Summary
Recently I have had a path-finding puzzle that has some complex constraints (currently, I don't have any solution for this one)
A 2D matrix represented the graph. The length of a path is the number of traversed cells.
One or more number sequences are to be found inside the matrix. Each sequence is scored with a value.
Maximum length of the path in the graph. The number of picked cells must not exceed this value.
At any given moment, you can only choose cells in a specific column or row.
On each turn, you need to switch between column and row and stay on
the same line as the last cell you picked. You have to move at right angles. (The direction is like the Snake game).
Always start with picking the first cell from the top row, then go
vertically down to pick the second cell, and then continue switching
between column and row as usual.
You can't choose the same cell twice. The resulting path must not contain duplicated
cells.
For example:
The task is to find the shortest path, if possible in the graph that contains one or more sequences with the highest total score and the path's length is not exceed the provided maximum length.
The picture below demonstrates the solved puzzle with the resulting path marked in red:
Here, we have a path 3A-10-9B. This path contains the given
sequence 3A-10-9B so, which earns 10pts. More complex graphs typically have longer paths containing various sequences at once.
More complex examples
Multiple Sequences
You can complete sequences in any order. The order in which the sequences are listed doesn't matter.
Wasted Moves
Sometimes we are forced to waste moves and choose different cells that don't belong to any sequence. Here are the rules:
Able to waste 1 or 2 moves before the first sequence.
Able to waste 1 or 2 moves between any neighboring sequences.
However, you cannot break sequences and waste moves in the middle of them.
Here, we must waste one move before the sequence 3A-9B and two moves between sequences 3A-9B and 72-D4. Also, notice how red lines between 3A and 9B as well as between 72 and D4 "cross" previously selected cells D4 and 9B, respectively. You can pick different cells from the same row or column multiple times.
Optimal Sequences
Sometimes, it is not possible to have a path that contains all of the provided sequences. In this case, choose the way which achieved the most significant score.
In the above example, we can complete either 9B-3A-72-D4 or 72-D4-3A but not both due to the maximum path length of 5 cells. We have chosen the sequence 9B-3A-72-D4 since it grants more score points than 72-D4-3A.
Unsolvable solution
The first sequence 3A-D4 can't be completed since the code matrix doesn't contain code D4 at all. The second sequence, 72-10, can't be completed for another reason: codes 72 and 10 aren't located in the same row or column anywhere in the matrix and, therefore, can't form a sequence.
Performance advice
One brute force way is to generate all possible paths in the code matrix, loop through them and choose the best one. This is the easiest but also the slowest approach. Solving larger matrices with larger maximum length of path might take dozens of minutes, if not hours.
Try to implement a faster algorithm that doesn’t iterate through all possible paths and can solve puzzles with the following parameters in less than 10 seconds:
Matrix size: 10x10
Number of sequences: 5
Average length of sequences: 4
Maximum path length: 12
At least one solution exists
For example:
Matrix:
41,0f,32,18,29,4b,55,3f,10,3a,
19,4f,57,43,3a,25,19,1e,5e,42,
13,5a,54,3c,1b,32,29,1c,15,30,
49,45,22,2e,25,51,2f,21,4c,37,
1a,5e,49,12,55,1e,49,19,43,2d,
34,26,53,48,49,60,32,3c,50,10,
0f,1e,30,3d,64,37,5b,5e,22,61,
4e,4f,15,5a,13,56,44,22,40,26,
43,2c,17,2b,1f,25,43,60,50,1f,
3c,2b,54,46,42,4d,32,46,30,24,
Sequences:
30, 26, 44, 32, 3c - 25pts
5a, 3c, 12, 1e, 4d - 10pts
1e, 5a, 12 - 10pts
4d, 1e - 5pts
32, 51, 2f, 49, 55, 42 - 30pts
Optimal solution
3f, 1c, 30, 26, 44, 32, 3c, 22, 5a, 12, 1e, 4d
Which contains
30, 26, 44, 32, 3c
5a, 12, 1e
1e, 4d
Conclusion
I am looking for any advice for this puzzle since I have no idea what keywords to look for. A pseudo-code or hints would be helpful for me, and I appreciate that. What has come to my mind is just Dijkstra:
For each sequence, since the order doesn't matter, I have to find all get all possible paths with every permutation, then find the highest score path that contains other input sequences
After that, choose the best of the best.
In this case, I doubt the performance will be the issue.
First step is to find if a required sequence exists.
