I would like to find out the number of different combinations of non-negative numbers(can be any number, it is not fixed) such that its total equals to the sum that is provided.
for example : I have 3 numbers and i want to find the different combinations of numbers such that the sum is 4. the value of num starts from 0. no negative numbers.
For 3 numbers that sum to 4, the combinations are
2 0 2
2 2 0
0 2 2
0 1 3
3 1 0
0 3 1
1 0 3
1 3 0
3 0 1
0 0 4
4 0 0
0 4 0
2 1 1
1 2 1
1 1 2
I saw this as an example : Finding the total number combinations for an integer using three numbers
But the problem is it only uses three numbers.
Any algorithm or code will be useful. Thanks.
You can view this as the number of ways to put s indistinguishable coins in n distinguishable jars. (In the example, s=4 and n=3).
As explained here, that is C(n+s-1,s-1), which gives 15 in the example.
If order does not matter and 0 counts too, like in example link, then
n=total+1
k=number-1
binomial(k+n-1,k) #combinations whith reptetitions
or
binomial(number+total-1,number-1)
If you represent number 5 as
1+1+1+1+1
and have to find number of sums sums whith 3 integers
You can see that you have to do 2 slices out of 6 calculating combinations whit repetitions.
Related
There is an existing array of size 64 that has values 6 values distributed as 0 1 2 3 4 5 0 1 2 3 4 5 0 1 2 3 4 5 ...
Please see the image for complete data.
The number of occurrence of 0 in the array is 11 times (at every 6th index), 1 is 11 times ... where as 4 and 5 occurs 10 times each.
There is a necessity to reduce the occurrence of any of these numbers [0 to 5] to a lesser number that could be any number from 0 to 10.
For example, it could be to reduce occurrence of 0 to 6 and 1 to 9.
I am looking for a solid idea to do this. Certainly all the numbers are to be evenly distributed and not something like 0 0 0 0 0 2 2 2 2 2 2 ...
I tried to find the index/position where the reduced value has to filled (64/occurrence of 0 or 2). But at times the index collide with each other and thus is not robust one.
From the example I quoted above, number of occurrence of 0 must be changed to 6 and occurrence of 1 to 9, the result after my algorithm is below -
New location to fill 0 = (Array size)/(new occurrence of 0) = 64/6 = ~10th index
New location to fill 1 = (Array size)/(new occurrence of 1) = 64/9 = ~7 index
For filling 6 0's and 9 1's, first the array is reset after which each of the values are filled to maintain balanced distribution.
After filling 6 0's, the array would be come like this:
Then, after filling 9 1's, the array would be come like this:
The index at 55 already has value 0 and apparently 8th 1 also index to 55 that creates a collision. So I believe, this algorithm to balance the distribution does not work.
How do I populate 6 's, 9 1's and rest of the numbers {2, 3, 4, 5} in the array in a balanced way?
I have an array that looks something like...
1 0 0 1 2 2 1 1 2 1 0
2 1 0 0 0 1 1 0 0 2 1
1 2 2 1 1 1 2 0 0 1 0
0 0 0 1 2 1 1 2 0 1 2
however my real array is (50x50).
I am relatively new to MATLAB and need to be able to count the amount of unique values in each row and column, for example there is four '1's in row-2 and three '0's in column-3. I need to be able to do this with my real array.
It would help even more if these quantities of unique values were in arrays of their own also.
PLEASE use simple language, or else i will get lost, for example if representing an array, don't call it x, but perhaps column_occurances_array... for me please :)
What I would do is iterate over each row of your matrix and calculate a histogram of occurrences for each row. Use histc to calculate the occurrences of each row. The thing that is nice about histc is that you are able to specify where the bins are to start accumulating. These correspond to the unique entries for each row of your matrix. As such, use unique to compute these unique entries.
