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Death is my excuse to stop working on Two-instruction madness!. And what do you know? Its extremely hard. It attempts to be harder than Malbolge (or Malbolge unshackled) without employing tricks like encryption. So, here are the instructions:

  • nop.
  • . takes the current value that the data pointer is pointing to, takes the accumulator, compares them and if the value is larger apply this: *d=(*d+a)/d

If its lower, apply this: *d=(*d+d) Else, terminate.

  • , does the same as a, but instead of compare, its comparing the parity. Same will do this: d=floor(*d+(d^2)/2)
  • ! a=(*d+d-a)/*(d+1)
  • / jumps back a+*d-d*2 instructions.
  • ( outputs ceil(*d-d/a) as an ascii character.
  • ) if there is a ( previously in the code, it will jump there and execute !!/.,.,, but if there isn't, the program jumps 1+a-*d+d instructions forwards, but if there is a .,/ or ( there, then it will jump back a+1 instructions.

And the number of every instruction has to be at most +-5 the average of all of the numbers of the instructions. I think this is Turing complete, but it's hard to prove anything when a esolang is this strict.