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I have a sorted array of N intervals of different length. I am plotting these intervals with alternating colors blue/green.
I am trying to find a method or algorithm to "downsample" the array of intervals to produce a visually similar plot, but with less elements.
Ideally I could write some function where I can pass the target number of output intervals as an argument. The output length only has to come close to the target.
input = [
[0, 5, "blue"],
[5, 6, "green"],
[6, 10, "blue"],
// ...etc
]
output = downsample(input, 25)
// [[0, 10, "blue"], ... ]
Below is a picture of what I am trying to accomplish. In this example the input has about 250 intervals, and the output about ~25 intervals. The input length can vary a lot.
Update 1:
Below is my original post which I initially deleted, because there were issues with displaying the equations and also I wasn't very confident if it really makes sense. But later, I figured that the optimisation problem that I described can be actually solved efficiently with DP (Dynamic programming).
So I did a sample C++ implementation. Here are some results:
Here is a live demo that you can play with in your browser (make sure browser support WebGL2, like Chrome or Firefox). It takes a bit to load the page.
Here is the C++ implementation: link
Update 2:
Turns out the proposed solution has the following nice property - we can easily control the importance of the two parts F1 and F2 of the cost function. Simply change the cost function to F(α)=F1 + αF2, where α >= 1.0 is a free parameter. The DP algorithm remains the same.
Here are some result for different α values using the same number of intervals N:
Live demo (WebGL2 required)
As can be seen, higher α means it is more important to cover the original input intervals even if this means covering more of the background in-between.
Original post
Even-though some good algorithms have already been proposed, I would like to propose a slightly unusual approach - interpreting the task as an optimisation problem. Although, I don't know how to efficiently solve the optimisation problem (or even if it can be solved in reasonable time at all), it might be useful to someone purely as a concept.
First, without loss of generality, lets declare the blue color to be background. We will be painting N green intervals on top of it (N is the number provided to the downsample() function in OP's description). The ith interval is defined by its starting coordinate 0 <= xi < xmax and width wi >= 0 (xmax is the maximum coordinate from the input).
Lets also define the array G(x) to be the number of green cells in the interval [0, x) in the input data. This array can easily be pre-calculated. We will use it to quickly calculate the number of green cells in arbitrary interval [x, y) - namely: G(y) - G(x).
We can now introduce the first part of the cost function for our optimisation problem:
The smaller F1 is, the better our generated intervals cover the input intervals, so we will be searching for xi, wi that minimise it. Ideally we want F1=0 which would mean that the intervals do not cover any of the background (which of course is not possible because N is less than the input intervals).
However, this function is not enough to describe the problem, because obviously we can minimise it by taking empty intervals: F1(x, 0)=0. Instead, we want to cover as much as possible from the input intervals. Lets introduce the second part of the cost function which corresponds to this requirement:
The smaller F2 is, the more input intervals are covered. Ideally we want F2=0 which would mean that we covered all of the input rectangles. However, minimising F2 competes with minimising F1.
Finally, we can state our optimisation problem: find xi, wi that minimize F=F1 + F2
How to solve this problem? Not sure. Maybe use some metaheuristic approach for global optimisation such as Simulated annealing or Differential evolution. These are typically easy to implement, especially for this simple cost function.
Best case would be to exist some kind of DP algorithm for solving it efficiently, but unlikely.
I would advise you to use Haar wavelet. That is a very simple algorithm which was often used to provide the functionality of progressive loading for big images on websites.
Here you can see how it works with 2D function. That is what you can use. Alas, the document is in Ukrainian, but code in C++, so readable:)
This document provides an example of 3D object:
Pseudocode on how to compress with Haar wavelet you can find in Wavelets for Computer Graphics: A Primer Part 1y.
You could do the following:
Write out the points that divide the whole strip into intervals as the array [a[0], a[1], a[2], ..., a[n-1]]. In your example, the array would be [0, 5, 6, 10, ... ].
Calculate double-interval lengths a[2]-a[0], a[3]-a[1], a[4]-a[2], ..., a[n-1]-a[n-3] and find the least of them. Let it be a[k+2]-a[k]. If there are two or more equal lengths having the lowest value, choose one of them randomly. In your example, you should get the array [6, 5, ... ] and search for the minimum value through it.
Swap the intervals (a[k], a[k+1]) and (a[k+1], a[k+2]). Basically, you need to assign a[k+1]=a[k]+a[k+2]-a[k+1] to keep the lengths, and to remove the points a[k] and a[k+2] from the array after that because two pairs of intervals of the same color are now merged into two larger intervals. Thus, the numbers of blue and green intervals decreases by one each after this step.
If you're satisfied with the current number of intervals, end the process, otherwise go to the step 1.
You performed the step 2 in order to decrease "color shift" because, at the step 3, the left interval is moved a[k+2]-a[k+1] to the right and the right interval is moved a[k+1]-a[k] to the left. The sum of these distances, a[k+2]-a[k] can be considered a measure of change you're introducing into the whole picture.
Main advantages of this approach:
It is simple.
It doesn't give a preference to any of the two colors. You don't need to assign one of the colors to be the background and the other to be the painting color. The picture can be considered both as "green-on-blue" and "blue-on-green". This reflects quite common use case when two colors just describe two opposite states (like the bit 0/1, "yes/no" answer) of some process extended in time or in space.
It always keeps the balance between colors, i.e. the sum of intervals of each color remains the same during the reduction process. Thus the total brightness of the picture doesn't change. It is important as this total brightness can be considered an "indicator of completeness" at some cases.
