Re: Balancing arrays matlab command

ellieandrogerxyzzy@xxxxxxxxxxxxxxxxxxxxxx (Roger Stafford) wrote in message <ellieandrogerxyzzy-0108071751340001@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>...
In article <f8r26s$drv$1@xxxxxxxxxxxxxxxxxx>, "Mark "
<medwar19.nospam@xxxxxxxxxxx> wrote:

Hi Roger,

Thanks so much for your help, it was 'repmat' that I needed, not
'meshgrid' as I was experimenting with.

So, I added your lines to my eggs problem to see what happens. After the
boxes are full a single swap usually get sufficent balance. I added a loop
to swap a second time if needed and that was always successful.
Interestingly I need to swap at least once around 50% of the time. I put
that down to the normal distribution of the rand command.

Thanks again for your help, I very much appreciate it.
Here's the solution in a test script:

% --- Balancing two egg boxes ---
MIN_MASS = 10;
MAX_MASS = 25;
EGGBOX_CAPACITY = 20;
MAX_MASS_TOLERANCE = 20;

TotalTests = 100;

for i = 1:TotalTests
% Start with random assortment of eggs between 10-25g
Eggs = rand(1,EGGBOX_CAPACITY * 2) * (MAX_MASS - MIN_MASS) + MIN_MASS;
% Put half the eggs in each box
Box1 = Eggs(1:EGGBOX_CAPACITY);
Box2 = Eggs(1 + EGGBOX_CAPACITY:EGGBOX_CAPACITY + EGGBOX_CAPACITY);

% Set up some vars
ix = 0;
Swaps = 0;
% Create a copy of the original data for comparison later
Box3 = Box1;
Box4 = Box2;
% Swap eggs until the egg boxes are within MAX_MASS_TOLERANCE of each
% other (limit number of swaps to 5 in case we fail)
while ix == 0 && Swaps < 5;
% Let w = initial difference between cassette masses
w = sum(Box3) - sum(Box4);
if abs(sum(Box3) - sum(Box4)) < MAX_MASS_TOLERANCE
% No swap necessary
ix = 0;
iy = 0;
break;
end
d = repmat(Box3,size(Box4,2),1)-repmat(Box4',1,size(Box3,2)); %
All x-y differences
[c,lx] = min(abs(d(:)-w/2)); % Get the closest match with w/2
[iy,ix] = ind2sub(size(d),lx); % Get the corresponding indices

if c > MAX_MASS_TOLERANCE
% Unable to get within tolerance with a swap
ix = -1;
iy = -1;
break;
end

if ix > 0
% perform swap in the boxes
t = Box3(ix);
Box3(ix) = Box4(iy);
Box4(iy) = t;
Swaps = Swaps + 1;
else
% no swop necessary
break;
end
if abs(sum(Box3) - sum(Box4)) > MAX_MASS_TOLERANCE
% Still not in tolerance after swap so force another go
ix = 0;
end
end
% save the egg data, original sums, sums after swap(s), and number of
% swaps needed
Dat(i,:) = [Eggs, Box1, Box2, sum(Box1) - sum(Box2), ix, iy,
sum(Box3) - sum(Box4), Swaps];
end

% Percentage of times that swaps were needed
PercentSwapped = (sum(Dat(:,end)) / TotalTests) * 100
------------------
It occurred to me that, in case you run into trouble making one or two
swaps alone do the trick, you could make the first step be a sorting of x
and y so that, if necessary, you could do the following. Swap the maximum
x's with the minimum y's, or visa versa, with a sufficient number of egg
pairs until just before that reverses the sign of the discrepancy. Then
with the remaining eggs you could do as before with one, possibly two,
more swaps which take into account all possible pairings. This way you
first remove any relatively large offset so that the final swaps have a
much larger probability of succeeding since they would be "centered"
better.

Also instead of just repeating a swap to get a second swap in this
second phase, you could compute all possible pairs of x's along with all
possible pairs of y's. If, for example, you have 20 eggs in each box,
that gets 190 x-pairs to be compared with 190 y-pairs in a 190 x 190
matrix. That's instead of a 20 x 20 matrix of single x-eggs compared with
single y-eggs. Then do as before with this larger matrix. It's not
really very much harder to do than for a single swap, though it involves a
larger matrix. With so many possible combinations (36100 in this case)
the likelihood of satisfying your tolerance should be quite high.

Of course all this software messing around with the egg locations
requires some careful indexing so that the programs manages to keep track
of the original location of all eggs to be swapped. This allows a final
swapping to be carried out that does not reflect all the computational
maneuvers that were required to arrive at it.

Roger Stafford

Indeed, as I know the mass of each egg after it's been put in a box I tried centering the mass of the eggs during loading by always adding the next egg to the lighter box. This usually meant that the boxes stayed within a tolerance of each other during loading, sometimes resulting in a space or two in one box when the other was full. I ran this load-sorting test along with a alternating-load strategy and found no difference in the number of swaps required (~50%).

My next challange is to modify the code for partially loaded boxes. I can see a case when the number of eggs is small where moving a single egg from one box to another will balance the boxes but a swop will not.

Thanks again for your help with my eggs.

Mark.
.

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