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Lecture 4 Calculations in Computers

Lecture 4 Calculations in Computers. Lecture 4: Calculations in Computer. In this lecture, we will study Nonnumeric calculations Logical calculations Logical Shift, Rotate Numeric calculations Calculations with Fixed Point Numbers Arithmetic Shift Addition and Subtraction

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Lecture 4 Calculations in Computers

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  1. Lecture 4Calculations in Computers Calculations in Computers

  2. Lecture 4:Calculations in Computer In this lecture, we will study • Nonnumeric calculations • Logical calculations • Logical Shift, Rotate • Numeric calculations • Calculations with Fixed Point Numbers • Arithmetic Shift • Addition and Subtraction • Multiplication • Division • Calculations with Floating Point Numbers • Conditional Sum Adder Calculations in Computers

  3. A B A nul = A nul = 10100001 00000000 nul B = 00000000 10100010 A nul v nul B A B = Calculations With Nonnumeric Data:Logical Calculations • Complement • Logical complement, Complement of numbers • AND • Masking out bits = 10100001 10100010 10100001 10100010 ^ 1111111100000000 = 10100001 00000000 Mask • OR • Merging data = 10100001 10100010 Calculations in Computers

  4. 0 0 0 x 0 x 110000011100001 1 100000111000010 0 0 0 1 0 0 110000011100001 100000111000010 0 Calculations With Nonnumeric Data:Logical Shifts Logical shifts • Change the location of data within a register • Moving a particular bit to the carry flag for the purpose of test • Shift Right/Left with/without Carry - always 0 bit is shifted in Shift Right with Carry Shift Left with Carry C Register x1100000111000010 C Register x 1100000111000010 Shift Right Shift Left 1100000111000010 1100000111000010 Always lost a bit by shifting a data, i.e., data cannot be preserved Calculations in Computers

  5. x 110000011100001 1 100000011100010 110000011100001 100000111000010 Calculations With Nonnumeric Data: Rotation Rotate • Similar to Logical Shift • Data cannot be restored after Logical Shifts, but data can be restored after Rotate • Rotate Right/Left with/without Carry Rotate Right with Carry Rotate Left with Carry C x1100000111000010 C x1100000111000010 0 x Rotate Right Rotate Left 1100000111000010 1100000111000010 0 1 Data is preserved Calculations in Computers

  6. A = an-1 an-2 … a1 a0 n-1 V(A) = S ai x 2i i=0 Generalization(n bits): V(ARm) = 2-m x V(A) V(ALm) = 2m x V(A) Calculations With Numeric Data: Arithmetic Shifts Arithmetic Shifts • Arithmetic Shift Right n bits is equivalent to division by 2n • Arithmetic Shift Left n bits is equivalent to multiplication by 2n • The bit shifted in is not always 0 A: an unsigned positive integer Right Shift: AR1 =0an-1 an-2 … a1 n-1 V(AR1) = 2-1 ( S ai x 2i ) = 2-1 x V(A) i=1 Left Shift: AL1 = an-2 … a1 a00 n-2 V(AL1) = 21 ( S ai x 2i ) = 21 x V(A) i=0 Calculations in Computers

  7. 1 0 0 01101000 Arithmetic Shift:Shift RightUnsigned Integer A = an-1 an-2 … a1 a0 n-1 V(A) = S ai x 2i i=0 AR1 =0an-1 an-2 … a1 n-1 V(AR1) = 2-1 ( S ai x 2i ) = 2-1 x V(A) i=1 Generalization: Right shift p bits AR1 =00…0an-1 an-2 … ap n-1 V(AR1) = 2-p ( S ai x 2i ) = 2-p x V(A) i=1 A = 011010001 Shift Right A 1 bit 011010001 Calculations in Computers

