Puzzler: Quickly tell if two absolute indices (a,b) are four connected for n x m matrix.
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function flag = isFourConnected(a,b,n,m)
%
% a,b: indices of interest a ~= b
% n,m: size of matrix of interest
%
% flag: True if indices a and b are four connected
% in a matrix of size n x m
%
%
% Your code here
Note, this code should use no toolboxes, and should be reasonably quick as this function will be called many times. Reasonably quick is up to debate as the rest of the code forms.
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채택된 답변
David Young
2011년 9월 1일
function flag = isFourConnected(a,b,n,m)
%
% a,b: indices of interest a ~= b
% n,m: size of matrix of interest
%
% flag: True if indices a and b are four connected
% in a matrix of size n x m
%
d = abs(a-b);
flag = d == n || (d == 1 && mod(min(a,b), n));
end
댓글 수: 3
추가 답변 (5개)
Fangjun Jiang
2011년 9월 1일
Circle-shifting neighbors are considered connected.
function flag = isFourConnected(a,b,n,m)
%
% a,b: indices of interest a ~= b
% n,m: size of matrix of interest
%
% flag: True if indices a and b are four connected
% in a matrix of size n x m
%
%
% Your code here
[x,y]=ind2sub([n,m],[a;b]);
xdiff=abs(x(1)-x(2));
ydiff=abs(y(1)-y(2));
flag = ((xdiff==0) && (ydiff==1) || (ydiff==m-1)) || ...
((ydiff==0) && (xdiff==1) || (xdiff==n-1));
A little test script. All other entries so far didn't pass this test.
clc;
TestVector={6,7,4,5
6,10,4,5
1,4,4,5
1,17,4,5};
for k=1:size(TestVector,1)
if isFourConnected(TestVector{k,:})~=true
disp(k);beep;
end
end
Walter Roberson
2011년 9월 1일
function flag = isFourConnected(a,b,n,m)
%
% a,b: indices of interest a ~= b
% n,m: size of matrix of interest
%
% flag: True if indices a and b are four connected
% in a matrix of size n x m
%
%
flag = abs(a-b)==n || (floor(a/n)==floor(b/n) && abs(a-b)==1);
댓글 수: 3
Oleg Komarov
2011년 9월 1일
I assume a,b,m,n always numeric and integer values > 1
function flag = isFourConnected(a,b,n,m)
% a,b : indices of interest a ~= b
% m,n : size of matrix of interest
% flag: True if indices a and b are four connected
% in a matrix of size n x m
d = a-b; flag = d == n || d == -n || (d == 1 && mod(a,n) ~= 1) || (d == -1 && mod(b,n) ~= 1);
댓글 수: 4
Oleg Komarov
2011년 9월 1일
Can't find any other valid solution to ensure bottom vs top not 4 conn except the ones already proposed.
Bruno Luong
2011년 9월 1일
function flag = isFourConnected(a,b,n,m)
% 10 arithmetic operations by pair
c = max(a,b);
d = min(a,b);
e = c - d;
flag = (e==1 & mod(d,n)) | (e==n & c>n);
댓글 수: 2
Walter Roberson
2011년 9월 1일
This might or might not be slightly faster:
c = sort([a,b]);
e = c(2)-c(1);
flag = (e==1 & mod(c(1),n)) | (e==m & c(2)>n);
Or if you prefer your original structure, then instead of max/min, you could use
c = max(a,b);
d = a + b - c;
Bruno Luong
2011년 9월 1일
I believe I had one redundant test in the earlier code:
function flag = isFourConnected(a,b,n,m)
% 8 arithmetic operations by pair
c = max(a,b);
d = min(a,b);
e = c - d;
flag = (e==1 & mod(d,n)) | (e==n);
Daniel Shub
2011년 9월 2일
I am not sure what to do about circle-shifting neighbors so I have two answers.
function flag = isFourConnected(a,b,n,m)
%
% a,b: indices of interest a ~= b
% n,m: size of matrix of interest
%
% flag: True if indices a and b are four connected
% in a matrix of size n x m
%
%
% Your code here
% Using ind2sub might be faster.
col = mod([a(:), b(:)]-1, n)+1;
row = ceil([a(:), b(:)]/n);
%[col, row] = ind2sub([n, m], [a(:), b(:)]);
flag = reshape(mod(abs(diff(col, 1, 2)), n-2)+mod(abs(diff(row, 1, 2)), m-2) == 1, size(a));
% if circle shifted points are not connected:
% flag = reshape(abs(diff(col, 1, 2))+abs(diff(row, 1, 2)) == 1, size(a));
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