第一个循环for j=k step-1 until 1 do，为什么j要从k到1，而不是1到k？

算法如下：Input: A sequence A1，A2，….An where each Ai is a k-tuple
(ai1，ai2，…，aik )
with aij an integer in the range 0 to m-1. (A convenient data structure for this sequence of k-tuples is an n*k array)
Output: A sequence B1，B2，….Bn which is a permutation of A1，A2，….An such that B<=B for 1<=iMethod. In transferring a k-tuple A to some bucket，only a pointer to A is actually moved. Thus A can be added to a bucket in fixed time rather than time bounded by k. We use a queue called QUEUE to hold the" current" sequence of elements. An array Qof m buckets is also used，where bucket Q[i] is intended to hold those k-tuples that have the integer i in the component currently under consideration, The algorithm is show as :
begin
place A，A，......，A in QUEUE;
for j=k step-1 until 1 do
begin
for l=0 until m-1 do make Q[l] empty;
while QUEUE not empty do
begin
let A be the first element in QUEUE;
move A from QUEUE to bucket Q[a]
end;
for l=0 until m-1 do
concatenate contents of Q[l] to the end of QUEUE
end
end

为什么要move A from QUEUE to bucket Q[a](这里似乎遗漏了一些下标)？bucket Q[a]的数据结构应该是怎样的？
为什么要for l=0 until m-1 do concatenate contents of Q[l] to the end of QUEUE？

i Method. In transferring a k-tuple A

搜一下：第一个循环for j=k step-1 until 1 do，为什么j要从k到1，而不是1到k？

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