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 Puzzle : Generating two unique numbers from an array of numbers (repetation not allowed) Rate Topic Display Mode Topic Options
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 Posted Wednesday, September 12, 2012 7:19 AM
 Grasshopper Group: General Forum Members Last Login: Friday, September 20, 2013 7:42 AM Points: 22, Visits: 1,102
 Hi All,(This is just for fun, so you may choose to ignore this topic.)Today I read a puzzle somewhere which is mentioned below.I thought why not solve it by some SQL query.Puzzle : Use the numbers 1,2,3,4,5,6,7,8,9 only once to fill in the blanks and complete the equation.(4 _ _ _ 6) - (_ _ _ _) = 33333As you see for the equation we have to generate 2 numbers, first a 3 digit number and second a 4 digit number and none of the digits in both these numbers can be repeated.So here is the query i tried and it worked fine.But i was wondering whether the negation joins i have used is the only way to achieve this?Is there any smarter/efficient way to crack this puzzle?`with a(num) as(select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9),b(num1,num2) as( select convert(bigint, '4' + convert(varchar,a.num) + convert(varchar,b.num) + convert(varchar,c.num) + '6') ,convert(bigint, convert(varchar,d.num) + convert(varchar,e.num) + convert(varchar,f.num) + convert(varchar,g.num)) from a join a as b on a.num <> b.num join a as c on a.num <> b.num and a.num <> c.num and b.num <> c.num join a as d on a.num <> b.num and a.num <> c.num and a.num <> d.num and b.num <> c.num and b.num <> d.num and c.num <> d.num join a as e on a.num <> b.num and a.num <> c.num and a.num <> d.num and a.num <> e.num and b.num <> c.num and b.num <> d.num and b.num <> e.num and c.num <> d.num and c.num <> e.num and d.num <> e.num join a as f on a.num <> b.num and a.num <> c.num and a.num <> d.num and a.num <> e.num and a.num <> f.num and b.num <> c.num and b.num <> d.num and b.num <> e.num and b.num <> f.num and c.num <> d.num and c.num <> e.num and c.num <> f.num and d.num <> e.num and d.num <> f.num and e.num <> f.num join a as g on a.num <> b.num and a.num <> c.num and a.num <> d.num and a.num <> e.num and a.num <> f.num and a.num <> g.num and b.num <> c.num and b.num <> d.num and b.num <> e.num and b.num <> f.num and b.num <> g.num and c.num <> d.num and c.num <> e.num and c.num <> f.num and c.num <> g.num and d.num <> e.num and d.num <> f.num and d.num <> g.num and e.num <> f.num and e.num <> g.num and f.num <> g.num)select * from b where num1-num2 = 33333order by 1,2`
Post #1357963
 Posted Wednesday, September 12, 2012 7:43 AM
 SSCrazy Group: General Forum Members Last Login: Wednesday, November 20, 2013 10:00 AM Points: 2,719, Visits: 4,724
 You don't really need last join as the last digit of the second number can only be '3' `with a(num) as(select 1 union select 2 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9),b(num1,num2) as( select convert(bigint, '4' + convert(varchar,a.num) + convert(varchar,b.num) + convert(varchar,c.num) + '6') ,convert(bigint, convert(varchar,d.num) + convert(varchar,e.num) + convert(varchar,f.num) + '3') from a join a as b on a.num <> b.num join a as c on a.num <> b.num and a.num <> c.num and b.num <> c.num join a as d on a.num <> b.num and a.num <> c.num and a.num <> d.num and b.num <> c.num and b.num <> d.num and c.num <> d.num join a as e on a.num <> b.num and a.num <> c.num and a.num <> d.num and a.num <> e.num and b.num <> c.num and b.num <> d.num and b.num <> e.num and c.num <> d.num and c.num <> e.num and d.num <> e.num join a as f on a.num <> b.num and a.num <> c.num and a.num <> d.num and a.num <> e.num and a.num <> f.num and b.num <> c.num and b.num <> d.num and b.num <> e.num and b.num <> f.num and c.num <> d.num and c.num <> e.num and c.num <> f.num and d.num <> e.num and d.num <> f.num and e.num <> f.num)select distinct *, num1-num2 from b where num1-num2 = 33333order by 1,2`Also, I thought that you shouldn't be able to use 4 and 6 as they were already used by the first number. In this case you will get only single possible result:`with a(num) as(select 1 union select 2 union select 5 union select 7 union select 8 union select 9),b(num1,num2) as( select convert(bigint, '4' + convert(varchar,a.num) + convert(varchar,b.num) + convert(varchar,c.num) + '6') ,convert(bigint, convert(varchar,d.num) + convert(varchar,e.num) + convert(varchar,f.num) + '3') from a join a as b on a.num <> b.num join a as c on a.num <> b.num and a.num <> c.num and b.num <> c.num join a as d on a.num <> b.num and a.num <> c.num and a.num <> d.num and b.num <> c.num and b.num <> d.num and c.num <> d.num join a as e on a.num <> b.num and a.num <> c.num and a.num <> d.num and a.num <> e.num and b.num <> c.num and b.num <> d.num and b.num <> e.num and c.num <> d.num and c.num <> e.num and d.num <> e.num join a as f on a.num <> b.num and a.num <> c.num and a.num <> d.num and a.num <> e.num and a.num <> f.num and b.num <> c.num and b.num <> d.num and b.num <> e.num and b.num <> f.num and c.num <> d.num and c.num <> e.num and c.num <> f.num and d.num <> e.num and d.num <> f.num and e.num <> f.num)select distinct *, num1-num2 from b where num1-num2 = 33333order by 1,2` _____________________________________________"The only true wisdom is in knowing you know nothing""O skol'ko nam otkrytiy chudnyh prevnosit microsofta duh!"(So many miracle inventions provided by MS to us...)How to post your question to get the best and quick help
Post #1357974
 Posted Wednesday, September 12, 2012 7:52 AM
 SSCrazy Group: General Forum Members Last Login: Wednesday, November 20, 2013 10:00 AM Points: 2,719, Visits: 4,724
 Also, you don't need to check if the same digits from different joins are not equal (eg. if checked a.num<>b.num in one "join" you don't need to check it again in other "joins"). Therefore all JOINs can be rewritten using "incremented" NOT IN:`with a(num) as(select 1 union select 2 union select 5 union select 7 union select 8 union select 9),b(num1,num2) as( select convert(bigint, '4' + convert(varchar,a.num) + convert(varchar,b.num) + convert(varchar,c.num) + '6') ,convert(bigint, convert(varchar,d.num) + convert(varchar,e.num) + convert(varchar,f.num) + '3') from a join a as b on a.num <> b.num join a as c on c.num NOT IN (a.num, b.num) join a as d on d.num NOT IN (a.num, b.num, c.num) join a as e on e.num NOT IN (a.num, b.num, c.num, d.num) join a as f on f.num NOT IN (a.num, b.num, c.num, d.num, e.num))select distinct *, num1-num2 from b where num1-num2 = 33333order by 1,2` _____________________________________________"The only true wisdom is in knowing you know nothing""O skol'ko nam otkrytiy chudnyh prevnosit microsofta duh!"(So many miracle inventions provided by MS to us...)How to post your question to get the best and quick help
Post #1357980
 Posted Wednesday, September 12, 2012 7:58 AM
 SSCoach Group: General Forum Members Last Login: Friday, November 01, 2013 1:55 PM Points: 15,442, Visits: 9,579
