I have an n × m grid and a collection of polyominos. I would like to know if it is possible to pack them into the grid: no overlapping or rotation is allowed.
I expect that like most packing problems this version is NP-hard and difficult to approximate, so I'm not expecting anything crazy, but an algorithm that could find reasonable packings on a grid around 25 × 25 and be fairly comprehensive around 10 × 10 would be great. (My tiles are mostly tetrominos -- four blocks -- but they could have 5–9+ blocks.)
I'll take whatever anyone has to offer: an algorithm, a paper, an existing program which can be adapted.
Here is a prototype-like SAT-solver approach, which tackles:
a-priori fixed polyomino patterns (see Constants / Input in code)
if rotations should be allowed, rotated pieces have to be added to the set
every polyomino can be placed 0-inf times
there is no scoring-mechanic besides:
the number of non-covered tiles is minimized!
Considering classic off-the-shelf methods for combinatorial-optimization (SAT, CP, MIP), this one will probably scale best (educated guess). It will also be very hard to beat when designing customized heuristics!
If needed, these slides provide some practical introduction to SAT-solvers in practice. Here we are using CDCL-based solvers which are complete (will always find a solution in finite time if there is one; will always be able to prove there is no solution in finite time if there is none; memory of course also plays a role!).
More complex (linear) per-tile scoring-functions are hard to incorporate in general. This is where a (M)IP-approach can be better. But in terms of pure search SAT-solving is much faster in general.
The N=25 problem with my polyomino-set takes ~ 1 second (and one could easily parallize this on multiple granularity-levels -> SAT-solver (threadings-param) vs. outer-loop; the latter will be explained later).
Of course the following holds:
as this is an NP-hard problem, there will be easy and non-easy instances
i did not do scientific benchmarks with many different sets of polyominos
it's to be expected that some sets are easier to solve than others
this is one possible SAT-formulation (not the most trivial!) of infinite many
each formulation has advantages and disadvantages
Idea
The general approach is creating a decision-problem and transforming it into CNF, which is then solved by highly efficient SAT-solvers (here: cryptominisat; CNF will be in DIMCAS-CNF format), which will be used as black-box solvers (no parameter-tuning!).
As the goal is to optimize the number of filled tiles and we are using a decision-problem, we need an outer-loop, adding a minimum tile-used constraint and try to solve it. If not successful, decrease this number. So in general we are calling the SAT-solver multiple times (from scratch!).
There are many different formulations / transformations to CNF possible. Here we use (binary) decision-variables X which indicate a placement. A placement is a tuple like polyomino, x_index, y_index (this index marks the top-left field of some pattern). There is a one-to-one mapping between the number of variables and the number of possible placements of all polyominos.
The core idea is: search in the space of all possible placement-combinations for one solution, which is not invalidating some constraints.
Additionally, we have decision-variables Y, which indicate a tile being filled. There are M*N such variables.
When having access to all possible placements, it's easy to calculate a collision-set for each tile-index (M*N). Given some fixed tile, we can check which placements can fill this one and constrain the problem to only select <=1 of those. This is active on X. In the (M)IP world this probably would be called convex-hull for the collisions.
n<=k-constraints are ubiquitous in SAT-solving and many different formulations are possible. Naive-encoding would need an exponential number of clauses in general which easily becomes infeasibly. Using new variables, there are many variable-clause trade-offs (see Tseitin-encoding) possible. I'm reusing one (old code; only reason why my code is python2-only) which worked good for me in the past. It's based on describing hardware-based counter-logic into CNF and provides good empirical- and theoretical performance (see paper). Of course there are many alternatives.
Additionally, we need to force the SAT-solver not to make all variables negative. We have to add constraints describing the following (that's one approach):
if some field is used: there has to be at least one placement active (poly + x + y), which results in covering this field!
this is a basic logical implication easily formulated as one potentially big logical or
Then only the core-loop is missing, trying to fill N fields, then N-1 until successful. This is again using the n<=k formulation mentioned earlier.
Code
This is python2-code, which needs the SAT-solver cryptominisat 5 in the directory the script is run from.
I'm also using tools from python's excellent scientific-stack.
