Had a small epiphany on processing based on one of the write-ups. For Lamda 6, needing >199 Rs.
The script to solve was:
(let ((triple (lambda (x) (concat x x x))))
(concat "solve lambdaman6 "
(triple (triple (triple "RRRRRRRR")))))
which becomes ICFP
B$ L! B. S3/,6%},!-"$!-!.[} B$ v! B$ v! B$ v! SLLLLLLLL L! B. v! B. v! v!
The 'solve Lambdaman6 ' is 'S3/,6%},!-"$!-!.[}'
The leftmost B$ L! means define function L!, aka triple() comes from L! B. v! B. v! v! thus L!=triple(x)=[v! v! v!], triple repeat
the second part of the first concat is
B$ v! B$ v! B$ v! SLLLLLLLL The usage B$ v! is L! thus
triple(triple(triple(SLLLLLLLL) )) thus where L is R we get
8Rs *3*3*3=27*8 Rs=216 Rs.
There are multiple ways to come to the same function substitution and variable load.
The ICFP 2024 is over. I was very disappointed with this year's contest as it required cryptic Caeser encryption and was based on Haskell processing for majority of challenges. Of 350 entrants only 100 could process the Lambda Extraction Haskell method. Hopefully next year they won't be so esoteric in their problems.
Oddly using
cd (fullfile(matlabroot,'toolbox/images/imdata/'))
gives a demo dir that includes cameraman.tif
With a few manipulations the expected results can be created.