# How to determine which grid cells a line segment passes through?

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Darcy 2015-7-16

Hi, I apologize if this question has been asked before but I have looked and cannot find the answer.
Similar to my previous question, lets say you have a grid with cells in the x-direction from [1:5] and in the y-direction from [1:5] (5 x 5 grid). Now lets say you have two points on the grid: (0.35, 2.65) and (4.2,4.73).
Is there a simple way to determine which grid cells the line connecting those two points passes through (and the distances of each line segment)?
It is very easily visually to see which cells the line passes through (see figure below). Clearly the line segment passes through cells B1, C1, C2, D2, D3, E3 and E4.
Determining the distances of each line segment through each cell would be more difficult by hand.
I have currently solved the problem by finding the slope and intercept of the line, then determining the intercepts of the line with the grid cell lines. Using this, I can determine the distance of each line segment. Then, to find which cell the line segment falls in, I have to find the centre point of each grid cell and the midpoint of each line segment and do a for loop to determine the minimum distance between each line segment and each cell centre.
Is there a more elegant way to solve this problem? Obviously when dealing with multiple lines and/or a large grid, this will become much more time intensive. ##### 2 个评论显示隐藏 1更早的评论
Walter Roberson 2015-7-17
This question deals with line segments, not individual points.

### 回答（4 个）

Star Strider 2015-7-17
This brings back memories of How do I store the Coordinates of Lines Intersecting a Grid? I refer you to it for the description and documentation.
I later updated the code for it but didn’t post it. The updated code is:
x = 0:5; % X-range
y = 0:25; % Y-range
lxmb = @(x,mb) mb(1).*x + mb(2); % Line equation: y = m*x+b
m = -5; % Slope (or slope array)
b = 15; % Intercept (or intercept array)
mb = [m b]; % Matrix of [slope intercept] values
L1 = lxmb(x,mb); % Calculate Line #1 = y(x,m,b)
hix = @(y,mb) [(y-mb(2))./mb(1); y]; % Calculate horizontal intercepts
vix = @(x,mb) [x; lxmb(x,mb)]; % Calculate vertical intercepts
hrz = hix(x(2:end),mb)'; % [X Y] Matrix of horizontal intercepts
vrt = vix(y(1:6),mb)'; % [X Y] Matrix of vertical intercepts
hvix = [hrz; vrt]; % Concatanated ‘hrz’ and ‘vrt’ arrays
exbd = find( (hvix(:,2) < 0) | (hvix(:,2) > 25) );
hvix(exbd,:) = [];
srtd = unique(hvix,'rows'); % Remove repeats and sort ascending by ‘x’
exL1 = find((L1 < 0) | (L1 > 25)); % Find ‘y’ values for ‘L1’ off grid
xp = x; % Create plotting x-vector for L1
xp(exL1) = []; % Eliminate out-of-bounds ‘y’ values from ‘x’
L1(exL1) = []; % Eliminate out-of-bounds ‘y’ values from ‘Li’
figure(1) % Draw grids & plot lines
plot(repmat(x,2,length(x)), [0 length(y)-1]) % Vertical gridlines
hold on
plot([0 length(x)-1], repmat(y,2,length(y))) % Horizontal gridlines
plot(xp, L1) % Plot more lines here (additional ‘plot’ statements)
hold off
axis equal
It’s been over a year since I wrote it so I would have to study it to remember what I did, but I did my best at the time to document it exhaustively with comments, so that should help. The output (grid intersections) is in the srtd array.
##### 1 个评论显示隐藏 无
Merisa Saber 2018-12-26
Do you have any idea how we can change it to 3 dimensional grid? I mean to determine which 3 dimensional cell a line in a 3D space goes through.

Walter Roberson 2015-7-17
##### 0 个评论显示隐藏 -1更早的评论

William Chamberlain 2018-10-17
Bresenham's line algorithm implementations:

Thank you guys.

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