func RabinMiller(p *big.Int) bool {
pdec := new(big.Int).Sub(p, big.NewInt(1)) // = p - 1
big2 := big.NewInt(2)
for i := 0; i < 20; i++ {
x := RandNumSmaller(p)
stg := new(big.Int).Exp(x, pdec, p) // = x^(p-1) mod p
if stg.Cmp(big1) != 0 {
return false
}
// test na Carmichaelova cisla (kontrola zda x^[(p-1)/2] je +1 nebo -1)
p2 := new(big.Int).Rsh(p, 1) // = (p - 1)/2
for {
stg.Exp(x, p2, p)
if stg.Cmp(pdec) == 0 {
break
}
if stg.Cmp(big1) != 0 {
return false
}
_, res := p2.Div(p2, big2)
if res.Cmp(big1) == 0 {
break
}
}
}
return true
}
// Zjisti a0, b0, z tak aby: a0*x + b0*y = z (mod n)
func Euklid(x, y, n *big.Int) (a0, b0, z *big.Int) {
a0 = big1
a := big0
b0 = big0
b := big1
g := new(big.Int)
t := new(big.Int) // tmp
if x.Cmp(y) < 0 {
x, y = y, x
a0, a, b0, b = b0, b, a0, a
}
for {
z = y
_, y = g.Div(x, y)
x = z
a0, a = a, _Euklid_sub(a0, t.Mul(a, g), n)
b0, b = b, _Euklid_sub(b0, t.Mul(b, g), n)
if y.Cmp(big0) == 0 {
break
}
}
return
}
func number_divide(x, y Obj) Obj {
xfx := (uintptr(unsafe.Pointer(x)) & fixnum_mask) == fixnum_tag
yfx := (uintptr(unsafe.Pointer(y)) & fixnum_mask) == fixnum_tag
if xfx && yfx {
i1 := int(uintptr(unsafe.Pointer(x))) >> fixnum_shift
i2 := int(uintptr(unsafe.Pointer(y))) >> fixnum_shift
// A good optimizer will combine the div and mod into
// one instruction.
r, m := i1/i2, i1%i2
if m == 0 && r > fixnum_min && r < fixnum_max {
return Make_fixnum(r)
} else {
return wrap(big.NewRat(int64(i1), int64(i2)))
}
}
if (!xfx && (uintptr(unsafe.Pointer(x))&heap_mask) != heap_tag) ||
(!yfx && (uintptr(unsafe.Pointer(y))&heap_mask) != heap_tag) {
panic("bad type")
}
if xfx {
x = wrap(big.NewInt(int64(fixnum_to_int(x))))
}
if yfx {
y = wrap(big.NewInt(int64(fixnum_to_int(y))))
//return wrap(z.Div(vx,vy))
}
switch vx := (*x).(type) {
case *big.Int:
var z *big.Int = big.NewInt(0)
switch vy := (*y).(type) {
case *big.Int:
return simpBig(z.Div(vx, vy))
case *big.Rat:
z := big.NewRat(1, 1)
z.SetInt(vx)
return simpRat(z.Quo(z, vy))
default:
panic("bad type")
}
case *big.Rat:
z := big.NewRat(1, 1)
switch vy := (*y).(type) {
case *big.Int:
z.SetInt(vy)
return simpRat(z.Quo(vx, z))
case *big.Rat:
return simpRat(z.Quo(vx, vy))
}
}
panic("bad type")
}
//encodes big.Int to base58 string
func Big2Base58(val *big.Int) Base58 {
answer := ""
valCopy := new(big.Int).Abs(val) //copies big.Int
if val.Cmp(big.NewInt(0)) <= 0 { //if it is less than 0, returns empty string
return Base58("")
}
tmpStr := ""
tmp := new(big.Int)
for valCopy.Cmp(big.NewInt(0)) > 0 { //converts the number into base58
tmp.Mod(valCopy, big.NewInt(58)) //takes modulo 58 value
valCopy.Div(valCopy, big.NewInt(58)) //divides the rest by 58
tmpStr += alphabet[tmp.Int64() : tmp.Int64()+1] //encodes
}
for i := (len(tmpStr) - 1); i > -1; i-- {
answer += tmpStr[i : i+1] //reverses the order
}
return Base58(answer) //returns
}
/**
* Find the next palindrome after the current number.
*/
func (factory *PalinFactory) Next() string {
var (
buffer bytes.Buffer
numberLen = len(factory.number)
oddLength = numberLen%2 > 0
// Greedily split the number into two halves. Greedily meaning for
// odd-length numbers, the middle digit belongs to both sides (e.g.,
// "987" splits into "98" and "87").
leftSide = factory.number[0 : (numberLen+1)/2]
rightSide = factory.number[numberLen/2:]
mirroredSide = Reverse(leftSide)
leftLength = len(leftSide)
)
// Before we create a palindrome, we can determine if the resulting
// palindrome will not be greater than the original number by mirroring the
// left side and comparing it to the right side.
//
// Using 123456 as an example, the palindrome will be 123321, which is less
// than 123456. If we mirror the left side, we get 321. The right side is
// 456. Since 321 <= 456, we know the palindrome will not be greater than
// the original number.
if !Greater(mirroredSide, rightSide) {
// Since it is not greater, we can increment the palindrome by
// incrementing the left side. This is essentially incrementing the
// middle digits of the palindrome, which results in the very next
// palindrome. Since we are also incrementing the left side, the
// resulting palindrome will also be greater than the original number.
leftSide = Increment(leftSide)
// One obvious optimization is to set leftSide here and then set it
// again in the next inner if {}, rather than calling leftValue.String()
// twice, but that seems to consistently be slower for some reason.
// If we introduced a new digit, this changes the evenness/oddness of
// the number, and we need to account for this when constructing the
// palindrome. Since we greedily split the number, going from odd to
// even means we need to drop a digit on the left.
//
// For example, the left side of 999 is "99". The left side of 1001 is
// "10". Notice both have two digits. However, if we increment "99"
// (as a left side), we get "100", so we need to divide by 10 to get the
// two digits that will be mirrored to form the palindrome 1001. This
// doesn't need to be done when going from odd to even, because the left
// side has one more digit (left side of 99 is "9", left side of 101 is
// "10").
if len(leftSide) != leftLength {
oddLength = !oddLength
if !oddLength {
var leftValue big.Int
leftValue.SetString(leftSide, 10)
leftValue.Div(&leftValue, big.NewInt(10))
leftSide = leftValue.String()
}
}
mirroredSide = Reverse(leftSide)
}
buffer.WriteString(leftSide)
// Mirror the left side onto the right to form a palindrome
if oddLength {
// When odd, the left and right sides "overlap" on the middle digit, so
// don't include in on the right
buffer.WriteString(mirroredSide[1:])
} else {
buffer.WriteString(mirroredSide)
}
return buffer.String()
}