paid for this, or it's part of your grades and it wouldn't be at all fair for us to do it all for you.
So we need you to do the work, and we will help you when you get stuck. That doesn't mean we will give you a step by step solution you can hand in!
Start by explaining where you are at the moment, and what the next step in the process is. Then tell us what you have tried to get that next step working, and what happened when you did.
And posting again because it didn't get through immediately never improves your chances of getting an answer: they stay at zero, only because they can't go down from there ...
"I have no idea what I did, but I'm taking full credit for it."  ThisOldTony
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Philip decided to supplement his income by participating in the popular game
show ‘Open the Boxes and keep the Best!’. The game is played in turns. At every turn, the
host shows Philip a box that can be opened by paying ci > 0. The box will contain a random
prize: in particular its value could be anything in {0, 1, 2, . . . , n}, uniformly at random.
At every step, Philip can stop playing and keep only one prize, the best found so far. Of course,
the costs he paid are not refunded. All of the boxes are known in advance, as well as the order
the host will present them. Help Philip find the optimal strategy and calculate his expected
payoff, which is the prize he keeps minus the total cost paid.
Goal: Given as input the number of boxes k, the number of different rewards n as well as
the cost ci (which is guaranteed to be integer) of every box:
1. Design a O(n^2· k) algorithm to find the expected optimal reward.
2. Improve the complexity to O(n · k).
Hint: One way is to use dynamic programming
can you help me to find a solution?





We are more than willing to help those that are stuck: but that doesn't mean that we are here to do it all for you! We can't do all the work, you are either getting paid for this, or it's part of your grades and it wouldn't be at all fair for us to do it all for you.
So we need you to do the work, and we will help you when you get stuck. That doesn't mean we will give you a step by step solution you can hand in!
Start by explaining where you are at the moment, and what the next step in the process is. Then tell us what you have tried to get that next step working, and what happened when you did.
"I have no idea what I did, but I'm taking full credit for it."  ThisOldTony
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Hello guys, I'm new here I was trying to learn more about algorithm by solving as much as I can from different sources anyway I found a question which makes me think about it every day and I couldn't find a proper solution for it check this link out and if you can give me ideas how to tackle this it will be great
infinite_recursive_shape_challenge
Thanks.





The ideas are all in the problem description.





I'm trying to convert a finite state machine back to a regular expression.
I'm looking for some help with one of two algorithms that i just don't have the math background to transfer to code. I've so far had no luck wrapping my head around it and it's frustrating.
Theory of Computation  Generating regular expression from finite automata  GeeksforGeeks[^]
I don't necessarily need actual code. I can deal with pseudo code. or pretty much any language (except java  for reasons having to do with the way their containers/collections work)
I'm trying to do this in C# though so some pointers would be helpful.
I know the basics of DFA and NFA machines enough to implement a regex engine. I simply cannot do this one thing.
When I was growin' up, I was the smartest kid I knew. Maybe that was just because I didn't know that many kids. All I know is now I feel the opposite.





For a coding project I'm working on I need to predict the amount of workers that are necessary to finish a task within a limited time frame T. Lets rephrase it a bit...
Given a collection of object X; that implements a blocking function that takes object Y as a parameter; from a queue of said Y objects; that does work for anywhere between TimeX1 < Time < TimeX2 ms.
Given the fact that the pool of Y objects is vastly larger then the X objects;
And that every time an object X finishes working on an object Y it takes another object Y from said queue.
And Given that the total amount of work time of all X objects combined cannot exceed MaxTime
What formula would predict the least amount of X objects necessary to work on the whole queue of Y objects?





