So I think this may be a hard one, may be it was only hard for me! So here is what it is, I will give a matrix, with some number of rows by 5 columns, to make it simpler I will not repeat any number in a single row, so all numbers in a single rows are unique. So you need to find up to 2, 3 or 4 numbers that are found in a single row that repeat in other rows and the amount of times they repeat in a given matrix. To make it a little better, you only need to display the rows/numbers that repeat at lease twice or more. Examples are the best way to show what to do:
%p is the given matrix p = [1 2 3 4 5; 2 3 4 5 1; 3 4 6 7 8; 1 2 4 3 5; 4 6 3 2 1]; %v is the amount of elements per row that are being checked to see if they repeat in another row v = 4; %Output: out =
1 2 3 4 4 1 2 3 5 3 1 2 4 5 3 1 3 4 5 3 2 3 4 5 3 %So the first to the fourth column are the four number that were found to repeat in rows of matrix p (any and all combination that can appear in the rows of the matrix p). The fifth column is the amount of times the that particular set of matrix out(?,1:4) appear in matrix p. for for row one in matrix out, 1,2,3 and 4 in any combination appear 4 time in the rows of matrix p.
So more examples...
p = [1 2 3 4 9; 2 3 1 4 8] v = 4;
out = 1 2 3 4 2 %so any combination of 1,2,3 and 4 appear 2 times in matrix p
and another:
p = [1 2 3 4 9; 2 3 1 4 8; 3 4 2 7 1] v = 3;
out = 1 2 3 3 1 2 4 3 1 3 4 3 2 3 4 3 %so this uses any combo of 3 values, so in this case out(?,1:3) are the number combos that can be found in matrix p, and the out(?,4) is the amount of times that those combos are found
last example:
p = [1 2 3 4 9; 2 3 1 4 8] v = 2;
out = 1 2 2 1 3 2 1 4 2 2 3 2 2 4 2 3 4 2 %so in this case only combos of out(?,1:2) are found and out(?,3) is the amounts of times that combo of out(?,1:2) is found in matrix p. so row four of matrix out (so out(4,:)...) shows that the combo of 2 and 3 are found 2 times in matrix p.
NOTE: only display any row number combos that occur at least 2 times or more!
I hope I gave enough instructions and explanation, if not please let me know and I will add to it! Thanks and good luck!
Return the 3n+1 sequence for n
6870 Solvers
Find the sum of the elements in the "second" diagonal
917 Solvers
256 Solvers
652 Solvers
3376 Solvers
Solution 288389
That's clever!