Instruction stringlengths 45 106 | input_code stringlengths 1 13.7k | output_code stringlengths 1 13.7k |
|---|---|---|
Translate the given Arturo code snippet into C without altering its behavior. | hailstone: function [n][
ret: @[n]
while [n>1][
if? 1 = and n 1 -> n: 1+3*n
else -> n: n/2
'ret ++ n
]
ret
]
print "Hailstone sequence for 27:"
print hailstone 27
maxHailstoneLength: 0
maxHailstone: 0
loop 2..1000 'x [
l: size hailstone x
if l>maxHailstoneLength [
maxHailstoneLength: l
maxHailstone: x
]
]
print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Write a version of this Arturo function in C# with identical behavior. | hailstone: function [n][
ret: @[n]
while [n>1][
if? 1 = and n 1 -> n: 1+3*n
else -> n: n/2
'ret ++ n
]
ret
]
print "Hailstone sequence for 27:"
print hailstone 27
maxHailstoneLength: 0
maxHailstone: 0
loop 2..1000 'x [
l: size hailstone x
if l>maxHailstoneLength [
maxHailstoneLength: l
maxHailstone: x
]
]
print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Produce a language-to-language conversion: from Arturo to C#, same semantics. | hailstone: function [n][
ret: @[n]
while [n>1][
if? 1 = and n 1 -> n: 1+3*n
else -> n: n/2
'ret ++ n
]
ret
]
print "Hailstone sequence for 27:"
print hailstone 27
maxHailstoneLength: 0
maxHailstone: 0
loop 2..1000 'x [
l: size hailstone x
if l>maxHailstoneLength [
maxHailstoneLength: l
maxHailstone: x
]
]
print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Convert the following code from Arturo to C++, ensuring the logic remains intact. | hailstone: function [n][
ret: @[n]
while [n>1][
if? 1 = and n 1 -> n: 1+3*n
else -> n: n/2
'ret ++ n
]
ret
]
print "Hailstone sequence for 27:"
print hailstone 27
maxHailstoneLength: 0
maxHailstone: 0
loop 2..1000 'x [
l: size hailstone x
if l>maxHailstoneLength [
maxHailstoneLength: l
maxHailstone: x
]
]
print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Port the provided Arturo code into C++ while preserving the original functionality. | hailstone: function [n][
ret: @[n]
while [n>1][
if? 1 = and n 1 -> n: 1+3*n
else -> n: n/2
'ret ++ n
]
ret
]
print "Hailstone sequence for 27:"
print hailstone 27
maxHailstoneLength: 0
maxHailstone: 0
loop 2..1000 'x [
l: size hailstone x
if l>maxHailstoneLength [
maxHailstoneLength: l
maxHailstone: x
]
]
print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Write the same algorithm in Java as shown in this Arturo implementation. | hailstone: function [n][
ret: @[n]
while [n>1][
if? 1 = and n 1 -> n: 1+3*n
else -> n: n/2
'ret ++ n
]
ret
]
print "Hailstone sequence for 27:"
print hailstone 27
maxHailstoneLength: 0
maxHailstone: 0
loop 2..1000 'x [
l: size hailstone x
if l>maxHailstoneLength [
maxHailstoneLength: l
maxHailstone: x
]
]
print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Write the same algorithm in Java as shown in this Arturo implementation. | hailstone: function [n][
ret: @[n]
while [n>1][
if? 1 = and n 1 -> n: 1+3*n
else -> n: n/2
'ret ++ n
]
ret
]
print "Hailstone sequence for 27:"
print hailstone 27
maxHailstoneLength: 0
maxHailstone: 0
loop 2..1000 'x [
l: size hailstone x
if l>maxHailstoneLength [
maxHailstoneLength: l
maxHailstone: x
]
]
print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Produce a functionally identical Python code for the snippet given in Arturo. | hailstone: function [n][
ret: @[n]
while [n>1][
if? 1 = and n 1 -> n: 1+3*n
else -> n: n/2
'ret ++ n
]
ret
]
print "Hailstone sequence for 27:"
print hailstone 27
maxHailstoneLength: 0
maxHailstone: 0
loop 2..1000 'x [
l: size hailstone x
if l>maxHailstoneLength [
maxHailstoneLength: l
maxHailstone: x
]
]
print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]
| def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))
|
Write a version of this Arturo function in Python with identical behavior. | hailstone: function [n][
ret: @[n]
while [n>1][
if? 1 = and n 1 -> n: 1+3*n
else -> n: n/2
'ret ++ n
]
ret
]
print "Hailstone sequence for 27:"
print hailstone 27
maxHailstoneLength: 0
maxHailstone: 0
loop 2..1000 'x [
l: size hailstone x
if l>maxHailstoneLength [
maxHailstoneLength: l
maxHailstone: x
]
]
print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]
| def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))
|
Convert the following code from Arturo to VB, ensuring the logic remains intact. | hailstone: function [n][
ret: @[n]
while [n>1][
if? 1 = and n 1 -> n: 1+3*n
else -> n: n/2
'ret ++ n
]
ret
]
print "Hailstone sequence for 27:"
print hailstone 27
maxHailstoneLength: 0
maxHailstone: 0
loop 2..1000 'x [
l: size hailstone x
if l>maxHailstoneLength [
maxHailstoneLength: l
maxHailstone: x
]
]