# PYTHON 2!
import math
import copy
import subprocess
import numpy as np
import matplotlib.pyplot as plt # plotting-only
import seaborn as sns # plotting-only
np.set_printoptions(linewidth=120) # more nice console-output
""" Constants / Input
Example: 5 tetrominoes; no rotation """
M, N = 25, 25
polyominos = [np.array([[1,1,1,1]]),
np.array([[1,1],[1,1]]),
np.array([[1,0],[1,0], [1,1]]),
np.array([[1,0],[1,1],[0,1]]),
np.array([[1,1,1],[0,1,0]])]
""" Preprocessing
Calculate:
A: possible placements
B: covered positions
C: collisions between placements
"""
placements = []
covered = []
for p_ind, p in enumerate(polyominos):
mP, nP = p.shape
for x in range(M):
for y in range(N):
if x + mP <= M: # assumption: no zero rows / cols in each p
if y + nP <= N: # could be more efficient
placements.append((p_ind, x, y))
cover = np.zeros((M,N), dtype=bool)
cover[x:x+mP, y:y+nP] = p
covered.append(cover)
covered = np.array(covered)
collisions = []
for m in range(M):
for n in range(N):
collision_set = np.flatnonzero(covered[:, m, n])
collisions.append(collision_set)
""" Helper-function: Cardinality constraints """
# K-ARY CONSTRAINT GENERATION
# ###########################
# SINZ, Carsten. Towards an optimal CNF encoding of boolean cardinality constraints.
# CP, 2005, 3709. Jg., S. 827-831.
def next_var_index(start):
next_var = start
while(True):
yield next_var
next_var += 1
class s_index():
def __init__(self, start_index):
self.firstEnvVar = start_index
def next(self,i,j,k):
return self.firstEnvVar + i*k +j
def gen_seq_circuit(k, input_indices, next_var_index_gen):
cnf_string = ''
s_index_gen = s_index(next_var_index_gen.next())
# write clauses of first partial sum (i.e. i=0)
cnf_string += (str(-input_indices[0]) + ' ' + str(s_index_gen.next(0,0,k)) + ' 0\n')
for i in range(1, k):
cnf_string += (str(-s_index_gen.next(0, i, k)) + ' 0\n')
# write clauses for general case (i.e. 0 < i < n-1)
for i in range(1, len(input_indices)-1):
cnf_string += (str(-input_indices[i]) + ' ' + str(s_index_gen.next(i, 0, k)) + ' 0\n')
cnf_string += (str(-s_index_gen.next(i-1, 0, k)) + ' ' + str(s_index_gen.next(i, 0, k)) + ' 0\n')
for u in range(1, k):
cnf_string += (str(-input_indices[i]) + ' ' + str(-s_index_gen.next(i-1, u-1, k)) + ' ' + str(s_index_gen.next(i, u, k)) + ' 0\n')
cnf_string += (str(-s_index_gen.next(i-1, u, k)) + ' ' + str(s_index_gen.next(i, u, k)) + ' 0\n')
cnf_string += (str(-input_indices[i]) + ' ' + str(-s_index_gen.next(i-1, k-1, k)) + ' 0\n')
# last clause for last variable
cnf_string += (str(-input_indices[-1]) + ' ' + str(-s_index_gen.next(len(input_indices)-2, k-1, k)) + ' 0\n')
return (cnf_string, (len(input_indices)-1)*k, 2*len(input_indices)*k + len(input_indices) - 3*k - 1)
def gen_at_most_n_constraints(vars, start_var, n):
constraint_string = ''
used_clauses = 0
used_vars = 0
index_gen = next_var_index(start_var)
circuit = gen_seq_circuit(n, vars, index_gen)
constraint_string += circuit[0]
used_clauses += circuit[2]
used_vars += circuit[1]
start_var += circuit[1]
return [constraint_string, used_clauses, used_vars, start_var]
def parse_solution(output):
# assumes there is one
vars = []
for line in output.split("\n"):
if line:
if line[0] == 'v':
line_vars = list(map(lambda x: int(x), line.split()[1:]))
vars.extend(line_vars)
return vars
def solve(CNF):
p = subprocess.Popen(["cryptominisat5.exe"], stdin=subprocess.PIPE, stdout=subprocess.PIPE)
result = p.communicate(input=CNF)[0]
sat_line = result.find('s SATISFIABLE')
if sat_line != -1:
# solution found!
vars = parse_solution(result)
return True, vars
else:
return False, None
""" SAT-CNF: BASE """
X = np.arange(1, len(placements)+1) # decision-vars
# 1-index for CNF
Y = np.arange(len(placements)+1, len(placements)+1 + M*N).reshape(M,N)
next_var = len(placements)+1 + M*N # aux-var gen
n_clauses = 0
cnf = '' # slow string appends
# int-based would be better
# <= 1 for each collision-set
for cset in collisions:
constraint_string, used_clauses, used_vars, next_var = \
gen_at_most_n_constraints(X[cset].tolist(), next_var, 1)
n_clauses += used_clauses
cnf += constraint_string
# if field marked: one of covering placements active
for x in range(M):
for y in range(N):
covering_placements = X[np.flatnonzero(covered[:, x, y])] # could reuse collisions
clause = str(-Y[x,y])
for i in covering_placements:
clause += ' ' + str(i)
clause += ' 0\n'
cnf += clause
n_clauses += 1
print('BASE CNF size')
print('clauses: ', n_clauses)
print('vars: ', next_var - 1)
""" SOLVE in loop -> decrease number of placed-fields until SAT """
print('CORE LOOP')
N_FIELD_HIT = M*N
while True:
print(' N_FIELDS >= ', N_FIELD_HIT)
# sum(y) >= N_FIELD_HIT
# == sum(not y) <= M*N - N_FIELD_HIT
cnf_final = copy.copy(cnf)
n_clauses_final = n_clauses
if N_FIELD_HIT == M*N: # awkward special case
constraint_string = ''.join([str(y) + ' 0\n' for y in Y.ravel()])
n_clauses_final += N_FIELD_HIT
else:
constraint_string, used_clauses, used_vars, next_var = \
gen_at_most_n_constraints((-Y).ravel().tolist(), next_var, M*N - N_FIELD_HIT)
n_clauses_final += used_clauses
n_vars_final = next_var - 1
cnf_final += constraint_string
cnf_final = 'p cnf ' + str(n_vars_final) + ' ' + str(n_clauses) + \
' \n' + cnf_final # header
status, sol = solve(cnf_final)
if status:
print(' SOL found: ', N_FIELD_HIT)
""" Print sol """
res = np.zeros((M, N), dtype=int)
counter = 1
for v in sol[:X.shape[0]]:
if v>0:
p, x, y = placements[v-1]
pM, pN = polyominos[p].shape
poly_nnz = np.where(polyominos[p] != 0)
x_inds, y_inds = x+poly_nnz[0], y+poly_nnz[1]
res[x_inds, y_inds] = p+1
counter += 1
print(res)
""" Plot """
# very very ugly code; too lazy
ax1 = plt.subplot2grid((5, 12), (0, 0), colspan=11, rowspan=5)
ax_p0 = plt.subplot2grid((5, 12), (0, 11))
ax_p1 = plt.subplot2grid((5, 12), (1, 11))
ax_p2 = plt.subplot2grid((5, 12), (2, 11))
ax_p3 = plt.subplot2grid((5, 12), (3, 11))
ax_p4 = plt.subplot2grid((5, 12), (4, 11))
ax_p0.imshow(polyominos[0] * 1, vmin=0, vmax=5)
ax_p1.imshow(polyominos[1] * 2, vmin=0, vmax=5)
ax_p2.imshow(polyominos[2] * 3, vmin=0, vmax=5)
ax_p3.imshow(polyominos[3] * 4, vmin=0, vmax=5)
ax_p4.imshow(polyominos[4] * 5, vmin=0, vmax=5)
ax_p0.xaxis.set_major_formatter(plt.NullFormatter())
ax_p1.xaxis.set_major_formatter(plt.NullFormatter())
ax_p2.xaxis.set_major_formatter(plt.NullFormatter())
ax_p3.xaxis.set_major_formatter(plt.NullFormatter())
ax_p4.xaxis.set_major_formatter(plt.NullFormatter())
ax_p0.yaxis.set_major_formatter(plt.NullFormatter())
ax_p1.yaxis.set_major_formatter(plt.NullFormatter())
ax_p2.yaxis.set_major_formatter(plt.NullFormatter())
ax_p3.yaxis.set_major_formatter(plt.NullFormatter())
ax_p4.yaxis.set_major_formatter(plt.NullFormatter())
mask = (res==0)
sns.heatmap(res, cmap='viridis', mask=mask, cbar=False, square=True, linewidths=.1, ax=ax1)
plt.tight_layout()
plt.show()
