Couple of points that are closer to each other in a list

Question

I have to do an algorithm, aprop, using divide and conquer to calculate the couple of points that are closer to each other in a list and also I have to calculate the complexity.

def aprop(a):
    longitudlista = len(a)
    if longitudlista <= 2:
        return a
    else:
        mig = longitudlista / 2
        pivot = (a[mig])

a= [1.2,2.9,3.1,4.0,5.7]

aprop(a)

Now I want that the algorithm return the couple of points that having the minimum difference of all the differences of all elements of the list using the pivot. How can i write this idea in code?

Because i have to divide the algorithm or also we can use a[:mig] and a[mig:] — Cristiano, Nov 30 '14 at 17:52
this is a standard algorithm homework question. I'm pretty sure you could Google the complete answer — kdopen, Nov 30 '14 at 17:53
But i dont want see the solution, i want to learn how can i do — Cristiano, Nov 30 '14 at 17:55

Mazdak · Answer 1 · 2014-11-30T18:35:06.027

1

The following code is General Implementation of the Divide and Conquer Scheme :

def divide_and_conquer(S, divide, combine): 
  if len(S) == 1: return S
  L, R = divide(S)
  A = divide_and_conquer(L, divide, combine)
  B = divide_and_conquer(R, divide, combine)
  return combine(A, B)

So based on the above algorithm you can use this :

from operator import sub

def divide_and_conquer(S, combine=[]):
    if len(S) == 2:
       combine.append(S)
    else :
       L, R = S[:(len(S)/2)+1],S[len(S)/2:]
       A = divide_and_conquer(L)
       B = divide_and_conquer(R)
       result=[abs(sub(i,j)) for i,j in combine]
       return combine[result.index(min(result))]

a= [1.2,2.9,3.1,4.0,5.7]

print divide_and_conquer(a)
[2.9, 3.1]

edited Nov 30 '14 at 18:35

answered Nov 30 '14 at 18:15

Mazdak

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Yeah i understand this but my result have to show 2.9 and 3.1 How can i do this? – Cristiano Nov 30 '14 at 18:21
1

@Cristiano i think you have a miss understanding about StackOverflow ,its not a magic place that do whole of your homeworks for you , and the peoples here want to learn and teach ! so its better to do it by your self ! in addition if you find this answer helpful you can tell this to community by voting or accepting the answer ! – Mazdak Nov 30 '14 at 18:42

score 0 · Accepted Answer · edited May 23 '17 at 10:26

Sort your array and check pairwise for the smallest distance.

The sort can be done using a divide and conquer algorithm, like merge_sort in O(nlog(n)) time.

For instance you could piggyback on merge sort like this :

# For convenience of representation, assume pairs to be defined as 
# tuples of form ( no_1, no_2 )

def closest_pair(array):
    l = len(array)
    pivot = l/2

    if len(array) == 2:
        # Simply returns the pair
        return (array[0], array[1])

    if len(array) == 3:
        # Returns the pair which has the smallest distance amongst the 3C2 pairs
        return min(itertools.combinations(array, r = 2), key = lambda x : abs(x[1] - x[0]) )


    left_closest = closest_pair(array[:pivot])
    right_closest = closest_pair(array[pivot:])
    split_pair = (array[pivot-1], array[pivot]) # Just incase the split_pair is the closest

    return min(left_closest, right_closest, split_pair, key = lambda x : abs(x[1] - x[0]) )

For your array [1.2,2.9,3.1,4.0,5.7]

>>> closest_pair([1.2,2.9,3.1,4.0,5.7])
(2.9, 3.1)

Side Note:

If you don't care about the divide and conquer implementation, you could simply use min using a key and itertools.combinations ( Like implemented above for the combine step / the base case of len(array) equal to 3. )

>>> a = [1.2,2.9,3.1,4.0,5.7]
>>> min(itertools.combinations(a, r = 2), key = lambda x : abs(x[1] - x[0])))
(2.9, 3.1)

Check this to learn more about :

itertools.combinations
min with a key parameter - Has been explained for max but the same applies for min too.

Hope this helps...

Couple of points that are closer to each other in a list

2 Answers2