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

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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!

Accepted Answer

Torsten on 9 Jun 2022
Edited: Torsten on 9 Jun 2022
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 = [];
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]];
Elapsed time is 0.003877 seconds.
b = 2×5
0 1 2 3 4 4 3 2 1 0
Torsten on 10 Jun 2022
Edited: Torsten on 10 Jun 2022
A better solution is the following:
m = min(a)
M = max(a)
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).

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More Answers (3)

KSSV on 9 Jun 2022
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:
Aly Osman
Aly Osman on 9 Jun 2022
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?

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Steven Lord
Steven Lord on 9 Jun 2022
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?

DGM on 10 Jun 2022
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.
DGM on 10 Jun 2022
Ah. It's good to explore. I'm just too eager to critique what I think are unhelpful homework tropes, that's all.

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