CSE 246: Computer Arithmetic Algorithms and Hardware Design Instructor: Prof. Chung-Kuan Cheng Winter 2004 Lecture 8

CSE 2462 Topics: Midterm Radix-4 SRT Division Division by a Constant Division by a Repeated Multiplication

CSE 2463 Midterm II.RNS to Decimal conversion (2|3|4)RNS(7|6|5) = ?  1 = 4  2 = 5  3 = 3

CSE 2464 Midterm III.One ’ s Complement Adder Loop back carryout of FA7 to the carry in of FA0 FA7FA6FA5FA4FA3FA2FA1FA0  The Delay is 8.7 because the maximum carry propagation is once cycle

CSE 2465 Midterm IV.Prefix Adders I.Ripple-carry Adder

CSE 2466 Midterm IV.Prefix Adders I.Prefix Adder

CSE 2467 Project Update  Come in to speak briefly about the final project Status Update 4 or 5 – 6:30 p.m. Tuesday or Thursday

CSE 2468 Radix-4 SRT Division  4s j-1 = q j d + s j where q j is in [-2,2] and s j-1 is in [-hd,+hd] h is less than or equal to 2/3 Therefore, s j-1 is in [-2d/3, 2d/3] And, 4s j-1 is in [-8d/3, 8d/3]  s shifts to the left by 2 bits

CSE 2469 Radix-4 SRT Division d/3 q j =1 q j =0 q j =2  The overlap regions of q j denote a choice still allowing for recursion Anything about 8/3 goes against our assumption and is therefore the infeasible region 4s j-1 d

CSE Radix-4 SRT Division  The value of q j determines the range it governs  For example, q j = /3 = 5/3 1 – 2/3 = 1/3 The range is 1/3 to 5/3

CSE Division by a Constant  Multiplication is O(log n) but division is linear…much slower Try to convert division to multiplication  Property: Given an odd number d m such that d*m = 2 n – 1  Ex. d = 3, m = 53*5 = 2 4 – 1 d = 7, m =97*9 = 2 6 – 1 d = 11, m = * 93 = E

CSE Division by a Constant  1/d = m/(2 n – 1)  1/(1-r) = 1+r+r 2 +r 3 + … = (1+r)(1+r 2 )(1+r 3 )(1+r 4 ) …  Example z/7 = mz/(2 n -1) log(n/6) operations m 1 2 n 1-2 -n = 2n2n m (1+2 -n )(1+2 -2n )(1+2 -4n ) z = z ( )( )( )

CSE Division by a Repeated Multiplication  q = z/d = (z/d)*(x 0 /x 0 )*(x 1 /x 1 ) … *(x k-1 /x k-1 )  Let xi = 2-di d 1 = dx o = d(2-d) = 1-(1-d) 2 d i+1 = d i x i = d i (2-d i ) = 1-(1-d i ) 2 1-d i +1 = (1-d i ) 2 … quadratic convergence  For k-bit operands, we need 2m-1 multiplications m 2 ’ s complement m = ceiling(log 2 k) with log 2 m extra bits for precision