1 Problem Statement
Pseudo-polynomialPartition
Given a set consisting of n integers [a1, a2, …an], you want to partition into two parts so that thesum of the two parts is equal. Suppose s = a1 + a2 …+ an. 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 - beginprint "cost time: ", totalreturn retreturn _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 sumif _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 Truefor 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] = Falsefor 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