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Comment by elevensies | Hacker News Reader

Well, here is my non-zero-handling answer to the first

  Z2
  I2  <-+
  J0,2 -+
  Z3
  I2  <-+
  I3    |
  J1,3 -+

And with zero handling (maybe)

    Z4
    I4
    Z2
    J0,2 -+
  +-J0,4  |
  | I2  <=+ 
  | J0,2 -+ 
  +>Z3    
    J1,3 -+
  +-J1,4  |
  | I2  <=+
  | I3    |
  | J1,3 -+ 
  +>exit

Edit * 2nd try

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My solution, looks pretty similar to your second.
I think this should work:
0 Z2 1 Z3 2 I4 3 J0,3 -> 5 // If r0 != 0, jump to add it to r2 4 J3,4 -> 9 // r0 == 0, force jump to check r1 5 I2 // add r0 6 I3 7 J0,3 -> 5 8 Z3 9 J1,3 -> 11 // If r1 != 0, jump to add it to r2 10 J3,4 -> exit // r1 == 0, force jump to exit 11 I2 // add r1 12 I3 13 J1,3 -> 11 14 exit
Phew that took a while. The simple case was easy enough. I guess exceptional candidates would be able to crack that out in an interview, otherwise they'd move you onto the balls.
Edited!

Couple corrections I think: - Register 4 should be set to zero before incrementing in instruction 2.
- inst. 4 should compare 0,4 , register 1 is unknown at this pt.
- inst. 3 should jump to 5

Yep 3->5 is a typo.
r4 could be zeroed, but I was lazy and assumed that as long as it's greater than zero it doesn't matter (since it's always compared to zero).
Instruction 4 should be cmp(3,4) to force a jump to 9.
EDIT: Although thinking about it, it doesn't make much difference. It has the same effect as doing cmp(0,4) since you know r4 > 0 and r0 == 0.
Thanks!

In your zero-handling program, you seem to suppose the first "J0,2" and "J1,3" commands take the arrow on equality, while the second "J0,2" and "J1,3" commands instead take the arrow on inequality.

oops! I changed it but I'm not sure if this is the right way to do it

My solution:
0: Z 2 1: Z 3 2: J 0 2 -> 4 3: J 1 3 -> 7 4: I 2 5: J 0 2 -> 4 6: J 1 3 -> 7 7: I 2 8: I 3 9: J 1 3 -> 7
Handles the zero case. Only uses 4 cells total.

Instruction 6 is a nop? if mem. 1 not-equals mem. 3, jump to 7, or else fall thru to 7?
I don't think you can depend on jumping in instruction 3, because you don't control the value of mem. 1, it's input. So if mem. 3 = 0, instructions 2 and 3 will always lead to instruction four.

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