Perfect Number series
N:1 =1
N:2 =34
N:3 =122
N:4 =1111
N:5 =11123
N:6 so on...??
Here N is the number of digits in an Integer.
Number is called perfect if it follow these Rules:-
Number(without leading zeros) does not contain any other zeros.
The sum of squares of all digits of Number is a perfect square.
Example:-For N=2 34
34:- 3^2+4^2 :- 25 which is a perfect square
How to Compute Next smallest N Digit Number of This series is there any algorithm or formula following this series ..??
package main
import (
"fmt"
"math"
"strconv"
)
func main() {
all := 1
for all < 21 {
fmt.Println(keepdo(0,0,all))
all++
}
}
func keepdo(i int,sum int, ma int) (bool, int, string){
if i>= ma {
return square_check(sum), sum, ""
}
try := 1
for try < 10 {
sumc := sum +try*try
check, s,ss := keepdo(i+1,sumc, ma)
if check {
return check, s, strconv.Itoa(try) + ss
}
try++
}
return false, sum, ""
}
func square_check(a int) bool {
var int_root int = int(math.Sqrt(float64(a)))
return (int_root * int_root) == a
}
This is a very naive solution.
Should work. Pls do let me know if any improvement
https://play.golang.org/p/drVPaE0R71I
Related
I am new to go and now evaluate a demo function about slice with Fibonacci sequence
package main
import "fmt"
func fbn(n int) []uint64 {
fbnSlice := make([]uint64, n)
fbnSlice[0] = 1
fbnSlice[1] = 1
for i := 2; i < n; i++ {
fbnSlice[i] = fbnSlice[i-1] + fbnSlice[i-2]
}
return fbnSlice
}
func main() {
fnbSlice := fbn(5)
fmt.Println(fnbSlice)
}
It will print "[1 1 2 3 5]"
My doubt is how the slice add it's len to 5，not the 5th num, thanks!
make([]uint64, n)
Will make a slice of length n, filled with zeros. Hence, fbn(5) will produce a slice of length 5.
I am drawing bar charts and i've come across a tricky problem. How to programmatically set the max value for the y axis label depending on the max value for a given series. So if you had a bar with a value of 7, you might want the y axis to go up to 10
My approach is not ideal but works like this:
Get a number to round, like 829
Count the number of digits (3)
Use a loop to convert to a string of 0s ("000")
Add a 1 to the start of the string then convert to a float (1000)
Find the difference (1000 - 829 = 171)
Get the first digit of the difference (1) and then add that to the first digit of the float, with the remaining set to zero ("900"), then convert to a number (900)
This means that 725 will see a y axis max label number of 800, and 829 of 900
My code works, but I feel like it's a piece of crap with a hacky approach
I have to code for big numbers. For example, if the float I want to find the max value for is >10000 then take the first two digits, and add 1000 to it. If >100,000 add 10,000
How can I improve here? I'm a little stuck, is my idea of converting to strings even right?!
