Instruction stringlengths 45 106 | input_code stringlengths 1 13.7k | output_code stringlengths 1 13.7k |
|---|---|---|
Can you help me rewrite this code in Java instead of MATLAB, keeping it the same logically? | function x = hailstone(n)
x = n;
while n > 1
if n ~= floor(n / 2) * 2
n = n * 3 + 1;
else
n = n / 2;
end
x(end + 1) = n;
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;
}
}
|
Produce a functionally identical Python code for the snippet given in MATLAB. | function x = hailstone(n)
x = n;
while n > 1
if n ~= floor(n / 2) * 2
n = n * 3 + 1;
else
n = n / 2;
end
x(end + 1) = n;
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)))
|
Maintain the same structure and functionality when rewriting this code in Python. | function x = hailstone(n)
x = n;
while n > 1
if n ~= floor(n / 2) * 2
n = n * 3 + 1;
else
n = n / 2;
end
x(end + 1) = n;
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)))
|
Can you help me rewrite this code in VB instead of MATLAB, keeping it the same logically? | function x = hailstone(n)
x = n;
while n > 1
if n ~= floor(n / 2) * 2
n = n * 3 + 1;
else
n = n / 2;
end
x(end + 1) = n;
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
|
Port the following code from MATLAB to VB with equivalent syntax and logic. | function x = hailstone(n)
x = n;
while n > 1
if n ~= floor(n / 2) * 2
n = n * 3 + 1;
else
n = n / 2;
end
x(end + 1) = n;
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
|
Can you help me rewrite this code in Go instead of MATLAB, keeping it the same logically? | function x = hailstone(n)
x = n;
while n > 1
if n ~= floor(n / 2) * 2
n = n * 3 + 1;
else
n = n / 2;
end
x(end + 1) = n;
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)
}
|
Convert this MATLAB block to Go, preserving its control flow and logic. | function x = hailstone(n)
x = n;
while n > 1
if n ~= floor(n / 2) * 2
n = n * 3 + 1;
else
n = n / 2;
end
x(end + 1) = n;
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)
}
|
Rewrite the snippet below in C so it works the same as the original Nim code. | proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."
var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
| #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 Nim snippet. | proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."
var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
| #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 Nim to C# with equivalent syntax and logic. | proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."
var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
| 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 Nim code snippet into C# without altering its behavior. | proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."
var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
| 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 Nim implementation. | proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."
var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
| #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 Nim to C++, same semantics. | proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."
var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Write the same algorithm in Java as shown in this Nim implementation. | proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."
var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
| 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;
}
}
|
Translate this program into Java but keep the logic exactly as in Nim. | proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."
var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
| 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 following Nim code into Python without altering its purpose. | proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."
var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
| 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 Nim to Python without modifying what it does. | proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."
var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
| 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")
|
Produce a language-to-language conversion: from Nim to VB, same semantics. | proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."
var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Produce a language-to-language conversion: from Nim to VB, same semantics. | proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."
var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
| 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
|
Can you help me rewrite this code in Go instead of Nim, keeping it the same logically? | proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."
var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Produce a language-to-language conversion: from Nim to Go, same semantics. | proc hailstone(n: int): seq[int] =
result = @[n]
var n = n
while n > 1:
if (n and 1) == 1:
n = 3 * n + 1
else:
n = n div 2
result.add n
when isMainModule:
import strformat, strutils
let h = hailstone(27)
echo &"Hailstone sequence for number 27 has {h.len} elements."
let first = h[0..3].join(", ")
let last = h[^4..^1].join(", ")
echo &"This sequence begins with {first} and ends with {last}."
var m, mi = 0
for i in 1..<100_000:
let n = hailstone(i).len
if n > m:
m = n
mi = i
echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
| 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)
}
|
Please provide an equivalent version of this OCaml code in C. | fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| (x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Change the programming language of this snippet from OCaml to C without modifying what it does. | fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| (x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Keep all operations the same but rewrite the snippet in C#. | fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| (x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Port the provided OCaml code into C# while preserving the original functionality. | fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| (x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Convert the following code from OCaml to C++, ensuring the logic remains intact. | fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| (x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Produce a language-to-language conversion: from OCaml to C++, same semantics. | fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| (x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Change the following OCaml code into Java without altering its purpose. | fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| (x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Maintain the same structure and functionality when rewriting this code in Java. | fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| (x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Change the programming language of this snippet from OCaml to Python without modifying what it does. | fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| (x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);
| def hailstone(n):
seq = [n]
while n > 1:
n = 3 * n + 1 if n & 1 else n // 2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert (len(h) == 112
and h[:4] == [27, 82, 41, 124]
and h[-4:] == [8, 4, 2, 1])
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")
|
Produce a language-to-language conversion: from OCaml to Python, same semantics. | fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| (x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);
| def hailstone(n):
seq = [n]
while n > 1:
n = 3 * n + 1 if n & 1 else n // 2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert (len(h) == 112
and h[:4] == [27, 82, 41, 124]
and h[-4:] == [8, 4, 2, 1])
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")
|
Produce a language-to-language conversion: from OCaml to VB, same semantics. | fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| (x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Change the programming language of this snippet from OCaml to VB without modifying what it does. | fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| (x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Translate the given OCaml code snippet into Go without altering its behavior. | fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| (x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Convert the following code from OCaml to Go, ensuring the logic remains intact. | fun hail (x = 1) = [1]
| (x rem 2 = 0) = x :: hail (x div 2)
| x = x :: hail (x * 3 + 1)
fun hailstorm
([], i, largest, largest_at) = (largest_at, largest)
| (x :: xs, i, largest, largest_at) =
let
val k = len (hail x)
in
if k > largest then
hailstorm (xs, i + 1, k, i)
else
hailstorm (xs, i + 1, largest, largest_at)
end
| (x :: xs) = hailstorm (x :: xs, 1, 0, 0)
;
val h27 = hail 27;
print "hailstone sequence for the number 27 has ";
print ` len (h27);
print " elements starting with ";
print ` sub (h27, 0, 4);
print " and ending with ";
print ` sub (h27, len(h27)-4, len h27);
println ".";
val biggest = hailstorm ` iota (100000 - 1);
print "The number less than 100,000 which has the longest ";
print "hailstone sequence is at element ";
print ` ref (biggest, 0);
print " and is of length ";
println ` ref (biggest, 1);
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Generate an equivalent C version of this Pascal code. | program ShowHailstoneSequence;
uses
SysUtils;
const
maxN = 10*1000*1000;
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;
IF n > 1 then
repeat
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
end.
