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Produce a language-to-language conversion: from Groovy to C++, same semantics.
def hailstone = { long start -> def sequence = [] while (start != 1) { sequence << start start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l } sequence << start }
#include <iostream> #include <vector> #include <utility> std::vector<int> hailstone(int i) { std::vector<int> v; while(true){ v.push_back(i); if (1 == i) break; i = (i % 2) ? (3 * i + 1) : (i / 2); } return v; } std::pair<int,int> find_longest_hailstone_seq(int n) { std::pair<int, int> maxseq(0, 0); int l; for(int i = 1; i < n; ++i){ l = hailstone(i).size(); if (l > maxseq.second) maxseq = std::make_pair(i, l); } return maxseq; } int main () { std::vector<int> h27; h27 = hailstone(27); int l = h27.size(); std::cout << "length of hailstone(27) is " << l; std::cout << " first four elements of hailstone(27) are "; std::cout << h27[0] << " " << h27[1] << " " << h27[2] << " " << h27[3] << std::endl; std::cout << " last four elements of hailstone(27) are " << h27[l-4] << " " << h27[l-3] << " " << h27[l-2] << " " << h27[l-1] << std::endl; std::pair<int,int> m = find_longest_hailstone_seq(100000); std::cout << "the longest hailstone sequence under 100,000 is " << m.first << " with " << m.second << " elements." <<std::endl; return 0; }
Rewrite the snippet below in Java so it works the same as the original Groovy code.
def hailstone = { long start -> def sequence = [] while (start != 1) { sequence << start start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l } sequence << start }
import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map; class Hailstone { public static List<Long> getHailstoneSequence(long n) { if (n <= 0) throw new IllegalArgumentException("Invalid starting sequence number"); List<Long> list = new ArrayList<Long>(); list.add(Long.valueOf(n)); while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; list.add(Long.valueOf(n)); } return list; } public static void main(String[] args) { List<Long> sequence27 = getHailstoneSequence(27); System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27); long MAX = 100000; { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = getHailstoneSequence(i).size(); if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = 1; long n = i; while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; count++; } if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; long highestCount = 1; Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>(); sequenceMap.put(Long.valueOf(1), Integer.valueOf(1)); List<Long> currentList = new ArrayList<Long>(); for (long i = 2; i < MAX; i++) { currentList.clear(); Long n = Long.valueOf(i); Integer count = null; while ((count = sequenceMap.get(n)) == null) { currentList.add(n); long nValue = n.longValue(); if ((nValue & 1) == 0) n = Long.valueOf(nValue / 2); else n = Long.valueOf(3 * nValue + 1); } int curCount = count.intValue(); for (int j = currentList.size() - 1; j >= 0; j--) sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount)); if (curCount > highestCount) { highestCount = curCount; highestNumber = i; } } System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } return; } }
Produce a functionally identical Java code for the snippet given in Groovy.
def hailstone = { long start -> def sequence = [] while (start != 1) { sequence << start start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l } sequence << start }
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 Groovy implementation into Python, maintaining the same output and logic.
def hailstone = { long start -> def sequence = [] while (start != 1) { sequence << start start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l } sequence << start }
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)))
Generate an equivalent Python version of this Groovy code.
def hailstone = { long start -> def sequence = [] while (start != 1) { sequence << start start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l } sequence << start }
def hailstone(n): seq = [n] while n>1: n = 3*n + 1 if n & 1 else n//2 seq.append(n) return seq if __name__ == '__main__': h = hailstone(27) assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1] print("Maximum length %i was found for hailstone(%i) for numbers <100,000" % max((len(hailstone(i)), i) for i in range(1,100000)))
Port the provided Groovy code into VB while preserving the original functionality.
def hailstone = { long start -> def sequence = [] while (start != 1) { sequence << start start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l } sequence << start }
Private Function hailstone(ByVal n As Long) As Collection Dim s As New Collection s.Add CStr(n), CStr(n) i = 0 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If s.Add CStr(n), CStr(n) Loop Set hailstone = s End Function Private Function hailstone_count(ByVal n As Long) Dim count As Long: count = 1 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If count = count + 1 Loop hailstone_count = count End Function Public Sub rosetta() Dim s As Collection, i As Long Set s = hailstone(27) Dim ls As Integer: ls = s.count Debug.Print "hailstone(27) = "; For i = 1 To 4 Debug.Print s(i); ", "; Next i Debug.Print "... "; For i = s.count - 4 To s.count - 1 Debug.Print s(i); ", "; Next i Debug.Print s(s.count) Debug.Print "length ="; ls Dim hmax As Long: hmax = 1 Dim imax As Long: imax = 1 Dim count As Integer For i = 2 To 100000# - 1 count = hailstone_count(i) If count > hmax Then hmax = count imax = i End If Next i Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements." End Sub
Rewrite this program in VB while keeping its functionality equivalent to the Groovy version.
def hailstone = { long start -> def sequence = [] while (start != 1) { sequence << start start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l } sequence << start }
Private Function hailstone(ByVal n As Long) As Collection Dim s As New Collection s.Add CStr(n), CStr(n) i = 0 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If s.Add CStr(n), CStr(n) Loop Set hailstone = s End Function Private Function hailstone_count(ByVal n As Long) Dim count As Long: count = 1 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If count = count + 1 Loop hailstone_count = count End Function Public Sub rosetta() Dim s As Collection, i As Long Set s = hailstone(27) Dim ls As Integer: ls = s.count Debug.Print "hailstone(27) = "; For i = 1 To 4 Debug.Print s(i); ", "; Next i Debug.Print "... "; For i = s.count - 4 To s.count - 1 Debug.Print s(i); ", "; Next i Debug.Print s(s.count) Debug.Print "length ="; ls Dim hmax As Long: hmax = 1 Dim imax As Long: imax = 1 Dim count As Integer For i = 2 To 100000# - 1 count = hailstone_count(i) If count > hmax Then hmax = count imax = i End If Next i Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements." End Sub
Write a version of this Groovy function in Go with identical behavior.
def hailstone = { long start -> def sequence = [] while (start != 1) { sequence << start start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l } sequence << start }
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 this program in Go while keeping its functionality equivalent to the Groovy version.
def hailstone = { long start -> def sequence = [] while (start != 1) { sequence << start start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l } sequence << start }
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 Haskell code in C.
import Data.List (maximumBy) import Data.Ord (comparing) collatz :: Int -> Int collatz n | even n = n `div` 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000] main :: IO () main = mapM_ putStrLn [ "Collatz sequence for 27: ", (show . hailstone) 27, "The number " <> show longestChain, "has the longest hailstone sequence", "for any number less then 100000. ", "The sequence has length: " <> (show . length . hailstone $ longestChain) ]
#include <stdio.h> #include <stdlib.h> int hailstone(int n, int *arry) { int hs = 1; while (n!=1) { hs++; if (arry) *arry++ = n; n = (n&1) ? (3*n+1) : (n/2); } if (arry) *arry++ = n; return hs; } int main() { int j, hmax = 0; int jatmax, n; int *arry; for (j=1; j<100000; j++) { n = hailstone(j, NULL); if (hmax < n) { hmax = n; jatmax = j; } } n = hailstone(27, NULL); arry = malloc(n*sizeof(int)); n = hailstone(27, arry); printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n", arry[0],arry[1],arry[2],arry[3], arry[n-4], arry[n-3], arry[n-2], arry[n-1], n); printf("Max %d at j= %d\n", hmax, jatmax); free(arry); return 0; }
Write the same algorithm in C as shown in this Haskell implementation.
