I'm using a library that allows computing large matrix operations on the GPU, gpu.js. What I'm trying to do is not really hard I don't think but I can't seem to begin to figure out the algorithm for it. Basically I have an array buffer stored as r,g,b,a repeated for each pixel, so a 4x4 image would be a an array of 64 values.
I want to output an an image X times larger than the input, but with "nearest neighbor" calculation, so every pixel just becomes a 2x2 square, or 3x3, etc.
So the operation is set up like this (note gpu.js requires arrays as inputs) and keep in mind it's iterating over the full size output, so I have to find the correct coordinates that will exist in the smaller sourceBuffer based on the current index in the outputBuffer (index is exposed by the lib as this.thread.x).
var pixelateMatrix = gpu.createKernel(function(inputBuffer, width, height, scale) {
var y = Math.floor(this.thread.x / (width[0] / scale[0]) * 4);
var x = this.thread.x % ((width[0] / scale[0]) * 4);
var remainder = this.thread.x % 4;
return inputBuffer[x * (width[0] * 4) + y * 4 + remainder];
}).setOutput([width * height * 4]);
This is what I tried but right now it's weirdly only outputting the current width of the screen as the value for each entry.
What's the correct algorithm for this? Normally I'm used to doing this kind of thing with a loop iterating over the source, but in this case I have to work with each pixel rgba value individually in a 1 dimentional array and I'm confused on how to do this.
Also obviously I need to do it with as few operations as possible.
Depends a bit if you store the item row or column major. Assuming you do row major.
Each row will look be r1 g2 b1 a1 r2 g2 b2 a2 .... then it will be followed by the next row and so on. You need to know how large is the image (at least how large each row is), i'll call this N. So the get component c from row x column y you need value_pos = x * (N * 4) + y * 4 + c. Everything here is 0 indexed. You can use the value for both reading and writing, just update N between the two since they are supposed to have different sizes.
Related
I have this image which is completely black except for a white object in the middle (it can be anything but it is always completely white). What I would like to do with nodeJs, is trace the boundary of the object (I would like to find all the white points which are next to black) in the image (performance is key!)
With pngjs I can read an image which gives me an array in which each pixels has 4 values (RGBA). Its a one dimensional array. So, suppose the image is 1000 x 1000 pixels it gives me an array of 1000 x 1000 x 4 = 4000000 entries.
The below expression converts x and y into an array index
var idx = (1000 * y + x) << 2;
data[idx] = 243;
data[idx + 1] = 16;
data[idx + 2] = 16;
Anyway, I could traverse the whole array and register the points where black changes into white, but as I said, performance is very important. I can imagine that some kind of smart iterative search algorithm exists that can follow the boundary somehow :)
Maybe someone knows a library that can help, or an article about how to do this would be great too!!
Check out chain codes like Freeman Code. You need 1 contour point to start with. So just iterate through your lines until you hit your object. Then you walk around your object until you reach your starting point. You will get a code that describes the direction you took for every step. This code can be used to calculate various object features or to just draw the contour of your object.
Btw if your obect is always white and your background is always black you don't have to process 4 channels. Red, green or blue channels contain the same information. Just use either one of them.
Specifically, I'm working in canvas with javascript.
Basically, I have objects which have boundaries that I want to avoid, but still surround with a bezier curve. However, I'm not even sure where to begin to write an algorithm that would move control points to avoid colliding.
The problem is in the image below, even if you're not familiar with music notation, the problem should still be fairly clear. The points of the curve are the red dots
Also, I have access to the bounding boxes of each note, which includes the stem.
So naturally, collisions must be detected between the bounding boxes and the curves (some direction here would be good, but I've been browsing and see that there's a decent amount of info on this). But what happens after collisions have been detected? What would have to happen to calculate control points locations to make something that looked more like:
Bezier approach
Initially the question is a broad one - perhaps even to broad for SO as there are many different scenarios that needs to be taken into consideration to make a "one solution that fits them all". It's a whole project in its self. Therefor I will present a basis for a solution which you can build upon - it's not a complete solution (but close to one..). I added some suggestions for additions at the end.
The basic steps for this solutions are:
Group the notes into two groups, a left and a right part.
The control points are then based on the largest angle from the first (end) point and distance to any of the other notes in that group, and the last end point to any point in the second group.
The resulting angles from the two groups are then doubled (max 90°) and used as basis to calculate the control points (basically a point rotation). The distance can be further trimmed using a tension value.
