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Convert the following code from Factor to PHP, ensuring the logic remains intact.
USING: arrays io kernel math math.ranges prettyprint sequences vectors ; IN: rosetta.hailstone : hailstone ( n -- seq ) [ 1vector ] keep [ dup 1 number= ] [ dup even? [ 2 / ] [ 3 * 1 + ] if 2dup swap push ] until drop ; <PRIVATE : main ( -- ) 27 hailstone dup dup "The hailstone sequence from 27:" print " has length " write length . " starts with " write 4 head [ unparse ] map ", " join print " ends with " write 4 tail* [ unparse ] map ", " join print 1 100000 [a,b) [ [ hailstone length ] keep 2array ] [ [ [ first ] bi@ > ] most ] map-reduce first2 "The hailstone sequence from " write pprint " has length " write pprint "." print ; PRIVATE> MAIN: main
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Can you help me rewrite this code in PHP instead of Factor, keeping it the same logically?
USING: arrays io kernel math math.ranges prettyprint sequences vectors ; IN: rosetta.hailstone : hailstone ( n -- seq ) [ 1vector ] keep [ dup 1 number= ] [ dup even? [ 2 / ] [ 3 * 1 + ] if 2dup swap push ] until drop ; <PRIVATE : main ( -- ) 27 hailstone dup dup "The hailstone sequence from 27:" print " has length " write length . " starts with " write 4 head [ unparse ] map ", " join print " ends with " write 4 tail* [ unparse ] map ", " join print 1 100000 [a,b) [ [ hailstone length ] keep 2array ] [ [ [ first ] bi@ > ] most ] map-reduce first2 "The hailstone sequence from " write pprint " has length " write pprint "." print ; PRIVATE> MAIN: main
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Maintain the same structure and functionality when rewriting this code in PHP.
: hail-next dup 1 and if 3 * 1+ else 2/ then ; : .hail begin dup . dup 1 > while hail-next repeat drop ; : hail-len 1 begin over 1 > while swap hail-next swap 1+ repeat nip ; 27 hail-len . cr 27 .hail cr : longest-hail 0 0 rot 1+ 1 do i hail-len 2dup < if nip nip i swap else drop then loop swap . ." has hailstone sequence length " . ; 100000 longest-hail
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Convert this Forth block to PHP, preserving its control flow and logic.
: hail-next dup 1 and if 3 * 1+ else 2/ then ; : .hail begin dup . dup 1 > while hail-next repeat drop ; : hail-len 1 begin over 1 > while swap hail-next swap 1+ repeat nip ; 27 hail-len . cr 27 .hail cr : longest-hail 0 0 rot 1+ 1 do i hail-len 2dup < if nip nip i swap else drop then loop swap . ." has hailstone sequence length " . ; 100000 longest-hail
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Can you help me rewrite this code in PHP instead of Fortran, keeping it the same logically?
program Hailstone implicit none integer :: i, maxn integer :: maxseqlen = 0, seqlen integer, allocatable :: seq(:) call hs(27, seqlen) allocate(seq(seqlen)) call hs(27, seqlen, seq) write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements" write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", & seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen) do i = 1, 99999 call hs(i, seqlen) if (seqlen > maxseqlen) then maxseqlen = seqlen maxn = i end if end do write(*,*) write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements" deallocate(seq) contains subroutine hs(number, length, seqArray) integer, intent(in) :: number integer, intent(out) :: length integer, optional, intent(inout) :: seqArray(:) integer :: n n = number length = 1 if(present(seqArray)) seqArray(1) = n do while(n /= 1) if(mod(n,2) == 0) then n = n / 2 else n = n * 3 + 1 end if length = length + 1 if(present(seqArray)) seqArray(length) = n end do end subroutine end program
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Maintain the same structure and functionality when rewriting this code in PHP.
program Hailstone implicit none integer :: i, maxn integer :: maxseqlen = 0, seqlen integer, allocatable :: seq(:) call hs(27, seqlen) allocate(seq(seqlen)) call hs(27, seqlen, seq) write(*,"(a,i0,a)") "Hailstone sequence for 27 has ", seqlen, " elements" write(*,"(a,4(i0,a),3(i0,a),i0)") "Sequence = ", seq(1), ", ", seq(2), ", ", seq(3), ", ", seq(4), " ...., ", & seq(seqlen-3), ", ", seq(seqlen-2), ", ", seq(seqlen-1), ", ", seq(seqlen) do i = 1, 99999 call hs(i, seqlen) if (seqlen > maxseqlen) then maxseqlen = seqlen maxn = i end if end do write(*,*) write(*,"(a,i0,a,i0,a)") "Longest sequence under 100000 is for ", maxn, " with ", maxseqlen, " elements" deallocate(seq) contains subroutine hs(number, length, seqArray) integer, intent(in) :: number integer, intent(out) :: length integer, optional, intent(inout) :: seqArray(:) integer :: n n = number length = 1 if(present(seqArray)) seqArray(1) = n do while(n /= 1) if(mod(n,2) == 0) then n = n / 2 else n = n * 3 + 1 end if length = length + 1 if(present(seqArray)) seqArray(length) = n end do end subroutine end program
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Write the same algorithm in PHP as shown in this Groovy implementation.
def hailstone = { long start -> def sequence = [] while (start != 1) { sequence << start start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l } sequence << start }
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Translate the given Groovy code snippet into PHP without altering its behavior.
def hailstone = { long start -> def sequence = [] while (start != 1) { sequence << start start = (start % 2l == 0l) ? start / 2l : 3l * start + 1l } sequence << start }
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Write the same code in PHP 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) ]
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Transform the following Haskell implementation into PHP, maintaining the same output 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) ]
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Generate an equivalent PHP version of this Icon code.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 end
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Rewrite this program in PHP while keeping its functionality equivalent to the Icon version.
procedure hailstone(n) while n > 1 do { suspend n n := if n%2 = 0 then n/2 else 3*n+1 } suspend 1 end
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Write the same code in PHP as shown below in J.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Write the same code in PHP as shown below in J.
hailseq=: -:`(1 3&p.)@.(2&|) ^:(1 ~: ]) ^:a:"0
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Can you help me rewrite this code in PHP instead of Julia, keeping it the same logically?
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Produce a language-to-language conversion: from Julia to PHP, same semantics.
function hailstonelength(n::Integer) len = 1 while n > 1 n = ifelse(iseven(n), n ÷ 2, 3n + 1) len += 1 end return len end @show hailstonelength(27); nothing @show findmax([hailstonelength(i) for i in 1:100_000]); nothing
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Write the same code in PHP as shown below in Lua.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Maintain the same structure and functionality when rewriting this code in PHP.
function hailstone( n, print_numbers ) local n_iter = 1 while n ~= 1 do if print_numbers then print( n ) end if n % 2 == 0 then n = n / 2 else n = 3 * n + 1 end n_iter = n_iter + 1 end if print_numbers then print( n ) end return n_iter; end hailstone( 27, true ) max_i, max_iter = 0, 0 for i = 1, 100000 do num = hailstone( i, false ) if num >= max_iter then max_i = i max_iter = num end end print( string.format( "Needed %d iterations for the number %d.\n", max_iter, max_i ) )
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Preserve the algorithm and functionality while converting the code from Mathematica to PHP.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 1 &]
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Preserve the algorithm and functionality while converting the code from Mathematica to PHP.
