Another German Yak ... with a suprise (RC #3)By Bill Graziano on 16 July 2002  Tags: Reader Challenges This Reader Challenge #3 solution is quite a bit different from the first solution I posted. And it's chock full of "sqlicious" goodness. And a suprise at the end!
The second solution I received was from Arnold Fribble (Arnold Fribble). And boy is it a doozy. I'll post it here and then walk through it.
CREATE TABLE #n5 (n int PRIMARY KEY) INSERT INTO #n5 SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 SELECT CASE WHEN Yak = Dane THEN 'Dane' WHEN Yak = English THEN 'English' WHEN Yak = German THEN 'German' WHEN Yak = Norwegian THEN 'Norwegian' WHEN Yak = Swede THEN 'Swede' END AS YakOwner FROM ( SELECT a1.n Dane, a2.n English, a3.n German, a4.n Norwegian, a5.n Swede, b1.n Blue, b2.n Green, b3.n Red, b4.n White, b5.n Yellow, c1.n Birds, c2.n Cats, c3.n Dog, c4.n Horse, c5.n Yak, d1.n Beer, d2.n Coffee, d3.n Milk, d4.n Tea, d5.n Water, e1.n Blend, e2.n BlueMaster, e3.n Dunhill, e4.n PallMall, e5.n Prince FROM #n5 a1, #n5 a2, #n5 a3, #n5 a4, #n5 a5,  nationalities #n5 b1, #n5 b2, #n5 b3, #n5 b4, #n5 b5,  house colors #n5 c1, #n5 c2, #n5 c3, #n5 c4, #n5 c5,  pets #n5 d1, #n5 d2, #n5 d3, #n5 d4, #n5 d5,  drinks #n5 e1, #n5 e2, #n5 e3, #n5 e4, #n5 e5  cigarettes WHERE a2.n NOT IN (a1.n) AND a3.n NOT IN (a1.n, a2.n) AND a4.n NOT IN (a1.n, a2.n, a3.n) AND a5.n NOT IN (a1.n, a2.n, a3.n, a4.n) AND b2.n NOT IN (b1.n) AND b3.n NOT IN (b1.n, b2.n) AND b4.n NOT IN (b1.n, b2.n, b3.n) AND b5.n NOT IN (b1.n, b2.n, b3.n, b4.n) AND c2.n NOT IN (c1.n) AND c3.n NOT IN (c1.n, c2.n) AND c4.n NOT IN (c1.n, c2.n, c3.n) AND c5.n NOT IN (c1.n, c2.n, c3.n, c4.n) AND d2.n NOT IN (d1.n) AND d3.n NOT IN (d1.n, d2.n) AND d4.n NOT IN (d1.n, d2.n, d3.n) AND d5.n NOT IN (d1.n, d2.n, d3.n, d4.n) AND e2.n NOT IN (e1.n) AND e3.n NOT IN (e1.n, e2.n) AND e4.n NOT IN (e1.n, e2.n, e3.n) AND e5.n NOT IN (e1.n, e2.n, e3.n, e4.n) ) AS perms WHERE English = Red AND Swede = Dog AND Dane = Tea AND Green = White  1 AND Green < White  alternate interpretation for previous line AND Coffee = Green AND PallMall = Birds AND Yellow = Dunhill AND Milk = 3 AND Norwegian = 1 AND ABS(Blend  Cats) = 1 AND ABS(Horse  Dunhill) = 1 AND BlueMaster = Beer AND German = Prince AND ABS(Norwegian  Blue) = 1 AND ABS(Water  Blend) = 1 DROP TABLE #n5 Yep, that's all there is to it. A temp table and a SELECT statement. So let's start breaking this down. First, he creates a temp table that hold the numbers 1 though 5. This is commonly called a Tally table and we've seen it before. Then he has a SELECT from a SELECT. I'm going to basically rebuild his query from scratch starting at the inside. The simplest piece looks like this: SELECT a1.n Dane, a2.n English FROM #n5 a1, #n5 a2 Note that I'm leaving off the part the generates the temp table and populates it. It generates the following result: Dane English   1 1 2 1 3 1 4 1 5 1 1 2 2 2 3 2 4 2 5 2 1 3 2 3 3 3 4 3 5 3 1 4 2 4 3 4 4 4 5 4 1 5 2 5 3 5 4 5 5 5 Which is 25 rows of possible solutions. He does a cross join of the temp table to itself. This joins each row in the table to every other row in the table  including itself. If there were nine rows in the table we'd have 81 rows in the result set. His naming of the columns is what maps it back to the problem. In this little sample set we can see every possible permutation of housing for the Dane and the Englishman. The first piece he adds is a WHERE clause to remove two people living in the same house. Modifying the query we get this: SELECT a1.n Dane, a2.n English FROM #n5 a1, #n5 a2 WHERE a2.n NOT IN (a1.n) The WHERE clause says I want all the possible permutations of this except where the English house (a2.n) is equal to the Danish house (a1.n). This would eliminate all the rows from the above result set where the two numbers are equal. Let's add a third parameter and see what that looks like. SELECT a1.n Dane, a2.n English, a3.n German FROM #n5 a1, #n5 a2, #n5 a3 WHERE a2.n NOT IN (a1.n) AND a3.n NOT IN (a1.n, a2.n) which returns the following result set: Dane English German    1 3 2 1 4 2 1 5 2 1 2 3 1 4 3 ... 55 rows omitted for ease of reading... Now let's pretend that one of our rules is that the Dane must live next to the Englishman. You might write that like this: SELECT a1.n Dane, a2.n English, a3.n German FROM #n5 a1, #n5 a2, #n5 a3 WHERE a2.n NOT IN (a1.n) AND a3.n NOT IN (a1.n, a2.n) AND ABS(a1.n  a2.n) = 1 And we're left with just the rows where the Dane and the Englishman are side by side. Dane English German    1 2 3 1 2 4 1 2 5 2 3 1 2 1 3 ... 19 rows omitted To make the whole thing easier to read he wraps it in a derived table so he can use field names. Rewriting this query to use a derived table gives us Select * From ( SELECT a1.n Dane, a2.n English, a3.n German FROM #n5 a1, #n5 a2, #n5 a3 WHERE a2.n NOT IN (a1.n) AND a3.n NOT IN (a1.n, a2.n) ) as perm WHERE ABS(Dane  English) = 1 which is much easier to read and returns the same result set as above. All that's really left is to add all the possible combinations and fill out the WHERE clause. At that point you get the query at the top of the page. And that brings us to the suprise. One of the rules from the challenge was that the "green house is on the left side of the white house." I interpreted this to mean next door on the left. You'll notice the WHERE clause in the full query also makes that assumption. There's a line below it that's commented out that has the more general assumption. Switching to the commented out line gives us SEVEN possible solutions. Removing the So that's the second solution. Great job Arnold. None of the others were drastically different than these two. If you'd like others to see your solution, please post it in the forums attached to one of these articles and we can all take a look.

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