RE: Oracle 10G

Since no one else tried it, here's a solution that immediately came to mind: The red hair remark tells that the oldest has no twin. The question do you have any children indicates that he did not have any older than 20. Table Create table kid_ages ( oldest smallint, middle smallint, youngest smallint) dataset o m y 12, 3, 1 9, 4, 1 9, 2, 2 etc. select * from kid_ages where oldest + middle + youngest = number of windows counted --EEM -----Original Message----- From: Daniel Morgan [mailto:damorgan@x.washington.edu] Sent: Wednesday, March 03, 2004 7:46 PM To: informix-list@iiug.org Subject: Re: Oracle 10G Obnoxio The Clown wrote: > Yeah, that's true, Oracle is so difficult and incomprehensible, you do need > to be a rocket scientist to work with it. And since NASA is not doing that > well at the moment... What is difficult for some ... is easy for others. The difficulty in the program is not the database. The course would be equally difficult taught with any RDBMS. The difficulty is that we teach problem solving skills: Not just syntax. Here's a problem given to students last night. As solvable in Informix as in Oracle: Go for it. ==================================================================== Two mathematicians (Boris and Vladimir) met accidently for the first time in 20 years. They greet each other and begin catching up on their respective lives. Boris asks Vladi "Do you have any children?" "Yes" replies Vladimir, "I have three." "How old are they?", asks Boris. "The product of their ages is 36 and the sum of their ages is equal to the number of windows on that building across the street." Boris looks at the building, counts the windows then says "Vladi, that still doesn't tell me the ages." "Ah, says Vladi, then I must tell you that the eldest has red hair." "Oh", says Boris, "now I know their ages." What are the ages of Boris' children? ==================================================================== Create a table, load it with data, and write a single SQL statement to produce the data set required to deduce the answer ... then deduce away! -- Daniel Morgan http://www.outreach.washington.edu/ext/certificates/oad/oad_crs.asp http://www.outreach.washington.edu/ext/certificates/aoa/aoa_crs.asp damorgan@x.washington.edu (replace 'x' with a 'u' to reply) sending to informix-list

Everett Mills wrote: > Since no one else tried it, here's a solution that immediately came to > mind: > > The red hair remark tells that the oldest has no twin. > The question do you have any children indicates that he did not have any > older than 20. > > Table > > Create table kid_ages > ( > oldest smallint, > middle smallint, > youngest smallint)> > dataset > o m y > 12, 3, 1 > 9, 4, 1 > 9, 2, 2 > etc. > > select * from kid_ages > where oldest + middle + youngest = number of windows counted> > --EEM Nice to see. Here's one solution from that 'other' group: STEP 1) CREATE TABLE ages (age NUMBWER); STEP 2) INSERT INTO ages VALUES (1); ... INSERT INTO ages VALUES (36);COMMIT; STEP 3) select a1, a2, a3 from (select c1.age a1, c2.age a2, c3.age a3, c1.age + c2.age + c3.age totals from ages c1, ages c2, ages c3 where c1.age * c2.age * c3.age = 36 and c1.age <= c2.age and c2.age <= c3.age) t1 where exists ( select * from ages c1, ages c2, ages c3 where c1.age * c2.age * c3.age = 36 and c1.age <= c2.age and c2.age <= c3.age and c1.age + c2.age + c3.age = t1.totals having count(*) > 1) and a3 > a1 and a3 > a2; -- Daniel Morgan http://www.outreach.washington.edu/ext/certificates/oad/oad_crs.asp http://www.outreach.washington.edu/ext/certificates/aoa/aoa_crs.asp damorgan@x.washington.edu (replace 'x' with a 'u' to reply)