  8. 0 0 0 11010001 Arithmetic Shift:Shift Left Unsigned Integer A = an-1 an-2 … a1 a0 n-1 V(A) = S ai x 2i i=0 AL1 =an-2 … a1a00 n-1 V(AL1) = 21 ( S ai x 2i ) = 21 x V(A) i=1 Generalization: Left shift p bits AR1 =an-p-1 … a000…0 n-1 V(AR1) = 2-p ( S ai x 2i ) = 2-p x V(A) i=1 A = 011010001 Shift Left A 1 bit 011010001 Calculations in Computers

  9. Representation Direction After Shift Comment 2’s Complement R AR1 = ban an-1 an-2 … a1b=an (if a0=1, truncation) L AL1 = an-1 an-2 … a1 a0b b=0 (if an-1=an, overflow) 1’s complement R AR1 = banan-1 an-2 … a1b=an (if a0=an, truncation) L AL1 = an-1 an-2 … a1 a0b b=an (if an-1=an, overflow) Calculations With Numeric Data:Arithmetic Shifts Arithmetic Shifts of a Signed Integer A A = an an-1 an-2 … a1 a0 with an: sign bit Sign plus Magnitude R AR1 = anb an-1 an-2 … a1b=0 (if a0=1, truncation) LAL1 = an an-2 … a1 a0bb=0 (if an-1=1, overflow) Calculations in Computers

  10. = +37/2 = +101/2 0 0 0 0 0 010010 0 0 100100 0 0 0 1 1 0 0 010010 0 100101 0 Arithmetic Shifts:Sign + Magnitude Shift R 1 bit: AR1 = anb an-1 an-2 … a1AR1 = A/2: b=0 (if a0=1, truncation) Shift L 1 bit: AL1 = an an-2 … a1 a0b AL1 = 2A: b=0 (if an-1=1, overflow) A = 0 0100100 = +36(a0=0 and an-1=0) Shift A 1 bit to the Right: Shift A 1 bit to the Left: 0 0100100 00100100 = +18 = A/2 = +72 = 2 x A C = 0 1100101 = +101(an-1=1) B = 0 0100101 = +37(b0=1) Shift B 1 bit to the Right: Shift C 1 bit to the Left: 0 0100101 01100101 = +18 = +74 Truncation Overflow Calculations in Computers

  11. Shift R 1 bit: AR1 = ban an-1 an-2 … a1A = A/2: b=an (if a0=1, truncation) Shift L 1 bit: AL1 = an-1 an-2 … a1 a0b A = 2A: b=0 (if an-1=an, overflow) 0 0 0 0 010110 0 101100 0 C = 0 1101101 = +109(an-1= an) 0 0 1 0 0 0 010110 1 101101 0 = B/2 = 2 x C Arithmetic Shifts:2’s Complement A = 0 0101100 = +44(a0=0 and an-1 = an) Shift A 1 bit to the Right: 00101100 Shift A 1 bit to the Left: 00101100 = +22 = A/2 = +88 = 2A B = 0 0101101 = +45(a0=1) Shift B to the Right 1 bit: Shift C to the Left 1 bit: 0 0101101 0 1101101 = +22 = - 22 Truncation Overflow Calculations in Computers

  12. 0 0 0 0 010110 0 101100 0 C = 0 1101101 = +109(an-1= an) B = 0 0101101 = +45(a0= an) Shift R 1 bit: AR1 = banan-1 an-2 … a1AR1= A/2:b=an (if a0=an, truncation) Shift L 1 bit: AL1 = an-1 an-2 … a1 a0b AL1 = 2A: b=an (if an-1=an, overflow) 0 0 1 0 0 0 010110 1 101101 0 = B/2 = 2 x C Arithmetic Shifts:1’s Complement A = 0 0101100 = +44(a0=an and an-1 = an) Shift A 1 bit to the Right: 00101100 Shift A 1 bit to the Left: 00101100 = +22 = A/2 = +88 = 2A Shift C to the Left 1 bit: Shift B to the Right 1 bit: 0 0101101 0 1101101 = +22 = - 37 Truncation Overflow Calculations in Computers