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Can you help me rewrite this code in C instead of Pascal, keeping it the same logically? | program ShowHailstoneSequence;
uses
SysUtils;
const
maxN = 10*1000*1000;
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;
IF n > 1 then
repeat
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
end.
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Translate this program into C# but keep the logic exactly as in Pascal. | program ShowHailstoneSequence;
uses
SysUtils;
const
maxN = 10*1000*1000;
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;
IF n > 1 then
repeat
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
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 Pascal function in C# with identical behavior. | program ShowHailstoneSequence;
uses
SysUtils;
const
maxN = 10*1000*1000;
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;
IF n > 1 then
repeat
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
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 the same code in C++ as shown below in Pascal. | program ShowHailstoneSequence;
uses
SysUtils;
const
maxN = 10*1000*1000;
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;
IF n > 1 then
repeat
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
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;
}
|
Port the following code from Pascal to C++ with equivalent syntax and logic. | program ShowHailstoneSequence;
uses
SysUtils;
const
maxN = 10*1000*1000;
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;
IF n > 1 then
repeat
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
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;
}
|
Keep all operations the same but rewrite the snippet in Java. | program ShowHailstoneSequence;
uses
SysUtils;
const
maxN = 10*1000*1000;
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;
IF n > 1 then
repeat
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
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 Pascal snippet. | program ShowHailstoneSequence;
uses
SysUtils;
const
maxN = 10*1000*1000;
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;
IF n > 1 then
repeat
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
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;
}
}
|
Convert the following code from Pascal to Python, ensuring the logic remains intact. | program ShowHailstoneSequence;
uses
SysUtils;
const
maxN = 10*1000*1000;
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;
IF n > 1 then
repeat
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
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])
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 Pascal block to Python, preserving its control flow and logic. | program ShowHailstoneSequence;
uses
SysUtils;
const
maxN = 10*1000*1000;
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;
IF n > 1 then
repeat
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
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])
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 VB but keep the logic exactly as in Pascal. | program ShowHailstoneSequence;
uses
SysUtils;
const
maxN = 10*1000*1000;
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;
IF n > 1 then
repeat
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
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
|
Change the following Pascal code into VB without altering its purpose. | program ShowHailstoneSequence;
uses
SysUtils;
const
maxN = 10*1000*1000;
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;
IF n > 1 then
repeat
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
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
|
Translate this program into Go but keep the logic exactly as in Pascal. | program ShowHailstoneSequence;
uses
SysUtils;
const
maxN = 10*1000*1000;
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;
IF n > 1 then
repeat
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
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)
}
|
Translate the given Pascal code snippet into Go without altering its behavior. | program ShowHailstoneSequence;
uses
SysUtils;
const
maxN = 10*1000*1000;
type
tiaArr = array[0..1000] of Uint64;
tIntArr = record
iaMaxPos : integer;
iaArr : tiaArr
end;
tpiaArr = ^tiaArr;
function HailstoneSeqCnt(n: UInt64): NativeInt;
begin
result := 0;
while not(ODD(n)) do
Begin
inc(result);
n := n shr 1;
end;
IF n > 1 then
repeat
repeat
n := (3*n+1) SHR 1;inc(result,2);
until NOT(Odd(n));
repeat
n := n shr 1; inc(result);
until odd(n);
until n = 1;
end;
procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr);
var
maxPos: NativeInt;
n: UInt64;
pArr : tpiaArr;
begin
with aHailstoneList do
begin
maxPos := 0;
pArr := @iaArr;
end;
n := aStartingNumber;
pArr^[maxPos] := n;
while n <> 1 do
begin
if odd(n) then
n := (3*n+1)
else
n := n shr 1;
inc(maxPos);
pArr^[maxPos] := n;
end;
aHailstoneList.iaMaxPos := maxPos;
end;
var
i,Limit: NativeInt;
lList: tIntArr;
lAverageLength:Uint64;
lMaxSequence: NativeInt;
lMaxLength,lgth: NativeInt;
begin
lList.iaMaxPos := 0;
GetHailstoneSequence(27, lList);
with lList do
begin
Limit := iaMaxPos;
writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1]));
write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..');
For i := iaMaxPos-3 to iaMaxPos-1 do
write(iaArr[i],',');
writeln(iaArr[iaMaxPos]);
end;
Writeln;
lMaxSequence := 0;
lMaxLength := 0;
i := 1;
limit := 10*i;
writeln(' Limit : number with max length | average length');
repeat
lAverageLength:= 0;
repeat
lgth:= HailstoneSeqCnt(i);
inc(lAverageLength, lgth);
if lgth >= lMaxLength then
begin
lMaxSequence := i;
lMaxLength := lgth+1;
end;
inc(i);
until i = Limit;
Writeln(Format(' %10d : %9d | %4d | %7.3f',
[limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit]));
limit := limit*10;
until Limit > maxN;
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 C translation of this Perl snippet without changing its computational steps. |
use warnings;
use strict;
my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n;
}
return @sequence;
}
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Change the following Perl code into C without altering its purpose. |
use warnings;
use strict;
my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n;
}
