1. Two’s and one’s complement arithmetic

2. 12 3 6 9 1 2 4 5 7 8 10 11 CLOCK ARITHMETIC

3. 12 3 6 9 1 2 4 5 7 8 10 11 2+3= ?

4. 12 3 6 9 1 2 4 5 7 8 10 11 2+3= 5; 9+4=?

5. 12 3 6 9 1 2 4 5 7 8 10 11 2+3= 5; 9+4=13=1; 7*3=?

6. 12 3 6 9 1 2 4 5 7 8 10 11 2+3= 5; 9+4=13=1; 7*3=21=9

7. 12=0 3 6 9 1 2 4 5 7 8 10 11 2+3= 5; 9+4=13=1; 7*3=21=9 You can add and multiply in clock arithmetic; but you subtract multiples of 12 to get the clock display (between 1 and 12) (or, better, between 0 and 11). In particular, you can perform accurate, real arithmetic on a clock (without correction), as long as you ensure your calculations don’t go beyond [0,11].

8. A base-16 clock 0 4 8 12 2 6 10 14 1 3 5 7 9 11 13 15 3+4 = ?; 11+13 = ?

9. 0 4 8 12 2 6 10 14 1 3 5 7 9 11 13 15 3+4 = 7; 11+13 = (24-16) = 8.

10. 0 4 8 12 2 6 10 14 1 3 5 7 9 11 13 15 It’s easy for a computer to do base-16 clock arithmetic if the numbers are encoded in binary. 0000 0001 0010 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

11. Whenever a sum in base-16 clock arithmetic is 16 or greater, it acquires a fifth bit. 13 or 1101 2 + 5 + 0101 2 18 10010 2 To subtract 16 in the binary form, || || simply drop the fifth bit 2 0010 2 (since 16 = 10000 2 ). So to do base 16 arithmetic, use a binary 4-bit register and drop the “carry” if it occurs in the 4 th bit.

12. A base-15 clock 0 4 8 2 6 10 13 1 3 5 7 9 11 12 14 3+4 = ?; 11+13 = ?

13. 3+4 = 7; 11+13 = 9 0 4 8 2 6 10 13 1 3 5 7 9 11 12 14

14. To implement base-15 arithmetic on a computer, again use the binary representation: 0 4 8 11 2 6 10 13 1 3 5 7 9 12 14 0000 0001 0010 0011 0100 0101 0110 01111000 1001 1010 1011 1100 1101 1110

15. Whenever a sum in base-15 clock arithmetic exceeds 15, it acquires a fifth bit. 13 or 1101 + 5 + 0101 18 10010 Subtract 15 || || 3 0011 The easy way to subtract 15 in binary is to subtract 16 and add 1. Subtract 16 by removing the fifth bit. Then to add 1 you can imagine that the fifth bit is “carried around” to the first bit. 1101 + 0101 10010 +1 ______ 0011 So you can do base-15 clock arithmetic with standard bit-adders, linking the 4 th - bit carry to the first bit.

16. So it’s easy to use standard bit-adders to do base-2 n clock arithmetic; simply drop the “carry” in the final bit. And it’s almost as easy to use standard adders to do base-[2 n -1] clock arithmetic; simply connect the “carry” in the final bit to the first bit. Next we will show how to interpret clock arithmetic in these two systems as signed arithmetic. We will not alter the computations or the hardware; we will simply interpret the numbers differently.

17. 12 3 6 9 1 2 4 5 7 8 10 11 Can you subtract on a standard clock? Are there negative numbers? 2 - 5 = ?

18. 12 3 6 9 1 2 4 5 7 8 10 11 Can you subtract on a standard clock? Are there negative numbers? 2 - 5 = 9 5

19. 0 3 6 9 1 2 4 5 7 8 10 11 (-1) is the solution to x + 1 = 0. So (-1) = 11. And (-2) = 10. (-1) (-2) Of course this respects the rule, because (-1) + 12 = 11; (-2) + 12 = 10.

20. 0 3 6 9 1 2 4 5 7 8 10 11 (-1) (-2) Therefore clock arithmetic replicates real arithmetic for numbers between (-6) and +5, if we identify 11 with -1, 10 with -2, …, 6 with -6. (-3) (-4) (-5) (-6) (0) (1) (2) (3) (4) (5) (-3) + 2 = -1 because 9 + 2 = 11 ; (-3) + 5 = 2 because 9 + 5 = 2 on the clock.