HailstoneF[n_] := NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 1 &]
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Change the following MATLAB code into PHP without altering its purpose.
function x = hailstone(n) x = n; while n > 1 if n ~= floor(n / 2) * 2 n = n * 3 + 1; else n = n / 2; end x(end + 1) = n; end
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Can you help me rewrite this code in PHP instead of MATLAB, keeping it the same logically?
function x = hailstone(n) x = n; while n > 1 if n ~= floor(n / 2) * 2 n = n * 3 + 1; else n = n / 2; end x(end + 1) = n; end
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Convert this Nim block to PHP, preserving its control flow and logic.
proc hailstone(n: int): seq[int] = result = @[n] var n = n while n > 1: if (n and 1) == 1: n = 3 * n + 1 else: n = n div 2 result.add n when isMainModule: import strformat, strutils let h = hailstone(27) echo &"Hailstone sequence for number 27 has {h.len} elements." let first = h[0..3].join(", ") let last = h[^4..^1].join(", ") echo &"This sequence begins with {first} and ends with {last}." var m, mi = 0 for i in 1..<100_000: let n = hailstone(i).len if n > m: m = n mi = i echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Write the same code in PHP as shown below in Nim.
proc hailstone(n: int): seq[int] = result = @[n] var n = n while n > 1: if (n and 1) == 1: n = 3 * n + 1 else: n = n div 2 result.add n when isMainModule: import strformat, strutils let h = hailstone(27) echo &"Hailstone sequence for number 27 has {h.len} elements." let first = h[0..3].join(", ") let last = h[^4..^1].join(", ") echo &"This sequence begins with {first} and ends with {last}." var m, mi = 0 for i in 1..<100_000: let n = hailstone(i).len if n > m: m = n mi = i echo &"\nFor numbers < 100_000, maximum length {m} was found for Hailstone({mi})."
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Translate the given OCaml code snippet into PHP without altering its behavior.
fun hail (x = 1) = [1] | (x rem 2 = 0) = x :: hail (x div 2) | x = x :: hail (x * 3 + 1) fun hailstorm ([], i, largest, largest_at) = (largest_at, largest) | (x :: xs, i, largest, largest_at) = let val k = len (hail x) in if k > largest then hailstorm (xs, i + 1, k, i) else hailstorm (xs, i + 1, largest, largest_at) end | (x :: xs) = hailstorm (x :: xs, 1, 0, 0) ; val h27 = hail 27; print "hailstone sequence for the number 27 has "; print ` len (h27); print " elements starting with "; print ` sub (h27, 0, 4); print " and ending with "; print ` sub (h27, len(h27)-4, len h27); println "."; val biggest = hailstorm ` iota (100000 - 1); print "The number less than 100,000 which has the longest "; print "hailstone sequence is at element "; print ` ref (biggest, 0); print " and is of length "; println ` ref (biggest, 1);
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Preserve the algorithm and functionality while converting the code from OCaml to PHP.
fun hail (x = 1) = [1] | (x rem 2 = 0) = x :: hail (x div 2) | x = x :: hail (x * 3 + 1) fun hailstorm ([], i, largest, largest_at) = (largest_at, largest) | (x :: xs, i, largest, largest_at) = let val k = len (hail x) in if k > largest then hailstorm (xs, i + 1, k, i) else hailstorm (xs, i + 1, largest, largest_at) end | (x :: xs) = hailstorm (x :: xs, 1, 0, 0) ; val h27 = hail 27; print "hailstone sequence for the number 27 has "; print ` len (h27); print " elements starting with "; print ` sub (h27, 0, 4); print " and ending with "; print ` sub (h27, len(h27)-4, len h27); println "."; val biggest = hailstorm ` iota (100000 - 1); print "The number less than 100,000 which has the longest "; print "hailstone sequence is at element "; print ` ref (biggest, 0); print " and is of length "; println ` ref (biggest, 1);
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Maintain the same structure and functionality when rewriting this code in PHP.
program ShowHailstoneSequence; uses SysUtils; const maxN = 10*1000*1000; type tiaArr = array[0..1000] of Uint64; tIntArr = record iaMaxPos : integer; iaArr : tiaArr end; tpiaArr = ^tiaArr; function HailstoneSeqCnt(n: UInt64): NativeInt; begin result := 0; while not(ODD(n)) do Begin inc(result); n := n shr 1; end; IF n > 1 then repeat repeat n := (3*n+1) SHR 1;inc(result,2); until NOT(Odd(n)); repeat n := n shr 1; inc(result); until odd(n); until n = 1; end; procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr); var maxPos: NativeInt; n: UInt64; pArr : tpiaArr; begin with aHailstoneList do begin maxPos := 0; pArr := @iaArr; end; n := aStartingNumber; pArr^[maxPos] := n; while n <> 1 do begin if odd(n) then n := (3*n+1) else n := n shr 1; inc(maxPos); pArr^[maxPos] := n; end; aHailstoneList.iaMaxPos := maxPos; end; var i,Limit: NativeInt; lList: tIntArr; lAverageLength:Uint64; lMaxSequence: NativeInt; lMaxLength,lgth: NativeInt; begin lList.iaMaxPos := 0; GetHailstoneSequence(27, lList); with lList do begin Limit := iaMaxPos; writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1])); write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..'); For i := iaMaxPos-3 to iaMaxPos-1 do write(iaArr[i],','); writeln(iaArr[iaMaxPos]); end; Writeln; lMaxSequence := 0; lMaxLength := 0; i := 1; limit := 10*i; writeln(' Limit  : number with max length | average length'); repeat lAverageLength:= 0; repeat lgth:= HailstoneSeqCnt(i); inc(lAverageLength, lgth); if lgth >= lMaxLength then begin lMaxSequence := i; lMaxLength := lgth+1; end; inc(i); until i = Limit; Writeln(Format(' %10d : %9d | %4d | %7.3f', [limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit])); limit := limit*10; until Limit > maxN; end.