The angle, doubling, distance, tension and padding offset will allow for fine-tuning to get the best over-all result. There might be special cases which need additional conditional checks but that is out of scope here to cover (it won't be a full key-ready solution but provide a good basis to work further upon).
A couple of snapshots from the process:
The main code in the example is split into two section, two loops that parses each half to find the maximum angle as well as the distance. This could be combined into a single loop and have a second iterator to go from right to middle in addition to the one going from left to middle, but for simplicity and better understand what goes on I split them into two loops (and introduced a bug in the second half - just be aware. I'll leave it as an exercise):
var dist1 = 0, // final distance and angles for the control points
dist2 = 0,
a1 = 0,
a2 = 0;
// get min angle from the half first points
for(i = 2; i < len * 0.5 - 2; i += 2) {
var dx = notes[i ] - notes[0], // diff between end point and
dy = notes[i+1] - notes[1], // current point.
dist = Math.sqrt(dx*dx + dy*dy), // get distance
a = Math.atan2(dy, dx); // get angle
if (a < a1) { // if less (neg) then update finals
a1 = a;
dist1 = dist;
}
}
if (a1 < -0.5 * Math.PI) a1 = -0.5 * Math.PI; // limit to 90 deg.
And the same with the second half but here we flip around the angles so they are easier to handle by comparing current point with end point instead of end point compared with current point. After the loop is done we flip it 180°:
// get min angle from the half last points
for(i = len * 0.5; i < len - 2; i += 2) {
var dx = notes[len-2] - notes[i],
dy = notes[len-1] - notes[i+1],
dist = Math.sqrt(dx*dx + dy*dy),
a = Math.atan2(dy, dx);
if (a > a2) {
a2 = a;
if (dist2 < dist) dist2 = dist; //bug here*
}
}
a2 -= Math.PI; // flip 180 deg.
if (a2 > -0.5 * Math.PI) a2 = -0.5 * Math.PI; // limit to 90 deg.
(the bug is that longest distance is used even if a shorter distance point has greater angle - I'll let it be for now as this is meant as an example. It can be fixed by reversing the iteration.).
The relationship I found works good is the angle difference between the floor and the point times two:
var da1 = Math.abs(a1); // get angle diff
var da2 = a2 < 0 ? Math.PI + a2 : Math.abs(a2);
a1 -= da1*2; // double the diff
a2 += da2*2;
Now we can simply calculate the control points and use a tension value to fine tune the result:
var t = 0.8, // tension
cp1x = notes[0] + dist1 * t * Math.cos(a1),
cp1y = notes[1] + dist1 * t * Math.sin(a1),
cp2x = notes[len-2] + dist2 * t * Math.cos(a2),
cp2y = notes[len-1] + dist2 * t * Math.sin(a2);
And voila:
ctx.moveTo(notes[0], notes[1]);
ctx.bezierCurveTo(cp1x, cp1y, cp2x, cp2y, notes[len-2], notes[len-1]);
ctx.stroke();
Adding tapering effect
To create the curve more visually pleasing a tapering can be added simply by doing the following instead:
Instead of stroking the path after the first Bezier curve has been added adjust the control points with a slight angle offset. Then continue the path by adding another Bezier curve going from right to left, and finally fill it (fill() will close the path implicit):
// first path from left to right
ctx.beginPath();
ctx.moveTo(notes[0], notes[1]); // start point
ctx.bezierCurveTo(cp1x, cp1y, cp2x, cp2y, notes[len-2], notes[len-1]);
// taper going from right to left
var taper = 0.15; // angle offset
cp1x = notes[0] + dist1*t*Math.cos(a1-taper);
cp1y = notes[1] + dist1*t*Math.sin(a1-taper);
cp2x = notes[len-2] + dist2*t*Math.cos(a2+taper);
cp2y = notes[len-1] + dist2*t*Math.sin(a2+taper);
// note the order of the control points
ctx.bezierCurveTo(cp2x, cp2y, cp1x, cp1y, notes[0], notes[1]);
ctx.fill(); // close and fill
Final result (with pseudo notes - tension = 0.7, padding = 10)
FIDDLE
Suggested improvements:
If both groups' distances are large, or angles are steep, they could probably be used as a sum to reduce tension (distance) or increase it (angle).