import Data.List (maximumBy) import Data.Ord (comparing) collatz :: Int -> Int collatz n | even n = n `div` 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000] main :: IO () main = mapM_ putStrLn [ "Collatz sequence for 27: ", (show . hailstone) 27, "The number " <> show longestChain, "has the longest hailstone sequence", "for any number less then 100000. ", "The sequence has length: " <> (show . length . hailstone $ longestChain) ]
#include <stdio.h> #include <stdlib.h> int hailstone(int n, int *arry) { int hs = 1; while (n!=1) { hs++; if (arry) *arry++ = n; n = (n&1) ? (3*n+1) : (n/2); } if (arry) *arry++ = n; return hs; } int main() { int j, hmax = 0; int jatmax, n; int *arry; for (j=1; j<100000; j++) { n = hailstone(j, NULL); if (hmax < n) { hmax = n; jatmax = j; } } n = hailstone(27, NULL); arry = malloc(n*sizeof(int)); n = hailstone(27, arry); printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n", arry[0],arry[1],arry[2],arry[3], arry[n-4], arry[n-3], arry[n-2], arry[n-1], n); printf("Max %d at j= %d\n", hmax, jatmax); free(arry); return 0; }
Translate the given Haskell code snippet into C# without altering its behavior.
import Data.List (maximumBy) import Data.Ord (comparing) collatz :: Int -> Int collatz n | even n = n `div` 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000] main :: IO () main = mapM_ putStrLn [ "Collatz sequence for 27: ", (show . hailstone) 27, "The number " <> show longestChain, "has the longest hailstone sequence", "for any number less then 100000. ", "The sequence has length: " <> (show . length . hailstone $ longestChain) ]
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 Haskell code snippet into C# without altering its behavior.
import Data.List (maximumBy) import Data.Ord (comparing) collatz :: Int -> Int collatz n | even n = n `div` 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000] main :: IO () main = mapM_ putStrLn [ "Collatz sequence for 27: ", (show . hailstone) 27, "The number " <> show longestChain, "has the longest hailstone sequence", "for any number less then 100000. ", "The sequence has length: " <> (show . length . hailstone $ longestChain) ]
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); } } }
Change the programming language of this snippet from Haskell to C++ without modifying what it does.
import Data.List (maximumBy) import Data.Ord (comparing) collatz :: Int -> Int collatz n | even n = n `div` 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000] main :: IO () main = mapM_ putStrLn [ "Collatz sequence for 27: ", (show . hailstone) 27, "The number " <> show longestChain, "has the longest hailstone sequence", "for any number less then 100000. ", "The sequence has length: " <> (show . length . hailstone $ longestChain) ]
#include <iostream> #include <vector> #include <utility> std::vector<int> hailstone(int i) { std::vector<int> v; while(true){ v.push_back(i); if (1 == i) break; i = (i % 2) ? (3 * i + 1) : (i / 2); } return v; } std::pair<int,int> find_longest_hailstone_seq(int n) { std::pair<int, int> maxseq(0, 0); int l; for(int i = 1; i < n; ++i){ l = hailstone(i).size(); if (l > maxseq.second) maxseq = std::make_pair(i, l); } return maxseq; } int main () { std::vector<int> h27; h27 = hailstone(27); int l = h27.size(); std::cout << "length of hailstone(27) is " << l; std::cout << " first four elements of hailstone(27) are "; std::cout << h27[0] << " " << h27[1] << " " << h27[2] << " " << h27[3] << std::endl; std::cout << " last four elements of hailstone(27) are " << h27[l-4] << " " << h27[l-3] << " " << h27[l-2] << " " << h27[l-1] << std::endl; std::pair<int,int> m = find_longest_hailstone_seq(100000); std::cout << "the longest hailstone sequence under 100,000 is " << m.first << " with " << m.second << " elements." <<std::endl; return 0; }
Write the same code in C++ as shown below in Haskell.
import Data.List (maximumBy) import Data.Ord (comparing) collatz :: Int -> Int collatz n | even n = n `div` 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000] main :: IO () main = mapM_ putStrLn [ "Collatz sequence for 27: ", (show . hailstone) 27, "The number " <> show longestChain, "has the longest hailstone sequence", "for any number less then 100000. ", "The sequence has length: " <> (show . length . hailstone $ longestChain) ]
#include <iostream> #include <vector> #include <utility> std::vector<int> hailstone(int i) { std::vector<int> v; while(true){ v.push_back(i); if (1 == i) break; i = (i % 2) ? (3 * i + 1) : (i / 2); } return v; } std::pair<int,int> find_longest_hailstone_seq(int n) { std::pair<int, int> maxseq(0, 0); int l; for(int i = 1; i < n; ++i){ l = hailstone(i).size(); if (l > maxseq.second) maxseq = std::make_pair(i, l); } return maxseq; } int main () { std::vector<int> h27; h27 = hailstone(27); int l = h27.size(); std::cout << "length of hailstone(27) is " << l; std::cout << " first four elements of hailstone(27) are "; std::cout << h27[0] << " " << h27[1] << " " << h27[2] << " " << h27[3] << std::endl; std::cout << " last four elements of hailstone(27) are " << h27[l-4] << " " << h27[l-3] << " " << h27[l-2] << " " << h27[l-1] << std::endl; std::pair<int,int> m = find_longest_hailstone_seq(100000); std::cout << "the longest hailstone sequence under 100,000 is " << m.first << " with " << m.second << " elements." <<std::endl; return 0; }
Can you help me rewrite this code in Java instead of Haskell, keeping it the same logically?
import Data.List (maximumBy) import Data.Ord (comparing) collatz :: Int -> Int collatz n | even n = n `div` 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000] main :: IO () main = mapM_ putStrLn [ "Collatz sequence for 27: ", (show . hailstone) 27, "The number " <> show longestChain, "has the longest hailstone sequence", "for any number less then 100000. ", "The sequence has length: " <> (show . length . hailstone $ longestChain) ]
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; } }
Port the provided Haskell code into Java while preserving the original functionality.
import Data.List (maximumBy) import Data.Ord (comparing) collatz :: Int -> Int collatz n | even n = n `div` 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000] main :: IO () main = mapM_ putStrLn [ "Collatz sequence for 27: ", (show . hailstone) 27, "The number " <> show longestChain, "has the longest hailstone sequence", "for any number less then 100000. ", "The sequence has length: " <> (show . length . hailstone $ longestChain) ]
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 Haskell snippet to Python and keep its semantics consistent.
import Data.List (maximumBy) import Data.Ord (comparing) collatz :: Int -> Int collatz n | even n = n `div` 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000] main :: IO () main = mapM_ putStrLn [ "Collatz sequence for 27: ", (show . hailstone) 27, "The number " <> show longestChain, "has the longest hailstone sequence", "for any number less then 100000. ", "The sequence has length: " <> (show . length . hailstone $ longestChain) ]
def hailstone(n): seq = [n] while n>1: n = 3*n + 1 if n & 1 else n//2 seq.append(n) return seq if __name__ == '__main__': h = hailstone(27) assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1] print("Maximum length %i was found for hailstone(%i) for numbers <100,000" % max((len(hailstone(i)), i) for i in range(1,100000)))
Write the same code in Python as shown below in Haskell.
import Data.List (maximumBy) import Data.Ord (comparing) collatz :: Int -> Int collatz n | even n = n `div` 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000] main :: IO () main = mapM_ putStrLn [ "Collatz sequence for 27: ", (show . hailstone) 27, "The number " <> show longestChain, "has the longest hailstone sequence", "for any number less then 100000. ", "The sequence has length: " <> (show . length . hailstone $ longestChain) ]
def hailstone(n): seq = [n] while n>1: n = 3*n + 1 if n & 1 else n//2 seq.append(n) return seq if __name__ == '__main__': h = hailstone(27) assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1] print("Maximum length %i was found for hailstone(%i) for numbers <100,000" % max((len(hailstone(i)), i) for i in range(1,100000)))
Write the same code in VB as shown below in Haskell.