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Preserve the algorithm and functionality while converting the code from Pascal to PHP.
program ShowHailstoneSequence; uses SysUtils; const maxN = 10*1000*1000; type tiaArr = array[0..1000] of Uint64; tIntArr = record iaMaxPos : integer; iaArr : tiaArr end; tpiaArr = ^tiaArr; function HailstoneSeqCnt(n: UInt64): NativeInt; begin result := 0; while not(ODD(n)) do Begin inc(result); n := n shr 1; end; IF n > 1 then repeat repeat n := (3*n+1) SHR 1;inc(result,2); until NOT(Odd(n)); repeat n := n shr 1; inc(result); until odd(n); until n = 1; end; procedure GetHailstoneSequence(aStartingNumber: NativeUint;var aHailstoneList: tIntArr); var maxPos: NativeInt; n: UInt64; pArr : tpiaArr; begin with aHailstoneList do begin maxPos := 0; pArr := @iaArr; end; n := aStartingNumber; pArr^[maxPos] := n; while n <> 1 do begin if odd(n) then n := (3*n+1) else n := n shr 1; inc(maxPos); pArr^[maxPos] := n; end; aHailstoneList.iaMaxPos := maxPos; end; var i,Limit: NativeInt; lList: tIntArr; lAverageLength:Uint64; lMaxSequence: NativeInt; lMaxLength,lgth: NativeInt; begin lList.iaMaxPos := 0; GetHailstoneSequence(27, lList); with lList do begin Limit := iaMaxPos; writeln(Format('sequence of %d has %d elements',[iaArr[0],Limit+1])); write(iaArr[0],',',iaArr[1],',',iaArr[2],',',iaArr[3],'..'); For i := iaMaxPos-3 to iaMaxPos-1 do write(iaArr[i],','); writeln(iaArr[iaMaxPos]); end; Writeln; lMaxSequence := 0; lMaxLength := 0; i := 1; limit := 10*i; writeln(' Limit  : number with max length | average length'); repeat lAverageLength:= 0; repeat lgth:= HailstoneSeqCnt(i); inc(lAverageLength, lgth); if lgth >= lMaxLength then begin lMaxSequence := i; lMaxLength := lgth+1; end; inc(i); until i = Limit; Writeln(Format(' %10d : %9d | %4d | %7.3f', [limit,lMaxSequence, lMaxLength,0.9*lAverageLength/Limit])); limit := limit*10; until Limit > maxN; end.
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Rewrite the snippet below in PHP so it works the same as the original Perl code.
use warnings; use strict; my @h = hailstone(27); print "Length of hailstone(27) = " . scalar @h . "\n"; print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n"; my ($max, $n) = (0, 0); for my $x (1 .. 99_999) { @h = hailstone($x); if (scalar @h > $max) { ($max, $n) = (scalar @h, $x); } } print "Max length $max was found for hailstone($n) for numbers < 100_000\n"; sub hailstone { my ($n) = @_; my @sequence = ($n); while ($n > 1) { if ($n % 2 == 0) { $n = int($n / 2); } else { $n = $n * 3 + 1; } push @sequence, $n; } return @sequence; }
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Write the same algorithm in PHP as shown in this Perl implementation.
use warnings; use strict; my @h = hailstone(27); print "Length of hailstone(27) = " . scalar @h . "\n"; print "[" . join(", ", @h[0 .. 3], "...", @h[-4 .. -1]) . "]\n"; my ($max, $n) = (0, 0); for my $x (1 .. 99_999) { @h = hailstone($x); if (scalar @h > $max) { ($max, $n) = (scalar @h, $x); } } print "Max length $max was found for hailstone($n) for numbers < 100_000\n"; sub hailstone { my ($n) = @_; my @sequence = ($n); while ($n > 1) { if ($n % 2 == 0) { $n = int($n / 2); } else { $n = $n * 3 + 1; } push @sequence, $n; } return @sequence; }
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Write the same algorithm in PHP as shown in this PowerShell implementation.
function Get-HailStone { param($n) switch($n) { 1 {$n;return} {$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)} {$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)} } } function Get-HailStoneBelowLimit { param($UpperLimit) for ($i = 1; $i -lt $UpperLimit; $i++) { [pscustomobject]@{ 'Number' = $i 'Count' = (Get-HailStone $i).count } } }
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Ensure the translated PHP code behaves exactly like the original PowerShell snippet.
function Get-HailStone { param($n) switch($n) { 1 {$n;return} {$n % 2 -eq 0} {$n; return Get-Hailstone ($n = $n / 2)} {$n % 2 -ne 0} {$n; return Get-Hailstone ($n = ($n * 3) +1)} } } function Get-HailStoneBelowLimit { param($UpperLimit) for ($i = 1; $i -lt $UpperLimit; $i++) { [pscustomobject]@{ 'Number' = $i 'Count' = (Get-HailStone $i).count } } }
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Translate the given Racket code snippet into PHP without altering its behavior.
#lang racket (define hailstone (let ([t (make-hasheq)]) (hash-set! t 1 '(1)) (λ(n) (hash-ref! t n (λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1))))))))) (define h27 (hailstone 27)) (printf "h(27) = ~s, ~s items\n" `(,@(take h27 4) ... ,@(take-right h27 4)) (length h27)) (define N 100000) (define longest (for/fold ([m #f]) ([i (in-range 1 (add1 N))]) (define h (hailstone i)) (if (and m (> (cdr m) (length h))) m (cons i (length h))))) (printf "for x<=~s, ~s has the longest sequence with ~s items\n" N (car longest) (cdr longest))
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Produce a language-to-language conversion: from Racket to PHP, same semantics.
#lang racket (define hailstone (let ([t (make-hasheq)]) (hash-set! t 1 '(1)) (λ(n) (hash-ref! t n (λ() (cons n (hailstone (if (even? n) (/ n 2) (+ (* 3 n) 1))))))))) (define h27 (hailstone 27)) (printf "h(27) = ~s, ~s items\n" `(,@(take h27 4) ... ,@(take-right h27 4)) (length h27)) (define N 100000) (define longest (for/fold ([m #f]) ([i (in-range 1 (add1 N))]) (define h (hailstone i)) (if (and m (> (cdr m) (length h))) m (cons i (length h))))) (printf "for x<=~s, ~s has the longest sequence with ~s items\n" N (car longest) (cdr longest))
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Generate an equivalent PHP version of this COBOL code.
identification division. program-id. hailstones. remarks. cobc -x hailstones.cob. data division. working-storage section. 01 most constant as 1000000. 01 coverage constant as 100000. 01 stones usage binary-long. 01 n usage binary-long. 01 storm usage binary-long. 01 show-arg pic 9(6). 01 show-default pic 99 value 27. 01 show-sequence usage binary-long. 01 longest usage binary-long occurs 2 times. 01 filler. 05 hail usage binary-long occurs 0 to most depending on stones. 01 show pic z(10). 01 low-range usage binary-long. 01 high-range usage binary-long. 01 range usage binary-long. 01 remain usage binary-long. 01 unused usage binary-long. procedure division. accept show-arg from command-line if show-arg less than 1 or greater than coverage then move show-default to show-arg end-if move show-arg to show-sequence move 1 to longest(1) perform hailstone varying storm from 1 by 1 until storm > coverage display "Longest at: " longest(2) " with " longest(1) " elements" goback. hailstone. move 0 to stones move storm to n perform until n equal 1 if stones > most then display "too many hailstones" upon syserr stop run end-if add 1 to stones move n to hail(stones) divide n by 2 giving unused remainder remain if remain equal 0 then divide 2 into n else compute n = 3 * n + 1 end-if end-perform add 1 to stones move n to hail(stones) if stones > longest(1) then move stones to longest(1) move storm to longest(2) end-if if storm equal show-sequence then display show-sequence ": " with no advancing perform varying range from 1 by 1 until range > stones move 5 to low-range compute high-range = stones - 4 if range < low-range or range > high-range then move hail(range) to show display function trim(show) with no advancing if range < stones then display ", " with no advancing end-if end-if if range = low-range and stones > 8 then display "..., " with no advancing end-if end-perform display ": " stones " elements" end-if . end program hailstones.