Full code here:
package main
import (
"fmt"
"strconv"
)
func main() {
myFloat := 899175.0
x := getMaxYAxisValueForChart(myFloat)
fmt.Println("The number to find the maximum value for is: ", myFloat)
fmt.Println("This should be the max value for the y axis: ", x)
}
func getMaxYAxisValueForChart(float float64) (YAxisMaximum float64) {
//Convert to string with no decimals
floatAsString := fmt.Sprintf("%.f", float)
//Get length of the string float
floatAsStringLength := len(floatAsString)
//For each digit in the string, make a zero-string
stringPowerTen := "0"
for i := 1; i < floatAsStringLength; i++ {
stringPowerTen += "0"
}
//Add a 1 to the 0 string to get the difference from the float
stringPowerTenWithOne := "1" + stringPowerTen
//Convert the number string to a float
convertStringPowerTenToFloat := ConvertStringsToFloat(stringPowerTenWithOne)
//Get the difference from the denominator from the numerator
difference := convertStringPowerTenToFloat - float
//We want to isolate the first digit to check how far the float is (100 is far from 1000) and then correct if so
floatAsStringDifference := fmt.Sprintf("%.f", difference)
runes := []rune(floatAsStringDifference)
floatAsStringDifferenceFirstDigit := string(runes[0])
//For the denominator we want to take away the difference that is rounded to the nearest ten, hundred etc
runes = []rune(stringPowerTen)
differenceLastDigitsAsString := ""
if difference < 10 {
differenceLastDigitsAsString = "1"
} else if difference < 30 && difference < 100 {
differenceLastDigitsAsString = "0"
} else {
differenceLastDigitsAsString = floatAsStringDifferenceFirstDigit + string(runes[1:])
}
//Convert the number difference string from total to a float
convertDifferenceStringPowerTenToFloat := ConvertStringsToFloat(differenceLastDigitsAsString)
YAxisMaximum = convertStringPowerTenToFloat - convertDifferenceStringPowerTenToFloat
//If float is less than 10,0000
if float < 10000 && (YAxisMaximum-float >= 500) {
YAxisMaximum = YAxisMaximum - 500
}
if float < 10000 && (YAxisMaximum-float < 500) {
YAxisMaximum = YAxisMaximum
}
//If number bigger than 10,000 then get the nearest 1,000
if float > 10000 {
runes = []rune(floatAsString)
floatAsString = string(runes[0:2])
runes = []rune(stringPowerTen)
stringPowerTen = string(runes[2:])
runes = []rune(stringPowerTenWithOne)
stringPowerTenWithOne = string(runes[0:(len(stringPowerTenWithOne) - 2)])
YAxisMaximum = ConvertStringsToFloat(floatAsString+stringPowerTen) + ConvertStringsToFloat(stringPowerTenWithOne)
}
if float > 10000 {
runes = []rune(floatAsString)
floatAsString = string(runes[0:2])
runes = []rune(stringPowerTen)
stringPowerTen = string(runes[:])
runes = []rune(stringPowerTenWithOne)
stringPowerTenWithOne = string(runes[0:(len(stringPowerTenWithOne))])
YAxisMaximum = ConvertStringsToFloat(floatAsString+stringPowerTen) + ConvertStringsToFloat(stringPowerTenWithOne)
}
return YAxisMaximum
}
func ConvertStringsToFloat(stringToConvert string) (floatOutput float64) {
floatOutput, Error := strconv.ParseFloat(stringToConvert, 64)
if Error != nil {
fmt.Println(Error)
}
return floatOutput
}
Here is the solution based off of Matt Timmermans answer, but converted to work in Go:
func testing(float float64) (YAxisMaximum float64) {
place := 1.0
for float >= place*10.0 {
place *= 10.0
}
return math.Ceil(float/place) * place
}
Wow, that's a pretty complicated procedure you have. This is how I would do it if the numbers aren't enormous. I don't know go, so I'm going to guess about how to write it in that language:
func getMaxYAxisValueForChart(float float64) {
place := 1.0;
while float >= place*10.0 {
place *= 10.0;
}
return math.Ceil(float/place) * place;
}
You can get the magnitude of a number using Math.Log10
int magnitude = (int)Math.Pow(10, (int)Math.Log10(value));
Use that to divide the number down, calculate ceiling and then scale it back up.
No strings, no while loops.
Take the length of the string and calculate that 10 to the power of that length
Or...better take the Log base 10, get the integer part, add 1 and then return that to the power of 10 :)
import (
"fmt"
"math"
)
//func PowerScale(x int) int64{
// return int64(math.Pow(10,float64(len((fmt.Sprintf("%d",x))))))
//}
func PowerScale(x int) int64 {
return int64(math.Pow(10,float64(int(math.Log10(float64(x))+1))))
}
func main() {
fmt.Println(PowerScale(829))
fmt.Println(PowerScale(7))
}
Since 829 is an int, or can be cast to, a pure integer solution :
func getMaxYAxisValueForChart(int int64) {
base := 10;
while int > base*10 {
base := 10 * base;
}
return int + (base - int) % base;
}
This question already has answers here:
How to generate a random string of a fixed length in Go?
