2022 PUZZLEUP

〇〇 · 发表于 2022-12-9 07:19

五个字母两行的情况，必须每行五个不同的字母，所以根据抽屉原则最多五列，也能达到五列，比如AA BB CC DD EE

〇〇 · 发表于 2022-12-9 07:31

三行的情况，在上述两行的基础上，第三行可以任选，所以五列可以与5种字母笛卡尔积，得到25列符合要求的解，如果再加1列，必有一个字母在某行出现6列，把这6列拎出来，去掉出现6列那行，剩下两行，根据上一回帖，它不符合要求，所以整个三行26列不符合要求，

〇〇 · 发表于 2022-12-9 09:08

上面一帖符合要求的25列还需要人工构造

〇〇 · 发表于 2022-12-9 15:45

用python构造5个字母2位总长度2行不同
import numpy as np
from communities.algorithms import bron_kerbosch

base=5
def bj(x,y):
      r=0
      while(x | y):
            r+=(x%base)!=(y%base)
            x//=base
            y//=base
      return r

def f_bk(rows,m):
  n=pow(base,rows)
  adj_matrix = np.zeros((n, n), dtype='int8')
  for i in range(n):
for j in range(n):
   cnt1=bj(i,j)
   if cnt1>=m: adj_matrix[ i][j]=1
  print(adj_matrix)
  communities = bron_kerbosch(adj_matrix)
  longest=0
  result=[]
  for i in communities:
if len(i)>longest:
   longest=len(i)
   result=i
  return (longest,result)

f_bk(2,2)

(5, {0, 18, 6, 24, 12})

〇〇 · 发表于 2022-12-9 15:51

本帖最后由〇〇于 2022-12-9 16:43 编辑

〇〇发表于 2022-12-9 15:45
用python构造5个字母2位总长度2行不同import numpy as npfrom communities.algorithms import bron_kerbosc ...

base=3
f_bk(3,2)
(9, {0, 4, 8, 10, 14, 15, 20, 21, 25})
select i,[i/pow(3,3-x)::int%3 for x in range(1,3+1)] AS a from unnest([0, 4, 8, 10, 14, 15, 20, 21, 25])t(i);
┌───────┬───────────┐
│ i │    a    │
│ int32 │  int32[]  │
├───────┼───────────┤
│    0 │ [0, 0, 0] │
│    4 │ [0, 1, 1] │
│    8 │ [0, 2, 2] │
│ 10 │ [1, 0, 1] │
│ 14 │ [1, 1, 2] │
│ 15 │ [1, 2, 0] │
│ 20 │ [2, 0, 2] │
│ 21 │ [2, 1, 0] │
│ 25 │ [2, 2, 1] │
└───────┴───────────┘

〇〇 · 发表于 2022-12-9 17:33

本帖最后由〇〇于 2022-12-9 17:58 编辑

按照前一帖构造了4个字母3行取2
with t(i) as(values
([0, 0, 0]),
([0, 1, 1]),
([0, 2, 2]),
([0, 3, 3]),
([1, 0, 1]),
([1, 1, 2]),
([1, 2, 3]),
([1, 3, 0]),
([2, 0, 2]),
([2, 1, 3]),
([2, 2, 0]),
([2, 3, 1])),
p as(
select a.i n1,b.i n2,count(*) from t a,t b , generate_series(1,3)q(x) where a.i[x]<>b.i[x] group by a.i,b.i having count(*)>=2)
select count(*) from p;
┌──────────────┐
│ count_star() │
│ int64    │
├──────────────┤
│       132 │
└──────────────┘
构造了5个字母3行取2
with t(i) as(values
([0, 0, 0]),
([0, 1, 1]),
([0, 2, 2]),
([0, 3, 3]),
([0, 4, 4]),
([1, 0, 1]),
([1, 1, 2]),
([1, 2, 3]),
([1, 3, 4]),
([1, 4, 0]),
([2, 0, 2]),
([2, 1, 3]),
([2, 2, 4]),
([2, 3, 0]),
([2, 4, 1]),
([3, 0, 3]),
([3, 1, 4]),
([3, 2, 0]),
([3, 3, 1]),
([3, 4, 2]),
([4, 0, 4]),
([4, 1, 0]),
([4, 2, 1]),
([4, 3, 2]),
([4, 4, 3])),
p as(
select a.i n1,b.i n2,count(*) from t a,t b , generate_series(1,3)q(x) where a.i[x]<>b.i[x] group by a.i,b.i having count(*)>=2)
select count(*) from p;
┌──────────────┐
│ count_star() │
│ int64    │
├──────────────┤
│       600 │
└──────────────┘