break
N_FIELD_HIT -= 1 # binary-search could be viable in some cases
# but beware the empirical asymmetry in SAT-solvers:
# finding solution vs. proving there is none!
Output console
BASE CNF size
('clauses: ', 31509)
('vars: ', 13910)
CORE LOOP
(' N_FIELDS >= ', 625)
(' N_FIELDS >= ', 624)
(' SOL found: ', 624)
[[3 2 2 2 2 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 2 2]
[3 2 2 2 2 1 1 1 1 1 1 1 1 2 2 2 2 2 2 1 1 1 1 2 2]
[3 3 3 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 1 1 1 1 2 2]
[2 2 3 1 1 1 1 1 1 1 1 2 2 2 2 1 1 1 1 2 2 2 2 2 2]
[2 2 3 3 3 2 2 2 2 2 2 2 2 2 2 1 1 1 1 2 2 2 2 2 2]
[1 1 1 1 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 2 2]
[1 1 1 1 3 3 3 2 2 1 1 1 1 2 2 2 2 2 2 2 2 1 1 1 1]
[2 2 1 1 1 1 3 2 2 2 2 2 2 2 2 1 1 1 1 2 2 2 2 2 2]
[2 2 2 2 2 2 3 3 3 2 2 2 2 1 1 1 1 2 2 2 2 2 2 2 2]
[2 2 2 2 2 2 2 2 3 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2]
[2 2 1 1 1 1 2 2 3 3 3 2 2 2 2 2 2 1 1 1 1 2 2 2 2]
[1 1 1 1 1 1 1 1 2 2 3 2 2 1 1 1 1 1 1 1 1 1 1 1 1]
[2 2 3 1 1 1 1 3 2 2 3 3 4 1 1 1 1 2 2 1 1 1 1 2 2]
[2 2 3 1 1 1 1 3 1 1 1 1 4 4 3 2 2 2 2 1 1 1 1 2 2]
[2 2 3 3 5 5 5 3 3 1 1 1 1 4 3 2 2 1 1 1 1 1 1 1 1]
[2 2 2 2 4 5 1 1 1 1 1 1 1 1 3 3 3 2 2 1 1 1 1 2 2]
[2 2 2 2 4 4 2 2 1 1 1 1 1 1 1 1 3 2 2 1 1 1 1 2 2]
[2 2 2 2 3 4 2 2 2 2 2 2 1 1 1 1 3 3 3 2 2 2 2 2 2]
[3 4 2 2 3 5 5 5 2 2 2 2 1 1 1 1 2 2 3 2 2 2 2 2 2]
[3 4 4 3 3 3 5 5 5 5 1 1 1 1 2 2 2 2 3 3 3 2 2 2 2]
[3 3 4 3 1 1 1 1 5 1 1 1 1 4 2 2 2 2 2 2 3 2 2 2 2]
[2 2 3 3 3 1 1 1 1 1 1 1 1 4 4 4 2 2 2 2 3 3 0 2 2]
[2 2 3 1 1 1 1 1 1 1 1 5 5 5 4 4 4 1 1 1 1 2 2 2 2]
[2 2 3 3 1 1 1 1 1 1 1 1 5 5 5 5 4 1 1 1 1 2 2 2 2]
[2 2 1 1 1 1 1 1 1 1 1 1 1 1 5 1 1 1 1 1 1 1 1 2 2]]
Output plot
One field cannot be covered in this parameterization!
