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Need help in building efficient exhaustive search algorithm
Imagine that you must open a locked door by inputting the correct 4-digit code on a keypad. After every keypress the lock evaluates the sequence of the last 4 digits inputted, i.e. by entering 123456 you have evaluated 3 codes: 1234, 2345 and 3456.
What is the shortest sequence of keypresses to evaluate all 10^4 different combinations?
Is there a method for traversing the entire space easy enough for a human to follow?
I have pondered this from time to time since a friend of mine had to brute force such a lock, to not having to spend the night outdoors in wintertime.
My feeble attempts at wrapping my head around it
With a code of length L=4 digits and an "alphabet" of digits of size D=10 the length of the optimal sequence cannot be shorter than D^L + L - 1. In simulations of smaller size than [L,D] = [4,10] I have obtained optimal results by semi-randomly searching the space. However I do not know if a solution exists for an arbitrary [L,D] pair and would not be able to remember the solution if I ever had to use it.
Lessons learned so far
When planning to spend the night at a friends house in another town, be sure to not arrive at 1 am if that person is going out to party and won't hear her cell phone.
I think you want a http://en.wikipedia.org/wiki/De_Bruijn_sequence - "a cyclic sequence of a given alphabet A with size k for which every possible subsequence of length n in A appears as a sequence of consecutive characters exactly once."
The link Evgeny provided should answer both of your quests. This answer is a bit offtopic, but you ask for a solution for humans.
In the real world you should probably rely more on Social engineering or heuristics, and after that on mathematics. I give a case on real life:
I went to visit an apartment and I found out that my cellphone was dead. Now way of contacting the person doing the visit. I was about to go back when I saw that the door used a keypad 0 - 9 and A B. I made several assumptions:
The code is 5 digits long. The length is pretty standard depending on the region you are in. I based this assumption on buildings I had access before (legally :D).
The code starts with numbers, then either A or B (based on my own building).
The keypad was not brand new. Conclusion, the numbers used in the code were a bit damaged. I knew with certainty which numbers were not in the code, and three of the four number in the code (given my previous assumptions)
By the amount of keys damaged I assumed the code didn't contain repeated keys (7 were damaged, it was clear A was used, B not used )
At the end I had 3 numbers which were in the code for sure, 2 candidates for the last number and I was sure A was at the end. On key was just slightly damaged compared to the others.
I just had to enumerate permutations starting with the candidate which seemed the more damaged, which give me 4! + 4! = 48 tries. Believe me, at the 5th try the door was opened. If I can give my 2 cents, the old put a key and open the door is still the most reliable method to restrict access to a building.
Related
I am working on a Wordle bot to calculate the best first (and subsequent) guess(es). I am assuming all 13,000 possible guesses are equally likely (for now), and I am running into a huge speed issue.
I can not think of a valid way to avoid a triple for loop, each with 13,000 elements. Obviously, this is ridiculously slow and would take about 20 hours on my laptop to compute the best first guess (I assume it would be faster for subsequent guesses due to fewer valid words).
The way I see it, I definitely need the first loop to test each word as a guess.
I need the second loop to find the colors/results given each possible answer with that guess.
Then, I need the third loop to see which guesses are valid given the guess and the colors.
Then, I find the average words remaining for each guess, and choose the one with the lowest average using a priority queue.
I think the first two loops are unavoidable, but maybe there's a way to get rid of the third?
Does anyone have any thoughts or suggestions?
I did a similar thing, for a list with 11,000 or so entries. It took 27 minutes.
I did it using a pregenerated (in one pass) list of the letters in a word,
as a bitfield (i.e. one 32 bit integer) and then did crude testing using
the AND instruction. if a word failed that, it exited the rest of the loop.
This is my first post on Stack Overflow, so please excuse my mistakes if I'm doing something wrong.
Ok so I'm trying to find an algorithm/function/something that can calculate how many times I have to do the same type of shuffle of 52 playing cards to get back to where I started.
The specific shuffle I'm using goes like this:
-You will have two piles.
-You have the deck with the back facing up. (Lets call this pile 1)
-You will now alternate between putting a card in the back of pile 1 Example: Let's say you have 4 cards in a pile, back facing up, going from 4 closest to the ground and 1 closest to the sky (Their order is 4,3,2,1. You take card 1 and put it beneath card 4 mening card 1 is now closest to the ground and card 4 is second closest, order is now 1,4,3,2. and putting one in pile 2. -Pile 2 will "stack downwards" meaning you will always put the new card at the bottom of that pile. (Back always facing up)
-The first card will always get put at the back of pile 1.
