It would be great if someone could point me towards an algorithm that would allow me to :
create a random square matrix, with entries 0 and 1, such that
every row and every column contain exactly two non-zero entries,
two non-zero entries cannot be adjacent,
all possible matrices are equiprobable.
Right now I manage to achieve points 1 and 2 doing the following : such a matrix can be transformed, using suitable permutations of rows and columns, into a diagonal block matrix with blocks of the form
1 1 0 0 ... 0
0 1 1 0 ... 0
0 0 1 1 ... 0
.............
1 0 0 0 ... 1
So I start from such a matrix using a partition of [0, ..., n-1] and scramble it by permuting rows and columns randomly. Unfortunately, I can't find a way to integrate the adjacency condition, and I am quite sure that my algorithm won't treat all the matrices equally.
Update
I have managed to achieve point 3. The answer was actually straight under my nose : the block matrix I am creating contains all the information needed to take into account the adjacency condition. First some properties and definitions:
a suitable matrix defines permutations of [1, ..., n] that can be build like so: select a 1 in row 1. The column containing this entry contains exactly one other entry equal to 1 on a row a different from 1. Again, row a contains another entry 1 in a column which contains a second entry 1 on a row b, and so on. This starts a permutation 1 -> a -> b ....
For instance, with the following matrix, starting with the marked entry
v
1 0 1 0 0 0 | 1
0 1 0 0 0 1 | 2
1 0 0 1 0 0 | 3
0 0 1 0 1 0 | 4
0 0 0 1 0 1 | 5
0 1 0 0 1 0 | 6
------------+--
1 2 3 4 5 6 |
we get permutation 1 -> 3 -> 5 -> 2 -> 6 -> 4 -> 1.
the cycles of such a permutation lead to the block matrix I mentioned earlier. I also mentioned scrambling the block matrix using arbitrary permutations on the rows and columns to rebuild a matrix compatible with the requirements.
But I was using any permutation, which led to some adjacent non-zero entries. To avoid that, I have to choose permutations that separate rows (and columns) that are adjacent in the block matrix. Actually, to be more precise, if two rows belong to a same block and are cyclically consecutive (the first and last rows of a block are considered consecutive too), then the permutation I want to apply has to move these rows into non-consecutive rows of the final matrix (I will call two rows incompatible in that case).
So the question becomes : How to build all such permutations ?
The simplest idea is to build a permutation progressively by randomly adding rows that are compatible with the previous one. As an example, consider the case n = 6 using partition 6 = 3 + 3 and the corresponding block matrix
1 1 0 0 0 0 | 1
0 1 1 0 0 0 | 2
1 0 1 0 0 0 | 3
0 0 0 1 1 0 | 4
0 0 0 0 1 1 | 5
0 0 0 1 0 1 | 6
------------+--
1 2 3 4 5 6 |
Here rows 1, 2 and 3 are mutually incompatible, as are 4, 5 and 6. Choose a random row, say 3.
We will write a permutation as an array: [2, 5, 6, 4, 3, 1] meaning 1 -> 2, 2 -> 5, 3 -> 6, ... This means that row 2 of the block matrix will become the first row of the final matrix, row 5 will become the second row, and so on.
Now let's build a suitable permutation by choosing randomly a row, say 3:
p = [3, ...]
The next row will then be chosen randomly among the remaining rows that are compatible with 3 : 4, 5and 6. Say we choose 4:
p = [3, 4, ...]
Next choice has to be made among 1 and 2, for instance 1:
p = [3, 4, 1, ...]
And so on: p = [3, 4, 1, 5, 2, 6].
Applying this permutation to the block matrix, we get:
1 0 1 0 0 0 | 3
0 0 0 1 1 0 | 4
1 1 0 0 0 0 | 1
0 0 0 0 1 1 | 5
0 1 1 0 0 0 | 2
0 0 0 1 0 1 | 6
------------+--
1 2 3 4 5 6 |
Doing so, we manage to vertically isolate all non-zero entries. Same has to be done with the columns, for instance by using permutation p' = [6, 3, 5, 1, 4, 2] to finally get
0 1 0 1 0 0 | 3
0 0 1 0 1 0 | 4
0 0 0 1 0 1 | 1
1 0 1 0 0 0 | 5
0 1 0 0 0 1 | 2
1 0 0 0 1 0 | 6
------------+--
6 3 5 1 4 2 |
So this seems to work quite efficiently, but building these permutations needs to be done with caution, because one can easily be stuck: for instance, with n=6 and partition 6 = 2 + 2 + 2, following the construction rules set up earlier can lead to p = [1, 3, 2, 4, ...]. Unfortunately, 5 and 6 are incompatible, so choosing one or the other makes the last choice impossible. I think I've found all situations that lead to a dead end. I will denote by r the set of remaining choices:
p = [..., x, ?], r = {y} with x and y incompatible
p = [..., x, ?, ?], r = {y, z} with y and z being both incompatible with x (no choice can be made)
p = [..., ?, ?], r = {x, y} with x and y incompatible (any choice would lead to situation 1)
p = [..., ?, ?, ?], r = {x, y, z} with x, y and z being cyclically consecutive (choosing x or z would lead to situation 2, choosing y to situation 3)
p = [..., w, ?, ?, ?], r = {x, y, z} with xwy being a 3-cycle (neither x nor y can be chosen, choosing z would lead to situation 3)
p = [..., ?, ?, ?, ?], r = {w, x, y, z} with wxyz being a 4-cycle (any choice would lead to situation 4)
p = [..., ?, ?, ?, ?], r = {w, x, y, z} with xyz being a 3-cycle (choosing w would lead to situation 4, choosing any other would lead to situation 4)
Now it seems that the following algorithm gives all suitable permutations:
As long as there are strictly more than 5 numbers to choose, choose randomly among the compatible ones.
If there are 5 numbers left to choose: if the remaining numbers contain a 3-cycle or a 4-cycle, break that cycle (i.e. choose a number belonging to that cycle).
If there are 4 numbers left to choose: if the remaining numbers contain three cyclically consecutive numbers, choose one of them.
If there are 3 numbers left to choose: if the remaining numbers contain two cyclically consecutive numbers, choose one of them.
I am quite sure that this allows me to generate all suitable permutations and, hence, all suitable matrices.
Unfortunately, every matrix will be obtained several times, depending on the partition that was chosen.
Intro
Here is some prototype-approach, trying to solve the more general task of
uniform combinatorial sampling, which for our approach here means: we can use this approach for everything which we can formulate as SAT-problem.
It's not exploiting your problem directly and takes a heavy detour. This detour to the SAT-problem can help in regards to theory (more powerful general theoretical results) and efficiency (SAT-solvers).
That being said, it's not an approach if you want to sample within seconds or less (in my experiments), at least while being concerned about uniformity.
Theory
The approach, based on results from complexity-theory, follows this work:
GOMES, Carla P.; SABHARWAL, Ashish; SELMAN, Bart. Near-uniform sampling of combinatorial spaces using XOR constraints. In: Advances In Neural Information Processing Systems. 2007. S. 481-488.
The basic idea:
formulate the problem as SAT-problem
add randomly generated xors to the problem (acting on the decision-variables only! that's important in practice)
this will reduce the number of solutions (some solutions will get impossible)
do that in a loop (with tuned parameters) until only one solution is left!
search for some solution is being done by SAT-solvers or #SAT-solvers (=model-counting)
if there is more than one solution: no xors will be added but a complete restart will be done: add random-xors to the start-problem!
The guarantees:
when tuning the parameters right, this approach achieves near-uniform sampling
this tuning can be costly, as it's based on approximating the number of possible solutions
empirically this can also be costly!
Ante's answer, mentioning the number sequence A001499 actually gives a nice upper bound on the solution-space (as it's just ignoring adjacency-constraints!)