(18 answers)
Closed 4 years ago.
I would like to generate number for invitation code.
I get a number of digit for this and i need to generate a string number according to this digit.
Example :
For 3 i need to generate a string number beetween 111 and 999
for 4 1111 to 9999
for 2 11 to 99
etc...
What is the best way to do that ?
I have some idea like making two empty string filling them with 1 for the first one according to x, with 9 for the second one according to X.
Then converting them to int and make a random between these two number, but i don't thik it's the optimal way to do it.
package main
import (
"fmt"
"math/rand"
"strconv"
"time"
)
func randInt(min int, max int) int {
return min + rand.Intn(max-min)
}
func main() {
x := 3
first := ""
second := ""
i := 0
for i < x {
first = first + "1"
i = i + 1
}
i = 0
for i < x {
second = second + "9"
i = i + 1
}
rand.Seed(time.Now().UTC().UnixNano())
fmt.Println(first)
fmt.Println(second)
firstInt, _ := strconv.Atoi(first)
secondInt, _ := strconv.Atoi(second)
fmt.Println(randInt(firstInt, secondInt))
}
regards
You can do something like this:
package main
import (
"fmt"
"math/rand"
"time"
)
func init() {
rand.Seed(time.Now().UnixNano())
}
var numbers = []rune("0123456789")
func GenerateNumberString(length int) string {
b := make([]rune, length)
for i := range b {
b[i] = numbers[rand.Intn(len(numbers))]
}
return string(b)
}
func main() {
fmt.Println(GenerateNumberString(2))
fmt.Println(GenerateNumberString(3))
fmt.Println(GenerateNumberString(4))
}
Try it here in the go playground.
How can I generate a stream of unique random number in Go?
I want to guarantee there are no duplicate values in array a using math/rand and/or standard Go library utilities.
func RandomNumberGenerator() *rand.Rand {
s1 := rand.NewSource(time.Now().UnixNano())
r1 := rand.New(s1)
return r1
}
rng := RandomNumberGenerator()
N := 10000
for i := 0; i < N; i++ {
a[i] = rng.Int()
}
There are questions and solutions on how to generate a series of random number in Go, for example, here.
But I would like to generate a series of random numbers that does not duplicate previous values. Is there a standard/recommended way to achieve this in Go?
My guess is to (1) use permutation or to (2) keep track of previously generated numbers and regenerate a value if it's been generated before.
But solution (1) sounds like overkill if I only want a few number and (2) sounds very time consuming if I end up generating a long series of random numbers due to collision, and I guess it's also very memory-consuming.
Use Case: To benchmark a Go program with 10K, 100K, 1M pseudo-random number that has no duplicates.
You should absolutely go with approach 2. Let's assume you're running on a 64-bit machine, and thus generating 63-bit integers (64 bits, but rand.Int never returns negative numbers). Even if you generate 4 billion numbers, there's still only a 1 in 4 billion chance that any given number will be a duplicate. Thus, you'll almost never have to regenerate, and almost never never have to regenerate twice.
Try, for example:
type UniqueRand struct {
generated map[int]bool
}
func (u *UniqueRand) Int() int {
for {
i := rand.Int()
if !u.generated[i] {
u.generated[i] = true
return i
}
}
}
I had similar task to pick elements from initial slice by random uniq index. So from slice with 10k elements get 1k random uniq elements.
Here is simple head on solution:
import (
"time"
"math/rand"
)
func getRandomElements(array []string) []string {
result := make([]string, 0)
existingIndexes := make(map[int]struct{}, 0)
randomElementsCount := 1000
for i := 0; i < randomElementsCount; i++ {
randomIndex := randomIndex(len(array), existingIndexes)
result = append(result, array[randomIndex])
}
return result
}
func randomIndex(size int, existingIndexes map[int]struct{}) int {
rand.Seed(time.Now().UnixNano())
for {
randomIndex := rand.Intn(size)
_, exists := existingIndexes[randomIndex]
if !exists {
existingIndexes[randomIndex] = struct{}{}
return randomIndex
}
}
}
I see two reasons for wanting this. You want to test a random number generator, or you want unique random numbers.