- SET found FALSE
- LOOP C1 over cells in first row
- CLEAR foundSequence
- ADD C1 to foundSequence
- LOOP C2 over cells is column containing C1
- IF C2 value == first value in sequence
- ADD C2 to foundSequence
- SET found TRUE
- break from LOOP C2
- IF found
- SET direction VERT
- LOOP V over remaining values in sequence
- TOGGLE direction
- SET found FALSE
- LOOP C2 over cells in same column or row ( depending on direction ) containing last cell in foundSequence
- IF C2 value == V
- ADD C2 to foundSequence
- SET found TRUE
- break from LOOP C2
- IF ! found
break out of LOOP V
- IF foundSequence == required sequence
- RETURN foundSequence
RETURN failed
Note: this doesn't find sequences that are feasible with "wasted moves". I would implement this first and get it working. Then, using the same ideas, it can be extended to allow wasted moves.
You have not specified an input format! I suggest a space delimited text files with lines beginning with 'm' containing matrix values and lines beginning 's' containing sequences, like this
m 3A 3A 10 9B
m 9B 72 3A 10
m 10 3A 3A 3A
m 3A 10 3A 9B
s 3A 10 9B
I have implemented the sequence finder in C++
std::vector<int> findSequence()
{
int w, h;
pA->size(w, h);
std::vector<int> foundSequence;
bool found = false;
bool vert = false;
// loop over cells in first row
for (int c = 0; c < w; c++)
{
foundSequence.clear();
found = false;
if (pA->cell(c, 0)->value == vSequence[0][0])
{
foundSequence.push_back(pA->cell(c, 0)->ID());
found = true;
}
while (found)
{
// found possible starting cell
// toggle search direction
vert = (!vert);
// start from last cell found
auto pmCell = pA->cell(foundSequence.back());
int c, r;
pA->coords(c, r, pmCell);
// look for next value in required sequence
std::string nextValue = vSequence[0][foundSequence.size()];
found = false;
if (vert)
{
// loop over cells in column
for (int r2 = 1; r2 < w; r2++)
{
if (pA->cell(c, r2)->value == nextValue)
{
foundSequence.push_back(pA->cell(c, r2)->ID());
found = true;
break;
}
}
}
else
{
// loop over cells in row
for (int c2 = 0; c2 < h; c2++)
{
if (pA->cell(c2, r)->value == nextValue)
{
foundSequence.push_back(pA->cell(c2, r)->ID());
found = true;
break;
}
}
}
if (!found) {
// dead end - try starting from next cell in first row
break;
}
if( foundSequence.size() == vSequence[0].size()) {
// success!!!
return foundSequence;
}
}
}
std::cout << "Cannot find sequence\n";
exit(1);
}
This outputs:
3A 3A 10 9B
9B 72 3A 10
10 3A 3A 3A
3A 10 3A 9B
row 0 col 1 3A
row 3 col 1 10
row 3 col 3 9B
You can check out the code for the complete application at https://github.com/JamesBremner/stackoverflow75410318
I have added the ability to find sequences that start elsewhere than the first row ( i.e. with "wasted moves" ). You can see the code in the github repo.
Here are the the results of a timing profile run on a 10 by 10 matrix - the algorithm finds 5 sequences in 0.6 milliseconds
Searching
41 0f 32 18 29 4b 55 3f 10 3a
19 4f 57 43 3a 25 19 1e 5e 42
13 5a 54 3c 1b 32 29 1c 15 30
49 45 22 2e 25 51 2f 21 4c 37
1a 5e 49 12 55 1e 49 19 43 2d
34 26 53 48 49 60 32 3c 50 10
0f 1e 30 3d 64 37 5b 5e 22 61
4e 4f 15 5a 13 56 44 22 40 26
43 2c 17 2b 1f 25 43 60 50 1f
3c 2b 54 46 42 4d 32 46 30 24
for sequence 4d 1e
Cannot find sequence starting in 1st row, using wasted moves
row 9 col 5 4d
row 4 col 5 1e
for sequence 30 26 44 32 3c
Cannot find sequence starting in 1st row, using wasted moves
Cannot find sequence
for sequence 5a 3c 12 1e 4d
Cannot find sequence starting in 1st row, using wasted moves
row 2 col 1 5a
row 2 col 3 3c
row 4 col 3 12
row 4 col 5 1e
row 9 col 5 4d
for sequence 1e 5a 12
Cannot find sequence starting in 1st row, using wasted moves
row 6 col 1 1e
row 4 col 5 1e
row 4 col 3 12
for sequence 32 51 2f 49 55 42
Cannot find sequence starting in 1st row, using wasted moves
row 2 col 5 32
row 3 col 5 51
