TimberLee's blog

Optimising DP using bitset in C++
By TimberLee, history, 8 years ago, In English

I recently read somewhere that some DP solutions like knapsack can be optimised and the overall complexity can be reduced by a factor of 32 using std::bitset in C++.

Can someone explain this optimisation and the kinds of DP on which this works ?

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8 years ago, # |
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Another Problem using this trick.

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8 years ago, # |
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COCI 2015/2016 Task UZASTOPNI http://hsin.hr/coci/contest1_tasks.pdf

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8 years ago, # |
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Also 685E - Travelling Through the Snow Queen's Kingdom in a recent CF round.
Basically, the idea is that you can use bit-wise operations on bitset to determine 32 times more values in one run compared to bool or int.
I think it can work on any DP where the output is boolean like can we reach this state or not.

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8 years ago, # |
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People are giving tasks but not a single one is describing the idea :(

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8 years ago, # |
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Hi !

Problem is: we have n numbers, calculate how many distinct numbers we can form by sum some of these numbers.

For example if we have set {17, 23, 40}, we can form {0, 17, 23, 40, 57, 63, 80}.

Dp solution is obvious:

dp[0] = 1;
for(int i = 0; i < n; i++)
    for(int j = maxv - 1; j >= a[i]; j--)  // maxv is some number bigger than sum of a[i]'s
        dp[j] |= dp[ j - a[i] ];
cout << count(dp, dp + maxv, 1) << '\n';

Now how to optimize it? bitset can store bits and do operations 32 times faster like this:

bitset<maxv> dp;
dp.set(0);
for(int i = 0; i < n; i++)
    dp |= dp << a[i];
cout << dp.count() << '\n';

Good problem (and hot!) to solve: SnackDown 2016 Online elimination round » Robot Walk.

My solution.

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    8 years ago, # ^ |
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    Can you tell how to solve knapsack using it? we can do the same thing with weights but how to store the value we are getting from that particular weight.

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      8 years ago, # ^ |
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      What is knapsack problem !?

      I saw several problems that called knapsack. Can you explain your problem?

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        8 years ago, # ^ |
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        [user:Arpa]The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible.

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    8 years ago, # ^ |
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    Can you explain why exactly using bitset is O(N / 32). I tried looking it up but couldn't find any exact reasons.

    I tried to read the library implementation of bitset(from usr/include/bitset) and it appears to be O(N) only(or O(maxv) from your code). This is the implementation on my system(G++ 4.9.2)

    void
      _M_do_or(const _Base_bitset<_Nw>& __x)
      {
	for (size_t __i = 0; __i < _Nw; __i++)
	  _M_w[__i] |= __x._M_w[__i];
      }

    And _Nw is defined in template<size_t _Nw> struct _Base_bitset so basically it makes O(maxv) operations only. The functions xor and and also seem to be implemented similarly. How is this faster than normal looping?

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      8 years ago, # ^ |
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        template<size_t _Nb>
    class bitset
    : private _Base_bitset<_GLIBCXX_BITSET_WORDS(_Nb)>

      So _Nw = _GLIBCXX_BITSET_WORDS(maxv) = maxv/32 (on a 32-bit system)

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        8 years ago, # ^ |
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        Ah nice I seemed to mix up the definitions of _Base_bitset and bitset. So now we basically have an N digit binary number split into groups of size 32 and dealt with individually. Is that understanding right?

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    4 years ago, # ^ |
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    https://www.spoj.com/problems/TUG/

    Can this be solved using this?

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5 years ago, # |
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Could someone show me some tutorials or blogs that about bitset optimization? Or you guys can give me some basic understanding and I will try to search for things.

Thanks in advanced <3 <3.

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