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我正在寻找一种算法来为一组团队生成时间表。例如，想象一个赛季，其中每支球队都互相比赛，一次是作为主队，另一次是作为客队在另一支球队的场地上比赛。

生成本赛季所有比赛的集合很容易，如果球队是球队的列表，则可以这样做：

set((x, y) for x in teams for y in teams if x != y)

但我也想按时间顺序排列游戏，使其满足有效游戏时间表的约束并且看起来“自然随机”。

约束是游戏列表应该可以分成多轮，每轮由 n / 2 场比赛组成（其中 n 是球队的数量），其中每支球队与另一支球队配对。

为了使赛程看起来更自然，两支球队不应在连续回合中两次面对对方。也就是说，如果（a，b）在一轮中进行，则游戏（b，a）不应该在下一轮中进行。

此外，每支球队都应该尽可能地每隔一轮比赛作为客队，另一轮作为主队。我认为不可能总是满足这个约束，所以拥有东西更好。例如，一支球队不应该先打 8 场主场比赛，然后再打 8 场客场比赛。

下面是我现在得到的。该算法的主要问题是它经常卡在while循环中。特别是当团队数量为 16 或更多时。它也非常低效，因为它建立在使用随机样本函数的基础上，并希望得到正确的结果：

from random import sample
def season_schedule_order(teams, pairs):
    n_games_per_round = len(teams) // 2
    last_pairs = set()
    while pairs:
        r_pairs = set(sample(pairs, n_games_per_round))
        # Check that each team is present once in the round.
        r_teams = set(x for (x, y) in r_pairs) | set(y for (x, y) in r_pairs)
        if r_teams != teams:
            continue
        # Check that two teams doesn't face each other again.
        rev_pairs = set((y, x) for (x, y) in r_pairs)
        if rev_pairs & last_pairs:
            continue
        pairs -= r_pairs
        for p in r_pairs:
            yield p
        last_pairs = r_pairs

teams = set(['aik', 'djurgarden', 'elfsborg', 'gais',
             'gefle', 'hacken', 'halmstad', 'helsingborg'])
pairs = set((x, y) for x in teams for y in teams if x != y)
for (ht, at) in season_schedule_order(teams, pairs):
    print '%-20s %-20s' % (ht, at)
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2 回答 2

2

我在这里找到了一种 我稍微适应的方法：

def round_robin(units, sets = None):
    """ Generates a schedule of "fair" pairings from a list of units """
    count = len(units)
    sets = sets or (count - 1)
    half = count / 2
    for turn in range(sets):
        left = units[:half]
        right = units[count - half - 1 + 1:][::-1]
        pairings = zip(left, right)
        if turn % 2 == 1:
            pairings = [(y, x) for (x, y) in pairings]
        units.insert(1, units.pop())
        yield pairings

teams = ['a', 'b', 'c', 'd']
print list(round_robin(teams, sets = len(teams) * 2 - 2))

现在我只需要把它变成 plpgsql。:)