〇〇 · 发表于 2022-12-10 11:55

本帖最后由〇〇于 2022-12-10 12:12 编辑

在https://gpt.chatapi.art/提问，结果是20,再问，又变成120

〇〇 · 发表于 2022-12-10 12:18

本帖最后由〇〇于 2022-12-10 12:22 编辑

看来是因为我没有贴全例子，引起歧义
用中文版问回答质量就更差了

solomon_007 · 发表于 2022-12-12 11:56

第 8 题

421 个

create or replace package body pkg_str_code is

  function str_diff_cnt (p_code1 in varchar2,p_code2 in varchar2)
  return pls_integer
  is
l_cnt pls_integer := 0;
l_ret pls_integer := 0;
  begin
for i in 1..5 loop
   if substr(p_code1,i,1) <> substr(p_code2,i,1) then
      l_cnt := l_cnt + 1;
   end if;
end loop;

if l_cnt > 1 then
   l_ret := 1;
else
   l_ret := 0;
end if;

return l_ret;

  end;

  procedure five_letter_code
  is
type aa_code is table of varchar2(5) index by pls_integer;
l_code aa_code;
l_result aa_code;

l_cnt pls_integer := 0;

  begin
with t(c) as (select chr(64+level) from dual connect by level<=5),
   s(n,str_code) as (select 1,c from t
                     union all
                     select n+1,
                           str_code||c
                     from s,t
                     where n < 5)
select str_code
   bulk collect into l_code
   from s
   where n = 5
   order by 1;

for i in 1..l_code.count loop

   if i=1 then
      l_result(i) := l_code(i);
   end if;

   l_cnt := 0;
   for j in 1..l_result.count loop

      if str_diff_cnt(l_result(j),l_code(i)) = 1 then
      l_cnt := l_cnt + 1;
      end if;

   end loop;

   if l_cnt = l_result.count then
      l_result(l_result.count + 1) := l_code(i);
   end if;

end loop;

dbms_output.put_line(l_result.count);

--for k in 1..l_result.count loop
--  dbms_output.put_line(l_result(k));
--end loop;

  end;

end;

solomon_007 · 发表于 2022-12-12 11:57

SQL> exec pkg_str_code.five_letter_code;

