Instruction stringlengths 45 106 | input_code stringlengths 1 13.7k | output_code stringlengths 1 13.7k |
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Convert the following code from Delphi to C#, ensuring the logic remains intact. | 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);
}
}
}
|
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 <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 C++ as shown below in Delphi. | 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 <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;
}
|
Can you help me rewrite this code in Java instead of Delphi, keeping it the same logically? | 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.
| 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;
}
}
|
Ensure the translated Java code behaves exactly like the original Delphi snippet. | 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.
| 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 Python as shown in this Delphi implementation. | 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.
| 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)))
|
Ensure the translated Python code behaves exactly like the original Delphi snippet. | 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.
| 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)))
|
Transform the following Delphi implementation into VB, maintaining the same output and logic. | 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.
| 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 VB code behaves exactly like the original Delphi snippet. | 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.
| 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 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.
| 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)
}
|
Generate a Go translation of this Delphi snippet without changing its computational steps. | 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.
| 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 the same algorithm in C as shown in this Elixir implementation. | defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with
end
end
Hailstone.run
| #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 Elixir code. | defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with
end
end
Hailstone.run
| #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 Elixir function in C# with identical behavior. | defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with
end
end
Hailstone.run
| 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 the given Elixir code snippet into C# without altering its behavior. | defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with
end
end
Hailstone.run
| 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 Elixir code into C++ while preserving the original functionality. | defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with
end
end
Hailstone.run
| #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;
}
|
Translate this program into C++ but keep the logic exactly as in Elixir. | defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with
end
end
Hailstone.run
| #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 Elixir. | defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with
end
end
Hailstone.run
| 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 an equivalent Java version of this Elixir code. | defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with
end
end
Hailstone.run
| 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 Elixir block to Python, preserving its control flow and logic. | defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with
end
end
Hailstone.run
| 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 Elixir function in Python with identical behavior. | defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with
end
end
Hailstone.run
| 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)))
|
Port the provided Elixir code into VB while preserving the original functionality. | defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with
end
end
Hailstone.run
| 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 VB as shown below in Elixir. | defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with
end
end
Hailstone.run
| 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
|
Preserve the algorithm and functionality while converting the code from Elixir to Go. | defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with
end
end
Hailstone.run
| 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 Elixir code into Go while preserving the original functionality. | defmodule Hailstone do
require Integer
def step(1) , do: 0
def step(n) when Integer.is_even(n), do: div(n,2)
def step(n) , do: n*3 + 1
def sequence(n) do
Stream.iterate(n, &step/1) |> Stream.take_while(&(&1 > 0)) |> Enum.to_list
end
def run do
seq27 = sequence(27)
len27 = length(seq27)
repr = String.replace(inspect(seq27, limit: 4) <> inspect(Enum.drop(seq27,len27-4)), "][", ", ")
IO.puts "Hailstone(27) has
{len, start} = Enum.map(1..100_000, fn(n) -> {length(sequence(n)), n} end) |> Enum.max
IO.puts "Longest sequence starting under 100000 begins with
end
end
Hailstone.run
| 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)
}
|
Ensure the translated C code behaves exactly like the original Erlang snippet. | -module(hailstone).
-import(io).
-export([main/0]).
hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).
| #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;
}
|
Produce a functionally identical C code for the snippet given in Erlang. | -module(hailstone).
-import(io).
-export([main/0]).
hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).
| #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 following code from Erlang to C# with equivalent syntax and logic. | -module(hailstone).
-import(io).
-export([main/0]).
hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).
| 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 Erlang code into C# while preserving the original functionality. | -module(hailstone).
-import(io).
-export([main/0]).
hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).
| 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 Erlang code in C++. | -module(hailstone).
-import(io).
-export([main/0]).
hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).
| #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;
}
|
Keep all operations the same but rewrite the snippet in C++. | -module(hailstone).
-import(io).
-export([main/0]).
hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).
| #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 Erlang to Java, same semantics. | -module(hailstone).
