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 time


def 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_clock
def 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

Pseudo-polynomial Partition Problem