-Repeat this process until all cards are in pile 2.
-Now take pile 2 and do the exact same thing you just did.
My question is: How many times do I have to repeat this process until I get back where I started?
Side notes:
- If this is a common way of shuffling cards and there already is a solution, please let me know.
- I'm still new to math and coding so if writing up an equation/algorithm/code for this is really easy then don't laugh at me pls ;<.
- Sorry if I'm asking this at the wrong place, I don't know how all this works.
- English isn't my main language and I'm not a native speaker either so please excuse any bad grammar and/or other grammatical errors.
I do however have a code that does all of this (Link here) but I'm unsure if it's the most effective way to do it, and it hasn't given a result yet so I don't even know if it works. If you wan't to give tips or suggestions on how to change it then please do, I would really appreciate it. It's done in scratch however because I can't write in any other languages... sorry...
Thanks in advance.
Any fixed shuffle is equivalent to a permutation; what you want to know is the order of that permutation. This can be computed by decomposing the permutation into cycles and then computing the least common multiple of the cycle lengths.
I'm not able to properly understand your algorithm, but here's an example of shuffling 8 elements and then finding the number of times that shuffle needs to be repeated to get back to an unshuffled state.
Suppose the sequence starts as 1,2,3,4,5,6,7,8 and after one shuffle, it's 3,1,4,5,2,8,7,6.
The number 1 goes to position 2, then 2 goes to position 5, then 5 goes to position 4, then 4 goes to position 3, then 3 goes to position 1. So the first cycle is (1 2 5 4 3).
The number 6 goes to position 8, then 8 goes to position 6. So the next cycle is (6 8).
The number 7 stays in position 7, so this is a trivial cycle (7).
The lengths of the cycles are 5, 2 and 1, so the least common multiple is 10. This shuffle takes 10 iterations to get back to the intitial state.
If you don't mind sitting down with pen and paper for a while, you should be able to follow this procedure for your own shuffling algorithm.
Edit: This question is not a duplicate of What is the optimal algorithm for the game 2048?
That question asks 'what is the best way to win the game?'
This question asks 'how can we work out the complexity of the game?'
They are completely different questions. I'm not interested in which steps are required to move towards a 'win' state - I'm interested in in finding out whether the total number of possible steps can be calculated.
I've been reading this question about the game 2048 which discusses strategies for creating an algorithm that will perform well playing the game.
The accepted answer mentions that:
the game is a discrete state space, perfect information, turn-based game like chess
which got me thinking about its complexity. For deterministic games like chess, its possible (in theory) to work out all the possible moves that lead to a win state and work backwards, selecting the best moves that keep leading towards that outcome. I know this leads to a large number of possible moves (something in the range of the number of atoms in the universe).. but is 2048 more or less complex?
Psudocode:
for the current arrangement of tiles
- work out the possible moves
- work out what the board will look like if the program adds a 2 to the board
- work out what the board will look like if the program adds a 4 to the board
- move on to working out the possible moves for the new state
At this point I'm thinking I will be here a while waiting on this to run...
So my question is - how would I begin to write this algorithm - what strategy is best for calculating the complexity of the game?
The big difference I see between 2048 and chess is that the program can select randomly between 2 and 4 when adding new tiles - which seems add a massive number of additional possible moves.
Ultimately I'd like the program to output a single figure showing the number of possible permutations in the game. Is this possible?!
Let's determine how many possible board configurations there are.
Each tile can be either empty, or contain a 2, 4, 8, ..., 512 or 1024 tile.
That's 12 possibilities per tile. There are 16 tiles, so we get 1612 = 248 possible board states - and this most likely includes a few unreachable ones.
Assuming we could store all of these in memory, we could work backwards from all board states that would generate a 2048 tile in the next move, doing a constant amount of work to link reachable board states to each other, which should give us a probabilistic best move for each state.
To store all bits in memory, let's say we'd need 4 bits per tile, i.e. 64 bits = 8 bytes per board state.
