How can I find all possible pairs within a range that result in the same average?

I'm trying to figure out how to output multiple pairs of values that happen to have the same average (added together and divided by 2) within a range. For example, if I set an array to be:

array= [0:1:100]

and I want all possible pairs that output an average of 2.

avg= 2

So for example, 2 pairs would be (0,2) and (2,2) and the avg. of both of these seprately is 2.

I'm sure I'll probably need a for loop, but I've read here on several function such as permutations or combinations, but I feel like this should be more simple. Obviously I'll have a lot more conditions involved, but the basic premise of getting this to work is somehow eluding me, so I'd appreciate a little nudge in the right direction here on how to go about this. Thank you!

[KSSV]

Average/ mean of (0,2) is 1.

Answer 1

[Torsten]

Do you know if there's any way that this can be done by just using for loops and nested for loops with if statements to get the same result?

Sure. A very slow version would be

mean_compare = 2.0;
a = 0:100;
n = numel(a);
ifound = 0;
b = [];
tic
for i = 1:n
    value1 = a(i);
    for j = 1:n
        value2 = a(j);
        if value1 + value2 == 2*mean_compare
            ifound = ifound + 1;
            b = [b,[value1;value2]];
        end
    end
end
toc

Elapsed time is 0.003877 seconds.

b

b = 2×5

0 1 2 3 4 4 3 2 1 0

[Aly Osman]

This is more what I was looking for. Thank you!

[Torsten]

A better solution is the following:

Let

m = min(a)

and

M = max(a)

Let

feasible_range = min(mean_compare - m, M - mean_compare)

Consider the points

union( (m,m) , (m-(1:feasible_range).',m+(1:feasible_range).') , (m+(1:feasible_range).',m-(1:feasible_range).') )

The intersection with

a x a

is the solution (at least if "a" is an array of integers).

Answer 2

[KSSV]

a = 0:10 ; 
b = permn(a,2) ; 
idx = mean(b,2)==2 ; 
iwant = b(idx,:)

You can download the function permn from the file exchange: https://in.mathworks.com/matlabcentral/fileexchange/7147-permn

[Aly Osman]

Do you know if there's any way that this can be done by just using for loops and nested for loops with if statements to get the same result?

[KSSV]

The function permn should have loops.... Check the function...

Answer 3

[Steven Lord]

As long as your arrays aren't that large just brute force it:

array= 0:1:100;
desiredAverage = 2;
ind = find(array+array.' == 2*desiredAverage)

ind = 5×1

5 105 205 305 405

[r, c] = ind2sub(numel(array)*[1 1], ind)

r = 5×1

5 4 3 2 1

c = 5×1

1 2 3 4 5

B = array([r, c]) % 5 pairs of numbers each of whose average is 2

B = 5×2

4 0 3 1 2 2 1 3 0 4

mean(B, 2) == desiredAverage % all true

ans = 5×1 logical array
   1
   1
   1
   1
   1

If you had to do this with a loop (because that's one of the requirements of your homework assignment, for instance) you only need one.

Hint: I'm thinking of two numbers whose average is 10. One of the numbers is 5. What's the other? How did you determine the answer?

What if instead I told you one of the numbers was 17. What's the other?

Answer 4

[DGM]

I know Steven hinted at it, but I'm just going to give an example. Since the assignment forces you to accept bad decisions (and you've already accepted them), I might as well demonstrate that the consequences aren't trivial.

candidatevalues = 0:1:100; % the set
desiredAverage = 2; % the target

companionvalues = 2*desiredAverage - candidatevalues;
isvalidpair = ismember(companionvalues,candidatevalues);

validpairs = [candidatevalues(isvalidpair); companionvalues(isvalidpair)].' % these are the pairs

validpairs = 5×2

0 4 1 3 2 2 3 1 4 0

mean(validpairs,2) % their average is as expected

ans = 5×1

2 2 2 2 2

Despite the use of ismember(), this naive algebra approach can be much faster than any of the examples given to meet the loop-based requirements. For a set size of 10E3 and a target mean of 25, on my hardware in R2019b, the 3-line example above is between 1400 to 35000 times as fast as the other examples. For a set size of 100, the smallest speed advantage is a factor of 5. Is my example ideal? I doubt it, but it's a remarkable amount better than what's been deemed acceptable.

I'm no compsci professor, but I'm sure there's a lesson to be found here regarding computational complexity.

[Aly Osman]

I see what you and Steven are saying and definitely the for loop method is slower. This is not for an assignment, it's for a personal project that I'm messing with. My project doesn't actually require the average or mean of any set of points, however the premise is similar to that in trying to output a pair of values that will provide the same output given a specific computation that they are involved in. So I am more looking for something where it could be x+y=10 and I want all possible pairs of x and y that sum to 10 within a range from say 0 to 10 so that could be x=1 y=9, x=2 y=8, etc. and with my knowledge of MATLAB, I wanted to do this with something I'm familiar with/more brute force (for loops) at first just to make sure the results and process is reasonable, than explore and implement some faster methods like you and Steven have included here which are much appreciated.

[DGM]

Ah. It's good to explore. I'm just too eager to critique what I think are unhelpful homework tropes, that's all.

[Aly Osman]

No problem at all! Thanks for looking out!