Problem 60591. ICFP2024 001: Lambdaman 6
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Had a small epiphany on processing based on one of the writeups. 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.
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