-import(io).
-export([main/0]).
hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).
| 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;
}
}
|
Can you help me rewrite this code in Java instead of Erlang, keeping it the same logically? | -module(hailstone).
-import(io).
-export([main/0]).
hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).
| 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 an equivalent Python version of this Erlang code. | -module(hailstone).
-import(io).
-export([main/0]).
hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).
| 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)))
|
Ensure the translated Python code behaves exactly like the original Erlang snippet. | -module(hailstone).
-import(io).
-export([main/0]).
hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).
| 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)))
|
Transform the following Erlang implementation into VB, maintaining the same output and logic. | -module(hailstone).
-import(io).
-export([main/0]).
hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).
| 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 functionally identical VB code for the snippet given in Erlang. | -module(hailstone).
-import(io).
-export([main/0]).
hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).
| 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 the snippet below in Go so it works the same as the original Erlang code. | -module(hailstone).
-import(io).
-export([main/0]).
hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).
| 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 Erlang code snippet into Go without altering its behavior. | -module(hailstone).
-import(io).
-export([main/0]).
hailstone(1) -> [1];
hailstone(N) when N band 1 == 1 -> [N|hailstone(N * 3 + 1)];
hailstone(N) when N band 1 == 0 -> [N|hailstone(N div 2)].
max_length(Start, Stop) ->
F = fun (N) -> {length(hailstone(N)), N} end,
Lengths = lists:map(F, lists:seq(Start, Stop)),
lists:max(Lengths).
main() ->
io:format("hailstone(4): ~w~n", [hailstone(4)]),
Seq27 = hailstone(27),
io:format("hailstone(27) length: ~B~n", [length(Seq27)]),
io:format("hailstone(27) first 4: ~w~n",
[lists:sublist(Seq27, 4)]),
io:format("hailstone(27) last 4: ~w~n",
[lists:nthtail(length(Seq27) - 4, Seq27)]),
io:format("finding maximum hailstone(N) length for 1 <= N <= 100000..."),
{Length, N} = max_length(1, 100000),
io:format(" done.~nhailstone(~B) length: ~B~n", [N, Length]).
| 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 F# to C with equivalent syntax and logic. | let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI
| #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 F# code snippet into C without altering its behavior. | let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI
| #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;
}
|
Maintain the same structure and functionality when rewriting this code in C#. | let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI
| 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 F# code into C# while preserving the original functionality. | let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI
| 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);
}
}
}
|
Ensure the translated C++ code behaves exactly like the original F# snippet. | let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI
| #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 this program in C++ while keeping its functionality equivalent to the F# version. | let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI
| #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 programming language of this snippet from F# to Java without modifying what it does. | let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI
| 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 this program in Java while keeping its functionality equivalent to the F# version. | let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI
| 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 F# to Python, ensuring the logic remains intact. | let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI
| 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])
max_length, n = max((len(hailstone(i)), i) for i in range(1, 100_000))
print(f"Maximum length {max_length} was found for hailstone({n}) "
f"for numbers <100,000")
|
Translate this program into Python but keep the logic exactly as in F#. | let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI
| 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])
max_length, n = max((len(hailstone(i)), i) for i in range(1, 100_000))
print(f"Maximum length {max_length} was found for hailstone({n}) "
f"for numbers <100,000")
|
Can you help me rewrite this code in VB instead of F#, keeping it the same logically? | let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI
| 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 the snippet below in VB so it works the same as the original F# code. | let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI
| 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 F# snippet. | let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI
| 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)
}
|
Generate a Go translation of this F# snippet without changing its computational steps. | let rec hailstone n = seq {
match n with
| 1 -> yield 1
| n when n % 2 = 0 -> yield n; yield! hailstone (n / 2)
| n -> yield n; yield! hailstone (n * 3 + 1)
}
let hailstone27 = hailstone 27 |> Array.ofSeq
assert (Array.length hailstone27 = 112)
assert (hailstone27.[..3] = [|27;82;41;124|])
assert (hailstone27.[108..] = [|8;4;2;1|])
let maxLen, maxI = Seq.max <| seq { for i in 1..99999 -> Seq.length (hailstone i), i}
printfn "Maximum length %d was found for hailstone(%d)" maxLen maxI
| 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 Factor function in C with identical behavior. |
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
: hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
: main ( -- )
27 hailstone dup dup
"The hailstone sequence from 27:" print
" has length " write length .
