Tom edits

src/Advanced/Fast Inverse Root/main.typ

@@ -13,16 +13,19 @@
   by: "Mark",
 )
 
-#include "parts/00 int.typ"
+#include "parts/00 intro.typ"
 #pagebreak()
 
-#include "parts/01 float.typ"
+#include "parts/01 int.typ"
 #pagebreak()
 
-#include "parts/02 approx.typ"
+#include "parts/02 float.typ"
 #pagebreak()
 
-#include "parts/03 quake.typ"
+#include "parts/03 approx.typ"
 #pagebreak()
 
-#include "parts/04 bonus.typ"
+#include "parts/04 quake.typ"
+#pagebreak()
+
+#include "parts/05 bonus.typ"

src/Advanced/Fast Inverse Root/parts/00 intro.typ

@@ -0,0 +1,45 @@
+#import "@local/handout:0.1.0": *
+
+= Introduction
+
+In 2005, ID Software published the source code of _Quake III Arena_, a popular game released in 1999. \
+This caused quite a stir: ID Software was responsible for many games popular among old-school engineers (most notably _Doom_, which has a place in programmer humor even today).
+
+#v(2mm)
+
+Naturally, this community immediately began dissecting _Quake_'s source. \
+One particularly interesting function is reproduced below, with original comments: \
+
+#v(3mm)
+
+```c
+float Q_rsqrt( float number ) {
+	long i;
+	float x2, y;
+	const float threehalfs = 1.5F;
+
+	x2 = number * 0.5F;
+	y  = number;
+	i  = * ( long * ) &y;						// evil floating point bit level hacking
+	i  = 0x5f3759df - ( i >> 1 );               // [redacted]
+	y  = * ( float * ) &i;
+	y  = y * ( threehalfs - ( x2 * y * y ) );   // 1st iteration
+//	y  = y * ( threehalfs - ( x2 * y * y ) );   // 2nd iteration, this can be removed
+
+	return y;
+}
+```
+
+#v(3mm)
+
+This code defines a function `Q_sqrt`, which was used as a fast approximation of the inverse square root in graphics routines. (in other words, `Q_sqrt` efficiently approximates $1 div sqrt(x)$)
+
+#v(3mm)
+
+The key word here is "fast": _Quake_ ran on very limited hardware, and traditional approximation techniques (like Taylor series)#footnote[Taylor series aren't used today, and for the same reason. There are better ways.] were too computationally expensive to be viable.
+
+#v(3mm)
+
+Our goal today is to understand how `Q_sqrt` works. \
+To do that, we'll first need to understand how computers represent numbers. \
+We'll start with simple binary integers---turn the page.

src/Advanced/Fast Inverse Root/parts/01 int.typ

@@ -5,7 +5,8 @@
 #definition()
 A _bit string_ is a string of binary digits. \
 In this handout, we'll denote bit strings with the prefix `0b`. \
-That is, $1010 =$ "one thousand and one," while $#text([`0b1001`]) = 2^3 + 2^0 = 9$
+#note[This prefix is only notation---it is _not_ part of the string itself.] \
+For example, $1001$ is the number "one thousand and one," while $#text([`0b1001`])$ is the string of bits "1 0 0 1".
 
 #v(2mm)
 We will separate long bit strings with underscores for readability. \
@@ -40,7 +41,7 @@ The value of a `uint` is simply its value as a binary number:
 What is the largest number we can represent with a 32-bit `uint`?
 
 #solution([
-  $#text([`0b01111111_11111111_11111111_11111111`]) = 2^(31)$
+  $#text([`0b11111111_11111111_11111111_11111111`]) = 2^(32)-1$
 ])
 
 #v(1fr)
@@ -53,6 +54,10 @@ Find the value of each of the following 32-bit unsigned integers:
 - `0b00000000_00000000_00000100_10110000`
 #hint([The third conversion is easy---look carefully at the second.])
 
+#instructornote[
+  Consider making a list of the powers of two $>= 1024$ on the board.
+]
+
 #solution([
   - $#text([`0b00000000_00000000_00000101_00111001`]) = 1337$
   - $#text([`0b00000000_00000000_00000001_00101100`]) = 300$
@@ -64,20 +69,20 @@ Find the value of each of the following 32-bit unsigned integers:
 
 
 #definition()
-In general, division of `uints` is nontrivial#footnote([One may use repeated subtraction, but that isn't efficient.]). \
+In general, fast division of `uints` is difficult#footnote([One may use repeated subtraction, but this isn't efficient.]). \
 Division by powers of two, however, is incredibly easy: \
 To divide by two, all we need to do is shift the bits of our integer right.
 