Here's another attempt at dynamic programming that's slightly different than Georgi Gerganov's, although the idea to try and formulate a dynamic program may have been inspired by his answer. Neither the implementation nor the concept is guaranteed to be sound but I did include a code sketch with a visual example :)
The search space in this case is not reliant on the total unit width but rather on the number of intervals. It's O(N * n^2) time and O(N * n) space, where N and n are the target and given number of (green) intervals, respectively, because we assume that any newly chosen green interval must be bound by two green intervals (rather than extend arbitrarily into the background).
The idea also utilises the prefix sum idea used to calculate runs with a majority element. We add 1 when we see the target element (in this case green) and subtract 1 for others (that algorithm is also amenable to multiple elements with parallel prefix sum tracking). (I'm not sure that restricting candidate intervals to sections with a majority of the target colour is always warranted but it may be a useful heuristic depending on the desired outcome. It's also adjustable -- we can easily adjust it to check for a different part than 1/2.)
Where Georgi Gerganov's program seeks to minimise, this dynamic program seeks to maximise two ratios. Let h(i, k) represent the best sequence of green intervals up to the ith given interval, utilising k intervals, where each is allowed to stretch back to the left edge of some previous green interval. We speculate that
h(i, k) = max(r + C*r1 + h(i-l, k-1))
where, in the current candidate interval, r is the ratio of green to the length of the stretch, and r1 is the ratio of green to the total given green. r1 is multiplied by an adjustable constant to give more weight to the volume of green covered. l is the length of the stretch.
JavaScript code (for debugging, it includes some extra variables and log lines):
function rnd(n, d=2){
let m = Math.pow(10,d)
return Math.round(m*n) / m;
}
function f(A, N, C){
let ps = [[0,0]];
let psBG = [0];
let totalG = 0;
A.unshift([0,0]);
for (let i=1; i<A.length; i++){
let [l,r,c] = A[i];
if (c == 'g'){
totalG += r - l;
let prevI = ps[ps.length-1][1];
let d = l - A[prevI][1];
let prevS = ps[ps.length-1][0];
ps.push(
[prevS - d, i, 'l'],
[prevS - d + r - l, i, 'r']
);
psBG[i] = psBG[i-1];
} else {
psBG[i] = psBG[i-1] + r - l;
}
}
//console.log(JSON.stringify(A));
//console.log('');
//console.log(JSON.stringify(ps));
//console.log('');
//console.log(JSON.stringify(psBG));
let m = new Array(N + 1);
m[0] = new Array((ps.length >> 1) + 1);
for (let i=0; i<m[0].length; i++)
m[0][i] = [0,0];
// for each in N
for (let i=1; i<=N; i++){
m[i] = new Array((ps.length >> 1) + 1);
for (let ii=0; ii<m[0].length; ii++)
m[i][ii] = [0,0];
// for each interval
for (let j=i; j<m[0].length; j++){
m[i][j] = m[i][j-1];
for (let k=j; k>i-1; k--){
// our anchors are the right
// side of each interval, k's are the left
let jj = 2*j;
let kk = 2*k - 1;
// positive means green
// is a majority
if (ps[jj][0] - ps[kk][0] > 0){
let bg = psBG[ps[jj][1]] - psBG[ps[kk][1]];
let s = A[ps[jj][1]][1] - A[ps[kk][1]][0] - bg;
let r = s / (bg + s);
let r1 = C * s / totalG;
let candidate = r + r1 + m[i-1][j-1][0];
if (candidate > m[i][j][0]){
m[i][j] = [
candidate,
ps[kk][1] + ',' + ps[jj][1],
bg, s, r, r1,k,m[i-1][j-1][0]
];
}
}
}
}
}
/*
for (row of m)
console.log(JSON.stringify(
row.map(l => l.map(x => typeof x != 'number' ? x : rnd(x)))));
*/
let result = new Array(N);
let j = m[0].length - 1;
for (let i=N; i>0; i--){
let [_,idxs,w,x,y,z,k] = m[i][j];
let [l,r] = idxs.split(',');
result[i-1] = [A[l][0], A[r][1], 'g'];
j = k - 1;
}
return result;
}
function show(A, last){
if (last[1] != A[A.length-1])
A.push(last);
let s = '';
let j;
for (let i=A.length-1; i>=0; i--){
let [l, r, c] = A[i];
let cc = c == 'g' ? 'X' : '.';
for (let j=r-1; j>=l; j--)
s = cc + s;
if (i > 0)
for (let j=l-1; j>=A[i-1][1]; j--)
s = '.' + s
}
for (let j=A[0][0]-1; j>=0; j--)
s = '.' + s
console.log(s);
return s;
}
function g(A, N, C){
const ts = f(A, N, C);
//console.log(JSON.stringify(ts));
show(A, A[A.length-1]);
show(ts, A[A.length-1]);
}
var a = [
[0,5,'b'],
[5,9,'g'],
[9,10,'b'],
[10,15,'g'],
[15,40,'b'],
[40,41,'g'],
[41,43,'b'],
[43,44,'g'],
[44,45,'b'],
[45,46,'g'],
[46,55,'b'],
[55,65,'g'],
[65,100,'b']
];
// (input, N, C)
g(a, 2, 2);
console.log('');
g(a, 3, 2);
console.log('');
g(a, 4, 2);
console.log('');
g(a, 4, 5);
I would suggest using K-means it is an algorithm used to group data(a more detailed explanation here: https://en.wikipedia.org/wiki/K-means_clustering and here https://scikit-learn.org/stable/modules/generated/sklearn.cluster.KMeans.html)
this would be a brief explanation of how the function should look like, hope it is helpful.
