Richard Suchenwirth 2002-04-27 - Periodic decimal fractions are numbers where a sequence of digits behind the decimal point (the period) is endlessly repeated, for example:
1/7 = 0.142857142857.. 1/3 = 0.3333..
The following routine tries to detect a period in a given number and returns the period (might be 0 for integers or non-strict periods like
1/2 = 0.5000..
or (implicitly) an empty string if no period could be detected - then the input number might be irrational (not representable by a nominator/divisor expression, e.g. sqrt(2)), or it has a period longer than 7 digits, which can not be confirmed at the maximum tcl_precision of 17, a limit imposed by the underlying double representation in C. For instance, 1/17 is certainly periodic, but the period is out of sight for Tcl's expr math..
proc period x {
set frac [expr {abs(double($x)-int($x))}]
if {!$frac || [string length $frac]<10} {return 0}
set digits [string range $frac 2 end]
foreach n {1 2 3 4 5 6 7} {
foreach offset {0 1 2 3 4 5 6} {
set try [string range $digits $offset [expr $offset+$n-1]]
if {[regexp .{0,$n}$try\($try)+.{0,$n}$ $digits]} {
return $try
}
}
}
}Arjen Markus As the fraction is contained in a string, there ought to be no problem with very long periods:
set long_fraction "0.123456789012345678901234567890"
As long, of course, as you avoid interpreting the string as a number! RS admits that he does, by letting expr compute the fractional part...
KBK In the spirit of Fraction math, let's do the period of a rational number whose numerator and denominator are integers. The following has no trouble finding out that 1/17 repeats with a period of 16 places. It also handles Arjen's problem nicely, if we remember that .12345678901234567890... = 137174210/1111111111
proc period2 { n d } {
set x 0
while { 1 } {
set r [expr { $n % $d }]
if { $r == 0 } {
return 0
} elseif { [info exists seen($n)] } {
return [expr { $x - $seen($n) }]
} else {
set seen($n) $x
incr x
set n [expr { 10 * $r }]
}
}
}
for { set i 1 } { $i <= 20 } { incr i } {
puts [list $i [period2 1 $i]]
}There is some interesting number theory here, but I haven't the time to get into it; I've typed this in in about 15 minutes while waiting for a test to run.
Arjen Markus Independently of Kevin, I came up with the following proc that will return the decimal string/number of such a division with any precision you like (also known as long division, if I have translated this correctly):
# decimalFraction --
# Determine the decimal fraction of num/denom by a given precision
#
# Arguments:
# num Numerator of the fraction
# denom Denominator of the fraction
# nodec Number of decimals required
# Result:
# The decimal fraction (as a string!)
# Notes:
# The method relies entirely on integer division and can achieve
# any precision required.
#
proc decimalFraction { num denom nodec } {
#
# The integer part first
#
set fract [expr {int($num/$denom)}]
set rem [expr {$num%$denom}]
set result "$fract"
append result "."
#
# Loop over the decimals
#
for { set i 0 } { $i < $nodec } { incr i } {
set rem [expr {$rem*10}]
if { $rem < $denom } {
append result "0"
} else {
set fract [expr {int($rem/$denom)}]
set rem [expr {$rem%$denom}]
append result "$fract"
}
}
return $result
}
#
# A few test cases
#
puts [decimalFraction 1 3 10]
puts [decimalFraction 1 5 10]
puts [decimalFraction 1 13 40]
puts [decimalFraction 1 97 200]
puts [decimalFraction 1 15 10]
puts [decimalFraction 1 17 40]
puts [decimalFraction 1 7 40]The number-theoretical aspects of this problem relate (probably among other things, but I am only an amateur :-) to Fermat's little theorem:
a^(p-1)-1 is divisible by p,
if p is a prime and a is not a multiple of p.
This allows us to put an upper limit to the period of any prime, say 127, which is one of my favourites: the period for 127 can not be more than 126, and otherwise it will be a factor of 126). The latter follows from the observation that:
a^gh-1 = m^h-1 = (m-1)*(1+m+m^2+...m^h) (where m=a^g)
Category Mathematics | Arts and crafts of Tcl-Tk programming