**Wesley Kerfoot**4 years ago

**1 changed files**with

**232 additions**and

**0 deletions**

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`#lang racket` |
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`(require data/bit-vector)` |
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```
``` |
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`; https://imgur.com/a/ClKK5Ac` |
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```
``` |
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`; See http://fabiensanglard.net/floating_point_visually_explained/ for an intuitive explanation` |
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```
``` |
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`(define (to-bin fl)` |
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` (~r fl` |
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` #:base 2` |
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` #:precision 52))` |
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```
``` |
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`(define (bit-vector->string bv)` |
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` (list->string` |
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` (for/list [(i (in-range (bit-vector-length bv)))]` |
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` (cond` |
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` [(bit-vector-ref bv i) #\1]` |
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` [else #\0]))))` |
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```
``` |
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`(define (bit-vector->posint bv)` |
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` (string->number` |
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` (format "#b~a"` |
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` (bit-vector->string bv))))` |
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```
``` |
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`(define (show-bv-slice bv start end)` |
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` (bit-vector->list` |
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` (bit-vector-copy bv start end)))` |
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```
``` |
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`(define (bool->number b)` |
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` (cond` |
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` [b 1]` |
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` [else 0]))` |
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```
``` |
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`(define (number->bool n)` |
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` (match n` |
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` [0 #f]` |
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` [1 #t]` |
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` [_ #f]))` |
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```
``` |
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`(define (sum xs)` |
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` (foldr + 0 xs))` |
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```
``` |
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`; conversion from base 10 functions` |
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```
``` |
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`;; Have to calculate the number of digits to remove from` |
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`;; the precision based on how far the decimal point needs` |
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`;; to be moved left,` |
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`;; or else maybe just do the calculation, and lop off digits from the right?` |
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```
``` |
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```
``` |
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`(define (int->binary n)` |
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` (let-values` |
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` ([(q r) (quotient/remainder n 2)])` |
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` (match q` |
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` [0 (list r)]` |
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` [_ (cons r (int->binary q))])))` |
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```
``` |
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`(define (real->binary-frac n [precision 0])` |
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` (define p` |
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` (* 2 n))` |
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` ` |
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` (displayln p)` |
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` (cond` |
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` [(= p 0.0) ""]` |
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` [(> precision 51) ""]` |
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` [(>= p 1)` |
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` (string-append "1"` |
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` (real->binary-frac` |
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` (sub1 p)` |
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` (add1 precision)))]` |
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` [(< p 1)` |
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` (string-append "0"` |
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` (real->binary-frac` |
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` p` |
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` (add1 precision)))]))` |
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```
``` |
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`; do the conversion from w.fff.. to binary` |
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`(define (real->bits whole fraction)` |
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` (list` |
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` (cond` |
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` [(> whole 0) 0]` |
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` [else 1])` |
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` (bit-vector->string` |
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` (list->bit-vector` |
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` (map number->bool` |
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` (reverse (int->binary whole)))))` |
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` ` |
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` (real->binary-frac fraction)))` |
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```
``` |
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```
``` |
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`; Conversion from base-2 functions` |
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```
``` |
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`(define (calculate-number bv)` |
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` (define sign (bv-sign bv))` |
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` (define mantissa (bv-mantissa bv))` |
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` (define exponent (bv-exponent bv))` |
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```
``` |
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` (displayln (format "Sign = ~a" (cond ((= 0 sign) "positive") (else "negative"))))` |
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` ` |
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` (displayln (format "Mantissa = ~a"` |
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` (exact->inexact` |
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` (calculate-mantissa mantissa))))` |
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` ` |
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` (displayln (format "Exponent = ~a" exponent))` |
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` ` |
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` (*` |
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` (expt -1 sign)` |
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` (calculate-mantissa mantissa)` |
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` (expt 2 exponent)))` |
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```
``` |
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`(define (exp-len bv)` |
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` (match (bit-vector-length bv)` |
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` [32 8]` |
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` [64 11]))` |
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```
``` |
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`(define (bv-mantissa bv)` |
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` (bit-vector-copy bv` |
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` (add1 (exp-len bv))` |
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` (bit-vector-length bv)))` |
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```
``` |
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```
``` |
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`;; Floating point numbers` |
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```
``` |
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`(define example` |
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` (string->bit-vector` |
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` ; 0.052 in binary` |
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` ;seeeeeeeeeeemmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm` |
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` "0011111110101010100111111011111001110110110010001011010000111001"))` |
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```
``` |
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`;; In this example, we are representing 0.052 as a 64 bit floating point number` |
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`;; The first bit is our sign` |