print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Convert this Arturo block to VB, preserving its control flow and logic. | hailstone: function [n][
ret: @[n]
while [n>1][
if? 1 = and n 1 -> n: 1+3*n
else -> n: n/2
'ret ++ n
]
ret
]
print "Hailstone sequence for 27:"
print hailstone 27
maxHailstoneLength: 0
maxHailstone: 0
loop 2..1000 'x [
l: size hailstone x
if l>maxHailstoneLength [
maxHailstoneLength: l
maxHailstone: x
]
]
print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Rewrite this program in Go while keeping its functionality equivalent to the Arturo version. | hailstone: function [n][
ret: @[n]
while [n>1][
if? 1 = and n 1 -> n: 1+3*n
else -> n: n/2
'ret ++ n
]
ret
]
print "Hailstone sequence for 27:"
print hailstone 27
maxHailstoneLength: 0
maxHailstone: 0
loop 2..1000 'x [
l: size hailstone x
if l>maxHailstoneLength [
maxHailstoneLength: l
maxHailstone: x
]
]
print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Translate the given Arturo code snippet into Go without altering its behavior. | hailstone: function [n][
ret: @[n]
while [n>1][
if? 1 = and n 1 -> n: 1+3*n
else -> n: n/2
'ret ++ n
]
ret
]
print "Hailstone sequence for 27:"
print hailstone 27
maxHailstoneLength: 0
maxHailstone: 0
loop 2..1000 'x [
l: size hailstone x
if l>maxHailstoneLength [
maxHailstoneLength: l
maxHailstone: x
]
]
print ["max hailstone sequence found (<100000): of length" maxHailstoneLength "for" maxHailstone]
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Preserve the algorithm and functionality while converting the code from AutoHotKey to C. |
List := varNum := 7
While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
Count := 1, List := varNum := 27
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
MaxNum := Max := 0
TimesToLoop := 100000
Offset := 70000
Loop, %TimesToLoop%
{
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...`n-------------------`n"
text .= "Current starting number: " Index "`n"
text .= "Current sequence count: " Count
text .= "`n-------------------`n"
text .= "Maximum starting number: " MaxNum "`n"
text .= "Maximum sequence count: " Max " <<"
ToolTip, %text%
Count := 1
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Convert the following code from AutoHotKey to C, ensuring the logic remains intact. |
List := varNum := 7
While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
Count := 1, List := varNum := 27
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
MaxNum := Max := 0
TimesToLoop := 100000
Offset := 70000
Loop, %TimesToLoop%
{
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...`n-------------------`n"
text .= "Current starting number: " Index "`n"
text .= "Current sequence count: " Count
text .= "`n-------------------`n"
text .= "Maximum starting number: " MaxNum "`n"
text .= "Maximum sequence count: " Max " <<"
ToolTip, %text%
Count := 1
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Can you help me rewrite this code in C# instead of AutoHotKey, keeping it the same logically? |
List := varNum := 7
While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
Count := 1, List := varNum := 27
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
MaxNum := Max := 0
TimesToLoop := 100000
Offset := 70000
Loop, %TimesToLoop%
{
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...`n-------------------`n"
text .= "Current starting number: " Index "`n"
text .= "Current sequence count: " Count
text .= "`n-------------------`n"
text .= "Maximum starting number: " MaxNum "`n"
text .= "Maximum sequence count: " Max " <<"
ToolTip, %text%
Count := 1
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Maintain the same structure and functionality when rewriting this code in C#. |
List := varNum := 7
While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
Count := 1, List := varNum := 27
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
MaxNum := Max := 0
TimesToLoop := 100000
Offset := 70000
Loop, %TimesToLoop%
{
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...`n-------------------`n"
text .= "Current starting number: " Index "`n"
text .= "Current sequence count: " Count
text .= "`n-------------------`n"
text .= "Maximum starting number: " MaxNum "`n"
text .= "Maximum sequence count: " Max " <<"
ToolTip, %text%
Count := 1
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Produce a language-to-language conversion: from AutoHotKey to C++, same semantics. |
List := varNum := 7
While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
Count := 1, List := varNum := 27
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
MaxNum := Max := 0
TimesToLoop := 100000
Offset := 70000
