Synacor Challenge

Please cf. to: synacor challenge website

== Synacor Challenge ==

In this challenge, your job is to use this architecture spec to create a virtual machine capable of running the included binary. Along the way, you will find codes; submit these to the challenge website to track your progress. Good luck!

== architecture ==

• three storage regions
  • memory with 15-bit address space storing 16-bit values
  • eight registers
  • an unbounded stack which holds individual 16-bit values
• all numbers are unsigned integers 0..32767 (15-bit)
• all math is modulo 32768; 32758 + 15 => 5

== binary format ==

• each number is stored as a 16-bit little-endian pair (low byte, high byte)
• numbers 0..32767 mean a literal value
• numbers 32768..32775 instead mean registers 0..7
• numbers 32776..65535 are invalid
• programs are loaded into memory starting at address 0
• address 0 is the first 16-bit value, address 1 is the second 16-bit value, etc

== execution ==

• After an operation is executed, the next instruction to read is immediately after the last argument of the current operation. If a jump was performed, the next operation is instead the exact destination of the jump.
• Encountering a register as an operation argument should be taken as reading from the register or setting into the register as appropriate.

== hints ==

• Start with operations 0, 19, and 21.
• Here's a code for the challenge website: mXGTwLElcNmB
• The program "9,32768,32769,4,19,32768" occupies six memory addresses and should:
  • Store into register 0 the sum of 4 and the value contained in register 1.
  • Output to the terminal the character with the ascii code contained in register 0.

== opcode listing ==

• halt: 0

  stop execution and terminate the program

• set: 1 a b

  set register a to the value of b

• push: 2 a

  push a onto the stack

• pop: 3 a

  remove the top element from the stack and write it into a; empty stack = error

• eq: 4 a b c

  set a to 1 if b is equal to c; set it to 0 otherwise

• gt: 5 a b c

  set a to 1 if b is greater than c; set it to 0 otherwise

• jmp: 6 a

  jump to a

• jt: 7 a b

  if a is nonzero, jump to b

• jf: 8 a b

  if a is zero, jump to b

• add: 9 a b c

  assign into a the sum of b and c (modulo 32768)

• mult: 10 a b c

  store into a the product of b and c (modulo 32768)

• mod: 11 a b c

  store into a the remainder of b divided by c

• and: 12 a b c

  stores into a the bitwise and of b and c

• or: 13 a b c

  stores into a the bitwise or of b and c

• not: 14 a b

  stores 15-bit bitwise inverse of b in a

• rmem: 15 a b

  read memory at address b and write it to a

• wmem: 16 a b

  write the value from b into memory at address a

• call: 17 a

  write the address of the next instruction to the stack and jump to a

• ret: 18

  remove the top element from the stack and jump to it; empty stack = halt

• out: 19 a

  write the character represented by ascii code a to the terminal

• in: 20 a

  read a character from the terminal and write its ascii code to a; it can be assumed that once input starts, it will continue until a newline is encountered; this means that you can safely read whole lines from the keyboard and trust that they will be fully read

• noop: 21

  no operation