21. 0 3 6 9 1 2 4 5 7 8 10 11 (-1) (-2) For example we can compute 3 – 5 = 3 +(-5) by adding the clock version of (-5) to 3. (-3) (-4) (-5) (-6) (0) (1) (2) (3) (4) (5) (-5) (+7) 3 – 5 = 3 + 7 = 10 = -2.

22. So the rules for signed clock arithmetic are: 1. Make sure your calculations don’t call for any numbers beyond [-6,5]. 2. Identify -1 with 11, -2 with 10, -3 with 9, -4 with 8, -5 with 7, -6 with 6. 3. Perform the signed addition [like 3 – 2 = 3 + (-2)] by doing the normal addition with clock arithmetic, using 0 through 11. (So if any answer lies outside [0,11], “repair” it by adding or subtracting 12.) 4. Identify 11 with -1. 10 with -2, 9 with -3, 8 with -4. 7 with -5, and 6 with -6. The hard parts, for computer implementation, are (i) making it easy to look at a coded negative number and identify its positive twin in steps 2 and 4. (ii) repairing any answer that lies out of the range in step 3. We have seen how the binary representation of the clock numbers makes (ii) very easy if the clock has 2 n or 2 n -1 numbers. Now we will show how 2’s complement and 1’s complement arithmetic, in the binary system, simplify (i).

23. Identify the negative numbers on the base-16 (=2 4 ) clock. 0 4 8 12 2 6 10 14 1 3 5 7 9 11 13 15 (0) (1) (2) (3) (4) (5) (6) (7)(-8) (-7) (-6) (-5) (-4) (-3) (-2) (-1)

24. 0 4 8 12 2 6 10 14 1 3 5 7 9 11 13 15 Now write the original clock numbers in binary form. (0) (1) (2) (3) (4) (5) (6) (7)(-8) (-7) (-6) (-5) (-4) (-3) (-2) (-1) 0000 0001 0010 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

25. 0 4 8 12 2 6 10 14 1 3 5 7 9 11 13 15 Recall our first objective: “(i) making it easy to identify negative numbers with their positive twins…” It’s easy to identify the negative numbers: their leading digit is “1”. (0) (1) (2) (3) (4) (5) (6) (7)(-8) (-7) (-6) (-5) (-4) (-3) (-2) (-1) 0000 0001 0010 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

26. 0 4 8 12 2 6 10 14 1 3 5 7 9 11 13 15 Recall our first objective: “(i) making it easy to identify negative numbers with their positive twins…” Get the positive twin by subtracting the number from 16 (10000 2 ). (0) (1) (2) (3) (4) (5) (6) (7)(-8) (-7) (-6) (-5) (-4) (-3) (-2) (-1) 0000 0001 0111 10000-1001=0111 10000-1100=0100 10000-1111=0001 0100

27. 0 4 8 12 2 6 10 14 1 3 5 7 9 11 13 15 Recall our first objective: “(i) making it easy to identify negative numbers with their positive twins…” Get the positive twin by subtracting the number from 16 (10000 2 ). (0) (1) (2) (3) (4) (5) (6) (7)(-8) (-7) (-6) (-5) (-4) (-3) (-2) (-1) 0000 0001 0010 0011 0100 0101 0110 0111 10000-1000=1000 10000-1001=0111 10000-1010=0110 10000-1011=0101 10000-1100=0100 10000-1101=0011 10000-1110=0010 10000-1111=0001

28. Recall our first objective: “(i) making it easy to identify negative numbers with their positive twins…”. If the binary number starts with 0, interpret it as a positive number or zero. If the binary number starts with 1, interpret it as the negative of [10000 2 minus the number]. This is called (the negative of) the 2’s complement of the number.

29. 0 4 8 12 2 6 10 14 1 3 5 7 9 11 13 15 (0) (1) (2) (3) (4) (5) (6) (7)(-8) (-7) (-6) (-5) (-4) (-3) (-2) (-1) 0000 0001 0010 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

30. CONCLUSION The 2’s complement system The 2’s complement system codes numbers with n bits. n = 4 in our example. Positive numbers from 0 to 2 n-1 -1 are written in standard binary notation, and are identified by that fact that their leading bit is zero. 0 to 7 in our example. Negative numbers from –2 n-1 to -1 are coded as their “2’s complement’, which is 2 n minus the absolute value of the number, and are identified by the fact that their leading bit is one. -8 to -1 in our example. Calculations are performed using ordinary arithmetic with the numbers binary coded, using an n-bit register, and dropping any “carry” in the n th bit. The user must avoid overflow, which occurs if any number larger than [2 n-1 -1] or lower that -2 n-1 is computed.