248 board states would then require 8*248 = 2251799813685248 bytes = 2048 TB (not to mention added overhead to keep track of the best boards). That's a bit beyond what a desktop computer these days has, although it might be possible to cleverly limit the number of boards required at any given time as to get down to something that will fit on, say, a 3 TB hard drive, or perhaps even in RAM.
For reference, chess has an upper bound of 2155 possible positions.
If we were to actually calculate, from the start, every possible move (in a breadth-first search-like manner), we'd get a massive number.
This isn't the exact number, but rather a rough estimate of the upper bound.
Let's make a few assumptions: (which definitely aren't always true, but, for the sake of simplicity)
There are always 15 open squares
You always have 4 moves (left, right, up, down)
Once the total sum of all tiles on the board reaches 2048, it will take the minimum number of combinations to get a single 2048 (so, if placing a 2 makes the sum 2048, the combinations will be 2 -> 4 -> 8 -> 16 -> ... -> 2048, i.e. taking 10 moves)
A 2 will always get placed, never a 4 - the algorithm won't assume this, but, for the sake of calculating the upper bound, we will.
We won't consider the fact that there may be duplicate boards generated during this process.
To reach 2048, there needs to be 2048 / 2 = 1024 tiles placed.
You start with 2 randomly placed tiles, then repeatedly make a move and another tile gets placed, so there's about 1022 'turns' (a turn consisting of making a move and a tile getting placed) until we get a sum of 2048, then there's another 10 turns to get a 2048 tile.
In each turn, we have 4 moves, and there can be one of two tiles placed in one of 15 positions (30 possibilities), so that's 4*30 = 120 possibilities.
This would, in total, give us 1201032 possible states.
If we instead assume a 4 will always get placed, we get 120519 states.
Calculating the exact number will likely involve working our way through all these states, which won't really be viable.
EDIT: Just to make sure someone is not breaking their head on the problem... I am not looking for the best optimal algorithm. Some heuristic that makes sense is fine.
I made a previous attempt at formulating this and realized I did not do a great job at it so I removed that question. I have taken another shot at formulating my problem. Please feel free to provide any constructive criticism that can help me improve this.
Input:
N people
k announcements that I can make
Distance that my voice can be heard (say 5 meters) i.e. I may decide to announce or not depending on the number of people within these 5 meters
Goal:
Maximize the total number of people who have heard my k announcements and (optionally) minimize the time in which I can finish announcing all k announcements
Constraints:
Once a person hears my announcement, he is be removed from the total i.e. if he had heard my first announcement, I do not count him even if he hears my second announcement
I can see the same person as well as the same set of people within my proximity
Example:
Let us consider 10 people numbered from 1 to 10 and the following pattern of arrival:
Time slot 1: 1 (payoff = 1)
Time slot 2: 2 3 4 5 (payoff = 4)
Time slot 3: 5 6 7 8 (payoff = 4 if no announcement was made previously in time slot 2, 3 if an announcement was made in time slot 2)
Time slot 4: 9 10 (payoff = 2)
and I am given 2 announcements to make. Now if I were an oracle, I would choose time slots 2 and time slots 3 because then 7 people would have heard (because 5 already heard my announcement in Time slot 2, I do not consider him anymore). I am looking for an online algorithm that will help me make these decisions on whether or not to make an announcement and if so based on what factors. Does anyone have any ideas on what algorithms can be used to solve this or a simpler version of this problem?
There should be an approach relying upon a max-flow algorithm. In essence, you're trying to push the maximum amount of messages from start->end. Though it would be multidimensional, you could have a super-sink, which connects to each value of t, then have each value of t connect to the people you can reach at this time and then have a super-sink. This way, you simply have to compute a max-flow (with the added constraint of no more than k shouts, which should be solvable with a bit of dynamic programming). It's a terrifically dirty way to solve it, but it should get the job done deterministically and without the use of heuristics.
I don't know that there is really a way to solve this or an algorithm to do it the way you have formulated it.
It seems like basically you are trying to reach the maximum number of people with exactly 2 announcements. But without knowing any information about the groups of people in advance, you can't really make any kind of intelligent decision about whether or not to use your first announcement. Your second one at least has the benefit of knowing when not to be used (i.e. if the group has no new members then you can know its not worth wasting the announcement). But it still has basically the same problem.