from sklearn.cluster import KMeans
import numpy as np
def downsample(input, cluster = 25):
# you will need to group your labels in a nmpy array as shown bellow
# for the sake of example I will take just a random array
X = np.array([[1, 2], [1, 4], [1, 0],[4, 2], [4, 4], [4, 0]])
# n_clusters will be the same as desired output
kmeans = KMeans(n_clusters= cluster, random_state=0).fit(X)
# then you can iterate through labels that was assigned to every entr of your input
# in our case the interval
kmeans_list = [None]*cluster
for i in range(0, X.shape[0]):
kmeans_list[kmeans.labels_[i]].append(X[i])
# after that you will basicly have a list of lists and every inner list will contain all points that corespond to a
# specific label
ret = [] #return list
for label_list in kmeans_list:
left = 10001000 # a big enough number to exced anything that you will get as an input
right = -left # same here
for entry in label_list:
left = min(left, entry[0])
right = max(right, entry[1])
ret.append([left,right])
return ret
I came across this question in one of the Job Interviews & i am unable to find the correct alogorithm of the solution so, i am posting this question here:
There is a robot who can move on a co-ordinate plane in eithr of the 2 ways:
Given that the robots current position is (x,y), The robot can move equal to the sum of x & y in either if the directon like so:
(x,y) -> (x+y, y)
(x,y) -> (x, x+y)
Now given a initial Point (x1, y1) and an destination point (x2, y2) you need to write a programme to check if the robot can ever reach the destination taking any number of moves.
Note: x1, y1 , x2 , y2 > 0
Explanation:
Suppose the robot's initial point is (2,3) and desintation is (7,5)
Result in this case is yes as the robot can take this path:
(2,3) -> (2, 2+3) => (2, 5)
(2,5) -> (2+5, 5) => (7,5)
Suppose the robot's initial point is (2,3) and desintation is (4,5)
Result in this case is No as no matter what path the robot takes it cannot reach (4,5)
A naive brute-force approach
One way would be to recursively explore every possible move until you reach the target.
Something to consider is that the robot can keep moving indefinitely (never reaching the target) so you need an end case so the function completes. Luckily the position is always increasing in the x and y axis, so when either the x-coordinate or y-coordinate is greater than the target, you can give up exploring that path.
So something like:
def can_reach_target(pos, target):
if pos == target:
return True
if pos[0] > target[0] or pos[1] > target[1]:
return False
return can_reach_target((pos[0], sum(pos)), target) or \
can_reach_target((sum(pos), pos[1]), target)
And it works:
>>> can_reach_target((2,3),(7,5))
True
>>> can_reach_target((2,3),(4,5))
False
A limitation is that this does not work for negative coordinates - not sure if this is a requirement, just let me know if it is and I will adapt the answer.
Bactracking
On the other hand, if negative co-ordinates are not allowed, then we can also approach this as Dave suggests. This is much more efficient, as the realisation is that there is one and only one way of the robot getting to each coordinate.
The method relies on being able to determine which way we stepped: either increasing the x-coordinate or the y-coordinate. We can determine which coordinate was last changed, by selecting the larger of the two. The following proof guarantees that this is the case.
The possibilities for a state change are:
1. (a, b) => (a+b, b) a x-coordinate change
and,
2. (a, b) => (a, a+b) a y-coordinate change
In case (1), the x-coordinate is now larger, since:
a > 0
a + b > b (add b to both sides)
and similarly, since b is also > 0, we can deduce that a+b is > a.
Now we can start from the target and ask: which coordinate led us here? And the answer is simple. If the x-coordinate is greater than the y-coordinate, subtract the y-coordinate from the x-coordinate, otherwise subtract the x-coordinate from the y-coordinate.
That is to say, for a coordinate, (x,y), if x > y, then we came from (x-y,y) otherwise (x,y-x).
The first code can now be adapted to:
def can_reach_target(pos, target):
if pos == target:
return True
if target[0] < pos[0] or target[1] < pos[1]:
return False
x, y = target
return can_reach_target(pos, (x-y,y) if x > y else (x,y-x))
which works as expected:
>>> can_reach_target((2,3),(7,5))
True
>>> can_reach_target((2,3),(4,5))
False
Timings
>>> timeit.timeit('brute_force((2,3),(62,3))',globals=locals(),number=10**5)
3.41243960801512
>>> timeit.timeit('backtracker((2,3),(62,3))',globals=locals(),number=10**5)
1.4046142909792252
>>> timeit.timeit('brute_force((2,3),(602,3))',globals=locals(),number=10**4)
3.518286211998202
>>> timeit.timeit('backtracker((2,3),(602,3))',globals=locals(),number=10**4)
1.4182081500184722
So you can see that the backtracker is nearly three times faster in both cases.
Go backwards. I'm assuming that the starting coordinates are positive. Say you want to know if a starting point of (a,b) is compatible with an end point of (x,y). One step back from (x,y) you were either at (x-y,y) or (x,y-x). If x > y choose the former, otherwise choose the latter.
I agree with Dave that going backwards is an efficient approach. If only positive coordinates are legal, then every coordinate has at most one valid parent. This lets you work backwards without a combinatorial explosion.