Loop, %TimesToLoop%
{
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...`n-------------------`n"
text .= "Current starting number: " Index "`n"
text .= "Current sequence count: " Count
text .= "`n-------------------`n"
text .= "Maximum starting number: " MaxNum "`n"
text .= "Maximum sequence count: " Max " <<"
ToolTip, %text%
Count := 1
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Write the same algorithm in C++ as shown in this AutoHotKey implementation. |
List := varNum := 7
While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
Count := 1, List := varNum := 27
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
MaxNum := Max := 0
TimesToLoop := 100000
Offset := 70000
Loop, %TimesToLoop%
{
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...`n-------------------`n"
text .= "Current starting number: " Index "`n"
text .= "Current sequence count: " Count
text .= "`n-------------------`n"
text .= "Maximum starting number: " MaxNum "`n"
text .= "Maximum sequence count: " Max " <<"
ToolTip, %text%
Count := 1
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Write the same code in Java as shown below in AutoHotKey. |
List := varNum := 7
While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
Count := 1, List := varNum := 27
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
MaxNum := Max := 0
TimesToLoop := 100000
Offset := 70000
Loop, %TimesToLoop%
{
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...`n-------------------`n"
text .= "Current starting number: " Index "`n"
text .= "Current sequence count: " Count
text .= "`n-------------------`n"
text .= "Maximum starting number: " MaxNum "`n"
text .= "Maximum sequence count: " Max " <<"
ToolTip, %text%
Count := 1
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Convert the following code from AutoHotKey to Java, ensuring the logic remains intact. |
List := varNum := 7
While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
Count := 1, List := varNum := 27
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
MaxNum := Max := 0
TimesToLoop := 100000
Offset := 70000
Loop, %TimesToLoop%
{
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...`n-------------------`n"
text .= "Current starting number: " Index "`n"
text .= "Current sequence count: " Count
text .= "`n-------------------`n"
text .= "Maximum starting number: " MaxNum "`n"
text .= "Maximum sequence count: " Max " <<"
ToolTip, %text%
Count := 1
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Convert this AutoHotKey snippet to Python and keep its semantics consistent. |
List := varNum := 7
While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
Count := 1, List := varNum := 27
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
MaxNum := Max := 0
TimesToLoop := 100000
Offset := 70000
Loop, %TimesToLoop%
{
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...`n-------------------`n"
text .= "Current starting number: " Index "`n"
text .= "Current sequence count: " Count
text .= "`n-------------------`n"
text .= "Maximum starting number: " MaxNum "`n"
text .= "Maximum sequence count: " Max " <<"
ToolTip, %text%
Count := 1
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max
| def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))
|
Produce a language-to-language conversion: from AutoHotKey to Python, same semantics. |
List := varNum := 7
While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
Count := 1, List := varNum := 27
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
MaxNum := Max := 0
TimesToLoop := 100000
Offset := 70000
Loop, %TimesToLoop%
{
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...`n-------------------`n"
text .= "Current starting number: " Index "`n"
text .= "Current sequence count: " Count
text .= "`n-------------------`n"
text .= "Maximum starting number: " MaxNum "`n"
text .= "Maximum sequence count: " Max " <<"
ToolTip, %text%
Count := 1
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max
| def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))
|
Produce a language-to-language conversion: from AutoHotKey to VB, same semantics. |
List := varNum := 7
While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
Count := 1, List := varNum := 27
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
MaxNum := Max := 0
TimesToLoop := 100000
Offset := 70000
Loop, %TimesToLoop%
{
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...`n-------------------`n"
text .= "Current starting number: " Index "`n"
text .= "Current sequence count: " Count
text .= "`n-------------------`n"
text .= "Maximum starting number: " MaxNum "`n"
text .= "Maximum sequence count: " Max " <<"
