Solve hanging lantern problem: Difference between revisions
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There are some columns of lanterns hanging from the ceiling. If you remove the lanterns one at a time, at each step removing the bottommost lantern from one column, how many legal sequences will let you take all of the lanterns down? |
There are some columns of lanterns hanging from the ceiling. If you remove the lanterns one at a time, at each step removing the bottommost lantern from one column, how many legal sequences will let you take all of the lanterns down? |
Revision as of 00:19, 19 June 2022
There are some columns of lanterns hanging from the ceiling. If you remove the lanterns one at a time, at each step removing the bottommost lantern from one column, how many legal sequences will let you take all of the lanterns down?
For example, there are some lanterns hanging like this:
🏮 🏮 🏮 🏮 🏮 🏮
If we number the lanterns like so:
1 2 4 3 5 6
You can take like this: [6,3,5,2,4,1] or [3,1,6,5,2,4]
But not like this: [6,3,2,4,5,1], because at that time 5 is under 4.
In total, there are 60 ways to take them down.
- Task
Input:
First an integer (n): the number of columns.
Then n integers: the number of lanterns in each column.
Output:
An integer: the number of sequences.
For example, the input of the example above could be:
3 1 2 3
And the output is:
60
- Optional task
Output all the sequences using this format:
[1,2,3,…] [2,1,3,…] ……
APL
<lang apl>lanterns ← { (!+/⍵) ÷ ×/!⍵ }</lang>
- Output:
lanterns 1 2 3 60 lanterns 1 3 3 140
Of course, for the simple sequences from 1, we can use iota to generate them instead of typing them out:
lanterns ⍳3 ⍝ same as lanterns 1 2 3 60 lanterns ⍳4 12600 lanterns ⍳5 37837800
BASIC
BASIC256
The result for n >= 5 is slow to emerge <lang freebasic>arraybase 1 n = 4 dim a(n) for i = 1 to a[?]
a[i] = i print "[ "; for j = 1 to i print a[j]; " "; next j print "] = "; getLantern(a)
next i end
function getLantern(arr)
res = 0 for i = 1 to arr[?] if arr[i] <> 0 then arr[i] -= 1 res += getLantern(arr) arr[i] += 1 end if next i if res = 0 then res = 1 return res
end function</lang>
- Output:
Same as FreeBASIC entry.
Commodore BASIC
The (1,2,3) example takes about 30 seconds to run on a stock C64; (1,2,3,4) takes about an hour and 40 minutes. Even on a 64 equipped with a 20MHz SuperCPU it takes about 5 minutes. <lang basic>100 PRINT CHR$(147);CHR$(18);"*** HANGING LANTERN PROBLEM ***" 110 INPUT "HOW MANY COLUMNS "; N 120 DIM NL(N-1):T=0 130 FOR I=0 TO N-1 140 : PRINT "HOW MANY LANTERNS IN COLUMN"I+1; 150 : INPUT NL(I):T=T+NL(I) 160 NEXT I 170 DIM I(T),R(T) 180 SP=0 190 GOSUB 300 200 PRINT R(0) 220 END 300 R(SP)=0 310 I(SP)=0 320 IF I(SP)=N THEN 420 330 IF NL(I(SP))=0 THEN 400 340 NL(I(SP))=NL(I(SP))-1 350 SP=SP+1 360 GOSUB 300 370 SP=SP-1 370 R(SP)=R(SP)+R(SP+1) 390 NL(I(SP))=NL(I(SP))+1 400 I(SP)=I(SP)+1 410 GOTO 320 420 IF R(SP)=0 THEN R(SP)=1 430 RETURN</lang>
- Output:
*** HANGING LANTERN PROBLEM *** HOW MANY COLUMNS ? 4 HOW MANY LANTERNS IN COLUMN 1 ? 1 HOW MANY LANTERNS IN COLUMN 2 ? 2 HOW MANY LANTERNS IN COLUMN 3 ? 3 HOW MANY LANTERNS IN COLUMN 4 ? 4 12600
FreeBASIC
<lang freebasic>Function getLantern(arr() As Uinteger) As Ulong
Dim As Ulong res = 0 For i As Ulong = 1 To Ubound(arr) If arr(i) <> 0 Then arr(i) -= 1 res += getLantern(arr()) arr(i) += 1 End If Next i If res = 0 Then res = 1 Return res
End Function
Dim As Uinteger n = 5 Dim As Uinteger a(n) 'Dim As Integer a(6) = {1,2,3,4,5,6} For i As Ulong = 1 To Ubound(a)
a(i) = i Print "[ "; For j As Ulong = 1 To i Print a(j); " "; Next j Print "] = "; getLantern(a())
Next i Sleep</lang>
- Output:
[ 1 ] = 1 [ 1 2 ] = 3 [ 1 2 3 ] = 60 [ 1 2 3 4 ] = 12600 [ 1 2 3 4 5 ] = 37837800
QBasic
The result for n >= 5 is slow to emerge <lang QBasic>FUNCTION getLantern (arr())
res = 0 FOR i = 1 TO UBOUND(arr) IF arr(i) <> 0 THEN arr(i) = arr(i) - 1 res = res + getLantern(arr()) arr(i) = arr(i) + 1 END IF NEXT i IF res = 0 THEN res = 1 getLantern = res
END FUNCTION
n = 4 DIM a(n) FOR i = 1 TO UBOUND(a)
a(i) = i PRINT "["; FOR j = 1 TO i PRINT a(j); " "; NEXT j PRINT "] = "; getLantern(a())
NEXT i END</lang>
- Output:
Same as FreeBASIC entry.
PureBasic
The result for n >= 5 is slow to emerge <lang PureBasic>;;The result For n >= 5 is slow To emerge Procedure getLantern(Array arr(1))
res.l = 0 For i.l = 1 To ArraySize(arr(),1) If arr(i) <> 0 arr(i) - 1 res + getLantern(arr()) arr(i) + 1 EndIf Next i If res = 0 res = 1 EndIf ProcedureReturn res
EndProcedure
OpenConsole() n.i = 4 Dim a.i(n) For i.l = 1 To ArraySize(a())
a(i) = i Print("[") For j.l = 1 To i Print(Str(a(j)) + " ") Next j PrintN("] = " + Str(getLantern(a())))
Next i Input() CloseConsole()</lang>
- Output:
Same as FreeBASIC entry.
