Finding close numbers in columns of data - sorting

Question

I have 3 columns of data (hours of the day)
C1 C2 C3
01 05 00
05 09 06
11 11 10
16 17 14
20 22 18
I need to be able to separate this into an n by 3 matrix where the three numbers on each row are +/-2 hours away from each other. (The range of each row must be <=4)
Each value in each column can only be used once, so if there are more than one combination that uses the same number then one of the combinations is ignored.
So the final result would be:
05 05 06 (Taken from the 2nd in C1, 1st in C2 and 2nd in C3)
11 09 10 (Taken from the 3rd in C1, 2nd in C2 and 3rd in C3)
16 17 18 (Taken from the 4th in C1, 4th in C2 and 5th in C3)
The data in each column must remain in that same column in the final matrix, for example the 16 found in C1 needs to be in the first column of the final matrix.
I'm really struggling to find a way to put this into code, can you help?

Answer 1

I managed to get this to almost work in MATLAB, I dont think its particularly efficient or clever but it does the job.
It relies quite heavily however on the value in the first column, treating it as a midpoint.
This means that the allowed points in columns 2 and 3 must be either +/-2.
The situation where the value in C1 is a lower limit (eg. 30) and the other 2 values are up to 4 greater than the one found in C1 is classed as invalid, although it is still technically a solution.
x = [1,5,0; 5,9,6; 11,11,10; 16,17,14; 20,22,18];
d=size(x); %Get the size in of x in the form [row,col]
rows=d(1); %Number Of Rows
cols=d(2); %Number Of Columns
clear d;
y = nan(rows,cols); %nan Matrix the same size as x used for the output
a = ones(1,cols); %Keep track of the current index in question in each column
c = 0; %Number of "matches" or rows that are valid in the output matrix
time = zeros(1,cols); %Keep track of the current values in each column
while(max(a)<rows+1) %For every row check that no index is invalid
time(1)=x(a(1),1); %Get the value in column1
b = 2; %column counter
skip=0; %Increment the counter for column 1 if this is true
while(b<cols+1&&~skip&&max(a)<rows+1) %For columns 2->cols, if we don't need to skip the value in column 1 and all the indexes are valid.
time(b)=x(a(b),b); %Get the value in column b at row a(b)
delta = time(b)-time(1); %work out the difference in value from the first column value that is selected
if(delta>2)
%Skip first column by 1
a(1)=a(1)+1; %Increment the counter for column 1
skip=1; %Return back to the first while loop
elseif(delta<-2)
%Skip b'th column by 1
a(b)=a(b)+1; %Increment the counter for column b
else
%Its valid
if(b==cols) %If at the last column and its valid
c=c+1; %Increment the match counter
y(c,:)=time(1:cols); %Set the c'th row of the output to what we've found
a=a+1; %Move onto next number in column 1
skip=1; %Start all over
else %Not at last column yet
b=b+1;
end
end
end
end
Final Result:
05 05 06
11 09 10
16 17 14
20 22 18
nan nan nan

Answer 2

It seems although you are talking of having three columns of values, their row and column has no meaning. Effectively you have just a list of numbers: 01, 05, 00, 05, 09, 06, 11, 11, 10, 16, 17, 14, 20, 22, 18.
Then you order them: 00, 01, 05, 05, 06, 09, 10, 11, 11, 14, 16, 17, 18, 20, 22
Then you take three and look at their distances: 00, 01, 05 = bad, for 01 and 05 are too far apart.
Next number. 01, 05, 05? No. Next number. 05, 05, 06? Yes. Continue after them. 09, 10, 11? Yes. Continue after them. 11, 14, 16? No. Next number. 14, 16, 17? Yes. You found a solution:
05, 05, 06
09, 10, 11
14, 16, 17