  13. 01101101 +) 00010110 10000011 carry bits 1 carry propagation xi yi ci-1 si ci 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 Calculation With Fixed Point Numbers: Addition X = xn xn-1 … x1 x0,Y = yn yn-1 … y1 y0 X + Y = S + cn Calculations in Computers

  14. If xn = yn then subtract the magnitudes If xn = yn, no overflow but if cn=1 then end-around carry Calculation With Fixed Point Numbers: Addition • Sign plus Magnitude • If xn = yn then add the magnitudes • 2’s complement • Add including sign bit • If xn = yn = 0 and cn-1 =1 then sn=1 thus overflow • If xn = yn = 1 and cn-1 =0 then sn=0 thus overflow • 1’s Complement • Add including sign bit • If xn = yn = 0 and cn-1 =1 then sn=1 thus overflow • If xn = yn = 1 and cn-1 =0 then sn=0 thus overflow , but if cn-1 = 1 no overflow • but if cn = 1 we need to add 1 to the result - end-around-carry Calculations in Computers

  15. +) X = 00100101 +) Y =00010011 +) X = 10111101 +) Y =11010011 +) X = 10100101 +) Y =10010011 +) X = 00111101 +) Y =01010011 cn-1 = 1 cn-1 = 1 Addition:Sign + Magnitude S = 0111000 0 S = 0111000 1 Overflow if cn-1=1 S = 0010000 1 S = 0010000 0 Calculations in Computers

  16. +) X = 00111101 = +61 +) Y = 01010011 = +83 +) X = 10101101 = -83 +) Y = 11000011 = -61 sn = 1 cn-1 = 1 sn = 0 cn-1 = 0 Addition:2’s Complement • Add including sign bit • If xn = yn = 0 and cn-1 =1 then sn=1 thus overflow • If xn = yn = 1 and cn-1 =0 then sn=0 thus overflow = -112 S = 10010000 = +112 S = 01110000 Overflow Overflow Calculations in Computers

  17. +) X = 10101101 = -82 +) Y = 11010011 = -44 +) X = 00010101 = +21 +) Y = 11010011 = -44 +) X = 00010101 = +21 +) Y = 11110011 = -12 +) X = 00111101 = +61 +) Y = 01010011 = +83 +) X = 10101101 = -82 +) Y = 11000011 = -60 • If xn = yn, no overflow but if cn=1 then end-around carry sn = 1 cn-1 = 1 sn = 0 cn-1 = 0 cn=1 cn=1 cn=0 End-around carry cn-1 = 1 Addition:1’s Complement • Add including sign bit • If xn = yn = 0 and cn-1 =1 then sn=1 thus overflow • If xn = yn = 1 and cn-1 =0 then sn=0 thus overflow , but if cn-1 = 1 no overflow • but if cn = 1 we need to add 1 to the result - end-around-carry = +112 S = 01110000 = -111 S = 10010000 Overflow Overflow S = 1 1101000 = -23 S = 00001000 S = 10000000 +) 1 Not overflow +) 1 1 0000001 = -126 0 0001001 = +9 Not overflow Not overflow Calculations in Computers

  18. Summer in Lake Powell, Arizona Calculations in Computers

  19. X - Y = D + bn xi yi bi-1 di bi 0 0 0 0 0 0 0 1 1 1 0 1 0 1 1 0 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 Calculation With Fixed Point Numbers: Subtraction Calculations in Computers

  20. Size of Z: (2n-1) x (2n-1) = 22n - 2n+1 + 1 22n-1 < Z < (22n - 1) 2n bits (n-1)-bit SL(X) (n-2)-bit SL(X) X Calculation With Fixed Point Numbers:Multiplication - Unsigned Numbers X(Multiplicand), Y(Multiplier) : Positive unsigned numbers X = xn-1 … x1 x0,Y = yn-1 … y1 y0 Z(Product) = X . Y We need a product register whose length is 2 times longer than the multiplicand and the multiplier. n-1 n-1 Z = X . Y = X. ( S yi . 2i) = S yi [ X . 2i] i=0 i=0 = yn-1 [ X . 2n-1 ] + yn-2 [ X . 2n-2 ] + … + y0 [ X . 20] Calculations in Computers