import Data.List (maximumBy) import Data.Ord (comparing) collatz :: Int -> Int collatz n | even n = n `div` 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000] main :: IO () main = mapM_ putStrLn [ "Collatz sequence for 27: ", (show . hailstone) 27, "The number " <> show longestChain, "has the longest hailstone sequence", "for any number less then 100000. ", "The sequence has length: " <> (show . length . hailstone $ longestChain) ]
Private Function hailstone(ByVal n As Long) As Collection Dim s As New Collection s.Add CStr(n), CStr(n) i = 0 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If s.Add CStr(n), CStr(n) Loop Set hailstone = s End Function Private Function hailstone_count(ByVal n As Long) Dim count As Long: count = 1 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If count = count + 1 Loop hailstone_count = count End Function Public Sub rosetta() Dim s As Collection, i As Long Set s = hailstone(27) Dim ls As Integer: ls = s.count Debug.Print "hailstone(27) = "; For i = 1 To 4 Debug.Print s(i); ", "; Next i Debug.Print "... "; For i = s.count - 4 To s.count - 1 Debug.Print s(i); ", "; Next i Debug.Print s(s.count) Debug.Print "length ="; ls Dim hmax As Long: hmax = 1 Dim imax As Long: imax = 1 Dim count As Integer For i = 2 To 100000# - 1 count = hailstone_count(i) If count > hmax Then hmax = count imax = i End If Next i Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements." End Sub
Rewrite the snippet below in VB so it works the same as the original Haskell code.
import Data.List (maximumBy) import Data.Ord (comparing) collatz :: Int -> Int collatz n | even n = n `div` 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000] main :: IO () main = mapM_ putStrLn [ "Collatz sequence for 27: ", (show . hailstone) 27, "The number " <> show longestChain, "has the longest hailstone sequence", "for any number less then 100000. ", "The sequence has length: " <> (show . length . hailstone $ longestChain) ]
Private Function hailstone(ByVal n As Long) As Collection Dim s As New Collection s.Add CStr(n), CStr(n) i = 0 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If s.Add CStr(n), CStr(n) Loop Set hailstone = s End Function Private Function hailstone_count(ByVal n As Long) Dim count As Long: count = 1 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If count = count + 1 Loop hailstone_count = count End Function Public Sub rosetta() Dim s As Collection, i As Long Set s = hailstone(27) Dim ls As Integer: ls = s.count Debug.Print "hailstone(27) = "; For i = 1 To 4 Debug.Print s(i); ", "; Next i Debug.Print "... "; For i = s.count - 4 To s.count - 1 Debug.Print s(i); ", "; Next i Debug.Print s(s.count) Debug.Print "length ="; ls Dim hmax As Long: hmax = 1 Dim imax As Long: imax = 1 Dim count As Integer For i = 2 To 100000# - 1 count = hailstone_count(i) If count > hmax Then hmax = count imax = i End If Next i Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements." End Sub
Rewrite this program in Go while keeping its functionality equivalent to the Haskell version.
import Data.List (maximumBy) import Data.Ord (comparing) collatz :: Int -> Int collatz n | even n = n `div` 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000] main :: IO () main = mapM_ putStrLn [ "Collatz sequence for 27: ", (show . hailstone) 27, "The number " <> show longestChain, "has the longest hailstone sequence", "for any number less then 100000. ", "The sequence has length: " <> (show . length . hailstone $ longestChain) ]
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 Haskell block to Go, preserving its control flow and logic.
import Data.List (maximumBy) import Data.Ord (comparing) collatz :: Int -> Int collatz n | even n = n `div` 2 | otherwise = 1 + 3 * n hailstone :: Int -> [Int] hailstone = takeWhile (1 /=) . iterate collatz longestChain :: Int longestChain = fst $ maximumBy (comparing snd) $ (,) <*> (length . hailstone) <$> [1 .. 100000] main :: IO () main = mapM_ putStrLn [ "Collatz sequence for 27: ", (show . hailstone) 27, "The number " <> show longestChain, "has the longest hailstone sequence", "for any number less then 100000. ", "The sequence has length: " <> (show . length . hailstone $ longestChain) ]
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 Icon block to C, preserving its control flow and logic.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 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; }
Write the same algorithm in C as shown in this Icon implementation.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 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; }
Ensure the translated C# code behaves exactly like the original Icon snippet.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 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); } } }
Transform the following Icon implementation into C#, maintaining the same output and logic.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 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); } } }
Produce a language-to-language conversion: from Icon to C++, same semantics.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 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; }
Generate an equivalent C++ version of this Icon code.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 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 Icon to Java with equivalent syntax and logic.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 end
import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map; class Hailstone { public static List<Long> getHailstoneSequence(long n) { if (n <= 0) throw new IllegalArgumentException("Invalid starting sequence number"); List<Long> list = new ArrayList<Long>(); list.add(Long.valueOf(n)); while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; list.add(Long.valueOf(n)); } return list; } public static void main(String[] args) { List<Long> sequence27 = getHailstoneSequence(27); System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27); long MAX = 100000; { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = getHailstoneSequence(i).size(); if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = 1; long n = i; while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; count++; } if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; long highestCount = 1; Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>(); sequenceMap.put(Long.valueOf(1), Integer.valueOf(1)); List<Long> currentList = new ArrayList<Long>(); for (long i = 2; i < MAX; i++) { currentList.clear(); Long n = Long.valueOf(i); Integer count = null; while ((count = sequenceMap.get(n)) == null) { currentList.add(n); long nValue = n.longValue(); if ((nValue & 1) == 0) n = Long.valueOf(nValue / 2); else n = Long.valueOf(3 * nValue + 1); } int curCount = count.intValue(); for (int j = currentList.size() - 1; j >= 0; j--) sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount)); if (curCount > highestCount) { highestCount = curCount; highestNumber = i; } } System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } return; } }
Write the same code in Java as shown below in Icon.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 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; } }
Generate an equivalent Python version of this Icon code.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 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)))
Port the provided Icon code into Python while preserving the original functionality.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 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)))
Port the provided Icon code into VB while preserving the original functionality.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 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 the given Icon code snippet into VB without altering its behavior.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 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
Please provide an equivalent version of this Icon code in Go.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 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 the following code from Icon to Go, ensuring the logic remains intact.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 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) }
Transform the following J implementation into C, maintaining the same output and logic.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
#include <stdio.h> #include <stdlib.h> int hailstone(int n, int *arry) { int hs = 1; while (n!=1) { hs++; if (arry) *arry++ = n; n = (n&1) ? (3*n+1) : (n/2); } if (arry) *arry++ = n; return hs; } int main() { int j, hmax = 0; int jatmax, n; int *arry; for (j=1; j<100000; j++) { n = hailstone(j, NULL); if (hmax < n) { hmax = n; jatmax = j; } } n = hailstone(27, NULL); arry = malloc(n*sizeof(int)); n = hailstone(27, arry); printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n", arry[0],arry[1],arry[2],arry[3], arry[n-4], arry[n-3], arry[n-2], arry[n-1], n); printf("Max %d at j= %d\n", hmax, jatmax); free(arry); return 0; }