 Here's what I came up with. Does the same thing the same way, just another way to write it:`WITH Numbers AS (SELECT * FROM ( VALUES ( '1'), ( '2'), ( '3'), ( '4'), ( '5'), ( '6'), ( '7'), ( '8'), ( '9') ) AS Nums (Number)) SELECT N1.Number + N2.Number + N3.Number + N4.Number + N5.Number, N6.Number + N7.Number + N8.Number + N9.Number FROM Numbers AS N1 CROSS APPLY (SELECT Number FROM Numbers AS N2 WHERE N2.Number != N1.Number) AS N2 CROSS APPLY (SELECT Number FROM Numbers AS N3 WHERE N3.Number NOT IN (N1.Number, N2.Number)) AS N3 CROSS APPLY (SELECT Number FROM Numbers AS N4 WHERE N4.Number NOT IN (N1.Number, N2.Number, N3.Number)) AS N4 CROSS APPLY (SELECT Number FROM Numbers AS N5 WHERE N5.Number NOT IN (N1.Number, N2.Number, N3.Number, N4.Number)) AS N5 CROSS APPLY (SELECT Number FROM Numbers AS N6 WHERE N6.Number NOT IN (N1.Number, N2.Number, N3.Number, N4.Number, N5.Number)) AS N6 CROSS APPLY (SELECT Number FROM Numbers AS N7 WHERE N7.Number NOT IN (N1.Number, N2.Number, N3.Number, N4.Number, N5.Number, N6.Number)) AS N7 CROSS APPLY (SELECT Number FROM Numbers AS N8 WHERE N8.Number NOT IN (N1.Number, N2.Number, N3.Number, N4.Number, N5.Number, N6.Number, N7.Number)) AS N8 CROSS APPLY (SELECT Number FROM Numbers AS N9 WHERE N9.Number NOT IN (N1.Number, N2.Number, N3.Number, N4.Number, N5.Number, N6.Number, N7.Number, N8.Number)) AS N9 WHERE N1.Number = '4' AND N5.Number = '6' AND CAST(N1.Number + N2.Number + N3.Number + N4.Number + N5.Number AS INT) - CAST(N6.Number + N7.Number + N8.Number + N9.Number AS INT) = 33333` - Gus "GSquared", RSVP, OODA, MAP, NMVP, FAQ, SAT, SQL, DNA, RNA, UOI, IOU, AM, PM, AD, BC, BCE, USA, UN, CF, ROFL, LOL, ETCProperty of The Thread"Nobody knows the age of the human race, but everyone agrees it's old enough to know better." - Anon
Post #1357989
 Posted Wednesday, September 12, 2012 8:06 AM
 SSCrazy Group: General Forum Members Last Login: Wednesday, November 20, 2013 10:00 AM Points: 2,719, Visits: 4,724
 As said before, you don't need to guess the last digit of the second number as it can be only 3!Saves one cross-apply... _____________________________________________"The only true wisdom is in knowing you know nothing""O skol'ko nam otkrytiy chudnyh prevnosit microsofta duh!"(So many miracle inventions provided by MS to us...)How to post your question to get the best and quick help
Post #1357995
 Posted Wednesday, September 12, 2012 8:22 AM
 SSCommitted Group: General Forum Members Last Login: 2 days ago @ 10:24 PM Points: 1,872, Visits: 4,127
 Eugene Elutin (9/12/2012)You don't really need last join as the last digit of the second number can only be '3' Wouldn't that be cheating?Because that way I can get the best performance`SELECT 41286 num1, 7953 num2`I had my solution that was very similar to yours (again) but using the different(<>) operators instead of the NOT IN.However, a better performance would be using the best datatype for the digits (char) since the beginning avoiding additional conversions. Luis C.Please don't trust me, test the solutions I give you before using them.Forum Etiquette: How to post data/code on a forum to get the best help
Post #1358004
 Posted Wednesday, September 12, 2012 8:23 AM
 Grasshopper Group: General Forum Members Last Login: Friday, September 20, 2013 7:42 AM Points: 22, Visits: 1,102
 Thank you Eugene and GSquared.Eugene, the NOT IN looks like a much better/easier/readable join.GSquared i love your signature ROFL,LOL,ETC .... I don't think it can get any better.
