Pseudo-polynomial Partition Problem

1       Problem Statement

Pseudo-polynomialPartition
Given a set consisting of n integers [a₁, a₂, …a_n], you want to partition into two parts so that thesum of the two parts is equal.  Suppose s =  a₁ + a₂ …+ a_n. The time complexity of your algorithm shouldbe O(ns) orbetter.  [Note: Due to the presence of the term s in the timecomplexity, such an algorithm is called pseudo polynomial algorithm.]

2       Theoretical Analysis

In this problem, I use part[i][j] to be true if asubset of {x1, x2,..,xj} sums to i and false otherwise. Thus, part[i][j] istrue if either p(i, j-1) is true or if part(i–xj, j-1) is true.

part(i, j) is false otherwise. And suppose s is thesum of the array.

Thus, the total time complexity is O(ns)

3       Experimental Analysis

3.1       Program Listing

 

I use python to write the program, if the O(ns) istoo small, the running time does not change obviously.

Thus, I choose “S”: 500, 1000, 1500, 2000, 2500

And, keep “N”: 6

 

3.2       Data Normalization Notes

 

I normalize the values by a constant of 679.91 It comes from theradio of average experimental result and average theoretical result.

 

 

 

 

 

3.3       Output Numerical Data

s

n

Experimental Result, in ns

Theoretical Result

Scaling Constant

Adjusted Theoretical Result

500

6

1699924.46

3000

　

2039723.46

1000

6

4694938.65

6000

　

4079446.91

1500

6

6446910.85

9000

　

6119170.37

2000

6

8310079.57

12000

　

8158893.83

2500

6

9443998.33

15000

　

10198617.3

　

　

6119170.37

9000

679.91

　

 

3.4       Graph

 

 

3.5       Graph Observations

 

From the graph, we can see the blue line, derivesfrom experiment, fits the red line which comes from theory.

 

 

4       Conclusions

 

From the experiment, we can come to a conclusionthat the time complexity of this code problem is O(ns).

# -*- coding: UTF-8 -*-# __author__ = 'Sengo'import timedef open_clock(func):    def _wrapper(*args, **kwargs):        begin = time.time()        ret = func(*args, **kwargs)        end = time.time()        total = end - begin        print "cost time: ", total        return ret    return _wrapper@open_clockdef find_partition(arr):    """    :param arr: A list of integers arr    :return: True if arr can be partitioned, and sum of two part is equal    """    _sum = sum(arr)    n = len(arr)    # Odd sum    if _sum % 2:        return False    # init part[_sum/2+1][n+1]    part = [[0] * (n+1) for _ in range(_sum/2+1)]    # init top row as True    for i in range(n+1):        part[0][i] = True    # init the leftmost column as False, expect part[0][0]    for i in range(1, _sum/2 + 1):        part[i][0] = False    for i in range(1, _sum/2 + 1):        for j in range(1, n + 1):            part[i][j] = part[i][j-1]            if i >= arr[j-1]:                part[i][j] = part[i][j] or part[i - arr[j-1]][j-1]    return part[_sum/2][n]if __name__ == '__main__':    a = [0, 1000, 0, 0, 0, 1000]    print find_partition(a)

参考 

http://www.geeksforgeeks.org/dynamic-programming-set-18-partition-problem/

https://en.wikipedia.org/wiki/Partition_problem