Please provide an equivalent version of this J code in C.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
#include <stdio.h> #include <stdlib.h> int hailstone(int n, int *arry) { int hs = 1; while (n!=1) { hs++; if (arry) *arry++ = n; n = (n&1) ? (3*n+1) : (n/2); } if (arry) *arry++ = n; return hs; } int main() { int j, hmax = 0; int jatmax, n; int *arry; for (j=1; j<100000; j++) { n = hailstone(j, NULL); if (hmax < n) { hmax = n; jatmax = j; } } n = hailstone(27, NULL); arry = malloc(n*sizeof(int)); n = hailstone(27, arry); printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n", arry[0],arry[1],arry[2],arry[3], arry[n-4], arry[n-3], arry[n-2], arry[n-1], n); printf("Max %d at j= %d\n", hmax, jatmax); free(arry); return 0; }
Write a version of this J function in C# with identical behavior.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Hailstone { class Program { public static List<int> hs(int n,List<int> seq) { List<int> sequence = seq; sequence.Add(n); if (n == 1) { return sequence; }else{ int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1; return hs(newn, sequence); } } static void Main(string[] args) { int n = 27; List<int> sequence = hs(n,new List<int>()); Console.WriteLine(sequence.Count + " Elements"); List<int> start = sequence.GetRange(0, 4); List<int> end = sequence.GetRange(sequence.Count - 4, 4); Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end)); int number = 0, longest = 0; for (int i = 1; i < 100000; i++) { int count = (hs(i, new List<int>())).Count; if (count > longest) { longest = count; number = i; } } Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest); } } }
Maintain the same structure and functionality when rewriting this code in C#.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Hailstone { class Program { public static List<int> hs(int n,List<int> seq) { List<int> sequence = seq; sequence.Add(n); if (n == 1) { return sequence; }else{ int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1; return hs(newn, sequence); } } static void Main(string[] args) { int n = 27; List<int> sequence = hs(n,new List<int>()); Console.WriteLine(sequence.Count + " Elements"); List<int> start = sequence.GetRange(0, 4); List<int> end = sequence.GetRange(sequence.Count - 4, 4); Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end)); int number = 0, longest = 0; for (int i = 1; i < 100000; i++) { int count = (hs(i, new List<int>())).Count; if (count > longest) { longest = count; number = i; } } Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest); } } }
Port the provided J code into C++ while preserving the original functionality.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
#include <iostream> #include <vector> #include <utility> std::vector<int> hailstone(int i) { std::vector<int> v; while(true){ v.push_back(i); if (1 == i) break; i = (i % 2) ? (3 * i + 1) : (i / 2); } return v; } std::pair<int,int> find_longest_hailstone_seq(int n) { std::pair<int, int> maxseq(0, 0); int l; for(int i = 1; i < n; ++i){ l = hailstone(i).size(); if (l > maxseq.second) maxseq = std::make_pair(i, l); } return maxseq; } int main () { std::vector<int> h27; h27 = hailstone(27); int l = h27.size(); std::cout << "length of hailstone(27) is " << l; std::cout << " first four elements of hailstone(27) are "; std::cout << h27[0] << " " << h27[1] << " " << h27[2] << " " << h27[3] << std::endl; std::cout << " last four elements of hailstone(27) are " << h27[l-4] << " " << h27[l-3] << " " << h27[l-2] << " " << h27[l-1] << std::endl; std::pair<int,int> m = find_longest_hailstone_seq(100000); std::cout << "the longest hailstone sequence under 100,000 is " << m.first << " with " << m.second << " elements." <<std::endl; return 0; }
Can you help me rewrite this code in C++ instead of J, keeping it the same logically?
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
#include <iostream> #include <vector> #include <utility> std::vector<int> hailstone(int i) { std::vector<int> v; while(true){ v.push_back(i); if (1 == i) break; i = (i % 2) ? (3 * i + 1) : (i / 2); } return v; } std::pair<int,int> find_longest_hailstone_seq(int n) { std::pair<int, int> maxseq(0, 0); int l; for(int i = 1; i < n; ++i){ l = hailstone(i).size(); if (l > maxseq.second) maxseq = std::make_pair(i, l); } return maxseq; } int main () { std::vector<int> h27; h27 = hailstone(27); int l = h27.size(); std::cout << "length of hailstone(27) is " << l; std::cout << " first four elements of hailstone(27) are "; std::cout << h27[0] << " " << h27[1] << " " << h27[2] << " " << h27[3] << std::endl; std::cout << " last four elements of hailstone(27) are " << h27[l-4] << " " << h27[l-3] << " " << h27[l-2] << " " << h27[l-1] << std::endl; std::pair<int,int> m = find_longest_hailstone_seq(100000); std::cout << "the longest hailstone sequence under 100,000 is " << m.first << " with " << m.second << " elements." <<std::endl; return 0; }
Generate an equivalent Java version of this J code.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map; class Hailstone { public static List<Long> getHailstoneSequence(long n) { if (n <= 0) throw new IllegalArgumentException("Invalid starting sequence number"); List<Long> list = new ArrayList<Long>(); list.add(Long.valueOf(n)); while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; list.add(Long.valueOf(n)); } return list; } public static void main(String[] args) { List<Long> sequence27 = getHailstoneSequence(27); System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27); long MAX = 100000; { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = getHailstoneSequence(i).size(); if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = 1; long n = i; while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; count++; } if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; long highestCount = 1; Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>(); sequenceMap.put(Long.valueOf(1), Integer.valueOf(1)); List<Long> currentList = new ArrayList<Long>(); for (long i = 2; i < MAX; i++) { currentList.clear(); Long n = Long.valueOf(i); Integer count = null; while ((count = sequenceMap.get(n)) == null) { currentList.add(n); long nValue = n.longValue(); if ((nValue & 1) == 0) n = Long.valueOf(nValue / 2); else n = Long.valueOf(3 * nValue + 1); } int curCount = count.intValue(); for (int j = currentList.size() - 1; j >= 0; j--) sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount)); if (curCount > highestCount) { highestCount = curCount; highestNumber = i; } } System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } return; } }
Convert this J block to Java, preserving its control flow and logic.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map; class Hailstone { public static List<Long> getHailstoneSequence(long n) { if (n <= 0) throw new IllegalArgumentException("Invalid starting sequence number"); List<Long> list = new ArrayList<Long>(); list.add(Long.valueOf(n)); while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; list.add(Long.valueOf(n)); } return list; } public static void main(String[] args) { List<Long> sequence27 = getHailstoneSequence(27); System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27); long MAX = 100000; { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = getHailstoneSequence(i).size(); if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = 1; long n = i; while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; count++; } if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; long highestCount = 1; Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>(); sequenceMap.put(Long.valueOf(1), Integer.valueOf(1)); List<Long> currentList = new ArrayList<Long>(); for (long i = 2; i < MAX; i++) { currentList.clear(); Long n = Long.valueOf(i); Integer count = null; while ((count = sequenceMap.get(n)) == null) { currentList.add(n); long nValue = n.longValue(); if ((nValue & 1) == 0) n = Long.valueOf(nValue / 2); else n = Long.valueOf(3 * nValue + 1); } int curCount = count.intValue(); for (int j = currentList.size() - 1; j >= 0; j--) sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount)); if (curCount > highestCount) { highestCount = curCount; highestNumber = i; } } System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } return; } }
Transform the following J implementation into Python, maintaining the same output and logic.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
def hailstone(n): seq = [n] while n>1: n = 3*n + 1 if n & 1 else n//2 seq.append(n) return seq if __name__ == '__main__': h = hailstone(27) assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1] print("Maximum length %i was found for hailstone(%i) for numbers <100,000" % max((len(hailstone(i)), i) for i in range(1,100000)))
Ensure the translated Python code behaves exactly like the original J snippet.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