ToolTip, %text%
Count := 1
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Produce a language-to-language conversion: from AutoHotKey to VB, same semantics. |
List := varNum := 7
While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
Count := 1, List := varNum := 27
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
MaxNum := Max := 0
TimesToLoop := 100000
Offset := 70000
Loop, %TimesToLoop%
{
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...`n-------------------`n"
text .= "Current starting number: " Index "`n"
text .= "Current sequence count: " Count
text .= "`n-------------------`n"
text .= "Maximum starting number: " MaxNum "`n"
text .= "Maximum sequence count: " Max " <<"
ToolTip, %text%
Count := 1
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Keep all operations the same but rewrite the snippet in Go. |
List := varNum := 7
While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
Count := 1, List := varNum := 27
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
MaxNum := Max := 0
TimesToLoop := 100000
Offset := 70000
Loop, %TimesToLoop%
{
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...`n-------------------`n"
text .= "Current starting number: " Index "`n"
text .= "Current sequence count: " Count
text .= "`n-------------------`n"
text .= "Maximum starting number: " MaxNum "`n"
text .= "Maximum sequence count: " Max " <<"
ToolTip, %text%
Count := 1
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Port the following code from AutoHotKey to Go with equivalent syntax and logic. |
List := varNum := 7
While ( varNum > 1 )
List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % List
Count := 1, List := varNum := 27
While ( varNum > 1 )
Count++ , List .= ", " ( varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 ) )
MsgBox % "Sequence:`n" List "`n`nCount: " Count
MaxNum := Max := 0
TimesToLoop := 100000
Offset := 70000
Loop, %TimesToLoop%
{
If ( TimesToLoop < ( varNum := Index := A_Index+Offset ) )
Break
text := "Processing...`n-------------------`n"
text .= "Current starting number: " Index "`n"
text .= "Current sequence count: " Count
text .= "`n-------------------`n"
text .= "Maximum starting number: " MaxNum "`n"
text .= "Maximum sequence count: " Max " <<"
ToolTip, %text%
Count := 1
While ( varNum > 1 )
Count++ , varNum := ( Mod(varNum,2) ? (varNum*3)+1 : varNum//2 )
If ( Count > Max )
Max := Count , MaxNum := Index
}
ToolTip
MsgBox % "Number: " MaxNum "`nCount: " Max
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Produce a functionally identical C code for the snippet given in AWK. |
function hailstone(v, verbose) {
n = 1;
u = v;
while (1) {
if (verbose) printf " "u;
if (u==1) return(n);
n++;
if (u%2 > 0 )
u = 3*u+1;
else
u = u/2;
}
}
BEGIN {
i = 27;
printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
ix=0;
m=0;
for (i=1; i<100000; i++) {
n = hailstone(i,0);
if (m<n) {
m=n;
ix=i;
}
}
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Write the same algorithm in C as shown in this AWK implementation. |
function hailstone(v, verbose) {
n = 1;
u = v;
while (1) {
if (verbose) printf " "u;
if (u==1) return(n);
n++;
if (u%2 > 0 )
u = 3*u+1;
else
u = u/2;
}
}
BEGIN {
i = 27;
printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
ix=0;
m=0;
for (i=1; i<100000; i++) {
n = hailstone(i,0);
if (m<n) {
m=n;
ix=i;
}
}
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Port the provided AWK code into C# while preserving the original functionality. |
function hailstone(v, verbose) {
n = 1;
u = v;
while (1) {
if (verbose) printf " "u;
if (u==1) return(n);
n++;
if (u%2 > 0 )
u = 3*u+1;
else
u = u/2;
}
}
BEGIN {
i = 27;
printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
ix=0;
m=0;
for (i=1; i<100000; i++) {
n = hailstone(i,0);
if (m<n) {
m=n;
ix=i;
}
}
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Port the provided AWK code into C# while preserving the original functionality. |
function hailstone(v, verbose) {
n = 1;
u = v;
while (1) {
if (verbose) printf " "u;
if (u==1) return(n);
n++;
if (u%2 > 0 )
u = 3*u+1;
else
u = u/2;
}
}
BEGIN {
i = 27;
printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
ix=0;
m=0;
for (i=1; i<100000; i++) {
n = hailstone(i,0);
if (m<n) {
m=n;
ix=i;
}
}
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Translate this program into C++ but keep the logic exactly as in AWK. |
function hailstone(v, verbose) {
n = 1;
u = v;
while (1) {
if (verbose) printf " "u;
if (u==1) return(n);
n++;
if (u%2 > 0 )
u = 3*u+1;
else
u = u/2;
}
}
BEGIN {
i = 27;
printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
ix=0;
m=0;
for (i=1; i<100000; i++) {
n = hailstone(i,0);
