Click here to monitor SSC
SQLServerCentral is supported by Red Gate Software Ltd.
 
Log in  ::  Register  ::  Not logged in
 
 
 
        
Home       Members    Calendar    Who's On


Add to briefcase 12345»»»

Puzzle : Generating two unique numbers from an array of numbers (repetation not allowed) Expand / Collapse
Author
Message
Posted Wednesday, September 12, 2012 7:19 AM
Grasshopper

GrasshopperGrasshopperGrasshopperGrasshopperGrasshopperGrasshopperGrasshopperGrasshopper

Group: General Forum Members
Last Login: Tuesday, January 21, 2014 3:19 AM
Points: 22, Visits: 1,104
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) - (_ _ _ _) = 33333

As 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 = 33333
order by 1,2

Post #1357963
Posted Wednesday, September 12, 2012 7:43 AM
SSCrazy

SSCrazySSCrazySSCrazySSCrazySSCrazySSCrazySSCrazySSCrazy

Group: General Forum Members
Last Login: Friday, July 4, 2014 3:55 AM
Points: 2,836, Visits: 5,062
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 = 33333
order 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 = 33333
order 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

SSCrazySSCrazySSCrazySSCrazySSCrazySSCrazySSCrazySSCrazy

Group: General Forum Members
Last Login: Friday, July 4, 2014 3:55 AM
Points: 2,836, Visits: 5,062
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 = 33333
order 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

SSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoach

Group: General Forum Members
Last Login: Friday, June 27, 2014 12:43 PM
Points: 15,444, Visits: 9,596
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, ETC
Property 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

SSCrazySSCrazySSCrazySSCrazySSCrazySSCrazySSCrazySSCrazy

Group: General Forum Members
Last Login: Friday, July 4, 2014 3:55 AM
Points: 2,836, Visits: 5,062
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


Hall of Fame

Hall of FameHall of FameHall of FameHall of FameHall of FameHall of FameHall of FameHall of FameHall of Fame

Group: General Forum Members
Last Login: Yesterday @ 10:33 PM
Points: 3,374, Visits: 7,296
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.
I am a great believer in luck, and I find the harder I work the more I have of it. Stephen Leacock

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

GrasshopperGrasshopperGrasshopperGrasshopperGrasshopperGrasshopperGrasshopperGrasshopper

Group: General Forum Members
Last Login: Tuesday, January 21, 2014 3:19 AM
Points: 22, Visits: 1,104
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

SSCrazySSCrazySSCrazySSCrazySSCrazySSCrazySSCrazySSCrazy

Group: General Forum Members
Last Login: Friday, July 4, 2014 3:55 AM
Points: 2,836, Visits: 5,062
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

SSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoach

Group: General Forum Members
Last Login: Friday, June 27, 2014 12:43 PM
Points: 15,444, Visits: 9,596
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, ETC
Property 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

SSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoachSSCoach

Group: General Forum Members
Last Login: Friday, June 27, 2014 12:43 PM
Points: 15,444, Visits: 9,596
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, ETC
Property of The Thread

"Nobody knows the age of the human race, but everyone agrees it's old enough to know better." - Anon
Post #1358019
« Prev Topic | Next Topic »

Add to briefcase 12345»»»

Permissions Expand / Collapse