Some other examples with a bigger set of patterns
Square M=N=61 (prime -> intuition: harder) where the base-CNF has 450.723 clauses and 185.462 variables. There is an optimal packing!
Non-square M,N =83,131 (double prime) where the base-CNF has 1.346.511 clauses and 553.748 variables. There is an optimal packing!
One approach could be using integer programming. I'll implement this using the python pulp package, though packages are available for pretty much any programming language.
The basic idea is to define a decision variable for every possible placement location for every tile. If a decision variable takes value 1, then its associated tile is placed there. If it takes value 0, then it is not placed there. The objective is therefore to maximize the sum of the decision variables times the number of squares in the variable's tile --- this corresponds to placing the maximum number of squares possible on the board.
My code implements two constraints:
Each tile can only be placed once (below we will relax this constraint)
Each square can have at most one tile on it
Here's the output for a set of five fixed tetrominoes on a 4x5 grid:
import itertools
import pulp
import string
def covered(tile, base):
return {(base[0] + t[0], base[1] + t[1]): True for t in tile}
tiles = [[(0,0), (1,0), (0,1), (0,2)],
[(0,0), (1,0), (2,0), (3,0)],
[(1,0), (0,1), (1,1), (2,0)],
[(0,0), (1,0), (0,1), (1,1)],
[(1,0), (0,1), (1,1), (2,1)]]
rows = 25
cols = 25
squares = {x: True for x in itertools.product(range(rows), range(cols))}
vars = list(itertools.product(range(rows), range(cols), range(len(tiles))))
vars = [x for x in vars if all([y in squares for y in covered(tiles[x[2]], (x[0], x[1])).keys()])]
x = pulp.LpVariable.dicts('tiles', vars, lowBound=0, upBound=1, cat=pulp.LpInteger)
mod = pulp.LpProblem('polyominoes', pulp.LpMaximize)
# Objective value is number of squares in tile
mod += sum([len(tiles[p[2]]) * x[p] for p in vars])
# Don't use any shape more than once
for tnum in range(len(tiles)):
mod += sum([x[p] for p in vars if p[2] == tnum]) <= 1
# Each square can be covered by at most one shape
for s in squares:
mod += sum([x[p] for p in vars if s in covered(tiles[p[2]], (p[0], p[1]))]) <= 1
# Solve and output
mod.solve()
out = [['-'] * cols for rep in range(rows)]
chars = string.ascii_uppercase + string.ascii_lowercase
numset = 0
for p in vars:
if x[p].value() == 1.0:
for off in tiles[p[2]]:
out[p[0] + off[0]][p[1] + off[1]] = chars[numset]
numset += 1
for row in out:
print(''.join(row))
It obtains the following optimal solution:
AAAB-
A-BBC
DDBCC
DD--C
If we allow repeats (comment out the constraint limiting to one copy of each shape), then we can completely tile the grid:
ABCDD
ABCDD
ABCEE
ABCEE
It worked near-instantaneously for a 10x10 grid:
ABCCDDEEFF
ABCCDDEEFF
ABGHHIJJKK
ABGHHIJJKK
LLGMMINOPP
LLGMMINOPP
QQRRSTNOUV
QQRRSTNOUV
WWXXSTYYUV
WWXXSTYYUV
The code obtains an optimal solution for the 25x25 grid in 100 seconds of runtime, though unfortunately there aren't enough letter and numbers for my output code to print the solution.
I don't know if its of any use to you but I coded up a small sketchy frame in Python. It doesn't place polyminos yet but the functions are there - checking for dead empty spaces is primitive, though and needs a better approach. Then again, maybe it is all rubbish...