A dominance/area factor could affect the distances. Dominance indicating where the most tallest parts are shifted at (does it lay more in the left or right side, and affects tension for each side accordingly). This could possibly/potentially be enough on its own but needs to be tested.
Taper angle offset should also have a relationship with the sum of distance. In some cases the lines crosses and does not look so good. Tapering could be replaced with a manual approach parsing Bezier points (manual implementation) and add a distance between the original points and the points for the returning path depending on array position.
Hope this helps!
Cardinal spline and filtering approach
If you're open to use a non-Bezier approach then the following can give an approximate curve above the note stems.
This solutions consists of 4 steps:
Collect top of notes/stems
Filter away "dips" in the path
Filter away points on same slope
Generate a cardinal spline curve
This is a prototype solution so I have not tested it against every possible combination there is. But it should give you a good starting point and basis to continue on.
The first step is easy, collect points representing the top of the note stem - for the demo I use the following point collection which slightly represents the image you have in the post. They are arranged in x, y order:
var notes = [60,40, 100,35, 140,30, 180,25, 220,45, 260,25, 300,25, 340,45];
which would be represented like this:
Then I created a simple multi-pass algorithm that filters away dips and points on the same slope. The steps in the algorithm are as follows:
While there is a anotherPass (true) it will continue, or until max number of passes set initially
The point is copied to another array as long as the skip flag isn't set
Then it will compare current point with next to see if it has a down-slope
If it does, it will compare the next point with the following and see if it has an up-slope
If it does it is considered a dip and the skip flag is set so next point (the current middle point) won't be copied
The next filter will compare slope between current and next point, and next point and the following.
If they are the same skip flag is set.
If it had to set a skip flag it will also set anotherPass flag.
If no points where filtered (or max passes is reached) the loop will end
The core function is as follows:
while(anotherPass && max) {
skip = anotherPass = false;
for(i = 0; i < notes.length - 2; i += 2) {
if (!skip) curve.push(notes[i], notes[i+1]);
skip = false;
// if this to next points goes downward
// AND the next and the following up we have a dip
if (notes[i+3] >= notes[i+1] && notes[i+5] <= notes[i+3]) {
skip = anotherPass = true;
}
// if slope from this to next point =
// slope from next and following skip
else if (notes[i+2] - notes[i] === notes[i+4] - notes[i+2] &&
notes[i+3] - notes[i+1] === notes[i+5] - notes[i+3]) {
skip = anotherPass = true;
}
}
curve.push(notes[notes.length-2], notes[notes.length-1]);
max--;
if (anotherPass && max) {
notes = curve;
curve = [];
}
}
The result of the first pass would be after offsetting all the points on the y-axis - notice that the dipping note is ignored:
After running through all necessary passes the final point array would be represented as this:
The only step left is to smoothen the curve. For this I have used my own implementation of a cardinal spline (licensed under MIT and can be found here) which takes an array with x,y points and smooths it adding interpolated points based on a tension value.
It won't generate a perfect curve but the result from this would be:
FIDDLE
There are ways to improve the visual result which I haven't addressed, but I will leave it to you to do that if you feel it's needed. Among those could be:
Find center of points and increase the offset depending on angle so it arcs more at top
The end points of the smoothed curve sometimes curls slightly - this can be fixed by adding an initial point right below the first point as well at the end. This will force the curve to have better looking start/end.
You could draw double curve to make a taper effect (thin beginning/end, thicker in the middle) by using the first point in this list on another array but with a very small offset at top of the arc, and then render it on top.
The algorithm was created ad-hook for this answer so it's obviously not properly tested. There could be special cases and combination throwing it off but I think it's a good start.
Known weaknesses:
It assumes the distance between each stem is the same for the slope detection. This needs to be replaced with a factor based comparison in case the distance varies within a group.
It compares the slope with exact values which may fail if floating point values are used. Compare with an epsilon/tolerance
I have animation follows this timing function: cubic-bezier(0.25, 0.1, 0.25, 1.0)
I want to mod this function so i just get the ending 40% of it. To make things easy lets just say I want the end 50% of the function. How can I do this.
So graphically this is what it is:
https://developer.mozilla.org/files/3429/cubic-bezier,ease.png
and I want to to make a cubic-bezier with parameters such that graphically we only see the top portion, so what we see from 0.5 to 1 (on the yaxist) porition of this graph, i want to make that same line but from 0 to 1.
Please help me how to make this function.
