# Numerical packing

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Hi all, here's a quick one. I've got an array and a limit:

A = [235 235 235 234 235 235 234 234 234 220 280 215 234]

lim = 1000

I want to group the elements of A so that:

1. The sum of each group's elements is below my lim

2. The group numbers of consecutive elements of A never decrease

3. A minimum number of groups is used. Ie, if a valid solution exists using 3 groups, all solutions using 4 or more groups are inferior.

4. The maximum of the sums within each group is minimized (clarified below)

As an example:

grps = zeros(size(A));

while any(grps==0)

openIdxs = find(grps==0);

mask = cumsum(A(openIdxs))<lim;

grps(openIdxs(mask)) = max(grps)+1;

end

The above returns:

grps =

1 1 1 1 2 2 2 2 3 3 3 3 4

This satisfies (1), (2) and (3), but note that the 4th group has only one element, and the sums of each group are:

grpSums = arrayfun(@(i)sum(A(grps==i)),1:max(grps))

grpSums =

939 938 949 234

Can anyone think of a nice way to "balance" the group sums so that the maximum group sum is minimised? Note that I want to keep the group count minimised (at 4 in this case)... I don't want to just make lots of groups of 1 element.

Bonus votes for simplicity of code over speed and/or exactness of solution... I don't mind if something is close to optimal but not quite there, and I'm only working with tiny arrays similar to the example above.

Clarification:

Point (4) refers to the grpSums variable calculated above. The command max(grpSums) in this example returns 949. If move some elements from group 3 to group 4:

grps = [1 1 1 1 2 2 2 2 3 3 4 4 4]

my grpSums becomes [939 938 734 449], making max(grpSums) give 939. An even better (and perhaps optimal for this case?) solution is:

grps = [1 1 1 2 2 2 3 3 3 3 4 4 4]

which gives grpSums = [705 704 922 729]. Note that it's not so much the range of numbers in grpSums that I care about - it's specifically its maximum.

##### 8 Comments

Sean de Wolski
on 19 Dec 2011

### Accepted Answer

Sven
on 29 Dec 2011

##### 3 Comments

Sean de Wolski
on 30 Dec 2011

Ans 0.55 seconds for A = [A;A];

Afetr that it's not obeying constraints.

### More Answers (4)

Walter Roberson
on 19 Dec 2011

##### 0 Comments

Seth DeLand
on 10 Mar 2014

Seeing how Optimization Toolbox has a mixed-integer solver in R2014a, I figured I would post this alternate solution that uses intlinprog to solve this problem. If you examine the output structure from intlinprog, the absolute gap is zero, meaning the solver has proven that this is the optimal solution to this problem.

A few things to note:

- This finds the same solution found in the brute force approach, with a max group size of 922.
- This code runs in a few hundredths of a second, likely much faster than brute force or global optimization methods for this problem.

%%Data

AA = [235 235 235 234 235 235 234 234 234 220 280 215 234];

lim = 1000; % maximum allowable in any group

ng = 4; % number of groups

%%Useful sizes

na = length(AA);

nx = ng*na;

nz = 1;

%%Variables

% There are two types of variables in this formulation:

% x: ng x na matrix of binary variables. x_i,j = 1 if A(i) is in group j.

% z: auxiliary variable used to keep track of maximum size of any group.

% The goal is to minimize z.

%%Objective function coefficients

f = zeros(nx + nz, 1);

f(end) = 1; % Minimize z, which is last in f

%%Sum of each group is less than limit

% A*x <= b

A = zeros(ng,nx+nz);

for ii = 1:na

A(:,ng*(ii-1)+1:ng*ii) = diag(repmat(AA(ii),ng,1));

end

b = repmat(lim,ng,1);

%%Minimax formulation

% Variable z changes the minimization problem into a minimax problem:

% min z s.t. Sum Ax - z <=0 (z is greater than or equal to size of each

% group)

Atmp = A;

Atmp(:,end) = -1;

A = [A; Atmp];

b = [b; zeros(ng,1)];

%%Groups of consecutive elements never increase (linear inequality)

Atmp = zeros(na-1,nx+nz);

for ii = 1:na-1

Atmp(ii,ng*(ii-1)+1:ng*ii) = 1:ng;

Atmp(ii,ng*ii+1:ng*(ii+1)) = -(1:ng)';

end

A = [A; Atmp];

b = [b; zeros(na-1,1)];

%%Equality constraint that assigns each member to exactly 1 group

Aeq = zeros(na,nx+nz);

for ii = 1:na

Aeq(ii,ng*(ii-1)+1:ng*ii) = 1;

end

beq = ones(na,1); % exaclty 1 in each group

%%Bounds

lb = zeros(nx+nz,1); % all variables positive

ub = ones(nx+nz,1); % upper bounds for binary variables

ub(end) = Inf; % no upper bound on z

%%Integer variables

intIdxs = 1:nx; % All x variables are binary integers

%%Solve

[xopt,fval,eflag,output] = intlinprog(f, intIdxs, A, b, Aeq, beq, lb, ub);

% If the problem is infeasible (negative eflag), might need to add more

% bins.

