How to resolve the algorithm Zeckendorf arithmetic step by step in the V (Vlang) programming language
How to resolve the algorithm Zeckendorf arithmetic step by step in the V (Vlang) programming language
Table of Contents
Problem Statement
This task is a total immersion zeckendorf task; using decimal numbers will attract serious disapprobation. The task is to implement addition, subtraction, multiplication, and division using Zeckendorf number representation. Optionally provide decrement, increment and comparitive operation functions. Like binary 1 + 1 = 10, note carry 1 left. There the similarity ends. 10 + 10 = 101, note carry 1 left and 1 right. 100 + 100 = 1001, note carry 1 left and 2 right, this is the general case. Occurrences of 11 must be changed to 100. Occurrences of 111 may be changed from the right by replacing 11 with 100, or from the left converting 111 to 100 + 100; 10 - 1 = 1. The general rule is borrow 1 right carry 1 left. eg: A larger example: Here you teach your computer its zeckendorf tables. eg. 101 * 1001: Lets try 1000101 divided by 101, so we can use the same table used for multiplication. Efficient algorithms for Zeckendorf arithmetic is interesting. The sections on addition and subtraction are particularly relevant for this task.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Zeckendorf arithmetic step by step in the V (Vlang) programming language
Source code in the v programming language
import strings
const (
dig = ["00", "01", "10"]
dig1 = ["", "1", "10"]
)
struct Zeckendorf {
mut:
d_val int
d_len int
}
fn new_zeck(xx string) Zeckendorf {
mut z := Zeckendorf{}
mut x := xx
if x == "" {
x = "0"
}
mut q := 1
mut i := x.len - 1
z.d_len = i / 2
for ; i >= 0; i-- {
z.d_val += int(x[i]-'0'[0]) * q
q *= 2
}
return z
}
fn (mut z Zeckendorf) a(ii int) {
mut i:=ii
for ; ; i++ {
if z.d_len < i {
z.d_len = i
}
j := (z.d_val >> u32(i*2)) & 3
if j in [0, 1] {
return
} else if j==2 {
if ((z.d_val >> (u32(i+1) * 2)) & 1) != 1 {
return
}
z.d_val += 1 << u32(i*2+1)
return
} else {// 3
z.d_val &= ~(3 << u32(i*2))
z.b((i + 1) * 2)
}
}
}
fn (mut z Zeckendorf) b(p int) {
mut pos := p
if pos == 0 {
z.inc()
return
}
if ((z.d_val >> u32(pos)) & 1) == 0 {
z.d_val += 1 << u32(pos)
z.a(pos / 2)
if pos > 1 {
z.a(pos/2 - 1)
}
} else {
z.d_val &= ~(1 << u32(pos))
z.b(pos + 1)
mut temp := 1
if pos > 1 {
temp = 2
}
z.b(pos - temp)
}
}
fn (mut z Zeckendorf) c(p int) {
mut pos := p
if ((z.d_val >> u32(pos)) & 1) == 1 {
z.d_val &= ~(1 << u32(pos))
return
}
z.c(pos + 1)
if pos > 0 {
z.b(pos - 1)
} else {
z.inc()
}
}
fn (mut z Zeckendorf) inc() {
z.d_val++
z.a(0)
}
fn (mut z1 Zeckendorf) plus_assign(z2 Zeckendorf) {
for gn := 0; gn < (z2.d_len+1)*2; gn++ {
if ((z2.d_val >> u32(gn)) & 1) == 1 {
z1.b(gn)
}
}
}
fn (mut z1 Zeckendorf) minus_assign(z2 Zeckendorf) {
for gn := 0; gn < (z2.d_len+1)*2; gn++ {
if ((z2.d_val >> u32(gn)) & 1) == 1 {
z1.c(gn)
}
}
for z1.d_len > 0 && ((z1.d_val>>u32(z1.d_len*2))&3) == 0 {
z1.d_len--
}
}
fn (mut z1 Zeckendorf) times_assign(z2 Zeckendorf) {
mut na := z2.copy()
mut nb := z2.copy()
mut nr := Zeckendorf{}
for i := 0; i <= (z1.d_len+1)*2; i++ {
if ((z1.d_val >> u32(i)) & 1) > 0 {
nr.plus_assign(nb)
}
nt := nb.copy()
nb.plus_assign(na)
na = nt.copy()
}
z1.d_val = nr.d_val
z1.d_len = nr.d_len
}
fn (z Zeckendorf) copy() Zeckendorf {
return Zeckendorf{z.d_val, z.d_len}
}
fn (z1 Zeckendorf) compare(z2 Zeckendorf) int {
if z1.d_val < z2.d_val {
return -1
} else if z1.d_val > z2.d_val {
return 1
} else {
return 0
}
}
fn (z Zeckendorf) str() string {
if z.d_val == 0 {
return "0"
}
mut sb := strings.new_builder(128)
sb.write_string(dig1[(z.d_val>>u32(z.d_len*2))&3])
for i := z.d_len - 1; i >= 0; i-- {
sb.write_string(dig[(z.d_val>>u32(i*2))&3])
}
return sb.str()
}
fn main() {
println("Addition:")
mut g := new_zeck("10")
g.plus_assign(new_zeck("10"))
println(g)
g.plus_assign(new_zeck("10"))
println(g)
g.plus_assign(new_zeck("1001"))
println(g)
g.plus_assign(new_zeck("1000"))
println(g)
g.plus_assign(new_zeck("10101"))
println(g)
println("\nSubtraction:")
g = new_zeck("1000")
g.minus_assign(new_zeck("101"))
println(g)
g = new_zeck("10101010")
g.minus_assign(new_zeck("1010101"))
println(g)
println("\nMultiplication:")
g = new_zeck("1001")
g.times_assign(new_zeck("101"))
println(g)
g = new_zeck("101010")
g.plus_assign(new_zeck("101"))
println(g)
}
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