" starts with " write 4 head [ unparse ] map ", " join print
" ends with " write 4 tail* [ unparse ] map ", " join print
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main
| #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 Factor code into C while preserving the original functionality. |
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
: hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
: main ( -- )
27 hailstone dup dup
"The hailstone sequence from 27:" print
" has length " write length .
" starts with " write 4 head [ unparse ] map ", " join print
" ends with " write 4 tail* [ unparse ] map ", " join print
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main
| #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 Factor version. |
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
: hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
: main ( -- )
27 hailstone dup dup
"The hailstone sequence from 27:" print
" has length " write length .
" starts with " write 4 head [ unparse ] map ", " join print
" ends with " write 4 tail* [ unparse ] map ", " join print
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main
| 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);
}
}
}
|
Write the same algorithm in C# as shown in this Factor implementation. |
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
: hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
: main ( -- )
27 hailstone dup dup
"The hailstone sequence from 27:" print
" has length " write length .
" starts with " write 4 head [ unparse ] map ", " join print
" ends with " write 4 tail* [ unparse ] map ", " join print
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main
| 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);
}
}
}
|
Rewrite this program in C++ while keeping its functionality equivalent to the Factor version. |
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
: hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
: main ( -- )
27 hailstone dup dup
"The hailstone sequence from 27:" print
" has length " write length .
" starts with " write 4 head [ unparse ] map ", " join print
" ends with " write 4 tail* [ unparse ] map ", " join print
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main
| #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;
}
|
Translate this program into C++ but keep the logic exactly as in Factor. |
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
: hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
: main ( -- )
27 hailstone dup dup
"The hailstone sequence from 27:" print
" has length " write length .
" starts with " write 4 head [ unparse ] map ", " join print
" ends with " write 4 tail* [ unparse ] map ", " join print
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main
| #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 programming language of this snippet from Factor to Java without modifying what it does. |
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
: hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
: main ( -- )
27 hailstone dup dup
"The hailstone sequence from 27:" print
" has length " write length .
" starts with " write 4 head [ unparse ] map ", " join print
" ends with " write 4 tail* [ unparse ] map ", " join print
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main
| 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 Java translation of this Factor snippet without changing its computational steps. |
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
: hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
: main ( -- )
27 hailstone dup dup
"The hailstone sequence from 27:" print
" has length " write length .
" starts with " write 4 head [ unparse ] map ", " join print
" ends with " write 4 tail* [ unparse ] map ", " join print
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main
| 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;
}
}
|
Change the programming language of this snippet from Factor to Python without modifying what it does. |
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
: hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
: main ( -- )
27 hailstone dup dup
"The hailstone sequence from 27:" print
" has length " write length .
" starts with " write 4 head [ unparse ] map ", " join print
" ends with " write 4 tail* [ unparse ] map ", " join print
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main
| 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])
max_length, n = max((len(hailstone(i)), i) for i in range(1, 100_000))
print(f"Maximum length {max_length} was found for hailstone({n}) "
f"for numbers <100,000")
|
Transform the following Factor implementation into Python, maintaining the same output and logic. |
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
: hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
: main ( -- )
27 hailstone dup dup
"The hailstone sequence from 27:" print
" has length " write length .