You're Testing A Random Number Generator
My first question is why? There's plenty of solid random number generators available. Don't write your own, it's basically dabbling in cryptography and that's never a good idea. Maybe you're testing a system that uses a random number generator to generate random output?
There's a problem: there's no guarantee random numbers are unique. They're random. There's always a possibility of collision. Testing that random output is unique is incorrect.
Instead, you want to test the results are distributed evenly. To do this I'll reference another answer about how to test a random number generator.
You Want Unique Random Numbers
From a practical perspective you don't need guaranteed uniqueness, but to make collisions so unlikely that it's not a concern. This is what UUIDs are for. They're 128 bit Universally Unique IDentifiers. There's a number of ways to generate them for particular scenarios.
UUIDv4 is basically just a 122 bit random number which has some ungodly small chance of a collision. Let's approximate it.
n = how many random numbers you'll generate
M = size of the keyspace (2^122 for a 122 bit random number)
P = probability of collision
P = n^2/2M
Solving for n...
n = sqrt(2MP)
Setting P to something absurd like 1e-12 (one in a trillion), we find you can generate about 3.2 trillion UUIDv4s with a 1 in a trillion chance of collision. You're 1000 times more likely to win the lottery than have a collision in 3.2 trillion UUIDv4s. I think that's acceptable.
Here's a UUIDv4 library in Go to use and a demonstration of generating 1 million unique random 128 bit values.
package main
import (
"fmt"
"github.com/frankenbeanies/uuid4"
)
func main() {
for i := 0; i <= 1000000; i++ {
uuid := uuid4.New().Bytes()
// use the uuid
}
}
you can generate a unique random number with len(12) using UnixNano in golang time package :
uniqueNumber:=time.Now().UnixNano()/(1<<22)
println(uniqueNumber)
it's always random :D
1- Fast positive and negative int32 unique pseudo random numbers in 296ms using std lib:
package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
const n = 1000000
rand.Seed(time.Now().UTC().UnixNano())
duplicate := 0
mp := make(map[int32]struct{}, n)
var r int32
t := time.Now()
for i := 0; i < n; {
r = rand.Int31()
if i&1 == 0 {
r = -r
}
if _, ok := mp[r]; ok {
duplicate++
} else {
mp[r] = zero
i++
}
}
fmt.Println(time.Since(t))
fmt.Println("len: ", len(mp))
fmt.Println("duplicate: ", duplicate)
positive := 0
for k := range mp {
if k > 0 {
positive++
}
}
fmt.Println(`n=`, n, `positive=`, positive)
}
var zero = struct{}{}
output:
296.0169ms
len: 1000000
duplicate: 118
n= 1000000 positive= 500000
2- Just fill the map[int32]struct{}:
for i := int32(0); i < n; i++ {
m[i] = zero
}
When reading it is not in order in Go:
for k := range m {
fmt.Print(k, " ")
}
And this just takes 183ms for 1000000 unique numbers, no duplicate (The Go Playground):
package main
import (
"fmt"
"time"
)
func main() {
const n = 1000000
m := make(map[int32]struct{}, n)
t := time.Now()
for i := int32(0); i < n; i++ {
m[i] = zero
}
fmt.Println(time.Since(t))
fmt.Println("len: ", len(m))
// for k := range m {
// fmt.Print(k, " ")
// }
}
var zero = struct{}{}
3- Here is the simple but slow (this takes 22s for 200000 unique numbers), so you may generate and save it to a file once:
package main
import "time"
import "fmt"
import "math/rand"
func main() {
dup := 0
t := time.Now()
const n = 200000
rand.Seed(time.Now().UTC().UnixNano())
var a [n]int32
var exist bool
for i := 0; i < n; {
r := rand.Int31()
exist = false
for j := 0; j < i; j++ {
if a[j] == r {
dup++
fmt.Println(dup)