Here's a sample implementation:
def get_path(source, destination):
path = [destination]
c,d = destination
while True:
if (c,d) == source:
return list(reversed(path))
if c > d:
c -= d
else:
d -= c
path.append((c,d))
if c < source[0] or d < source[1]:
return None
print(get_path((1,1), (1,1)))
print(get_path((2,3), (7,5)))
print(get_path((2,3), (4,5)))
print(get_path((1,1), (6761, 1966)))
print(get_path((4795, 1966), (6761, 1966)))
Result:
[(1, 1)]
[(2, 3), (2, 5), (7, 5)]
None
[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (6, 5), (11, 5), (16, 5), (21, 5), (26, 5), (31, 5), (36, 5), (41, 5), (46, 5), (46, 51), (46, 97), (143, 97), (143, 240), (383, 240), (623, 240), (863, 240), (863, 1103), (863, 1966), (2829, 1966), (4795, 1966), (6761, 1966)]
[(4795, 1966), (6761, 1966)]
Appendix: some observations I made along the way that might be useful for finding an O(1) solution:
(a,b) is reachable from (1,1) if and only if a and b are coprime.
If a and b have a common factor, then all children of (a,b) also have that common factor. Equivalently, if there is a path from (a,b) to (c,d), then there is also a path from (n*a, n*b) to (n*c, n*d), for any positive integer n.
if a and b are coprime and aren't (1,1), then there are infinitely many coprime coordinates that are unreachable from (a,b). By choosing (a,b) as a starting point, you're effectively limiting yourself to some sub-branch of the tree formed by (1,1). You can never reach any of the sibling branches of (a,b), where infinitely many coordinates reside.
A recursive funtion should work fine for that. You even got the number of possibilities.
def find_if_possible(x,y,x_obj,y_obj,max_depth):
if(max_depth < 0):
return 0
elif(x == x_obj and y == y_obj):
return 1
elif(x>x_obj or y>y_obj):
return 0
else:
return(sum(find_if_possible(x+y,y,x_obj,y_obj,max_depth-1),find_if_possible(x,y+x,x_obj,y_obj,max_depth-1))
I am modelling a particle in 3D space.
{0} The particle starts at time t0 from a known position P0 with a velocity V0. The velocity is computed using its known previous position of P-1 at t-1.
{1} The particle is targeted to go to P1 at t1 with a known velocity of V1.
{..} The particle moves as fast as it can, without jerks (C1 continuous) bound by a set of constraints that clamp the acceleration along x, y and z independently. The maximum acceleration/deceleration along x, y and z are known and are Xa, Ya, and Za. The max rate of change of acceleration along x, y and z are defined by Xr, Yr, and Zr.
{n} After an unknown number of time steps it reaches Pn at some time (say tn) with a velocity of Vn.
{n+1} It moves to Pn+1 at tn+1.
The problem I have is to compute the minimum time for the transition from P0 to Pn and to generate the intermediate positions and velocity directions thereof. A secondary goal is to accelerate smoothly instead of applying acceleration that results in jerks.
Current Approach:
find the dimension {x, y or z} that will take the longest to align from start P0 to end Pn. This will be the critical dimension and will determine the total time. This is fairly straightforward and I can write something to this effect.
interpolate smoothly without jitters from P0 to Pn in all dimensions such that the velocity at Pn is as expected. I am not sure, how to approach this.
Any inputs/physics engines that already do this will be useful. It is a commercial project and I cannot put dependencies on large 3rd party libraries with restrictive licenses.
Note: Particle at P0 and Pn have little or no acceleration.
If I understand correctly, you have a point (P0, V0), with V0 = P0 - P-1, and a point (Pn, Vn), with Vn = Pn - Pn-1, and you want to find the fewest intermediate points by adjusting the acceleration at each time step.
Let's define the acceleration at ti: Ai = Vi - Vi-1, with abs(Ai) <= mA. Here, since the problem is axis-independant, abs is the member-wise absolute instead of the norm (or vector magnitude), and mA is the maximum acceleration vector, positive in each dimension. Let's also consider that Pn > P0 (member-wise).
From that, we get Vi = Vi-1 + Ai and so Pi = Pi-1 + Vi-1 + Ai.
If you need to go from some point to another in the fastest way possible, the obvious thing to do, whatever the initial velocity, is accelerate as much as possible until you reach the goal. However, since your problem is discrete and you have a terminal velocity Vn, using that method will probably lead too far and with a different terminal velocity.
However, you can do the same thing in reverse, starting from the end point. And if you start simultaneously from both points, you will make two paths crossing each other in each dimension (not necessarily crossing in 3D, but, in each dimension, the relative direction of both paths changes at some "crossing" point).
Let's take a one-dimensional example. (P0, V0) = (0, -2) and (Pn, Vn) = (35, -1), and mA = 1.
The first path, with Ai = mA, goes like this:
(0, -2) -> (-1, -1) -> (-1, 0) -> (0, 1) -> (2, 2) ->
(5, 3) -> (9, 4) -> (14, 5) -> (20, 6) -> (27, 7) -> ...
The second path, with Ai = -mA but in reverse, goes like this:
(35, -1) <- (36, 0) <- (36, 1) <- (35, 2) <- (33, 3) <-
(30, 4) <- (26, 5) <- (21, 6) <- (15, 7) <- ...
You can see the paths cross with the same velocity somewhere between 20 and 21. That gives you the fastest acceleration and deceleration parts of the path you need, but the two parts aren't connected. However, it's easy to connect them by finding the closest points of same velocity; let's call these points Pq and Pr. Here, Pq = (20, 6) and Pr = (21, 6). Since that velocity is calculated between current and previous points, take the point before Pq (Pq-1, or (14, 5) in the example) and the point Pr, and try connecting them.