The drawbacks:
inefficient for large problems (in general; not necessarily compared to the alternatives like MCMC and co.)
need to change / reduce parameters to produce samples
those reduced parameters lose the theoretical guarantees
but empirically: good results are still possible!
Parameters:
In practice, the parameters are:
N: number of xors added
L: minimum number of variables part of one xor-constraint
U: maximum number of variables part of one xor-constraint
N is important to reduce the number of possible solutions. Given N constant, the other variables of course also have some effect on that.
Theory says (if i interpret correctly), that we should use L = R = 0.5 * #dec-vars.
This is impossible in practice here, as xor-constraints hurt SAT-solvers a lot!
Here some more scientific slides about the impact of L and U.
They call xors of size 8-20 short-XORS, while we will need to use even shorter ones later!
Implementation
Final version
Here is a pretty hacky implementation in python, using the XorSample scripts from here.
The underlying SAT-solver in use is Cryptominisat.
The code basically boils down to:
Transform the problem to conjunctive normal-form
as DIMACS-CNF
Implement the sampling-approach:
Calls XorSample (pipe-based + file-based)
Call SAT-solver (file-based)
Add samples to some file for later analysis
Code: (i hope i did warn you already about the code-quality)
from itertools import count
from time import time
import subprocess
import numpy as np
import os
import shelve
import uuid
import pickle
from random import SystemRandom
cryptogen = SystemRandom()
""" Helper functions """
# 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)
# K=2 clause GENERATION
# #####################
def gen_at_most_2_constraints(vars, start_var):
constraint_string = ''
used_clauses = 0
used_vars = 0
index_gen = next_var_index(start_var)
circuit = gen_seq_circuit(2, 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 gen_at_least_2_constraints(vars, start_var):
k = len(vars) - 2
vars = [-var for var in vars]
constraint_string = ''
used_clauses = 0
used_vars = 0
index_gen = next_var_index(start_var)
circuit = gen_seq_circuit(k, 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]
# Adjacency conflicts
# ###################
def get_all_adjacency_conflicts_4_neighborhood(N, X):
conflicts = set()
for x in range(N):
for y in range(N):
if x < (N-1):
conflicts.add(((x,y),(x+1,y)))
if y < (N-1):
conflicts.add(((x,y),(x,y+1)))
cnf = '' # slow string appends
for (var_a, var_b) in conflicts:
var_a_ = X[var_a]
var_b_ = X[var_b]
cnf += '-' + var_a_ + ' ' + '-' + var_b_ + ' 0 \n'
return cnf, len(conflicts)
# Build SAT-CNF
#############
def build_cnf(N, verbose=False):
var_counter = count(1)
N_CLAUSES = 0
X = np.zeros((N, N), dtype=object)
for a in range(N):
for b in range(N):
X[a,b] = str(next(var_counter))
# Adjacency constraints
CNF, N_CLAUSES = get_all_adjacency_conflicts_4_neighborhood(N, X)
# k=2 constraints
NEXT_VAR = N*N+1
for row in range(N):
constraint_string, used_clauses, used_vars, NEXT_VAR = gen_at_most_2_constraints(X[row, :].astype(int).tolist(), NEXT_VAR)
N_CLAUSES += used_clauses
CNF += constraint_string
constraint_string, used_clauses, used_vars, NEXT_VAR = gen_at_least_2_constraints(X[row, :].astype(int).tolist(), NEXT_VAR)
N_CLAUSES += used_clauses
CNF += constraint_string
for col in range(N):
constraint_string, used_clauses, used_vars, NEXT_VAR = gen_at_most_2_constraints(X[:, col].astype(int).tolist(), NEXT_VAR)
N_CLAUSES += used_clauses
CNF += constraint_string
constraint_string, used_clauses, used_vars, NEXT_VAR = gen_at_least_2_constraints(X[:, col].astype(int).tolist(), NEXT_VAR)
N_CLAUSES += used_clauses
CNF += constraint_string
# build final cnf
CNF = 'p cnf ' + str(NEXT_VAR-1) + ' ' + str(N_CLAUSES) + '\n' + CNF
return X, CNF, NEXT_VAR-1
# External tools
# ##############
def get_random_xor_problem(CNF_IN_fp, N_DEC_VARS, N_ALL_VARS, s, min_l, max_l):
# .cnf not part of arg!
p = subprocess.Popen(['./gen-wff', CNF_IN_fp,
str(N_DEC_VARS), str(N_ALL_VARS),
str(s), str(min_l), str(max_l), 'xored'],
stdin=subprocess.PIPE, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
result = p.communicate()
os.remove(CNF_IN_fp + '-str-xored.xor') # file not needed
return CNF_IN_fp + '-str-xored.cnf'
def solve(CNF_IN_fp, N_DEC_VARS):
seed = cryptogen.randint(0, 2147483647) # actually no reason to do it; but can't hurt either
p = subprocess.Popen(["./cryptominisat5", '-t', '4', '-r', str(seed), CNF_IN_fp], stdin=subprocess.PIPE, stdout=subprocess.PIPE)
result = p.communicate()[0]
sat_line = result.find('s SATISFIABLE')
if sat_line != -1:
# solution found!
vars = parse_solution(result)[:N_DEC_VARS]
# forbid solution (DeMorgan)
negated_vars = list(map(lambda x: x*(-1), vars))
with open(CNF_IN_fp, 'a') as f:
f.write( (str(negated_vars)[1:-1] + ' 0\n').replace(',', ''))
# assume solve is treating last constraint despite not changing header!
# solve again
seed = cryptogen.randint(0, 2147483647)
p = subprocess.Popen(["./cryptominisat5", '-t', '4', '-r', str(seed), CNF_IN_fp], stdin=subprocess.PIPE, stdout=subprocess.PIPE)
result = p.communicate()[0]
sat_line = result.find('s SATISFIABLE')
if sat_line != -1:
os.remove(CNF_IN_fp) # not needed anymore
return True, False, None
else:
return True, True, vars
else:
return False, False, None
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
# Core-algorithm
# ##############
def xorsample(X, CNF_IN_fp, N_DEC_VARS, N_VARS, s, min_l, max_l):
start_time = time()
while True:
# add s random XOR constraints to F
xored_cnf_fp = get_random_xor_problem(CNF_IN_fp, N_DEC_VARS, N_VARS, s, min_l, max_l)
state_lvl1, state_lvl2, var_sol = solve(xored_cnf_fp, N_DEC_VARS)
print('------------')
if state_lvl1 and state_lvl2:
print('FOUND')
d = shelve.open('N_15_70_4_6_TO_PLOT')
d[str(uuid.uuid4())] = (pickle.dumps(var_sol), time() - start_time)
d.close()
return True
else:
if state_lvl1:
print('sol not unique')
else:
print('no sol found')
print('------------')
""" Run """
N = 15
N_DEC_VARS = N*N
X, CNF, N_VARS = build_cnf(N)
with open('my_problem.cnf', 'w') as f:
f.write(CNF)
counter = 0
while True:
print('sample: ', counter)
xorsample(X, 'my_problem', N_DEC_VARS, N_VARS, 70, 4, 6)
counter += 1
Output will look like (removed some warnings):
------------
no sol found
------------
------------
no sol found
------------
------------
no sol found
------------
------------
sol not unique
------------
------------
FOUND
Core: CNF-formulation
We introduce one variable for every cell of the matrix. N=20 means 400 binary-variables.
Adjancency:
Precalculate all symmetry-reduced conflicts and add conflict-clauses.
Basic theory:
a -> !b
<->
!a v !b (propositional logic)
Row/Col-wise Cardinality:
This is tough to express in CNF and naive approaches need an exponential number
of constraints.
