Friday, March 4, 2016

Bin Packing Problem (Minimize number of used Bins) - GeeksforGeeks


Bin Packing Problem (Minimize number of used Bins) - GeeksforGeeks
Given n items of different weights and bins each of capacity c, assign each item to a bin such that number of total used bins is minimized. It may be assumed that all items have weights smaller than bin capacity.
Input:  wieght[]       = {4, 8, 1, 4, 2, 1}
        Bin Capacity c = 10
Output: 2
We need minimum 2 bins to accommodate all items
First bit contains {4, 4, 2} and second bin {8, 2}
We can always find a lower bound on minimum number of bins required. The lower bound can be given as :
   Min no. of bins  >=  Ceil ((Total Weight) / (Bin Capacity))
  
In the above examples, lower bound for first example is “ceil(4 + 8 + 1 + 4 + 2 + 1)/10″ = 2 and lower bound in second example is “ceil(9 + 8 + 2 + 2 + 5 + 4)/10″ = 3.
This problem is a NP Hard problem and finding an exact minimum number of bins takes exponential time. 
Online Algorithms

These algorithms are for Bin Packing problems where items arrive one at a time (in unknown order), each must be put in a bin, before considering the next item.
1. Next Fit:
When processing next item, check if it fits in the same bin as the last item. Use a new bin only if it does not.
nt nextFit(int weight[], int n, int c)
{
   // Initialize result (Count of bins) and remaining
   // capacity in current bin.
   int res = 0, bin_rem = c;
 
   // Place items one by one
   for (int i=0; i<n; i++)
   {
       // If this item can't fit in current bin
       if (weight[i] > bin_rem)
       {
          res++;  // Use a new bin
          bin_rem = c - weight[i];
       }
       else
         bin_rem -= weight[i];
   }
   return res;
}
Next Fit is a simple algorithm. It requires only O(n) time and O(1) extra space to process n items.
Next Fit is 2 approximate, i.e., the number of bins used by this algorithm is bounded by twice of optimal. Consider any two adjacent bins. The sum of items in these two bins must be > c; otherwise, NextFit would have put all the items of second bin into the first. The same holds for all other bins. Thus, at most half the space is wasted, and so Next Fit uses at most 2M bins if M is optimal.


2. First Fit:
When processing the next item, see if it fits in the same bin as the last item. Start a new bin only if it does not.
int firstFit(int weight[], int n, int c)
{
    // Initialize result (Count of bins)
    int res = 0;
 
    // Create an array to store remaining space in bins
    // there can be at most n bins
    int bin_rem[n];
 
    // Place items one by one
    for (int i=0; i<n; i++)
    {
        // Find the first bin that can accommodate
        // weight[i]
        int j;
        for (j=0; j<res; j++)
        {
            if (bin_rem[j] >= weight[i])
            {
                bin_rem[j] = bin_rem[j] - weight[i];
                break;
            }
        }
 
        // If no bin could accommodate weight[i]
        if (j==res)
        {
            bin_rem[res] = c - weight[i];
            res++;
        }
    }
    return res;
}
The above implementation of First Fit requires O(n2) time, but First Fit can be implemented in O(n Log n) time using Self-Balancing Binary Search Trees.
If M is the optimal number of bins, then First Fit never uses more than 1.7M bins. So First Fit is better than Next Fit in terms of upper bound on number of bins.


3. Best Fit:
The idea is to places the next item in the *tightest* spot. That is, put it in the bin so that smallest empty space is left.
int bestFit(int weight[], int n, int c)
{
    // Initialize result (Count of bins)
    int res = 0;
 
    // Create an array to store remaining space in bins
    // there can be at most n bins
    int bin_rem[n];
 
    // Place items one by one
    for (int i=0; i<n; i++)
    {
        // Find the best bin that ca\n accomodate
        // weight[i]
        int j;
 
        // Initialize minimum space left and index
        // of best bin
        int min = c+1, bi = 0;
 
        for (j=0; j<res; j++)
        {
            if (bin_rem[j] >= weight[i] &&
                    bin_rem[j] - weight[i] < min)
            {
                bi = j;
                min = bin_rem[j] - weight[i];
            }
        }
 
        // If no bin could accommodate weight[i],
        // create a new bin
        if (min==c+1)
        {
            bin_rem[res] = c - weight[i];
            res++;
        }
        else // Assign the item to best bin
            bin_rem[bi] -= weight[i];
    }
    return res;
}
Offline Algorithms

In the offline version, we have all items upfront. Unfortunately offline version is also NP Complete, but we have a better approximate algorithm for it. First Fit Decreasing uses at most (4M + 1)/3 bins if the optimal is M.


4. First Fit Decreasing:
A trouble with online algorithms is that packing large items is difficult, especially if they occur late in the sequence. We can circumvent this by *sorting* the input sequence, and placing the large items first. With sorting, we get First Fit Decreasing and Best Fit Decreasing, as offline analogs of online First Fit and Best Fit.
int firstFitDec(int weight[], int n, int c)
{
    // First sort all weights in decreasing order
    sort(weight, weight+n, std::greater<int>());
 
    // Now call first fit for sorted items
    return firstFit(weight, n, c);
}
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