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Hi ppolymorphe,
Thank you for your comment.
What you suggest changes the acceleration... I would like to change the speed and find the lowest permissible speed without changing the acceleration.
The idea is to keep the same dynamics for all the movements.
Doing what you suggest would end with a near to infinite acceleration in really short movements... reducing the speed would keep movements soft and get the same result...
But I've not been able to understand why I can't find it directly, if you take a look at the code snippet I've pasted into the thread with Ralf you will see that at the end, after a first approximation, I'm searching for a suitable speed decreasing it by one (terrible).
Thank you for your post.
Joan.
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Joan M wrote: What you suggest changes the acceleration... I would like to change the speed and find the lowest permissible speed without changing the acceleration.
I suggest to change the acceleration duration, not the rate.
In a car, if it takes 60 secondes to accelerate from 0 to 100km/h, then plateau and decelerate.
it will take 15 secondes to reach 25km/h, 30 secondes to reach 50km/h, 45 secondes to reach 75km/h.
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
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Of course, Patrice, you are right ... because the acceleration (or the deceleration) represents the angle of a right-angled triangle.
The Trapezoid consists of 3 parts : a triangle for the acceleration, a rectangle and finally a triangle for the deceleration.
There are 2 Problems in this discussion :
1st : the distance between both points don't allow to reach the max. speed with the given acceleration. In this case we get only 2 triangles with a reduced peak - but all the time the same dynamic
2nd : the movement consists of 2 parts with different max.-speed. In this case we get a triangle for the acceleration, a rectangle, a triangle for the 2nd acceleration or deceleration, the 2nd rectangle and finally again triangle for the deceleration.
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Hi ppolymorphe,
First post updated with the real problem explained, a piece of code and the new question.
If you could take a look at it...
Thank you very much for your time! 
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Code looks not so bad.
Can you give example values of a test that work and one that fail.
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
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Of course Patrice:
Notice that the variables I've written that start with u_ are values entered by the user and the ones that start with c_ are values calculated by the algorithm.
This set of parameters doesn't reach the user vmax and require 37 iterations (ugliest code in the world) to fins the right one:
u_vmax 3000 mm/s
u_v0 2200 mm/s <--------- the important different one
u_vf 0 mm/s
u_a 2000 mm/s^2
u_x0 3000 mm
u_xf 6000 mm
c_dist 3000 mm
c_vmax 2901,250001 mm
This set of parameters works without reaching the vmax as the dynamics doesn't allow the motor to reach it:
u_vmax 3000 mm/s
u_v0 0 mm/s <--------- the important different one
u_vf 0 mm/s
u_a 2000 mm/s^2
u_x0 3000 mm
u_xf 6000 mm
c_dist 3000 mm
c_vmax 2449,489743 mm
This set of parameters work as the vmax entered by the user is reached and we have some part of the movement at constant speed:
u_vmax 3000 mm/s
u_v0 2600 mm/s <--------- the important different one
u_vf 0 mm/s
u_a 2000 mm/s^2
u_x0 3000 mm
u_xf 6000 mm
c_dist 3000 mm
c_vmax 3000 mm
Thank you!
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You should change
LimitIterations = 1000
to
LimitIterations = vmax
because for very short moves, you will need more than 1000 iterations to get the answer.
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
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Yes, but this is not acceptable...
That code will run in a very small processor in a realtime deterministic system and I can't afford spending this amount of time in each cycle.
That's why I need to find the vmax calculating and not by decreasing the value little by little...
I can see two possible approaches:
A) graphical approach with trigonometry trying to find the intersection point of the two straight lines in a time/velocity graph.
B) trying to find the solution via quadratic equations.
In case none of them work, then, at least I would implement a sort of binary search to minimize the searching time for vmax, but I will work to avoid that.
Thank you for your suggestion though. 
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c) dichotomy search
Expect 12 iterations for a vmax between 0 and 4000.
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
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Yes, that would be an option, but I would prefer to solve it mathematically and with only one cycle.
Anyway, that binary (or dichotomy) search would do it more or less well.
Usually the dichotomy search is used to find exact values, in this case I should search for two states in each cycle:
vmax + 1 is not OK
vmax is ok
But that's the last resort if all the other calculations fail.
Thank you!
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Solved... updated the initial post to show the solution.
Thank you Patrice!
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I have to write a program that, given an array of n numbers, finds the longest contiguous subarray whose minimum is the first element of the subarray and maximum the last element of the subarray. For example, with [5, 18, 2, 12, 4, 7, 13, -2, -5, 12] the answer should be [2, 12, 4, 7, 13] (2 <= of all the elements of the subarray, 13 >= of all the elements of the subarray). If possible, the time complexity in the worst case has to be equal to O(n). If it isn't possible, could you explain me why? I already tried to find the minimums and then the maximums, but doesn't work with arrays like [-1,12,1,2,3,4]. I also tried to split the array when there is an element followed by a smaller element and then merge the subarrays, but it doesn't work for arrays like [2,15,6,10,3,20].
modified 5-Feb-17 6:11am.
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Member 10566089 wrote: I have to write a program that
OK, go ahead then.
So what exactly is your question? What do you need help with?
The difficult we do right away...
...the impossible takes slightly longer.
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I can't find an algorithm that solves this problem in O(n). Do you have any idea?
modified 5-Feb-17 13:06pm.
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Member 10566089 wrote: I can't find an algorithm that solves this problem in O(n). Do you have any idea?
You have not been told it is possible, you are asked if it is possible or not. If you don't find an algorithm, the answer is 'no'.