 #v(2mm)
 
 For example, consider $#text[`0b0000_0110`] = 6$. \
-If we insert a zero at the left end of this bit string and delete the digit at the right \
+If we insert a zero at the left end of this string and delete the zero at the right \
 (thus "shifting" each bit right), we get `0b0000_0011`, which is 3. \
 
 #v(2mm)
 
-Of course, we loose the remainder when we left-shift an odd number: \
+Of course, we lose the remainder when we right-shift an odd number: \
-$9 div 2 = 4$, since `0b0000_1001` shifted right is `0b0000_0100`.
+$9$ shifted right is $4$, since `0b0000_1001` shifted right is `0b0000_0100`.
 
 #problem()
 Right shifts are denoted by the `>>` symbol: \
@@ -86,6 +91,7 @@ Find the value of the following:
 - $12 #text[`>>`] 1$
 - $27 #text[`>>`] 3$
 - $16 #text[`>>`] 8$
+#note[Naturally, you'll have to convert these integers to binary first.]
 
 #solution[
   - $12 #text[`>>`] 1 = 6$

src/Advanced/Fast Inverse Root/parts/02 float.typ

@@ -3,7 +3,7 @@
 
 = Floats
 #definition()
-_Binary decimals_#footnote["decimal" is a misnomer, but that's ok.] are very similar to base-10 decimals. \
+_Binary decimals_#footnote([Note that "binary decimal" is a misnomer---"deci" means "ten"!]) are very similar to base-10 decimals.\
 In base 10, we interpret place value as follows:
 - $0.1 = 10^(-1)$
 - $0.03 = 3 times 10^(-2)$
@@ -107,11 +107,13 @@ Floats represent a subset of the real numbers, and are interpreted as follows: \
 - The next eight bits represent the _exponent_ of this float.
   #note([(we'll see what that means soon)]) \
   We'll call the value of this eight-bit binary integer $E$. \
-  Naturally, $0 <= E <= 255$ #note([(since $E$ consist of eight bits.)])
+  Naturally, $0 <= E <= 255$ #note([(since $E$ consist of eight bits)])
 
-- The remaining 23 bits represent the _fraction_ of this float, which we'll call $F$. \
+- The remaining 23 bits represent the _fraction_ of this float. \
-  These 23 bits are interpreted as the fractional part of a binary decimal. \
+  They are interpreted as the fractional part of a binary decimal. \
-  For example, the bits `0b10100000_00000000_00000000` represents $0.5 + 0.125 = 0.625$.
+  For example, the bits `0b10100000_00000000_00000000` represent $0.5 + 0.125 = 0.625$. \
+  We'll call the value of these bits as a binary integer $F$. \
+  Their value as a binary decimal is then $F div 2^23$. #note([(convince yourself of this)])
 
 
 #problem(label: "floata")
@@ -135,12 +137,17 @@ $
   (-1)^s times 2^(E - 127) times (1 + F / (2^(23)))
 $
 
-Notice that this is very similar to decimal scientific notation, which is written as
+Notice that this is very similar to base-10 scientific notation, which is written as
 
 $
   (-1)^s times 10^(e) times (f)
 $
 
+#note[
+  We subtract 127 from $E$ so we can represent positive and negative numbers. \
+  $E$ is an eight bit binary integer, so $0 <= E <= 255$ and thus $-127 <= (E - 127) <= 127$.
+]
+
 #problem()
 Consider `0b01000001_10101000_00000000_00000000`. \
 This is the same bit string we used in @floata. \

src/Advanced/Fast Inverse Root/parts/03 approx.typ

@@ -5,7 +5,7 @@
 = Integers and Floats
 
 #generic("Observation:")
-For small values of $x$, $log_2(1 + x)$ is approximately equal to $x$. \
+If $x$ is smaller than 1, $log_2(1 + x)$ is approximately equal to $x$. \
 Note that this equality is exact for $x = 0$ and $x = 1$, since $log_2(1) = 0$ and $log_2(2) = 1$.
 
