Part One
根据电脑两两连接的信息，类似a<->b, c<->b，推断出三个电脑互相连接的场景
首先拆分出两两相连的情况，为了方便计算，强行让左侧小于右侧，再通过自关联求出三个电脑互联的场景
with origin AS (
SELECT least(split_part(line, '-', 1), split_part(line, '-', 2)) as l,
greatest(split_part(line, '-', 1), split_part(line, '-', 2)) as r
FROM lance_input
)
SELECT count(*)
FROM (
SELECT distinct a.l, a.r, c.r
FROM origin a
JOIN origin b
ON a.r = b.l
JOIN origin c
ON a.l = c.l
AND b.r = c.r
WHERE left(a.l, 1) = 't'
OR left(a.r, 1) = 't'
OR left(c.r, 1) = 't'
) t;
Part Two
一个LAN中可能不止三台电脑，求出一个LAN中最多的电脑组合
那么如何确定一个LAN中每台电脑都互联呢？上一题中3台电脑就需要关联3次，来确保每台都互联。电脑越多，需要关联的次数越多，计算量太大。
目前已知的条件，可以看作是有很多2台电脑的LAN，2台电脑LAN组成3台的时候，比如{a,c}与{c,f}，只需要确保a和f互联，同理可推导出两个N-1的LAN，组成N的LAN，只需要校验a1与bn的互联即可。
{a1, xx, xx, xx, xx, xx}
{xx, xx, xx, xx, xx, bn}
{a1, xx, xx, xx, xx, xx, bn}
with recursive origin AS (
SELECT least(split_part(line, '-', 1), split_part(line, '-', 2)) as l,
greatest(split_part(line, '-', 1), split_part(line, '-', 2)) as r
FROM lance_input
), all_paths AS (
SELECT l, r, array[l, r] as path
FROM origin
UNION ALL
SELECT *
FROM (
WITH all_paths_inner AS (select * from all_paths)
SELECT a.l, c.r, a.path || c.r as path
FROM all_paths_inner a
JOIN all_paths_inner b
ON a.path[2:] = b.path[1:array_length(b.path,1) - 1]
JOIN origin c
ON a.l = c.l
AND b.r = c.r
) t
)
SELECT * FROM all_paths;
cu | wb | {cu,iv,ll,oh,oi,oj,qu,qw,uu,vb,vx,wb}
gy | xx | {gy,jw,jx,kb,lo,mu,pa,pk,ty,vr,xo,xx}
gy | xx | {gy,jx,kb,lo,mu,no,pa,pk,ty,vr,xo,xx}
ix | xm | {ix,jy,ls,ly,or,ow,pf,qq,qy,rk,ue,xm}
ix | xm | {ix,jy,ls,ly,or,ow,pf,qq,rk,ue,xe,xm}
ch | zk | {ch,cz,di,gb,ht,ku,lu,tw,vf,vt,wo,xz,zk}
(247547 rows)