# [Tutor] problem solving with lists

Dennis Lee Bieber wlfraed at ix.netcom.com
Thu Mar 24 13:26:58 EDT 2022

```On Thu, 24 Mar 2022 11:41:45 -0400, Dennis Lee Bieber
<wlfraed at ix.netcom.com> declaimed the following:

>
>of a string.
>
>	Strings are indexable -- the syntax looks just like what you'd use with
>a list of 1-char strings).
>

Alternatively -- don't use LIST or STRING... Since for purposes of the
problem itself, there is no difference between

"abcd" and "dbac" (and similarly, there is no difference between
"abcd", "efgh", "ijkl", "mnop" and "ijkl", "abcd", "mnop", "efgh")

you should be thinking of using SET as a data structure.

You state you are trying to come up with an algorithm, but your
description is always at the level of implementation at the low-level of
the language itself -- and you have already defined a one-use
implementation.

An algorithm would be something like:

{
/w/ 	<= a week
/nw/	<= number of weeks
/g/		<= a group
/ng/	<= number of groups
/pg/	<= player in group
/sg/	<= size of group
/p/		<= a player
}
for each /w/ in /nw/ weeks
for each /g/ in /ng/ groups
for each /pg/ in  /sg/ players-in-group
select /p/ from (pg * g) players
such that /p/ is only selected once in the current week
AND
any pair of /p/ in /pg/ does not appear in any prior
week's groups

At an implementation level -- .combinations() does most of the
for each /pg/ ...
select /p/ ...
in that it will return /sg/ size groups of /p/, and within the group, there
are no duplicates. You are still responsible for testing that they are
unique across the groups in each week (that means rejecting many of the
groups that .combinations() will provide). If you use set logic, you don't
even have to index the members within each group.

--
Wulfraed                 Dennis Lee Bieber         AF6VN
wlfraed at ix.netcom.com    http://wlfraed.microdiversity.freeddns.org/

```