if (m<n) {
m=n;
ix=i;
}
}
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Port the provided AWK code into C++ while preserving the original functionality. |
function hailstone(v, verbose) {
n = 1;
u = v;
while (1) {
if (verbose) printf " "u;
if (u==1) return(n);
n++;
if (u%2 > 0 )
u = 3*u+1;
else
u = u/2;
}
}
BEGIN {
i = 27;
printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
ix=0;
m=0;
for (i=1; i<100000; i++) {
n = hailstone(i,0);
if (m<n) {
m=n;
ix=i;
}
}
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Write the same code in Java as shown below in AWK. |
function hailstone(v, verbose) {
n = 1;
u = v;
while (1) {
if (verbose) printf " "u;
if (u==1) return(n);
n++;
if (u%2 > 0 )
u = 3*u+1;
else
u = u/2;
}
}
BEGIN {
i = 27;
printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
ix=0;
m=0;
for (i=1; i<100000; i++) {
n = hailstone(i,0);
if (m<n) {
m=n;
ix=i;
}
}
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Produce a functionally identical Java code for the snippet given in AWK. |
function hailstone(v, verbose) {
n = 1;
u = v;
while (1) {
if (verbose) printf " "u;
if (u==1) return(n);
n++;
if (u%2 > 0 )
u = 3*u+1;
else
u = u/2;
}
}
BEGIN {
i = 27;
printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
ix=0;
m=0;
for (i=1; i<100000; i++) {
n = hailstone(i,0);
if (m<n) {
m=n;
ix=i;
}
}
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Transform the following AWK implementation into Python, maintaining the same output and logic. |
function hailstone(v, verbose) {
n = 1;
u = v;
while (1) {
if (verbose) printf " "u;
if (u==1) return(n);
n++;
if (u%2 > 0 )
u = 3*u+1;
else
u = u/2;
}
}
BEGIN {
i = 27;
printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
ix=0;
m=0;
for (i=1; i<100000; i++) {
n = hailstone(i,0);
if (m<n) {
m=n;
ix=i;
}
}
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
| def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))
|
Maintain the same structure and functionality when rewriting this code in Python. |
function hailstone(v, verbose) {
n = 1;
u = v;
while (1) {
if (verbose) printf " "u;
if (u==1) return(n);
n++;
if (u%2 > 0 )
u = 3*u+1;
else
u = u/2;
}
}
BEGIN {
i = 27;
printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
ix=0;
m=0;
for (i=1; i<100000; i++) {
n = hailstone(i,0);
if (m<n) {
m=n;
ix=i;
}
}
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
| def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))
|
Rewrite the snippet below in VB so it works the same as the original AWK code. |
function hailstone(v, verbose) {
n = 1;
u = v;
while (1) {
if (verbose) printf " "u;
if (u==1) return(n);
n++;
if (u%2 > 0 )
u = 3*u+1;
else
u = u/2;
}
}
BEGIN {
i = 27;
printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
ix=0;
m=0;
for (i=1; i<100000; i++) {
n = hailstone(i,0);
if (m<n) {
m=n;
ix=i;
}
}
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Write the same algorithm in VB as shown in this AWK implementation. |
function hailstone(v, verbose) {
n = 1;
u = v;
while (1) {
if (verbose) printf " "u;
if (u==1) return(n);
n++;
if (u%2 > 0 )
u = 3*u+1;
else
u = u/2;
}
}
BEGIN {
i = 27;
printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
ix=0;
m=0;
for (i=1; i<100000; i++) {
n = hailstone(i,0);
if (m<n) {
m=n;
ix=i;
}
}
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Port the following code from AWK to Go with equivalent syntax and logic. |
function hailstone(v, verbose) {
n = 1;
u = v;
while (1) {
if (verbose) printf " "u;
if (u==1) return(n);
n++;
if (u%2 > 0 )
u = 3*u+1;
else
u = u/2;
}
}
BEGIN {
i = 27;
printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
ix=0;
m=0;
for (i=1; i<100000; i++) {
n = hailstone(i,0);
if (m<n) {
m=n;
ix=i;
}
}
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Translate the given AWK code snippet into Go without altering its behavior. |
function hailstone(v, verbose) {
n = 1;
u = v;
while (1) {
if (verbose) printf " "u;
if (u==1) return(n);
n++;
if (u%2 > 0 )
u = 3*u+1;
else
u = u/2;
}
}
BEGIN {
i = 27;
printf("hailstone(%i) has %i elements\n",i,hailstone(i,1));
ix=0;
m=0;
for (i=1; i<100000; i++) {
n = hailstone(i,0);
if (m<n) {
m=n;
ix=i;
}
}
printf("longest hailstone sequence is %i and has %i elements\n",ix,m);
}