  21. (22n-1 - 1) < (22n - 2n+1 + 1) < (22n - 1) Calculation With Fixed Point Numbers:Size of the Register for Product • Size of registers for multiplicand and multiplier = n bits • Largest integer multiplicand and multiplier = 2n-1 • Largest product • (2n-1) x (2n-1) = 22n - 2 x 2n + 1 • = 22n - 2n+1 + 1 • Therefore, product register size must be 2n bits Calculations in Computers

  22. Add if X y0=1 X.21 y1=1 X.22 y2=1 … … X.2n-2 yn-2=1 +) X.2n-1 yn-1=1 Z Calculation With Fixed Point Numbers:Multiplication - Unsigned Numbers Example - Unsigned positive integer Multiplication Rule X = 01101010 = +106 Y = 00100101 = +37 X +) 01101010 y0=1 X.21+) 011010100y1=0 X.22 +) 0110101000 y2=1 +) 1010010010 X.23+) 01101010000 y3=0 X.24+)011010100000 y4=0 X.25+) 0110101000000 y5=1 a+). 0111101010010 X.26+) 01101010000000y6=0 X.27+) 011010100000000y7=0 Z +) 0111101010010 =+3922 Calculations in Computers

  23. n-bit SL(X) (n-1)-bit SL(X) (n-2)-bit SL(X) X X.20 y0 X . 21 y1 … … X . 2n-2 yn-2 +) X . 2n-1 yn-1 - ) X . 2n yn Z Calculation With Fixed Point Numbers:Multiplication - Signed 2’s Complement X = xn xn-1 … x1 x0,Y = yn yn-1 … y1 y0 n-1 Z = X . Y = X [ - yn . 2n + S yi . 2i] i=0 = - yn .[X . 2n ]+ yn-1[ X . 2n-1 ] + yn-2[ X . 2n-2 ] + … + y0[ X . 20] Calculations in Computers

  24. X.20 y0 X . 21 y1 … … X . 2n-2 yn-2 +) X . 2n-1 yn-1 - ) X . 2n yn Z Calculation With Fixed Point Numbers:Multiplication - Signed 2’s Complement X = 0 0101010 = +42 Y = 1 1110101 = - 11 X.20+)00101010 y0=1:Add X.21+)001010100y1=0 X.22+) 0010101000 y2=1:Add X.20+)0011010010 X.23+)00101010000y3=0 X.24+) 001010100000 y4=1:Add X.23+)001101110010 y3=0 X.25+) 0010101000000 y5=1:Add X.23+)0100010110010 y3=0 X.26+) 00101010000000y6=1 X.23+)01001100110010 y3=0 X.27-) 001010100000000y7=1:Sub X.2)=+) 110101100000000 + 2’s Compl X.27-) 111111000110010 =- 462 y7=1 Calculations in Computers

  25. n-bit SL(X) (n-1)-bit SL(X) (n-2)-bit SL(X) X X X.20 yn X.20 y0 X . 21 y1 … … X . 2n-2 yn-2 +) X . 2n-1 yn-1 - ) X . 2n yn Z Calculation With Fixed Point Numbers:Multiplication - Signed 1’s Complement X = xn xn-1 … x1 x0,Y = yn yn-1 … y1 y0 n-1 Z = X . Y = X [ yn . (1 - 2n )+ S yi . 2i] i=0 = - yn .[X . 2n ]+ yn-1[ X . 2n-1 ]+ yn-2[ X . 2n-2 ]+ … + y0[ X . 20]+ yn .[ X ] Calculations in Computers