" starts with " write 4 head [ unparse ] map ", " join print
" ends with " write 4 tail* [ unparse ] map ", " join print
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main
| 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])
max_length, n = max((len(hailstone(i)), i) for i in range(1, 100_000))
print(f"Maximum length {max_length} was found for hailstone({n}) "
f"for numbers <100,000")
|
Convert this Factor snippet to VB and keep its semantics consistent. |
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
: hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
: main ( -- )
27 hailstone dup dup
"The hailstone sequence from 27:" print
" has length " write length .
" starts with " write 4 head [ unparse ] map ", " join print
" ends with " write 4 tail* [ unparse ] map ", " join print
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main
| 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 the snippet below in VB so it works the same as the original Factor code. |
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
: hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
: main ( -- )
27 hailstone dup dup
"The hailstone sequence from 27:" print
" has length " write length .
" starts with " write 4 head [ unparse ] map ", " join print
" ends with " write 4 tail* [ unparse ] map ", " join print
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main
| 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 programming language of this snippet from Factor to Go without modifying what it does. |
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
: hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
: main ( -- )
27 hailstone dup dup
"The hailstone sequence from 27:" print
" has length " write length .
" starts with " write 4 head [ unparse ] map ", " join print
" ends with " write 4 tail* [ unparse ] map ", " join print
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main
| 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)
}
|
Ensure the translated Go code behaves exactly like the original Factor snippet. |
USING: arrays io kernel math math.ranges prettyprint sequences vectors ;
IN: rosetta.hailstone
: hailstone ( n -- seq )
[ 1vector ] keep
[ dup 1 number= ]
[
dup even? [ 2 / ] [ 3 * 1 + ] if
2dup swap push
] until
drop ;
<PRIVATE
: main ( -- )
27 hailstone dup dup
"The hailstone sequence from 27:" print
" has length " write length .
" starts with " write 4 head [ unparse ] map ", " join print
" ends with " write 4 tail* [ unparse ] map ", " join print
1 100000 [a,b)
[ [ hailstone length ] keep 2array ]
[ [ [ first ] bi@ > ] most ] map-reduce
first2
"The hailstone sequence from " write pprint
" has length " write pprint "." print ;
PRIVATE>
MAIN: main
| 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 the same algorithm in C as shown in this Forth implementation. | : hail-next
dup 1 and if 3 * 1+ else 2/ then ;
: .hail
begin dup . dup 1 > while hail-next repeat drop ;
: hail-len
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr
27 .hail cr
: longest-hail
0 0 rot 1+ 1 do
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail
| #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 Forth block to C, preserving its control flow and logic. | : hail-next
dup 1 and if 3 * 1+ else 2/ then ;
: .hail
begin dup . dup 1 > while hail-next repeat drop ;
: hail-len
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr
27 .hail cr
: longest-hail
0 0 rot 1+ 1 do
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail
| #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;
}
|
Preserve the algorithm and functionality while converting the code from Forth to C#. | : hail-next
dup 1 and if 3 * 1+ else 2/ then ;
: .hail
begin dup . dup 1 > while hail-next repeat drop ;
: hail-len
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr
27 .hail cr
: longest-hail
0 0 rot 1+ 1 do
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail
| 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 Forth to C#, same semantics. | : hail-next
dup 1 and if 3 * 1+ else 2/ then ;
: .hail
begin dup . dup 1 > while hail-next repeat drop ;
: hail-len
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr
27 .hail cr
: longest-hail
0 0 rot 1+ 1 do
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail
| 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 Forth to C++, ensuring the logic remains intact. | : hail-next
dup 1 and if 3 * 1+ else 2/ then ;
: .hail
begin dup . dup 1 > while hail-next repeat drop ;
: hail-len
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr
27 .hail cr
: longest-hail
0 0 rot 1+ 1 do
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail
| #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;
}
|
Keep all operations the same but rewrite the snippet in C++. | : hail-next
dup 1 and if 3 * 1+ else 2/ then ;
: .hail
begin dup . dup 1 > while hail-next repeat drop ;
: hail-len
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr
27 .hail cr
: longest-hail
0 0 rot 1+ 1 do
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail
| #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;