We use some adder-circuit based encoding (SINZ, Carsten. Towards an optimal CNF encoding of boolean cardinality constraints) which introduces new auxiliary-variables.
Remark:
sum(var_set) <= k
<->
sum(negated(var_set)) >= len(var_set) - k
These SAT-encodings can be put into exact model-counters (for small N; e.g. < 9). The number of solutions equals Ante's results, which is a strong indication for a correct transformation!
There are also interesting approximate model-counters (also heavily based on xor-constraints) like approxMC which shows one more thing we can do with the SAT-formulation. But in practice i have not been able to use these (approxMC = autoconf; no comment).
Other experiments
I did also build a version using pblib, to use more powerful cardinality-formulations
for the SAT-CNF formulation. I did not try to use the C++-based API, but only the reduced pbencoder, which automatically selects some best encoding, which was way worse than my encoding used above (which is best is still a research-problem; often even redundant-constraints can help).
Empirical analysis
For the sake of obtaining some sample-size (given my patience), i only computed samples for N=15. In this case we used:
N=70 xors
L,U = 4,6
I also computed some samples for N=20 with (100,3,6), but this takes a few mins and we reduced the lower bound!
Visualization
Here some animation (strengthening my love-hate relationship with matplotlib):
Edit: And a (reduced) comparison to brute-force uniform-sampling with N=5 (NXOR,L,U = 4, 10, 30):
(I have not yet decided on the addition of the plotting-code. It's as ugly as the above one and people might look too much into my statistical shambles; normalizations and co.)
Theory
Statistical analysis is probably hard to do as the underlying problem is of such combinatoric nature. It's even not entirely obvious how that final cell-PDF should look like. In the case of N=odd, it's probably non-uniform and looks like a chess-board (i did brute-force check N=5 to observe this).
One thing we can be sure about (imho): symmetry!
Given a cell-PDF matrix, we should expect, that the matrix is symmetric (A = A.T).
This is checked in the visualization and the euclidean-norm of differences over time is plotted.
We can do the same on some other observation: observed pairings.
For N=3, we can observe the following pairs:
0,1
0,2
1,2
Now we can do this per-row and per-column and should expect symmetry too!
Sadly, it's probably not easy to say something about the variance and therefore the needed samples to speak about confidence!
Observation
According to my simplified perception, current-samples and the cell-PDF look good, although convergence is not achieved yet (or we are far away from uniformity).
The more important aspect are probably the two norms, nicely decreasing towards 0.
(yes; one could tune some algorithm for that by transposing with prob=0.5; but this is not done here as it would defeat it's purpose).
Potential next steps
Tune parameters
Check out the approach using #SAT-solvers / Model-counters instead of SAT-solvers
Try different CNF-formulations, especially in regards to cardinality-encodings and xor-encodings
XorSample is by default using tseitin-like encoding to get around exponentially grow
for smaller xors (as used) it might be a good idea to use naive encoding (which propagates faster)
XorSample supports that in theory; but the script's work differently in practice
Cryptominisat is known for dedicated XOR-handling (as it was build for analyzing cryptography including many xors) and might gain something by naive encoding (as inferring xors from blown-up CNFs is much harder)
More statistical-analysis
Get rid of XorSample scripts (shell + perl...)
Summary
The approach is very general
This code produces feasible samples
It should be not hard to prove, that every feasible solution can be sampled
Others have proven theoretical guarantees for uniformity for some params
does not hold for our params
Others have empirically / theoretically analyzed smaller parameters (in use here)
(Updated test results, example run-through and code snippets below.)
You can use dynamic programming to calculate the number of solutions resulting from every state (in a much more efficient way than a brute-force algorithm), and use those (pre-calculated) values to create equiprobable random solutions.
Consider the example of a 7x7 matrix; at the start, the state is:
0,0,0,0,0,0,0
meaning that there are seven adjacent unused columns. After adding two ones to the first row, the state could be e.g.:
0,1,0,0,1,0,0
with two columns that now have a one in them. After adding ones to the second row, the state could be e.g.:
0,1,1,0,1,0,1
After three rows are filled, there is a possibility that a column will have its maximum of two ones; this effectively splits the matrix into two independent zones:
1,1,1,0,2,0,1 -> 1,1,1,0 + 0,1
These zones are independent in the sense that the no-adjacent-ones rule has no effect when adding ones to different zones, and the order of the zones has no effect on the number of solutions.
In order to use these states as signatures for types of solutions, we have to transform them into a canonical notation. First, we have to take into account the fact that columns with only 1 one in them may be unusable in the next row, because they contain a one in the current row. So instead of a binary notation, we have to use a ternary notation, e.g.:
2,1,1,0 + 0,1
where the 2 means that this column was used in the current row (and not that there are 2 ones in the column). At the next step, we should then convert the twos back into ones.
Additionally, we can also mirror the seperate groups to put them into their lexicographically smallest notation:
2,1,1,0 + 0,1 -> 0,1,1,2 + 0,1
Lastly, we sort the seperate groups from small to large, and then lexicographically, so that a state in a larger matrix may be e.g.:
0,0 + 0,1 + 0,0,2 + 0,1,0 + 0,1,0,1
Then, when calculating the number of solutions resulting from each state, we can use memoization using the canonical notation of each state as a key.
Creating a dictionary of the states and the number of solutions for each of them only needs to be done once, and a table for larger matrices can probably be used for smaller matrices too.
Practically, you'd generate a random number between 0 and the total number of solutions, and then for every row, you'd look at the different states you could create from the current state, look at the number of unique solutions each one would generate, and see which option leads to the solution that corresponds with your randomly generated number.
Note that every state and the corresponding key can only occur in a particular row, so you can store the keys in seperate dictionaries per row.
TEST RESULTS
A first test using unoptimized JavaScript gave very promising results. With dynamic programming, calculating the number of solutions for a 10x10 matrix now takes a second, where a brute-force algorithm took several hours (and this is the part of the algorithm that only needs to be done once). The size of the dictionary with the signatures and numbers of solutions grows with a diminishing factor approaching 2.5 for each step in size; the time to generate it grows with a factor of around 3.
