create table emp (
  ename varchar2(100),
  mgrname varchar2(100
);
      

        select 'TRUE' from tcemp where ename='SCOTT' and mgrname='KING'
      

        with tcemp as （select ename,mgrname from tcemp 
union select tcemp.ename,emp.mgrname from tcemp, emp where tcemp.mgrname = emp.ename) 
select 'TRUE' from tcempwhere ename = 'SCOTT' and mgrname = 'KING';
      

        select 'TRUE' from (  select ename from emp  connect by prior mgrname = ename  
start with ename = 'SCOTT') where ename = 'KING';
      

        select e1.ename from emp e1, emp e2 where e2.path like e1.path || '%' and e2.ename='FORD'
      

        select e1.ename from emp e1, emp e2 where e1.path like e2.path || '%' and e2.ename='JONES'
      

        select e1.ename from emp e1, emp e2 where e2.path > e1.path and e2.path < e1.path || 'Z' 
and e2.ename='FORD'
      

        select e1.ename from emp e1 where e2.path>e1.path and e2.path<e1.path || 'Z'
      

        select e1.ename from emp where e1.path in ('1.1','1')
      

        select p2.emp from Personnel p1, Personnel p2where p1.lft between p2.lft and 
p2.rgtand p1.emp = 'Chuck'
      

每個(gè)節(jié)點(diǎn)[lft,rgt]在二維圓錐形里有它的子節(jié)點(diǎn)邊界y>=lft&x<=rgt。且右區(qū)間邊界總是小于左區(qū)間，所有節(jié)點(diǎn)均不允許超過(guò)對(duì)角線y=x。

另一種方式看這幅圖片應(yīng)注意父節(jié)點(diǎn)的子類(lèi)中的一個(gè)孩子節(jié)點(diǎn)，無(wú)論何時(shí)一系列定義在圓錐形孩子y>=clft&x<=crgt的所有點(diǎn)是父節(jié)點(diǎn)y>=plft&x<=prgt的一個(gè)子集。這個(gè)子集與平面上的圓錐形的關(guān)系是一個(gè)偏序。

現(xiàn)在我們知道遵照樹(shù)節(jié)點(diǎn)的兩個(gè)制約因素，我將確切地描述如何在xy平面上放置他們。

映射(The Mapping)

---

樹(shù)根的選擇完全是隨意的：我們假定根節(jié)點(diǎn)為區(qū)間[0,1]。在我們幾何圖案的解釋中，所有樹(shù)節(jié)點(diǎn)屬于xy平面上的正方形單元下部的三角形。

我們會(huì)通過(guò)歸納來(lái)進(jìn)一步詳細(xì)的描述映射 。對(duì)于每個(gè)樹(shù)節(jié)點(diǎn)，讓我們首先在xy平面定義兩個(gè)重要的點(diǎn)。深度優(yōu)先會(huì)聚點(diǎn)是對(duì)角線與通過(guò)節(jié)點(diǎn)的垂直線之間的一個(gè)交叉點(diǎn)。例如，節(jié)點(diǎn)<x=1,y=1/2>的深度優(yōu)先會(huì)聚點(diǎn)為<x=1,y=1>。廣度優(yōu)先會(huì)聚點(diǎn)是對(duì)角線與通過(guò)這點(diǎn)的水平線之間的交叉點(diǎn)。例如， 點(diǎn)<x=1,y=1/2>的廣度優(yōu)先點(diǎn)為<x=1/2,y=1/2>。

現(xiàn)在，為每個(gè)父節(jié)點(diǎn)，我們定義首個(gè)孩子的位置為一個(gè)父親點(diǎn)和深度優(yōu)先會(huì)聚點(diǎn)之間中點(diǎn)的一半。那么，每個(gè)兄弟節(jié)點(diǎn)被定義為一個(gè)前兄弟點(diǎn)和廣度優(yōu)先會(huì)聚點(diǎn)中點(diǎn)的一半：

例如，節(jié)點(diǎn)2.1的位置在x=1/2, y=3/8。

現(xiàn)在映射定義了，很顯然我們正使用密集型域：它既不是有理數(shù)，也不是實(shí)數(shù)，而是一對(duì)分?jǐn)?shù)(當(dāng)然，盡管兩個(gè)前者已足夠)。

有趣的是，父節(jié)點(diǎn)"1.2"的子樹(shù)后代是節(jié)點(diǎn)"1.1"子樹(shù)的一個(gè)向下縮小的復(fù)制品。 同樣的，節(jié)點(diǎn)1.1的子樹(shù)是節(jié)點(diǎn)"1."樹(shù)的一個(gè)向下縮小的復(fù)制品，一個(gè)帶自相似性的結(jié)構(gòu)被稱(chēng)為分形圖。

規(guī)范化(Normalization)

---

接著，我們注意到x和y并沒(méi)有完全獨(dú)立。假如知道他們的和，就能知道x和y兩者是什么？給出有理數(shù)的分子和分母代表節(jié)點(diǎn)坐標(biāo)的和，我們能計(jì)算x和y坐標(biāo)追溯到：

        function x_numer( numer integer, denom integer )
RETURN integer IS
ret_num integer;
ret_den integer;
BEGIN
ret_num := numer+1;
ret_den := denom*2;
while floor(ret_num/2) = ret_num/2 loop
ret_num := ret_num/2;
ret_den := ret_den/2;
end loop;
RETURN ret_num;
END;

function x_denom( numer integer, denom integer )
...
RETURN ret_den;
END;
      