VBA
- See Visual Basic
Visual Basic
Note: Integer may overflow if the input number is too big. To solve this problem, simply change Integer to Long or Variant for Decimal.
Recursive version
- Main code
<lang vb> Dim n As Integer, c As Integer Dim a() As Integer
Private Sub Command1_Click()
Dim res As Integer If c < n Then Label3.Caption = "Please input completely.": Exit Sub res = getLantern(a()) Label3.Caption = "Result:" + Str(res)
End Sub
Private Sub Text1_Change()
If Val(Text1.Text) <> 0 Then n = Val(Text1.Text) ReDim a(1 To n) As Integer End If
End Sub
Private Sub Text2_KeyPress(KeyAscii As Integer)
If KeyAscii = Asc(vbCr) Then If Val(Text2.Text) = 0 Then Exit Sub c = c + 1 If c > n Then Exit Sub a(c) = Val(Text2.Text) List1.AddItem Str(a(c)) Text2.Text = "" End If
End Sub
Function getLantern(arr() As Integer) As Integer
Dim res As Integer, i As Integer For i = 1 To n If arr(i) <> 0 Then arr(i) = arr(i) - 1 res = res + getLantern(arr()) arr(i) = arr(i) + 1 End If Next i If res = 0 Then res = 1 getLantern = res
End Function</lang>
- Form code
<lang vb> VERSION 5.00 Begin VB.Form Form1
Caption = "Get Lantern" ClientHeight = 4410 ClientLeft = 120 ClientTop = 465 ClientWidth = 6150 LinkTopic = "Form1" ScaleHeight = 4410 ScaleWidth = 6150 StartUpPosition = 3 Begin VB.CommandButton Command1 Caption = "Start" Height = 495 Left = 2040 TabIndex = 5 Top = 3000 Width = 1935 End Begin VB.ListBox List1 Height = 1320 Left = 360 TabIndex = 4 Top = 1440 Width = 5175 End Begin VB.TextBox Text2 Height = 855 Left = 3360 TabIndex = 1 Top = 480 Width = 2175 End Begin VB.TextBox Text1 Height = 855 Left = 360 TabIndex = 0 Top = 480 Width = 2175 End Begin VB.Label Label3 Height = 495 Left = 2040 TabIndex = 6 Top = 3720 Width = 2295 End Begin VB.Label Label2 Caption = "Number Each" Height = 375 Left = 3960 TabIndex = 3 Top = 120 Width = 1695 End Begin VB.Label Label1 Caption = "Total" Height = 255 Left = 960 TabIndex = 2 Top = 120 Width = 1455 End
End Attribute VB_Name = "Form1" Attribute VB_GlobalNameSpace = False Attribute VB_Creatable = False Attribute VB_PredeclaredId = True Attribute VB_Exposed = False</lang>
Math solution
Reimplemented "getLantern" function above
<lang vb>Function getLantern(arr() As Integer) As Integer
Dim tot As Integer, res As Integer Dim i As Integer For i = 1 To n tot = tot + arr(i) Next i res = factorial(tot) For i = 1 To n res = res / factorial(arr(i)) Next i getLantern = res
End Function
Function factorial(num As Integer) As Integer
Dim i As Integer factorial = 1 For i = 2 To n factorial = factorial * i Next i
End Function</lang>
Yabasic
The result for n >= 5 is slow to emerge <lang yabasic>n = 4 dim a(n) for i = 1 to arraysize(a(),1)
a(i) = i print "[ "; for j = 1 to i print a(j), " "; next j print "] = ", getLantern(a())
next i
sub getLantern(arr())
local res, i res = 0 for i = 1 to arraysize(arr(),1) if arr(i) <> 0 then arr(i) = arr(i) - 1 res = res + getLantern(arr()) arr(i) = arr(i) + 1 fi next i if res = 0 res = 1 return res
end sub</lang>
- Output:
Same as FreeBASIC entry.
J
Translation of APL:
<lang J>lanterns=: Template:(!+/y) % */!y<</lang>
Example use:
<lang J> lanterns 1 2 3 60
lanterns 1 3 3
140 </lang>
Also, a pedantic version where we must manually count how many values we are providing the computer:
<lang J>pedantic=: {{
assert. ({. = #@}.) y lanterns }.y
}}</lang>
And, in the spirit of providing unnecessary but perhaps pleasant (for some) overhead, we'll throw in an unnecessary comma between this count and the relevant values:
<lang J> pedantic 3, 1 2 3 60
pedantic 3, 1 3 3
140</lang>
If we wanted to impose even more overhead, we could insist that the numbers be read from a file where tabs, spaces and newlines are all treated equivalently. For that, we must specify the file name and implement some parsing:
<lang J>yetmoreoverhead=: {{
pedantic ({.~ 1+{.) _ ". rplc&(TAB,' ',LF,' ') fread y
}}</lang>
Examples of this approach are left as an exercise for the user (note: do not use commas with this version, unless you modify the code to treat them as whitespace).