Sounds like a simple maths problem.
To guarantee that the queue will be processed in time, you have to assume the worstcase scenario  that each Y takes TimeX2 ms to process.
Therefore, the total time to process the queue will be:
TimeX2 * Count(Y) / Count(X) You want that value to be less than or equal to the MaxTime , so:
MaxTime >= TimeX2 * Count(Y) / Count(X)
Count(X) >= TimeX2 * Count(Y) / MaxTime Therefore, the minimum number of X s required will be the ceiling of the number of Y s multiplied by the maximum time to process a Y divided by the maximum time allowed.
NB: If you were willing to accept that the queue might not be fully processed within the time limit, and you had more information about the distribution of processing times, then you might be able to bring that number down a bit. You could also consider an adaptive approach, where you increase or decrease the number of X s over time based on the number of Y s waiting to be processed and the amount of time remaining.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
 Homer





It would have been simple if every X could handle only one Y.
But every time an X finished doing work on a Y he takes another one.
And X is a precious resource, so I can't just go all out with X's. and these damn Y just keep coming...





That's already covered.
Try putting some numbers in  eg:
TimeX2 = 10MaxTime = 100Count(Y) = 20TimeX2 * Count(Y) / MaxTime = 2
If each X was only handling one Y, you'd need 20 Xs to process the queue.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
 Homer





You need an average time for x; or an average time for EACH x.
You need a Y arrival rate. If you're just processing what the "night shift" produced, you still need some idea of quantity.
With Y type stacks / queues and an x type timer factory, you have a simulation.
The Master said, 'Am I indeed possessed of knowledge? I am not knowing. But if a mean person, who appears quite emptylike, ask anything of me, I set it forth from one end to the other, and exhaust it.'
― Confucian Analects





We know how much y's would enter the queue beforehand, once the collection of Xs start working there's no stopping unless we are past the MaxTime limit. EACH X should not work more than given MaxTime. How many Xs do I supply to work on a given amount of Ys. When each X takes another Y from the queue as soon as he finishes with a previous Y?





N = (Y * T1) / (X * T2) where T2 "is less than MaxTime".
The Master said, 'Am I indeed possessed of knowledge? I am not knowing. But if a mean person, who appears quite emptylike, ask anything of me, I set it forth from one end to the other, and exhaust it.'
― Confucian Analects





I have think out a sorting algorithm.
It's comparison count is fewer about 12% than gcc quick sort. (By 1000 item sort test 1000 times average)
If I am scholar, it is a case of dissertation.
But I'm not so. (I am smartphone application developer.)
(I had ever made a article about previous version algorithm and submitted to Cornell University and it was refused. It said to go to forum like this.)
The characteristic is:
.Variation of merge sort, it would be classified.
.It is inplace. But different from "Inplace merge sort" (Normally merge sort is not inplace.)
I think it may be fastest sorting ever. But I have now no way to publish.
Is there someone who knows the good way to announce it to the people who can evaluate it?
(The program list is a little long to include in this message.
I have not made explanation document about it. I am still testing and modifying it.)





Member 14560162 wrote: But I have now no way to publish.
Is there someone who knows the good way to announce it to the people who can evaluate it?
Have you considered publishing it as an article here?
Submit a new Article[^]
They are designed for code and algorithms, plus the explanation of them.
Here's a couple of mine, to give you the idea:
Using struct and class  what's that all about?[^]
List<T>  Is it really as efficient as you probably think?[^]
You'll reach up to 14,000,000 software developers, and get feedback (positive and / or negative) fairly quickly once it's moderated.
Sent from my Amstrad PC 1640
Never throw anything away, Griff
Bad command or file name. Bad, bad command! Sit! Stay! Staaaay...
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Thank you for answer.
I had not reached article page before you indicate.
I try to write article.
But the submit will be need times because I have written nothing and I must write it between other work.





No rush  take your time!
Articles tabn=ke me a fair amount of time to write as well  often several times longer than the code they are based on.
Sent from my Amstrad PC 1640
Never throw anything away, Griff
Bad command or file name. Bad, bad command! Sit! Stay! Staaaay...
AntiTwitter: @DalekDave is now a follower!





You can mark the article as a work in progress and simply save it for further editing without having to publish in an unfinished state (which would be bad thing to do).