If Pq >= Pr >= Pq - 2mA, then you can connect them directly by taking Pq-1 unchanged, and Pr with Vr = Pr - Pq-1.
Else, take Pq-2 and Pr-1 (where Vr-1 = Vr - mA, because it's in reverse) and try connecting those by adding intermediate points. Since these points have a velocity difference of mA, you can search only for intermediate points with the same velocity Vs such that Vq-2 <= Vs <= Vr-1.
If you still can't find a solution, then take Pq-3 and Pr-2 and repeat the process with more intermediate points.
In the example I took, Pq < Pr, so we have to try with Pq-2 = (9, 4) and Pr-1 = (26, 5). We can connect those with a sequence of 3 points, for example (9, 4) -> (13, 4) -> (17, 4) -> (21, 4) -> (26, 5).
In any case, this method will give you the smallest amount of intermediate points, meaning the fastest path between P0 and Pn.
If you then want to reduce jerk, then you can forget the points calculated previously and do an interpolation with the number of points you now know to be minimal.
After playing around with some ideas, I came up with another solution, more accurate and probably faster, if done correctly, than that of my previous answer. It is however quite complicated and requires quite a bit of maths, although not very complex maths. Moreover, this is a work in progress: I am still investigating some areas. Nonetheless, from what I've tried, it does already produce very good results.
The problem
Definitions and goal
Throughout this answer, p[n] refers to the position of the nth point, v[n] to its velocity, a[n] to its acceleration, and j[n] to its jerk (the derivative of acceleration). The velocity of the nth point depends only on its position and that of the previous point. Similarly for acceleration and jerk, but with the points velocity and acceleration, respectively.
We have a start point and an end point, respectively p[0] and p[n], both with associated velocities v[0] and v[n]. The goal is to place n-1 points in between, with an arbitrary n, such that, along the X, Y, and Z axes, the absolute values of acceleration and jerk at any of these points (and at p[n]) are below some limits, respectively aMaxX, aMaxY, and aMaxZ for acceleration, and jMaxX, jMaxY, and jMaxZ for jerk.
What we want to find is the values of p[i] for all i ∈ [1; n-1]. Because p[i] = p[i-1] + v[i], this is the same as finding v[i]. By the same reasoning, with v[i] = v[i-1] + a[i] and a[i] = a[i-1] + j[i], it is also the same as finding a[i] or j[i].
a[0] and a[n+1] are assumed to be zero.
Observations and simplifications
Because the problem's constraints are independant of the dimension, we can solve for each of the three dimensions separately, as long as the number of points obtained in each case is the same. Therefore, I am only going to solve the one-dimensional version of the problem, using aMax and jMax, irrespective of the axis.
*[WIP]* Determine the worst case to solve first, then solve the other ones, knowing the number of points.
The actual positions of the two given points are irrelevant, what matters is the relative distance between them, which we can define as P = p[n] - p[0]. Let's also define the ranges R = [1; n] and R* = [1; n+1].
Because of the discrete nature of the problem, we can obtain the following equations. Note that ∑{i∈R}(x[i]) is the sum of all x[i] for i∈R.
Ⓐ ∑{i∈R}(v[i]) = P
Ⓑ ∑{i∈R}(a[i]) = v[n] - v[0]
Ⓧ ∑{i∈R*}(j[i]) = 0
Ⓧ comes from the assumption that a[0] = a[n+1] = 0.
From Ⓐ and v[i] = v[i-1] + a[i], i∈R, we can deduce:
Ⓒ ∑{i∈R}((n+1-i)*a[i]) = P - n*v[0]
By the same logic, from Ⓑ, Ⓒ, and a[i] = a[i-1] + j[i], i∈R, we can deduce:
Ⓨ ∑{i∈R}((n+1-i)*j[i]) = v[n] - v[0]
Ⓩ ∑{i∈R}(T[n+1-i]*j[i]) = P - n*v[0]
Here, T[n] is the nth triangular number, defined by T[n] = n*(n+1)/2.
The equations Ⓧ, Ⓨ, and Ⓩ are the relevant ones for the next parts.
The approach
In order to minimize n, we can start with a small value of n (1, 2?) and find a solution. Then, if max{i∈R}(abs(a[i])) > aMax or max{i∈R}(abs(j[i])) > jMax, we can increment n and repeat the process.
*[WIP]* Find a lower bound for n to avoid unnecessary calculations from small values of n. Or estimate the correct value of n and pinpoint it by testing solutions.
Finding a solution requires finding the values of j[i] for all i∈R*. I have yet to find an optimal form for j[i], but defining j*[i], r[i] and s[i] such that
j[i] = j*[i] + r[i]v[0] + s[i]v[n]
works quite well.
*[WIP]* Find a better form for j[i]
By doing that, we transform our n-1 unknowns (j[i], i∈R, note that j[n+1] = -∑{i∈R}(j[i])) into 3(n-1) easier to find unknowns. Here are a few things we can deduce right now from Ⓧ, Ⓨ, and Ⓩ.
∑{i∈R*}(r[i]) = 0
∑{i∈R*}(s[i]) = 0
∑{i∈R}((n+1-i)*r[i]) = -1
∑{i∈R}((n+1-i)*s[i]) = 1
∑{i∈R}(T(n+1-i)*r[i]) = -n
∑{i∈R}(T(n+1-i)*s[i]) = 0
As a reminder, here are Ⓧ, Ⓨ, and Ⓩ.