Finally, enumerating solutions might be approached recursively:
<lang J>showlanterns=: {{
arrange=. ($ $ (* +/\)@,) y $&>1 echo 'lantern ids:' echo rplc&(' 0';' ')"1 ' ',.":|:arrange echo cols=. <@-.&0"1 arrange recur=: <@Template:Todo=. ("0 1 echo 'all lantern removal sequences:' echo >a:-.~ -.&0 each;0 recur cols
}}</lang>
Example use:
<lang J> showlanterns 1 2 1 lantern ids:
1 2 4 3
all lantern removal sequences: 1 3 2 4 1 3 4 2 1 4 3 2 3 1 2 4 3 1 4 2 3 2 1 4 3 2 4 1 3 4 1 2 3 4 2 1 4 1 3 2 4 3 1 2 4 3 2 1</lang>
Julia
<lang ruby>""" rosettacode.org /wiki/Lantern_Problem """
using Combinatorics
function lanternproblem(verbose = true)
println("Input number of columns, then column heights in sequence:") inputs = [parse(Int, i) for i in split(readline(), r"\s+")] n = popfirst!(inputs) println("\nThere are ", multinomial(BigInt.(inputs)...), " ways to take these ", n, " columns down.") if verbose idx, fullmat = 0, zeros(Int, n, maximum(n)) for col in 1:size(fullmat, 2), row in 1:size(fullmat, 1) if inputs[col] >= row fullmat[row, col] = (idx += 1) end end show(stdout, "text/plain", map(n -> n > 0 ? "$n " : " ", fullmat)) println("\n") takedownways = unique(permutations(reduce(vcat, [fill(i, m) for (i, m) in enumerate(inputs)]))) for way in takedownways print("[") mat = copy(fullmat) for (i, col) in enumerate(way) row = findlast(>(0), @view mat[:, col]) print(mat[row, col], i == length(way) ? "]\n" : ", ") mat[row, col] = 0 end end end
end
lanternproblem() lanternproblem() lanternproblem(false)
</lang>
- Output:
Input number of columns, then column heights in sequence: 3 1 2 3 There are 60 ways to take these 3 columns down. 3×3 Matrix{String}: "1 " "2 " "4 " " " "3 " "5 " " " " " "6 " [1, 3, 2, 6, 5, 4] [1, 3, 6, 2, 5, 4] [1, 3, 6, 5, 2, 4] [1, 3, 6, 5, 4, 2] [1, 6, 3, 2, 5, 4] [1, 6, 3, 5, 2, 4] [1, 6, 3, 5, 4, 2] [1, 6, 5, 3, 2, 4] [1, 6, 5, 3, 4, 2] [1, 6, 5, 4, 3, 2] [3, 1, 2, 6, 5, 4] [3, 1, 6, 2, 5, 4] [3, 1, 6, 5, 2, 4] [3, 1, 6, 5, 4, 2] [3, 2, 1, 6, 5, 4] [3, 2, 6, 1, 5, 4] [3, 2, 6, 5, 1, 4] [3, 2, 6, 5, 4, 1] [3, 6, 1, 2, 5, 4] [3, 6, 1, 5, 2, 4] [3, 6, 1, 5, 4, 2] [3, 6, 2, 1, 5, 4] [3, 6, 2, 5, 1, 4] [3, 6, 2, 5, 4, 1] [3, 6, 5, 1, 2, 4] [3, 6, 5, 1, 4, 2] [3, 6, 5, 2, 1, 4] [3, 6, 5, 2, 4, 1] [3, 6, 5, 4, 1, 2] [3, 6, 5, 4, 2, 1] [6, 1, 3, 2, 5, 4] [6, 1, 3, 5, 2, 4] [6, 1, 3, 5, 4, 2] [6, 1, 5, 3, 2, 4] [6, 1, 5, 3, 4, 2] [6, 1, 5, 4, 3, 2] [6, 3, 1, 2, 5, 4] [6, 3, 1, 5, 2, 4] [6, 3, 1, 5, 4, 2] [6, 3, 2, 1, 5, 4] [6, 3, 2, 5, 1, 4] [6, 3, 2, 5, 4, 1] [6, 3, 5, 1, 2, 4] [6, 3, 5, 1, 4, 2] [6, 3, 5, 2, 1, 4] [6, 3, 5, 2, 4, 1] [6, 3, 5, 4, 1, 2] [6, 3, 5, 4, 2, 1] [6, 5, 1, 3, 2, 4] [6, 5, 1, 3, 4, 2] [6, 5, 1, 4, 3, 2] [6, 5, 3, 1, 2, 4] [6, 5, 3, 1, 4, 2] [6, 5, 3, 2, 1, 4] [6, 5, 3, 2, 4, 1] [6, 5, 3, 4, 1, 2] [6, 5, 3, 4, 2, 1] [6, 5, 4, 1, 3, 2] [6, 5, 4, 3, 1, 2] [6, 5, 4, 3, 2, 1] Input number of columns, then column heights in sequence: 3 1 3 3 There are 140 ways to take these 3 columns down. 3×3 Matrix{String}: "1 " "2 " "5 " " " "3 " "6 " " " "4 " "7 " [1, 4, 3, 2, 7, 6, 5] [1, 4, 3, 7, 2, 6, 5] [1, 4, 3, 7, 6, 2, 5] [1, 4, 3, 7, 6, 5, 2] [1, 4, 7, 3, 2, 6, 5] [1, 4, 7, 3, 6, 2, 5] [1, 4, 7, 