Groups of pairwise combinations where each member appears only once

Question

I have a list of unique tuples each containing 2 elements from 1 to 10. A total number of elements in a list is 45. I would like to divide them into 10 groups each of them containing only numbers from 1 to 10.

I have tried solve my problem using this answer: python get groups of combinations that each member appear only once

python:

from itertools import combinations, chain
l = ['A','B','C','D','E', 'F', 'G','H','I','J']
c = list(combinations(l,2))
[set(i) for i in list(combinations(c,5)) if (len(set(l) & set(chain(*i))) == len(l))]

But I get repetitions, like so:

[{('A', 'B'), ('C', 'D'), ('E', 'F'), ('G', 'H'), ('I', 'J')},
 {('A', 'B'), ('C', 'D'), ('E', 'F'), ('G', 'I'), ('H', 'J')},...]

Akins

Could you explain your requirements with an example? Also, 45 tuples can't be divided into 10 equal groups, so what will the division be? – Cindycine

Of course you are right, in that case I could have 9 groups 5 elements each. Each element is a tuple of 2 letters, all letters should be unique. So letters from A to J appear only once. Is it possible? Thanks – Akins

I suggest you provide the desired outcome for e.g. ['A', 'B', 'C', 'D'] and explain why repetitions like above are bad. – Knave

A practical example should hopefully clarify this issue. Imagine we have 10 players which play a chess game that consists of 9 rounds. Each player should play against all others players, but only once over the course of 9 rounds. So in round 1, we have 5 pairs: players 1 and 2, 3 and 4, 5 and 6, 7 and 8, 9 and 10 playing. In round 2, we have another 5 pairs. They could be 1 and 3, 2 and 4, 5 and 7, 6 and 9, 8 and 10. – Akins

Answer 1

not 10 pairs , but there are 945 such pair that satisy your conditions

what i did was, get all the permutation of number, create a dictionary which keeps all combinations as key.

now for each permutation element, I have taken them in pair of 2 ie [1,2,3,4] is [(1,2),(2,3),(3,4)]

this will create a list

now for all such list and their element, I have compared them from the dictionary whether they are present in the dictionary or not.

ps. this is a long and time-consuming solution, with graph theory we can reduce size considerably.

from itertools import combinations, permutations
l=['A','B','C','D','E','F','G','H','I','J']
c=list(permutations(l))
d=list(combinations(l,2))
from collections import defaultdict

dic = defaultdict(int)

for i in d:
    dic[i]=1


new =[]
for i in c:
    tmp=[]
    for j in range(1,len(i),2):
        tmp.append((i[j-1],i[j]))
    new.append(tmp)


final =[]

for i in new:
    flag =True
    count =0 
    for j in i:
        try:
            if dic[j]:
                count+=1
                pass

        except:
            flag=False
            count=0
            break
    if flag and count==5:
        final.append(i)

final2 = [tuple(sorted(i)) for i in final]   

solution = list(set(final2))
print(solution)

there will be 945 such values pair that exist this way

Piglet