Sample solutions and discussion
Perl 'Medium' Quiz of the Week 2005-1
- Written by Shlomi Fish
The problem that was posted was:
<<<<<<<<<<<<<<<
You are given N sport teams (where N is even) who wish to compete against
each other. Each team should have a match against any other team once and
only once. Moreover, the tournament should take place in N-1 days, where
in
every day, each team plays once against some other team. (for a total of
N*(N-1)/2 matches, which is the number of possible matches).
Your mission is to write a Perl program that will allocate such a
schedule.
The function in question would be called allocate_schedule() and will
receive a single scalar parameter, which is the number of the teams. It
will return a reference to an array of day allocations. Each day
allocation
will be a reference to an array that for each index will specify the index
of the team that the team with the index in that array will compete
against
that day.
So for example:
allocate_schedule(2)
will return:
[[1, 0]];
allocate_schedule(4)
may return:
[[1,0,3,2],[2,3,0,1],[3,2,1,0]];
You can try solving the problem for an N that is a whole power of 2, as
this problem is easier.
>>>>>>>>>>>>>>>
1. The story behind the quiz (and a wrong solution):
When I took the course "Introduction to Data Structures and Algorithms"
in the Technion, we were given as our first computer assignment to solve
the problem for an N that was a whole power of 2, using the "Divide and
Conquer" method.
I thought I had a right solution which I'll explain below, that can work for
every N. However, it was not a divide and conquer one, so my lecturer
instructed me not to use it. Then when I talked to a fellow programmer he
told me that when he was at the Chess club, they used to perform such
full-coverage tournaments, by splitting the teams into two, and performing
a circular allocation for them. This gave me an idea for a divide and conquer
where I use this for N+2 days, and then recurse into both teams for the extra
N/2+1, and I implemented this.
When I proposed this quiz a few days ago, I tried to implement the solution
I had in mind, and discovered that it fails for all N's that are not powers
of 2 (like 6).
The solution goes like that:
On the (i-1)'th day, pair off teams 0 and teams i, teams 2*i and teams 3*i,
and so on, cyclically. If the greater common divisor of i and N is greater
than 1, also do so in offsets up to the gcd (exclusive). I.e: 1 and i+1,
2*i+1 and 3*i+1, etc.
The reason it doesn't work is because it's possible N is divided by i an odd
number of times. For example, if we take 6, and on day 1 pair 0 and
2, then 4 has no-one to compete against.
This solution still works for an N that is a power of 2, but not in the
general
case. Trying to figure out a different solution that works for every even
N took me several days of thinking, and one full day of implementation. But
as you'll soon see, it was an exceptionally complicated solution, and there
are considerably simpler solutions available.
2. Daniel Martin and I submitted automated test suites to test the validity of
the results produced by a correct solver:
http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2412
http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2411
3. Pr. Martin also submitted a very nice recursive solution:
http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2413
What he did was for 2*N where N is even, split the teams into half,
generate two schedules, and then put them side by side, and then perform
a cyclical allocation for the other days. For 2*N-2, he allocated two N
schedules, and then paired every team that competes against the extra "ghost"
team, to compete against the team from the other bracket instead. Then
he did a cyclical assignment only this time, without pairing off teams at
relatives offsets of 0. (which were already allocated previously).
I really like his solution, and was able to improve upon it slightly:
http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2423
4. Zsban Ambrus submitted a very short (but correct) solution which he
implemented in Ruby:
http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2408
I translated it to Perl:
http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2416
And Pr. Martin continued with a mathematical analysis of it:
http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2418
5. Jeffrey M. Vinocur submitted his own solution:
http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2409
Pr. Greg Matheson conjectured that it was in essence the same as Pr. Ambrus
solution and I was able to present a variation of Ambrus' code that produced
the same results as Vinocur's solution (without making use of the extra
arrays he kept):
http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2421
6. My solution was presented in all of its glory^W complexity here:
http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2415
To quote Linus Torvalds, it can be used to frighten small children. I handled
the 2*(2*N+1) case, by dividing according to a non-two prime divisor, which
resulted in a very complicated, but still correct solution.
7. Some solutions were sent by:
Walt Mankowski -
http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2422
Greg Matheson -
http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2410
Rod Adams -
http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2414
I did not took too close a look at them, but at least some of them
seem similar to Pr. Vinocur's solution.
8. Tor Fuglerud sent a solution that was not entirely to spec:
http://article.gmane.org/gmane.comp.lang.perl.qotw.discuss/2417
9. The obligatory benchmark:
Benchmarked on producing solutions for all numbers between 2 and 200:
Daniel Martin (modified) - 03.48
Daniel Martin (modified) - 03.42
Greg Matheson - 02.38
Jeffrey Vinocur - 00.94
Shlomi Fish - 12.00
Walt Mankowski - 01.20
Zsban Ambrus - 01.53
This is on a Mandrake Linux 10.1 Running on a Pentium 2.4 GHz machine.
I expected Pr. Ambrus' solution to be the fastest of the three, mine to
be the slowest, and Pr. Martin's to be somewhere in between, so the results
are not surprising. It also seems Pr. Vinocur's solution is the fastest. While
it was shown to be equivalent to Ambrus' solution, it is possibly faster,
because it doesn't make a heavy use of the modulo operation, which is costy.
Note that Pr. Ambrus solution was optimized, by putting one of the
for { push ... } loops into a map { } call.
The solution of Rob Adams was eliminated from the test, after witnessing that
it was extremely slow for large numbers for some reason.
10. Some more solutions for the Power of 2 case.
I can mention some more solutions, that work only on the power of 2 case. One
of them, is my cyclical allocation and recurse solution. Another one proposed
by a friend from the Technion who also studied the course in my semester, is
an
almost pure divide-only solution in which the teams are split into two, and at
some days compete against other, and the others against themselves.
And finally, Daniel Martin proposed here his own solution to this case using
XOR operations:
return mapr {
my($a) = $_;
mapr {$a^$_} [0..$n-1]
} [1..$n-1];
And naturally, all of the more generalized solutions work in this case.
11. Finale
I hope you had fun solving and discussing this quiz. So far it seems the
MJD-less quizzing has been a good idea, and there was already an easier quiz
posted, and more to come in the future.
---------------------------------------------------------------------
Shlomi Fish shlomif-ik1l9ssToec+JF/nGntIXQ@xxxxxxxxxxxxxxxx
Homepage: http://www.shlomifish.org/
Knuth is not God! It took him two days to build the Roman Empire.
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