3, 6, 5, 2] [1, 4, 7, 6, 3, 2, 5] [1, 4, 7, 6, 3, 5, 2] [1, 4, 7, 6, 5, 3, 2] [1, 7, 4, 3, 2, 6, 5] [1, 7, 4, 3, 6, 2, 5] [1, 7, 4, 3, 6, 5, 2] [1, 7, 4, 6, 3, 2, 5] [1, 7, 4, 6, 3, 5, 2] [1, 7, 4, 6, 5, 3, 2] [1, 7, 6, 4, 3, 2, 5] [1, 7, 6, 4, 3, 5, 2] [1, 7, 6, 4, 5, 3, 2] [1, 7, 6, 5, 4, 3, 2] [4, 1, 3, 2, 7, 6, 5] [4, 1, 3, 7, 2, 6, 5] [4, 1, 3, 7, 6, 2, 5] [4, 1, 3, 7, 6, 5, 2] [4, 1, 7, 3, 2, 6, 5] [4, 1, 7, 3, 6, 2, 5] [4, 1, 7, 3, 6, 5, 2] [4, 1, 7, 6, 3, 2, 5] [4, 1, 7, 6, 3, 5, 2] [4, 1, 7, 6, 5, 3, 2] [4, 3, 1, 2, 7, 6, 5] [4, 3, 1, 7, 2, 6, 5] [4, 3, 1, 7, 6, 2, 5] [4, 3, 1, 7, 6, 5, 2] [4, 3, 2, 1, 7, 6, 5] [4, 3, 2, 7, 1, 6, 5] [4, 3, 2, 7, 6, 1, 5] [4, 3, 2, 7, 6, 5, 1] [4, 3, 7, 1, 2, 6, 5] [4, 3, 7, 1, 6, 2, 5] [4, 3, 7, 1, 6, 5, 2] [4, 3, 7, 2, 1, 6, 5] [4, 3, 7, 2, 6, 1, 5] [4, 3, 7, 2, 6, 5, 1] [4, 3, 7, 6, 1, 2, 5] [4, 3, 7, 6, 1, 5, 2] [4, 3, 7, 6, 2, 1, 5] [4, 3, 7, 6, 2, 5, 1] [4, 3, 7, 6, 5, 1, 2] [4, 3, 7, 6, 5, 2, 1] [4, 7, 1, 3, 2, 6, 5] [4, 7, 1, 3, 6, 2, 5] [4, 7, 1, 3, 6, 5, 2] [4, 7, 1, 6, 3, 2, 5] [4, 7, 1, 6, 3, 5, 2] [4, 7, 1, 6, 5, 3, 2] [4, 7, 3, 1, 2, 6, 5] [4, 7, 3, 1, 6, 2, 5] [4, 7, 3, 1, 6, 5, 2] [4, 7, 3, 2, 1, 6, 5] [4, 7, 3, 2, 6, 1, 5] [4, 7, 3, 2, 6, 5, 1] [4, 7, 3, 6, 1, 2, 5] [4, 7, 3, 6, 1, 5, 2] [4, 7, 3, 6, 2, 1, 5] [4, 7, 3, 6, 2, 5, 1] [4, 7, 3, 6, 5, 1, 2] [4, 7, 3, 6, 5, 2, 1] [4, 7, 6, 1, 3, 2, 5] [4, 7, 6, 1, 3, 5, 2] [4, 7, 6, 1, 5, 3, 2] [4, 7, 6, 3, 1, 2, 5] [4, 7, 6, 3, 1, 5, 2] [4, 7, 6, 3, 2, 1, 5] [4, 7, 6, 3, 2, 5, 1] [4, 7, 6, 3, 5, 1, 2] [4, 7, 6, 3, 5, 2, 1] [4, 7, 6, 5, 1, 3, 2] [4, 7, 6, 5, 3, 1, 2] [4, 7, 6, 5, 3, 2, 1] [7, 1, 4, 3, 2, 6, 5] [7, 1, 4, 3, 6, 2, 5] [7, 1, 4, 3, 6, 5, 2] [7, 1, 4, 6, 3, 2, 5] [7, 1, 4, 6, 3, 5, 2] [7, 1, 4, 6, 5, 3, 2] [7, 1, 6, 4, 3, 2, 5] [7, 1, 6, 4, 3, 5, 2] [7, 1, 6, 4, 5, 3, 2] [7, 1, 6, 5, 4, 3, 2] [7, 4, 1, 3, 2, 6, 5] [7, 4, 1, 3, 6, 2, 5] [7, 4, 1, 3, 6, 5, 2] [7, 4, 1, 6, 3, 2, 5] [7, 4, 1, 6, 3, 5, 2] [7, 4, 1, 6, 5, 3, 2] [7, 4, 3, 1, 2, 6, 5] [7, 4, 3, 1, 6, 2, 5] [7, 4, 3, 1, 6, 5, 2] [7, 4, 3, 2, 1, 6, 5] [7, 4, 3, 2, 6, 1, 5] [7, 4, 3, 2, 6, 5, 1] [7, 4, 3, 6, 1, 2, 5] [7, 4, 3, 6, 1, 5, 2] [7, 4, 3, 6, 2, 1, 5] [7, 4, 3, 6, 2, 5, 1] [7, 4, 3, 6, 5, 1, 2] [7, 4, 3, 6, 5, 2, 1] [7, 4, 6, 1, 3, 2, 5] [7, 4, 6, 1, 3, 5, 2] [7, 4, 6, 1, 5, 3, 2] [7, 4, 6, 3, 1, 2, 5] [7, 4, 6, 3, 1, 5, 2] [7, 4, 6, 3, 2, 1, 5] [7, 4, 6, 3, 2, 5, 1] [7, 4, 6, 3, 5, 1, 2] [7, 4, 6, 3, 5, 2, 1] [7, 4, 6, 5, 1, 3, 2] [7, 4, 6, 5, 3, 1, 2] [7, 4, 6, 5, 3, 2, 1] [7, 6, 1, 4, 3, 2, 5] [7, 6, 1, 4, 3, 5, 2] [7, 6, 1, 4, 5, 3, 2] [7, 6, 1, 5, 4, 3, 2] [7, 6, 4, 1, 3, 2, 5] [7, 6, 4, 1, 3, 5, 2] [7, 6, 4, 1, 5, 3, 2] [7, 6, 4, 3, 1, 2, 5] [7, 6, 4, 3, 1, 5, 2] [7, 6, 4, 3, 2, 1, 5] [7, 6, 4, 3, 2, 5, 1] [7, 6, 4, 3, 5, 1, 2] [7, 6, 4, 3, 5, 2, 1] [7, 6, 4, 5, 1, 3, 2] [7, 6, 4, 5, 3, 1, 2] [7, 6, 4, 5, 3, 2, 1] [7, 6, 5, 1, 4, 3, 2] [7, 6, 5, 4, 1, 3, 2] [7, 6, 5, 4, 3, 1, 2] [7, 6, 5, 4, 3, 2, 1] Input number of columns, then column heights in sequence: 9 1 2 3 4 5 6 7 8 9 There are 65191584694745586153436251091200000 ways to take these 9 columns down.
Pascal
A console application in Free Pascal, created with the Lazarus IDE.