def hailstone(n): seq = [n] while n>1: n = 3*n + 1 if n & 1 else n//2 seq.append(n) return seq if __name__ == '__main__': h = hailstone(27) assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1] print("Maximum length %i was found for hailstone(%i) for numbers <100,000" % max((len(hailstone(i)), i) for i in range(1,100000)))
Port the provided J code into VB while preserving the original functionality.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
Private Function hailstone(ByVal n As Long) As Collection Dim s As New Collection s.Add CStr(n), CStr(n) i = 0 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If s.Add CStr(n), CStr(n) Loop Set hailstone = s End Function Private Function hailstone_count(ByVal n As Long) Dim count As Long: count = 1 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If count = count + 1 Loop hailstone_count = count End Function Public Sub rosetta() Dim s As Collection, i As Long Set s = hailstone(27) Dim ls As Integer: ls = s.count Debug.Print "hailstone(27) = "; For i = 1 To 4 Debug.Print s(i); ", "; Next i Debug.Print "... "; For i = s.count - 4 To s.count - 1 Debug.Print s(i); ", "; Next i Debug.Print s(s.count) Debug.Print "length ="; ls Dim hmax As Long: hmax = 1 Dim imax As Long: imax = 1 Dim count As Integer For i = 2 To 100000# - 1 count = hailstone_count(i) If count > hmax Then hmax = count imax = i End If Next i Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements." End Sub
Transform the following J implementation into VB, maintaining the same output and logic.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
Private Function hailstone(ByVal n As Long) As Collection Dim s As New Collection s.Add CStr(n), CStr(n) i = 0 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If s.Add CStr(n), CStr(n) Loop Set hailstone = s End Function Private Function hailstone_count(ByVal n As Long) Dim count As Long: count = 1 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If count = count + 1 Loop hailstone_count = count End Function Public Sub rosetta() Dim s As Collection, i As Long Set s = hailstone(27) Dim ls As Integer: ls = s.count Debug.Print "hailstone(27) = "; For i = 1 To 4 Debug.Print s(i); ", "; Next i Debug.Print "... "; For i = s.count - 4 To s.count - 1 Debug.Print s(i); ", "; Next i Debug.Print s(s.count) Debug.Print "length ="; ls Dim hmax As Long: hmax = 1 Dim imax As Long: imax = 1 Dim count As Integer For i = 2 To 100000# - 1 count = hailstone_count(i) If count > hmax Then hmax = count imax = i End If Next i Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements." End Sub
Please provide an equivalent version of this J code in Go.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
package main import "fmt" func hs(n int, recycle []int) []int { s := append(recycle[:0], n) for n > 1 { if n&1 == 0 { n = n / 2 } else { n = 3*n + 1 } s = append(s, n) } return s } func main() { seq := hs(27, nil) fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n", len(seq), seq[0], seq[1], seq[2], seq[3], seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1]) var maxN, maxLen int for n := 1; n < 100000; n++ { seq = hs(n, seq) if len(seq) > maxLen { maxN = n maxLen = len(seq) } } fmt.Printf("hs(%d): %d elements\n", maxN, maxLen) }
Produce a functionally identical Go code for the snippet given in J.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
package main import "fmt" func hs(n int, recycle []int) []int { s := append(recycle[:0], n) for n > 1 { if n&1 == 0 { n = n / 2 } else { n = 3*n + 1 } s = append(s, n) } return s } func main() { seq := hs(27, nil) fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n", len(seq), seq[0], seq[1], seq[2], seq[3], seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1]) var maxN, maxLen int for n := 1; n < 100000; n++ { seq = hs(n, seq) if len(seq) > maxLen { maxN = n maxLen = len(seq) } } fmt.Printf("hs(%d): %d elements\n", maxN, maxLen) }
Convert this Julia block to C, preserving its control flow and logic.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
#include <stdio.h> #include <stdlib.h> int hailstone(int n, int *arry) { int hs = 1; while (n!=1) { hs++; if (arry) *arry++ = n; n = (n&1) ? (3*n+1) : (n/2); } if (arry) *arry++ = n; return hs; } int main() { int j, hmax = 0; int jatmax, n; int *arry; for (j=1; j<100000; j++) { n = hailstone(j, NULL); if (hmax < n) { hmax = n; jatmax = j; } } n = hailstone(27, NULL); arry = malloc(n*sizeof(int)); n = hailstone(27, arry); printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n", arry[0],arry[1],arry[2],arry[3], arry[n-4], arry[n-3], arry[n-2], arry[n-1], n); printf("Max %d at j= %d\n", hmax, jatmax); free(arry); return 0; }
Rewrite the snippet below in C so it works the same as the original Julia code.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
#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 Julia code into C# without altering its purpose.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Hailstone { class Program { public static List<int> hs(int n,List<int> seq) { List<int> sequence = seq; sequence.Add(n); if (n == 1) { return sequence; }else{ int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1; return hs(newn, sequence); } } static void Main(string[] args) { int n = 27; List<int> sequence = hs(n,new List<int>()); Console.WriteLine(sequence.Count + " Elements"); List<int> start = sequence.GetRange(0, 4); List<int> end = sequence.GetRange(sequence.Count - 4, 4); Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end)); int number = 0, longest = 0; for (int i = 1; i < 100000; i++) { int count = (hs(i, new List<int>())).Count; if (count > longest) { longest = count; number = i; } } Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest); } } }
Maintain the same structure and functionality when rewriting this code in C#.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Hailstone { class Program { public static List<int> hs(int n,List<int> seq) { List<int> sequence = seq; sequence.Add(n); if (n == 1) { return sequence; }else{ int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1; return hs(newn, sequence); } } static void Main(string[] args) { int n = 27; List<int> sequence = hs(n,new List<int>()); Console.WriteLine(sequence.Count + " Elements"); List<int> start = sequence.GetRange(0, 4); List<int> end = sequence.GetRange(sequence.Count - 4, 4); Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end)); int number = 0, longest = 0; for (int i = 1; i < 100000; i++) { int count = (hs(i, new List<int>())).Count; if (count > longest) { longest = count; number = i; } } Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest); } } }
Keep all operations the same but rewrite the snippet in C++.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
#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 this Julia block to C++, preserving its control flow and logic.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
#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 Julia.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
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 Julia implementation into Java, maintaining the same output and logic.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
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 Julia to Python, same semantics.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
def hailstone(n): seq = [n] while n>1: n = 3*n + 1 if n & 1 else n//2 seq.append(n) return seq if __name__ == '__main__': h = hailstone(27) assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1] print("Maximum length %i was found for hailstone(%i) for numbers <100,000" % max((len(hailstone(i)), i) for i in range(1,100000)))
Ensure the translated Python code behaves exactly like the original Julia snippet.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
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)))
Generate an equivalent VB version of this Julia code.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
Private Function hailstone(ByVal n As Long) As Collection Dim s As New Collection s.Add CStr(n), CStr(n) i = 0 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If s.Add CStr(n), CStr(n) Loop Set hailstone = s End Function Private Function hailstone_count(ByVal n As Long) Dim count As Long: count = 1 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If count = count + 1 Loop hailstone_count = count End Function Public Sub rosetta() Dim s As Collection, i As Long Set s = hailstone(27) Dim ls As Integer: ls = s.count Debug.Print "hailstone(27) = "; For i = 1 To 4 Debug.Print s(i); ", "; Next i Debug.Print "... "; For i = s.count - 4 To s.count - 1 Debug.Print s(i); ", "; Next i Debug.Print s(s.count) Debug.Print "length ="; ls Dim hmax As Long: hmax = 1 Dim imax As Long: imax = 1 Dim count As Integer For i = 2 To 100000# - 1 count = hailstone_count(i) If count > hmax Then hmax = count imax = i End If Next i Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements." End Sub
Rewrite this program in VB while keeping its functionality equivalent to the Julia version.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