Now, I would use arrayfun to iterate over all of your rows in your matrix, and this will produce a cell array. Each element in this cell array will give you the counts for each unique value for each row. Therefore, assuming your matrix of values is stored in A, you would simply do:
vals = arrayfun(#(x) [unique(A(x,:)); histc(A(x,:), unique(A(x,:)))], 1:size(A,1), 'uni', 0);
Now, if we want to display all of our counts, use celldisp. Using your example, and with the above code combined with celldisp, this is what I get:
vals{1} =
0 1 2
3 5 3
vals{2} =
0 1 2
5 4 2
vals{3} =
0 1 2
3 5 3
vals{4} =
0 1 2
4 4 3
What the above display is saying is that for the first row, you have 3 zeros, 5 ones and 3 twos. The second row has 5 zeros, 4 ones and 2 twos and so on. These are just for the rows. If you want to do these for columns, you have to modify your code slightly to operate along columns:
vals = arrayfun(#(x) [unique(A(:,x)) histc(A(:,x), unique(A(:,x)))].', 1:size(A,2), 'uni', 0);
By using celldisp, this is what we get:
vals{1} =
0 1 2
1 2 1
vals{2} =
0 1 2
2 1 1
vals{3} =
0 2
3 1
vals{4} =
0 1
1 3
vals{5} =
0 1 2
1 1 2
vals{6} =
1 2
3 1
vals{7} =
1 2
3 1
vals{8} =
0 1 2
2 1 1
vals{9} =
0 2
3 1
vals{10} =
1 2
3 1
vals{11} =
0 1 2
2 1 1
This means that in the first column, we see 1 zero, 2 ones and 1 two, etc. etc.
I absolutely agree with rayryeng! However, here is some code which might be easier to understand for you as a beginner. It is without cell arrays or arrayfuns and quite self-explanatory:
%% initialize your array randomly for demonstration:
numRows = 50;
numCols = 50;
yourArray = round(10*rand(numRows,numCols));
%% do some stuff of what you are asking for
% find all occuring numbers in yourArray
occVals = unique(yourArray(:));
% now you could sort them just for convinience
occVals = sort(occVals);
% now we could create a matrix occMat_row of dimension |occVals| x numRows
% where occMat_row(i,j) represents how often the ith value occurs in the
% jth row, analoguesly occMat_col:
occMat_row = zeros(length(occVals),numRows);
occMat_col = zeros(length(occVals),numCols);
for k = 1:length(occVals)
occMat_row(k,:) = sum(yourArray == occVals(k),2)';
occMat_col(k,:) = sum(yourArray == occVals(k),1);
end
I have a table of rows which consist of zeros and numbers like this:
A B C D E F G H I J K L M N
0 0 0 4 3 1 0 1 0 2 0 0 0 0
0 1 0 1 4 0 0 0 0 0 1 0 0 0
9 5 7 9 10 7 2 3 6 4 4 0 1 0
I want to calculate an average of the numbers including zeros, but starting from the first nonzero value and put it into column after tables end. E.g. for the first row first value is 4, so average - 11/11; for the second - 7/13; the last one is 67/14.
How could I using excel formulas do this? Probably OFFSET with nested IF?
This still needs to be entered as an array formula (ctrl-shift-enter) but it isn't volatile:
=AVERAGE(INDEX(($A2:$O2),MATCH(TRUE,$A2:$O2<>0,0)):$O2)
or, depending on location:
=AVERAGE(INDEX(($A2:$O2);MATCH(TRUE;$A2:$O2<>0;0)):$O2)
The sum is the same no matter how many 0's you include, so all you need to worry about is what to divide it by, which you could determine using nested IFs, or take a cue from this: https://superuser.com/questions/671435/excel-formula-to-get-first-non-zero-value-in-row-and-return-column-header
Thank you, Scott Hunter, for good reference.
I solved the problem using a huge formula, and I think it's a bit awkward.
Here it is:
=AVERAGE(INDIRECT(CELL("address";INDEX(A2:O2;MATCH(TRUE;INDEX(A2:O2<>0;;);0)));TRUE):O2)
Imagine the following binary image exemplified by the matrix below. This is a simplified version of the images I'll be working with:
0 1 0 1
0 1 1 1
0 0 0 1
0 1 1 1
I want to construct a graph that will represent the randomness of each column. My thought is to develop a random index = the total transitions between each value in the column / by the total possible transitions. In the matrix above, each column could have a total possible of 3 transitions.
For the example above:
Column 1 would have a random index of 0% (0/3)
Column 2 would have a random index of 66.7% (2/3)
Column 3 = 100% (3/3)
Column 4 = 0% (0/3) even though they are 1's and not 0's. Doesn't matter, I just want the transitions.
