Re: Listing all Permutations using recursive function
[
Date Prev][
Date Next][
Thread Prev][
Thread Next]
[
Date Index]
[
Thread Index]
- Subject: Re: Listing all Permutations using recursive function
- From: Rici Lake <lua@...>
- Date: 2005年8月16日 17:05:33 -0500
On 16-Aug-05, at 3:05 PM, Szilard wrote:
(This program was mangled in the mail; I tried to clean it up)
A={}
function Combinations(n,m)
for i = 1, n do
A[m] = i
if (m<n) then
Combinations(n,m+1)
else
B=A[1]
for j=2,n do
B=B ..", ".. A[j]
end
print(B)
end
end
end
Combinations(4,1)
You would have been better off to have said
local B = A[1]
since B is clearly a local temporary variable, but even better would be
to make use of the standard library:
else
print(table.concat(A, ', '))
end
As a little note about optimization, in case you wanted to go beyond 4
elements, the program spends a significant amount of its time
repeatedly converting integers to strings in order to concatenate them.
Here's a slightly rewritten version, which shows how Lua's scoping
rules work:
function Combinations(n)
local a, vals = {}, {}
for i = 1, n do
vals[i] = tostring(i)
end
local function aux(m)
for i = 1, n do
a[m] = vals[i]
if m < n then
aux(m + 1)
else
print(table.concat(a))
end
end
end
aux(1)
end
I know that wasn't what you were asking, but I couldn't resist.
How can I solve the similar problem when I need only the permutations?
There's a nice example in Programming in Lua, page 73 (available at
<http://www.lua.org/pil/9.3.html>
The example goes on to demonstrate some interesting features of Lua:
coroutines and generator functions.