Post #1358006
 Posted Wednesday, September 12, 2012 8:39 AM
 SSCrazy Group: General Forum Members Last Login: Wednesday, November 20, 2013 10:00 AM Points: 2,719, Visits: 4,724
 Luis Cazares (9/12/2012)Eugene Elutin (9/12/2012)You don't really need last join as the last digit of the second number can only be '3' Wouldn't that be cheating?....Cheating? No, it's a common sense. There is no other number you can deduct from 6 to get 3. There is no need to guess that.Regarding performance... I don't think it is relevant to this kind of task at all. I guess finding the more "elegant" looking solution is what OP is after. Saying that, removing number of casting will make it look more elegant:`;with a (num)as( select * from (values ('1'),('2'),('5'),('7'),('8'),('9')) a(num) ),b as( select '4' + a.num + b.num + c.num + '6' as num1 ,d.num + e.num + f.num + '3' as num2 from a join a as b on a.num <> b.num join a as c on c.num NOT IN (a.num, b.num) join a as d on d.num NOT IN (a.num, b.num, c.num) join a as e on e.num NOT IN (a.num, b.num, c.num, d.num) join a as f on f.num NOT IN (a.num, b.num, c.num, d.num, e.num))select * from b where cast(num1 as int) - cast(num2 as int)= 33333` _____________________________________________"The only true wisdom is in knowing you know nothing""O skol'ko nam otkrytiy chudnyh prevnosit microsofta duh!"(So many miracle inventions provided by MS to us...)How to post your question to get the best and quick help
Post #1358013
 Posted Wednesday, September 12, 2012 8:42 AM
 SSCoach Group: General Forum Members Last Login: Friday, November 01, 2013 1:55 PM Points: 15,442, Visits: 9,579
 Luis Cazares (9/12/2012)Eugene Elutin (9/12/2012)You don't really need last join as the last digit of the second number can only be '3' Wouldn't that be cheating?Because that way I can get the best performance`SELECT 41286 num1, 7953 num2`I had my solution that was very similar to yours (again) but using the different(<>) operators instead of the NOT IN.However, a better performance would be using the best datatype for the digits (char) since the beginning avoiding additional conversions.(emphasis added)That's why I did mine that way. Makes the concatenation, etc., more efficient, at the cost of the final Where clause being slightly more costly.I also kept the rules for what fixed numbers go in what columns in the Where clause, instead of in the various From objects, because that way it can be modified for related queries more easily. If, for example, the puzzle were modified to "second number must start with '79'" instead of "first number must start with '4' and end with '6'", then a simple modification of the Where clause would make my version work.For related puzzles, you could assign a parameter to each column. If not null, then fixed value for that column. Final value could also be a parameter.And, yeah, I really do think extensibility like that with every piece of code I write. - Gus "GSquared", RSVP, OODA, MAP, NMVP, FAQ, SAT, SQL, DNA, RNA, UOI, IOU, AM, PM, AD, BC, BCE, USA, UN, CF, ROFL, LOL, ETCProperty of The Thread"Nobody knows the age of the human race, but everyone agrees it's old enough to know better." - Anon
Post #1358017
 Posted Wednesday, September 12, 2012 8:44 AM
 SSCoach Group: General Forum Members Last Login: Friday, November 01, 2013 1:55 PM Points: 15,442, Visits: 9,579
 Shaun-884394 (9/12/2012)...GSquared i love your signature ROFL,LOL,ETC .... I don't think it can get any better.Thanks! The really funny part is, there was a noticable increase in people assuming my posts were credible, when I added all that to my sig. - Gus "GSquared", RSVP, OODA, MAP, NMVP, FAQ, SAT, SQL, DNA, RNA, UOI, IOU, AM, PM, AD, BC, BCE, USA, UN, CF, ROFL, LOL, ETCProperty of The Thread"Nobody knows the age of the human race, but everyone agrees it's old enough to know better." - Anon
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