return @sequence;
}
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Port the following code from Perl to C# with equivalent syntax and logic. |
use warnings;
use strict;
my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n;
}
return @sequence;
}
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Please provide an equivalent version of this Perl code in C#. |
use warnings;
use strict;
my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n;
}
return @sequence;
}
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Can you help me rewrite this code in C++ instead of Perl, keeping it the same logically? |
use warnings;
use strict;
my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n;
}
return @sequence;
}
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Preserve the algorithm and functionality while converting the code from Perl to C++. |
use warnings;
use strict;
my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n;
}
return @sequence;
}
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Write a version of this Perl function in Java with identical behavior. |
use warnings;
use strict;
my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n;
}
return @sequence;
}
| 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 Perl implementation into Java, maintaining the same output and logic. |
use warnings;
use strict;
my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n;
}
return @sequence;
}
| 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 Perl implementation into Python, maintaining the same output and logic. |
use warnings;
use strict;
my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n;
}
return @sequence;
}
| def hailstone(n):
seq = [n]
while n > 1:
n = 3 * n + 1 if n & 1 else n // 2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert (len(h) == 112
and h[:4] == [27, 82, 41, 124]
and h[-4:] == [8, 4, 2, 1])
max_length, n = max((len(hailstone(i)), i) for i in range(1, 100_000))
print(f"Maximum length {max_length} was found for hailstone({n}) "
f"for numbers <100,000")
|
Can you help me rewrite this code in Python instead of Perl, keeping it the same logically? |
use warnings;
use strict;
my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n;
}
return @sequence;
}
| def hailstone(n):
seq = [n]
while n > 1:
n = 3 * n + 1 if n & 1 else n // 2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert (len(h) == 112
and h[:4] == [27, 82, 41, 124]
and h[-4:] == [8, 4, 2, 1])
max_length, n = max((len(hailstone(i)), i) for i in range(1, 100_000))
print(f"Maximum length {max_length} was found for hailstone({n}) "
f"for numbers <100,000")
|
Rewrite the snippet below in VB so it works the same as the original Perl code. |
use warnings;
use strict;
my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n;
}
return @sequence;
}
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Change the following Perl code into VB without altering its purpose. |
use warnings;
use strict;
my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n;
}
return @sequence;
}
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Maintain the same structure and functionality when rewriting this code in Go. |
use warnings;
use strict;
my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n;
}
return @sequence;
}
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Convert the following code from Perl to Go, ensuring the logic remains intact. |
use warnings;
use strict;
my @h = hailstone(27);
print "Length of hailstone(27) = " . scalar @h . "\n";
print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n";
my ($max, $n) = (0, 0);
for my $x (1 .. 99_999) {
@h = hailstone($x);
if (scalar @h > $max) {
($max, $n) = (scalar @h, $x);
}
}
print "Max length $max was found for hailstone($n) for numbers < 100_000\n";
sub hailstone {
my ($n) = @_;
my @sequence = ($n);
while ($n > 1) {
if ($n % 2 == 0) {
$n = int($n / 2);
} else {
$n = $n * 3 + 1;
}
push @sequence, $n;
}
return @sequence;
}
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Produce a functionally identical C code for the snippet given in PowerShell. | function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Maintain the same structure and functionality when rewriting this code in C. | function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Preserve the algorithm and functionality while converting the code from PowerShell to C#. | function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Transform the following PowerShell implementation into C#, maintaining the same output and logic. | function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}
| using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Hailstone
{
class Program
{
public static List<int> hs(int n,List<int> seq)
{
List<int> sequence = seq;
sequence.Add(n);
if (n == 1)
{
return sequence;
}else{
int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1;
return hs(newn, sequence);
}
}
static void Main(string[] args)
{
int n = 27;
List<int> sequence = hs(n,new List<int>());
Console.WriteLine(sequence.Count + " Elements");
List<int> start = sequence.GetRange(0, 4);
List<int> end = sequence.GetRange(sequence.Count - 4, 4);
Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end));
int number = 0, longest = 0;
for (int i = 1; i < 100000; i++)
{
int count = (hs(i, new List<int>())).Count;
if (count > longest)
{
longest = count;
number = i;
}
}
Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest);
}
}
}
|
Convert the following code from PowerShell to C++, ensuring the logic remains intact. | function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Convert the following code from PowerShell to C++, ensuring the logic remains intact. | function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Translate this program into Java but keep the logic exactly as in PowerShell. | function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}
| 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;
}
}
|
Maintain the same structure and functionality when rewriting this code in Java. | function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Produce a language-to-language conversion: from PowerShell to Python, same semantics. | function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}
| def hailstone(n):
seq = [n]
while n > 1:
