Find the longest arithmetic progression inside a sequence

问题

Suppose I have a sequence of increasing numbers, and I want to find the length of longest arithmetic progression within the sequence. Longest arithmetic progression means an increasing sequence with common difference, such as [2, 4, 6, 8] or [3, 6, 9, 12].

For example, for [5, 10, 14, 15, 17], [5, 10, 15] is the longest arithmetic progression, with length 3;

for [10, 12, 13, 20, 22, 23, 30], [10, 20, 30] is the longest arithmetic progression with length 3;

for [7, 10, 12, 13, 15, 20, 21], [10, 15, 20] or [7, 10, 13] are the longest arithmetic progressions with length 3.

This site https://prismoskills.appspot.com/lessons/Dynamic_Programming/Chapter_22_-_Longest_arithmetic_progression.jsp offers some insight into the problem, i.e. by looping around j and consider every 3 elements. I intend to use this algorithm in Python, and my code is as follows:

def length_of_AP(L):
n = len(L)
Table = [[0 for _ in range(n)] for _ in range(n)]
length_of_AP = 2

# initialise the last column of the table as all i and (n-1) pairs have lenth 2
for i in range(n):
        Table[i][n-1] =2

# loop around the list and i, k such that L[i] + L[k] = 2 * L[j]
for j in range(n - 2, 0, -1):
        i = j - 1
        k = j + 1
        while i >= 0 and k < n:
                difference = (L[i] + L[k]) - 2 * L[j]
                if difference < 0:
                        k = k + 1
                else:
                        if difference > 0:
                                i = i - 1
                        else:
                                Table[i][j] = Table[j][k] + 1
                                length_of_AP = max(length_of_AP, Table[i][j])
                                k = k + 1
                                i = i - 1
return length_of_AP

This function works fine with [1, 3, 4, 5, 7, 8, 9], but it doesn't work for [5, 10, 14, 15, 20, 25, 26, 27, 28, 30, 31], where I am supposed to get 6 but I got 4. I can see the reason being that 25, 26, 27, 28 inside the list may be a distracting factor for my function. How do I change my function so that it gives me the result desired.

Any help may be appreciated.

回答1:

Following your link and running second sample, it looks like the code actually find proper LAP

5, 10, 15, 20, 25, 30,

but fails to find proper length. I didn't spend too much time analyzing the code but the piece

    // Any 2-letter series is an AP
    // Here we initialize only for the last column of lookup because
    // all i and (n-1) pairs form an AP of size 2  
    for (int i=0; i<n; i++)
        lookup[i][n-1] = 2;

looks suspicious to me. It seems that you need to initialize whole lookup table with 2 instead of just last column and if I do so, it starts to get correct length on your sample as well.

So get rid of the "initialise" loop and change your 3rd line to following code:

# initialise whole table with 2 as all (i, j) pairs have length 2    
Table = [[2 for _ in range(n)] for _ in range(n)]

Moreover their

Sample Execution:
Max AP length = 6
3, 5, 7, 9, 11, 13, 15, 17,

Contains this bug as well and actually prints correct sequence only because of sheer luck. If I modify the sortedArr to

int sortedArr[] = new int[] {3, 4, 5, 7, 8, 9, 11, 13, 14, 15, 16, 17, 18,  112, 113, 114, 115, 116, 117, 118};

I get following output

Max AP length = 7
112, 113, 114, 115, 116, 117, 118,

which is obviously wrong as original 8-items long sequence 3, 5, 7, 9, 11, 13, 15, 17, is still there.

回答2:

Did you try it?
Here's a quick brute force implementation, for small datasets it should run fast enough:

def gen(seq):
    diff = ((b-a, a) for a, b in it.combinations(sorted(seq), 2))
    for d, n in diff:
        k = []
        while n in seq:
            k.append(n)
            n += d
        yield (d, k)

def arith(seq):
    return max(gen(seq), key=lambda x: len(x[1]))

In [1]: arith([7, 10, 12, 13, 15, 20, 21])
Out[1]: (3, [7, 10, 13])
In [2]: %timeit arith([7, 10, 12, 13, 15, 20, 21])
10000 loops, best of 3: 23.6 µs per loop
In [3]: seq = {random.randrange(1000) for _ in range(100)}
In [4]: arith(seq)
Out[4]: (171, [229, 400, 571, 742, 913])
In [5]: %timeit arith(seq)
100 loops, best of 3: 3.79 ms per loop
In [6]: seq = {random.randrange(1000000) for _ in range(1000)}
In [7]: arith(seq)
Out[7]: (81261, [821349, 902610, 983871])
In [8]: %timeit arith(seq)
1 loop, best of 3: 434 ms per loop

来源：https://stackoverflow.com/questions/43151904/find-the-longest-arithmetic-progression-inside-a-sequence

标签

python

math

Sequence