RACKET Scheme prog

list has two equal elements recursively in scheme

These are the elements initially pointed by REF and REST. If they are the same, we return #t. If not, we can make a recursive call on the list from which we have removed the second element (in Java, we change REST to [login to view URL] and by doing so we skip the second element pointed by REST). If the recursive call returns #t, the whole function returns #t. This recursive call will recursively compare the first element of the list with all subsequent elements by removing the elements that are not equal to the first element until the end of the list is reached. Otherwise (if the first recursive call fails), we need to make one more recursive call in which we delete the first element of the list (in Java, we change REF to [login to view URL], therefore skipping the first element). Obviously, the first element does not have a match if the first recursive call failed.

Examples

(two ‘(7 7)) returns #t

(two ‘(5 3 3 4)) returns #t

(two ‘(11 3 5 11)) returns #t

(two ‘(1 2 3 4 5 6 10)) returns #f

(two ‘(24 14 11 4 6)) returns #f

The whole solution must be packed in one recursive function two which must look as follows: (define two (lambda (list)

(cond

...

)))

In other words, you have to define a single COND statement. Nested COND statements are not

allowed.

Inside COND, you can use ONLY the following constructs: - null?

- car

- cdr

- cons

- else

- =

- #t

- #f

- two

- list

Java

Project ID: #32192739

About the project