于 2011-05-06T16:36:07.463 回答
1 
REQUIREMENTS for the BALANCED ROUND ROBIN algorithm
The requirements of the algorithm can be defined by these four rules:
 1) All versus all
 Each team must meet exactly once, and once only, the other teams in the division league. 
 If the division is composed of n teams, the championship takes place in the n-1 rounds.
2) Alternations HOME / AWAY rule
The sequence of alternations HOME / AWAY matches for every teams in the division league, should be retained if possible. 
For any team in the division league at most once in the sequence of consecutive matches HAHA, occurs the BREAK of the rhythm, i.e. HH or AA match in the two consecutive rounds.
3) The rule of the last slot number
The team with the highest slot number must always be positioned in the last row of the grid. 
For each subsequent iteration the highest slot number of grid alternates left and right position; left column (home) and right (away).
The system used to compose the league schedule is "counter-clockwise circuit." 
In the construction of matches in one round of the championship, a division with an even number of teams. 
If in a division is present odd number of teams, it will be inserted a BYE/Dummy team in the highest slot number of grid/ring.
4) HH and AA are non-terminal and not initial
 Cadence HH or AA must never happen at the beginning or at the end of the of matches for any team in the division.
 Corrective inversion RULE performs only once, in the bottom line in the RING, LeftRight redundant inversion flip-flop RULE, so we will never obtain in the last two rounds CC or FF.
Round Robin ALGORITHM
The algorithm that satisfies rule (1) is obtained with a simple algorithm Round Robin:
where every successive round is obtained applying to the slot numbers ring a   "counterclockwise" rotation.
To satisfy the rule (2) we must "improve", the simple round robin algorithm by performing balancing of Home and Away sequence. 
It is performed by applying several "counterclockwise" rotations in the slot numbers ring in order to obtain acceptable combinations of slot positions for the next round.

The number of rotations required in the ring is (n / 2 -1).
So we will get that in the two successive rounds, almost all the teams playing at home in the previous round, will play away from home in the next round.
DATA STRUCTURE
n_teams: (4,6,8,..,20)
n_rounds:  n_teams -1;

环——电路环就是这样的数据结构，即一种特定类型的序列，可以执行并在算法上正式定义：环元素之间的（逆时针）旋转操作和INSERT_LAST_TEAM操作，环内容定义了每个特定轮次的所有匹配项在比赛日程中 Ring.Length 是一个偶数，等于 n_teams Left_ring 子序列：元素从 1 到 n/2 的 Ring 子序列。Right_ring 子序列：元素从 n/2+1 到 n 的环子序列。匹配 [i] = Ring_element [ i ] : Ring_element [ n – i + 1 ]; 其中 i = 1 到 n/2 MATCH 是由两个 TEAMS 组成的有序对，(Home_team : Away_team)。第一轮主场：客场 01:07 02:06 03:05 04:08

Next_Iteration is obtained by applying these rules:
a)   perform (n/2 – 1) times * ANTI CLOCKWISE ROTATIONs  
b)   all right column elements, below the last team, will be shifted upwards 
c)   INSERT_LAST_TEAM 
(Insert_Last_Element_into_ Ring_onLeft (n) or 
 Insert_Last_Element_into_Ring_onRight(n), alternatively)
d.1) Last Line LeftRight redundant swap RULE 
eventually have to swap elements in the last row once again
d.1) last team left/right - flip/flop



In the following example it will be explained how the Ring elements of 8 teams 
are gradually transformed in five steps, 
from from the SECOND ROUND into the THIRD ROUND.
If we start from from the SECOND ROUND:
05 :04
06 :03
07 :02
08 :01
a. After we perform THREE Anti-clockwise rotations, (n =8, 3 = 8/2 -1)
we will get the situation like this:
02 :01
03 :08
04 :07
05 :06
b. When we apply the LAST SLOT RULE: 
the right column elements (06,07) below last team (08) will be shifted upwards
02 :01
03
04 :07
05 :06
c. And now we will apply the LAST SLOT RULE-bottom right, 
we will get the situation which describes the THIRD ROUND: 
(08 must be moved to the bottom right position)
02 :01
03 :07
04 :06
05 :08
d.1 Now it will be checked if for this iteration number (i.e. round) 
depending on CODE SIX or ZERO Cadence, we eventually have to 
swap elements in the last row redundantly,
Left/Right swaping
d.2 And at the end we will apply the LAST TEAM SLOT ROULE
swap left & right elements, 
if in the previous iteration last team was positioned on the right, 
in this iteration it should be positioned on the left bottom position, 
so the last line elements will be swapped, else do nothing.
THIRD ROUND:
02 :01
03 :07
04 :06
05 :08

阅读更多

这是 Perl 中的代码片段。

于 2012-08-19T06:02:43.420 回答