On 09/09/2015 18:59, Oscar Benjamin wrote:
On 9 September 2015 at 12:05, Francesco Loffredo via Tutor
<tutor@python.org> wrote:
A quick solution is to add one "dummy" letter to the pool of the OP's
golfers.
I used "!" as the dummy one. This way, you end up with 101 triples, 11 of
which contain the dummy player.
But this is better than the 25-item pool, that resulted in an incomplete set
of triples (for example, A would never play with Z)
So, in your real-world problem, you will have 11 groups of 2 people instead
of 3. Is this a problem?
import pprint, itertools
pool = "abcdefghijklmnopqrstuvwxyz!"
def maketriples(tuplelist):
final = []
used = set()
for a, b, c in tuplelist:
if ( ((a,b) in used) or ((b,c) in used) or ((a,c) in used) or ((b,a)
in used) or ((c,b) in used) or ((c,a) in used) ):
continue
else:
final.append((a, b, c))
used.add((a,b))
used.add((a,c))
used.add((b,c))
used.add((b,a))
used.add((c,a))
used.add((c,b))
return final
combos = list(itertools.combinations(pool, 3))
print("combos contains %s triples." % len(combos))
triples = maketriples(combos)
print("maketriples(combos) contains %s triples." % len(triples))
pprint.pprint(triples)
I don't think the code above works. For n=27 it should count 117
(according to the formula I showed) but instead it comes up with 101.
I tried it with a smaller n by setting pool to range(1, 9+1) meaning
that n=9. The output is:
combos contains 84 triples.
maketriples(combos) contains 8 triples.
[(1, 2, 3),
(1, 4, 5),
(1, 6, 7),
(1, 8, 9),
(2, 4, 6),
(2, 5, 7),
(3, 4, 7),
(3, 5, 6)]
However I can construct a set of 12 triples containing each number
exactly 4 times which is the exact Steiner triple system:
1 6 8
1 2 3
1 5 9
1 4 7
2 6 7
2 4 9
2 5 8
3 5 7
3 6 9
3 8 4
4 5 6
7 8 9
This is the number of triples predicted by the formula: 9*(9-1)/6 = 12
--
Oscar
That's very interesting! This takes me to my question to Tutors:
what's wrong with the above code?
Francesco
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