Solve hanging lantern problem
You are encouraged to solve this task according to the task description, using any language you may know.
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:
[a,b,c,…] [b,a,c,…] ……
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.
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.
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.
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
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
Phix
fast analytical count only
with javascript_semantics include mpfr.e function get_lantern(sequence s) mpz {z,f} = mpz_inits(2) mpz_fac_ui(z,sum(s)) for d in s do mpz_fac_ui(f,d) mpz_fdiv_q(z,z,f) end for return mpz_get_str(z) end function for t in apply(tagset(9),tagset)&{{1,3,3},{10,14,12}} do printf(1,"%v = %s\n",{t,get_lantern(t)}) end for
- Output:
{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
full solution
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
Rather than take the number of columns as an explicit argument, this program infers the number from the size of the array of columns passed in.
Sequence as columns
The verbose mode of this version outputs the sequence of columns to remove lanterns from, rather than numbering the lanterns individually as in the description:
<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 1 2 3 60 $ 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 lanterns
This longer version numbers the lanterns as in the example in the task description.
<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 4
1 2 4 7
3 5 8
6 9
10
There are 12600 possible takedown sequences:
[1,3,2,6,5,4,10,9,8,7]
[1,3,2,6,5,10,4,9,8,7]
[1,3,2,6,5,10,9,4,8,7]
[1,3,2,6,5,10,9,8,4,7]
[1,3,2,6,5,10,9,8,7,4]
...
[10,9,8,7,6,5,3,4,1,2]
[10,9,8,7,6,5,3,4,2,1]
[10,9,8,7,6,5,4,1,3,2]
[10,9,8,7,6,5,4,3,1,2]
[10,9,8,7,6,5,4,3,2,1]
VBA
- See Visual Basic
Visual Basic
- 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
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>
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]