How can I find the maximum sum of a sub-sequence u

问题:

I'm re-reading Skiena's Algorithm Design Manual to catch up on some stuff I've forgotten since school, and I'm a little baffled by his descriptions of Dynamic Programming. I've looked it up on Wikipedia and various other sites, and while the descriptions all make sense, I'm having trouble figuring out specific problems myself. Currently, I'm working on problem 3-5 from the Skiena book. (Given an array of n real numbers, find the maximum sum in any contiguous subvector of the input.) I have an O(n^2) solution, such as described in this answer. But I'm stuck on the O(N) solution using dynamic programming. It's not clear to me what the recurrence relation should be.

I see that the subsequences form a set of sums, like so:

S = {a,b,c,d}

a    a+b    a+b+c    a+b+c+d
     b      b+c      b+c+d
            c        c+d
                     d

What I don't get is how to pick which one is the greatest in linear time. I've tried doing things like keeping track of the greatest sum so far, and if the current value is positive, add it to the sum. But when you have larger sequences, this becomes problematic because there may be stretches of negative numbers that would decrease the sum, but a later large positive number may bring it back to being the maximum.

I'm also reminded of summed area tables. You can calculate all the sums using only the cumulative sums: a, a+b, a+b+c, a+b+c+d, etc. (For example, if you need b+c, it's just (a+b+c) - (a).) But don't see an O(N) way to get it.

Can anyone explain to me what the O(N) dynamic programming solution is for this particular problem? I feel like I almost get it, but that I'm missing something.

回答1:

You should take a look to this pdf back in the school in http://castle.eiu.edu here it is:

The explanation of the following pseudocode is also int the pdf.

回答2:

There is a solution like, first sort the array in to some auxiliary memory, then apply Longest Common Sub-Sequence method to the original array and the sorted array, with sum(not the length) of common sub-sequence in the 2 arrays as the entry into the table (Memoization). This can also solve the problem

Total running time is O(nlogn)+O(n^2) => O(n^2) Space is O(n) + O(n^2) => O(n^2)

This is not a good solution when memory comes into picture. This is just to give a glimpse on how problems can be reduced to one another.

回答3:

My understanding of DP is about "making a table". In fact, the original meaning "programming" in DP is simply about making tables.

The key is to figure out what to put in the table, or modern terms: what state to track, or what's the vertex key/value in DAG (ignore these terms if they sound strange to you).

How about choose dp[i] table being the largest sum ending at index i of the array, for example, the array being [5, 15, -30, 10]

The second important key is "optimal substructure", that is to "assume" dp[i-1] already stores the largest sum for sub-sequences ending at index i-1, that's why the only step at i is to decide whether to include a[i] into the sub-sequence or not

dp[i] = max(dp[i-1], dp[i-1] + a[i])

The first term in max is to "not include a[i]", the second term is to "include a[i]". Notice, if we don't include a[i], the largest sum so far remains dp[i-1], which comes from the "optimal substructure" argument.

So the whole program looks like this (in Python):

a = [5,15,-30,10]

dp = [0]*len(a)
dp[0] = max(0,a[0]) # include a[0] or not

for i in range(1,len(a)):
    dp[i] = max(dp[i-1], dp[i-1]+a[i]) # for sub-sequence, choose to add or not     


 print(dp, max(dp)) 

The result: largest sum of sub-sequence should be the largest item in dp table, after i iterate through the array a. But take a close look at dp, it holds all the information.

Since it only goes through items in array a once, it's a O(n) algorithm.

This problem seems silly, because as long as a[i] is positive, we should always include it in the sub-sequence, because it will only increase the sum. This intuition matches the code

dp[i] = max(dp[i-1], dp[i-1] + a[i])

So the max. sum of sub-sequence problem is easy, and doesn't need DP at all. Simply,

sum = 0
for v in a:
     if  v >0
         sum += v

However, what about largest sum of "continuous sub-array" problem. All we need to change is just a single line of code

dp[i] = max(dp[i-1]+a[i], a[i])    

The first term is to "include a[i] in the continuous sub-array", the second term is to decide to start a new sub-array, starting a[i].

In this case, dp[i] is the max. sum continuous sub-array ending with index-i.

This is certainly better than a naive approach O(n^2)*O(n), to for j in range(0,i): inside the i-loop and sum all the possible sub-arrays.

One small caveat, because the way dp[0] is set, if all items in a are negative, we won't select any. So for the max sum continuous sub-array, we change that to

dp[0] = a[0]