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Change the programming language of this snippet from BBC_Basic to C without modifying what it does. | seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Keep all operations the same but rewrite the snippet in C. | seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Write the same algorithm in C# as shown in this BBC_Basic implementation. | seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Can you help me rewrite this code in C# instead of BBC_Basic, keeping it the same logically? | seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Can you help me rewrite this code in C++ instead of BBC_Basic, keeping it the same logically? | seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Produce a language-to-language conversion: from BBC_Basic to C++, same semantics. | seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Preserve the algorithm and functionality while converting the code from BBC_Basic to Java. | seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Convert this BBC_Basic block to Java, preserving its control flow and logic. | seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Write a version of this BBC_Basic function in Python with identical behavior. | seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1
| def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))
|
Write the same code in Python as shown below in BBC_Basic. | seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1
| def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))
|
Convert the following code from BBC_Basic to VB, ensuring the logic remains intact. | seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Change the following BBC_Basic code into VB without altering its purpose. | seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Write the same code in Go as shown below in BBC_Basic. | seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Port the following code from BBC_Basic to Go with equivalent syntax and logic. | seqlen% = FNhailstone(27, TRUE)
PRINT '"Sequence length = "; seqlen%
maxlen% = 0
FOR number% = 2 TO 100000
seqlen% = FNhailstone(number%, FALSE)
IF seqlen% > maxlen% THEN
maxlen% = seqlen%
maxnum% = number%
ENDIF
NEXT
PRINT "The number with the longest hailstone sequence is " ; maxnum%
PRINT "Its sequence length is " ; maxlen%
END
DEF FNhailstone(N%, S%)
LOCAL L%
IF S% THEN PRINT N%;
WHILE N% <> 1
IF N% AND 1 THEN N% = 3 * N% + 1 ELSE N% DIV= 2
IF S% THEN PRINT N%;
L% += 1
ENDWHILE
= L% + 1
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Produce a functionally identical C code for the snippet given in Clojure. | (defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Translate this program into C but keep the logic exactly as in Clojure. | (defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Write the same algorithm in C# as shown in this Clojure implementation. | (defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Keep all operations the same but rewrite the snippet in C#. | (defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Please provide an equivalent version of this Clojure code in C++. | (defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Please provide an equivalent version of this Clojure code in C++. | (defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Rewrite the snippet below in Java so it works the same as the original Clojure code. | (defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Rewrite the snippet below in Java so it works the same as the original Clojure code. | (defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Generate a Python translation of this Clojure snippet without changing its computational steps. | (defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))
| def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))
|
Maintain the same structure and functionality when rewriting this code in Python. | (defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))
| def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))
|
Translate the given Clojure code snippet into VB without altering its behavior. | (defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Port the following code from Clojure to VB with equivalent syntax and logic. | (defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Convert this Clojure block to Go, preserving its control flow and logic. | (defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Translate this program into Go but keep the logic exactly as in Clojure. | (defn hailstone-seq [n]
{:pre [(pos? n)]}
(lazy-seq
(cond (= n 1) '(1)
(even? n) (cons n (hailstone-seq (/ n 2)))
:else (cons n (hailstone-seq (+ (* n 3) 1))))))
(let [hseq (hailstone-seq 27)]
(-> hseq count (= 112) assert)
(->> hseq (take 4) (= [27 82 41 124]) assert)
(->> hseq (drop 108) (= [8 4 2 1]) assert))
(let [{max-i :num, max-len :len}
(reduce #(max-key :len %1 %2)
(for [i (range 1 100000)]
{:num i, :len (count (hailstone-seq i))}))]
(println "Maximum length" max-len "was found for hailstone(" max-i ")."))
