5/15/2019 Square Root - Direct Method Square Root - Direct Method - PowerPoint PPT Presentation

5/15/2019 Square Root - Direct Method Square Root - Direct Method In IEEE floating point standard a real number is � To help the calculation of represented as : −1 � × � × 2 � the representation of R must be changed into � = −1 � × �′ × 2 �� � � = 0 where In 32-bit representation : � � = Int � 2 � � ∈ 0, 1 � � ∈ 0, 4 � ∈ 1, 2 � ∈ 0, 2 Or � ∈ −126, ⋯ , 127 � = −127 if � = −1 � × � � ∈ 0, 2 � � × 2 � � with Then normal sub-normal Square Root February 2017 Square Root February 2017 Square Root - Direct Method Square Root - Direct Method � = −1 � × � × 2 � Let R be a positive real number ( ) � = 0 �′ ∈ 0, 1 �′ However, when the calculation of may lead to a lost of precision � We seek to calculate the positive real number � = −1 � × � × 2 � � � �′ = 0 �′ = 0 Therefore if �′ ∈ 0, 1 �′ �′ ×2 � and if is decreased and is � is easy to calculate ... except when is odd 1, 4 2 until it can fit within Square Root February 2017 Square Root February 2017 1

5/15/2019 Square Root - Direct Method Square Root - Direct Method - Then, the problem can be stated as : � ≤ � < � - + 2 $- � � - = " # $% × 2 $% � - Let such as Given a positive real number � ∈ 1, 4 %&' we seek to calculate such as � � = � � � � � ∈ 1, 2 � - � -$, is obtained from At each iteration � ! � � Let be an approximation of coded on n +1 bits � - = � -$, + # $- 2 $- # $- ∈ 0, 1 ! � ≤ � < � ! + 2 $! � � ! = " # $% × 2 $% � ! 2 $- 0 such as is denoted - %&' � ! We propose to calculate digit-by-digit Square Root February 2017 Square Root February 2017 Square Root - Direct Method Square Root - Direct Method -$, � ≤ � < � - + 2 $- � � - � -$, = " # $% × 2 $% Let such as � < � - + 2 $- � − � - %&' � ≤ � < � -$, + 2 $(-$,) � � Let Δ - = � − � - � 0 ≤ � − � - � -$, 0 ≤ Δ - < 2 $- 2� - + 2 $- � � - < 2 yet and � - + 2 $- ≤ 2 � - 2 $, 2 $- $- 0≤Δ - < 4×2 � -$, then 0 ≤ Δ ' < 4 o 0 0 0 0 0 ? o At each iteration the upper bound 2 ' 2 $� Δ - of is divided by 2 Square Root February 2017 Square Root February 2017 2

5/15/2019 Square Root - Direct Method Square Root - Direct Method zero Implementation � Δ - = � − � - detect Δ - = � − � -$, + # $- 2 $- � 2 - = 2 2 -$, −# $- � -$, +2 $� 0 -$, +2 $-$, 2 - = 2 2 -$, −# $- � -$, -$, + # $- 2 $- -$, + # $- 2 $, 0 � � - = � - = � D k-1 -$, � + 2# $- 2 $- � -$, + # $- 2 $- 2 $- << 1 Δ - = � − � -$, subtract - = 2 $- = 2 $, 0 0 -$, Δ - = Δ -$, − # $- 2 $- 2� -$, + 2 $- { or X k-1 or 2 - Δ - = 2 - Δ -$, − # $- 2� -$, + 2 $- Let 2 - = 2 - Δ - >> 2 W k-1 >> 1 0 2 - = 22 -$, − # $- 2� -$, + 2 $- such as 0 ≤ 2 - # $- = last 1 weight Square Root February 2017 Square Root February 2017 { Square Root - Direct Method Square Root - Direct Method - Improvement radix 4 0 2 - = 22 -$, − # $- 2� -$, + 2 $- � - = � -$� + 2# $-3, + # $- 2 $- # $- = such as 0 ≤ 2 - 1 � - { Iteration scheme : # $-3, # $- 2 $, � -$� 2 - = 2 2 -$, − # $- � -$, + 2 $-$, ? ? 0 0 0 � - = � -$, + # $- 2 $- 0 # $- = such as 0 ≤ 2 - 2 ' 2 $� 1 Square Root February 2017 Square Root February 2017 3

5/15/2019 Square Root - Direct Method - Improvement radix 4 2 - = 4 2 -$� # $-3, = 0, # $- = 0 � - = � -$� -$� −1 2 � -$� +1 × 2 $-$, # $-3, = 0, # $- = 1 2 - = 4 2 � � - = � -$� + 1 × 2 $- -$� − � -$� +2 × 2 $-$, 2 - = 4 2 # $-3, = 1, # $- = 0 � - = � -$� + 2 × 2 $- -$� − 3 2 � -$� + 3× 2 $-$, 2 - = 4 2 � # $-3, = 1, # $- = 1 � - = � -$� + 3 × 2 $- Square Root February 2017 Square Root - Direct Method - Improvement zero Implementation detect 2 - = 4 2 -$� −1 2 $-$, � � -$� +1×2 � - = � -$� D k-2 D k-1 $-$, << 2 − � -$� +2×2 << 1 subtract -$� −1 2 -$� +1×2 $-$, 2 - = 4 2 � � −3 2 $-$, � � -$� +3×2 or � - = � -$� + 1 × 2 $- X k-2 X k-1 ×1×2 $� or or -$� +2×2 $-$, 2 - = 4 2 -$� − � $� or ×2×2 � - = � -$� + 2 × 2 $- >> 2 or W k-1 W k-2 ×3×2 $� >> 1 >> 2 -$� −3 2 2 - = 4 2 � � -$� +3×2 $-$, � - = � -$� + 3 × 2 $- last weight Square Root February 2017 4