Ⓧ ∑{i∈R*}(j[i]) + j[n+1] = 0
Ⓨ ∑{i∈R}((n+1-i)*j[i]) = v[n] - v[0]
Ⓩ ∑{i∈R}(T[n+1-i]*j[i]) = P - n*v[0]
The goal now is to find adequate special cases to help us determine these unknowns.
The special cases
v[0] = v[n] = 0
By playing with values of jerk, I observed that taking all of j[i], i∈R* as part of a parabola yields excellent results for minimizing both jerk and acceleration. Although it isn't the best possible fit, I haven't found better yet.
The intuition behind values of jerk coming from a parabola is that, if the values of position are to follow a polynomial, then its degree must be at least 5, and can be 5. This is easier to understand if you think about the values of velocity following a 4th degree polynomial. The constraints that v[0] and v[n] are set, a[0] = a[n+1] = 0, and that its integral over [0; n] must equal P, this polynomial must have a degree of at least 4. This holds for both continuous and dicrete cases. Finally, it seems that taking the smallest degree leads to a smoother jerk as well as making it easier to calculate.
Here is an example of a continuous case where the position is in purple, the velocity in blue, the acceleration in yellow and the jerk in red.
In case you want to play with this, here is how to define the position curve in terms of n, p[0], p[n], v[0], and v[n] (the other ones are simply derivatives).
a = (-3(v[n]+v[0]) + 6(p[n]-p[0])) / n^5
b = (n(7v[n]+8v[0]) - 15(p[n]-p[0])) / n^4
c = (-n(4v[n]+6v[0]) + 10(p[n]-p[0])) / n^3
p[x] = ax^5 + bx^4 + cx^3 + v[0]x + p[0]
If v[0] = v[n] = 0, then j[i] = j*[i], i∈R*. That means that the values j*[i] follow a quadratic polynomial. So we want to find α, β, and γ such that Ⓟ holds.
Ⓟ j*[i] = αi^2 + βi + γ, i∈R*
From Ⓧ, Ⓨ, and Ⓩ follow these equations.
α*∑{i∈R*}(i^2) + β*∑{i∈R*}(i) + c*∑{i∈R*}(1) = 0
α*∑{i∈R}((n+1-i)*i^2) + β*∑{i∈R}((n+1-i)*i) + c*∑{i∈R}(n+1-i) = 0
α*∑{i∈R}(T(n+1-i)*i^2) + β*∑{i∈R}(T(n+1-i)*i) + c*∑{i∈R}(T(n+1-i)) = P
Solving this system gives α, β, and γ, which can be used with Ⓟ to calculate j*[i], i∈R*. Note that j*[i] = j*[n+2-i], so only the upper half of the calculations need to be done.
v[0] = v[n] = 1/n
If v[0] = v[n] = 1/n, then j[i] = 0, i∈R*. This means that Ⓠ holds.
Ⓠ r[i] + s[i] = -n*j[i], i∈R*
v[0] = 0, j[i∈L] = J, j[h] = 0, j[i∈U] = -J
L and U are respectively the lower and upper halves of R*, and h is the value in between, if n+1 is odd. In other words:
if n is odd:
L = [1; (n+1)/2]
U = [(n+3)/2; n+1]
if n is even:
L = [1; n/2]
h = n/2+1
U = [n/2+2; n]
This special case corresponds to the maximum overall acceleration between p[0] and p[n] while minimizing abs(j[i]), i∈R*. Here, Ⓩ gives us the following equation.
∑{i∈R}(T[n+1-i]*j[i]) = P
∑{i∈L}(T[n+1-i])*j[1] + ∑{i∈U}(T[n+1-i])*j[n+1] = P
j[1] = P / [ ∑{i∈L}(T[n+1-i]) - ∑{i∈U}(T[n+1-i]) ]
This gives j[1], and so every j[i], i∈R*. We can then calculate v[n] using Ⓨ.
Putting the pieces together
Each special case gives us, for some values of v[0], v[n] and P, a relation of the form
αj*[i] + βr[i] + γs[i] = δ.
By treating three special cases (assuming they are not similar, meaning the do not give the same relation), we have a system of three equations that, once solved, gives the values of j*[i], r[i] and s[i] for all i∈R*.
As a result, we can calculate, for each value of n, values of j[i] depending on v[0], v[n] and P. They can be precalculated, which means testing them for any value of n can be very fast. Thereby, we can very quicklyt find a good estimate for the fewest amount of points needed in the trajectory, as well as a good approximation of the best trajectory possible, as long as we have precalculated values up to a sufficiently large value of n.
Answer
I suggest you to take following function :
X(n) = Xstart + Vxstart n+ (-6xstart+3Vxstart+6xend-3Vxend+c/2) n^2 + (8xstart+3Vxstart-8xend+5Vxend-c) n^3 + (-3Xstart-Vxstart+3xend-2Vxend+c/2) n^4
(for each coordinate X,Y,Z)
Here are some graphs of what this gives, I took c=3 for each samples:
For xstart=1, vstart=1, xend=3, vstart=-2, this gives :
X(n)= 1 + n + 16 n^2 -25 n^3 + 10 n^4
For xstart = -4, vstart =-4, xend = 4, vend = 0, this gives :
(-4 -4n +61n^2 -78n^3 + 29yn^4)
where c is a number from 0.1 to 5, it is up to you to decide, the higher c will be, the faster the function will go to that point (but it might have to turn back if c > 4). (See graphs below).
The polynomial comes from following calculation : where a=x0,b=v0,c=xe,d=v2,e=the magic constant
Explanation
Based on Nelfeal's answer, my idea was to try to solve the given problem with polynomials.
We can change the problem as to define a new Axis which goes in the P[last]-P[0], to have the problem reduced to dimension 1.