  26. X.20 yn X.20 y0 X . 21 y1 … … X . 2n-2 yn-2 +) X . 2n-1 yn-1 - ) X . 2n yn Z Calculation With Fixed Point Numbers:Multiplication - Signed 1’s Complement X = 0 0101010 = +42 Y = 1 1110101 = - 10 X.20 +) 00101010 y7=1:Add X.20+) 00101010 y0=1:Add X.20+)01010100 y0=1 X.21+)001010100y1=0 X.22+) 0010101000 y2=1:Add X.20+)0011111100 X.23+)00101010000 y3=0 X.24+) 001010100000 y4=1:Add X.23+)001110011100 y3=0 X.25+) 0010101000000 y5=1:Add X.23+)0100011011100 y3=0 X.26+) 00101010000000y6=1 X.23+)01001101011100 y3=0 X.27-) 001010100000000y7=1:Sub X.2)=+) 110101011111111 + 1’s Comp X.27-) 111111001011011 =- 420 y7=1 Calculations in Computers

  27. Time Out • 매 주말 만사 제쳐 놓고 등산을 즐기는 초등학교 때부터 단짝으로 지내는 중년 신사 두 사람이 있는데, 이번 주말에도 어김없이 등산을 하고 있었다. • 이 때 저 아래 산을 끼고 도는 국도에는 하나의 장례차 행렬이 지나가고 있었다. • 이를 본 이들 중 한 사람이 등산모를 벗어 가슴에 얹고 정중하게 예의를 갖추었다. • 장례 행렬이 산 모퉁이를 사라지자 이를 보고 있던 다른 신사가 말했다. “ 우리가 지금까지 오랫동안 절친한 친구로 지내 왔는데, 자네가 그렇게 예의 바른 사람인줄은 몰랐네” • “자네는 모르겠지만 저 장례 행렬은 40년을 같이 살아 온 우리 집사람의 장례 행렬인데 그 정도의 예의는 갖추어야 하지 않겠는가?” 예의 바른 신사가 대답했다. Calculations in Computers

  28. Calculation With Fixed Point Numbers: Division 125 - ) 7 118 1 -) 7 111 2 -) 7 104 3 -) 7 97 4 -) 7 90 5 … 20 -) 7 13 16 -) 7 6 17 -) 7 - 1 +) 7restore 6 remainder Addition(+,-) 19 times -) 00125 -) 00700 -- 00575 00000 +) 00700restore -) 00125 -) 00070 -) 00055 00010 -) 00070 -- 00025 00010 +) 00070restore -)00055 -) 00007 -) 00048 00011 ... -) 00013 -) 00007 00017 -) 00006 -) 00007 -- 00001 00017 +) 00007 restore -) 00006 remainder Addition(+,-) 14 times 125 / 7 Calculations in Computers

  29. Calculation With Fixed Point Numbers:Division - Unsigned Numbers X, Y : Positive unsigned numbers X = xn-1 … x1 x0,Y = yn-1 … y1 y0 X(Dividend) / Y(Divisor) = Q(Quotient) + R(Remainder) Size of Q: Initial subtraction after aligning the MSB of Dividend with the LSB of Divisor. Size of Q is 2n-1 bits. However, if we want to deal with signed numbers, the size of Q should be 2n bits. Calculations in Computers

  30. Calculation With Fixed Point Numbers:Division - Unsigned Numbers 10 / 5 = 2(q3q2q1q0) + 0 Restoring Division X = 1010, Y = 0101 Initial X = X(4)0001010 0001010 X(4) -) 0101000 -Y.23 1100010 X(3) <0 q3 = 0 +) 0101000 Restore X(3) 0001010 X(3) -) 0010100 -Y.22 1110110 X(2) < 0 q2 = 0 +) 0010100 Restore X(2) 0001010 X(2) -) 0001010 -Y.21 0000000 X(1) > 0 q1 = 1 -) 0000101 -Y.20 1111011 X(0) < 0 q0 = 0 +) 0000101 +Y.20 Restore 0000000 X(0) R Non-restoring Division X = 1010, Y = 0101 Initial X=X(4) = 0001010 0001010 X(4) -) 0101000 -Y.23 1100010 X(3) < 0 q3 = 0 +) 0010100 +Y.22 1110110 X(2) < 0 q2 = 0 +) 0001010 +Y.21 0000000 X(1) > 0 q1 = 1 -) 0000101 -Y.20 1111011 X(0) < 0 q0 = 0 +) 0000101 +Y.20Restore 0000000 X(0) R Calculations in Computers