%%Post-process

x = xopt(1:end-1); % Extract x from solution vector

grpMax = xopt(end); % grpMax is z

x = round(reshape(x,ng,na));

optBins = zeros(size(AA)); % assigned bins

for ii = 1:na

optBins(ii) = find(x(:,ii));

end

binVals = A(1:ng,:)*xopt; % grpMax is z

The found solution:

binVals =

705.0000

704.0000

922.0000

729.0000

Dr. Seis
on 19 Dec 2011

UGLY/BRUTE-FORCE: I assumed you were trying to minimize the standard deviation of your group sum AND the difference between your lim and mean group sum. I also assumed that there would always be at least 2 groups. I check all possible group sizes/combinations. Here goes:

A = [235 235 235 234 235 235 234 234 234 220 280 215 234];

lim = 1000;

intmax = prod(A);

% WEIGHT

% >0.5 weight STD of sum more heavily

% <0.5 weight difference of lim and MEAN of sum more heavily

weight = 0.5;

best_combo = zeros(1,length(A));

best_fit = intmax;

for i = 1 : length(A)-1

B = nchoosek(1:length(A)-1,i);

for j = 1 : nchoosek(length(A)-1,i)

C = (i+1)*ones(1,length(A));

for k = sort(1 : i,'descend')

C(1:length(A)<=B(j,k))=k;

end

fit = (std(arrayfun(@(k)sum(A(C==k)),1:i+1))*(weight))^2 + ...

((lim-mean(arrayfun(@(k)sum(A(C==k)),1:i+1)))*(1-weight))^2;

if sum(arrayfun(@(k)sum(A(C==k)),1:i+1)>lim) == 0 && fit < best_fit

best_fit = fit;

best_combo = C;

end

end

end

grpsum=arrayfun(@(k)sum(A(best_combo==k)),1:max(best_combo));

This results in:

best_combo =

1 1 1 2 2 2 3 3 3 3 4 4 4

grpsum =

705 704 922 729

##### 2 Comments

Sean de Wolski
on 20 Dec 2011

Alright, here goes!! accumarray is certainly a part of it, as is ga() the awesome genetic algorithm function. So yes, Global Optimization Toolbox is required :)

main.m:

%12/19/2011

%

%Data

A = [235 235 235 234 235 235 234 234 234 220 280 215 234]'; %column vector (because they're cool)

lim = 1000; %upper limit.

szA = size(A);

% Get bounds and number of variables:

[bounds, n] = getValidGroup(A,lim);

% Engine:

ObjectiveFunction = @(x)binWeight(A,index2groups(x,szA),lim); %get the weight

[x,fval] = ga(ObjectiveFunction,n,[],[],[],[],ones(1,n),bounds(2,:),[],1:n);

index = index2groups(x,szA);

fprintf('Groups: %s \n',num2str(index));

fprintf('Weight is: %i!\n',binWeight(A,index,lim));

binWeight.m:

function weight = binWeight(A,index,lim)

%

%data nx1 vector of values

%index = nx1 vector of groupings

%lim = threshold of group sum

%

groups = accumarray(index,A); %sum bins

groups(groups>lim) = lim*10; %lim is soft constraint with stiff penalty.

weight = max(groups); %worst one

getValidGroup.m:

function [bounds cnt] = getValidGroup(A,lim)

%

%inputs:

%A - nx1 vector of values

%lim - the group limit

%

%outputs:

%bounds - 2x(cnt) lower/upper bounds

%cnt - number of bins

%

cnt = 0;

szA = size(A);

index = zeros(szA);

A2 = A;

while ~isempty(A)

cnt = cnt+1;

index(cnt) = find(cumsum(A)<lim,1,'last');

A(1:index(cnt)) = [];

end

index = index(1:cnt);

index(end) = find(cumsum(flipud(A2))<lim,1,'last');

bounds = 1:cnt; %lower bounds

bounds(2,:) = index; %upper bounds

index2groups.m:

function groups = index2groups(index,szA)

%index is a vector cumulative max values

index = index(:);

index = cumsum([1;index(1:end-1)]);

groups = zeros(szA(1),1);

groups(index) = 1; %min(index,szA(1))

groups = cumsum(groups);

And executing main.m:

Optimization terminated: average change in the penalty fitness value less than options.TolFun and constraint violation is less than options.TolCon. Groups: 1112223333444 Weight is: 922!

I have to say this is one of the most fun/educational/rewarding MATLAB problems I've ever worked on. Thanks!

##### 2 Comments

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