AAAAA
AAABB
AAACC
AAADD
AAAEE
AABAB
AABBA
AABCD
AABDC
AACAC
AACBD
AACCA
AACDB
AADAD
AADBC
AADCB
AADDA
AAEAE
AAEEA
ABAAB
ABABA
ABACD
ABADC
ABBAA
ABBBB
ABBCC
ABBDD
ABBEE
ABCAD
ABCBC
ABCCB
ABCDA
ABDAC
ABDBD
ABDCA
ABDDB
ABEBE
ABEEB
ACAAC
ACABD
ACACA
ACADB
ACBAD
ACBBC
ACBCB
ACBDA
ACCAA
ACCBB
ACCCC
ACCDD
ACCEE
ACDAB
ACDBA
ACDCD
ACDDC
ACECE
ACEEC
ADAAD
ADABC
ADACB
ADADA
ADBAC
ADBBD
ADBCA
ADBDB
ADCAB
ADCBA
ADCCD
ADCDC
ADDAA
ADDBB
ADDCC
ADDDD
ADDEE
ADEDE
ADEED
AEAAE
AEAEA
AEBBE
AEBEB
AECCE
AECEC
AEDDE
AEDED
AEEAA
AEEBB
AEECC
AEEDD
AEEEE
BAAAB
BAABA
BAACD
BAADC
BABAA
BABBB
BABCC
BABDD
BABEE
BACAD
BACBC
BACCB
BACDA
BADAC
BADBD
BADCA
BADDB
BAEBE
BAEEB
BBAAA
BBABB
BBACC
BBADD
BBAEE
BBBAB
BBBBA
BBBCD
BBBDC
BBCAC
BBCBD
BBCCA
BBCDB
BBDAD
BBDBC
BBDCB
BBDDA
BBEAE
BBEEA
BCAAD
BCABC
BCACB
BCADA
BCBAC
BCBBD
BCBCA
BCBDB
BCCAB
BCCBA
BCCCD
BCCDC
BCDAA
BCDBB
BCDCC
BCDDD
BCDEE
BCEDE
BCEED
BDAAC
BDABD
BDACA
BDADB
BDBAD
BDBBC
BDBCB
BDBDA
BDCAA
BDCBB
BDCCC
BDCDD
BDCEE
BDDAB
BDDBA
BDDCD
BDDDC
BDECE
BDEEC
BEABE
BEAEB
BEBAE
BEBEA
BECDE
BECED
BEDCE
BEDEC
BEEAB
BEEBA
BEECD
BEEDC
CAAAC
CAABD
CAACA
CAADB
CABAD
CABBC
CABCB
CABDA
CACAA
CACBB
CACCC
CACDD
CACEE
CADAB
CADBA
CADCD
CADDC
CAECE
CAEEC
CBAAD
CBABC
CBACB
CBADA
CBBAC
CBBBD
CBBCA
CBBDB
CBCAB
CBCBA
CBCCD
CBCDC
CBDAA
CBDBB
CBDCC
CBDDD
CBDEE
CBEDE
CBEED
CCAAA
CCABB
CCACC
CCADD
CCAEE
CCBAB
CCBBA
CCBCD
CCBDC
CCCAC
CCCBD
CCCCA
CCCDB
CCDAD
CCDBC
CCDCB
CCDDA
CCEAE
CCEEA
CDAAB
CDABA
CDACD
CDADC
CDBAA
CDBBB
CDBCC
CDBDD
CDBEE
CDCAD
CDCBC
CDCCB
CDCDA
CDDAC
CDDBD
CDDCA
CDDDB
CDEBE
CDEEB
CEACE
CEAEC
CEBDE
CEBED
CECAE
CECEA
CEDBE
CEDEB
CEEAC
CEEBD
CEECA
CEEDB
DAAAD
DAABC
DAACB
DAADA
DABAC
DABBD
DABCA
DABDB
DACAB
DACBA
DACCD
DACDC
DADAA
DADBB
DADCC
DADDD
DADEE
DAEDE
DAEED
DBAAC
DBABD
DBACA
DBADB
DBBAD
DBBBC
DBBCB
DBBDA
DBCAA
DBCBB
DBCCC
DBCDD
DBCEE
DBDAB
DBDBA
DBDCD
DBDDC
DBECE
DBEEC
DCAAB
DCABA
DCACD
DCADC
DCBAA
DCBBB
DCBCC
DCBDD
DCBEE
DCCAD
DCCBC
DCCCB
DCCDA
DCDAC
DCDBD
DCDCA
DCDDB
DCEBE
DCEEB
DDAAA
DDABB
DDACC
DDADD
DDAEE
DDBAB
DDBBA
DDBCD
DDBDC
DDCAC
DDCBD
DDCCA
DDCDB
DDDAD
DDDBC
DDDCB
DDDDA
DDEAE
DDEEA
DEADE
DEAED
DEBCE
DEBEC
DECBE
DECEB
DEDAE
DEDEA
DEEAD
DEEBC
DEECB
DEEDA
EAAAE
EAAEA
EABBE
EABEB
EACCE
EACEC
EADDE
EADED
EAEAA
EAEBB
EAECC
EAEDD
EAEEE
EBABE
EBAEB
EBBAE
EBBEA
EBCDE
EBCED
EBDCE
EBDEC
EBEAB
EBEBA
EBECD
EBEDC
ECACE
ECAEC
ECBDE
ECBED
ECCAE
ECCEA
ECDBE
ECDEB
ECEAC
ECEBD
ECECA
ECEDB
EDADE
EDAED
EDBCE
EDBEC
EDCBE
EDCEB
EDDAE
EDDEA
EDEAD
EDEBC
EDECB
EDEDA
EEAAA
EEABB
EEACC
EEADD
EEAEE
EEBAB
EEBBA
EEBCD
EEBDC
EECAC
EECBD
EECCA
EECDB
EEDAD
EEDBC
EEDCB
EEDDA
EEEAE
EEEEA

PL/SQL procedure successfully completed

SQL> exec pkg_str_code.five_letter_code;

421

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