If you want only a section of a cubic curve, with t from 0 to 1, there are "simple" formulae to determine what the new coordinates need to be. I say simple because it's pretty straight forward to implement, but if you also want to know why the implementation actually works, that generally requires diving into maths, and some people consider that scary.
(The end result of the section on matrix splitting pretty much gives you the new coordinates for an arbitrary split-point without needing to read the explanation of why that works)
Let's take your example curve: first, we need to figure out what the curve's original coordinates are. We go with a guess of (0,0)-(0.4,0.25)-(0.2,1)-(1,1). We then want to split that curve up at t=0.4, so we ignore all of section 7 except for the final bit that tells us how to derive new coordinates. For any splitting point t=z (where z is somewhere between 0 and 1` we'll have two new sets of coordinates. One for the curve "before" the splitting point, and one for "after" the splitting point. We want the latter, so we pick:
So we just plug in 0.4 for z and off we go. Our new first point is 0.064 * P4 - 3 * 0.096 * P3 + 3 * 0.144 * P2 + 0.216 * P1 = 0.2944 (which we need to evaluate twice. Once for our x values, and one for our y values). We do the same for P2, P3 and P4 (although our fourth point is of course still the same so we don't need to bother. It was (1,1) and is still (1,1) after the split).
So, let's implement that in javascript:
function split(options) {
var z = options.z,
cz = z-1,
z2 = z*z,
cz2 = cz*cz,
z3 = z2*z,
cz3 = cz2*cz,
x = options.x,
y = options.y;
var left = [
x[0],
y[0],
z*x[1] - cz*x[0],
z*y[1] - cz*y[0],
z2*x[2] - 2*z*cz*x[1] + cz2*x[0],
z2*y[2] - 2*z*cz*y[1] + cz2*y[0],
z3*x[3] - 3*z2*cz*x[2] + 3*z*cz2*x[1] - cz3*x[0],
z3*y[3] - 3*z2*cz*y[2] + 3*z*cz2*y[1] - cz3*y[0]];
var right = [
z3*x[3] - 3*z2*cz*x[2] + 3*z*cz2*x[1] - cz3*x[0],
z3*y[3] - 3*z2*cz*y[2] + 3*z*cz2*y[1] - cz3*y[0],
z2*x[3] - 2*z*cz*x[2] + cz2*x[1],
z2*y[3] - 2*z*cz*y[2] + cz2*y[1],
z*x[3] - cz*x[2],
z*y[3] - cz*y[2],
x[3],
y[3]];
return { left: left, right: right};
}
Done deal. This function will give us two subcurves (called left and right, both Number[8] arrays in x1/y1/x2/y2/... ordering) that are mathematically identical to our original curve if taken together, except modeled as two new t=[0,1] intervals, for any splitting point t=z with z between 0 and 1. Our work is done forever.
Given an array of integers heights I would like to split these into n sets, each with equal totalHeight (sum of the values in set), or as close to as possible. There must be a fixed distance, gap, between each value in a set. Sets do not have to have the same number of values.
For example, supposing:
heights[0, 1, 2, 3, 4, 5] = [120, 78, 110, 95, 125, 95]
n = 3
gaps = 10
Possible arrangements would be:
a[0, 1], b[2, 3], c[4, 5] giving totalHeight values of
a = heights[0] + gap + heights[1] = 120 + 10 + 78 = 208
b = heights[2] + gap + heights[3] = 110 + 10 + 95 = 215
c = heights[4] + gap + heights[5] = 125 + 10 + 95 = 230
a[0], b[1, 2, 3], c[4, 5] giving totalHeight values of
a = heights[0] = 120
b = heights[1] + gap + heights[2] + gap + heights[3] = 303
c = heights[4] + gap + heights[5] = 125 + 10 + 95 = 230
And so on. I want to find the combination that gives the most evenly-sized sets. So in this example the first combination is better since it gives an overall error of:
max - min = 230 - 208 = 22
Whereas the second combination gives an error of 183. I'm trying to do this in JavaScript, but I'm just looking for some sort of outline of an algorithm. Pseudo code or whatever would be great. Any help would be highly appreciated.
MY POOR ATTEMPTS: Obviously one way of solving this would be to just try every possible combination. That would be horrible though once heights gets large.
Another method I tried is to get the expected mean height of the sets, calculated as the sum of the values in height / n. Then I tried to fill each set individually by getting as close to this average as possible. It works alright in some cases, but it's too slow.