import functools
import itertools
M = 4 # x
N = 5 # y
field = [[9999]*(N+1)]+[[9999]+[0]*N+[9999] for _ in range(M)]+[[9999]*(N+1)]
def field_rd(p2d):
return field[p2d[0]+1][p2d[1]+1]
def field_add(p2d,val):
field[p2d[0]+1][p2d[1]+1] += val
def add2d(p,k):
return p[0]+k[0],p[1]+k[1]
def norm(polymino_2d):
x0,y0 = min(x for x,y in polymino_2d),min(y for x,y in polymino_2d)
return tuple(sorted(map(lambda p: add2d(p,(-x0,-y0)), polymino_2d)))
def create_cutoff(occupied):
"""Receive a polymino and create the outer area of squares which could be cut off by a placement of this polymino"""
cutoff = set(itertools.chain.from_iterable(map(lambda p: add2d(p,(x,y)),occupied) for (x,y) in [(-1,0),(1,0),(0,-1),(0,1)])) #(-1,-1),(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1)]))
return tuple(cutoff.difference(occupied))
def is_occupied(p2d):
return field_rd(p2d) == 0
def is_cutoff(p2d):
return not is_occupied(p2d) and all(map(is_occupied,map(lambda p: add2d(p,p2d),[(-1,0),(1,0),(0,-1),(0,1)])))
def polym_colliding(p2d,occupied):
return any(map(is_occupied,map(lambda p: add2d(p,p2d),occupied)))
def polym_cutoff(p2d,cutoff):
return any(map(is_cutoff,map(lambda p: add2d(p,p2d),cutoff)))
def put(p2d,occupied,polym_nr):
for p in occupied:
field_add(add2d(p2d,p),polym_nr)
def remove(p2d,occupied,polym_nr):
for p in polym:
field_add(add2d(p2d,p),-polym_nr)
def place(p2d,polym_nr):
"""Try to place a polymino at point p2d. If it fits without cutting off unreachable single cells return True else False"""
occupied = polym[polym_nr][0]
if polym_colliding(p2d,occupied):
return False
put(p2d,occupied,polym_nr)
cutoff = polym[polym_nr][1]
if polym_cutoff(p2d,cutoff):
remove(p2d,occupied,polym_nr)
return False
return True
def NxM_array(N,M):
return [[0]*N for _ in range(M)]
def generate_all_polyminos(n):
"""Create all polyminos with size n"""
def gen_recur(polymino,i,result):
if i > 1:
new_pts = set(itertools.starmap(add2d,itertools.product(polymino,[(-1,0),(1,0),(0,-1),(0,1)])))
new_pts = new_pts.difference(polymino)
for p in new_pts:
gen_recur(polymino.union({p}),i-1,result)
else:
result.add(norm(polymino))
#---------------------------------------
all_polyminos = set()
gen_recur({(0,0)},n,all_polyminos)
return all_polyminos
print("All possible Tetris blocks (all orientations): ",generate_all_polyminos(4))
Let's say I have a number of base 3, 1211. How could I check this number is divisible by 2 without converting it back to base 10?
Update
The original problem is from TopCoder
The digits 3 and 9 share an interesting property. If you take any multiple of 3 and sum its digits, you get another multiple of 3. For example, 118*3 = 354 and 3+5+4 = 12, which is a multiple of 3. Similarly, if you take any multiple of 9 and sum its digits, you get another multiple of 9. For example, 75*9 = 675 and 6+7+5 = 18, which is a multiple of 9. Call any digit for which this property holds interesting, except for 0 and 1, for which the property holds trivially.
A digit that is interesting in one base is not necessarily interesting in another base. For example, 3 is interesting in base 10 but uninteresting in base 5. Given an int base, your task is to return all the interesting digits for that base in increasing order. To determine whether a particular digit is interesting or not, you need not consider all multiples of the digit. You can be certain that, if the property holds for all multiples of the digit with fewer than four digits, then it also holds for multiples with more digits. For example, in base 10, you would not need to consider any multiples greater than 999.
Notes
- When base is greater than 10, digits may have a numeric value greater than 9. Because integers are displayed in base 10 by default, do not be alarmed when such digits appear on your screen as more than one decimal digit. For example, one of the interesting digits in base 16 is 15.
Constraints
- base is between 3 and 30, inclusive.
This is my solution:
class InterestingDigits {
public:
vector<int> digits( int base ) {
vector<int> temp;
for( int i = 2; i <= base; ++i )
if( base % i == 1 )
temp.push_back( i );
return temp;
}
};
The trick was well explained here : https://math.stackexchange.com/questions/17242/how-does-base-of-a-number-relate-to-modulos-of-its-each-individual-digit
Thanks,
Chan
If your number k is in base three, then you can write it as
k = a0 3^n + a1 3^{n-1} + a2 3^{n-2} + ... + an 3^0
where a0, a1, ..., an are the digits in the base-three representation.
