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15-122: Principles of Imperative Computation Recitation 3 Section C September 5, 2012 Office Hours Tuesdays from 7:00 to 9:00 PM, GHC 4122/26 Come with questions about class, homework, life. . . TA Carsten Josh Jonathan Kyle Alex Nivedita Bill Klas Ryan Room GHC 6505 GHC 4102 GHC 5709 GHC 4122/26 ” ” ” ” ” Time Thu 1:00 - 3:00 Wed 6:00 - 8:00 Wed 1:30 - 2:30, Fri 1:30 - 2:30 Fri 2:30-4:30 Tue 4:30-5:30, Thurs 4:30-5:30 Wed 3:30-4:30, Fri 3:30 - 4:30 Tue 3:00 - 4:00, Thu 3:00 - 4:00 Thu 7:30 - 9:30 Mon 3:00 - 4:00 CA Arjun Elliot Kathy Ling-Yi Maria Room GHC 5201 ” ” ” ” Time Mon 6:00 - 8:00 Sun 6:00 - 8:00 Thu 10:00 - midnight Wed 8:30 - 10:30 Tue 6:00 -8:00 Binary Representation Our familiar numerical system is known as base-10. 10 unique digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 348210 = (3 × 103 ) + (4 × 102 ) + (8 × 101 ) + (2 × 100 ) Computers use a base-2 numeral system. 2 unique digits: 0, 1 2710 = 110112 = (1 × 24 ) + (1 × 23 ) + (0 × 22 ) + (1 × 21 ) + (1 × 20 ) Conversion - Decimal to Binary: Let n be a number in base-10 with binary representation nk nk−1 . . . n1 n0 . Method 1: Division with remainder i 0 1 ··· k−1 k n 2i Method 2: Power Table 2k nk remainder n0 n1 ··· nk−1 nk n n 2 ··· n 2k−1 n 2k 2k−1 ··· i ··· 20 Test 2k ≤ n nk 2k + 2k−1 ≤ n nk−1 .. . ··· ni nk .. 2k + nk−1 . n0 ( 2k−1 + · · · + 2i ≤ n ··· i 0 i=1 ni 2 ) + 2 = n Pk Modular Arithmetic Suppose we have 4-bit integers. Then there are 24 = 16 numbers we can represent: 0000, 0001, . . . , 1110, 1111. That’s the interval [0, 15]. Let’s do some simple math: + 1 0 1 0 1 1 1 0 1 0 1 1 (10) (5) (15) - 01 10 1 10 0 0 1 1 1 1 1 1 0 (13) (7) Awesome. Things seem to work as expected. (6) But what if our operations go outside [0, 15]? 11 + 1 0 0 1 1 0 0 1 1 1 0 1 (13) (6) We’re dealing with 4-bit ints; there’s no room for the fifth bit. (19) 11 So we just chop it off and look at the 4 lowest bits. + (1) 0 0 1 1 0 0 1 1 1 0 1 (13) (6) (3) We’re working mod 24 , so our values can’t be greater than 15. In general, if you have k bits, the highest number you can represent is 2k − 1. The following properties are true for arithmetic on integers modulo some number n: Addition Multiplication x+y =y+x x∗y =y∗x (commutativity) (x + y) + z = x + (y + z) (x*y)*z = x*(y*z) (associativity) x+0=x x∗1=x (application of identity) x + (−x) = 0 (additive inverse) x ∗ (y + z) = x ∗ y + x ∗ z (distributivity) −(−x) = x (cancellation) x ∗ 0 = 0 (annihilation) 2’s Complement Arithmetic Looking again at our 4-bit numbers, we decided that they span [0, 15]. But what about negative numbers? Let’s look at the 4 bits a little differently. Let the most significant (leftmost) bit carry a negative weight. Then n3 n2 n1 n0 = (n3 ∗ (−1) ∗ 23 ) + (n2 ∗ 22 ) + (n1 ∗ 21 ) + (n0 ∗ 20 ). What are the numbers we have? 0000 (0) 1111 (-1) 0001 (1) 1110 (-2) 0010 (2) 1101 (-3) 0011 (3) 1100 (-4) 0100 (4) 1011 (-5) 0101 (5) 1010 (-6) 0110 (6) 1001 (-7) 0111 (7) 1000 (-8) Things to note: 1. -0 = 0 : we didn’t lose this property when changing the way we interpret things. 2. min int = -8, max int = 7 : the min negative number has a greater magnitude than the max positive number 3. −x = (∼ x) + 1 : To get −x from x, you flip all the bits of x and then add 1. 4. −min int = min int : the negative of the minimum expressible number is itself. Hexadecimal Back to representations - hexadecimal, or base-16, is another numerical system computers use. 16 unique digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, a, b, c, d, e, f If n = nk nk−1 . . . n1 n0 is a hex number, then n = (nk × 16k ) + (nk−1 × 16k−1 ) + . . . + (n1 × 161 ) + (n0 × 160 ) 16 is kind of a clunky number to work with. Thankfully, we’ve got a nicer way of looking at these hex numbers. Each hex digit can range from 0 to 16. That’s the span of 4 bits. Let’s explore that idea: hex 2 3 d 1 9 4 f 6 binary 0010 0011 1101 0001 1001 0100 1111 0110 Then 23d194f616 = 001000111101000110010100111101102