Private Function hailstone(ByVal n As Long) As Collection Dim s As New Collection s.Add CStr(n), CStr(n) i = 0 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If s.Add CStr(n), CStr(n) Loop Set hailstone = s End Function Private Function hailstone_count(ByVal n As Long) Dim count As Long: count = 1 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If count = count + 1 Loop hailstone_count = count End Function Public Sub rosetta() Dim s As Collection, i As Long Set s = hailstone(27) Dim ls As Integer: ls = s.count Debug.Print "hailstone(27) = "; For i = 1 To 4 Debug.Print s(i); ", "; Next i Debug.Print "... "; For i = s.count - 4 To s.count - 1 Debug.Print s(i); ", "; Next i Debug.Print s(s.count) Debug.Print "length ="; ls Dim hmax As Long: hmax = 1 Dim imax As Long: imax = 1 Dim count As Integer For i = 2 To 100000# - 1 count = hailstone_count(i) If count > hmax Then hmax = count imax = i End If Next i Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements." End Sub
Generate a Go translation of this Julia snippet without changing its computational steps.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
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 Go version of this Julia code.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
package main import "fmt" func hs(n int, recycle []int) []int { s := append(recycle[:0], n) for n > 1 { if n&1 == 0 { n = n / 2 } else { n = 3*n + 1 } s = append(s, n) } return s } func main() { seq := hs(27, nil) fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n", len(seq), seq[0], seq[1], seq[2], seq[3], seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1]) var maxN, maxLen int for n := 1; n < 100000; n++ { seq = hs(n, seq) if len(seq) > maxLen { maxN = n maxLen = len(seq) } } fmt.Printf("hs(%d): %d elements\n", maxN, maxLen) }
Write the same code in C as shown below in Lua.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
#include <stdio.h> #include <stdlib.h> int hailstone(int n, int *arry) { int hs = 1; while (n!=1) { hs++; if (arry) *arry++ = n; n = (n&1) ? (3*n+1) : (n/2); } if (arry) *arry++ = n; return hs; } int main() { int j, hmax = 0; int jatmax, n; int *arry; for (j=1; j<100000; j++) { n = hailstone(j, NULL); if (hmax < n) { hmax = n; jatmax = j; } } n = hailstone(27, NULL); arry = malloc(n*sizeof(int)); n = hailstone(27, arry); printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n", arry[0],arry[1],arry[2],arry[3], arry[n-4], arry[n-3], arry[n-2], arry[n-1], n); printf("Max %d at j= %d\n", hmax, jatmax); free(arry); return 0; }
Generate a C translation of this Lua snippet without changing its computational steps.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
#include <stdio.h> #include <stdlib.h> int hailstone(int n, int *arry) { int hs = 1; while (n!=1) { hs++; if (arry) *arry++ = n; n = (n&1) ? (3*n+1) : (n/2); } if (arry) *arry++ = n; return hs; } int main() { int j, hmax = 0; int jatmax, n; int *arry; for (j=1; j<100000; j++) { n = hailstone(j, NULL); if (hmax < n) { hmax = n; jatmax = j; } } n = hailstone(27, NULL); arry = malloc(n*sizeof(int)); n = hailstone(27, arry); printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n", arry[0],arry[1],arry[2],arry[3], arry[n-4], arry[n-3], arry[n-2], arry[n-1], n); printf("Max %d at j= %d\n", hmax, jatmax); free(arry); return 0; }
Maintain the same structure and functionality when rewriting this code in C#.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Hailstone { class Program { public static List<int> hs(int n,List<int> seq) { List<int> sequence = seq; sequence.Add(n); if (n == 1) { return sequence; }else{ int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1; return hs(newn, sequence); } } static void Main(string[] args) { int n = 27; List<int> sequence = hs(n,new List<int>()); Console.WriteLine(sequence.Count + " Elements"); List<int> start = sequence.GetRange(0, 4); List<int> end = sequence.GetRange(sequence.Count - 4, 4); Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end)); int number = 0, longest = 0; for (int i = 1; i < 100000; i++) { int count = (hs(i, new List<int>())).Count; if (count > longest) { longest = count; number = i; } } Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest); } } }
Write the same algorithm in C# as shown in this Lua implementation.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Hailstone { class Program { public static List<int> hs(int n,List<int> seq) { List<int> sequence = seq; sequence.Add(n); if (n == 1) { return sequence; }else{ int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1; return hs(newn, sequence); } } static void Main(string[] args) { int n = 27; List<int> sequence = hs(n,new List<int>()); Console.WriteLine(sequence.Count + " Elements"); List<int> start = sequence.GetRange(0, 4); List<int> end = sequence.GetRange(sequence.Count - 4, 4); Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end)); int number = 0, longest = 0; for (int i = 1; i < 100000; i++) { int count = (hs(i, new List<int>())).Count; if (count > longest) { longest = count; number = i; } } Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest); } } }
Please provide an equivalent version of this Lua code in C++.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
#include <iostream> #include <vector> #include <utility> std::vector<int> hailstone(int i) { std::vector<int> v; while(true){ v.push_back(i); if (1 == i) break; i = (i % 2) ? (3 * i + 1) : (i / 2); } return v; } std::pair<int,int> find_longest_hailstone_seq(int n) { std::pair<int, int> maxseq(0, 0); int l; for(int i = 1; i < n; ++i){ l = hailstone(i).size(); if (l > maxseq.second) maxseq = std::make_pair(i, l); } return maxseq; } int main () { std::vector<int> h27; h27 = hailstone(27); int l = h27.size(); std::cout << "length of hailstone(27) is " << l; std::cout << " first four elements of hailstone(27) are "; std::cout << h27[0] << " " << h27[1] << " " << h27[2] << " " << h27[3] << std::endl; std::cout << " last four elements of hailstone(27) are " << h27[l-4] << " " << h27[l-3] << " " << h27[l-2] << " " << h27[l-1] << std::endl; std::pair<int,int> m = find_longest_hailstone_seq(100000); std::cout << "the longest hailstone sequence under 100,000 is " << m.first << " with " << m.second << " elements." <<std::endl; return 0; }
Change the programming language of this snippet from Lua to C++ without modifying what it does.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
#include <iostream> #include <vector> #include <utility> std::vector<int> hailstone(int i) { std::vector<int> v; while(true){ v.push_back(i); if (1 == i) break; i = (i % 2) ? (3 * i + 1) : (i / 2); } return v; } std::pair<int,int> find_longest_hailstone_seq(int n) { std::pair<int, int> maxseq(0, 0); int l; for(int i = 1; i < n; ++i){ l = hailstone(i).size(); if (l > maxseq.second) maxseq = std::make_pair(i, l); } return maxseq; } int main () { std::vector<int> h27; h27 = hailstone(27); int l = h27.size(); std::cout << "length of hailstone(27) is " << l; std::cout << " first four elements of hailstone(27) are "; std::cout << h27[0] << " " << h27[1] << " " << h27[2] << " " << h27[3] << std::endl; std::cout << " last four elements of hailstone(27) are " << h27[l-4] << " " << h27[l-3] << " " << h27[l-2] << " " << h27[l-1] << std::endl; std::pair<int,int> m = find_longest_hailstone_seq(100000); std::cout << "the longest hailstone sequence under 100,000 is " << m.first << " with " << m.second << " elements." <<std::endl; return 0; }
Generate an equivalent Java version of this Lua code.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map; class Hailstone { public static List<Long> getHailstoneSequence(long n) { if (n <= 0) throw new IllegalArgumentException("Invalid starting sequence number"); List<Long> list = new ArrayList<Long>(); list.add(Long.valueOf(n)); while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; list.add(Long.valueOf(n)); } return list; } public static void main(String[] args) { List<Long> sequence27 = getHailstoneSequence(27); System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27); long MAX = 100000; { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = getHailstoneSequence(i).size(); if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = 1; long n = i; while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; count++; } if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; long highestCount = 1; Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>(); sequenceMap.put(Long.valueOf(1), Integer.valueOf(1)); List<Long> currentList = new ArrayList<Long>(); for (long i = 2; i < MAX; i++) { currentList.clear(); Long n = Long.valueOf(i); Integer count = null; while ((count = sequenceMap.get(n)) == null) { currentList.add(n); long nValue = n.longValue(); if ((nValue & 1) == 0) n = Long.valueOf(nValue / 2); else n = Long.valueOf(3 * nValue + 1); } int curCount = count.intValue(); for (int j = currentList.size() - 1; j >= 0; j--) sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount)); if (curCount > highestCount) { highestCount = curCount; highestNumber = i; } } System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } return; } }
