Attempt at Lua
I've tried to implement the iterative solution from wikipedia, but it doesn't really work, but the time i'm spending on it is up, so I hope this inspires someone to adapt it.
It does parse everything well, including empty columns.
Extra goodie: it does pretty printing of the stacks as in the visual representation in the question.
-- Input "rod1,rod2,rod3" where rod? = a - seperated list of numbers, representing the disks.
p,q,r=io.read():match'([^,]*),([^,]*),([^,]*)'
print(p,q,r)
i=table.insert
u=unpack
function gen(t)
return function(v)i(t,tonumber(v)) end
end
function basic(t,n)
for k,v in pairs(t) do
print(k,"----")
for kk,vv in pairs(v) do print("\t",kk,vv) end
end
print'================'
end
function pretty(t,n)
local out={}
for k=1,n do out[k]={} end
for k=1,n do -- K is each row
local line=out[k]
for l=1,3 do -- L is each rod
local d=t[l][k]
if d~=1e9 then -- TODO Check if metahack necesarry
line[#line+1]=(" "):rep(n-d+1)
line[#line+1]=("="):rep(d)
line[#line+1]="|"
line[#line+1]=("="):rep(d)
line[#line+1]=(" "):rep(n-d+1)
line[#line+1]=" "
else
line[#line+1]=(" "):rep(2*n+4)
end
end
out[k]=table.concat(line)
end
for k=n,1,-1 do
io.write(out[k],"\n")
end
end
function T(f,...)
w=0
for k=1,3 do
l=({...})[k]
w=#l==0 and w or f(w,u(l))
end
return w
end
Stat=pretty
t={{},{},{}} --rods 1 - 3, discs ordered 1 = bottom
for k,v in pairs{p,q,r}do -- loop over strings
v:gsub('%d+',gen(t[k])) -- add decimal to rod
end
n=T(math.max,t[1],t[2],t[3]) -- Biggest disc = number of discs
--for k=1,3 do c=1*t[k][1] if n==c then A=k elseif m==c then C=k else B=k end end -- Rod where the biggest disc is (A)
for k=1,3 do setmetatable(t[k],{__index = function() return 1e9 end}) c=t[k] if c[#c]==1 then one=k end end -- locate smallest disc, and set index for nonexistant discs to 1e9
-- Locate second biggest disc (B), smallest stack = C -> move C to B
-- Algorithm:
-- uneven : move to the left, even: move to the right
-- move smallest, then move non-smallest.
-- repeat until done
--
-- For an even number of disks:
--
-- * make the legal move between pegs A and B
-- * make the legal move between pegs A and C
-- * make the legal move between pegs B and C
-- * repeat until complete
--
-- For an odd number of disks:
--
-- * make the legal move between pegs A and C
-- * make the legal move between pegs A and B
-- * make the legal move between pegs B and C
-- * repeat until complete
--
-- In each case, a total of 2n-1 moves are made.
d={{2,3,1},{3,1,2}}
s=d[math.fmod(n,2)+1] -- sense of movement -1 left (uneven # of discs), 1 right (even # of discs)
Stat(t,n)
for qqq=1,10 do
-- move smallest
d=s[one]
print(one,d)
if #t[d]==0 then print("skip rod",d,"next rod",s[d]) d=s[d] end-- if rod is empty, move to next in same direction
table.insert(t[d],table.remove(t[one])) --TODO Problem
print("Moved",one,"to",d)
one=d -- track the small disc
Stat(t,n)
if #t[d]==n then break end -- destination stack reached number of discs, break off.
-- find next valid move (compare the two non-previous-destination rod) to see which has the smallest disc, move disc to other rod.
z=0
for k=1,3 do
print("-- k="..k)
if k~=one then
if z>0 then
if t[k][#t[k]] > t[z][#t[z]] then -- disc at rod z (source) is smaller than at k (destination)
d=k -- destination = k
print("-- t["..k.."]>t["..z.."], d="..d..", z="..z)
else -- disc at rod z (source) is bigger than at k (destination
d,z=z,k -- switch destination and source, so d will be z, and z will be the current rod
print("-- t["..k.."]
