You’re right. Thanks for pointing out the error; I don’t know how I managed to report this incorrectly, but I’ve added some notes to the Errata page for this chapter.

(Coincidentally, someone else sent me an email a couple of days ago with a list of errors and typos – including this one, and they included a pl/sql block to demonstrate arithmetic as well.)

Good Day

while reading your fabulous book “cost based oracle”

In chapter 13 on pg 383 in the second last paragraph you said

“for every row in t1, Oracle has located the first relevant row in the second sorted
data set, and scanned from there to the end of the sorted data set. That’s
10,000 rows checked for the first row in t1; 9998 rows checked for the second row in t1;
9996 rows checked for the third row in t1 and so on”

and here you are talking about the access predicate (t2.n2>=t1.n2-1)

Now the the column n2 in both t1 and t2 have pattern like 0,0,1,1,2,2 ……1000,1000 ….

As per access predicate first two row in t1.n2 is -1, -1 (n2-1) hence we are doing 10,000 for the first two rows in t2
then there is 0, 0 (n2-1) for the third and fourth rows for t1.n2 hence again we are doing 10,000 for the third and fourth rows..

should not be oracle is doing 10,000 check in t2 for the first four rows in t1;
9998 for the 5th and 6th rows; 9996 for the 7th and 8th rows ……

Am I missing something?

Thanks