These are the number of solutions, states, signatures (total size of the dictionaries), and maximum number of signatures per row (largest dictionary per row) that are created:
size unique solutions states signatures max/row
4x4 2 9 6 2
5x5 16 73 26 8
6x6 722 514 107 40
7x7 33,988 2,870 411 152
8x8 2,215,764 13,485 1,411 596
9x9 179,431,924 56,375 4,510 1,983
10x10 17,849,077,140 218,038 13,453 5,672
11x11 2,138,979,146,276 801,266 38,314 14,491
12x12 304,243,884,374,412 2,847,885 104,764 35,803
13x13 50,702,643,217,809,908 9,901,431 278,561 96,414
14x14 9,789,567,606,147,948,364 33,911,578 723,306 238,359
15x15 2,168,538,331,223,656,364,084 114,897,838 1,845,861 548,409
16x16 546,386,962,452,256,865,969,596 ... 4,952,501 1,444,487
17x17 155,420,047,516,794,379,573,558,433 12,837,870 3,754,040
18x18 48,614,566,676,379,251,956,711,945,475 31,452,747 8,992,972
19x19 17,139,174,923,928,277,182,879,888,254,495 74,818,773 20,929,008
20x20 6,688,262,914,418,168,812,086,412,204,858,650 175,678,000 50,094,203
(Additional results obtained with C++, using a simple 128-bit integer implementation. To count the states, the code had to be run using each state as a seperate signature, which I was unable to do for the largest sizes. )
EXAMPLE
The dictionary for a 5x5 matrix looks like this:
row 0: 00000 -> 16 row 3: 101 -> 0
1112 -> 1
row 1: 20002 -> 2 1121 -> 1
00202 -> 4 1+01 -> 0
02002 -> 2 11+12 -> 2
02020 -> 2 1+121 -> 1
0+1+1 -> 0
row 2: 10212 -> 1 1+112 -> 1
12012 -> 1
12021 -> 2 row 4: 0 -> 0
12102 -> 1 11 -> 0
21012 -> 0 12 -> 0
02121 -> 3 1+1 -> 1
01212 -> 1 1+2 -> 0
The total number of solutions is 16; if we randomly pick a number from 0 to 15, e.g. 13, we can find the corresponding (i.e. the 14th) solution like this:
state: 00000
options: 10100 10010 10001 01010 01001 00101
signature: 00202 02002 20002 02020 02002 00202
solutions: 4 2 2 2 2 4
This tells us that the 14th solution is the 2nd solution of option 00101. The next step is:
state: 00101
options: 10010 01010
signature: 12102 02121
solutions: 1 3
This tells us that the 2nd solution is the 1st solution of option 01010. The next step is:
state: 01111
options: 10100 10001 00101
signature: 11+12 1112 1+01
solutions: 2 1 0
This tells us that the 1st solution is the 1st solution of option 10100. The next step is:
state: 11211
options: 01010 01001
signature: 1+1 1+1
solutions: 1 1
This tells us that the 1st solutions is the 1st solution of option 01010. The last step is:
state: 12221
options: 10001
And the 5x5 matrix corresponding to randomly chosen number 13 is:
0 0 1 0 1
0 1 0 1 0
1 0 1 0 0
0 1 0 1 0
1 0 0 0 1
And here's a quick'n'dirty code example; run the snippet to generate the signature and solution count dictionary, and generate a random 10x10 matrix (it takes a second to generate the dictionary; once that is done, it generates random solutions in half a millisecond):
function signature(state, prev) {
var zones = [], zone = [];
for (var i = 0; i < state.length; i++) {
if (state[i] == 2) {
if (zone.length) zones.push(mirror(zone));
zone = [];
}
else if (prev[i]) zone.push(3);
else zone.push(state[i]);
}
if (zone.length) zones.push(mirror(zone));
zones.sort(function(a,b) {return a.length - b.length || a - b;});
return zones.length ? zones.join("2") : "2";
function mirror(zone) {
var ltr = zone.join('');
zone.reverse();
var rtl = zone.join('');
return (ltr < rtl) ? ltr : rtl;
}
}
function memoize(n) {
var memo = [], empty = [];
for (var i = 0; i <= n; i++) memo[i] = [];
for (var i = 0; i < n; i++) empty[i] = 0;
memo[0][signature(empty, empty)] = next_row(empty, empty, 1);
return memo;
function next_row(state, prev, row) {
if (row > n) return 1;
var solutions = 0;
for (var i = 0; i < n - 2; i++) {
if (state[i] == 2 || prev[i] == 1) continue;
for (var j = i + 2; j < n; j++) {
if (state[j] == 2 || prev[j] == 1) continue;
var s = state.slice(), p = empty.slice();
++s[i]; ++s[j]; ++p[i]; ++p[j];
var sig = signature(s, p);
var sol = memo[row][sig];
if (sol == undefined)
memo[row][sig] = sol = next_row(s, p, row + 1);
solutions += sol;
}
}
return solutions;
}
}
function random_matrix(n, memo) {
var matrix = [], empty = [], state = [], prev = [];
for (var i = 0; i < n; i++) empty[i] = state[i] = prev[i] = 0;
var total = memo[0][signature(empty, empty)];
var pick = Math.floor(Math.random() * total);
document.write("solution " + pick.toLocaleString('en-US') +
" from a total of " + total.toLocaleString('en-US') + "<br>");
for (var row = 1; row <= n; row++) {
var options = find_options(state, prev);
for (var i in options) {
var state_copy = state.slice();
for (var j in state_copy) state_copy[j] += options[i][j];
var sig = signature(state_copy, options[i]);
var solutions = memo[row][sig];
if (pick < solutions) {
matrix.push(options[i].slice());
prev = options[i].slice();
state = state_copy.slice();
break;
}
else pick -= solutions;
}
}
return matrix;
function find_options(state, prev) {
var options = [];
for (var i = 0; i < n - 2; i++) {
if (state[i] == 2 || prev[i] == 1) continue;
for (var j = i + 2; j < n; j++) {
if (state[j] == 2 || prev[j] == 1) continue;
var option = empty.slice();
++option[i]; ++option[j];
options.push(option);
}
}
return options;
}
}
var size = 10;
var memo = memoize(size);
var matrix = random_matrix(size, memo);
for (var row in matrix) document.write(matrix[row] + "<br>");
The code snippet below shows the dictionary of signatures and solution counts for a matrix of size 10x10. I've used a slightly different signature format from the explanation above: the zones are delimited by a '2' instead of a plus sign, and a column which has a one in the previous row is marked with a '3' instead of a '2'. This shows how the keys could be stored in a file as integers with 2×N bits (padded with 2's).
function signature(state, prev) {
var zones = [], zone = [];
for (var i = 0; i < state.length; i++) {
if (state[i] == 2) {
if (zone.length) zones.push(mirror(zone));
zone = [];
}
else if (prev[i]) zone.push(3);
else zone.push(state[i]);
}
if (zone.length) zones.push(mirror(zone));
zones.sort(function(a,b) {return a.length - b.length || a - b;});
return zones.length ? zones.join("2") : "2";
function mirror(zone) {
var ltr = zone.join('');
zone.reverse();
var rtl = zone.join('');
return (ltr < rtl) ? ltr : rtl;
}
}
function memoize(n) {
var memo = [], empty = [];
for (var i = 0; i <= n; i++) memo[i] = [];
for (var i = 0; i < n; i++) empty[i] = 0;
memo[0][signature(empty, empty)] = next_row(empty, empty, 1);
return memo;
function next_row(state, prev, row) {
if (row > n) return 1;
var solutions = 0;
for (var i = 0; i < n - 2; i++) {
if (state[i] == 2 || prev[i] == 1) continue;
for (var j = i + 2; j < n; j++) {
if (state[j] == 2 || prev[j] == 1) continue;
var s = state.slice(), p = empty.slice();
++s[i]; ++s[j]; ++p[i]; ++p[j];
var sig = signature(s, p);
var sol = memo[row][sig];
if (sol == undefined)
memo[row][sig] = sol = next_row(s, p, row + 1);
solutions += sol;
}
}
return solutions;
}
}
var memo = memoize(10);
for (var i in memo) {
document.write("row " + i + ":<br>");
for (var j in memo[i]) {
document.write(""" + j + "": " + memo[i][j] + "<br>");
}
}
Just few thoughts. Number of matrices satisfying conditions for n <= 10:
3 0
4 2
5 16
6 722
7 33988
8 2215764
9 179431924
10 17849077140
Unfortunatelly there is no sequence with these numbers in OEIS.
There is one similar (A001499), without condition for neighbouring one's. Number of nxn matrices in this case is 'of order' as A001499's number of (n-1)x(n-1) matrices. That is to be expected since number
of ways to fill one row in this case, position 2 one's in n places with at least one zero between them is ((n-1) choose 2). Same as to position 2 one's in (n-1) places without the restriction.
I don't think there is an easy connection between these matrix of order n and A001499 matrix of order n-1, meaning that if we have A001499 matrix than we can construct some of these matrices.
With this, for n=20, number of matrices is >10^30. Quite a lot :-/
This solution use recursion in order to set the cell of the matrix one by one. If the random walk finish with an impossible solution then we rollback one step in the tree and we continue the random walk.
