Tutorial 1, Answers

1.   (.1) q           (.6) d            (.10) o
     (.2) u           (.7) i            (.11) w
     (.3) f           (.8) k            (.12) p
     (.4) a           (.9) j            (.13) n
     (.5) c
   

2. A compiler might translate f = -p*r + (a/b)*a*a - b into

      mul $10, $3, $4    # $10 = p*r
      sub $2, $0, $10    # $2 = -p*r
      div $10, $5, $6    # $10 = (a/b)
      mul $10, $10, $5   # $10 = (a/b)*a
      mul $10, $10, $5   # $10 = (a/b)*a*a
      add $2, $2, $10    # $2 = -p*r + (a/b)*a*a
      sub $2, $2, $6     # $2 = f = -p*r + (a/b)*a*a - b
   

3. To save the trouble of commenting each instruction, I just say that the program stores the value 1,2,3 into memory locations starting from 0x10000000. Then it loads the same set of values and add 1 to each, and saves them (2,3,4, respectively) in the next three locations in memory.
4. The binary encoding is

        Decimal     Hexadecimal       Symbolic
     35 2 5   0      0x8c450000     lw $5, 0($2) 
      0 5 3 5 0 32   0x00a32820     add $5, $5, $3
     43 2 5  12      0xac45000c     sw $5, 12($2)