My script isn't working, can someone please! help me? Further explanation in thread.

조회 수: 2 (최근 30일)
The Legend
The Legend 2020년 1월 13일
편집: The Legend 2020년 1월 13일
function M = initializeBoard(N, p)
for i = 1 : N
for j = 1 : N
x = rand(1);
if (x < p)
M(i,j) = 1;
else
M(i,j) = 0;
end
end
end
end
Imagine that this functions creates a N x N matrix A (M).
This matrix consists of 1's and 0's, and the distribution of these depends on p, as can be seen:
if (x < p)
M(i,j) = 1;
else
M(i,j) = 0;
end
function pathFound = findPath(N, p)
A = initializeBoard(N, p);
scanPath = bwlabel(A,4);
[rowNum, ~] = size(scanPath);
stats = regionprops(scanPath, 'PixelList');
pathFound = false;
for iblob = 1:length(stats)
if ((min(stats(iblob).PixelList(:,2)) == 1) && (max(stats(iblob).PixelList(:,2)) == rowNum))
scanPath(scanPath == iblob) = 10*iblob;
pathFound = true;
else
pathFound = false;
end
end
end
This function should check if there's a path between bottom and top of the matrix, and if there is, return a 'true' value.
clear; close all; clc;
pSteps = 20;
M = 100;
N = 200;
probs = [];
ps = [];
for i = 1:pSteps
p = (1/pSteps) * i;
fprintf('p is at %.2f\n', p);
paths = 0;
for m = 1:M
if findPath(N, p) == 1
paths = paths + 1;
end
end
prob = paths / 50;
probs = [probs, prob];
ps = [ps, p];
end
plot(ps, probs)
The script above checks for 20 equally distributed values between 0 and 1 what the probability is of finding a path in a matrix with size N x N.
The output I get from N = 100, M = 100, pStep = 20 (0.05 - 1.00):
loool.png
The output I get from N = 200, M = 100, pStep = 20 (0.05 - 1.00):
aaaakkrzooi.png
However, the output I get from N = 200 and N = 100 is very different from the output I am supposed to get:
pcimg.png
My Question: Does anyone know where I made a mistake / why do my graphs not look like the one shown above at all?
  댓글 수: 2
The Legend
The Legend 2020년 1월 13일
편집: The Legend 2020년 1월 13일
Uhh, I don't think there are equations?
If p is very small, matrix A will contain only a few 1s. If p is close to 1 almost all elements will be 1. These two situations are the two phases.

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