The algorithm is efficient and i think that the generated data are highly equiprobable.
package rndsqmatrix;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.stream.IntStream;
public class RndSqMatrix {
/**
* Generate a random matrix
* #param size the size of the matrix
* #return the matrix encoded in 1d array i=(x+y*size)
*/
public static int[] generate(final int size) {
return generate(size, new int[size * size], new int[size],
new int[size]);
}
/**
* Build a matrix recursivly with a random walk
* #param size the size of the matrix
* #param matrix the matrix encoded in 1d array i=(x+y*size)
* #param rowSum
* #param colSum
* #return
*/
private static int[] generate(final int size, final int[] matrix,
final int[] rowSum, final int[] colSum) {
// generate list of valid positions
final List<Integer> positions = new ArrayList();
for (int y = 0; y < size; y++) {
if (rowSum[y] < 2) {
for (int x = 0; x < size; x++) {
if (colSum[x] < 2) {
final int p = x + y * size;
if (matrix[p] == 0
&& (x == 0 || matrix[p - 1] == 0)
&& (x == size - 1 || matrix[p + 1] == 0)
&& (y == 0 || matrix[p - size] == 0)
&& (y == size - 1 || matrix[p + size] == 0)) {
positions.add(p);
}
}
}
}
}
// no valid positions ?
if (positions.isEmpty()) {
// if the matrix is incomplete => return null
for (int i = 0; i < size; i++) {
if (rowSum[i] != 2 || colSum[i] != 2) {
return null;
}
}
// the matrix is complete => return it
return matrix;
}
// random walk
Collections.shuffle(positions);
for (int p : positions) {
// set '1' and continue recursivly the exploration
matrix[p] = 1;
rowSum[p / size]++;
colSum[p % size]++;
final int[] solMatrix = generate(size, matrix, rowSum, colSum);
if (solMatrix != null) {
return solMatrix;
}
// rollback
matrix[p] = 0;
rowSum[p / size]--;
colSum[p % size]--;
}
// we can't find a valid matrix from here => return null
return null;
}
public static void printMatrix(final int size, final int[] matrix) {
for (int y = 0; y < size; y++) {
for (int x = 0; x < size; x++) {
System.out.print(matrix[x + y * size]);
System.out.print(" ");
}
System.out.println();
}
}
public static void printStatistics(final int size, final int count) {
final int sumMatrix[] = new int[size * size];
for (int i = 0; i < count; i++) {
final int[] matrix = generate(size);
for (int j = 0; j < sumMatrix.length; j++) {
sumMatrix[j] += matrix[j];
}
}
printMatrix(size, sumMatrix);
}
public static void checkAlgorithm() {
final int size = 8;
final int count = 2215764;
final int divisor = 122;
final int sumMatrix[] = new int[size * size];
for (int i = 0; i < count/divisor ; i++) {
final int[] matrix = generate(size);
for (int j = 0; j < sumMatrix.length; j++) {
sumMatrix[j] += matrix[j];
}
}
int total = 0;
for(int i=0; i < sumMatrix.length; i++) {
total += sumMatrix[i];
}
final double factor = (double)total / (count/divisor);
System.out.println("Factor=" + factor + " (theory=16.0)");
}
public static void benchmark(final int size, final int count,
final boolean parallel) {
final long begin = System.currentTimeMillis();
if (!parallel) {
for (int i = 0; i < count; i++) {
generate(size);
}
} else {
IntStream.range(0, count).parallel().forEach(i -> generate(size));
}
final long end = System.currentTimeMillis();
System.out.println("rate="
+ (double) (end - begin) / count + "ms/matrix");
}
public static void main(String[] args) {
checkAlgorithm();
benchmark(8, 10000, true);
//printStatistics(8, 2215764/36);
printStatistics(8, 2215764);
}
}
The output is:
Factor=16.0 (theory=16.0)
rate=0.2835ms/matrix
552969 554643 552895 554632 555680 552753 554567 553389
554071 554847 553441 553315 553425 553883 554485 554061
554272 552633 555130 553699 553604 554298 553864 554028
554118 554299 553565 552986 553786 554473 553530 554771
554474 553604 554473 554231 553617 553556 553581 553992
554960 554572 552861 552732 553782 554039 553921 554661
553578 553253 555721 554235 554107 553676 553776 553182
553086 553677 553442 555698 553527 554850 553804 553444
Here is a very fast approach of generating the matrix row by row, written in Java:
public static void main(String[] args) throws Exception {
int n = 100;
Random rnd = new Random();
byte[] mat = new byte[n*n];
byte[] colCount = new byte[n];
//generate row by row
for (int x = 0; x < n; x++) {
//generate a random first bit
int b1 = rnd.nextInt(n);
while ( (x > 0 && mat[(x-1)*n + b1] == 1) || //not adjacent to the one above
(colCount[b1] == 2) //not in a column which has 2
) b1 = rnd.nextInt(n);
//generate a second bit, not equal to the first one
int b2 = rnd.nextInt(n);
while ( (b2 == b1) || //not the same as bit 1
(x > 0 && mat[(x-1)*n + b2] == 1) || //not adjacent to the one above
(colCount[b2] == 2) || //not in a column which has 2
(b2 == b1 - 1) || //not adjacent to b1
(b2 == b1 + 1)
) b2 = rnd.nextInt(n);
//fill the matrix values and increment column counts
mat[x*n + b1] = 1;
mat[x*n + b2] = 1;
colCount[b1]++;
colCount[b2]++;
}
String arr = Arrays.toString(mat).substring(1, n*n*3 - 1);
System.out.println(arr.replaceAll("(.{" + n*3 + "})", "$1\n"));
}
It essentially generates each a random row at a time. If the row will violate any of the conditions, it is generated again (again randomly). I believe this will satisfy condition 4 as well.
Adding a quick note that it will spin forever for N-s where there is no solutions (like N=3).
I have a deck of 24 cards - 8 red, 8 blue and 8 yellow cards.
red |1|2|3|4|5|6|7|8|
yellow |1|2|3|4|5|6|7|8|
blue |1|2|3|4|5|6|7|8|
I can take 3 of cards (same numbers, straight, straigh flush), whereas each of the type is scored differently.
My question is, how to calculate maximal possible score (find optimal groups) for a game in progress, where some cards are already missing.
for example:
red |1|2|3|4|5|6|7|8|
yellow |1|2|3| |5| |7|8|
blue |1|2| |4|5|6| |8|
The score for a three-of-a-kind is:
1-1-1 20
2-2-2 30
3-3-3 40
4-4-4 50
5-5-5 60
6-6-6 70
7-7-7 80
8-8-8 90
The score for a straight is:
1-2-3 10
2-3-4 20
3-4-5 30
4-5-6 40
5-6-7 50
6-7-8 60
The score for a straight flush is:
1-2-3 50
2-3-4 60
3-4-5 70
4-5-6 80
5-6-7 90
6-7-8 100
A solution which recursively tries every combination would go like this:
Start looking at combinations that have a red 8 as the highest card: three-of-a-kind r8-y8-b8, straight flush r6-r7-r8, and every possible straight *6-*7-r8. For each of these, remove the cards from the set, and recurse to check combinations with the yellow 8, then blue 8, then red 7, yellow 7, blue 7, red 6 ... until you've checked everything except the 2's and 1's; then add three-of-a-kind 2-2-2 and 1-1-1 if available. At each step, check which recursion returns the maximum score, and return this maximum.
Let's look at what happens in each of these steps. Say we're looking at combinations with red 8; we have available cards like:
red ...|6|7|8|
yellow ...|6| |8|
blue ...| |7|8|
First, use three-of-a-kind r8-y8-b8, if possible. Create a copy of the available cards, remove the 8's, and recurse straight to the 7's:
score = 90 + max_score(cards_copy, next = red 7)
(Trying the three-of-a-kind should only be done when the current card is red, to avoid duplicate solutions.)
