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Comparison between adjacent elements in sequences

I need a comparison index between elements in a sequence. This index is a factor between the sum of all absolute comparisons between adjacent elements in a sequence and the highest value that a sequence with its length can have.

For example, the sequences s1 = [0, 1, 2]and s2 = [0, 2, 1]have absolute comparisons [1, 1]and, [2, 1]respectively. There is no other combination of sequences with a length of 3 with a higher sum of absolute comparisons than 3. Thus, the comparison index should be 2/3 and 3/3 for s1 and s2.

These sequences always have integers from 0to length - 1and can have non-adjacent repeating elements, such as [0, 1, 0, 1, 0]. These sequences have all integers between their lowest and highest element values.

I need a function to calculate the maximum value of absolute comparisons that a sequence with a given length can have. The function that I wrote (highest) returns incorrect results. I wrote this code:

    def aux_pairs(seq):
        n = 2
        return [seq[i:i + n] for i in range(len(seq) - (n - 1))]

    def comparisons_sum(seq):
        return sum([abs(el[-1] - el[0]) for el in aux_pairs(seq)])

    def highest(seq):
        card = len(seq)
        pair = [0, card - 1]
        r = (pair * (card / 2))
        if card & 1 == 1:
            r.append(0)
        return comparisons_sum(r)

    def comparison_index(seq):
        return comparisons_sum(seq) / float(highest(seq))
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1 answer

The easiest way to create a list is simply:

def comparisons(seq):
    return [abs(a-b) for a, b in zip(seq, seq[1:])]

As for your comparison, the highest value will always be the maximum, followed by the minimum, repeated. For example: for length 4:

[3, 0, 3, 0]

. (length-1) ( length-1). , (length-1)**2.

, , , 3 3, [0, 2, 0] ( [2, 2], 4)?

, 0 length-1, (: [0, 1, 0]) . , ( n 0 n-1, ).

, .

0 len-1, , 0 len-1, "", , :

from collections import deque
from itertools import permutations
from operator import itemgetter

def comparisons(seq):
    return [abs(a-b) for a, b in zip(seq, seq[1:])]

def best_order(n):
    temp = deque([0, n-1])
    low = 1
    high = n-2
    while low < high:
        left = temp[0]
        right = temp[-1]
        if left < right:
            temp.append(low)
            temp.appendleft(high)
        else:
            temp.append(high)
            temp.appendleft(low)
        low += 1
        high -= 1
    if len(temp) < n:
        temp.append(low)
    return list(temp)

def brute_force(n):
    getcomp = itemgetter(2)
    return max([(list(a), comparisons(a), sum(comparisons(a))) for a in permutations(range(n))], key=getcomp)

for i in range(2, 6):
    print("Algorithmic:", best_order(i), comparisons(best_order(i)), sum(comparisons(best_order(i))))
    print("Brute Force:", *brute_force(i))

:

Algorithmic: [0, 1] [1] 1
Brute Force: [0, 1] [1] 1
Algorithmic: [0, 2, 1] [2, 1] 3
Brute Force: [0, 2, 1] [2, 1] 3
Algorithmic: [2, 0, 3, 1] [2, 3, 2] 7
Brute Force: [1, 3, 0, 2] [2, 3, 2] 7
Algorithmic: [3, 0, 4, 1, 2] [3, 4, 3, 1] 11
Brute Force: [1, 3, 0, 4, 2] [2, 3, 4, 2] 11

, .

:

from collections import deque

def comparisons(seq):
    return [abs(a-b) for a, b in zip(seq, seq[1:])]

def best_order(seq):
    pool = deque(sorted(seq))
    temp = deque([pool.popleft(), pool.pop()])
    try:
        while pool:
            if temp[0] < temp[-1]:
                temp.append(pool.popleft())
                temp.appendleft(pool.pop())
            else:
                temp.append(pool.pop())
                temp.appendleft(pool.popleft())
    except IndexError:
        pass
    return list(temp)

i = [0, 1, 2, 3, 4, 5, 6, 0, 0, 1, 1, 2, 2]
print("Algorithmic:", best_order(i), comparisons(best_order(i)), sum(comparisons(best_order(i))))
for n in range(2, 6):
    i = list(range(n))
    print("Algorithmic:", best_order(i), comparisons(best_order(i)), sum(comparisons(best_order(i))))

:

Algorithmic: [2, 1, 3, 0, 5, 0, 6, 0, 4, 1, 2, 1, 2] [1, 2, 3, 5, 5, 6, 6, 4, 3, 1, 1, 1] 38
Algorithmic: [0, 1] [1] 1
Algorithmic: [0, 2, 1] [2, 1] 3
Algorithmic: [2, 0, 3, 1] [2, 3, 2] 7
Algorithmic: [3, 0, 4, 1, 2] [3, 4, 3, 1] 11

, .

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