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Port the provided Common_Lisp code into C while preserving the original functionality. | (defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Generate an equivalent C version of this Common_Lisp code. | (defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Translate the given Common_Lisp code snippet into C# without altering its behavior. | (defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Maintain the same structure and functionality when rewriting this code in C#. | (defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Port the following code from Common_Lisp to C++ with equivalent syntax and logic. | (defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Change the following Common_Lisp code into C++ without altering its purpose. | (defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Rewrite the snippet below in Java so it works the same as the original Common_Lisp code. | (defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Convert this Common_Lisp snippet to Java and keep its semantics consistent. | (defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Write a version of this Common_Lisp function in Python with identical behavior. | (defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))
| def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))
|
Produce a functionally identical Python code for the snippet given in Common_Lisp. | (defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))
| def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))
|
Convert the following code from Common_Lisp to VB, ensuring the logic remains intact. | (defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Write the same algorithm in VB as shown in this Common_Lisp implementation. | (defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Ensure the translated Go code behaves exactly like the original Common_Lisp snippet. | (defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Produce a functionally identical Go code for the snippet given in Common_Lisp. | (defun hailstone (len)
(loop for x = len
then (if (evenp x)
(/ x 2)
(+ 1 (* 3 x)))
collect x until (= x 1)))
(defun max-hailstone-start (limit mx curr)
(declare (xargs :mode :program))
(if (zp limit)
(mv mx curr)
(let ((new-mx (len (hailstone limit))))
(if (> new-mx mx)
(max-hailstone-start (1- limit) new-mx limit)
(max-hailstone-start (1- limit) mx curr)))))
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Port the following code from D to C with equivalent syntax and logic. | import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Convert the following code from D to C, ensuring the logic remains intact. | import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Convert this D block to C#, preserving its control flow and logic. | import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Generate an equivalent C# version of this D code. | import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Convert this D block to C++, preserving its control flow and logic. | import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Change the following D code into C++ without altering its purpose. | import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Please provide an equivalent version of this D code in Java. | import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Produce a language-to-language conversion: from D to Java, same semantics. | import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Preserve the algorithm and functionality while converting the code from D to Python. | import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}
| def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))
|
Convert this D block to Python, preserving its control flow and logic. | import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}
| def hailstone(n):
seq = [n]
while n>1:
n = 3*n + 1 if n & 1 else n//2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
max((len(hailstone(i)), i) for i in range(1,100000)))
|
Translate this program into VB but keep the logic exactly as in D. | import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Produce a language-to-language conversion: from D to VB, same semantics. | import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Generate an equivalent Go version of this D code. | import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Transform the following D implementation into Go, maintaining the same output and logic. | import std.stdio, std.algorithm, std.range, std.typecons;