Then, use straight flush r6-r7-r8, if possible. Create a copy of the available cards, remove r6, r7 and r8, and recurse to yellow 8:
score = 100 + max_score(cards_copy, next = yellow 8)
Then, use every possible non-flush straight containing red 8; in the example, those are r6-b7-r8, y6-r7-r8 and y6-b7-r8 (there could be up to nine). For each of these, create a copy of the available cards, remove the three cards and recurse to yellow 8:
score = 60 + max_score(cards_copy, next = yellow 8)
Then, finally, recurse without using red 8: create a copy of the available cards, remove red 8 and recurse to yellow 8:
score = max_score(cards_copy, next = yellow 8)
You then calculate which of these options has the greatest score (with the score returned by its recursion added), and return that maximum score.
A quick test in JavaScript shows that for a full set of 24 cards, the algorithm goes through 30 million recursions to find the maximum score 560, and becomes quite slow. However, as soon as 3 higher-value cards have been removed, the number of recursions falls below one million and it takes around 1 second, and with 6 higher-value cards removed, it falls below 20,000 and returns almost instantly.
For almost-complete sets, you could pre-compute the maximum scores, and only calculate the score once a certain number of cards have been removed. A lot of sets will be duplicates anyway; removing r6-r7-r8 will result in the same maximum score as removing y6-y7-y8; removing r6-y7-b8 is a duplicate of removing b6-y7-r8... So first you change the input to a canonical version, and then you look up the pre-computed score. E.g. using pre-computed scores for all sets with 3 or 6 cards removed would require storing 45,340 scores.
As a code example, here's the JavaScript code I tested the algorithm with:
function clone(array) { // copy 2-dimensional array
var copy = [];
array.forEach(function(item) {copy.push(item.slice())});
return copy;
}
function max_score(cards, suit, rank) {
suit = suit || 0; rank = rank || 7; // start at red 8
var max = 0;
if (rank < 2) { // try 3-of-a-kind for rank 1 and 2
if (cards[0][0] && cards[1][0] && cards[2][0]) max += 20;
if (cards[0][1] && cards[1][1] && cards[2][1]) max += 30;
return max;
}
var next_rank = suit == 2 ? rank - 1: rank;
var next_suit = (suit + 1) % 3;
max = max_score(clone(cards), next_suit, next_rank); // try skipping this card
if (! cards[suit][rank]) return max;
if (suit == 0 && cards[1][rank] && cards[2][rank]) { // try 3-of-a-kind
var score = rank * 10 + 20 + max_score(clone(cards), 0, rank - 1);
if (score > max) max = score;
}
for (var i = 0; i < 3; i++) { // try all possible straights
if (! cards[i][rank - 2]) continue;
for (var j = 0; j < 3; j++) {
if (! cards[j][rank - 1]) continue;
var copy = clone(cards);
copy[j][rank - 1] = 0; copy[i][rank - 2] = 0;
var score = rank * 10 - 10 + max_score(copy, next_suit, next_rank);
if (i == suit && j == suit) score += 40; // straight is straight flush
if (score > max) max = score;
}
}
return max;
}
document.write(max_score([[1,1,1,1,1,0,1,1], [1,1,1,1,1,1,1,0], [1,1,1,0,1,1,1,1]]));
An obvious way to speed up the algorithm is to use a 24-bit pattern instead of a 3x8 bit array to represent the cards; that way the array cloning is no longer necessary, and most of the code is turned into bit manipulation. In JavaScript, it's about 8 times faster:
function max_score(cards, suit, rank) {
suit = suit || 0; rank = rank || 7; // start at red 8
var max = 0;
if (rank < 2) { // try 3-of-a-kind for rank 1 and 2
if ((cards & 65793) == 65793) max += 20; // 65793 = rank 1 of all suits
if ((cards & 131586) == 131586) max += 30; // 131586 = rank 2 of all suits
return max;
}
var next_rank = suit == 2 ? rank - 1: rank;
var next_suit = (suit + 1) % 3;
var this_card = 1 << rank << suit * 8;
max = max_score(cards, next_suit, next_rank); // try skipping this card
if (! (cards & this_card)) return max;
if (suit == 0 && cards & this_card << 8 && cards & this_card << 16) { // try 3oaK
var score = rank * 10 + 20 + max_score(cards, 0, rank - 1);
if (score > max) max = score;
}
for (var i = 0; i < 3; i++) { // try all possible straights
var mid_card = 1 << rank - 1 << i * 8;
if (! (cards & mid_card)) continue;
for (var j = 0; j < 3; j++) {
var low_card = 1 << rank - 2 << j * 8;
if (! (cards & low_card)) continue;
var cards_copy = cards - mid_card - low_card;
var score = rank * 10 - 10 + max_score(cards_copy, next_suit, next_rank);
if (i == suit && j == suit) score += 40; // straight is straight flush
if (score > max) max = score;
}
}
return max;
}
document.write(max_score(parseInt("111101110111111111011111", 2)));
// B Y R
// 876543218765432187654321
The speed for almost-complete sets can be further improved by using the observation that if a straight flush for all three suits can be be made for the current rank, then this is always the best option. This reduces the number of recursions drastically, because nine cards can be skipped at once. This check should be added immediately after trying 3-of-a-kind for rank 1 and 2:
if (suit == 0) { // try straight flush for all suits
var flush3 = 460551 << rank - 2; // 460551 = rank 1, 2 and 3 of all suits
if ((cards & flush3) == flush3) {
max = rank * 30 + 90;
if (rank > 2) max += max_score(cards - flush3, 0, rank - 3);
return max;
}
}
Given a sequence of N integers where 1 <= N <= 500 and the numbers are between 1 and 50. In a step any two adjacent equal numbers x x can be replaced with a single x + 1. What is the maximum number achievable by such steps.
For example if given 2 3 1 1 2 2 then the maximum possible is 4:
2 3 1 1 2 2 ---> 2 3 2 2 2 ---> 2 3 3 2 ---> 2 4 2.
It is evident that I should try to do better than the maximum number available in the sequence. But I can't figure out a good algorithm.
Each substring of the input can make at most one single number (invariant: the log base two of the sum of two to the power of each entry). For every x, we can find the set of substrings that can make x. For each x, this is (1) every occurrence of x (2) the union of two contiguous substrings that can make x - 1. The resulting algorithm is O(N^2)-time.
An algorithm could work like this:
Convert the input to an array where every element has a frequency attribute, collapsing repeated consecutive values in the input into one single node. For example, this input:
1 2 2 4 3 3 3 3
Would be represented like this:
{val: 1, freq: 1} {val: 2, freq: 2} {val: 4, freq: 1} {val: 3, freq: 4}
Then find local minima nodes, like the node (3 3 3 3) in 1 (2 2) 4 (3 3 3 3) 4, i.e. nodes whose neighbours both have higher values. For those local minima that have an even frequency, "lift" those by applying the step. Repeat this until no such local minima (with even frequency) exist any more.
Start of the recursive part of the algorithm:
At both ends of the array, work inwards to "lift" values as long as the more inner neighbour has a higher value. With this rule, the following:
1 2 2 3 5 4 3 3 3 1 1
will completely resolve. First from the left side inward:
1 4 5 4 3 3 3 1 1
Then from the right side:
1 4 6 3 2
Note that when there is an odd frequency (like for the 3s above), there will be a "remainder" that cannot be incremented. The remainder should in this rule always be left on the outward side, so to maximise the potential towards the inner part of the array.
