Lab1: Exercise 8

[BertLisser]

-- Because there is only one possibility where 3 people are correct according to the 
-- teacher the boy with 3 accusers will be found guilty. So we check for each person how 
-- many accusers he has untill we find the one with 3 accusers.
checkGuilty :: Boy -> [Boy] -> [Boy]
checkGuilty x xs = if length (accusers x) == 3 then [x] 
                             else checkGuilty (head xs) (tail xs)

The remark if the puzzle is well-designed... is not an assumption. So checkGuilty must be changed in:

checkGuilty :: Boy -> [Boy] -> [Boy]
checkGuilty x [] = []
checkGuilty x xs = if length (accusers x) == 3 then [x]++checkGuilty (head xs) (tail xs)  
                             else checkGuilty (head xs) (tail xs)

Simpler interface will be:

checkGuilty ::[Boy] -> [Boy]
checkGuilty [] =  ...
checkGuilty (x:xs) =  ...

Well done!