Types

Challenge 1

Write a program that increments the first character of a string.

C1 Solution

⊂+1⊢⟜(↘1)

or the idiomatic solution

⍜(⊢|+1)

Why?

Once you understand the idiomatic solution, there is no going back. I solved this idiomatically but then I struggled with the intended solution. I will admit that I just kinda forgot about on and made this horror

back join dip(add 1) dip first by drop 1
˜⊂ ⊙(+ 1) ⊙⊢ ⊸↘ 1

But let's look at what we're trying to accomplish, step by step. Let's assume we want to turn bingo into cingo.

# What does on look like with drop 1?
    on (drop 1) "bingo"
    ⟜(↘ 1) "bingo"
"ingo"
"bingo"

# And how is it different from by?
    by drop 1 "bingo"
    ⊸↘ 1 "bingo"
"bingo"
"ingo"

# We want to operate on "bingo" first
# Therefore we choose `on`
    first on (drop 1) "bingo"
    ⊢ ⟜(↘ 1) "bingo"
"ingo"
@b

# We want to change @b into @c
    add 1 first on (drop 1) "bingo"
    + 1 ⊢ ⟜(↘ 1) "bingo"
"ingo"
@c

# Finally we join it
    join add 1 first on (drop 1) "bingo"
    ⊂ + 1 ⊢ ⟜(↘ 1) "bingo"
"cingo"

But what about the idiomatic solution?

We have discussed under in chapter 2 but let's recap

What under does is transform something, apply a function to the transformation and then undo the transformation.

What does under(first|add1) do to "bingo"?

• - transforms bingo into @b via first
  • + You can think of this as dropping ingo
• - add 1 to @b making @c
• - reverses the transformation
  • + "Undrops" ingo back onto @c making cingo