}
|
Generate a Java translation of this Forth snippet without changing its computational steps. | : hail-next
dup 1 and if 3 * 1+ else 2/ then ;
: .hail
begin dup . dup 1 > while hail-next repeat drop ;
: hail-len
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr
27 .hail cr
: longest-hail
0 0 rot 1+ 1 do
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail
| 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 Forth implementation into Java, maintaining the same output and logic. | : hail-next
dup 1 and if 3 * 1+ else 2/ then ;
: .hail
begin dup . dup 1 > while hail-next repeat drop ;
: hail-len
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr
27 .hail cr
: longest-hail
0 0 rot 1+ 1 do
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail
| 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 Forth snippet to Python and keep its semantics consistent. | : hail-next
dup 1 and if 3 * 1+ else 2/ then ;
: .hail
begin dup . dup 1 > while hail-next repeat drop ;
: hail-len
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr
27 .hail cr
: longest-hail
0 0 rot 1+ 1 do
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail
| 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])
max_length, n = max((len(hailstone(i)), i) for i in range(1, 100_000))
print(f"Maximum length {max_length} was found for hailstone({n}) "
f"for numbers <100,000")
|
Change the programming language of this snippet from Forth to Python without modifying what it does. | : hail-next
dup 1 and if 3 * 1+ else 2/ then ;
: .hail
begin dup . dup 1 > while hail-next repeat drop ;
: hail-len
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr
27 .hail cr
: longest-hail
0 0 rot 1+ 1 do
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail
| 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])
max_length, n = max((len(hailstone(i)), i) for i in range(1, 100_000))
print(f"Maximum length {max_length} was found for hailstone({n}) "
f"for numbers <100,000")
|
Convert this Forth block to VB, preserving its control flow and logic. | : hail-next
dup 1 and if 3 * 1+ else 2/ then ;
: .hail
begin dup . dup 1 > while hail-next repeat drop ;
: hail-len
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr
27 .hail cr
: longest-hail
0 0 rot 1+ 1 do
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail
| 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
|
Transform the following Forth implementation into VB, maintaining the same output and logic. | : hail-next
dup 1 and if 3 * 1+ else 2/ then ;
: .hail
begin dup . dup 1 > while hail-next repeat drop ;
: hail-len
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr
27 .hail cr
: longest-hail
0 0 rot 1+ 1 do
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail
| 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 programming language of this snippet from Forth to Go without modifying what it does. | : hail-next
dup 1 and if 3 * 1+ else 2/ then ;
: .hail
begin dup . dup 1 > while hail-next repeat drop ;
: hail-len
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr
27 .hail cr
: longest-hail
0 0 rot 1+ 1 do
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail
| 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)
}
|
Convert this Forth block to Go, preserving its control flow and logic. | : hail-next
dup 1 and if 3 * 1+ else 2/ then ;
: .hail
begin dup . dup 1 > while hail-next repeat drop ;
: hail-len
1 begin over 1 > while swap hail-next swap 1+ repeat nip ;
27 hail-len . cr
27 .hail cr
: longest-hail
0 0 rot 1+ 1 do
i hail-len 2dup < if
nip nip i swap
else drop then
loop
swap . ." has hailstone sequence length " . ;
100000 longest-hail
| 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 Fortran to C# with equivalent syntax and logic. | program Hailstone
implicit none
integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program
| 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#. | program Hailstone
implicit none
integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program
| 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 Fortran code in C++. | program Hailstone
implicit none
integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program
| #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;
}
|
Keep all operations the same but rewrite the snippet in C++. | program Hailstone
implicit none
integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program
| #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 a version of this Fortran function in C with identical behavior. | program Hailstone
implicit none
integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program
| #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;
}
|
Preserve the algorithm and functionality while converting the code from Fortran to C. | program Hailstone
implicit none
integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program
| #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 the snippet below in Java so it works the same as the original Fortran code. | program Hailstone
implicit none
integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program
| 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 Fortran function in Java with identical behavior. | program Hailstone
implicit none
integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program
| 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 this program in Python while keeping its functionality equivalent to the Fortran version. | program Hailstone
implicit none
integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program
| 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 Fortran code snippet into Python without altering its behavior. | program Hailstone
implicit none
integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program
| 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)))
|
Transform the following Fortran implementation into VB, maintaining the same output and logic. | program Hailstone
implicit none
integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program
| 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 programming language of this snippet from Fortran to VB without modifying what it does. | program Hailstone
implicit none
integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program
| 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
|
Preserve the algorithm and functionality while converting the code from Fortran to PHP. | program Hailstone
implicit none
integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program
| function hailstone($n,$seq=array()){
$sequence = $seq;
$sequence[] = $n;
if($n == 1){
return $sequence;
}else{
$n = ($n%2==0) ? $n/2 : (3*$n)+1;
return hailstone($n, $sequence);
}
}
$result = hailstone(27);
echo count($result) . ' Elements.<br>';
echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>';
$maxResult = array(0);
for($i=1;$i<=100000;$i++){
$result = count(hailstone($i));
if($result > max($maxResult)){
$maxResult = array($i=>$result);
}
}
foreach($maxResult as $key => $val){
echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val;
}
|
Can you help me rewrite this code in PHP instead of Fortran, keeping it the same logically? | program Hailstone
implicit none
integer :: i, maxn
integer :: maxseqlen = 0, seqlen
integer, allocatable :: seq(:)
call hs(27, seqlen)
allocate(seq(seqlen))
call hs(27, seqlen, seq)
write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements"
write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", &
seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen)
do i = 1, 99999
call hs(i, seqlen)
if (seqlen > maxseqlen) then
maxseqlen = seqlen
maxn = i
end if
end do
write(*,*)
write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements"
deallocate(seq)
contains
subroutine hs(number, length, seqArray)
integer, intent(in) :: number
integer, intent(out) :: length
integer, optional, intent(inout) :: seqArray(:)
integer :: n
n = number
length = 1
if(present(seqArray)) seqArray(1) = n
do while(n /= 1)
if(mod(n,2) == 0) then
n = n / 2
else
n = n * 3 + 1
end if
length = length + 1
if(present(seqArray)) seqArray(length) = n
end do
end subroutine
end program
| function hailstone($n,$seq=array()){
$sequence = $seq;
$sequence[] = $n;
if($n == 1){
return $sequence;
}else{
$n = ($n%2==0) ? $n/2 : (3*$n)+1;
return hailstone($n, $sequence);
}
}
$result = hailstone(27);
echo count($result) . ' Elements.<br>';
echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>';
$maxResult = array(0);
for($i=1;$i<=100000;$i++){
$result = count(hailstone($i));
if($result > max($maxResult)){
$maxResult = array($i=>$result);
}
}
foreach($maxResult as $key => $val){
echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val;
}
|
Write a version of this Groovy function in C with identical behavior. | def hailstone = { long start ->
def sequence = []
while (start != 1) {
sequence << start
start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l
}
sequence << start
}
| #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;
}
|
Ensure the translated C code behaves exactly like the original Groovy snippet. | def hailstone = { long start ->
def sequence = []
while (start != 1) {
sequence << start
start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l
}
sequence << start
}
| #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 Groovy code. | def hailstone = { long start ->
def sequence = []
while (start != 1) {
sequence << start
start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l
}
sequence << start
}
| 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 Groovy code. | def hailstone = { long start ->
def sequence = []
while (start != 1) {
sequence << start
start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l
}
sequence << start
}
| 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);
}
}
}
|
Write the same code in C++ as shown below in Groovy. | def hailstone = { long start ->
def sequence = []
while (start != 1) {
sequence << start
start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l
}
sequence << start
}
| #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;
}
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