Javascript, 5 Distinct characters, 9 Bytes

~[]/~-~[]

posted about 2 years ago

CC BY-SA 4.0

2y ago

AkshatJ‭

21 reputation 0 1 2 1

dc, 3 unique bytes, 138 bytes

This challenge is pretty easy if you aim for just 4 unique bytes. For example, you could push the stack depth repeatedly to get ascending integers, set the precision to something positive, divide and then print.

zzzzk/p

And surely it can't go any lower, right? After all, we have to increase the precision (k), print (p, n, f), do a floating-point operation (v, ^, /, .), and still get a number from somewhere...

Luckily though, there is actually a narrow path forward thanks to a and x. a pops a number off the stack, casts it to a long, and pushes its LSB back as a single-character string. Using x we can then execute that string as a command. This leaves us to choose a third command which should give us some helpful numbers that we can feed into a. The only options that can give us a range of numbers are z as well as the hex digits A through F. I haven't found a way to tame z, since it buries every intermediate result you get, so let's just see what useful characters we can get if we choose one of the hex digits.

At first, this might not look too promising, but if we choose F as our third command, we can use the i to change the input base to 15 and basically get a reroll on our commands. 🤞

The ? doesn't help us, since we don't get any input, the / however is game-changing. By carefully placing the right constants on the stack before and after the base change we can later use / to efficiently calculate the ASCII value of any command. We use this to first set the precision to 15 and then sum, divide and print three pre-placed Fs (as in F k F F F + / n) to do the actual calculation.

Finally, here is an annotated version of the complete program:

F F F         # three 15s for calculating 0.5
FFFFFFFFFFFFF # "+" numerator
FFFFFF        # "+" denominator 1
F             # argument for k
FFFFFFFFFFF   # "k" numerator
FFFFF         # "k" denominator 1
F             # argument for i
FFFFFFax      # execute "i" to switch base
FFFFFax       # "k" division 1
F             # "k" denominator 2
FFFFFax       # "k" division 2
ax            # execute "k" to set precedence
FFFFFax       # "+" division 1
F             # "+" denominator 2
FFFFFax       # "+" division 2
ax            # execute "+" to add two 15s
FFFFFax       # divide 15 by 30
FFFFFFFFFFFF  # "n" numerator 1
FFF           # "n" denominator 1
FFFFFax       # "n" division 1
FFFF          # "n" denominator 2
FFFFFax       # "n" division 2
ax            # execute "n" to print the result

We can get rid of any whitespace by placing xs between constants since they act as no-ops on numbers.

FxFxFxFFFFFFFFFFFFFxFFFFFFxFxFFFFFFFFFFFxFFFFFxFxFFFFFFaxFFFFFaxFxFFFFFaxaxFFFFFaxFxFFFFFaxaxFFFFFaxFFFFFFFFFFFFxFFFxFFFFFaxFFFFxFFFFFaxax

Try it online!

posted about 2 years ago

CC BY-SA 4.0

orthoplex‭

191 reputation 0 13 19 1

Python 3, 7 distinct characters, 11 bytes

True/-~True

Try it online!

posted about 2 years ago

CC BY-SA 4.0

m90‭

221 reputation 0 9 22 1