  31. Calculation With Fixed Point Numbers:Division - 2’s Complement 10 / 5 = 2(q4 q3q2q1q0) + 0 Instead of subtraction, 2’s compliment of addition Non-restoring Division X = 10 = 01010>0, Y = - 5 = 11011<0 Initial X = 00001010 000001010 X(5) +) 110110000 +Y.24 110111010 X(4) < 0 q4 = 1 sign bit +) 000101000 -Y.23 + 2’s complement of Y.23 111100010 X(3) < 0 q3 = 1 +) 000010100 -Y.22 + 2’s complement of Y.22 111110110 X(2) < 0 q2 = 1 +) 000001010 -Y.21 + 2’s complement of Y.21 000000000 X(1) > 0 q1 = 0 +) 111111011 +Y.20 111111011 X(0) < 0 q0 =1 +) 000000101 -Y.20 + 2’s complement of Y 000000000 X(0) R = 0 Q = 1 1101 + 0 0001 = 1 1110 (since q4 is 1, i.e. Q is negative add 1 to the resulting Q) Calculations in Computers

  32. Calculation With Fixed Point Numbers:Division - 1’s Complement 10 / 5 = 2(q4 q3q2q1q0) + 0 Instead of subtraction, 1’s compliment of addition Non-restoring Division X = 10 = 01010, Y = - 5 = 11010 Initial X = 00001010 000001010 X(5) +) 110101111 +Y.24 110111001 X(4) < 0 q4 = 1 sign bit +) 000101000 -Y.23 111100001 X(3) < 0 q3 = 1 +) 000010100 -Y.22 111110101 X(2) < 0 q2 = 1 +) 000001010 -Y.21 111111111 X(1) = 0 q1 = 0 +) 111111010 +Y.20 1 111111001 end-around carry +) 000000001 111111010 X(0) < 0 q0 = 1 +) 000000101 -Y.20 111111111 X(0) R = 0 Q = 1 1101 Calculations in Computers

  33. Calculations in Computers

  34. Calculation With Floating Point Numbers F1 = A1 . B1 , F2 = A2 . B2, #: add or subtract, !: multiply or divide F1 # F2 = [B1 . RA1] # [B2 . RA2] = [B1 # B2 . RA2-A1 ] . RA1 (A1 > A2) or = [B1 . RA1-A2 # B2 ] . RA2 (A1 < A2) F1 ! F2 = [B1! B2] . RA1 #A2 Calculations in Computers

  35. No need to worry about carry propagation between code symbols, simply add code symbols. |1/2|7 = |2/4|7 = |3/6|7 = |4/8|7 = |5/10|7 = |6/12|7 = |7/14|7 2/4 3/6 4/1 5/3 6/5 0/0 Chinese Remainder Representation: Addition M1=2, m2=3, m3=5, m4=7 M=210 si = | ai + bi |mi 3: 1 0 3 3 +) 12: 0 0 2 5 15: |1|2 |0|3 |5|5 |8|7 = (1 0 0 1) Verification: N1=105, N2=70, N3=42, N4=30 | (105|1/105|2) + (70|0/70|3) + (42|0/42|5) + (30|1/30|7) |210 = |(105|1/1|2) + 0 + 0 + (30 |1/2|7 ) |210 = | 105 + (30|4/1|7) |210 = | 105 + 120 | 210 = | 225 |210 = 15 Calculations in Computers

  36. Subtraction by Adding mi-complement: di = | ai + bi’ |mi bi’ = mi - bi if bi = 0 = 0 if bi = 0 (1 0 4 2) = | (|105|1/105|2+0 + (42|4/42|5) + (30|2/30|7) |210 = | 105 + 0 + (42|2/1|5) + (30|2/2|7) |210 = |105 + 84 + 30 |210 = |219|210 = 9 Chinese Remainder Representation: Subtraction Subtraction of a positive number: di = | ai - bi |mi 12: 0 0 2 5 -) 3: 1 0 3 3 |-1|2 |0|3 |-1|5 |2|7 = (1 0 4 2) 12: 0 0 2 5 +) -3: 1 0 2 4 |1|2 |0|3 |4|5 |9|7 = (1 0 4 2) Calculations in Computers