exist = true
break
}
}
if !exist {
a[i] = r
i++
}
}
fmt.Println(time.Since(t))
}
Temporary workaround based on #joshlf's answer
type UniqueRand struct {
generated map[int]bool //keeps track of
rng *rand.Rand //underlying random number generator
scope int //scope of number to be generated
}
//Generating unique rand less than N
//If N is less or equal to 0, the scope will be unlimited
//If N is greater than 0, it will generate (-scope, +scope)
//If no more unique number can be generated, it will return -1 forwards
func NewUniqueRand(N int) *UniqueRand{
s1 := rand.NewSource(time.Now().UnixNano())
r1 := rand.New(s1)
return &UniqueRand{
generated: map[int]bool{},
rng: r1,
scope: N,
}
}
func (u *UniqueRand) Int() int {
if u.scope > 0 && len(u.generated) >= u.scope {
return -1
}
for {
var i int
if u.scope > 0 {
i = u.rng.Int() % u.scope
}else{
i = u.rng.Int()
}
if !u.generated[i] {
u.generated[i] = true
return i
}
}
}
Client side code
func TestSetGet2(t *testing.T) {
const N = 10000
for _, mask := range []int{0, -1, 0x555555, 0xaaaaaa, 0x333333, 0xcccccc, 0x314159} {
rng := NewUniqueRand(2*N)
a := make([]int, N)
for i := 0; i < N; i++ {
a[i] = (rng.Int() ^ mask) << 1
}
//Benchmark Code
}
}
I imported the math library in my program, and I was trying to find the minimum of three numbers in the following way:
v1[j+1] = math.Min(v1[j]+1, math.Min(v0[j+1]+1, v0[j]+cost))
where v1 is declared as:
t := "stackoverflow"
v1 := make([]int, len(t)+1)
However, when I run my program I get the following error:
./levenshtein_distance.go:36: cannot use int(v0[j + 1] + 1) (type int) as type float64 in argument to math.Min
I thought it was weird because I have another program where I write
fmt.Println(math.Min(2,3))
and that program outputs 2 without complaining.
so I ended up casting the values as float64, so that math.Min could work:
v1[j+1] = math.Min(float64(v1[j]+1), math.Min(float64(v0[j+1]+1), float64(v0[j]+cost)))
With this approach, I got the following error:
./levenshtein_distance.go:36: cannot use math.Min(int(v1[j] + 1), math.Min(int(v0[j + 1] + 1), int(v0[j] + cost))) (type float64) as type int in assignment
so to get rid of the problem, I just casted the result back to int
I thought this was extremely inefficient and hard to read:
v1[j+1] = int(math.Min(float64(v1[j]+1), math.Min(float64(v0[j+1]+1), float64(v0[j]+cost))))
I also wrote a small minInt function, but I think this should be unnecessary because the other programs that make use of math.Min work just fine when taking integers, so I concluded this has to be a problem of my program and not the library per se.
Is there anything that I'm doing terrible wrong?
Here's a program that you can use to reproduce the issues above, line 36 specifically:
package main
import (
"math"
)
func main() {
LevenshteinDistance("stackoverflow", "stackexchange")
}
func LevenshteinDistance(s string, t string) int {
if s == t {
return 0
}
if len(s) == 0 {
return len(t)
}
if len(t) == 0 {
return len(s)
}
v0 := make([]int, len(t)+1)
v1 := make([]int, len(t)+1)
for i := 0; i < len(v0); i++ {
v0[i] = i
}
for i := 0; i < len(s); i++ {
v1[0] = i + 1
for j := 0; j < len(t); j++ {
cost := 0
if s[i] != t[j] {
cost = 1
}
v1[j+1] = int(math.Min(float64(v1[j]+1), math.Min(float64(v0[j+1]+1), float64(v0[j]+cost))))
}
for j := 0; j < len(v0); j++ {
v0[j] = v1[j]
}
}
return v1[len(t)]
}
Until Go 1.18 a one-off function was the standard way; for example, the stdlib's sort.go does it near the top of the file:
func min(a, b int) int {
if a < b {
return a
}
return b
}
You might still want or need to use this approach so your code works on Go versions below 1.18!