31. How about negative numbers on a base-15 clock? 0 4 8 2 6 10 13 1 3 5 7 9 11 12 14 4-7=?

32. How about negative numbers on a base-15 clock? 0 4 8 2 6 10 13 1 3 5 7 9 11 12 14 4-7=12 7

33. The negative numbers are as shown: 0 4 8 11 2 6 10 13 1 3 5 7 9 12 14 (0) (1) (2) (3) (4) (5) (6) (7) (-7) (-6) (-5) (-4) (-3) (-2) (-1)

34. Now write the original numbers in binary form. Note that 0 (which is equivalent to 15) has two representations now, as 0000 and as 1111. 0 4 8 11 2 6 10 13 1 3 5 7 9 12 14 (0) (1) (2) (3) (4) (5) (6) (7) (-7) (-6) (-5) (-4) (-3) (-2) (-1) 0000 0001 0010 0011 0100 0101 0110 01111000 1001 1010 1011 1100 1101 1110 1111

35. Recall our first objective: “(i) making it easy to identify negative numbers with their positive twins…” Again it’s easy to identify the negative numbers: their leading digit is “1”. (Sometimes we call 1111 the “negative zero.”) 0 4 8 11 2 6 10 13 1 3 5 7 9 12 14 (0) (1) (2) (3) (4) (5) (6) (7) (-7) (-6) (-5) (-4) (-3) (-2) (-1) 0000 0001 0010 0011 0100 0101 0110 01111000 1001 1010 1011 1100 1101 1110 1111

36. 0 4 8 11 2 6 10 13 1 3 5 7 9 12 14 Recall our first objective: “(i) making it easy to identify negative numbers with their positive twins…” Get the positive twin by subtracting the number from 15, i.e. (1111 2 ). (0) (1) (2) (3) (4) (5) (6) (7) (-7) (-6) (-5) (-4) (-3) (-2) (-1) 0000 0001 01111111-1000=0111 1111-1011=0100 1111-1110=0001 0100 1111

37. 0 4 8 11 2 6 10 13 1 3 5 7 9 12 14 (0) (1) (2) (3) (4) (5) (6) (7) (-7) (-6) (-5) (-4) (-3) (-2) (-1) 0000 0001 0010 0011 0100 0101 0110 01111000 1001 1010 1011 1100 1101 1110 Recall our first objective: “(i) making it easy to identify negative numbers with their positive twins…” Get the positive twin by subtracting the number from 15, i.e. (1111 2 ). This is too easy. Change every 1 to 0 and every 0 to 1.

38. Recall our first objective: “(i) making it easy to identify negative numbers with their positive twins…”. If the binary number starts with 0, interpret it as a positive number or zero. If the binary number starts with 1, interpret it as the negative of the number obtained by changing each bit. This the (the negative of) the 1’s complement of the number.

39. 0 4 8 11 2 6 10 13 1 3 5 7 9 12 14 (0) (1) (2) (3) (4) (5) (6) (7) (-7) (-6) (-5) (-4) (-3) (-2) (-1) 0000 0001 0010 0011 0100 0101 0110 01111000 1001 1010 1011 1100 1101 1110 1111

40. CONCLUSION The 1’s complement system The 1’s complement system codes numbers with n bits. n = 4 in our example Positive numbers from 0 to (2 n-1 -1) are written in standard binary notation, and are identified by that fact that their leading bit is zero. 0 to 7 in our example Negative numbers from –(2 n-1 -1 ) to -1 are coded as their “1’s complement’, which means every bit is changed from 0 to 1 or from 1 to 0; and are identified by the fact that their leading bit is one. -7 to -1 in our example As a result, 0000 is identified with 1111, which is sometimes called “negative zero.” Calculations are performed using ordinary arithmetic with the numbers binary coded, using an n-bit register, except that any “carry” generated in the n th bit of a sum is carried around and added to the first bit. The user must avoid overflow, which occurs if any number larger than (2 n-1 -1) or lower that –(2 n-1 -1) is computed.