  37. mi-complement (0 0 2 5) (0 0 2 5) = | (105|0/105|2 + (70|0/70|3) + (42|2/42|5) + (30|5/30|7) |210 = | 0 + 0 + 42 + (30|5/2|7) |210 = | 42 + (30|6/1|7) |210 = | 222 |210 =12 Chinese Remainder Representation: Multiplication Product P: pi = | ai x bi |mi 3: 1 0 3 3 x) 4: 0 1 4 4 0 0 |12|5 |12|7 = (0 0 2 5) Multiplication of a negative number by mi-complement 4: 0 1 4 4 x) -3: 1 0 2 4 |0|2 |0|3 |8|5 |16|7 = (0 0 3 2) Calculations in Computers

  38. Calculation With Chinese Remainder Representations • Very compact code - mi-remainders • Allows to calculate with smaller numbers • Faster with smaller numbers • mi-complement arithmetic is possible • Difficult to compare values of numbers • No carries to propagate between mi-remainders (although we have to consider the carry propagation within an mi-remainder) • Division is extremely difficult Calculations in Computers

  39. log2n selection time 30 0 0 0 1 1 1 1 0 c level +) 177 1 0 1 1 0 0 0 1 c s c s c s c s c s c s c s c s 0 1 00 0 1 1 00 1 010 1 0 1 0 1001 101 1 10 1 0 10 10 1 1 Conditional Sum Adder • Generate 2 pairs of sum and carry bits for each bit with 1 and 0 carry-in • 1 bit Add time • Results are examined to select the correct sum and carry bits by examining • in the 2nd level, adjacent bit pairs, • in the 3rd level, adjacent 2-bit pairs, • in the 4th level, adjacent 4-bit pairs, and so on to select correct sum and carry bits => 0 10 0 1 1 1 0 0 10 1 0 1 1 10 0 01 1 2 0 1 1 0 0 0 1 1 0 1 01 1 1 1 3 0 1 1 0 0 1 1 1 1 4 Calculations in Computers

  40. a0 b0 a1 b1 a2 b2 an bn . . . CSA CSA CSA CSA s0 s1 s2 sn Addition with CR RepresentationUsing CSA S = A + B A = (a0, a1, a2, … , an) and B = (b0, b1, b2, … , bn) • Each of ai’s and bi’s is represented with p bits, where p is the number of bits needed to represent Max{mi} • Thus if q is the number of bits required to represent A or B, where q>>p which makes the CR representation faster • Each of ai’s and bi’s is represented in any binary number representation Calculations in Computers

  41. r=2 • 1964 S. Winograd(IBM) • Fan-in must be considered: Fan-in = r • Number of levels = logr 2n • Theoritical lower bound T = logr 2n t, t=block processing delay r=4 • Theoritical Lower bound for CR representation • A(N): largest remainder • d: base used for representing remainders • n: number of digit when representing A(N) in base d n = logd A(N) • add time T = logr 2 logd A(N) t Theoritical Lower Bound of Add Time • 1962 U. Offman(USSR) • For n-bit numbers, T = f(log2 n) Calculations in Computers

  42. Fan-in = r, but each processing block must take a pair of r/2 digits from both operands } r T > 1-bit add time + log r/2 logd A(N) t r/2 r/2 r/2 . . . . . . For n-bit binary remainder T > 1-bit add time + log r/2 (n/ r/2 ) t Binary n=32, r=3 r=5 r=7 CSA 6t 4t 3t Winograd(LB) 4t 3t 3t Achievable Speed of Addition Conditional Sum Adder with Chinese Remainder representation A(N) = Max {mi} i Calculations in Computers

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