At this point the remaining local minima have odd frequencies. Applying the step to such a node will always leave a "remainder" (like above) with the original value. This remaining node can appear anywhere, but it only makes sense to look at solutions where this remainder is on the left side or the right side of the lift (not in the middle). So for example:
4 1 1 1 1 1 2 3 4
Can resolve to one of these:
4 2 2 1 2 3 4
Or:
4 1 2 2 2 3 4
The 1 in either second or fourth position, is the above mentioned "remainder". Obviously, the second way of resolving is more promising in this example. In general, the choice is obvious when on one side there is a value that is too high to merge with, like the left-most 4 is too high for five 1 values to get to. The 4 is like a wall.
When the frequency of the local minimum is one, there is nothing we can do with it. It actually separates the array in a left and right side that do not influence each other. The same is true for the remainder element discussed above: it separates the array into two parts that do not influence each other.
So the next step in the algorithm is to find such minima (where the choice is obvious), apply that kind of step and separate the problem into two distinct problems which should be solved recursively (from the top). So in the last example, the following two problems would be solved separately:
4
2 2 3 4
Then the best of both solutions will count as the overall solution. In this case that is 5.
The most challenging part of the algorithm is to deal with those local minima for which the choice of where to put the remainder is not obvious. For instance;
3 3 1 1 1 1 1 2 3
This can go to either:
3 3 2 2 1 2 3
3 3 1 2 2 2 3
In this example the end result is the same for both options, but in bigger arrays it would be less and less obvious. So here both options have to be investigated. In general you can have many of them, like 2 in this example:
3 1 1 1 2 3 1 1 1 1 1 3
Each of these two minima has two options. This seems like to explode into too many possibilities for larger arrays. But it is not that bad. The algorithm can take opposite choices in neighbouring minima, and go alternating like this through the whole array. This way alternating sections are favoured, and get the most possible value drawn into them, while the other sections are deprived of value. Now the algorithm turns the tables, and toggles all choices so that the sections that were previously favoured are now deprived, and vice versa. The solution of both these alternatives is derived by resolving each section recursively, and then comparing the two "grand" solutions to pick the best one.
Snippet
Here is a live JavaScript implementation of the above algorithm.
Comments are provided which hopefully should make it readable.
"use strict";
function Node(val, freq) {
// Immutable plain object
return Object.freeze({
val: val,
freq: freq || 1, // Default frequency is 1.
// Max attainable value when merged:
reduced: val + (freq || 1).toString(2).length - 1
});
}
function compress(a) {
// Put repeated elements in a single node
var result = [], i, j;
for (i = 0; i < a.length; i = j) {
for (j = i + 1; j < a.length && a[j] == a[i]; j++);
result.push(Node(a[i], j - i));
}
return result;
}
function decompress(a) {
// Expand nodes into separate, repeated elements
var result = [], i, j;
for (i = 0; i < a.length; i++) {
for (j = 0; j < a[i].freq; j++) {
result.push(a[i].val);
}
}
return result;
}
function str(a) {
return decompress(a).join(' ');
}
function unstr(s) {
s = s.replace(/\D+/g, ' ').trim();
return s.length ? compress(s.split(/\s+/).map(Number)) : [];
}
/*
The function merge modifies an array in-place, performing a "step" on
the indicated element.
The array will get an element with an incremented value
and decreased frequency, unless a join occurs with neighboring
elements with the same value: then the frequencies are accumulated
into one element. When the original frequency was odd there will
be a "remainder" element in the modified array as well.
*/
function merge(a, i, leftWards, stats) {
var val = a[i].val+1,
odd = a[i].freq % 2,
newFreq = a[i].freq >> 1,
last = i;
// Merge with neighbouring nodes of same value:
if ((!odd || !leftWards) && a[i+1] && a[i+1].val === val) {
newFreq += a[++last].freq;
}
if ((!odd || leftWards) && i && a[i-1].val === val) {
newFreq += a[--i].freq;
}
// Replace nodes
a.splice(i, last-i+1, Node(val, newFreq));
if (odd) a.splice(i+leftWards, 0, Node(val-1));
// Update statistics and trace: this is not essential to the algorithm
if (stats) {
stats.total_applied_merges++;
if (stats.trace) stats.trace.push(str(a));
}
return i;
}
/* Function Solve
Parameters:
a: The compressed array to be reduced via merges. It is changed in-place
and should not be relied on after the call.
stats: Optional plain object that will be populated with execution statistics.
Return value:
The array after the best merges were applied to achieve the highest
value, which is stored in the maxValue custom property of the array.
*/
function solve(a, stats) {
var maxValue, i, j, traceOrig, skipLeft, skipRight, sections, goLeft,
b, choice, alternate;
if (!a.length) return a;
if (stats && stats.trace) {
traceOrig = stats.trace;
traceOrig.push(stats.trace = [str(a)]);
}
// Look for valleys of even size, and "lift" them
for (i = 1; i < a.length - 1; i++) {
if (a[i-1].val > a[i].val && a[i].val < a[i+1].val && (a[i].freq % 2) < 1) {
// Found an even valley
i = merge(a, i, false, stats);
if (i) i--;
}
}
// Check left-side elements with always increasing values
for (i = 0; i < a.length-1 && a[i].val < a[i+1].val; i++) {
if (a[i].freq > 1) i = merge(a, i, false, stats) - 1;
};
// Check right-side elements with always increasing values, right-to-left
for (j = a.length-1; j > 0 && a[j-1].val > a[j].val; j--) {
if (a[j].freq > 1) j = merge(a, j, true, stats) + 1;
};
// All resolved?
if (i == j) {
while (a[i].freq > 1) merge(a, i, true, stats);
a.maxValue = a[i].val;
} else {
skipLeft = i;
skipRight = a.length - 1 - j;
// Look for other valleys (odd sized): they will lead to a split into sections
sections = [];
for (i = a.length - 2 - skipRight; i > skipLeft; i--) {
if (a[i-1].val > a[i].val && a[i].val < a[i+1].val) {
// Odd number of elements: if more than one, there
// are two ways to merge them, but maybe
// one of both possibilities can be excluded.
goLeft = a[i+1].val > a[i].reduced;
if (a[i-1].val > a[i].reduced || goLeft) {
if (a[i].freq > 1) i = merge(a, i, goLeft, stats) + goLeft;
// i is the index of the element which has become a 1-sized valley
// Split off the right part of the array, and store the solution
sections.push(solve(a.splice(i--), stats));
}
}
}
if (sections.length) {
// Solve last remaining section
sections.push(solve(a, stats));
sections.reverse();
// Combine the solutions of all sections into one
maxValue = sections[0].maxValue;
for (i = sections.length - 1; i >= 0; i--) {
maxValue = Math.max(sections[i].maxValue, maxValue);
}
} else {
// There is no more valley that can be resolved without branching into two
// directions. Look for the remaining valleys.
sections = [];
b = a.slice(0); // take copy
for (choice = 0; choice < 2; choice++) {
if (choice) a = b; // restore from copy on second iteration
alternate = choice;
for (i = a.length - 2 - skipRight; i > skipLeft; i--) {
if (a[i-1].val > a[i].val && a[i].val < a[i+1].val) {
// Odd number of elements
alternate = !alternate
i = merge(a, i, alternate, stats) + alternate;
sections.push(solve(a.splice(i--), stats));
}
}
// Solve last remaining section
sections.push(solve(a, stats));
}
sections.reverse(); // put in logical order
// Find best section:
maxValue = sections[0].maxValue;
for (i = sections.length - 1; i >= 0; i--) {
maxValue = Math.max(sections[i].maxValue, maxValue);
}
for (i = sections.length - 1; i >= 0 && sections[i].maxValue < maxValue; i--);
// Which choice led to the highest value (choice = 0 or 1)?
choice = (i >= sections.length / 2)
// Discard the not-chosen version
sections = sections.slice(choice * sections.length/2);
}
// Reconstruct the solution from the sections.
