Author Topic: Power function  (Read 177 times)

B+

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Power function
« on: 24. July 2018, 04:39:24 »
what? Naalaa has no power (function)! well here is one that offers some rough estimates:
Code: [Select]

' Power pack the short version.txt
' written for Naalaa 6 by bplus posted 2018-07-23
' extracted from large test of fractions, random number functions, string... in Power test.txt

' The main purpose of this code: to demo a power function for real numbers,

' I had an idea for how real numbers to the power of real numbers might be done ie x# ^ y# = ?
' This means that not only can you take SQR of a number, you can get cube or cube root, quartic, 5th 6th... roots and any multiple

' It came from this simple idea
' 2 ^ 3.5 = 2 ^ 3 * 2 ^ .5 = 8 * Sqr(2)
' 3 ^ 3.5 = 3 ^ 3 * 3 ^ .5 = 27 * Sqr(3)

' so 2 ^ 3.25 = 2 ^ 3 * 2 ^ .25
' what is 2 ^ .25 ?  It is sqr(sqr(2)) !

' likewise 2 ^ 3.125 = 2 ^ 3 * 2 ^ 1/8
' what is 2 ^ 1/8 ? It is sqr(sqr(sqr(2))) !

' any decimal can be written as a sum of fraction powers of 2, as any integer can be written in powers of 2.
' in binary expansions
' 1/2 = .1       or .5 base 10
' 1/4 = .01      or .25 base 10
' 1/8 = .001     or .125 base 10
' 1/16 = .0001   or .0625 base 10

' So with binary expansion of decimal, we can SQR and multiply our way to an estimate
' of any real number to the power of another real number using binary expansion of
' the decimal parts as long as we stay in Naalaa integer limits and are mindful of precision.

constant:
wW 800
wH 720
hidden:
set window 100, 20, wW, wH

do
    wln "And now for the main event! We test the power#(x#, pow#) function!"
    write "(nothing) quits, Please enter a real (float) decimal number raise to some power. x# = "
    x# = rln#()
    if x# = 0.0 then break
    write "(nothing) quits, Please enter a real (float) decimal power. pow# = "
    pw# = rln#()
    if pw# = 0.0 then break
    result# = power#(x#, pw#)
    wln result#, " is what we estimate for x# raised to power of pow#"
    wln
loop
wln
wln "Far from precise of course, but this code is clear proof of concept!"
wln " OMG, it worked!!!"
wait keydown


' A power function for real numbers (small ones but still!)
' x# to the power of pow#
function power#(x#, pow#)
'this sub needs 2 other subs
'rite$
'bExpand20$

    'this is going to follow covertReal2Fraction$, first deal with integer part if any
    r$ = str$(pow#)
    s$[] = split(r$, ".")
    integer$ = s$[0]
    build# = 1.0
    if integer$ <> "0"
        p = int(integer$)
        for i = 1 to p
            build# = build# * x#
        next
    endif
    'that takes care of integer part,
    'now for the fraction part convert decimal to fraction
    n$ = s$[1]
    ld = len(n$)
    while rite$(n$, 1) = "0"
        n$ = left$(n$, ld - 1)
        ld = len(n$)
    wend
    denom = 10
    for i = 2 to ld
        denom = denom * 10   
    next
    numer = int(n$)
    'OK for bExpand20$ don't have to simplify and that saves us having to extract n and d again from n/d
    bs$ = bExpand20$(numer, denom)
    'at moment we haven't taken any sqr of x
    runningXSQR# = x#
    'run through all the 0's and 1's in the bianry expansion of the fraction part of the power float
    for i = 1 to len(bs$)
        'this is the matching sqr of the sqr of the sqr... of x#
        runningXSQR# = sqr#(runningXSQR#)
        'for every 1 in the expansion, multiple our build with the running sqr of ... sqr of x
        if mid$(bs$, i - 1, 1) = "1" then build# = build# * runningXSQR#
    next
    'our build# should be a estimate or x# to power of pow#
    return build#
endfunc

'write a series of 1s and 0s that represent the decimal fraction n/d in binary 20 places long
function bExpand20$(nOver, d)
    ' b for base
    b# = 0.5
    ' r for remainder
    r# = float(nOver)/float(d)
    ' s for string$ 0's and 1's that we will build and return for function value
    s$ = ""
    ' f for flag to stop
    f = 0
    ' c for count to track how far we are, don't want to go past 20
    c = 0
    while f = 0
        if r# < b#
            s$ = s$ + "0"
        else
            s$ = s$ + "1"
            if r# > b#
                r# = r# - b#
            else
                f = 1
            endif
        endif
        c = c + 1
        if c >= 20 then f = 1
        b# = b# * 0.5
    wend
    return s$
endfunc

'not right$
function rite$(stringy$, amount)
    return mid$(stringy$, len(stringy$) - amount, amount)
endfunc
« Last Edit: 24. July 2018, 17:34:23 by B+ »

B+

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Re: Power function
« Reply #1 on: 24. July 2018, 14:15:08 »
Oh apparently Naalaa does have a power function pow#(a#, b#) a closely kept secret I guess.

