RosettaCodeData/Task/Stern-Brocot-sequence/AutoHotkey/stern-brocot-sequence.ahk
2016-12-05 22:15:40 +01:00

77 lines
2.2 KiB
AutoHotkey

Found := FindOneToX(100), FoundList := ""
Loop, 10
FoundList .= "First " A_Index " found at " Found[A_Index] "`n"
MsgBox, 64, Stern-Brocot Sequence
, % "First 15: " FirstX(15) "`n"
. FoundList
. "First 100 found at " Found[100] "`n"
. "GCDs of all two consecutive members are " (GCDsUpToXAreOne(1000) ? "" : "not ") "one."
return
class SternBrocot
{
__New()
{
this[1] := 1
this[2] := 1
this.Consider := 2
}
InsertPair()
{
n := this.Consider
this.Push(this[n] + this[n - 1], this[n])
this.Consider++
}
}
; Show the first fifteen members of the sequence. (This should be: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3,
; 5, 2, 5, 3, 4)
FirstX(x)
{
SB := new SternBrocot()
while SB.MaxIndex() < x
SB.InsertPair()
Loop, % x
Out .= SB[A_Index] ", "
return RTrim(Out, " ,")
}
; Show the (1-based) index of where the numbers 1-to-10 first appears in the sequence.
; Show the (1-based) index of where the number 100 first appears in the sequence.
FindOneToX(x)
{
SB := new SternBrocot(), xRequired := x, Found := []
while xRequired > 0 ; While the count of numbers yet to be found is > 0.
{
Loop, 2 ; Consider the second last member and then the last member.
{
n := SB[i := SB.MaxIndex() - 2 + A_Index]
; If number (n) has not been found yet, and it is less than the maximum number to
; find (x), record the index (i) and decrement the count of numbers yet to be found.
if (Found[n] = "" && n <= x)
Found[n] := i, xRequired--
}
SB.InsertPair() ; Insert the two members that will be checked next.
}
return Found
}
; Check that the greatest common divisor of all the two consecutive members of the series up to
; the 1000th member, is always one.
GCDsUpToXAreOne(x)
{
SB := new SternBrocot()
while SB.MaxIndex() < x
SB.InsertPair()
Loop, % x - 1
if GCD(SB[A_Index], SB[A_Index + 1]) > 1
return 0
return 1
}
GCD(a, b) {
while b
b := Mod(a | 0x0, a := b)
return a
}