n = 3 * n + 1 if n & 1 else n // 2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert (len(h) == 112
and h[:4] == [27, 82, 41, 124]
and h[-4:] == [8, 4, 2, 1])
max_length, n = max((len(hailstone(i)), i) for i in range(1, 100_000))
print(f"Maximum length {max_length} was found for hailstone({n}) "
f"for numbers <100,000")
|
Rewrite this program in Python while keeping its functionality equivalent to the PowerShell version. | function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}
| def hailstone(n):
seq = [n]
while n > 1:
n = 3 * n + 1 if n & 1 else n // 2
seq.append(n)
return seq
if __name__ == '__main__':
h = hailstone(27)
assert (len(h) == 112
and h[:4] == [27, 82, 41, 124]
and h[-4:] == [8, 4, 2, 1])
max_length, n = max((len(hailstone(i)), i) for i in range(1, 100_000))
print(f"Maximum length {max_length} was found for hailstone({n}) "
f"for numbers <100,000")
|
Produce a language-to-language conversion: from PowerShell to VB, same semantics. | function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Convert this PowerShell snippet to VB and keep its semantics consistent. | function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}
| Private Function hailstone(ByVal n As Long) As Collection
Dim s As New Collection
s.Add CStr(n), CStr(n)
i = 0
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
s.Add CStr(n), CStr(n)
Loop
Set hailstone = s
End Function
Private Function hailstone_count(ByVal n As Long)
Dim count As Long: count = 1
Do While n <> 1
If n Mod 2 = 0 Then
n = n / 2
Else
n = 3 * n + 1
End If
count = count + 1
Loop
hailstone_count = count
End Function
Public Sub rosetta()
Dim s As Collection, i As Long
Set s = hailstone(27)
Dim ls As Integer: ls = s.count
Debug.Print "hailstone(27) = ";
For i = 1 To 4
Debug.Print s(i); ", ";
Next i
Debug.Print "... ";
For i = s.count - 4 To s.count - 1
Debug.Print s(i); ", ";
Next i
Debug.Print s(s.count)
Debug.Print "length ="; ls
Dim hmax As Long: hmax = 1
Dim imax As Long: imax = 1
Dim count As Integer
For i = 2 To 100000# - 1
count = hailstone_count(i)
If count > hmax Then
hmax = count
imax = i
End If
Next i
Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements."
End Sub
|
Port the provided PowerShell code into Go while preserving the original functionality. | function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Preserve the algorithm and functionality while converting the code from PowerShell to Go. | function Get-HailStone {
param($n)
switch($n) {
1 {$n;return}
{$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)}
{$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)}
}
}
function Get-HailStoneBelowLimit {
param($UpperLimit)
for ($i = 1; $i -lt $UpperLimit; $i++) {
[pscustomobject]@{
'Number' = $i
'Count' = (Get-HailStone $i).count
}
}
}
| package main
import "fmt"
func hs(n int, recycle []int) []int {
s := append(recycle[:0], n)
for n > 1 {
if n&1 == 0 {
n = n / 2
} else {
n = 3*n + 1
}
s = append(s, n)
}
return s
}
func main() {
seq := hs(27, nil)
fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n",
len(seq), seq[0], seq[1], seq[2], seq[3],
seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1])
var maxN, maxLen int
for n := 1; n < 100000; n++ {
seq = hs(n, seq)
if len(seq) > maxLen {
maxN = n
maxLen = len(seq)
}
}
fmt.Printf("hs(%d): %d elements\n", maxN, maxLen)
}
|
Produce a language-to-language conversion: from Racket to C, same semantics. | #lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
| #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. | #lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
| #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 language-to-language conversion: from Racket to C#, same semantics. | #lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
| 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 Racket code. | #lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
| 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 Racket implementation. | #lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
| #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;
}
|
Convert the following code from Racket to C++, ensuring the logic remains intact. | #lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Please provide an equivalent version of this Racket code in Java. | #lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
| 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 Racket to Java without modifying what it does. | #lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
| 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 code in Python as shown below in Racket. | #lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
| 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")
|
Rewrite the snippet below in Python so it works the same as the original Racket code. | #lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
| 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")
|
Generate a VB translation of this Racket snippet without changing its computational steps. | #lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
| 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
|
Maintain the same structure and functionality when rewriting this code in VB. | #lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
| 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 Racket code. | #lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
| 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)
}
|
Rewrite the snippet below in Go so it works the same as the original Racket code. | #lang racket
(define hailstone
(let ([t (make-hasheq)])
(hash-set! t 1 '(1))
(λ(n) (hash-ref! t n
(λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1)))))))))
(define h27 (hailstone 27))
(printf "h(27) = ~s, ~s items\n"
`(,@(take h27 4) ... ,@(take-right h27 4))
(length h27))
(define N 100000)
(define longest
(for/fold ([m #f]) ([i (in-range 1 (add1 N))])
(define h (hailstone i))
(if (and m (> (cdr m) (length h))) m (cons i (length h)))))
(printf "for x<=~s, ~s has the longest sequence with ~s items\n"
N (car longest) (cdr longest))
| 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 COBOL to C with equivalent syntax and logic. | identification division.
program-id. hailstones.
remarks. cobc -x hailstones.cob.
data division.
working-storage section.
01 most constant as 1000000.
