@loop: subleq @tmp, @IN subleq @tmp, @IN subleq @OUT, @tmp subleq @tmp, @tmp, @loop

Not a solution but I felt that it would fit here. If your solution starts by taking an input like this subleq @input @IN...

@loop: subleq @A, @IN subleq @B, @IN subleq @A, @B subleq @OUT, @A subleq @B, @B subleq @A, @A, @loop @A: .data 0 @B: .d...

This is a solution with least cycles (989) on Steam at present. Some highlights: Specialized instruction sequences to mu...

Well, the solution is easy subleq @OUT @IN This solution does read 5 bytes. BUT there is a solution that can read only...

@start: subleq @count, @IN @loop: subleq @count, @n_one, @continue @done: subleq @OUT, @zero subleq @count, @count, @sta...

@loop: subleq @tmp, @IN subleq @tmp2, @tmp, @done subleq @count, @one @loop_reset: subleq @tmp2, @tmp2 subleq @tmp, @tmp...

@loop: subleq @tmp2, @IN subleq @tmp, @tmp2, @done subleq @sum, @tmp subleq @tmp2, @tmp2 subleq @tmp, @tmp, @loop @done...

@start: subleq @tmp, @IN subleq @remainder, @tmp subleq @tmp, @tmp subleq @tmp, @IN subleq @divisor, @tmp @loop: subleq...

Here is my solution for the self hosting problem, inspired by CakeEaterGames ' version. It takes 893 cycles and 193 byte...

Simple solution: @loop: subleq @tmp, @IN subleq @OUT, @tmp subleq @tmp, @tmp, @loop @tmp: .data 0

149 cycles, 62 bytes @loop: subleq @tmp, @IN subleq @tmp2, @tmp2 subleq @tmp2, @tmp, @dequeue @queue: subleq @stack, @tm...

@program: .data 0, @IN, 3, 0, 8, 6, @OUT, 0, -48, 0, 0, 0 If a command is guaranteed to output a positive number to @OUT...

@l0: .data 0,0,@l2+3,0,0,0,0,0,0,0,0,0,0,0,0,0 @l1: .data 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 @l2: .data 0,0,0,0,0,@l3+...

Very similar to the "stack memory" puzzle, just with handling lists' zero terminators: @start: subleq @tmp, @IN subleq @...

@start: subleq @a, @IN subleq @b, @IN subleq @count, @n_one subleq @count, @b @loop: subleq @count, @one, @next subleq @...

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