北师大版三年级上册数学柯克曼女生问题

柯克曼女生问题
有一个学校有15个女生，她们每天要做三人行的散步，要使每个女生在一周内的每天做三人行散步时，与其她同学在组成三人小组同行时，彼此只有一次相遇在同一小组，应怎样安排？
这个问题是英国数学家柯克曼（1806～1895）于1850年提出，下面介绍一位英国牧师Andrew Frost的解答。

设15位女生用下面15个符号表示：x, a
1, a
2
, b
1
, b
2
, c
1
, c
2
, d
1
, d
2
, e
1
,
e 2, f
1
, f
2
, g
1
, g
2
;将它们排成七行，每天五个三人行小组（共十五人），使ｘ处
于七行中的最前一位置上：(x,a
1,a
2
); (x,b
1
,b
2
); (x,c
1
,c
2
); (x,d
1
,d
2
); (x,e
1
,e
2
);
(x,f
1,f
2
); (x,g
1
,g
2
).
于是只须分配14个元素，再每一行中，后继三人行小组，即对有下标的七个元素a，b，c，d，e，f，g进行三元素组合，填入每行，但每个字母只许出项两次。

即
Sunday: (x,a,a), (b,d,f), (b,e,g), (c,d,g), (c,e,f);
Monday: (x,b,b), (a,b,e), (a,f,g), (c,d,g), (c,e,f);
Tuesday: (x,c,c), (a,d,e), (a,f,g), (b,d,f),(b,e,g);
Wednsday:(x,d,d), (a,b,c), (a,f,g), (b,e,g),(c,e,f);
Thursday: (x,e,e), (a,b,c), (a,f,g), (b,d,f), (c,d,g)
Friday: (x,f,f), (a,b,c), (a,d,e), (b,e,g), (c,d,g);
Saturday:(x,g,g), (a,b,c), (a,d,e), (b,d,f), (c,e,f)
现在来填下标，如果在同一行中，可以有两个相同字母，例如在第三行中
bdf,beg中，b出现两次，可标上不同的脚标b
1,b
2
;若每一个“三人行”，有两个
脚标已定，则在同一行，别的三人行组不能再用；若不是由两种原则定出脚标，就定为１。

得到解：
Sunday: (x,a
1,a
2
), (b
1
,d
1
,f
1
), (b
2
,e
1
,g
1
), (c
1
,d
2
,g
2
), (c
2
,e
2
,f
2
);
Monday: (x,b
1,b
2
), (a
1
,b
2
,e
2
), (a
2
,f
2
,g
2
), (c
1
,d
1
,g
1
), (c
2
,e
1
,f
1
);
Tuesday: (x,c
1,c
2
), (a
1
,d
1
,e
1
), (a2,f
1
,g
1
), (b
1
,d
2
,f
2
),(b
2
,e
2
,g
2
);
Wednsday:(x,d
1,d
2
), (a
1
,b
2
,c
2
), (a
2
,f
2
,g1), (b
2
,e
1
,g
2
),(c1,e
2
,f
1
);
Thursday: (x,e
1,e
2
), (a
1
,b
1
,c
1
), (a
2
,f
1
,g
2
), (b
2
,d
1
,f
2
), (c
2
,d
2
,g
1
)
Friday: (x,f
1,f
2
), (a1,b
2
,c
1
), (a
2
,d
2
,e
1
), (b
1
,e
2
,g1), (c
2
,d
1
,g
2
);
Saturday:(x,g
1,g
2
), (a
1
,b
1
,c
2
), (a
2
,d
1
,e
2
), (b
2
,d
2
,f
1
), (c
1
,e
1
,f
2
)。