Here is why I thought it didn't have a power function (plus the keyword does not highlight when typed because I tried that before rolling my own):
« Last Edit: 24. July 2018, 17:34:33 by B+ »

B+

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Re: Power function
« Reply #2 on: 24. July 2018, 17:33:18 »
Wow in QB64, this power estimator is on par with the ^ operator!
Code: [Select]
_TITLE "Power Function 2 by bplus"
'QB64 X 64 version 1.2 20180228/86  from git b301f92

' started 2018-07-23  Naalaa has no power function (or operator), so I wrote a power function for it.
' ''Power pack the short version.txt
' ''written for Naalaa 6 by bplus posted 2018-07-23
' ''extracted from large test of fractions, random number functions, string... in Power test.txt
' 2018-07-24 Power Function 2 split is replaced with two much smaller and very handy string functions.

' OMG the crazy thing worked! It produced decent estimates of roots given the limitations of precision...
' Now I want to see how well it works with far greater precision available. So here we are, looking to see
' how this function compares to the regualar ^ operator in QB64.

'from Naalaa comments:
' The main purpose of this code: to demo a power function for real numbers,

' I had an idea for how real numbers to the power of real numbers might be done ie x ^ y = ?
' This means that not only can you take SQR of a number, you can get cube or cube root, quartic, 5th 6th... roots and any multiple

' It came from this simple idea
' 2 ^ 3.5 = 2 ^ 3 * 2 ^ .5 = 8 * Sqr(2)
' 3 ^ 3.5 = 3 ^ 3 * 3 ^ .5 = 27 * Sqr(3)

' so 2 ^ 3.25 = 2 ^ 3 * 2 ^ .25
' what is 2 ^ .25 ?  It is sqr(sqr(2)) !

' likewise 2 ^ 3.125 = 2 ^ 3 * 2 ^ 1/8
' what is 2 ^ 1/8 ? It is sqr(sqr(sqr(2))) !

' any decimal can be written as a sum of fraction powers of 2 ie 1/2^n, as any integer can be written in powers of 2.
' in binary expansions
' 1/2 = .1       or .5 base 10
' 1/4 = .01      or .25 base 10
' 1/8 = .001     or .125 base 10
' 1/16 = .0001   or .0625 base 10

' So with binary expansion of decimal, we can SQR and multiply our way to an estimate
' of any real number to the power of another real number using binary expansion of
' the decimal parts as long as we stay in Naalaa integer limits and are mindful of precision.

CONST wW = 800
CONST wH = 600
SCREEN _NEWIMAGE(wW, wH, 32)
_SCREENMOVE 360, 60
_DEFINE A-Z AS _FLOAT

DO
    PRINT "Testing the power(x, pow) function:"
    INPUT "(nothing) quits, Please enter a real number to raise to some power. x = "; x
    IF x = 0 THEN EXIT DO
    INPUT "(nothing) quits, Please enter a real number for the power. pow = ", pw
    IF pw = 0 THEN EXIT DO
    result = power(x, pw)
    PRINT result; " is what we estimate for"; x; " raised to power of"; pw
    PRINT x ^ pw; " is what the ^ operator gives us."
    PRINT
LOOP
PRINT
PRINT "This is matching the ^ operator very well! This code is clear proof of concept!"
PRINT " OMG, it worked!!!"
SLEEP


' A power function for real numbers (small ones but still!)
' x to the power of pow
FUNCTION power## (x AS _FLOAT, pow AS _FLOAT)
    'this sub needs: bExpand60$, leftOf$, rightOf$

    DIM build AS _FLOAT
    r$ = "0" + STR$(pow) 'in case pow starts with decimal
    integer$ = leftOf$(r$, ".")
    build = 1.0
    IF integer$ <> "0" THEN
        p = VAL(integer$)
        FOR i = 1 TO p
            build = build * x
        NEXT
    END IF
    'that takes care of integer part

    n$ = rightOf$(r$, ".")
    IF n$ = "" THEN power = build: EXIT SUB

    'remove 0's to right of main digits
    ld = LEN(n$)
    WHILE RIGHT$(n$, 1) = "0"
        n$ = LEFT$(n$, ld - 1)
        ld = LEN(n$)
    WEND

    'note: we are pretending that the ^ operator is not available, so this is hand made integer power
    denom& = 10
    FOR i = 2 TO ld
        denom& = denom& * 10
    NEXT

    'OK for bExpand60$ don't have to simplify fraction and that saves us having to extract n and d again from n/d
    bs$ = bExpand60$(VAL(n$), denom&)