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Please provide an equivalent version of this COBOL code in PHP.
identification division. program-id. hailstones. remarks. cobc -x hailstones.cob. data division. working-storage section. 01 most constant as 1000000. 01 coverage constant as 100000. 01 stones usage binary-long. 01 n usage binary-long. 01 storm usage binary-long. 01 show-arg pic 9(6). 01 show-default pic 99 value 27. 01 show-sequence usage binary-long. 01 longest usage binary-long occurs 2 times. 01 filler. 05 hail usage binary-long occurs 0 to most depending on stones. 01 show pic z(10). 01 low-range usage binary-long. 01 high-range usage binary-long. 01 range usage binary-long. 01 remain usage binary-long. 01 unused usage binary-long. procedure division. accept show-arg from command-line if show-arg less than 1 or greater than coverage then move show-default to show-arg end-if move show-arg to show-sequence move 1 to longest(1) perform hailstone varying storm from 1 by 1 until storm > coverage display "Longest at: " longest(2) " with " longest(1) " elements" goback. hailstone. move 0 to stones move storm to n perform until n equal 1 if stones > most then display "too many hailstones" upon syserr stop run end-if add 1 to stones move n to hail(stones) divide n by 2 giving unused remainder remain if remain equal 0 then divide 2 into n else compute n = 3 * n + 1 end-if end-perform add 1 to stones move n to hail(stones) if stones > longest(1) then move stones to longest(1) move storm to longest(2) end-if if storm equal show-sequence then display show-sequence ": " with no advancing perform varying range from 1 by 1 until range > stones move 5 to low-range compute high-range = stones - 4 if range < low-range or range > high-range then move hail(range) to show display function trim(show) with no advancing if range < stones then display ", " with no advancing end-if end-if if range = low-range and stones > 8 then display "..., " with no advancing end-if end-perform display ": " stones " elements" end-if . end program hailstones.
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Ensure the translated PHP code behaves exactly like the original REXX snippet.
options replace format comments java crossref savelog symbols binary do start = 27 hs = hailstone(start) hsCount = hs.words say 'The number' start 'has a hailstone sequence comprising' hsCount 'elements' say ' its first four elements are:' hs.subword(1, 4) say ' and last four elements are:' hs.subword(hsCount - 3) hsMax = 0 hsCountMax = 0 llimit = 100000 loop x_ = 1 to llimit - 1 hs = hailstone(x_) hsCount = hs.words if hsCount > hsCountMax then do hsMax = x_ hsCountMax = hsCount end end x_ say 'The number' hsMax 'has the longest hailstone sequence in the range 1 to' llimit - 1 'with a sequence length of' hsCountMax catch ex = Exception ex.printStackTrace end return method hailstone(hn = long) public static returns Rexx signals IllegalArgumentException hs = Rexx('') if hn <= 0 then signal IllegalArgumentException('Invalid start point. Must be a positive integer greater than 0') loop label n_ while hn > 1 hs = hs' 'hn if hn // 2 \= 0 then hn = hn * 3 + 1 else hn = hn % 2 end n_ hs = hs' 'hn return hs.strip
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Write the same algorithm in PHP as shown in this REXX implementation.
options replace format comments java crossref savelog symbols binary do start = 27 hs = hailstone(start) hsCount = hs.words say 'The number' start 'has a hailstone sequence comprising' hsCount 'elements' say ' its first four elements are:' hs.subword(1, 4) say ' and last four elements are:' hs.subword(hsCount - 3) hsMax = 0 hsCountMax = 0 llimit = 100000 loop x_ = 1 to llimit - 1 hs = hailstone(x_) hsCount = hs.words if hsCount > hsCountMax then do hsMax = x_ hsCountMax = hsCount end end x_ say 'The number' hsMax 'has the longest hailstone sequence in the range 1 to' llimit - 1 'with a sequence length of' hsCountMax catch ex = Exception ex.printStackTrace end return method hailstone(hn = long) public static returns Rexx signals IllegalArgumentException hs = Rexx('') if hn <= 0 then signal IllegalArgumentException('Invalid start point. Must be a positive integer greater than 0') loop label n_ while hn > 1 hs = hs' 'hn if hn // 2 \= 0 then hn = hn * 3 + 1 else hn = hn % 2 end n_ hs = hs' 'hn return hs.strip
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Change the following Ruby code into PHP without altering its purpose.
def hailstone(n) seq = [n] until n == 1 n = n.even? ? n // 2 : n * 3 + 1 seq << n end seq end max_len = (1...100_000).max_by{|n| hailstone(n).size } max = hailstone(max_len) puts ([max_len, max.size, max.max, max.first(4), max.last(4)]) twenty_seven = hailstone(27) puts ([twenty_seven.size, twenty_seven.first(4), max.last(4)])
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Convert the following code from Ruby to PHP, ensuring the logic remains intact.
def hailstone(n) seq = [n] until n == 1 n = n.even? ? n // 2 : n * 3 + 1 seq << n end seq end max_len = (1...100_000).max_by{|n| hailstone(n).size } max = hailstone(max_len) puts ([max_len, max.size, max.max, max.first(4), max.last(4)]) twenty_seven = hailstone(27) puts ([twenty_seven.size, twenty_seven.first(4), max.last(4)])
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Can you help me rewrite this code in PHP instead of Scala, keeping it the same logically?
object HailstoneSequence extends App { def hailstone(n: Int): Stream[Int] = n #:: (if (n == 1) Stream.empty else hailstone(if (n % 2 == 0) n / 2 else n * 3 + 1)) val nr = args.headOption.map(_.toInt).getOrElse(27) val collatz = hailstone(nr) println(s"Use the routine to show that the hailstone sequence for the number: $nr.") println(collatz.toList) println(s"It has ${collatz.length} elements.") println println( "Compute the number < 100,000, which has the longest hailstone sequence with that sequence's length.") val (n, len) = (1 until 100000).map(n => (n, hailstone(n).length)).maxBy(_._2) println(s"Longest hailstone sequence length= $len occurring with number $n.") }
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Please provide an equivalent version of this Scala code in PHP.