This solution avoids recursion and calculates the result mathematically. As noted in the Picat solution, the result is a multinomial coefficient, e.g. with columns of length 3, 6, 4 the result is (3 + 6 + 4)!/(3!*6!*4!). <lang pascal> program LanternProblem; uses SysUtils;
// Calculate multinomial coefficient, e.g. input array [3, 6, 4] // would give (3 + 6 + 4)! / (3!*6!*4!). // Result is calculated as a product of binomial coefficients. function Multinomial( a : array of integer) : UInt64; var
n, i, j, k : integer; b : array of integer; // sorted copy of ionput row : array of integer; // start of row in Pascal's triangle
begin
result := 1; // in case of trivial input n := Length( a); if (n <= 1) then exit;
// Copy caller's array to local array in descending order SetLength( b, n); for j := 0 to n - 1 do begin k := j; while (k > 0) and (b[k - 1] < a[j]) do begin b[k] := b[k - 1]; dec(k); end; b[k] := a[j]; end;
// Zero entries don't affect the result, so remove them while (n > 0) and (b[n - 1] = 0) do dec(n); if (n <= 1) then exit;
// Non-trivial input, do the calculation by means of Pascal's triangle SetLength( row, b[1] + 1); row[0] := 1; for k := 1 to b[1] do row[k] := 0; for i := 1 to b[0] + b[1] do begin for k := b[1] downto 1 do inc( row[k], row[k - 1]); end; result := row[b[1]]; // first binomial coefficient
// Since b[1] >= b[2] >= b[3] ... there are always enough valid terms // in the row to allow calculation of the next binomial coefficient. for j := 2 to n - 1 do begin for i := 1 to b[j] do begin for k := b[1] downto 1 do inc( row[k], row[k - 1]); end; result := result*row[b[j]]; // multiply by next binomial coefficient end;
end;
// Prompt user for non-negative integer. // Avoid raising exception when user input isn't an integer. function UserInt( const prompt : string) : integer; var
userInput : string; inputOK : boolean;
begin
repeat Write( prompt, ' '); ReadLn(userInput); inputOK := SysUtils.TryStrToInt( userInput, result) and (result >= 0); if not inputOK then WriteLn( 'Try again'); until inputOK;
end;
// Main routine var
nrCols, j : integer; counts : array of integer;
begin
repeat nrCols := UserInt( 'Number of columns (0 to quit)?'); if nrCols = 0 then exit; SetLength( counts, nrCols); for j := 0 to nrCols - 1 do counts[j] := UserInt( SysUtils.Format('How many in column %d?', [j + 1])); // column numbers 1-based for user Write( 'Columns are '); for j := 0 to nrCols - 1 do Write(' ', counts[j]); WriteLn( ', number of ways = ', Multinomial(counts)); until false;
end. </lang>
- Output:
Number of columns (0 to quit)? 3 How many in column 1? 1 How many in column 2? 2 How many in column 3? 3 Columns are 1 2 3, number of ways = 60 [input omitted from now on] Columns are 1 2 3 4, number of ways = 12600 Columns are 1 2 3 4 5, number of ways = 37837800 Columns are 1 2 3 4 5 6, number of ways = 2053230379200 Columns are 1 2 3 4 5 6 7, number of ways = 2431106898187968000 Columns are 1 3 3, number of ways = 140
Perl
<lang perl>#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Solve_hanging_lantern_problem use warnings;
$_ = 'a bc def';
my $answer = ; find($_, ); print "count = @{[ $answer =~ tr/\n// ]}\n", $answer;
sub find
{ my ($in, $found) = @_; find( $` . $', $found . $& ) while $in =~ /\w\b/g; $in =~ /\w/ or $answer .= '[' . $found =~ s/\B/,/gr . "]\n"; }</lang>
- Output:
count = 60 [a,c,b,f,e,d] [a,c,f,b,e,d] [a,c,f,e,b,d] [a,c,f,e,d,b] [a,f,c,b,e,d] [a,f,c,e,b,d] [a,f,c,e,d,b] [a,f,e,c,b,d] [a,f,e,c,d,b] [a,f,e,d,c,b] [c,a,b,f,e,d] [c,a,f,b,e,d] [c,a,f,e,b,d] [c,a,f,e,d,b] [c,b,a,f,e,d] [c,b,f,a,e,d] [c,b,f,e,a,d] [c,b,f,e,d,a] [c,f,a,b,e,d] [c,f,a,e,b,d] [c,f,a,e,d,b] [c,f,b,a,e,d] [c,f,b,e,a,d] [c,f,b,e,d,a] [c,f,e,a,b,d] [c,f,e,a,d,b] [c,f,e,b,a,d] [c,f,e,b,d,a] [c,f,e,d,a,b] [c,f,e,d,b,a] [f,a,c,b,e,d] [f,a,c,e,b,d] [f,a,c,e,d,b] [f,a,e,c,b,d] [f,a,e,c,d,b] [f,a,e,d,c,b] [f,c,a,b,e,d] [f,c,a,e,b,d] [f,c,a,e,d,b] [f,c,b,a,e,d] [f,c,b,e,a,d] [f,c,b,e,d,a] [f,c,e,a,b,d] [f,c,e,a,d,b] [f,c,e,b,a,d] [f,c,e,b,d,a] [f,c,e,d,a,b] [f,c,e,d,b,a] [f,e,a,c,b,d] [f,e,a,c,d,b] [f,e,a,d,c,b] [f,e,c,a,b,d] [f,e,c,a,d,b] [f,e,c,b,a,d] [f,e,c,b,d,a] [f,e,c,d,a,b] [f,e,c,d,b,a] [f,e,d,a,c,b] [f,e,d,c,a,b] [f,e,d,c,b,a]
Phix
with javascript_semantics include mpfr.e function get_lantern(mpz z, sequence s, bool bJustCount=true, integer na=-1) if bJustCount then sequence l = apply(s,length) mpz_fac_ui(z,sum(l)) mpz f = mpz_init() for d in l do mpz_fac_ui(f,d) mpz_fdiv_q(z,z,f) end for return 0 end if if na=-1 then na = sum(apply(s,length)) end if if na=0 then mpz_set_si(z,1) return {""} end if s = deep_copy(s) sequence res = {} for i=1 to length(s) do if length(s[i]) then integer si = s[i][$] s[i] = s[i][1..$-1] mpz z2 = mpz_init() object r = get_lantern(z2, s, false, na-1) for k=1 to length(r) do res = append(res,si&r[k]) end for mpz_add(z,z,z2) s[i] &= si end if end for return res end function procedure test(sequence s, bool bJustCount=true) mpz z = mpz_init() object r = get_lantern(z,s,bJustCount) string sj = join(s,", "), sz = mpz_get_str(z) if bJustCount then printf(1,"%s = %s\n",{sj,sz}) else string rj = join_by(r,1,10,",") printf(1,"%s = %s:\n%s\n",{sj,sz,rj}) end if end procedure test({"1"},false) test({"1","23"},false) test({"1","23","456"},false) test({"1","234","567"}) test({"1234567890","ABCDEFGHIJKLMN","OPQRSTUVWXYZ"}) sequence s = {"1", "23", "456", "7890", "ABCDE", "FGHIJK", "LMNOPQR", "STUVWXYZ", "abcdefghi"} for i=1 to 9 do test(s[1..i]) end for
- Output:
1 = 1: 1 1, 23 = 3: 132,312,321 1, 23, 456 = 60: 132654,136254,136524,136542,163254,163524,163542,165324,165342,165432 312654,316254,316524,316542,321654,326154,326514,326541,361254,361524 361542,362154,362514,362541,365124,365142,365214,365241,365412,365421 613254,613524,613542,615324,615342,615432,631254,631524,631542,632154 632514,632541,635124,635142,635214,635241,635412,635421,651324,651342 651432,653124,653142,653214,653241,653412,653421,654132,654312,654321 1, 234, 567 = 140 1234567890, ABCDEFGHIJKLMN, OPQRSTUVWXYZ = 2454860399191200 1 = 1 1, 23 = 3 1, 23, 456 = 60 1, 23, 456, 7890 = 12600 1, 23, 456, 7890, ABCDE = 37837800 1, 23, 456, 7890, ABCDE, FGHIJK = 2053230379200 1, 23, 456, 7890, ABCDE, FGHIJK, LMNOPQR = 2431106898187968000 1, 23, 456, 7890, ABCDE, FGHIJK, LMNOPQR, STUVWXYZ = 73566121315513295589120000 1, 23, 456, 7890, ABCDE, FGHIJK, LMNOPQR, STUVWXYZ, abcdefghi = 65191584694745586153436251091200000
Picat
<lang Picat>main =>
run_lantern().