01 coverage constant as 100000.
01 stones usage binary-long.
01 n usage binary-long.
01 storm usage binary-long.
01 show-arg pic 9(6).
01 show-default pic 99 value 27.
01 show-sequence usage binary-long.
01 longest usage binary-long occurs 2 times.
01 filler.
05 hail usage binary-long
occurs 0 to most depending on stones.
01 show pic z(10).
01 low-range usage binary-long.
01 high-range usage binary-long.
01 range usage binary-long.
01 remain usage binary-long.
01 unused usage binary-long.
procedure division.
accept show-arg from command-line
if show-arg less than 1 or greater than coverage then
move show-default to show-arg
end-if
move show-arg to show-sequence
move 1 to longest(1)
perform hailstone varying storm
from 1 by 1 until storm > coverage
display "Longest at: " longest(2) " with " longest(1) " elements"
goback.
hailstone.
move 0 to stones
move storm to n
perform until n equal 1
if stones > most then
display "too many hailstones" upon syserr
stop run
end-if
add 1 to stones
move n to hail(stones)
divide n by 2 giving unused remainder remain
if remain equal 0 then
divide 2 into n
else
compute n = 3 * n + 1
end-if
end-perform
add 1 to stones
move n to hail(stones)
if stones > longest(1) then
move stones to longest(1)
move storm to longest(2)
end-if
if storm equal show-sequence then
display show-sequence ": " with no advancing
perform varying range from 1 by 1 until range > stones
move 5 to low-range
compute high-range = stones - 4
if range < low-range or range > high-range then
move hail(range) to show
display function trim(show) with no advancing
if range < stones then
display ", " with no advancing
end-if
end-if
if range = low-range and stones > 8 then
display "..., " with no advancing
end-if
end-perform
display ": " stones " elements"
end-if
.
end program hailstones.
| #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 a C translation of this COBOL snippet without changing its computational steps. | identification division.
program-id. hailstones.
remarks. cobc -x hailstones.cob.
data division.
working-storage section.
01 most constant as 1000000.
01 coverage constant as 100000.
01 stones usage binary-long.
01 n usage binary-long.
01 storm usage binary-long.
01 show-arg pic 9(6).
01 show-default pic 99 value 27.
01 show-sequence usage binary-long.
01 longest usage binary-long occurs 2 times.
01 filler.
05 hail usage binary-long
occurs 0 to most depending on stones.
01 show pic z(10).
01 low-range usage binary-long.
01 high-range usage binary-long.
01 range usage binary-long.
01 remain usage binary-long.
01 unused usage binary-long.
procedure division.
accept show-arg from command-line
if show-arg less than 1 or greater than coverage then
move show-default to show-arg
end-if
move show-arg to show-sequence
move 1 to longest(1)
perform hailstone varying storm
from 1 by 1 until storm > coverage
display "Longest at: " longest(2) " with " longest(1) " elements"
goback.
hailstone.
move 0 to stones
move storm to n
perform until n equal 1
if stones > most then
display "too many hailstones" upon syserr
stop run
end-if
add 1 to stones
move n to hail(stones)
divide n by 2 giving unused remainder remain
if remain equal 0 then
divide 2 into n
else
compute n = 3 * n + 1
end-if
end-perform
add 1 to stones
move n to hail(stones)
if stones > longest(1) then
move stones to longest(1)
move storm to longest(2)
end-if
if storm equal show-sequence then
display show-sequence ": " with no advancing
perform varying range from 1 by 1 until range > stones
move 5 to low-range
compute high-range = stones - 4
if range < low-range or range > high-range then
move hail(range) to show
display function trim(show) with no advancing
if range < stones then
display ", " with no advancing
end-if
end-if
if range = low-range and stones > 8 then
display "..., " with no advancing
end-if
end-perform
display ": " stones " elements"
end-if
.
end program hailstones.
| #include <stdio.h>
#include <stdlib.h>
int hailstone(int n, int *arry)
{
int hs = 1;
while (n!=1) {
hs++;
if (arry) *arry++ = n;
n = (n&1) ? (3*n+1) : (n/2);
}
if (arry) *arry++ = n;
return hs;
}
int main()
{
int j, hmax = 0;
int jatmax, n;
int *arry;
for (j=1; j<100000; j++) {
n = hailstone(j, NULL);
if (hmax < n) {
hmax = n;
jatmax = j;
}
}
n = hailstone(27, NULL);
arry = malloc(n*sizeof(int));
n = hailstone(27, arry);
printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n",
arry[0],arry[1],arry[2],arry[3],
arry[n-4], arry[n-3], arry[n-2], arry[n-1], n);
printf("Max %d at j= %d\n", hmax, jatmax);
free(arry);
return 0;
}
|
Convert the following code from COBOL to C#, ensuring the logic remains intact. | identification division.
program-id. hailstones.
remarks. cobc -x hailstones.cob.
data division.
working-storage section.
01 most constant as 1000000.
01 coverage constant as 100000.
01 stones usage binary-long.
01 n usage binary-long.
01 storm usage binary-long.
01 show-arg pic 9(6).
01 show-default pic 99 value 27.
01 show-sequence usage binary-long.
01 longest usage binary-long occurs 2 times.
01 filler.
05 hail usage binary-long
occurs 0 to most depending on stones.
01 show pic z(10).
01 low-range usage binary-long.
01 high-range usage binary-long.
01 range usage binary-long.