NOTE: If it helps, I would be happy to have symmetric sets. So for example, with sets (a, b c), a = b. Or with five sets (a, b, c, d, e), a = b and c = d. I think this would be even more difficult to implement but I could be wrong.
EDIT: For anyone who may be interested, the best I could come up with was the following algorithm:
Sort heights in descending order.
Create n sets.
Put the first n values from heights into the first slot of each set. i.e. put the n largest values at the start of each set. Remove the values from heights as they are added.
While heights.count > 0
Find the smallest totalHeight (including gap) in each of the n sets.
Add the next value in heights to this set (and remove the value from heights).
Then there's some little algorithm at the end where each set can make x number of swaps with the other sets, if the totalHeight gets closer to the average. I'm keeping x small because this process could go on forever.
It's not terrible, but obviously not perfect.
Seems like it is NP-complete and reducible to Subset sum problem or more precisely to Partition problem.
Your second approach - finding the mean (height / n), then attempting to fill sets with a mean as close as possible seems like a good practical approach. You say this is too slow... the following implementation is O(n*m log n) where m is the maximum number of elements allowed in a set. If m can be very large, then this could be quite slow, however if m is constrained to be within a certain range then it could approach O(n log n) which is about as fast as you are going to get.
Find the mean height of all values. h_mean = Sum(h) / n; O(n).
Sort all heights. O(n log n).
Examine the highest and lowest height.
Add the value which is furthest from the mean to a new set.
Remove this value from the sorted heights.
Repeat for max_number allowed in set = 1 .. m (m < n / 2)
{
Repeat:
{
If the set mean is higher than the mean.
Add the lowest value from the sorted heights.
Remove this value from the sorted heights.
If the set mean is lower than the mean
Add the highest value from the sorted heights.
Remove this value from the sorted heights.
Recalculate the set mean (taking account of the gap).
If the new set mean is further from the h_mean than the last OR
If the set has too many elements
break
}
Until all numbers are used.
Keep track of the standard deviations for this assignment.
If this assignment is the best so far, keep it.
}
This isn't going to give a provably optimal solution, but it's simple and that has a lot going for it...
Note, in this algorithm, all sets have the same number of elements, m. You repeat an iteration for different values of m, say, 2, 3, 4 (note that m should be a factor of N). Each set ends up with approximately m * mean_height for total height.
You may ask, well what if N is prime?
Then clearly, one set will value short on total value.
Does this mean this algorithm is useless?
Not at all. It's simple and it should produce a good first attempt at a solution. You may wish to use this algorithm first, then refine the first result using optimization techniques (such as selective swapping of heights being sets).
I'm looking for possibly efficient algorithm to detect "win" situation in a gomoku (five-in-a-row) game, played on a 19x19 board. Win situation happens when one of the players manages to get five and NO MORE than five "stones" in a row (horizontal, diagonal or vertical).
I have the following data easily accessible:
previous moves ("stones") of both players stored in a 2d array (can be also json notation object), with variables "B" and "W" to difference players from each other,
"coordinates" of the incoming move (move.x, move.y),
number of moves each player did
I'm doing it in javascript, but any solution that doesn't use low-level stuff like memory allocation nor higher-level (python) array operations would be good.
I've found similiar question ( Detect winning game in nought and crosses ), but solutions given there only refer to small boards (5x5 etc).
A simple to understand solution without excessive loops (only pseudocode provided, let me know if you need more explanation):
I assume your 2-d array runs like this:
board = [
[...],
[...],
[...],
...
];
I.e. the inner arrays represent the horizontal rows of the board.
I also assume that the array is populated by "b", "w", and "x", representing black pieces, white pieces, and empty squares, respectively.
My solution is somewhat divide-and-conquer, so I've divided it into the 3 cases below. Bear with me, it may seem more complex than simply running multiple nested loops at first, but the concept is easy to understand, read, and with the right approach, quite simple to code.
Horizontal lines
Let's first consider the case of detecting a win situation ONLY if the line is horizontal - this is the easiest. First, join a row into a single string, using something like board[0].join(""). Do this for each row. You end up with an array like this:
rows = [
"bxwwwbx...",
"xxxwbxx...",
"wwbbbbx...",
...
]
Now join THIS array, but inserting an "x" between elements to separate each row: rows.join("x").