object HailstoneSequence extends App { def hailstone(n: Int): Stream[Int] = n #:: (if (n == 1) Stream.empty else hailstone(if (n % 2 == 0) n / 2 else n * 3 + 1)) val nr = args.headOption.map(_.toInt).getOrElse(27) val collatz = hailstone(nr) println(s"Use the routine to show that the hailstone sequence for the number: $nr.") println(collatz.toList) println(s"It has ${collatz.length} elements.") println println( "Compute the number < 100,000, which has the longest hailstone sequence with that sequence's length.") val (n, len) = (1 until 100000).map(n => (n, hailstone(n).length)).maxBy(_._2) println(s"Longest hailstone sequence length= $len occurring with number $n.") }
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Generate an equivalent PHP version of this Swift code.
func hailstone(var n:Int) -> [Int] { var arr = [n] while n != 1 { if n % 2 == 0 { n /= 2 } else { n = (3 * n) + 1 } arr.append(n) } return arr } let n = hailstone(27) println("hailstone(27): \(n[0...3]) ... \(n[n.count-4...n.count-1]) for a count of \(n.count).") var longest = (n: 1, len: 1) for i in 1...100_000 { let new = hailstone(i) if new.count > longest.len { longest = (i, new.count) } } println("Longest sequence for numbers under 100,000 is with \(longest.n). Which has \(longest.len) items.")
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Maintain the same structure and functionality when rewriting this code in PHP.
func hailstone(var n:Int) -> [Int] { var arr = [n] while n != 1 { if n % 2 == 0 { n /= 2 } else { n = (3 * n) + 1 } arr.append(n) } return arr } let n = hailstone(27) println("hailstone(27): \(n[0...3]) ... \(n[n.count-4...n.count-1]) for a count of \(n.count).") var longest = (n: 1, len: 1) for i in 1...100_000 { let new = hailstone(i) if new.count > longest.len { longest = (i, new.count) } } println("Longest sequence for numbers under 100,000 is with \(longest.n). Which has \(longest.len) items.")
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Transform the following Tcl implementation into PHP, maintaining the same output and logic.
proc hailstone n { while 1 { lappend seq $n if {$n == 1} {return $seq} set n [expr {$n & 1 ? $n*3+1 : $n/2}] } } set h27 [hailstone 27] puts "h27 len=[llength $h27]" puts "head4 = [lrange $h27 0 3]" puts "tail4 = [lrange $h27 end-3 end]" set maxlen [set max 0] for {set i 1} {$i<100000} {incr i} { set l [llength [hailstone $i]] if {$l>$maxlen} {set maxlen $l;set max $i} } puts "max is $max, with length $maxlen"
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Write the same algorithm in PHP as shown in this Tcl implementation.
proc hailstone n { while 1 { lappend seq $n if {$n == 1} {return $seq} set n [expr {$n & 1 ? $n*3+1 : $n/2}] } } set h27 [hailstone 27] puts "h27 len=[llength $h27]" puts "head4 = [lrange $h27 0 3]" puts "tail4 = [lrange $h27 end-3 end]" set maxlen [set max 0] for {set i 1} {$i<100000} {incr i} { set l [llength [hailstone $i]] if {$l>$maxlen} {set maxlen $l;set max $i} } puts "max is $max, with length $maxlen"
function hailstone($n,$seq=array()){ $sequence = $seq; $sequence[] = $n; if($n == 1){ return $sequence; }else{ $n = ($n%2==0) ? $n/2 : (3*$n)+1; return hailstone($n, $sequence); } } $result = hailstone(27); echo count($result) . ' Elements.<br>'; echo 'Starting with : ' . implode(",",array_slice($result,0,4)) .' and ending with : ' . implode(",",array_slice($result,count($result)-4)) . '<br>'; $maxResult = array(0); for($i=1;$i<=100000;$i++){ $result = count(hailstone($i)); if($result > max($maxResult)){ $maxResult = array($i=>$result); } } foreach($maxResult as $key => $val){ echo 'Number < 100000 with longest Hailstone seq.: ' . $key . ' with length of ' . $val; }
Convert this C snippet to Rust and keep its semantics consistent.
#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; }
fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); }
Please provide an equivalent version of this C++ code in Rust.
#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; }
fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); }
Preserve the algorithm and functionality while converting the code from C# to Rust.
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); } } }
fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); }
Write the same algorithm in Rust as shown in this C# implementation.
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); } } }
fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); }
Write a version of this Go function in Rust with identical behavior.
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) }
fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); }
Convert this Rust snippet to Python and keep its semantics consistent.
fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); }
def hailstone(n): seq = [n] while n>1: n = 3*n + 1 if n & 1 else n//2 seq.append(n) return seq if __name__ == '__main__': h = hailstone(27) assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1] print("Maximum length %i was found for hailstone(%i) for numbers <100,000" % max((len(hailstone(i)), i) for i in range(1,100000)))
Translate this program into VB but keep the logic exactly as in Rust.
fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); }
Private Function hailstone(ByVal n As Long) As Collection Dim s As New Collection s.Add CStr(n), CStr(n) i = 0 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If s.Add CStr(n), CStr(n) Loop Set hailstone = s End Function Private Function hailstone_count(ByVal n As Long) Dim count As Long: count = 1 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If count = count + 1 Loop hailstone_count = count End Function Public Sub rosetta() Dim s As Collection, i As Long Set s = hailstone(27) Dim ls As Integer: ls = s.count Debug.Print "hailstone(27) = "; For i = 1 To 4 Debug.Print s(i); ", "; Next i Debug.Print "... "; For i = s.count - 4 To s.count - 1 Debug.Print s(i); ", "; Next i Debug.Print s(s.count) Debug.Print "length ="; ls Dim hmax As Long: hmax = 1 Dim imax As Long: imax = 1 Dim count As Integer For i = 2 To 100000# - 1 count = hailstone_count(i) If count > hmax Then hmax = count imax = i End If Next i Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements." End Sub
Write the same code in VB as shown below in Rust.
fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); }
Private Function hailstone(ByVal n As Long) As Collection Dim s As New Collection s.Add CStr(n), CStr(n) i = 0 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If s.Add CStr(n), CStr(n) Loop Set hailstone = s End Function Private Function hailstone_count(ByVal n As Long) Dim count As Long: count = 1 Do While n <> 1 If n Mod 2 = 0 Then n = n / 2 Else n = 3 * n + 1 End If count = count + 1 Loop hailstone_count = count End Function Public Sub rosetta() Dim s As Collection, i As Long Set s = hailstone(27) Dim ls As Integer: ls = s.count Debug.Print "hailstone(27) = "; For i = 1 To 4 Debug.Print s(i); ", "; Next i Debug.Print "... "; For i = s.count - 4 To s.count - 1 Debug.Print s(i); ", "; Next i Debug.Print s(s.count) Debug.Print "length ="; ls Dim hmax As Long: hmax = 1 Dim imax As Long: imax = 1 Dim count As Integer For i = 2 To 100000# - 1 count = hailstone_count(i) If count > hmax Then hmax = count imax = i End If Next i Debug.Print "The longest hailstone sequence under 100,000 is"; imax; "with"; hmax; "elements." End Sub
Convert the following code from Java to Rust, ensuring the logic remains intact.