      function y_numer( numer integer, denom integer )
RETURN integer IS
num integer;
den integer;
BEGIN
num := x_numer(numer, denom);
den := x_denom(numer, denom);
while den < denom loop
num := num*2;
den := den*2;
end loop;
num := numer - num;
while floor(num/2) = num/2 loop
num := num/2;
den := den/2;
end loop;
RETURN num;
END;

function y_denom( numer integer, denom integer )
...
RETURN den;
END;
    

      select x_number(39,32)||'/'||x_denom(39,32),
y_number(39,32)||'/'||y_denom(39,32) from dual
5/8  19/32

select 5/8+19/32,39/32 from dual
1.21875 1.21875
    

      function parent_numer( numer integer, denom integer )
RETURN integer IS
ret_num integer;
ret_den integer;
BEGIN
if numer=3 then
return NULL;
end if;
ret_num := (numer-1)/2;
ret_den := denom/2;
while floor((ret_num-1)/4) = (ret_num-1)/4 loop
ret_num := (ret_num+1)/2;
ret_den := ret_den/2;
end loop;
RETURN ret_num;
END;

function parent_denom( numer integer, denom integer )
...
RETURN ret_den;
END;
    

      select parent_numer(27,32)||'/'||parent_denom(27,32) from dual
7/8
    

      function sibling_number( numer integer, denom integer )
RETURN integer IS
  ret_num integer;
  ret_den integer;
  ret integer;
BEGIN
  if numer=3 then
    return NULL;
  end if;
  ret_num := (numer-1)/2;
  ret_den := denom/2;
  ret := 1;
while floor((ret_num-1)/4) = (ret_num-1)/4 loop
  if ret_num=1 and ret_den=1 then
  return ret;
end if;
ret_num := (ret_num+1)/2;
ret_den := ret_den/2;
ret := ret+1;
end loop;
RETURN ret;
END;
    

      select sibling_number(7,8) from dual;
             1
    

      function path( numer integer, denom integer )
RETURN varchar2 IS
BEGIN
if numer is NULL then
return '';
end if;
RETURN path(parent_numer(numer, denom),
parent_denom(numer, denom))
|| '.' || sibling_number(numer, denom);
END;

select path(15,16) from dual

.2.1.1
    

      function distance( num1 integer, den1 integer, num2 integer, den2 integer )
RETURN integer IS
BEGIN
 if num1 is NULL then
 return -999999;
end if;
 if num1=num2 and den1=den2 then
  return 0;
end if;
 RETURN 1+distance(parent_numer(num1, den1), parent_denom(num1, den1), num2,den2);
END;

select distance(27,32,3,4) from dual
 2 
    

      function child_numer
( num integer, den integer, child integer )
RETURN integer IS
BEGIN
RETURN num*power(2, child)+3-power(2, child);
END;

function child_denom
( num integer, den integer, child integer )
RETURN integer IS
BEGIN
RETURN den*power(2, child);
END;

select child_numer(3,2,3) || '/' ||
child_denom(3,2,3) from dual

19/16

    

      function path_numer( path varchar2 )
RETURN integer IS
 num integer;
 den integer;
 postfix varchar2(1000);
 sibling varchar2(100);
BEGIN
   num := 1;
   den := 1;
   postfix := '.' || path || '.';
while length(postfix) > 1 loop
  sibling := substr(postfix, 2, instr(postfix,'.',2)-2);
  postfix := substr(postfix, instr(postfix,'.',2), length(postfix) -instr(postfix,'.',2)+1);
  num := child_numer(num,den,to_number(sibling));
  den := child_denom(num,den,to_number(sibling));
  end loop;
 RETURN num;
END;

 function path_denom( path varchar2 ) ...
  RETURN den;
 END;

select path_numer('2.1.3') || '/' || path_denom('2.1.3') from dual
 51/64 
    

      create table emps (
name varchar2(30),
numer integer,
denom integer
)

alter table emps
ADD CONSTRAINT uk_name UNIQUE (name) USING INDEX
(CREATE UNIQUE INDEX name_idx on emps(name))
ADD CONSTRAINT UK_node
UNIQUE (numer, denom) USING INDEX
(CREATE UNIQUE INDEX node_idx on emps(numer, denom))
    

      insert into emps values ('KING',
path_numer('1'),path_denom('1'));
insert into emps values ('JONES',
path_numer('1.1'),path_denom('1.1'));
insert into emps values ('SCOTT',
path_numer('1.1.1'),path_denom('1.1.1'));
insert into emps values ('ADAMS',
path_numer('1.1.1.1'),path_denom('1.1.1.1'));
insert into emps values ('FORD',
path_numer('1.1.2'),path_denom('1.1.2'));
insert into emps values ('SMITH',
path_numer('1.1.2.1'),path_denom('1.1.2.1'));
insert into emps values ('BLAKE',
path_numer('1.2'),path_denom('1.2'));
insert into emps values ('ALLEN',
path_numer('1.2.1'),path_denom('1.2.1'));
insert into emps values ('WARD',
path_numer('1.2.2'),path_denom('1.2.2'));
insert into emps values ('MARTIN',
path_numer('1.2.3'),path_denom('1.2.3'));
insert into emps values ('TURNER',
path_numer('1.2.4'),path_denom('1.2.4'));
insert into emps values ('CLARK',
path_numer('1.3'),path_denom('1.3'));
insert into emps values ('MILLER',
path_numer('1.3.1'),path_denom('1.3.1'));
commit;
    

      create or replace
view hierarchy as
select name, numer, denom,
y_numer(numer,denom) numer_left,
y_denom(numer,denom) denom_left,
x_numer(numer,denom) numer_right,
x_denom(numer,denom) denom_right,
path (numer,denom) path,
distance(numer,denom,3,2) depth
from emps