Starting with Go 1.18, you can write a generic min function which is just as efficient at run time as the hand-coded single-type version, but works with any type with < and > operators:
func min[T constraints.Ordered](a, b T) T {
if a < b {
return a
}
return b
}
func main() {
fmt.Println(min(1, 2))
fmt.Println(min(1.5, 2.5))
fmt.Println(min("Hello", "世界"))
}
There's been discussion of updating the stdlib to add generic versions of existing functions, but if that happens it won't be until a later version.
math.Min(2, 3) happened to work because numeric constants in Go are untyped. Beware of treating float64s as a universal number type in general, though, since integers above 2^53 will get rounded if converted to float64.
There is no built-in min or max function for integers, but it’s simple to write your own. Thanks to support for variadic functions we can even compare more integers with just one call:
func MinOf(vars ...int) int {
min := vars[0]
for _, i := range vars {
if min > i {
min = i
}
}
return min
}
Usage:
MinOf(3, 9, 6, 2)
Similarly here is the max function:
func MaxOf(vars ...int) int {
max := vars[0]
for _, i := range vars {
if max < i {
max = i
}
}
return max
}
For example,
package main
import "fmt"
func min(x, y int) int {
if x < y {
return x
}
return y
}
func main() {
t := "stackoverflow"
v0 := make([]int, len(t)+1)
v1 := make([]int, len(t)+1)
cost := 1
j := 0
v1[j+1] = min(v1[j]+1, min(v0[j+1]+1, v0[j]+cost))
fmt.Println(v1[j+1])
}
Output:
1
Though the question is quite old, maybe my package imath can be helpful for someone who does not like reinventing a bicycle. There are few functions, finding minimal of two integers: ix.Min (for int), i8.Min (for int8), ux.Min (for uint) and so on. The package can be obtained with go get, imported in your project by URL and functions referred as typeabbreviation.FuncName, for example:
package main
import (
"fmt"
"<Full URL>/go-imath/ix"
)
func main() {
a, b := 45, -42
fmt.Println(ix.Min(a, b)) // Output: -42
}
As the accepted answer states, with the introduction of generics in go 1.18 it's now possible to write a generic function that provides min/max for different numeric types (there is not one built into the language). And with variadic arguments we can support comparing 2 elements or a longer list of elements.
func Min[T constraints.Ordered](args ...T) T {
min := args[0]
for _, x := range args {
if x < min {
min = x
}
}
return min
}
func Max[T constraints.Ordered](args ...T) T {
max := args[0]
for _, x := range args {
if x > max {
max = x
}
}
return max
}
example calls:
Max(1, 2) // 2
Max(4, 5, 3, 1, 2) // 5
Could use https://github.com/pkg/math:
import (
"fmt"
"github.com/pkg/math"
)
func main() {
a, b := 45, -42
fmt.Println(math.Min(a, b)) // Output: -42
}
Since the issue has already been resolved, I would like to add a few words. Always remember that the math package in Golang operates on float64. You can use type conversion to cast int into a float64. Keep in mind to account for type ranges. For example, you cannot fit a float64 into an int16 if the number exceeds the limit for int16 which is 32767. Last but not least, if you convert a float into an int in Golang, the decimal points get truncated without any rounding.
If you want the minimum of a set of N integers you can use (assuming N > 0):
import "sort"
func min(set []int) int {
sort.Slice(set, func(i, j int) bool {
return set[i] < set[j]
})
return set[0]
}
Where the second argument to min function is your less function, that is, the function that decides when an element i of the passed slice is less than an element j
Check it out here in Go Playground: https://go.dev/play/p/lyQYlkwKrsA