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; } }
fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); }
Please provide an equivalent version of this Java code in Rust.
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; } }
fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); }
Change the following Go code into Rust without altering its purpose.
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) }
fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); }
Translate this program into Rust but keep the logic exactly as in C.
#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; }
fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); }
Transform the following C++ implementation into Rust, maintaining the same output and logic.
#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; }
fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); }
Write the same code in Python as shown below in Rust.
fn hailstone(start : u32) -> Vec<u32> { let mut res = Vec::new(); let mut next = start; res.push(start); while next != 1 { next = if next % 2 == 0 { next/2 } else { 3*next+1 }; res.push(next); } res } fn main() { let test_num = 27; let test_hailseq = hailstone(test_num); println!("For {} number of elements is {} ", test_num, test_hailseq.len()); let fst_slice = test_hailseq[0..4].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); let last_slice = test_hailseq[test_hailseq.len()-4..].iter() .fold("".to_owned(), |acc, i| { acc + &*(i.to_string()).to_owned() + ", " }); println!(" hailstone starting with {} ending with {} ", fst_slice, last_slice); let max_range = 100000; let mut max_len = 0; let mut max_seed = 0; for i_seed in 1..max_range { let i_len = hailstone(i_seed).len(); if i_len > max_len { max_len = i_len; max_seed = i_seed; } } println!("Longest sequence is {} element long for seed {}", max_len, max_seed); }
def hailstone(n): seq = [n] while n>1: n = 3*n + 1 if n & 1 else n//2 seq.append(n) return seq if __name__ == '__main__': h = hailstone(27) assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1] print("Maximum length %i was found for hailstone(%i) for numbers <100,000" % max((len(hailstone(i)), i) for i in range(1,100000)))
Write a version of this Ada function in C# with identical behavior.
with Ada.Text_IO; use Ada.Text_IO; procedure Mfact is function MultiFact (num : Natural; deg : Positive) return Natural is Result, N : Integer := num; begin if N = 0 then return 1; end if; loop N := N - deg; exit when N <= 0; Result := Result * N; end loop; return Result; end MultiFact; begin for deg in 1..5 loop Put("Degree"& Integer'Image(deg) &":"); for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop; New_line; end loop; end Mfact;
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Maintain the same structure and functionality when rewriting this code in C#.
with Ada.Text_IO; use Ada.Text_IO; procedure Mfact is function MultiFact (num : Natural; deg : Positive) return Natural is Result, N : Integer := num; begin if N = 0 then return 1; end if; loop N := N - deg; exit when N <= 0; Result := Result * N; end loop; return Result; end MultiFact; begin for deg in 1..5 loop Put("Degree"& Integer'Image(deg) &":"); for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop; New_line; end loop; end Mfact;
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Write the same code in C as shown below in Ada.
with Ada.Text_IO; use Ada.Text_IO; procedure Mfact is function MultiFact (num : Natural; deg : Positive) return Natural is Result, N : Integer := num; begin if N = 0 then return 1; end if; loop N := N - deg; exit when N <= 0; Result := Result * N; end loop; return Result; end MultiFact; begin for deg in 1..5 loop Put("Degree"& Integer'Image(deg) &":"); for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop; New_line; end loop; end Mfact;
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Produce a functionally identical C code for the snippet given in Ada.
with Ada.Text_IO; use Ada.Text_IO; procedure Mfact is function MultiFact (num : Natural; deg : Positive) return Natural is Result, N : Integer := num; begin if N = 0 then return 1; end if; loop N := N - deg; exit when N <= 0; Result := Result * N; end loop; return Result; end MultiFact; begin for deg in 1..5 loop Put("Degree"& Integer'Image(deg) &":"); for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop; New_line; end loop; end Mfact;
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Port the following code from Ada to C++ with equivalent syntax and logic.
with Ada.Text_IO; use Ada.Text_IO; procedure Mfact is function MultiFact (num : Natural; deg : Positive) return Natural is Result, N : Integer := num; begin if N = 0 then return 1; end if; loop N := N - deg; exit when N <= 0; Result := Result * N; end loop; return Result; end MultiFact; begin for deg in 1..5 loop Put("Degree"& Integer'Image(deg) &":"); for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop; New_line; end loop; end Mfact;
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Convert this Ada block to C++, preserving its control flow and logic.
with Ada.Text_IO; use Ada.Text_IO; procedure Mfact is function MultiFact (num : Natural; deg : Positive) return Natural is Result, N : Integer := num; begin if N = 0 then return 1; end if; loop N := N - deg; exit when N <= 0; Result := Result * N; end loop; return Result; end MultiFact; begin for deg in 1..5 loop Put("Degree"& Integer'Image(deg) &":"); for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop; New_line; end loop; end Mfact;
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Preserve the algorithm and functionality while converting the code from Ada to Go.
with Ada.Text_IO; use Ada.Text_IO; procedure Mfact is function MultiFact (num : Natural; deg : Positive) return Natural is Result, N : Integer := num; begin if N = 0 then return 1; end if; loop N := N - deg; exit when N <= 0; Result := Result * N; end loop; return Result; end MultiFact; begin for deg in 1..5 loop Put("Degree"& Integer'Image(deg) &":"); for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop; New_line; end loop; end Mfact;
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Keep all operations the same but rewrite the snippet in Go.
with Ada.Text_IO; use Ada.Text_IO; procedure Mfact is function MultiFact (num : Natural; deg : Positive) return Natural is Result, N : Integer := num; begin if N = 0 then return 1; end if; loop N := N - deg; exit when N <= 0; Result := Result * N; end loop; return Result; end MultiFact; begin for deg in 1..5 loop Put("Degree"& Integer'Image(deg) &":"); for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop; New_line; end loop; end Mfact;
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Rewrite this program in Java while keeping its functionality equivalent to the Ada version.
with Ada.Text_IO; use Ada.Text_IO; procedure Mfact is function MultiFact (num : Natural; deg : Positive) return Natural is Result, N : Integer := num; begin if N = 0 then return 1; end if; loop N := N - deg; exit when N <= 0; Result := Result * N; end loop; return Result; end MultiFact; begin for deg in 1..5 loop Put("Degree"& Integer'Image(deg) &":"); for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop; New_line; end loop; end Mfact;
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Port the following code from Ada to Java with equivalent syntax and logic.
