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) To avoid getting into a loop, the sample unconditionally ends the conversation whether the message was processed successfully or not This is not what you would want to do in a real application After inserting the message contents into the Inventory table, the response message is sent back to the ManufacturingService and the dialog is ended The response message contains either a message indicating success or the error text if the insert failed The @dialog variable used for the SEND and END DIALOG was obtained from the received message, so the message will be routed back to the calling dialog Now that a service is ready to process messages, let s implement a stored procedure to send the AddItem message to this service This stored procedure will begin a dialog to the InventoryService and send a message on the dialog The target service name for this.

Because the Employees table contains nine rows, the result set will contain 81 rows. And here's the ANSI SQL:1989 syntax you would use for the same task: SELECT E1.FirstName, E1.LastName AS emp1, E2.FirstName, E2.LastName AS emp2 FROM dbo.Employees AS E1, dbo.Employees AS E2;

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A good way to start solving this puzzle is to rst list all groups of three integers whose product is 36, then calculate each group s sum: 1 + 1 + 36 = 38 1 + 2 + 18 = 21 1 + 3 + 12 = 16 1 + 4 + 9 = 14 1 + 6 + 6 = 13 2 + 2 + 9 = 13 2 + 3 + 6 = 11 3 + 3 + 4 = 10 M knows the sum of the kids ages (equal to the number of the bus N pointed to). Notice that all sums arise in one way except for the sum 13, which arises from two different groups of three integers. Had the bus number been something other than 13, M would have immediately known the answer. Because M said that he doesn t have suf cient information to solve the puzzle, the bus number must have been 13. Now, the question remains, which of the two age variations is the correct one Notice that in both cases (1, 6, 6 and 2, 2, 9), there are twins. The additional piece of information N provided was, One of my kids was born before we bought the house. The implication is that one of the kids is older than the other two, so of the two variations, the correct one is 2, 2, 9. Now, how would the solution change if N s additional piece of information had been that one of his kids was born after he bought the house In this case, one of the kids is younger than the other two, so the correct answer would be 1, 6, 6. Interestingly, you can solve this puzzle with a T-SQL query, like so:

empname VARCHAR(25) NOT NULL, salary MONEY NOT NULL, lvl INT NOT NULL, UNIQUE CLUSTERED(lvl, empid) ) AS BEGIN DECLARE @lvl AS INT; SET @lvl = 0;

AS(SELECT 0 AS c UNION ALL SELECT 0), AS(SELECT 0 AS c FROM L0 AS a CROSS JOIN L0 AS b), AS(SELECT 0 AS c FROM L1 AS a CROSS JOIN L1 AS b), AS(SELECT TOP(36) 0 AS c FROM L2 AS a CROSS JOIN L2 AS b), AS(SELECT ROW_NUMBER() OVER(ORDER BY (SELECT 0)) AS n FROM L3), Divisors AS ( SELECT C1.n AS age1, C2.n AS age2, C3.n AS age3, COUNT(*) OVER(PARTITION BY C1.n + C2.n + C3.n) AS cnt FROM Nums AS C1 CROSS JOIN Nums AS C2 CROSS JOIN Nums AS C3 WHERE C1.n * C2.n * C3.n = 36 AND C1.n <= C2.n AND C2.n <= C3.n )

For cross joins only, I prefer using a comma (as opposed to using the CROSS JOIN keywords) because it allows for shorter code. I also find the older syntax to be more natural and readable. The optimizer will produce the same plan for both, so you shouldn't have any concerns about performance. As you will see later on, I will give a different recommendation for inner joins. Now let's look at more sophisticated uses of cross joins. In 4, I presented a powerful key technique to generate duplicates. Recall that I used an auxiliary table of numbers (Nums) as follows to generate the requested number of duplicates of each row: SELECT ... FROM T1, Nums WHERE n <= <num_of_dups>

-- Insert root node to @Subs INSERT INTO @Subs(empid, mgrid, empname, salary, lvl) SELECT empid, mgrid, empname, salary, @lvl FROM dbo.Employees WHERE empid = @mgrid; WHILE @@rowcount > 0 BEGIN SET @lvl = @lvl + 1; -- While prev level had rows -- Increment level counter

SELECT age1, age2, age3 FROM Divisors WHERE cnt > 1 AND age3 > age2; -- One born before others (before house);

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