Generate a Java translation of this Lua snippet without changing its computational steps.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map; class Hailstone { public static List<Long> getHailstoneSequence(long n) { if (n <= 0) throw new IllegalArgumentException("Invalid starting sequence number"); List<Long> list = new ArrayList<Long>(); list.add(Long.valueOf(n)); while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; list.add(Long.valueOf(n)); } return list; } public static void main(String[] args) { List<Long> sequence27 = getHailstoneSequence(27); System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27); long MAX = 100000; { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = getHailstoneSequence(i).size(); if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = 1; long n = i; while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; count++; } if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; long highestCount = 1; Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>(); sequenceMap.put(Long.valueOf(1), Integer.valueOf(1)); List<Long> currentList = new ArrayList<Long>(); for (long i = 2; i < MAX; i++) { currentList.clear(); Long n = Long.valueOf(i); Integer count = null; while ((count = sequenceMap.get(n)) == null) { currentList.add(n); long nValue = n.longValue(); if ((nValue & 1) == 0) n = Long.valueOf(nValue / 2); else n = Long.valueOf(3 * nValue + 1); } int curCount = count.intValue(); for (int j = currentList.size() - 1; j >= 0; j--) sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount)); if (curCount > highestCount) { highestCount = curCount; highestNumber = i; } } System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } return; } }
Keep all operations the same but rewrite the snippet in Python.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
def hailstone(n): seq = [n] while n>1: n = 3*n + 1 if n & 1 else n//2 seq.append(n) return seq if __name__ == '__main__': h = hailstone(27) assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1] print("Maximum length %i was found for hailstone(%i) for numbers <100,000" % max((len(hailstone(i)), i) for i in range(1,100000)))
Translate this program into Python but keep the logic exactly as in Lua.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
def hailstone(n): seq = [n] while n>1: n = 3*n + 1 if n & 1 else n//2 seq.append(n) return seq if __name__ == '__main__': h = hailstone(27) assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1] print("Maximum length %i was found for hailstone(%i) for numbers <100,000" % max((len(hailstone(i)), i) for i in range(1,100000)))
Change the programming language of this snippet from Lua to VB without modifying what it does.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
Private Function hailstone(ByVal n As Long) As Collection Dim s As New Collection s.Add CStr(n), CStr(n) i = 0 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If s.Add CStr(n), CStr(n) Loop Set hailstone = s End Function Private Function hailstone_count(ByVal n As Long) Dim count As Long: count = 1 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If count = count + 1 Loop hailstone_count = count End Function Public Sub rosetta() Dim s As Collection, i As Long Set s = hailstone(27) Dim ls As Integer: ls = s.count Debug.Print "hailstone(27) = "; For i = 1 To 4 Debug.Print s(i); ", "; Next i Debug.Print "... "; For i = s.count - 4 To s.count - 1 Debug.Print s(i); ", "; Next i Debug.Print s(s.count) Debug.Print "length ="; ls Dim hmax As Long: hmax = 1 Dim imax As Long: imax = 1 Dim count As Integer For i = 2 To 100000# - 1 count = hailstone_count(i) If count > hmax Then hmax = count imax = i End If Next i Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements." End Sub
Maintain the same structure and functionality when rewriting this code in VB.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
Private Function hailstone(ByVal n As Long) As Collection Dim s As New Collection s.Add CStr(n), CStr(n) i = 0 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If s.Add CStr(n), CStr(n) Loop Set hailstone = s End Function Private Function hailstone_count(ByVal n As Long) Dim count As Long: count = 1 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If count = count + 1 Loop hailstone_count = count End Function Public Sub rosetta() Dim s As Collection, i As Long Set s = hailstone(27) Dim ls As Integer: ls = s.count Debug.Print "hailstone(27) = "; For i = 1 To 4 Debug.Print s(i); ", "; Next i Debug.Print "... "; For i = s.count - 4 To s.count - 1 Debug.Print s(i); ", "; Next i Debug.Print s(s.count) Debug.Print "length ="; ls Dim hmax As Long: hmax = 1 Dim imax As Long: imax = 1 Dim count As Integer For i = 2 To 100000# - 1 count = hailstone_count(i) If count > hmax Then hmax = count imax = i End If Next i Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements." End Sub
Please provide an equivalent version of this Lua code in Go.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
package main import "fmt" func hs(n int, recycle []int) []int { s := append(recycle[:0], n) for n > 1 { if n&1 == 0 { n = n / 2 } else { n = 3*n + 1 } s = append(s, n) } return s } func main() { seq := hs(27, nil) fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n", len(seq), seq[0], seq[1], seq[2], seq[3], seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1]) var maxN, maxLen int for n := 1; n < 100000; n++ { seq = hs(n, seq) if len(seq) > maxLen { maxN = n maxLen = len(seq) } } fmt.Printf("hs(%d): %d elements\n", maxN, maxLen) }
Write the same algorithm in Go as shown in this Lua implementation.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
package main import "fmt" func hs(n int, recycle []int) []int { s := append(recycle[:0], n) for n > 1 { if n&1 == 0 { n = n / 2 } else { n = 3*n + 1 } s = append(s, n) } return s } func main() { seq := hs(27, nil) fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n", len(seq), seq[0], seq[1], seq[2], seq[3], seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1]) var maxN, maxLen int for n := 1; n < 100000; n++ { seq = hs(n, seq) if len(seq) > maxLen { maxN = n maxLen = len(seq) } } fmt.Printf("hs(%d): %d elements\n", maxN, maxLen) }
Keep all operations the same but rewrite the snippet in C.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 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; }
Ensure the translated C code behaves exactly like the original Mathematica snippet.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 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; }
Preserve the algorithm and functionality while converting the code from Mathematica to C#.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 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); } } }
Produce a language-to-language conversion: from Mathematica to C#, same semantics.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 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); } } }
Write a version of this Mathematica function in C++ with identical behavior.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 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; }
Generate a C++ translation of this Mathematica snippet without changing its computational steps.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 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; }
Port the following code from Mathematica to Java with equivalent syntax and logic.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 1 &]
import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map; class Hailstone { public static List<Long> getHailstoneSequence(long n) { if (n <= 0) throw new IllegalArgumentException("Invalid starting sequence number"); List<Long> list = new ArrayList<Long>(); list.add(Long.valueOf(n)); while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; list.add(Long.valueOf(n)); } return list; } public static void main(String[] args) { List<Long> sequence27 = getHailstoneSequence(27); System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27); long MAX = 100000; { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = getHailstoneSequence(i).size(); if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = 1; long n = i; while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; count++; } if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; long highestCount = 1; Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>(); sequenceMap.put(Long.valueOf(1), Integer.valueOf(1)); List<Long> currentList = new ArrayList<Long>(); for (long i = 2; i < MAX; i++) { currentList.clear(); Long n = Long.valueOf(i); Integer count = null; while ((count = sequenceMap.get(n)) == null) { currentList.add(n); long nValue = n.longValue(); if ((nValue & 1) == 0) n = Long.valueOf(nValue / 2); else n = Long.valueOf(3 * nValue + 1); } int curCount = count.intValue(); for (int j = currentList.size() - 1; j >= 0; j--) sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount)); if (curCount > highestCount) { highestCount = curCount; highestNumber = i; } } System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } return; } }