with Ada.Text_IO; use Ada.Text_IO; procedure Mfact is function MultiFact (num : Natural; deg : Positive) return Natural is Result, N : Integer := num; begin if N = 0 then return 1; end if; loop N := N - deg; exit when N <= 0; Result := Result * N; end loop; return Result; end MultiFact; begin for deg in 1..5 loop Put("Degree"& Integer'Image(deg) &":"); for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop; New_line; end loop; end Mfact;
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Translate this program into Python but keep the logic exactly as in Ada.
with Ada.Text_IO; use Ada.Text_IO; procedure Mfact is function MultiFact (num : Natural; deg : Positive) return Natural is Result, N : Integer := num; begin if N = 0 then return 1; end if; loop N := N - deg; exit when N <= 0; Result := Result * N; end loop; return Result; end MultiFact; begin for deg in 1..5 loop Put("Degree"& Integer'Image(deg) &":"); for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop; New_line; end loop; end Mfact;
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Change the following Ada code into Python without altering its purpose.
with Ada.Text_IO; use Ada.Text_IO; procedure Mfact is function MultiFact (num : Natural; deg : Positive) return Natural is Result, N : Integer := num; begin if N = 0 then return 1; end if; loop N := N - deg; exit when N <= 0; Result := Result * N; end loop; return Result; end MultiFact; begin for deg in 1..5 loop Put("Degree"& Integer'Image(deg) &":"); for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop; New_line; end loop; end Mfact;
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Ensure the translated VB code behaves exactly like the original Ada snippet.
with Ada.Text_IO; use Ada.Text_IO; procedure Mfact is function MultiFact (num : Natural; deg : Positive) return Natural is Result, N : Integer := num; begin if N = 0 then return 1; end if; loop N := N - deg; exit when N <= 0; Result := Result * N; end loop; return Result; end MultiFact; begin for deg in 1..5 loop Put("Degree"& Integer'Image(deg) &":"); for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop; New_line; end loop; end Mfact;
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Generate a VB translation of this Ada snippet without changing its computational steps.
with Ada.Text_IO; use Ada.Text_IO; procedure Mfact is function MultiFact (num : Natural; deg : Positive) return Natural is Result, N : Integer := num; begin if N = 0 then return 1; end if; loop N := N - deg; exit when N <= 0; Result := Result * N; end loop; return Result; end MultiFact; begin for deg in 1..5 loop Put("Degree"& Integer'Image(deg) &":"); for num in 1..10 loop Put(Integer'Image(MultiFact(num,deg))); end loop; New_line; end loop; end Mfact;
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Transform the following Arturo implementation into C, maintaining the same output and logic.
multifact: function [n deg][ if? n =< deg -> n else -> n * multifact n-deg deg ] loop 1..5 'i [ prints ["Degree" i ":"] loop 1..10 'j [ prints [multifact j i " "] ] print "" ]
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Keep all operations the same but rewrite the snippet in C.
multifact: function [n deg][ if? n =< deg -> n else -> n * multifact n-deg deg ] loop 1..5 'i [ prints ["Degree" i ":"] loop 1..10 'j [ prints [multifact j i " "] ] print "" ]
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Write a version of this Arturo function in C# with identical behavior.
multifact: function [n deg][ if? n =< deg -> n else -> n * multifact n-deg deg ] loop 1..5 'i [ prints ["Degree" i ":"] loop 1..10 'j [ prints [multifact j i " "] ] print "" ]
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Produce a language-to-language conversion: from Arturo to C#, same semantics.
multifact: function [n deg][ if? n =< deg -> n else -> n * multifact n-deg deg ] loop 1..5 'i [ prints ["Degree" i ":"] loop 1..10 'j [ prints [multifact j i " "] ] print "" ]
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Ensure the translated C++ code behaves exactly like the original Arturo snippet.
multifact: function [n deg][ if? n =< deg -> n else -> n * multifact n-deg deg ] loop 1..5 'i [ prints ["Degree" i ":"] loop 1..10 'j [ prints [multifact j i " "] ] print "" ]
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Transform the following Arturo implementation into C++, maintaining the same output and logic.
multifact: function [n deg][ if? n =< deg -> n else -> n * multifact n-deg deg ] loop 1..5 'i [ prints ["Degree" i ":"] loop 1..10 'j [ prints [multifact j i " "] ] print "" ]
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Port the provided Arturo code into Java while preserving the original functionality.
multifact: function [n deg][ if? n =< deg -> n else -> n * multifact n-deg deg ] loop 1..5 'i [ prints ["Degree" i ":"] loop 1..10 'j [ prints [multifact j i " "] ] print "" ]
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Rewrite this program in Java while keeping its functionality equivalent to the Arturo version.
multifact: function [n deg][ if? n =< deg -> n else -> n * multifact n-deg deg ] loop 1..5 'i [ prints ["Degree" i ":"] loop 1..10 'j [ prints [multifact j i " "] ] print "" ]
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Change the programming language of this snippet from Arturo to Python without modifying what it does.
multifact: function [n deg][ if? n =< deg -> n else -> n * multifact n-deg deg ] loop 1..5 'i [ prints ["Degree" i ":"] loop 1..10 'j [ prints [multifact j i " "] ] print "" ]
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Port the provided Arturo code into Python while preserving the original functionality.
multifact: function [n deg][ if? n =< deg -> n else -> n * multifact n-deg deg ] loop 1..5 'i [ prints ["Degree" i ":"] loop 1..10 'j [ prints [multifact j i " "] ] print "" ]
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Rewrite the snippet below in VB so it works the same as the original Arturo code.
multifact: function [n deg][ if? n =< deg -> n else -> n * multifact n-deg deg ] loop 1..5 'i [ prints ["Degree" i ":"] loop 1..10 'j [ prints [multifact j i " "] ] print "" ]
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Change the following Arturo code into VB without altering its purpose.
multifact: function [n deg][ if? n =< deg -> n else -> n * multifact n-deg deg ] loop 1..5 'i [ prints ["Degree" i ":"] loop 1..10 'j [ prints [multifact j i " "] ] print "" ]
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Convert this Arturo snippet to Go and keep its semantics consistent.
multifact: function [n deg][ if? n =< deg -> n else -> n * multifact n-deg deg ] loop 1..5 'i [ prints ["Degree" i ":"] loop 1..10 'j [ prints [multifact j i " "] ] print "" ]
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Write the same algorithm in Go as shown in this Arturo implementation.
multifact: function [n deg][ if? n =< deg -> n else -> n * multifact n-deg deg ] loop 1..5 'i [ prints ["Degree" i ":"] loop 1..10 'j [ prints [multifact j i " "] ] print "" ]
package main import "fmt" func multiFactorial(n, k int) int { r := 1 for ; n > 1; n -= k { r *= n } return r } func main() { for k := 1; k <= 5; k++ { fmt.Print("degree ", k, ":") for n := 1; n <= 10; n++ { fmt.Print(" ", multiFactorial(n, k)) } fmt.Println() } }
Port the following code from AutoHotKey to C with equivalent syntax and logic.