row 3 col 6 2f
row 4 col 6 49
row 4 col 4 55
row 9 col 4 42
raven::set::cRunWatch code timing profile
Calls Mean (secs) Total Scope
5 0.00059034 0.0029517 findSequence
I want to compute the mean and standard derivation of sub region that is created by a window (dashed line) and center at identified pixel-red color( called local mean and standard derivation). This is figure to describe it
We can do it by convolution image with a mask. However, it takes long time because I only care the mean and standard derivation of a server points, while convolution computes for whole point in image. Could you have a faster way to resolve it that only compute the mean and standard derivation at identified pixel? I am doing it by matlab. This is my code by convolution function
I=[18 36 70 33 64 40 62 76 71 37 5
82 49 86 45 96 29 74 7 60 56 45
25 32 55 48 25 30 12 82 95 77 8
24 18 78 74 19 57 67 59 16 46 78
28 9 59 2 29 11 7 31 75 15 25
83 26 96 8 82 26 85 12 11 28 19
81 64 78 70 26 33 17 72 81 16 54
75 39 78 34 59 31 77 31 61 81 89
89 84 29 99 79 25 26 35 65 56 76
93 90 45 7 61 13 34 24 11 34 92
88 82 91 81 100 4 88 70 85 8 19];
identified_position=[30 36 84 90] %indices of pixel 78, 48,72 60
mask=1/9.*ones(3,3);
mean_all=imfilter(I,mask,'same');
%Mean of identified pixels
mean_all(identified_position)
% Compute the variance
std_all=stdfilt(I,ones(3));
%std of identified pixels
std_all(identified_position)
This is the comparison code
function compare_mean(dimx,dimy)
I=randi(100,[dimx,dimy]);
rad=3;
identified_position=randi(max(I(:)),[1,5]);% Get 5 random position
function way1()
mask=ones(rad,rad);
mask=mask./sum(mask(:));
mean_all=conv2(I,mask,'same');
mean_out =mean_all(identified_position);
end
function way2()
box_size = rad; %// Edit your window size here (an odd number is preferred)
bxr = floor(box_size/2); %// box radius
%// Get neighboring indices and those elements for all identified positions
off1 = bsxfun(#plus,[-bxr:bxr]',[-bxr:bxr]*size(I,1)); %//'#neighborhood offsets
idx = bsxfun(#plus,off1(:),identified_position); %// all absolute offsets
I_selected_neigh = I(idx); %// all offsetted elements
mean_out = mean(I_selected_neigh,1); %// mean output
end
way2()
time_way1=#()way1();timeit(time_way1)
time_way2=#()way2();timeit(time_way2)
end
Sometime the way2 has error is
Subscript indices must either be real positive integers or logicals.
Error in compare_mean/way2 (line 18)
I_selected_neigh = I(idx); %// all offsetted elements
Error in compare_mean (line 22)
way2()
Discussion & Solution Codes
Given I as the input image, identified_position as the linear indices of the selected points and bxsz as the window/box size, the approach listed next must be pretty efficient -
%// Get XY coordinates
[X,Y] = ind2sub(size(I),identified_position);
pts = [X(:) Y(:)];
%// Parameters
bxr = (bxsz-1)/2;
Isz = size(I);
%// XY coordinates of neighboring elements
[offx,offy] = ndgrid(-bxr:bxr,-bxr:bxr);
x_idx = bsxfun(#plus,offx(:),pts(:,1)'); %//'
y_idx = bsxfun(#plus,offy(:),pts(:,2)'); %//'
%// Outside image boundary elements
invalids = x_idx>Isz(1) | x_idx<1 | y_idx>Isz(2) | y_idx<1;
%// All neighboring indices
all_idx = (y_idx-1)*size(I,1) + x_idx;
all_idx(invalids) = 1;
%// All neighboring elements
all_vals = I(all_idx);
all_vals(invalids) = 0;
mean_out = mean(all_vals,1); %// final mean output
stdfilts = stdfilt(all_vals,ones(bxsz^2,1))
std_out = stdfilts(ceil(size(stdfilts,1)/2),:) %// final stdfilt output
Basically, it gets all the neighbouring indices for all identified positions in one go with bsxfun and thus, gets all those neighbouring elements. Those selected elements are then used to get the mean and stdfilt outputs. The whole idea is to keep the memory requirement minimum and at the same time doing everything in a vectorized fashion within those selected elements. Hopefully, this must be faster!