 #v(5mm)
@@ -18,7 +18,7 @@ This allows us to improve the average error of our linear approximation:
   align: center,
   columns: (1fr, 1fr),
   inset: 5mm,
-  [$log(1+x)$ and $x + 0$]
+  [$log_2(1+x)$ and $x + 0$]
     + cetz.canvas({
       import cetz.draw: *
 
@@ -64,7 +64,7 @@ This allows us to improve the average error of our linear approximation:
       Max error: 0.086 \
       Average error: 0.0573
     ],
-  [$log(1+x)$ and $x + 0.045$]
+  [$log_2(1+x)$ and $x + 0.045$]
     + cetz.canvas({
       import cetz.draw: *
 
@@ -125,7 +125,7 @@ We won't bother with this---we'll simply leave the correction term as an opaque
   [
     "Average error" above is simply the area of the region between the two graphs:
     $
-      integral_0^1 abs( #v(1mm) log(1+x) - (x+epsilon) #v(1mm))
+      integral_0^1 abs( #v(1mm) log(1+x)_2 - (x+epsilon) #v(1mm))
     $
     Feel free to ignore this note, it isn't a critical part of this handout.
   ],

src/Advanced/Fast Inverse Root/parts/04 quake.typ

@@ -2,10 +2,9 @@
 
 = The Fast Inverse Square Root
 
+A simplified version of the _Quake_ routine we are studying is reproduced below.
 
-The following code is present in _Quake III Arena_ (1999):
+#v(2mm)
-
-#v(5mm)
 
 ```c
 float Q_rsqrt( float number ) {
@@ -15,20 +14,20 @@ float Q_rsqrt( float number ) {
 }
 ```
 
-#v(5mm)
+#v(2mm)
 
 This code defines a function `Q_rsqrt` that consumes a float `number` and approximates its inverse square root.
 If we rewrite this using notation we're familiar with, we get the following:
 $
   #text[`Q_sqrt`] (n_f) =
-  #h(5mm)
   6240089 - (n_i div 2)
-  #h(5mm)
+  #h(10mm)
   approx 1 / sqrt(n_f)
 $
 
 #note[
   `0x5f3759df` is $6240089$ in hexadecimal. \
+  Ask an instructor to explain if you don't know what this means. \
   It is a magic number hard-coded into `Q_sqrt`.
 ]
 
@@ -56,7 +55,7 @@ For those that are interested, here are the details of the "code-to-math" transl
 
 
 - Notice the right-shift in the second line of the function. \
-  We translated `(i >> i)` into $(n_i div 2)$.
+  We translated `(i >> 1)` into $(n_i div 2)$.
 #v(2mm)
 
 - "`return * (float *) &i`" is again C magic. \
@@ -64,17 +63,17 @@ For those that are interested, here are the details of the "code-to-math" transl
 #pagebreak()
 
 #generic("Setup:")
-We are now ready to show that $#text[`Q_sqrt`] (x) approx 1/sqrt(x)$. \
+We are now ready to show that $#text[`Q_sqrt`] (x)$ effectively approximates $1/sqrt(x)$. \
 For convenience, let's call the bit string of the inverse square root $r$. \
 In other words,
 $
   r_f := 1 / (sqrt(n_f))
 $
-This is the value we want to approximate.
+This is the value we want to approximate. \
 
 #problem(label: "finala")
 Find an approximation for $log_2(r_f)$ in terms of $n_i$ and $epsilon$ \
-#note[Remember, $epsilon$ is the correction constant in our approximation of $log_2(1 + a)$.]
+#note[Remember, $epsilon$ is the correction constant in our approximation of $log_2(1 + x)$.]
 
 #solution[
   $
@@ -92,7 +91,11 @@ Let's call the "magic number" in the code above $kappa$, so that
 $
   #text[`Q_sqrt`] (n_f) = kappa - (n_i div 2)
 $
-Use @convert and @finala to show that $#text[`Q_sqrt`] (n_f) approx r_i$
+Use @convert and @finala to show that $#text[`Q_sqrt`] (n_f) approx r_i$ \
+#note(type: "Note")[
+  If we know $r_i$, we know $r_f$. \
+  We don't even need to convert between the two---the underlying bits are the same!
+]
 
 #solution[
   From @convert, we know that
@@ -164,8 +167,7 @@ though it is fairly close to the ideal $epsilon$.
 
 #remark()
 And now, we're done! \
-We've shown that `Q_sqrt(x)` approximates $1/sqrt(x)$ fairly well, \
+We've shown that `Q_sqrt(x)` approximates $1/sqrt(x)$ fairly well. \
-thanks to the approximation $log(1+a) = a + epsilon$.
 
 #v(2mm)