Loop, 5 { Output .= "Degree " (i := A_Index) ": " Loop, 10 Output .= MultiFact(A_Index, i) (A_Index = 10 ? "`n" : ", ") } MsgBox, % Output MultiFact(n, d) { Result := n while 1 < n -= d Result *= n return, Result }
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Ensure the translated C code behaves exactly like the original AutoHotKey snippet.
Loop, 5 { Output .= "Degree " (i := A_Index) ": " Loop, 10 Output .= MultiFact(A_Index, i) (A_Index = 10 ? "`n" : ", ") } MsgBox, % Output MultiFact(n, d) { Result := n while 1 < n -= d Result *= n return, Result }
#include <stdio.h> #define HIGHEST_DEGREE 5 #define LARGEST_NUMBER 10 int multifact(int n, int deg){ return n <= deg ? n : n * multifact(n - deg, deg); } int multifact_i(int n, int deg){ int result = n; while (n >= deg + 1){ result *= (n - deg); n -= deg; } return result; } int main(void){ int i, j; for (i = 1; i <= HIGHEST_DEGREE; i++){ printf("\nDegree %d: ", i); for (j = 1; j <= LARGEST_NUMBER; j++){ printf("%d ", multifact(j, i)); } } }
Produce a functionally identical C# code for the snippet given in AutoHotKey.
Loop, 5 { Output .= "Degree " (i := A_Index) ": " Loop, 10 Output .= MultiFact(A_Index, i) (A_Index = 10 ? "`n" : ", ") } MsgBox, % Output MultiFact(n, d) { Result := n while 1 < n -= d Result *= n return, Result }
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Port the following code from AutoHotKey to C# with equivalent syntax and logic.
Loop, 5 { Output .= "Degree " (i := A_Index) ": " Loop, 10 Output .= MultiFact(A_Index, i) (A_Index = 10 ? "`n" : ", ") } MsgBox, % Output MultiFact(n, d) { Result := n while 1 < n -= d Result *= n return, Result }
namespace RosettaCode.Multifactorial { using System; using System.Linq; internal static class Program { private static void Main() { Console.WriteLine(string.Join(Environment.NewLine, Enumerable.Range(1, 5) .Select( degree => string.Join(" ", Enumerable.Range(1, 10) .Select( number => Multifactorial(number, degree)))))); } private static int Multifactorial(int number, int degree) { if (degree < 1) { throw new ArgumentOutOfRangeException("degree"); } var count = 1 + (number - 1) / degree; if (count < 1) { throw new ArgumentOutOfRangeException("number"); } return Enumerable.Range(0, count) .Aggregate(1, (accumulator, index) => accumulator * (number - degree * index)); } } }
Generate a C++ translation of this AutoHotKey snippet without changing its computational steps.
Loop, 5 { Output .= "Degree " (i := A_Index) ": " Loop, 10 Output .= MultiFact(A_Index, i) (A_Index = 10 ? "`n" : ", ") } MsgBox, % Output MultiFact(n, d) { Result := n while 1 < n -= d Result *= n return, Result }
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Port the following code from AutoHotKey to C++ with equivalent syntax and logic.
Loop, 5 { Output .= "Degree " (i := A_Index) ": " Loop, 10 Output .= MultiFact(A_Index, i) (A_Index = 10 ? "`n" : ", ") } MsgBox, % Output MultiFact(n, d) { Result := n while 1 < n -= d Result *= n return, Result }
#include <algorithm> #include <iostream> #include <iterator> int main(void) { for (int g = 1; g < 10; g++) { int v[11], n=0; generate_n(std::ostream_iterator<int>(std::cout, " "), 10, [&]{n++; return v[n]=(g<n)? v[n-g]*n : n;}); std::cout << std::endl; } return 0; }
Convert this AutoHotKey snippet to Java and keep its semantics consistent.
Loop, 5 { Output .= "Degree " (i := A_Index) ": " Loop, 10 Output .= MultiFact(A_Index, i) (A_Index = 10 ? "`n" : ", ") } MsgBox, % Output MultiFact(n, d) { Result := n while 1 < n -= d Result *= n return, Result }
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Port the provided AutoHotKey code into Java while preserving the original functionality.
Loop, 5 { Output .= "Degree " (i := A_Index) ": " Loop, 10 Output .= MultiFact(A_Index, i) (A_Index = 10 ? "`n" : ", ") } MsgBox, % Output MultiFact(n, d) { Result := n while 1 < n -= d Result *= n return, Result }
public class MultiFact { private static long multiFact(long n, int deg){ long ans = 1; for(long i = n; i > 0; i -= deg){ ans *= i; } return ans; } public static void main(String[] args){ for(int deg = 1; deg <= 5; deg++){ System.out.print("degree " + deg + ":"); for(long n = 1; n <= 10; n++){ System.out.print(" " + multiFact(n, deg)); } System.out.println(); } } }
Transform the following AutoHotKey implementation into Python, maintaining the same output and logic.
Loop, 5 { Output .= "Degree " (i := A_Index) ": " Loop, 10 Output .= MultiFact(A_Index, i) (A_Index = 10 ? "`n" : ", ") } MsgBox, % Output MultiFact(n, d) { Result := n while 1 < n -= d Result *= n return, Result }
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Rewrite this program in Python while keeping its functionality equivalent to the AutoHotKey version.
Loop, 5 { Output .= "Degree " (i := A_Index) ": " Loop, 10 Output .= MultiFact(A_Index, i) (A_Index = 10 ? "`n" : ", ") } MsgBox, % Output MultiFact(n, d) { Result := n while 1 < n -= d Result *= n return, Result }
>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m)) >>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)])) 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>>
Rewrite the snippet below in VB so it works the same as the original AutoHotKey code.
Loop, 5 { Output .= "Degree " (i := A_Index) ": " Loop, 10 Output .= MultiFact(A_Index, i) (A_Index = 10 ? "`n" : ", ") } MsgBox, % Output MultiFact(n, d) { Result := n while 1 < n -= d Result *= n return, Result }
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next
Ensure the translated VB code behaves exactly like the original AutoHotKey snippet.
Loop, 5 { Output .= "Degree " (i := A_Index) ": " Loop, 10 Output .= MultiFact(A_Index, i) (A_Index = 10 ? "`n" : ", ") } MsgBox, % Output MultiFact(n, d) { Result := n while 1 < n -= d Result *= n return, Result }
Function multifactorial(n,d) If n = 0 Then multifactorial = 1 Else For i = n To 1 Step -d If i = n Then multifactorial = n Else multifactorial = multifactorial * i End If Next End If End Function For j = 1 To 5 WScript.StdOut.Write "Degree " & j & ": " For k = 1 To 10 If k = 10 Then WScript.StdOut.Write multifactorial(k,j) Else WScript.StdOut.Write multifactorial(k,j) & " " End If Next WScript.StdOut.WriteLine Next