run_lantern() =>
N = read_int(), A = [], foreach(_ in 1..N) A := A ++ [read_int()] end, println(A), println(lantern(A)), nl.
table lantern(A) = Res =>
Arr = copy_term(A), Res = 0, foreach(I in 1..Arr.len) if Arr[I] != 0 then Arr[I] := Arr[I] - 1, Res := Res + lantern(Arr), Arr[I] := Arr[I] + 1 end end, if Res == 0 then Res := 1 end.</lang>
Some tests: <lang Picat>main =>
A = [1,2,3], println(lantern(A)), foreach(N in 1..8) println(1..N=lantern(1..N)) end, nl.</lang>
- Output:
60 [1] = 1 [1,2] = 3 [1,2,3] = 60 [1,2,3,4] = 12600 [1,2,3,4,5] = 37837800 [1,2,3,4,5,6] = 2053230379200 [1,2,3,4,5,6,7] = 2431106898187968000 [1,2,3,4,5,6,7,8] = 73566121315513295589120000
The sequence of lantern(1..N) is the OEIS sequence A022915 ("Multinomial coefficients (0, 1, ..., n)! = C(n+1,2)!/(0!*1!*2!*...*n!)").
Python
Recursive version
<lang python> def getLantern(arr):
res = 0 for i in range(0, n): if arr[i] != 0: arr[i] -= 1 res += getLantern(arr) arr[i] += 1 if res == 0: res = 1 return res
a = [] n = int(input()) for i in range(0, n):
a.append(int(input()))
print(getLantern(a)) </lang>
Math solution
<lang python> import math n = int(input()) a = [] tot = 0 for i in range(0, n):
a.append(int(input())) tot += a[i]
res = math.factorial(tot) for i in range(0, n):
res /= math.factorial(a[i])
print(int(res)) </lang>
Raku
Note: All of these solutions accept the list of column sizes as command-line arguments and infer the number of columns from the number of sizes provided, rather than requiring that a count be supplied as an extra distinct parameter.
Directly computing the count
If all we need is the count, then we can compute that directly:
<lang perl6>unit sub MAIN(*@columns);
sub postfix:<!>($n) { [*] 1..$n }
say [+](@columns)! / [*](@columns»!);</lang>
- Output:
$ raku lanterns.raku 1 2 3 60
Sequence as column numbers
If we want to list all of the sequences, we have to do some more work. This version outputs the sequences as lists of column numbers (assigned from 1 to N left to right); at each step the bottommost lantern from the numbered column is removed.
<lang perl6>unit sub MAIN(*@columns, :v(:$verbose)=False);
my @sequences = @columns
. pairs . map({ (.key+1) xx .value }) . flat . permutations . map( *.join(',') ) . unique;
if ($verbose) {
say "There are {+@sequences} possible takedown sequences:"; say "[$_]" for @sequences;
} else {
say +@sequences;
} </lang>
- Output:
$ raku lanterns.raku --verbose 1 2 3 There are 60 possible takedown sequences: [1,2,2,3,3,3] [1,2,3,2,3,3] [1,2,3,3,2,3] [1,2,3,3,3,2] [1,3,2,2,3,3] [1,3,2,3,2,3] ... [3,3,2,2,3,1] [3,3,2,3,1,2] [3,3,2,3,2,1] [3,3,3,1,2,2] [3,3,3,2,1,2] [3,3,3,2,2,1]
Sequence as lantern numbers
If we want individually-numbered lanterns in the sequence instead of column numbers, as in the example given in the task description, that requires yet more work:
<lang perl6>unit sub MAIN(*@columns, :v(:$verbose)=False);
my @sequences = @columns
. pairs . map({ (.key+1) xx .value }) . flat . permutations . map( *.join(',') ) . unique;
if ($verbose) {
my @offsets = |0,|(1..@columns).map: { [+] @columns[0..$_-1] }; my @matrix; for ^@columns.max -> $i { for ^@columns -> $j { my $value = $i < @columns[$j] ?? ($i+@offsets[$j]+1) !! Nil; @matrix[$j][$i] = $value if $value;; print "\t" ~ ($value // " "); } say ; } say "There are {+@sequences} possible takedown sequences:"; for @sequences».split(',') -> @seq { my @work = @matrix».clone; my $seq = '['; for @seq -> $col { $seq ~= @work[$col-1].pop ~ ','; } $seq ~~ s/','$/]/; say $seq; }
} else {
say +@sequences;
}</lang>
- Output:
$ raku lanterns.raku -v 1 2 3 1 2 4 3 5 6 There are 60 possible takedown sequences: [1,3,2,6,5,4] [1,3,6,2,5,4] [1,3,6,5,2,4] ... [6,5,4,1,3,2] [6,5,4,3,1,2] [6,5,4,3,2,1]
Wren
Version 1
The result for n == 5 is slow to emerge. <lang ecmascript>var lantern // recursive function lantern = Fn.new { |n, a|
var count = 0 for (i in 0...n) { if (a[i] != 0) { a[i] = a[i] - 1 count = count + lantern.call(n, a) a[i] = a[i] + 1 } } if (count == 0) count = 1 return count
}
System.print("Number of permutations for n (<= 5) groups and lanterns per group [1..n]:") var n = 0 for (i in 1..5) {
var a = (1..i).toList n = n + 1 System.print("%(a) => %(lantern.call(n, a))")
}</lang>
- Output:
Number of permutations for n (<= 5) groups and lanterns per group [1..n]: [1] => 1 [1, 2] => 3 [1, 2, 3] => 60 [1, 2, 3, 4] => 12600 [1, 2, 3, 4, 5] => 37837800
Version 2
Alternatively, using library methods. <lang ecmascript>import "./perm" for Perm import "./big" for BigInt
var listPerms = Fn.new { |a, rowSize|
var lows = List.filled(a.count, 0) var sum = 0 var mlist = [] for (i in 0...a.count) { sum = sum + a[i] lows[i] = sum mlist = mlist + [i+1] * a[i] } var n = Perm.countDistinct(sum, a) System.print("\nList of %(n) permutations for %(a.count) groups and lanterns per group %(a):") var count = 0 for (p in Perm.listDistinct(mlist)) { var curr = lows.toList var perm = List.filled(sum, 0) for (i in 0...sum) { perm[i] = curr[p[i]-1] curr[p[i]-1] = curr[p[i]-1] - 1 } System.write("%(perm) ") count = count + 1 if (count % rowSize == 0) System.print() } if (count % rowSize != 0) System.print()
}
System.print("Number of permutations for the lanterns per group shown:") var n = 0 for (i in 1..9) {
var a = (1..i).toList n = n + i System.print("%(a) => %(BigInt.multinomial(n, a))")
} var a = [1, 3, 3] System.print("%(a) => %(BigInt.multinomial(7, a))") a = [10, 14, 12] System.print("%(a) => %(BigInt.multinomial(36, a))") listPerms.call([1, 2, 3], 4) listPerms.call([1, 3, 3], 3)</lang>
- Output:
Number of permutations for the lanterns per group shown: [1] => 1 [1, 2] => 3 [1, 2, 3] => 60 [1, 2, 3, 4] => 12600 [1, 2, 3, 4, 5] => 37837800 [1, 2, 3, 4, 5, 6] => 2053230379200 [1, 2, 3, 4, 5, 6, 7] => 2431106898187968000 [1, 2, 3, 4, 5, 6, 7, 8] => 73566121315513295589120000 [1, 2, 3, 4, 5, 6, 7, 8, 9] => 65191584694745586153436251091200000 [1, 3, 3] => 140 [10, 14, 12] => 2454860399191200 List of 60 permutations for 3 groups and lanterns per group [1, 2, 3]: [1, 3, 2, 6, 5, 4] [1, 3, 6, 2, 5, 4] [1, 3, 6, 5, 2, 4] [1, 3, 6, 5, 4, 2] [1, 6, 3, 2, 5, 4] [1, 6, 3, 5, 2, 4] [1, 6, 3, 5, 4, 2] [1, 6, 5, 3, 2, 4] [1, 6, 5, 3, 4, 2] [1, 6, 5, 4, 3, 2] [3, 1, 2, 6, 5, 4] [3, 1, 6, 2, 5, 4] [3, 1, 6, 5, 2, 4] [3, 1, 6, 5, 4, 2] [3, 2, 1, 6, 5, 4] [3, 2, 6, 1, 5, 4] [3, 2, 6, 5, 1, 4] [3, 2, 6, 5, 4, 1] [3, 6, 2, 1, 5, 4] [3, 6, 2, 5, 1, 4] [3, 6, 2, 5, 4, 1] [3, 6, 1, 2, 5, 4] [3, 6, 1, 5, 2, 4] [3, 6, 1, 5, 4, 2] [3, 6, 5, 1, 2, 4] [3, 6, 5, 1, 4, 2] [3, 6, 5, 2, 1, 4] [3, 6, 5, 2, 4, 1] [3, 6, 5, 4, 2, 1] [3, 6, 5, 4, 1, 2] [6, 3, 2, 1, 5, 4] [6, 3, 2, 5, 1, 4] [6, 3, 2, 5, 4, 1] [6, 3, 1, 2, 5, 4] [6, 3, 1, 5, 2, 4] [6, 3, 1, 5, 4, 2] [6, 3, 5, 1, 2, 4] [6, 3, 5, 1, 4, 2] [6, 3, 5, 2, 1, 4] [6, 3, 5, 2, 4, 1] [6, 3, 5, 4, 2, 1] [6, 3, 5, 4, 1, 2] [6, 1, 3, 2, 5, 4] [6, 1, 3, 5, 2, 4] [6, 1, 3, 5, 4, 2] [6, 1, 5, 3, 2, 4] [6, 1, 5, 3, 4, 2] [6, 1, 5, 4, 3, 2] [6, 5, 3, 1, 2, 4] [6, 5, 3, 1, 4, 2] [6, 5, 3, 2, 1, 4] [6, 5, 3, 2, 4, 1] [6, 5, 3, 4, 2, 1] [6, 5, 3, 4, 1, 2] [6, 5, 1, 3, 2, 4] [6, 5, 1, 3, 4, 2] [6, 5, 1, 4, 3, 2] [6, 5, 4, 1, 3, 2] [6, 5, 4, 3, 1, 2] [6, 5, 4, 3, 2, 1] List of 140 permutations for 3 groups and