We can think about the problem in continuous mathematics instead of discrete mathematics (eg use functions instead of sequences), and go back to the discrete world which is just a special case of the continuous.
We can change the unit for time and space so that the time is 1 and the distance is 1, so that the problem is simplified to
Find a function 𝒇 which satisfies the following :
𝒇(0) = 0 and 𝒇(1) = 1
𝒇'(0) = 0 and 𝒇'(1) = 0
For x∈ℝ |𝒇''(x)| < c, where c is the max speed
We have
P(X) = ∑{i∈ℕ} Ai Xi
P'(X) = ∑{i∈ℕ} (i+1) Ai+1 Xi
P''(X) = ∑{i∈ℕ} (i+2)(i+1) Ai+2 Xi
We need :
P(0) = 0
P(1) = 1
P'(0) = 0
P'(1) = 0
-c <= P''(x) <= c
Thus it means :
a0 = 0 (from 1.)
a1 = 0 (from 3.)
P(1) = ∑{i∈ℕ} Ai = 1
P'(1) = ∑{i∈ℕ} (i+1) Ai = 0
P''(x) = ∑{i∈ℕ} (i+2)(i+1) Ai Xi in [-c,c]
The third equation is the most complex one, and can be simplified by saying that P(1) = c.
We will have c vary to see what changes.
After inverting a 3x3 matrix, we get following result :
P(x) = (c/2+6) x^2 - (c+8) x^3 + (c/2+3) x^4
For c=0.15, this gives :
For c=1, this gives:
For c=4, we see a bounce back :
If we take c from 0.1 to 6, we get following 3d graph :
Note that we have solved this for polynomals of degree 4, but you might do the same things to higher degrees (up to 10 if you want to) to get more possibilities in your functions.
Given a set of closed regions [a,b] where a and b are integers I need to find another set of regions that cover the same numbers but are disjoint.
I suppose it is possible to do naively by iterating through the set several times, but I am looking for a recommendation of a good algorithm for this. Please help.
EDIT:
to clarify, the resulting regions cannot be larger than the original ones, I have to come up with disjoint regions that are contained by the original ones. In other words, I need to split the original regions on the boundaries where they overlap.
example:
3,8
1,4
7,9
11,14
result:
1,3
3,4
4,7
7,8
8,9
11,14
Just sort all endpoints left to right (remember their type: start or end). Swype left to right. Keep a counter starting at 0. Whenever you come across a start: increment the counter. When you come across an end: decrement (note that the counter is always at least 0).
Keep track of the last two points. If the counter is greater than zero - and the last two points are different (prevent empty ranges) - add the interval between the last two points.
Pseudocode:
points = all interval endpoints
sort(points)
previous = points[0]
counter = 1
for(int i = 1; i < #points; i++) {
current = points[i]
if (current was start point)
counter++
else
counter--
if (counter > 0 and previous != current)
add (previous, current) to output
previous = current
}
(This is a modification of an answer that I posted earlier today which I deleted after I discovered it had a logic error. I later realized that I could modify Vincent van der Weele's elegant idea of using parenthesis depth to fix the bug)
On Edit: Modified to be able to accept intervals of length 0
Call an interval [a,a] of length 0 essential if a doesn't also appear as an endpoint of any interval of length > 0. For example, in [1,3], [2,2], [3,3], [4,4] the 0-length intervals [2,2] and [4,4] are essential but [3,3] isn't.
Inessential 0-length intervals are redundant thus need not appear in the final output. When the list of intervals is initially scanned (loading the basic data structures) points corresponding to 0-length intervals are recorded, as are endpoint of intervals of length > 0. When the scan is completed, two instances of each point corresponding to essential 0-length intervals are added into the list of endpoints, which is then sorted. The resulting data structure is a multiset where the only repetitions correspond to essential 0-length intervals.
For every endpoint in the interval define the pdelta (parentheses delta) of the endpoint as the number of times that point appears as a left endpoint minus the number of times it appears as a right endpoint. Store these in a dictionary keyed by the endpoints.
[a,b] where a,b are the first two elements of the list of endpoints is the first interval in the list of disjoint intervals. Define the parentheses depth of b to be the sum of pdelta[a] and pdelta[b]. We loop through the rest of the endpoints as follows:
In each pass through the loop look at the parenthesis depth of b. If it is not 0 than b is still needed for one more interval. Let a = b and let the new p be the next value in the list. Adjust the parentheses depth be the pdelta of the new b and add [a,b] to the list of disjoint intervals. Otherwise (if the parenthesis depth of b is 0) let the next [a,b] be the next two points in the list and adjust the parenthesis depth accordingly.
Here is a Python implementation:
def disjointify(intervals):
if len(intervals) == 0: return []
pdelta = {}
ends = set()
disjoints = []
onePoints = set() #onePoint intervals
for (a,b) in intervals:
if a == b:
onePoints.add(a)
if not a in pdelta: pdelta[a] = 0
else:
ends.add(a)
ends.add(b)
pdelta[a] = pdelta.setdefault(a,0) + 1
pdelta[b] = pdelta.setdefault(b,0) - 1
onePoints.difference_update(ends)
ends = list(ends)
for a in onePoints:
ends.extend([a,a])
ends.sort()
a = ends[0]
b = ends[1]
pdepth = pdelta[a] + pdelta[b]
i = 1
disjoints.append((a,b))
while i < len(ends) - 1:
if pdepth != 0:
a = b
b = ends[i+1]
pdepth += pdelta[b]
i += 1
else:
a = ends[i+1]
b = ends[i+2]
pdepth += (pdelta[a] + pdelta[b])
i += 2
disjoints.append((a,b))
return disjoints
Sample output which illustrates various edge cases:
>>> example = [(1,1), (1,4), (2,2), (4,4),(5,5), (6,8), (7,9), (10,10)]
>>> disjointify(example)
[(1, 2), (2, 2), (2, 4), (5, 5), (6, 7), (7, 8), (8, 9), (10, 10)]
>>> disjointify([(1,1), (2,2)])
[(1, 1), (2, 2)]
(I am using Python tuples to represent the closed intervals even though it has the minor drawback of looking like the standard mathematical notation for open intervals).