    'at moment we haven't taken any sqr of x
    runningXSQR = x

    'run through all the 0's and 1's in the bianry expansion, bs$, the fraction part of the power float
    FOR i = 1 TO LEN(bs$)
        'this is the matching sqr of the sqr of the sqr... of x
        runningXSQR = SQR(runningXSQR)
        'for every 1 in the expansion, multiple our build with the running sqr of ... sqr of x
        IF MID$(bs$, i, 1) = "1" THEN build = build * runningXSQR
    NEXT

    'our build should now be an estimate or x to power of pow
    power = build
END FUNCTION

'write a series of 1s and 0s that represent the decimal fraction n/d in binary 60 places long
FUNCTION bExpand60$ (nOver&, d&)
    DIM b AS _FLOAT, r AS _FLOAT
    ' b for base
    b = 0.5
    ' r for remainder
    r = nOver& / d&
    ' s for string$ 0's and 1's that we will build and return for function value
    s$ = ""
    ' f for flag to stop
    f% = 0
    ' c for count to track how far we are, don't want to go past 20
    c% = 0
    WHILE f% = 0
        IF r < b THEN
            s$ = s$ + "0"
        ELSE
            s$ = s$ + "1"
            IF r > b THEN
                r = r - b
            ELSE
                f% = 1
            END IF
        END IF
        c% = c% + 1
        IF c% >= 60 THEN f% = 1
        b = b * 0.5
    WEND
    bExpand60$ = s$
END FUNCTION

FUNCTION leftOf$ (source$, of$)
    posOf = INSTR(source$, of$)
    IF posOf > 0 THEN leftOf$ = MID$(source$, 1, posOf - 1)
END FUNCTION

FUNCTION rightOf$ (source$, of$)
    posOf = INSTR(source$, of$)
    IF posOf > 0 THEN rightOf$ = MID$(source$, posOf + LEN(of$))
END FUNCTION


John

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Re: Power function
« Reply #3 on: 27. July 2018, 06:18:36 »
Quote
' It came from this simple idea
' 2 ^ 3.5 = 2 ^ 3 * 2 ^ .5 = 8 * Sqr(2)
' 3 ^ 3.5 = 3 ^ 3 * 3 ^ .5 = 27 * Sqr(3)

Good luck finding a language that allows the results to be an expression rather than a variable.

B+

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Re: Power function
« Reply #4 on: 27. July 2018, 15:52:42 »
Quote
' It came from this simple idea
' 2 ^ 3.5 = 2 ^ 3 * 2 ^ .5 = 8 * Sqr(2)
' 3 ^ 3.5 = 3 ^ 3 * 3 ^ .5 = 27 * Sqr(3)

Good luck finding a language that allows the results to be an expression rather than a variable.

Hi John,

The point of those 2 little examples was to show that 2 ^ 3.5 for instance can be calculated without the ^ operator.
The SQR(2) part of course is further evaluated down to a single value. 8 * 1.4142.. = 11.3137...

x to an integer power like x ^ n  where n is positive integer (and not real number with decimals) is easy peasy without ^

p = 1
for i = 1 to n
   p = p * x
next

result: p = x ^ n when n is positive integer

Integer powers are easy to do without the ^ operator but how can we do a power that is a real number (AKA float or decimal) (and positive).

This became the first most crude estimate for x ^ power for the return value of function for the integer part only.

For the decimal part of the power, I took it as a fraction .75 = 75/100
and expanded that fraction to binary 1's and 0's.

Then I ran down the expansion a 0 or 1 at a time, keeping a running value of the sqr(sqr... (sqr(x))) for each step.
If the expansion had a 1 in it, then the return result was multiplied by that running value.
Such that by the time you reach the end of the expansion, you have a very decent estimate of x ^ power without using the ^.

From the time BASIC started using the ^ operator there had to be a way to calculate that expression because of course you can't use ^ to calculate an expression using the ^ operator. I don't know if this is the way it was done but it is the way it could have been done if the SQR() function was available.


   

John

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Re: Power function
« Reply #5 on: 28. July 2018, 01:22:45 »
Sorry!

I get what you meant now.

Code: [Select]
' 2 ^ 3.5 = 2 ^ 3 * 2 ^ .5 = 8 * Sqr(2)
' 3 ^ 3.5 = 3 ^ 3 * 3 ^ .5 = 27 * Sqr(3)

PRINT FORMAT("%g",2 ^ 3.5),"\n"
PRINT FORMAT("%g",2 ^ 3 * 2 ^ .5), "\n"
PRINT FORMAT("%g",8 * Sqr(2)), "\n"
PRINTNL
PRINT FORMAT("%g",3 ^ 3.5),"\n"
PRINT FORMAT("%g",3 ^ 3 * 3 ^ .5),"\n"
PRINT FORMAT("%g",27 * Sqr(3)),"\n"


jrs@jrs-laptop:~/sb/examples/test$ scriba bpwr.sb
11.3137
11.3137
11.3137

46.7654
46.7654
46.7654
jrs@jrs-laptop:~/sb/examples/test$