Write the same code in Java as shown below in Mathematica.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 1 &]
import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map; class Hailstone { public static List<Long> getHailstoneSequence(long n) { if (n <= 0) throw new IllegalArgumentException("Invalid starting sequence number"); List<Long> list = new ArrayList<Long>(); list.add(Long.valueOf(n)); while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; list.add(Long.valueOf(n)); } return list; } public static void main(String[] args) { List<Long> sequence27 = getHailstoneSequence(27); System.out.println("Sequence for 27 has " + sequence27.size() + " elements: " + sequence27); long MAX = 100000; { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = getHailstoneSequence(i).size(); if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 1, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; int highestCount = 1; for (long i = 2; i < MAX; i++) { int count = 1; long n = i; while (n != 1) { if ((n & 1) == 0) n = n / 2; else n = 3 * n + 1; count++; } if (count > highestCount) { highestCount = count; highestNumber = i; } } System.out.println("Method 2, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } { long highestNumber = 1; long highestCount = 1; Map<Long, Integer> sequenceMap = new HashMap<Long, Integer>(); sequenceMap.put(Long.valueOf(1), Integer.valueOf(1)); List<Long> currentList = new ArrayList<Long>(); for (long i = 2; i < MAX; i++) { currentList.clear(); Long n = Long.valueOf(i); Integer count = null; while ((count = sequenceMap.get(n)) == null) { currentList.add(n); long nValue = n.longValue(); if ((nValue & 1) == 0) n = Long.valueOf(nValue / 2); else n = Long.valueOf(3 * nValue + 1); } int curCount = count.intValue(); for (int j = currentList.size() - 1; j >= 0; j--) sequenceMap.put(currentList.get(j), Integer.valueOf(++curCount)); if (curCount > highestCount) { highestCount = curCount; highestNumber = i; } } System.out.println("Method 3, number " + highestNumber + " has the longest sequence, with a length of " + highestCount); } return; } }
Write a version of this Mathematica function in Python with identical behavior.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 1 &]
def hailstone(n): seq = [n] while n>1: n = 3*n + 1 if n & 1 else n//2 seq.append(n) return seq if __name__ == '__main__': h = hailstone(27) assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1] print("Maximum length %i was found for hailstone(%i) for numbers <100,000" % max((len(hailstone(i)), i) for i in range(1,100000)))
Generate a Python translation of this Mathematica snippet without changing its computational steps.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 1 &]
def hailstone(n): seq = [n] while n>1: n = 3*n + 1 if n & 1 else n//2 seq.append(n) return seq if __name__ == '__main__': h = hailstone(27) assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1] print("Maximum length %i was found for hailstone(%i) for numbers <100,000" % max((len(hailstone(i)), i) for i in range(1,100000)))
Generate a VB translation of this Mathematica snippet without changing its computational steps.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 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
Port the provided Mathematica code into VB while preserving the original functionality.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 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
Rewrite the snippet below in Go so it works the same as the original Mathematica code.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 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) }
Translate the given Mathematica code snippet into Go without altering its behavior.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 1 &]
package main import "fmt" func hs(n int, recycle []int) []int { s := append(recycle[:0], n) for n > 1 { if n&1 == 0 { n = n / 2 } else { n = 3*n + 1 } s = append(s, n) } return s } func main() { seq := hs(27, nil) fmt.Printf("hs(27): %d elements: [%d %d %d %d ... %d %d %d %d]\n", len(seq), seq[0], seq[1], seq[2], seq[3], seq[len(seq)-4], seq[len(seq)-3], seq[len(seq)-2], seq[len(seq)-1]) var maxN, maxLen int for n := 1; n < 100000; n++ { seq = hs(n, seq) if len(seq) > maxLen { maxN = n maxLen = len(seq) } } fmt.Printf("hs(%d): %d elements\n", maxN, maxLen) }
Produce a language-to-language conversion: from MATLAB to C, same semantics.
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
#include <stdio.h> #include <stdlib.h> int hailstone(int n, int *arry) { int hs = 1; while (n!=1) { hs++; if (arry) *arry++ = n; n = (n&1) ? (3*n+1) : (n/2); } if (arry) *arry++ = n; return hs; } int main() { int j, hmax = 0; int jatmax, n; int *arry; for (j=1; j<100000; j++) { n = hailstone(j, NULL); if (hmax < n) { hmax = n; jatmax = j; } } n = hailstone(27, NULL); arry = malloc(n*sizeof(int)); n = hailstone(27, arry); printf("[ %d, %d, %d, %d, ...., %d, %d, %d, %d] len=%d\n", arry[0],arry[1],arry[2],arry[3], arry[n-4], arry[n-3], arry[n-2], arry[n-1], n); printf("Max %d at j= %d\n", hmax, jatmax); free(arry); return 0; }
Generate an equivalent C version of this MATLAB code.
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
#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 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
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 MATLAB to C#, ensuring the logic remains intact.
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
using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Hailstone { class Program { public static List<int> hs(int n,List<int> seq) { List<int> sequence = seq; sequence.Add(n); if (n == 1) { return sequence; }else{ int newn = (n % 2 == 0) ? n / 2 : (3 * n) + 1; return hs(newn, sequence); } } static void Main(string[] args) { int n = 27; List<int> sequence = hs(n,new List<int>()); Console.WriteLine(sequence.Count + " Elements"); List<int> start = sequence.GetRange(0, 4); List<int> end = sequence.GetRange(sequence.Count - 4, 4); Console.WriteLine("Starting with : " + string.Join(",", start) + " and ending with : " + string.Join(",", end)); int number = 0, longest = 0; for (int i = 1; i < 100000; i++) { int count = (hs(i, new List<int>())).Count; if (count > longest) { longest = count; number = i; } } Console.WriteLine("Number < 100000 with longest Hailstone seq.: " + number + " with length of " + longest); } } }
Maintain the same structure and functionality when rewriting this code in C++.
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
#include <iostream> #include <vector> #include <utility> std::vector<int> hailstone(int i) { std::vector<int> v; while(true){ v.push_back(i); if (1 == i) break; i = (i % 2) ? (3 * i + 1) : (i / 2); } return v; } std::pair<int,int> find_longest_hailstone_seq(int n) { std::pair<int, int> maxseq(0, 0); int l; for(int i = 1; i < n; ++i){ l = hailstone(i).size(); if (l > maxseq.second) maxseq = std::make_pair(i, l); } return maxseq; } int main () { std::vector<int> h27; h27 = hailstone(27); int l = h27.size(); std::cout << "length of hailstone(27) is " << l; std::cout << " first four elements of hailstone(27) are "; std::cout << h27[0] << " " << h27[1] << " " << h27[2] << " " << h27[3] << std::endl; std::cout << " last four elements of hailstone(27) are " << h27[l-4] << " " << h27[l-3] << " " << h27[l-2] << " " << h27[l-1] << std::endl; std::pair<int,int> m = find_longest_hailstone_seq(100000); std::cout << "the longest hailstone sequence under 100,000 is " << m.first << " with " << m.second << " elements." <<std::endl; return 0; }
Write the same code in C++ as shown below in 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
#include <iostream> #include <vector> #include <utility> std::vector<int> hailstone(int i) { std::vector<int> v; while(true){ v.push_back(i); if (1 == i) break; i = (i % 2) ? (3 * i + 1) : (i / 2); } return v; } std::pair<int,int> find_longest_hailstone_seq(int n) { std::pair<int, int> maxseq(0, 0); int l; for(int i = 1; i < n; ++i){ l = hailstone(i).size(); if (l > maxseq.second) maxseq = std::make_pair(i, l); } return maxseq; } int main () { std::vector<int> h27; h27 = hailstone(27); int l = h27.size(); std::cout << "length of hailstone(27) is " << l; std::cout << " first four elements of hailstone(27) are "; std::cout << h27[0] << " " << h27[1] << " " << h27[2] << " " << h27[3] << std::endl; std::cout << " last four elements of hailstone(27) are " << h27[l-4] << " " << h27[l-3] << " " << h27[l-2] << " " << h27[l-1] << std::endl; std::pair<int,int> m = find_longest_hailstone_seq(100000); std::cout << "the longest hailstone sequence under 100,000 is " << m.first << " with " << m.second << " elements." <<std::endl; return 0; }
Generate a Java translation of this MATLAB snippet without changing its computational steps.
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; } }