01 remain usage binary-long.
01 unused usage binary-long.
procedure division.
accept show-arg from command-line
if show-arg less than 1 or greater than coverage then
move show-default to show-arg
end-if
move show-arg to show-sequence
move 1 to longest(1)
perform hailstone varying storm
from 1 by 1 until storm > coverage
display "Longest at: " longest(2) " with " longest(1) " elements"
goback.
hailstone.
move 0 to stones
move storm to n
perform until n equal 1
if stones > most then
display "too many hailstones" upon syserr
stop run
end-if
add 1 to stones
move n to hail(stones)
divide n by 2 giving unused remainder remain
if remain equal 0 then
divide 2 into n
else
compute n = 3 * n + 1
end-if
end-perform
add 1 to stones
move n to hail(stones)
if stones > longest(1) then
move stones to longest(1)
move storm to longest(2)
end-if
if storm equal show-sequence then
display show-sequence ": " with no advancing
perform varying range from 1 by 1 until range > stones
move 5 to low-range
compute high-range = stones - 4
if range < low-range or range > high-range then
move hail(range) to show
display function trim(show) with no advancing
if range < stones then
display ", " with no advancing
end-if
end-if
if range = low-range and stones > 8 then
display "..., " with no advancing
end-if
end-perform
display ": " stones " elements"
end-if
.
end program hailstones.
| 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 COBOL code in C#. | identification division.
program-id. hailstones.
remarks. cobc -x hailstones.cob.
data division.
working-storage section.
01 most constant as 1000000.
01 coverage constant as 100000.
01 stones usage binary-long.
01 n usage binary-long.
01 storm usage binary-long.
01 show-arg pic 9(6).
01 show-default pic 99 value 27.
01 show-sequence usage binary-long.
01 longest usage binary-long occurs 2 times.
01 filler.
05 hail usage binary-long
occurs 0 to most depending on stones.
01 show pic z(10).
01 low-range usage binary-long.
01 high-range usage binary-long.
01 range usage binary-long.
01 remain usage binary-long.
01 unused usage binary-long.
procedure division.
accept show-arg from command-line
if show-arg less than 1 or greater than coverage then
move show-default to show-arg
end-if
move show-arg to show-sequence
move 1 to longest(1)
perform hailstone varying storm
from 1 by 1 until storm > coverage
display "Longest at: " longest(2) " with " longest(1) " elements"
goback.
hailstone.
move 0 to stones
move storm to n
perform until n equal 1
if stones > most then
display "too many hailstones" upon syserr
stop run
end-if
add 1 to stones
move n to hail(stones)
divide n by 2 giving unused remainder remain
if remain equal 0 then
divide 2 into n
else
compute n = 3 * n + 1
end-if
end-perform
add 1 to stones
move n to hail(stones)
if stones > longest(1) then
move stones to longest(1)
move storm to longest(2)
end-if
if storm equal show-sequence then
display show-sequence ": " with no advancing
perform varying range from 1 by 1 until range > stones
move 5 to low-range
compute high-range = stones - 4
if range < low-range or range > high-range then
move hail(range) to show
display function trim(show) with no advancing
if range < stones then
display ", " with no advancing
end-if
end-if
if range = low-range and stones > 8 then
display "..., " with no advancing
end-if
end-perform
display ": " stones " elements"
end-if
.
end program hailstones.
| 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 COBOL code. | identification division.
program-id. hailstones.
remarks. cobc -x hailstones.cob.
data division.
working-storage section.
01 most constant as 1000000.
01 coverage constant as 100000.
01 stones usage binary-long.
01 n usage binary-long.
01 storm usage binary-long.
01 show-arg pic 9(6).
01 show-default pic 99 value 27.
01 show-sequence usage binary-long.
01 longest usage binary-long occurs 2 times.
01 filler.
05 hail usage binary-long
occurs 0 to most depending on stones.
01 show pic z(10).
01 low-range usage binary-long.
01 high-range usage binary-long.
01 range usage binary-long.
01 remain usage binary-long.
01 unused usage binary-long.
procedure division.
accept show-arg from command-line
if show-arg less than 1 or greater than coverage then
move show-default to show-arg
end-if
move show-arg to show-sequence
move 1 to longest(1)
perform hailstone varying storm
from 1 by 1 until storm > coverage
display "Longest at: " longest(2) " with " longest(1) " elements"
goback.
hailstone.
move 0 to stones
move storm to n
perform until n equal 1
if stones > most then
display "too many hailstones" upon syserr
stop run
end-if
add 1 to stones
move n to hail(stones)
divide n by 2 giving unused remainder remain
if remain equal 0 then
divide 2 into n
else
compute n = 3 * n + 1
end-if
end-perform
add 1 to stones
move n to hail(stones)
if stones > longest(1) then
move stones to longest(1)
move storm to longest(2)
end-if
if storm equal show-sequence then
display show-sequence ": " with no advancing
perform varying range from 1 by 1 until range > stones
move 5 to low-range
compute high-range = stones - 4
if range < low-range or range > high-range then
move hail(range) to show
display function trim(show) with no advancing
if range < stones then
display ", " with no advancing
end-if
end-if
if range = low-range and stones > 8 then
display "..., " with no advancing
end-if
end-perform
display ": " stones " elements"
end-if
.
end program hailstones.
