hackerbase/doc/utils.md

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Utility Modules
===============
To ensure there are no external dependencies (including chicken eggs),
these modules re-implement any basic procedures which are required for
any algorithms used. And some advanced yet generic functionality as
well.
The modules are listed in alphabetical order.
### Set (List)
(import util-set-list)
This module implements linear-time set with custom comparator.
(make-lset [comparator])
* ```comparator``` - comparison procedure, defaults to ```equal?```
Creates new lset with given comparator.
(lset? v)
* ```v``` - any value
Returns ```#t``` if given value is a lset.
(lset-empty? ls)
* ```ls``` - lset instance
Returns ```#t``` if given lset contains no elements.
(lset-member? ls el)
* ```ls``` - lset instance
* ```el``` - element to check
Returns ```#t``` if given element ```el``` is contained in the lset
```ls``` using its equality comparator.
(lset-count ls)
* ```ls``` - lset instance
Returns the number of elements in given lset.
(lset-add ls [el ...])
* ```ls``` - lset instance
* ```el``` - element(s) to add
Adds given element(s) ```el``` to the lset instance ```ls``` returning
a new lset instance.
(lset-remove ls el)
* ```ls``` - lset instance
* ```el``` - element to remove
If given lset instance ```ls``` contains the element ```el```
provided, it is removed and the new lset is returned. If the element
is not contained in ```ls```, it is returned intact.
(list->lset lst [comparator])
* ```lst``` - list to convert
* ```comparator``` - equality comparison procedure, defaults to ```equal?```
Returns a new lset with the comparator provided containing all
elements in given list.
(lset->list ls)
* ```ls``` - lset instance
Returns the list of elements in given lset.
(lset-merge ls1 ls2)
* ```ls1``` - lset instance
* ```ls2``` - lset instance
Returns a new lset instance with all elements in both lset instances given.
(lset-intersect ls1 ls2)
* ```ls1``` - lset instance
* ```ls2``` - lset instance
Returns a new lset instance containing elements present both in
```ls1``` and ```ls2```.
(lset-subtract ls1 ls2)
Returns a new lset instance from ```ls1``` with all elements in
```ls2``` removed from it.
(lset=? ls1 ls2)
* ```ls1``` - lset instance
* ```ls2``` - lset instance
Returns true if the sets contain exactly the same values.