A final remark: referring to the result as a collection of disjoint interval might not be accurate since some of these intervals have nonempty albeit 1-point intersections
I think there should be an algorithm for this out there - probably in a field like bioinformatics (the problem reminds me a bit of sequence alignment) so I hope someone can help me out here.
The problem is as follows: Assume I have classified some data into two different classes X and Y. The result of this may look something like this: ..XXX Y XXX.. Further assume that we have some domain knowledge about those classes and know that it's extremely unlikely to have less than a certain number of instances in a row (ie it's unlikely that there are less than 4 Xs or Ys in a sequence - preferably I could use a different threshold per class but that's not a must). So if we use this domain knowledge it's "obvious" that we'd like to replace the single Y in the middle with a X.
So the algorithm should take a sequence of classified instances and the thresholds for the classes (or 1 threshold for all if it simplifies the problem) and try to find a sequence that fulfills the property (no sequences of classes shorter than the given threshold). Obviously there can be an extremely large number of correct solutions (eg in the above example we could also replace all X with a Y) so I think a reasonable optimization criterium would be to minimize the number of replacements.
I don't need an especially efficient algorithm here since the number of instances will be rather small (say < 4k) and we'll only have two classes. Also since this is obviously only a heuristic I'm fine with some inaccuracies if they vastly simplify the algorithm.
A very similar problem to this can be solved as a classic dynamic programming shortest path problem. We wish to find the sequence which minimises some notion of cost. Penalise each character in the sequence that is different from the corresponding character in the original sequence. Penalise each change of character in the sequence, so penalise each change from X to Y and vice versa.
This is not quite what you want because the penalty for YYYXYYY is the same as the penalty for YXXXXXXY - one penalty for YX and one for XY - however it may be a good approximation because e.g. if the base sequence says YYY....YXY....YY then it will be cheaper to change the central X to a Y than to pay the cost of XY and YX - and you can obviously fiddle with the different cost penalties to get something that looks plausible.
Now think of each position in the sequence as being two points, one above the other, one point representing "X goes here" and one representing "Y goes here". You can link points with lines of cost depending on whether the corresponding character is X or Y in the original sequence, and whether the line joins an X with an X or an X with a Y or so on. Then work out the shortest path from left to right using a dynamic program that works out the best paths terminating in X and Y at position i+1, given knowledge of the cost of the best paths terminating in X and Y at position i.
If you really want to penalise short lived changes more harshly than long lived changes you can probably do so by increasing the number of points in the path-finding representation - you would have points that correspond to "X here and the most recent Y was 3 characters ago". But depending on what you want for a penalty you might end up with an incoveniently large number of points at each character.
You can use dynamic programming as in the following pseudocode sketch (for simplicity, this code assumes the threshold is 3 Xs or Ys in a row, rather than 4):
min_switch(s):
n = len(s)
optx = array(4, n, infinity) // initialize all values to infinity
opty = array(4, n, infinity) // initialize all values to infinity
if s[0] == 'X':
optx[1][0] = 0
opty[1][0] = 1
else:
optx[1][0] = 1
opty[1][0] = 0
for i in {1, n - 1}:
x = s[i]
if x == 'X':
optx[1][i] = opty[3][i - 1]
optx[2][i] = optx[1][i - 1]
optx[3][i] = min(optx[2][i - 1], optx[3][i - 1])
opty[1][i] = 1 + min(optx[1][i - 1], optx[2][i - 1], optx[3][i - 1])
opty[2][i] = 1 + opty[1][i - 1]
opty[3][i] = 1 + min(opty[2][i - 1], opty[3][i - 1])
else:
optx[1][i] = 1 + min(opty[1][i - 1], opty[2][i - 1], opty[3][i - 1])
optx[2][i] = 1 + opty[1][i - 1]
optx[3][i] = 1 + min(opty[2][i - 1], opty[3][i - 1])
opty[1][i] = optx[3][i - 1]
opty[2][i] = opty[1][i - 1]
opty[3][i] = min(opty[2][i - 1], opty[3][i - 1])
return min(optx[3][n - 1], opty[3][n - 1])
The above code essentially computes the lowest cost of creating a smooth sequence up to the ith character storing the optimal value for all relevant numbers of consecutive Xs or Ys in a row (1, 2, or 3 in a row). More formally
opt[i][0][k] stores the smallest
cost to convert the string s[0...k]
into a smooth sequence then ends in
i consecutive Xs. Runs of 3 or more
are accounted for in opt[3][0][k].
opt[0][j][k] stores the smallest
cost to convert the string s[0...k]
into a smooth sequence then ends in
j consecutive Ys. Runs of 3 or more
are accounted for in opt[0][3][k].
It is straightforward to convert this to an algorithm that returns the sequence as well as the optimal cost.
Note that some of the cases in the above code are probably unnecessary, it's just a straightforward recurrence derived from the constraints.