auto hailstone(uint n) pure nothrow {
auto result = [n];
while (n != 1) {
n = (n & 1) ? (n * 3 + 1) : (n / 2);
result ~= n;
}
return result;
}
void main() {
enum M = 27;
immutable h = M.hailstone;
writeln("hailstone(", M, ")= ", h[0 .. 4], " ... " , h[$ - 4 .. $]);
writeln("Length hailstone(", M, ")= ", h.length);
enum N = 100_000;
immutable p = iota(1, N)
.map!(i => tuple(i.hailstone.length, i))
.reduce!max;
writeln("Longest sequence in [1,", N, "]= ",p[1]," with len ",p[0]);
}
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Write a version of this Delphi function in C with identical behavior. | program ShowHailstoneSequence;
uses SysUtils, Generics.Collections;
procedure GetHailstoneSequence(aStartingNumber: Integer; aHailstoneList: TList<Integer>);
var
n: Integer;
begin
aHailstoneList.Clear;
aHailstoneList.Add(aStartingNumber);
n := aStartingNumber;
while n <> 1 do
begin
if Odd(n) then
n := (3 * n) + 1
else
n := n div 2;
aHailstoneList.Add(n);
end;
end;
var
i: Integer;
lList: TList<Integer>;
lMaxSequence: Integer;
lMaxLength: Integer;
begin
lList := TList<Integer>.Create;
try
GetHailstoneSequence(27, lList);
Writeln(Format('27: %d elements', [lList.Count]));
Writeln(Format('[%d,%d,%d,%d ... %d,%d,%d,%d]',
[lList[0], lList[1], lList[2], lList[3],
lList[lList.Count - 4], lList[lList.Count - 3], lList[lList.Count - 2], lList[lList.Count - 1]]));
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
for i := 1 to 100000 do
begin
GetHailstoneSequence(i, lList);
if lList.Count > lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lList.Count;
end;
end;
Writeln(Format('Longest sequence under 100,000: %d with %d elements', [lMaxSequence, lMaxLength]));
finally
lList.Free;
end;
Readln;
end.
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Please provide an equivalent version of this Delphi code in C. | program ShowHailstoneSequence;
uses SysUtils, Generics.Collections;
procedure GetHailstoneSequence(aStartingNumber: Integer; aHailstoneList: TList<Integer>);
var
n: Integer;
begin
aHailstoneList.Clear;
aHailstoneList.Add(aStartingNumber);
n := aStartingNumber;
while n <> 1 do
begin
if Odd(n) then
n := (3 * n) + 1
else
n := n div 2;
aHailstoneList.Add(n);
end;
end;
var
i: Integer;
lList: TList<Integer>;
lMaxSequence: Integer;
lMaxLength: Integer;
begin
lList := TList<Integer>.Create;
try
GetHailstoneSequence(27, lList);
Writeln(Format('27: %d elements', [lList.Count]));
Writeln(Format('[%d,%d,%d,%d ... %d,%d,%d,%d]',
[lList[0], lList[1], lList[2], lList[3],
lList[lList.Count - 4], lList[lList.Count - 3], lList[lList.Count - 2], lList[lList.Count - 1]]));
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
for i := 1 to 100000 do
begin
GetHailstoneSequence(i, lList);
if lList.Count > lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lList.Count;
end;
end;
Writeln(Format('Longest sequence under 100,000: %d with %d elements', [lMaxSequence, lMaxLength]));
finally
lList.Free;
end;
Readln;
end.
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Rewrite this program in C# while keeping its functionality equivalent to the Delphi version. | program ShowHailstoneSequence;
uses SysUtils, Generics.Collections;
procedure GetHailstoneSequence(aStartingNumber: Integer; aHailstoneList: TList<Integer>);
var
n: Integer;
begin
aHailstoneList.Clear;
aHailstoneList.Add(aStartingNumber);
n := aStartingNumber;
while n <> 1 do
begin
if Odd(n) then
n := (3 * n) + 1
else
n := n div 2;
aHailstoneList.Add(n);
end;
end;
var
i: Integer;
lList: TList<Integer>;
lMaxSequence: Integer;
lMaxLength: Integer;
begin
lList := TList<Integer>.Create;
try
GetHailstoneSequence(27, lList);
Writeln(Format('27: %d elements', [lList.Count]));
Writeln(Format('[%d,%d,%d,%d ... %d,%d,%d,%d]',
[lList[0], lList[1], lList[2], lList[3],
lList[lList.Count - 4], lList[lList.Count - 3], lList[lList.Count - 2], lList[lList.Count - 1]]));
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
for i := 1 to 100000 do
begin
GetHailstoneSequence(i, lList);
if lList.Count > lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lList.Count;
end;
end;
Writeln(Format('Longest sequence under 100,000: %d with %d elements', [lMaxSequence, lMaxLength]));
finally
lList.Free;
end;
Readln;
end.
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.