a = [].concat.apply([], sections);
a.maxValue = maxValue;
}
if (traceOrig) stats.trace = traceOrig;
return a;
}
function randomValues(len) {
var a = [];
for (var i = 0; i < len; i++) {
// 50% chance for a 1, 25% for a 2, ... etc.
a.push(Math.min(/\.1*/.exec(Math.random().toString(2))[0].length,5));
}
return a;
}
// I/O
var inputEl = document.querySelector('#inp');
var randEl = document.querySelector('#rand');
var lenEl = document.querySelector('#len');
var goEl = document.querySelector('#go');
var outEl = document.querySelector('#out');
goEl.onclick = function() {
// Get the input and structure it
var a = unstr(inputEl.value),
stats = {
total_applied_merges: 0,
trace: a.length < 100 ? [] : undefined
};
// Apply algorithm
a = solve(a, stats);
// Output results
var output = {
value: a.maxValue,
compact: str(a),
total_applied_merges: stats.total_applied_merges,
trace: stats.trace || 'no trace produced (input too large)'
};
outEl.textContent = JSON.stringify(output, null, 4);
}
randEl.onclick = function() {
// Get input (count of numbers to generate):
len = lenEl.value;
// Generate
var a = randomValues(len);
// Output
inputEl.value = a.join(' ');
// Simulate click to find the solution immediately.
goEl.click();
}
// Tests
var tests = [
' ', '',
'1', '1',
'1 1', '2',
'2 2 1 2 2', '3 1 3',
'3 2 1 1 2 2 3', '5',
'3 2 1 1 2 2 3 1 1 1 1 3 2 2 1 1 2', '6',
'3 1 1 1 3', '3 2 1 3',
'2 1 1 1 2 1 1 1 2 1 1 1 1 1 2', '3 1 2 1 4 1 2',
'3 1 1 2 1 1 1 2 3', '4 2 1 2 3',
'1 4 2 1 1 1 1 1 1 1', '1 5 1',
];
var res;
for (var i = 0; i < tests.length; i+=2) {
var res = str(solve(unstr(tests[i])));
if (res !== tests[i+1]) throw 'Test failed: ' + tests[i] + ' returned ' + res + ' instead of ' + tests[i+1];
}
Enter series (space separated):<br>
<input id="inp" size="60" value="2 3 1 1 2 2"><button id="go">Solve</button>
<br>
<input id="len" size="4" value="30"><button id="rand">Produce random series of this size and solve</button>
<pre id="out"></pre>
As you can see the program produces a reduced array with the maximum value included. In general there can be many derived arrays that have this maximum; only one is given.
An O(n*m) time and space algorithm is possible, where, according to your stated limits, n <= 500 and m <= 58 (consider that even for a billion elements, m need only be about 60, representing the largest element ± log2(n)). m is representing the possible numbers 50 + floor(log2(500)):
Consider the condensed sequence, s = {[x, number of x's]}.
If M[i][j] = [num_j,start_idx] where num_j represents the maximum number of contiguous js ending at index i of the condensed sequence; start_idx, the index where the sequence starts or -1 if it cannot join earlier sequences; then we have the following relationship:
M[i][j] = [s[i][1] + M[i-1][j][0], M[i-1][j][1]]
when j equals s[i][0]
j's greater than s[i][0] but smaller than or equal to s[i][0] + floor(log2(s[i][1])), represent converting pairs and merging with an earlier sequence if applicable, with a special case after the new count is odd:
When M[i][j][0] is odd, we do two things: first calculate the best so far by looking back in the matrix to a sequence that could merge with M[i][j] or its paired descendants, and then set a lower bound in the next applicable cells in the row (meaning a merge with an earlier sequence cannot happen via this cell). The reason this works is that:
if s[i + 1][0] > s[i][0], then s[i + 1] could only possibly pair with the new split section of s[i]; and
if s[i + 1][0] < s[i][0], then s[i + 1] might generate a lower j that would combine with the odd j from M[i], potentially making a longer sequence.
At the end, return the largest entry in the matrix, max(j + floor(log2(num_j))), for all j.
JavaScript code (counterexamples would be welcome; the limit on the answer is set at 7 for convenient visualization of the matrix):
function f(str){
var arr = str.split(/\s+/).map(Number);
var s = [,[arr[0],0]];
for (var i=0; i<arr.length; i++){
if (s[s.length - 1][0] == arr[i]){
s[s.length - 1][1]++;
} else {
s.push([arr[i],1]);
}
}
var M = [new Array(8).fill([0,0])],
best = 0;
for (var i=1; i<s.length; i++){
M[i] = new Array(8).fill([0,i]);
var temp = s[i][1],
temp_odd,
temp_start,
odd = false;
for (var j=s[i][0]; temp>0; j++){
var start_idx = odd ? temp_start : M[i][j-1][1];
if (start_idx != -1 && M[start_idx - 1][j][0]){
temp += M[start_idx - 1][j][0];
start_idx = M[start_idx - 1][j][1];
}
if (!odd){
M[i][j] = [temp,start_idx];
temp_odd = temp;
} else {
M[i][j] = [temp_odd,-1];
temp_start = start_idx;
}
if (!odd && temp & 1 && temp > 1){
odd = true;
temp_start = start_idx;
}
best = Math.max(best,j + Math.floor(Math.log2(temp)));
temp >>= 1;
temp_odd >>= 1;
}
}
return [arr, s, best, M];
}
// I/O
var button = document.querySelector('button');
var input = document.querySelector('input');
var pre = document.querySelector('pre');
button.onclick = function() {
var val = input.value;
var result = f(val);
var text = '';
for (var i=0; i<3; i++){
text += JSON.stringify(result[i]) + '\n\n';
}
for (var i in result[3]){
text += JSON.stringify(result[3][i]) + '\n';
}
pre.textContent = text;
}
<input value ="2 2 3 3 2 2 3 3 5">
<button>Solve</button>
<pre></pre>
Here's a brute force solution:
function findMax(array A, int currentMax)
for each pair (i, i+1) of indices for which A[i]==A[i+1] do
currentMax = max(A[i]+1, currentMax)
replace A[i],A[i+1] by a single number A[i]+1
currentMax = max(currentMax, findMax(A, currentMax))
end for
return currentMax
Given the array A, let currentMax=max(A[0], ..., A[n])
print findMax(A, currentMax)
The algorithm terminates because in each recursive call the array shrinks by 1.
It's also clear that it is correct: we try out all possible replacement sequences.
The code is extremely slow when the array is large and there's lots of options regarding replacements, but actually works reasonbly fast on arrays with small number of replaceable pairs. (I'll try to quantify the running time in terms of the number of replaceable pairs.)
A naive working code in Python:
def findMax(L, currMax):
for i in range(len(L)-1):
if L[i] == L[i+1]:
L[i] += 1
del L[i+1]
currMax = max(currMax, L[i])
currMax = max(currMax, findMax(L, currMax))
L[i] -= 1
L.insert(i+1, L[i])
return currMax
# entry point
if __name__ == '__main__':
L1 = [2, 3, 1, 1, 2, 2]
L2 = [2, 3, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2]
print findMax(L1, max(L1))
print findMax(L2, max(L2))
The result of the first call is 4, as expected.
The result of the second call is 5 as expected; the sequence that gives the result: 2,3,1,1,2,2,2,2,2,2,2,2, -> 2,3,1,1,3,2,2,2,2,2,2 -> 2,3,1,1,3,3,2,2,2,2, -> 2,3,1,1,3,3,3,2,2 -> 2,3,1,1,3,3,3,3 -> 2,3,1,1,4,3, -> 2,3,1,1,4,4 -> 2,3,1,1,5