Benchmarking
Benchmarking Code
dx = 10000; %// x-dimension of input image
dy = 10000; %// y-dimension of input image
npts = 1000; %// number of points
I=randi(100,[dx,dy]); %// create input image of random intensities
identified_position=randi(max(I(:)),[1,npts]);
rad=5; %// blocksize (rad x rad)
%// Run the approaches fed with the inputs
func1 = #() way1(I,identified_position,rad); %// original approach
time1 = timeit(func1);
clear func1
func2 = #() way2(I,identified_position,rad); %// proposed approach
time2 = timeit(func2);
clear func2
disp(['Input size: ' num2str(dx) 'x' num2str(dy) ' & Points: ' num2str(npts)])
disp(['With Original Approach: Elapsed Time = ' num2str(time1) '(s)'])
disp(['With Proposed Approach: Elapsed Time = ' num2str(time2) '(s)'])
disp(['**Speedup w/ Proposed Approach : ' num2str(time1/time2) 'x!**'])
Associated function codes
%// OP's stated approach
function mean_out = way1(I,identified_position,rad)
mask=ones(rad,rad);
mask=mask./sum(mask(:));
mean_all=conv2(I,mask,'same');
mean_out =mean_all(identified_position);
return;
function mean_out = way2(I,identified_position,rad)
%//.... code from proposed approach stated earlier until mean_out %//
Runtime results
Input size: 10000x10000 & Points: 1000
With Original Approach: Elapsed Time = 0.46394(s)
With Proposed Approach: Elapsed Time = 0.00049403(s)
**Speedup w/ Proposed Approach : 939.0778x!**
I'm looking for a faster way to do decrease the value of certain numbers in a vector in Matlab, for example I've this vector:
Vector a=[1 21 35 44 45 67 77 83 93 100]
Then I have to remove the elements 35,45,77, so:
RemoveVector b=[3,5,7]
RemoveElements c=[35,45,77]
After remove the elements, the should be:
Vector=[1 21 43 65 80 90 97]
Note that besides remove the element, all the next elements decrease their values in 1, I've this code in Matlab:
a(:,b) = [];
b = fliplr(b);
for i=1:size(a,2)
for j=1:size(c,2)
if(a(1,i)>=c(1,j))
a(1,i) = a(1,i) -1;
end
end
end
But is too slow, m0=2.8*10^-3 seconds, there is a faster algorithm? I believe with matrix operations could be faster and elegant.
#Geoff has a good overall approach, but the adjustment can be done in O(n) not O(n*k):
adjustment = zeros(size(a));
adjustment(b(:)) = 1;
a = a - cumsum(adjustment);
a(b(:)) = [];
I think prior to removing the elements from a whose indices are given in b, the code could do all the decrementing first
% copy a
c = a;
% iterate over each index in b
for k=1:length(b)
% for all elements in c that follow the index in b (so b(k)+1…end)
% subtract one
c(b(k)+1:end) = c(b(k)+1:end) - 1;
end
% now remove the elements that correspond to the indices in b
c(b) = [];
Try the above and see what happens!
Thank so much to Geoff and Ben for yours answer, I've proved both answers by this way:
tic
a=[1 21 35 44 45 67 77 83 93 100];
b=[3 5 7];
%Code by Geoff
c = a;
for k=1:length(b)
% for all elements in c that follow the index in b (so b(k)+1…end)
% subtract one
c(b(k)+1:end) = c(b(k)+1:end) - 1;
end
c(b) = [];
m1 = toc;
and
tic
a=[1 21 35 44 45 67 77 83 93 100];
b=[3 5 7];
%Code by Ben
adjustment = zeros(size(a));
adjustment(b(:)) = 1;
a = a - cumsum(adjustment);
a(b(:)) = [];
m2 = toc;
The results in my machine were m1=1.2648*10^-4 seconds and m2=7.426*10^-5 seconds, the second code is faster, my first code gives m0 = 2.8*10^-3 seconds .
suppose I have the following matrix
a =
76 NaN 122 NaN
78 NaN 123 NaN
84 NaN 124 54
77 NaN 126 58
82 45 129 62
90 50 135 45
76 63 133 66
79 52 122 49
88 56 140 24
Is there any way to calculate exponential moving average for each column, disregarding the first NaN values? For instance, if I use a 3 days exponential factor, I would expect to get a matrix starting with 2 NaN values in the 1st column, 6 NaN values in the 2nd column,2 NaN values in the 3rd column and 4 NaN values in the 4th column. Any suggestion? Thank you in advance
Just use filter on the whole matrix, which will pass through the NaN's as appropriate. If you want to "infect" edge values with NaN as well, add some extras at the top edge, then trim the result:
kernel = [1 1 1].'; % Any 3-element kernel, as column vector
a2 = [repmat(NaN, 2, 4); a]; % Add extra NaN's at the start, to avoid partial answers
xtemp = filter(kernel, 1, a2);
x = xtemp(3:end, :);