| #include <iostream>
#include <vector>
#include <utility>
std::vector<int> hailstone(int i)
{
std::vector<int> v;
while(true){
v.push_back(i);
if (1 == i) break;
i = (i % 2) ? (3 * i + 1) : (i / 2);
}
return v;
}
std::pair<int,int> find_longest_hailstone_seq(int n)
{
std::pair<int, int> maxseq(0, 0);
int l;
for(int i = 1; i < n; ++i){
l = hailstone(i).size();
if (l > maxseq.second) maxseq = std::make_pair(i, l);
}
return maxseq;
}
int main () {
std::vector<int> h27;
h27 = hailstone(27);
int l = h27.size();
std::cout << "length of hailstone(27) is " << l;
std::cout << " first four elements of hailstone(27) are ";
std::cout << h27[0] << " " << h27[1] << " "
<< h27[2] << " " << h27[3] << std::endl;
std::cout << " last four elements of hailstone(27) are "
<< h27[l-4] << " " << h27[l-3] << " "
<< h27[l-2] << " " << h27[l-1] << std::endl;
std::pair<int,int> m = find_longest_hailstone_seq(100000);
std::cout << "the longest hailstone sequence under 100,000 is " << m.first
<< " with " << m.second << " elements." <<std::endl;
return 0;
}
|
Port the following code from COBOL to C++ with equivalent syntax and logic. | identification division.
program-id. hailstones.
remarks. cobc -x hailstones.cob.
data division.
working-storage section.
01 most constant as 1000000.
01 coverage constant as 100000.
01 stones usage binary-long.
01 n usage binary-long.
01 storm usage binary-long.
01 show-arg pic 9(6).
01 show-default pic 99 value 27.
01 show-sequence usage binary-long.
01 longest usage binary-long occurs 2 times.
01 filler.
05 hail usage binary-long
occurs 0 to most depending on stones.
01 show pic z(10).
01 low-range usage binary-long.
01 high-range usage binary-long.
01 range usage binary-long.
01 remain usage binary-long.
01 unused usage binary-long.
procedure division.
accept show-arg from command-line
if show-arg less than 1 or greater than coverage then
move show-default to show-arg
end-if
move show-arg to show-sequence
move 1 to longest(1)
perform hailstone varying storm
from 1 by 1 until storm > coverage
display "Longest at: " longest(2) " with " longest(1) " elements"
goback.
hailstone.
move 0 to stones
move storm to n
perform until n equal 1
if stones > most then
display "too many hailstones" upon syserr
stop run
end-if
add 1 to stones
move n to hail(stones)
divide n by 2 giving unused remainder remain
if remain equal 0 then
divide 2 into n
else
compute n = 3 * n + 1
end-if
end-perform
add 1 to stones
move n to hail(stones)
if stones > longest(1) then
move stones to longest(1)
move storm to longest(2)
end-if
if storm equal show-sequence then
display show-sequence ": " with no advancing
perform varying range from 1 by 1 until range > stones
move 5 to low-range
compute high-range = stones - 4
if range < low-range or range > high-range then
move hail(range) to show
display function trim(show) with no advancing
if range < stones then
display ", " with no advancing
end-if
end-if
if range = low-range and stones > 8 then
display "..., " with no advancing
end-if
end-perform
display ": " stones " elements"
end-if
.
end program hailstones.
| #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 Java. | identification division.
program-id. hailstones.
remarks. cobc -x hailstones.cob.
data division.
working-storage section.
01 most constant as 1000000.
01 coverage constant as 100000.
01 stones usage binary-long.
01 n usage binary-long.
01 storm usage binary-long.
01 show-arg pic 9(6).
01 show-default pic 99 value 27.
01 show-sequence usage binary-long.
01 longest usage binary-long occurs 2 times.
01 filler.
05 hail usage binary-long
occurs 0 to most depending on stones.
01 show pic z(10).
01 low-range usage binary-long.
01 high-range usage binary-long.
01 range usage binary-long.
01 remain usage binary-long.
01 unused usage binary-long.
procedure division.
accept show-arg from command-line
if show-arg less than 1 or greater than coverage then
move show-default to show-arg
end-if
move show-arg to show-sequence
move 1 to longest(1)
perform hailstone varying storm
from 1 by 1 until storm > coverage
display "Longest at: " longest(2) " with " longest(1) " elements"
goback.
hailstone.
move 0 to stones
move storm to n
perform until n equal 1
if stones > most then
display "too many hailstones" upon syserr
stop run
end-if
add 1 to stones
move n to hail(stones)
divide n by 2 giving unused remainder remain
if remain equal 0 then
divide 2 into n
else
compute n = 3 * n + 1
end-if
end-perform
add 1 to stones
move n to hail(stones)
if stones > longest(1) then
move stones to longest(1)
move storm to longest(2)
end-if
if storm equal show-sequence then
display show-sequence ": " with no advancing
perform varying range from 1 by 1 until range > stones
move 5 to low-range
compute high-range = stones - 4
if range < low-range or range > high-range then
move hail(range) to show
display function trim(show) with no advancing
if range < stones then
display ", " with no advancing
end-if
end-if
if range = low-range and stones > 8 then
display "..., " with no advancing
end-if
end-perform
display ": " stones " elements"
end-if
.
end program hailstones.