Now you have one long string representing your board, and it's simply a matter of applying a regexp to find consecutive "w" or "b" of exactly 5 length: superString.test(/(b{5,5})|(w{5,5})/). If the test returns true you have a win situation. If not, let's move on to vertical lines.
Vertical lines
You want to reuse the above code, so create a function testRows for it. Testing for vertical lines is exactly the same process, but you want to transpose the board, so that rows become columns and columns become rows. Then you apply the same testRows function. Transposing can be done by copying values into a new 2-d array, or by writing a simple getCol function and using that within testRows.
Diagonal lines
Again, we want to reuse the `testRows' function. A diagonal such as this:
b x x x x
x b x x x
x x b x x
x x x b x
x x x x b
Can be converted to a vertical such as this:
b x x x x
b x x x
b x x
b x
b
By shifting row i by i positions. Now it's a matter of transposing and we are back at testing for horizontals. You'll need to do the same for diagonals that go the other way, but this time shift row i by length - 1 - i positions, or in your case, 18 - i positions.
Functional javascript
As a side note, my solution fits nicely with functional programming, which means that it can be quite easily coded if you have functional programming tools with you, though it's not necessary. I recommend using underscore.js as it's quite likely you'll need basic tools like map, reduce and filter in many different game algorithms. For example, my section on testing horizontal lines can be written in one line of javascript with the use of map:
_(board).map(function (row) {return row.join("")}).join("x").test(/(b{5,5})|(w{5,5})/);
Even though this is a really old question I want to provide my answer because I took a deeper look into this problem today and solved it in a much (much) more efficient way.
I'm using a bit board, which is used in most of the board games and engines (chess engines) due to the efficiency, to represent my field.
You can do everything you need in this game with bitwise operations.
A bit can just have 2 states (0 and 1) however what we need are 3 states e.g. p1, p2 or empty.
To solve this problem we're going to have 2 boards instead, one for each player.
Another problem is that Gomoku has a lot of fields (19x19) and there is no number type that has that many bits to represent the field.
We will use an array of numbers to represent each line and just use the first lsb 15bits of it.
Vertical rows
A simplified board of player 1 could look like this
000000
101100
001000
011000
000000
Lets say we want to detect 3 in a row. We take the first 3 rows(0-2) and took at them.
000000
001100
101000
With the & (AND) operator you can check if there is a 1 in every row.
var result = field[player][0] & field[player][1] & field[player][2];
In this case the result will be 0 which means no winner. Lets continue... The next step is to take rows 1-3
101100
001000
011000
Apply the AND again and that we will get is 001000. We don't have to care what number this is, just if it's 0 or not. (result != 0)
Horizontal rows
Ok now we can detect vertical rows. To detect the horizontal rows we need to save another 2 boards, again one for each player. But we need to invert x and y axis. Then we can do the same check again to detect horizontal lines. Your array would then be:
//[player][hORv][rows]
var field[2][2][19];
Diagonals :)
The trickiest part are the diagonals of course but with a simple trick you can do the same check as above. A simple board:
000000
010000
001000
000100
000000
Basically we do the same as above but before we do that we need to shift the rows. Lets say we're at row 1-3.
010000
001000
000100
The first row stays as it is. Then you shift the second row one to the left and the third 2 to the left.
var r0 = field[0][0][i];
var r1 = field[0][0][i+1] << 1;
var r2 = field[0][0][i+2] << 2;
What you will get is:
010000
010000
010000
Apply AND you can have your win detection. To get the other diagonal direction just do it again, but instead of shifting to the left <<, shift to the right >>
I hopes this helps someone.
untested:
int left = max(0, move.x-5), right = min(width-1, move.x+5), top = max(0, move.y-5), bottom = min(width-1, move.y+5);
// check the primary diagonal (top-left to bottom-right)
for (int x = left, y = top; x <= right && y <= bottom; x++, y++) {
for (int count = 0; x <= right && y <= bottom && stones[x][y] == lastPlayer; x++, y++, count++) {
if (count >= 5) return true;
}
}
// check the secondary diagonal (top-right to bottom-left)
// ...
// check the horizontal
// ...
// check the vertical
// ...
return false;
alternatively, if you don't like the nested loops (untested):
// check the primary diagonal (top-left to bottom-right)
int count = 0, maxCount = 0;
for (int x = left, y = top; x <= right && y <= bottom; x++, y++) {
if (count < 5) {
count = stones[x][y] == lastPlayer ? count + 1 : 0;
} else {
return true;
}
}