Math and Comparison

Challenge 1

Write a program that for arguments A, B, and C, computes (A + B) × C.

C1 Solution

×+

Why?

It's helpful to look at the arguments and work backwards from that. When we have arguments A B C notice that A B comes first. Let's set A B C to 1 2 3.

+ 1 2 3
# We see + with its arguments 1 2
# We have (+ 1 2) 3
# or (+ A B) C
# output: 3 3
# or
# output: (A+B) C

So, putting an add at the front results in D C where D = A + B. We now just have to use a mul function, or mul D C. Remember that we read from right to left and therefore we consider add to be on the front with mul behind it.

Challenge 2

Write a program that calculates the equation √(A² + B), where A is the first argument and B is the second.

C2 Solution

√+˙×

Alternatively we could use pow 2

√+ⁿ2

Why?

Again, this is an exercise in reading right to left and getting a feeling for how arguments are used/consumed.

Let's see how we would work backwards. Let's use 3 16 for A B.

˙× 3 16
# output: 9 16
# We now have A² B as arguments 

+ 9 16
# output: 25
# We now have the argument C, where C = A²+B

√ 25 
# output: 5
# We now have sqrt(C)

What about the idiomatic solution?

⍜˙×+

I am not going to go through under here in great detail. There will be a thorough discussion of inverses later.

What under does is transform something, apply a function to the transformation and then undo the transformation. Here is what happens in the idiomatic solution:

• under transforms the first argument, A, by squaring it
• under then adds together A² + B
  • → let's refer to this result as C
• under computes the inverse of self mul which is sqrt
• under applies this inverse to the first argument, C