Performance 带顺序限制的置换
让Performance 带顺序限制的置换,performance,matlab,permutation,Performance,Matlab,Permutation,让L成为对象列表。此外,让C成为一组约束条件,例如: C(1)=t1位于t2之前,其中t1和t2属于L C(2)=t3位于t2之后,其中t3和t2属于L 如何(在MATLAB中)找到不违反C中约束的置换集 我的第一个解决方案很幼稚: orderings = perms(L); toBeDeleted = zeros(1,size(orderings,1)); for ii = 1:size(orderings,1) for jj = 1:size
L
成为对象列表。此外,让C
成为一组约束条件,例如:
C(1)=t1位于t2之前,其中
和t1
属于t2
L
C(2)=t3位于t2之后,其中
和t3
属于t2
L
C
中约束的置换集
我的第一个解决方案很幼稚:
orderings = perms(L);
toBeDeleted = zeros(1,size(orderings,1));
for ii = 1:size(orderings,1)
for jj = 1:size(constraints,1)
idxA = find(orderings(ii,:) == constraints(jj,1));
idxB = find(orderings(ii,:) == constraints(jj,2));
if idxA > idxB
toBeDeleted(ii) = 1;
end
end
end
其中约束
是一组约束(每个约束位于两个元素的行上,指定第一个元素位于第二个元素之前)
我想知道是否存在更简单(更有效)的解决方案
提前谢谢。我想说,到目前为止,这是一个非常好的解决方案 不过,我看到了一些优化。以下是我的变体:
% INITIALIZE
NN = 9;
L = rand(1,NN-1);
while numel(L) ~= NN;
L = unique( randi(100,1,NN) ); end
% Some bogus constraints
constraints = [...
L(1) L(2)
L(3) L(6)
L(3) L(5)
L(8) L(4)];
% METHOD 0 (your original method)
tic
orderings = perms(L);
p = size(orderings,1);
c = size(constraints,1);
toKeep = true(p,1);
for perm = 1:p
for constr = 1:c
idxA = find(orderings(perm,:) == constraints(constr,1));
idxB = find(orderings(perm,:) == constraints(constr,2));
if idxA > idxB
toKeep(perm) = false;
end
end
end
orderings0 = orderings(toKeep,:);
toc
% METHOD 1 (your original, plus a few optimizations)
tic
orderings = perms(L);
p = size(orderings,1);
c = size(constraints,1);
toKeep = true(p,1);
for perm = 1:p
for constr = 1:c
% break on first condition breached
if toKeep(perm)
% find only *first* entry
toKeep(perm) = ...
find(orderings(perm,:) == constraints(constr,1), 1) < ...
find(orderings(perm,:) == constraints(constr,2), 1);
else
break
end
end
end
orderings1 = orderings(toKeep,:);
toc
% METHOD 2
tic
orderings = perms(L);
p = size(orderings,1);
c = size(constraints,1);
toKeep = true(p,1);
for constr = 1:c
% break on first condition breached1
if any(toKeep)
% Vectorized search for constraint values
[i1, j1] = find(orderings == constraints(constr,1));
[i2, j2] = find(orderings == constraints(constr,2));
% sort by rows
[i1, j1i] = sort(i1);
[i2, j2i] = sort(i2);
% Check if columns meet condition
toKeep = toKeep & j1(j1i) < j2(j2i);
else
break
end
end
orderings2 = orderings(toKeep,:);
toc
% Check for equality
all(orderings2(:) == orderings1(:))
然而,整个方法有一个基本缺陷,即IMHO;直接使用perms
。由于内存限制(NN<10
,如help perms
中所述),这本身就造成了限制
我强烈怀疑,当您将定制的
烫发
组合在一起时,您可以在时间和内存方面获得更好的性能。幸运的是,perms
不是内置的,所以您可以先将该代码复制粘贴到自定义函数中 无论何时在向量或矩阵中查找某些内容,都应该使用find或ismember.Sure。建议的解决方案是伪代码(perms指令除外)。然而,我认为这不是可能的最佳解决方案,因为我必须迭代所有排序和所有约束。但是,我已经更新了代码(现在不是伪代码)。这是最好的。这个问题与。
Elapsed time is 17.911469 seconds. % your method
Elapsed time is 10.477549 seconds. % your method + optimizations
Elapsed time is 2.184242 seconds. % vectorized outer loop
ans =
1
ans =
1