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
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Show what you have done so far. We will not write it for you.
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
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int l, l_max, u_max, u;
int A[N+1] = l_max= l =1;
u_max= u =1;
int i;
for (i=2;i<N;i++)
{
if (A[i] >= A[u])
u = i;
else if (A[i] < A[l])
{
if (u-l > u_max-l_max)
{
l_max = l;
u_max = u;
}
l = u = i;
}
}
if (u-l > u_max - l_max)
{
l_max = l;
u_max = u;
}
printf("%d %d", l_max,u_max);
But doesn't work with arrays like {1,10,5,5,5,5,1,9}
public static void main(String[] args) {
int [] arr=
int N=arr.length;
LinkedList <Subarray> subarrays=new LinkedList<Subarray> ();
Subarray s=new Subarray();
s.setStart(0);
s.setEnd(0);
for (int i=1;i<N;i++)
{
if (arr[i] < arr[i-1])
{
s.setEnd(i-1);
subarrays.add(s);
s=new Subarray();
s.setStart(i);
s.setEnd(i);
}
}
s.setEnd(N-1);
subarrays.add(s);
Iterator<Subarray> iterator = subarrays.iterator();
s=iterator.next();
Subarray nxt;
while (iterator.hasNext()) {
nxt=iterator.next();
if (arr[nxt.getEnd()]>=arr[s.getEnd()] && arr[s.getStart()]<=arr[nxt.getStart()])
nxt.setStart(s.getStart());
s=nxt;
}
iterator = subarrays.iterator();
s=iterator.next();
while (iterator.hasNext())
{
nxt=iterator.next();
if ((nxt.getEnd()-nxt.getStart()) > (s.getEnd()-s.getStart())){
s=nxt;
}
}
System.out.print("+s.getStart()+""+s.getEnd());
}
}
But doesn't work with {2,15,6,10,3,20}
public static void main(String[] args) {
int [] input=
int max = input.length - 1;
int[] aux = new int[input.length];
for(int i = input.length; i>0; i--){
if(input[i-1] > input[max])
{
max = i - 1;
}
aux[i-1] = max;
}
int l = 0;
int u = 0;
max = 0;
int lMax = 0;
int uMax = 0;
for (int i=0;i<aux.length;i++)
System.out.print(aux[i]+" ");
for(int i = 0; i< input.length; i++){
System.out.println(""+input[i]+",l: "+l+",u: "+u);
if (input[i] >= input[l] && aux[l] >= i)
{
u = i;
}
else
{
if(u-l > max)
{
lMax = l;
uMax = u;
max = u-l;
}
l = i;
u = i;
}
}
if(u-l > max){
lMax = l;
uMax = u;
}
System.out.println("l "+l+" u "+u);
}
But doesn't work with {1,2,3,4,5,15,4,-1,18}
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Why do you have 3 different programs ?
Are you rewriting a new program every time the previous one don't work ?
Advice: Take a sheet of paper and simulate how your program should behave and how variables evolve over execution. And validate that you find the right answer.
If the simulation don't endup with the answer, algorithm is wrong.
if the simulation is ok, use the debugger to check that the program conform to simulation, if not, it will help you to find the reason.
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
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Yes, I wrote the code and discovered that there is a type of array for which the algorithm gives the wrong answer.
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Hi,
SOLVED.
Thanks.
modified 12-Apr-17 15:00pm.
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(Accounts payable and receivable) "Aging" has to do with "Invoice dates" and "Due Dates" and payment terms.
Your concept of "aging" makes no sense; particularly when you start talking in terms of "years".
And when you start using the wrong term for a given accounting concept, you are in for a lot of pain (through misunderstandings).
"(I) am amazed to see myself here rather than there ... now rather than then".
― Blaise Pascal
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Excuse my lack of understanding of the accounting "Aging" method but somehow my accounting person is calling it "Aging". I hope I was able to explain and communicate my requirement on what I'm trying to achieve. I would appreciate some help with this.
Thanks.
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You might be "forecasting"; but nailing one small amount to a given month is meaningless unless it has some significance to it, like "Mother's Day".
If anything, the accounting person may also be "depreciating" or modeling the life cycle of a marketing campaign or product.
Maybe it's all on a "need to know" basis...
"(I) am amazed to see myself here rather than there ... now rather than then".
― Blaise Pascal
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public static int LastIndexOf<T>(this IReadOnlyList<T> s, IReadOnlyList<T> t, int startIndex,
IEqualityComparer<T> comparer) where T : IEquatable<T>
{
Validate(s, t, startIndex);
if (t.Count == 0) return 0;
if (s.Count == 0 || s.Count < t.Count) return -1;
if (comparer == null) comparer = EqualityComparer<T>.Default;
if (t.Count == 1) return LastIndexOf(s, t[0], startIndex, comparer);
var table = BuildTable(t, comparer);
var i = 0;
while (startIndex - i >= 0)
{
if (comparer.Equals(t[t.Count - i - 1], s[startIndex - i]))
{
if (i == t.Count - 1)
return startIndex - t.Count + 1;
i++;
}
else
{
if (table[i] > -1)
{
startIndex -= i;
i = table[i];
}
else
{
startIndex--;
i = 0;
}
}
}
return -1;
}
Full source code can be seen here[^].
The method works like String.LastIndexOf which searches string from the last character. The BuildTable method is exactly same with normal KMP algorithm.
I'm not sure if this is the best solution though.
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I would have just reversed the 2 strings and compared those.
(KMP: Keep Me Partying).
"(I) am amazed to see myself here rather than there ... now rather than then".
― Blaise Pascal
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