lanterns per group [1, 3, 3]: [1, 4, 3, 2, 7, 6, 5] [1, 4, 3, 7, 2, 6, 5] [1, 4, 3, 7, 6, 2, 5] [1, 4, 3, 7, 6, 5, 2] [1, 4, 7, 3, 2, 6, 5] [1, 4, 7, 3, 6, 2, 5] [1, 4, 7, 3, 6, 5, 2] [1, 4, 7, 6, 3, 2, 5] [1, 4, 7, 6, 3, 5, 2] [1, 4, 7, 6, 5, 3, 2] [1, 7, 4, 3, 2, 6, 5] [1, 7, 4, 3, 6, 2, 5] [1, 7, 4, 3, 6, 5, 2] [1, 7, 4, 6, 3, 2, 5] [1, 7, 4, 6, 3, 5, 2] [1, 7, 4, 6, 5, 3, 2] [1, 7, 6, 4, 3, 2, 5] [1, 7, 6, 4, 3, 5, 2] [1, 7, 6, 4, 5, 3, 2] [1, 7, 6, 5, 4, 3, 2] [4, 1, 3, 2, 7, 6, 5] [4, 1, 3, 7, 2, 6, 5] [4, 1, 3, 7, 6, 2, 5] [4, 1, 3, 7, 6, 5, 2] [4, 1, 7, 3, 2, 6, 5] [4, 1, 7, 3, 6, 2, 5] [4, 1, 7, 3, 6, 5, 2] [4, 1, 7, 6, 3, 2, 5] [4, 1, 7, 6, 3, 5, 2] [4, 1, 7, 6, 5, 3, 2] [4, 3, 1, 2, 7, 6, 5] [4, 3, 1, 7, 2, 6, 5] [4, 3, 1, 7, 6, 2, 5] [4, 3, 1, 7, 6, 5, 2] [4, 3, 2, 1, 7, 6, 5] [4, 3, 2, 7, 1, 6, 5] [4, 3, 2, 7, 6, 1, 5] [4, 3, 2, 7, 6, 5, 1] [4, 3, 7, 2, 1, 6, 5] [4, 3, 7, 2, 6, 1, 5] [4, 3, 7, 2, 6, 5, 1] [4, 3, 7, 1, 2, 6, 5] [4, 3, 7, 1, 6, 2, 5] [4, 3, 7, 1, 6, 5, 2] [4, 3, 7, 6, 1, 2, 5] [4, 3, 7, 6, 1, 5, 2] [4, 3, 7, 6, 2, 1, 5] [4, 3, 7, 6, 2, 5, 1] [4, 3, 7, 6, 5, 2, 1] [4, 3, 7, 6, 5, 1, 2] [4, 7, 3, 2, 1, 6, 5] [4, 7, 3, 2, 6, 1, 5] [4, 7, 3, 2, 6, 5, 1] [4, 7, 3, 1, 2, 6, 5] [4, 7, 3, 1, 6, 2, 5] [4, 7, 3, 1, 6, 5, 2] [4, 7, 3, 6, 1, 2, 5] [4, 7, 3, 6, 1, 5, 2] [4, 7, 3, 6, 2, 1, 5] [4, 7, 3, 6, 2, 5, 1] [4, 7, 3, 6, 5, 2, 1] [4, 7, 3, 6, 5, 1, 2] [4, 7, 1, 3, 2, 6, 5] [4, 7, 1, 3, 6, 2, 5] [4, 7, 1, 3, 6, 5, 2] [4, 7, 1, 6, 3, 2, 5] [4, 7, 1, 6, 3, 5, 2] [4, 7, 1, 6, 5, 3, 2] [4, 7, 6, 3, 1, 2, 5] [4, 7, 6, 3, 1, 5, 2] [4, 7, 6, 3, 2, 1, 5] [4, 7, 6, 3, 2, 5, 1] [4, 7, 6, 3, 5, 2, 1] [4, 7, 6, 3, 5, 1, 2] [4, 7, 6, 1, 3, 2, 5] [4, 7, 6, 1, 3, 5, 2] [4, 7, 6, 1, 5, 3, 2] [4, 7, 6, 5, 1, 3, 2] [4, 7, 6, 5, 3, 1, 2] [4, 7, 6, 5, 3, 2, 1] [7, 4, 3, 2, 1, 6, 5] [7, 4, 3, 2, 6, 1, 5] [7, 4, 3, 2, 6, 5, 1] [7, 4, 3, 1, 2, 6, 5] [7, 4, 3, 1, 6, 2, 5] [7, 4, 3, 1, 6, 5, 2] [7, 4, 3, 6, 1, 2, 5] [7, 4, 3, 6, 1, 5, 2] [7, 4, 3, 6, 2, 1, 5] [7, 4, 3, 6, 2, 5, 1] [7, 4, 3, 6, 5, 2, 1] [7, 4, 3, 6, 5, 1, 2] [7, 4, 1, 3, 2, 6, 5] [7, 4, 1, 3, 6, 2, 5] [7, 4, 1, 3, 6, 5, 2] [7, 4, 1, 6, 3, 2, 5] [7, 4, 1, 6, 3, 5, 2] [7, 4, 1, 6, 5, 3, 2] [7, 4, 6, 3, 1, 2, 5] [7, 4, 6, 3, 1, 5, 2] [7, 4, 6, 3, 2, 1, 5] [7, 4, 6, 3, 2, 5, 1] [7, 4, 6, 3, 5, 2, 1] [7, 4, 6, 3, 5, 1, 2] [7, 4, 6, 1, 3, 2, 5] [7, 4, 6, 1, 3, 5, 2] [7, 4, 6, 1, 5, 3, 2] [7, 4, 6, 5, 1, 3, 2] [7, 4, 6, 5, 3, 1, 2] [7, 4, 6, 5, 3, 2, 1] [7, 1, 4, 3, 2, 6, 5] [7, 1, 4, 3, 6, 2, 5] [7, 1, 4, 3, 6, 5, 2] [7, 1, 4, 6, 3, 2, 5] [7, 1, 4, 6, 3, 5, 2] [7, 1, 4, 6, 5, 3, 2] [7, 1, 6, 4, 3, 2, 5] [7, 1, 6, 4, 3, 5, 2] [7, 1, 6, 4, 5, 3, 2] [7, 1, 6, 5, 4, 3, 2] [7, 6, 4, 3, 1, 2, 5] [7, 6, 4, 3, 1, 5, 2] [7, 6, 4, 3, 2, 1, 5] [7, 6, 4, 3, 2, 5, 1] [7, 6, 4, 3, 5, 2, 1] [7, 6, 4, 3, 5, 1, 2] [7, 6, 4, 1, 3, 2, 5] [7, 6, 4, 1, 3, 5, 2] [7, 6, 4, 1, 5, 3, 2] [7, 6, 4, 5, 1, 3, 2] [7, 6, 4, 5, 3, 1, 2] [7, 6, 4, 5, 3, 2, 1] [7, 6, 1, 4, 3, 2, 5] [7, 6, 1, 4, 3, 5, 2] [7, 6, 1, 4, 5, 3, 2] [7, 6, 1, 5, 4, 3, 2] [7, 6, 5, 4, 1, 3, 2] [7, 6, 5, 4, 3, 1, 2] [7, 6, 5, 4, 3, 2, 1] [7, 6, 5, 1, 4, 3, 2]
XPL0
<lang XPL0>char N, Column, Sequences, I, Lanterns;
proc Tally(Level); char Level, Col; [for Col:= 0 to N-1 do
if Column(Col) > 0 then [Column(Col):= Column(Col)-1; if Level = Lanterns-1 then Sequences:= Sequences+1 else Tally(Level+1); Column(Col):= Column(Col)+1; ];
];
[Sequences:= 0; Lanterns:= 0; N:= IntIn(0); Column:= Reserve(N); for I:= 0 to N-1 do
[Column(I):= IntIn(0); Lanterns:= Lanterns + Column(I); ];
Tally(0); IntOut(0, Sequences); ]</lang>
- Output:
5 1 3 5 2 4 37837800