Can I draw a boundary around all the 1 values and then have MATLAB sum all of the boundaries?
To calculate what you are suggesting you can just do:
sum( diff(A) ~= 0 )
The diff(A) will take the forward difference down the columns and the sum will count the number of non-zero changes. So if you do this you will get:
ans =
0 2 3 0
Let your image be defined as
im = [ 0 1 0 1
0 1 1 1
0 0 0 1
0 1 1 1 ];
The random index you want can be computed as
result = sum(diff(im)~=0) / (size(im,1)-1);
Explanation: diff computes the difference between consecutive elemtents down each column. The result is compared against zero (~=0), and all nonzero values within each row are added (with sum). Finally, the result is divided by the maximum number os transitions, which is the number of rows minus 1 (size(im,1)-1)
Equivalently, you could use xor between consecutive rows:
result = sum(xor(im(1:end-1,:), im(2:end,:))) / (size(im,1)-1)
actual problem is like this which I got from an Online competition. I solved it but my solution, which is in C, couldn't produce answer in time for large numbers. I need to solve it in C.
Given below is a word from the English dictionary arranged as a matrix:
MATHE
ATHEM
THEMA
HEMAT
EMATI
MATIC
ATICS
Tracing the matrix is starting from the top left position and at each step move either RIGHT or DOWN, to reach the bottom right of the matrix. It is assured that any such tracing generates the same word. How many such tracings can be possible for a given word of length m+n-1 written as a matrix of size m * n?
1 ≤ m,n ≤ 10^6
I have to print the number of ways S the word can be traced as explained in the problem statement. If the number is larger than 10^9+7, I have to print S mod (10^9 + 7).
In the testcases, m and n can be very large.
Imagine traversing the matrix, whatever path you choose you need to take exatcly n+m-2 steps to make the word, among of which n-1 are down and m-1 are to the right, their order may change but the numbers n-1 and m-1 remain same. So the problem got reduced to only select n-1 positions out of n+m-2, so the answer is
C(n+m-2,n-1)=C(n+m-2,m-1)
How to calculate C(n,r) for this problem:
You must be knowing how to multiply two numbers in modular arithmetics, i.e.
(a*b)%mod=(a%mod*b%mod)%mod,
now to calculate C(n,r) you also need to divide, but division in modular arithmetic can be performed by using modular multiplicative inverse of the number i.e.
((a)*(a^-1))%mod=1
Ofcourse a^-1 in modular arithmetic need not equal to 1/a, and can be computed using Extended Euclidean Algorithm, as in your case mod is a prime number therefore
(a^(-1))=a^(mod-2)%mod
a^(mod-2) can be computed efficiently using repetitive squaring method.
I would suggest a dynamic programming approach for this problem since calculation of factorials of large numbers shall involve a lot of time, especially since you have multiple queries.
Starting from a small matrix (say 2x1), keep finding solutions for bigger matrices. Note that this solution works since in finding the solution for bigger matrix, you can use the value calculated for smaller matrices and speed up your calculation.
The complexity of the above soltion IMO is polynomial in M and N for an MxN matrix.
Use Laplace's triangle, incorrectly named also "binomial"
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 1 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 1 1 0 0
1 2 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 1 1 1 0
1 2 3 0 0
1 3 0 0 0
1 0 0 0 0
0 0 0 0 0
1 1 1 1 1
1 2 3 4 0
1 3 6 0 0
1 4 0 0 0
1 0 0 0 0
1 1 1 1 1
1 2 3 4 5
1 3 6 10 0
1 4 10 0 0
1 5 0 0 0
1 1 1 1 1
1 2 3 4 5
1 3 6 10 15
1 4 10 20 0
1 5 15 0 0
1 1 1 1 1
1 2 3 4 5
1 3 6 10 15
1 4 10 20 35
1 5 15 35 0
1 1 1 1 1
1 2 3 4 5
1 3 6 10 15
1 4 10 20 35
1 5 15 35 70
Got it? Notice, that elements could be counted as binomial members. The diag members are here: C^1_2, C^2_4,C^3_6,C^4_8, and so on. Choose which you need.