| import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
class Hailstone {
public static List<Long> getHailstoneSequence(long n) {
if (n <= 0)
throw new IllegalArgumentException("Invalid starting sequence number");
List<Long> list = new ArrayList<Long>();
list.add(Long.valueOf(n));
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
list.add(Long.valueOf(n));
}
return list;
}
public static void main(String[] args) {
List<Long> sequence27 = getHailstoneSequence(27);
System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27);
long MAX = 100000;
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = getHailstoneSequence(i).size();
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
int highestCount = 1;
for (long i = 2; i < MAX; i++) {
int count = 1;
long n = i;
while (n != 1) {
if ((n & 1) == 0)
n = n / 2;
else
n = 3 * n + 1;
count++;
}
if (count > highestCount) {
highestCount = count;
highestNumber = i;
}
}
System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
{
long highestNumber = 1;
long highestCount = 1;
Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>();
sequenceMap.put(Long.valueOf(1), Integer.valueOf(1));
List<Long> currentList = new ArrayList<Long>();
for (long i = 2; i < MAX; i++) {
currentList.clear();
Long n = Long.valueOf(i);
Integer count = null;
while ((count = sequenceMap.get(n)) == null) {
currentList.add(n);
long nValue = n.longValue();
if ((nValue & 1) == 0)
n = Long.valueOf(nValue / 2);
else
n = Long.valueOf(3 * nValue + 1);
}
int curCount = count.intValue();
for (int j = currentList.size() - 1; j >= 0; j--)
sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount));
if (curCount > highestCount) {
highestCount = curCount;
highestNumber = i;
}
}
System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount);
}
return;
}
}
|
Produce a language-to-language conversion: from COBOL to Java, same semantics. | identification division.
program-id. hailstones.
remarks. cobc -x hailstones.cob.
data division.
working-storage section.
01 most constant as 1000000.
01 coverage constant as 100000.
01 stones usage binary-long.
01 n usage binary-long.
01 storm usage binary-long.
01 show-arg pic 9(6).
01 show-default pic 99 value 27.
01 show-sequence usage binary-long.
01 longest usage binary-long occurs 2 times.
01 filler.
05 hail usage binary-long
occurs 0 to most depending on stones.
01 show pic z(10).
01 low-range usage binary-long.
01 high-range usage binary-long.
01 range usage binary-long.
01 remain usage binary-long.
01 unused usage binary-long.
procedure division.
accept show-arg from command-line
if show-arg less than 1 or greater than coverage then
move show-default to show-arg
end-if
move show-arg to show-sequence
move 1 to longest(1)
perform hailstone varying storm
from 1 by 1 until storm > coverage
display "Longest at: " longest(2) " with " longest(1) " elements"
goback.
hailstone.
move 0 to stones
move storm to n
perform until n equal 1
if stones > most then
display "too many hailstones" upon syserr
stop run
end-if
add 1 to stones
move n to hail(stones)
divide n by 2 giving unused remainder remain
if remain equal 0 then
divide 2 into n
else
compute n = 3 * n + 1
end-if
end-perform
add 1 to stones
move n to hail(stones)
if stones > longest(1) then
move stones to longest(1)
move storm to longest(2)
end-if
if storm equal show-sequence then
display show-sequence ": " with no advancing
perform varying range from 1 by 1 until range > stones
move 5 to low-range
compute high-range = stones - 4
if range < low-range or range > high-range then
move hail(range) to show
display function trim(show) with no advancing
if range < stones then
display ", " with no advancing
end-if
end-if
if range = low-range and stones > 8 then
display "..., " with no advancing
end-if
end-perform
display ": " stones " elements"
end-if
.
end program hailstones.
| 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 COBOL block to Python, preserving its control flow and logic. | identification division.
program-id. hailstones.
remarks. cobc -x hailstones.cob.
data division.
working-storage section.
01 most constant as 1000000.
01 coverage constant as 100000.
01 stones usage binary-long.
01 n usage binary-long.
01 storm usage binary-long.
01 show-arg pic 9(6).
01 show-default pic 99 value 27.
01 show-sequence usage binary-long.
01 longest usage binary-long occurs 2 times.
01 filler.
05 hail usage binary-long
occurs 0 to most depending on stones.
01 show pic z(10).
01 low-range usage binary-long.
01 high-range usage binary-long.
01 range usage binary-long.
01 remain usage binary-long.
01 unused usage binary-long.
procedure division.
accept show-arg from command-line
if show-arg less than 1 or greater than coverage then
move show-default to show-arg
end-if
move show-arg to show-sequence
move 1 to longest(1)
perform hailstone varying storm
from 1 by 1 until storm > coverage
display "Longest at: " longest(2) " with " longest(1) " elements"
goback.
hailstone.
move 0 to stones
move storm to n
perform until n equal 1
if stones > most then
display "too many hailstones" upon syserr
stop run
end-if
add 1 to stones
move n to hail(stones)
divide n by 2 giving unused remainder remain
if remain equal 0 then
divide 2 into n
else
compute n = 3 * n + 1
end-if
end-perform
add 1 to stones
move n to hail(stones)
if stones > longest(1) then
move stones to longest(1)
move storm to longest(2)
end-if
if storm equal show-sequence then
display show-sequence ": " with no advancing
perform varying range from 1 by 1 until range > stones
move 5 to low-range
compute high-range = stones - 4
if range < low-range or range > high-range then
move hail(range) to show
display function trim(show) with no advancing
if range < stones then
display ", " with no advancing
end-if
end-if
if range = low-range and stones > 8 then
display "..., " with no advancing
end-if
end-perform
display ": " stones " elements"
end-if
.
end program hailstones.
| 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")
|
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