To see if the number is divisible by two, you're interested in whether the number, modulo 2, is equal to zero. Well, k mod 2 is given by
k mod 2 = (a0 3^n + a1 3^{n-1} + a2 3^{n-2} + ... + an 3^0) mod 2
= (a0 3^n) mod 2 + (a1 3^{n-1}) mod 2 + ... + an (3^0) mod 2
= (a0 mod 2) (3^n mod 2) + ... + (an mod 2) (3^0 mod 2)
The trick here is that 3^i = 1 (mod 2), so this expression is
k mod 2 = (a0 mod 2) + (a1 mod 2) + ... + (an mod 2)
In other words, if you sum up the digits of the ternary representation and get that this value is divisible by two, then the number itself must be divisible by two. To make this even cooler, since the only ternary digits are 0, 1, and 2, this is equivalent to asking whether the number of 1s in the ternary representation is even!
More generally, though, if you have a number in base m, then that number is divisible by m - 1 iff the sum of the digits is divisible by m. This is why you can check if a number in base 10 is divisible by 9 by summing the digits and seeing if that value is divisible by nine.
You can always build a finite automaton for any base and any divisor:
Normally to compute the value n of a string of digits in base b
you iterate over the digits and do
n = (n * b) + d
for each digit d.
Now if you are interested in divisibility you do this modulo m instead:
n = ((n * b) + d) % m
Here n can take at most m different values. Take these as states of a finite automaton, and compute the transitions depending on the digit d according to that formula. The accepting state is the one where the remainder is 0.
For your specific case we have
n == 0, d == 0: n = ((0 * 3) + 0) % 2 = 0
n == 0, d == 1: n = ((0 * 3) + 1) % 2 = 1
n == 0, d == 2: n = ((0 * 3) + 2) % 2 = 0
n == 1, d == 0: n = ((1 * 3) + 0) % 2 = 1
n == 1, d == 1: n = ((1 * 3) + 1) % 2 = 0
n == 1, d == 2: n = ((1 * 3) + 2) % 2 = 1
which shows that you can just sum the digits 1 modulo 2 and ignore any digits 0 or 2.
Add all the digits together (or even just count the ones) - if the answer is odd, the number is odd; if it's even, the nmber is even.
How does that work? Each digit from the number contributes 0, 1 or 2 times (1, 3, 9, 27, ...). A 0 or a 2 adds an even number, so no effect on the oddness/evenness (parity) of the number as a whole. A 1 adds one of the powers of 3, which is always odd, and so flips the parity). And we start from 0 (even). So by counting whether the number of flips is odd or even we can tell whether the number itself is.
I'm not sure on what CPU you have a number in base-3, but the normal way to do this is to perform a modulus/remainder operation.
if (n % 2 == 0) {
// divisible by 2, so even
} else {
// odd
}
How to implement the modulus operator is going to depend on how you're storing your base-3 number. The simplest to code will probably be to implement normal pencil-and-paper long division, and get the remainder from that.
0 2 2 0
_______
2 ⟌ 1 2 1 1
0
---
1 2
1 1
-----
1 1
1 1
-----
0 1 <--- remainder = 1 (so odd)
(This works regardless of base, there are "tricks" for base-3 as others have mentioned)
Same as in base 10, for your example:
1. Find the multiple of 2 that's <= 1211, that's 1210 (see below how to achieve it)
2. Substract 1210 from 1211, you get 1
3. 1 is < 10, thus 1211 isn't divisible by 2
how to achieve 1210:
1. starts with 2
2. 2 + 2 = 11
3. 11 + 2 = 20
4. 20 + 2 = 22
5. 22 + 2 = 101
6. 101 + 2 = 110
7. 110 + 2 = 112
8. 112 + 2 = 121
9. 121 + 2 = 200
10. 200 + 2 = 202
... // repeat until you get the biggest number <= 1211
it's basically the same as base 10 it's just the round up happens on 3 instead of 10.