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`;; The next 11 bits are our exponent` |
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`;; The next 52 bits are our mantissa (also called the significand or fraction)` |
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```
``` |
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`;; Starting with the sign, if it is 1 it is negative, otherwise positive` |
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```
``` |
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`(define (bv-sign bv)` |
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` (cond` |
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` [(bit-vector-ref bv 0) -1]` |
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` [else 0]))` |
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```
``` |
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`;; The exponent (next 11 bits) is represented in a biased form, meaning there is a subtraction that occurs` |
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`;; So for 0.052, the exponent is -5` |
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`;; 01111111010 = 1018 in binary` |
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`;; the bias is 1023, so we do 1018 - (2^10-1) = 1028 - 1023 = -5` |
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```
``` |
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`(define (bv-exponent bv)` |
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` ; bias is basically half the range of the exp minus 1` |
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` (define bias` |
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` (sub1` |
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` (expt 2` |
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` (sub1 (exp-len bv)))))` |
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` ; subtract bias from exponent` |
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` (-` |
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` (bit-vector->posint` |
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` (bit-vector-copy bv 1 (add1 (exp-len bv))))` |
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` bias))` |
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```
``` |
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`;; The mantissa (next 52 bits) is usually represented in a *normalized* form, meaning 1.xxx... (52 bits for a 64 bit float)` |
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`;; The mantissa can be calculated in decimal using a summation, e.g. b1 / 2^1 + b2 / 2^2 + ... (b1 and b2 are bits)` |
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```
``` |
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`(define (calculate-mantissa n)` |
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` (define bits` |
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` (map bool->number (bit-vector->list n)))` |
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` (define powers` |
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` (map add1 (range (length bits))))` |
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` ; add 1 for the implicit 1.xxx` |
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` ; sum of bits divided by increasing powers of 2` |
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` ; basically each "place" in the binary digits` |
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` (add1 (sum (map` |
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` (lambda (b p)` |
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` (/ b (expt 2 p)))` |
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` bits powers))))` |
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```
``` |
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`;; s m exp` |
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`;; Putting that together, you get (-1)^0 * 1.664 * 2^(-5) = 0.052` |
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```
``` |
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`;; Keep in mind that the computer does not do this conversion every time it calculates something` |
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`;; There are various algorithms for adding/multiplying binary floating point numbers efficiently (which I won't get into)` |
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```
``` |
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`;; You may ask why there is always an implicit leading 1. in the mantissa/significand. The answer is that it's` |
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`;; somewhat arbitrary. There are things called subnormal, or denormalized numbers, which can change this.` |
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```
``` |
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`;; From wikipedia:` |
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```
``` |
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`;; In a denormal number, since the exponent is the least that it can be,` |
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`;; zero is the leading significand digit (0.m1m2m3...mp−2mp−1)` |
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`;; allowing the representation of numbers closer to zero than the smallest normal number.` |
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```
``` |
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`;; ` |
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```
``` |
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`;; Other fun things about floating point numbers` |
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```
``` |
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`;; You may also notice that as the exponent gets larger and larger, the range of numbers between a given whole number` |
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`;; and the next one increases.` |
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```
``` |
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`;; There is something called "epsilon" which essentially tells you which number is the upper bound on any rounding error` |
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`;; For example, on my machine 2.0 + 2.220446049250313e-16 = 2.0` |
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```
``` |
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`;; Why? because 2.220446049250313e-16 (or anything smaller) is going to simply get rounded off.` |
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`;; This number basically tells you the limit of the precision for your floats on a given machine` |
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`;; It ends up being useful for various numerical algorithms that you probably don't need to care about.` |
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```
``` |
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`;; It is important to understand that floating point intervals have an inherent limit to the range of numbers` |
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```
``` |
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`;; NaN` |
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`;; NaNs are represented with an exponent that is all 1s, and a mantissa that is anything except all 0s` |
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`;; NaN == NaN is always false. This implies there is more than one NaN. Some software will actually use this` |
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`;; as a way of encoding error codes.` |
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```
``` |
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`;; Infinity is represented with a mantissa of all 0s and an exponent of all 1s` |
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`;; We can have -/+ Infinity because of this` |
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```
``` |
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`;; E.g.` |
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`;; NaN = 0111111111111000000000000000000000000000000000000000000000000000` |
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`;;-Inf = 1111111111110000000000000000000000000000000000000000000000000000` |
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```
``` |
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`;; (note that my code does not properly handle infinity or NaNs)` |
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```
``` |
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`;; Decimal floating point` |
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`;; There is an entire separate standard for this but all in Decimal, not Binary! Conceptually, you could do this` |
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`;; with any base. There are even hexadecimal floating point number systems.` |
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```
``` |
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`;; If you need to deal with anything that must be exact, use rationals. If you need performance, use floats.` |
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`;; The problem with using floats is that some numbers can only be approximated, not perfectly accurately represented.` |
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`;; This is true of any base, not just base 2. It is also true of irrational numbers like pi.` |
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```
``` |
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`;; There are things called "Minifloats" which are only 16 bits or smaller, and are non-standard, but useful` |
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`;; E.g. in graphics where you don't care too much about precision but performance matters a lot` |
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```
``` |
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`(displayln` |
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` (exact->inexact (calculate-number example)))` |

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