The only real way to solve this is to use knowledge about the type of data or the desired outcome to make guesses. If you know that groups average 100 people with a standard deviation of 10, then you could just refuse to announce if less than 90 people are present. Or, if you know you need to reach at least 100 people with two announcements, you could choose never to announce to less than 50 at once. Obviously those approaches risk never announcing at all if the actual data does not meet what you would expect. But that's always going to be a risk, since you could get 1 person in the first group and then 0 in all of the rest, no matter what you do.
Or, you could try more clearly defining the problem, I have a hard time figuring out how to relate this to computers.
Lets start my trying to solve the simplest possible variant of the problem: Lets assume N people and K timeslots, but only one possible announcement. Lets also assume that each person will only ever stay for one timeslot and that each person who hasn't yet shown up has an equally probable chance of showing up at any future timeslot.
Given these simplifications, at each timeslot you look at the payoff of announcing at the current timeslot and compare to the chance of a future timeslot having a higher payoff, eg, lets assume 4 people 3 timeslots:
Timeslot 1: Person 1 shows up, so you know you could get a payoff of 1 by announcing, but then you have 3 people to show up in 2 remaining timeslots, so at least one of those timeslots is guaranteed to have 2 people, so don't announce..
So at each timeslot, you can calculate the chance that a later timeslot will have a higher payoff than the current by treating the remaining (N) people and (K) timeslots as being N independent random numbers each from 1..k, and calculate the chance of at least one value k being hit more than or equal to the current-payoff times. (Similar to the Birthday problem, but for more than 1 collision) and then you need to decide hwo much to discount based on expected variances. (bird in the hand, etc)
Generalization of this solution to the original problem is left as an exercise for the reader.
Here is the facts first.
In the game of bridge there are 4
players named North, South, East and
West.
All 52 cards are dealt with 13 cards
to each player.
There is a Honour counting systems.
Ace=4 points, King=3 points, Queen=2
points and Jack=1 point.
I'm creating a "Card dealer" with constraints where for example you might say that the hand dealt to north has to have exactly 5 spades and between 13 to 16 Honour counting points, the rest of the hands are random.
How do I accomplish this without affecting the "randomness" in the best way and also having effective code?
I'm coding in C# and .Net but some idea in Pseudo code would be nice!
Since somebody already mentioned my Deal 3.1, I'd like to point out some of the optimizations I made in that code.
First of all, to get the most flexibly constraints, I wanted to add a complete programming language to my dealer, so you could generate whole libraries of constraints with different types of evaluators and rules. I used Tcl for that language, because I was already learning it for work, and, in 1994 when Deal 0.0 was released, Tcl was the easiest language to embed inside a C application.
Second, I needed the constraint language to run fairly fast. The constraints are running deep inside the loop. Quite a lot of code in my dealer is little optimizations with lookup tables and the like.
One of the most surprising and simple optimizations was to not deal cards to a seat until a constraint is checked on that seat. For example, if you want north to match constraint A and south to match constraint B, and your constraint code is:
match constraint A to north
match constraint B to south
Then only when you get to the first line do you fill out the north hand. If it fails, you reject the complete deal. If it passes, next fill out the south hand and check its constraint. If it fails, throw out the entire deal. Otherwise, finish the deal and accept it.
I found this optimization when doing some profiling and noticing that most of the time was spent in the random number generator.
There is one fancy optimization, which can work in some instances, call "smart stacking."
deal::input smartstack south balanced hcp 20 21
This generates a "factory" for the south hand which takes some time to build but which can then very quickly fill out the one hand to match this criteria. Smart stacking can only be applied to one hand per deal at a time, because of conditional probability problems. [*]
Smart stacking takes a "shape class" - in this case, "balanced," a "holding evaluator", in this case, "hcp", and a range of values for the holding evaluator. A "holding evaluator" is any evaluator which is applied to each suit and then totaled, so hcp, controls, losers, and hcp_plus_shape, etc. are all holding evalators.
For smartstacking to be effective, the holding evaluator needs to take a fairly limited set of values. How does smart stacking work? That might be a bit more than I have time to post here, but it's basically a huge set of tables.
One last comment: If you really only want this program for bidding practice, and not for simulations, a lot of these optimizations are probably unnecessary. That's because the very nature of practicing makes it unworthy of the time to practice bids that are extremely rare. So if you have a condition which only comes up once in a billion deals, you really might not want to worry about it. :)
[Edit: Add smart stacking details.]
Okay, there are exactly 8192=2^13 possible holdings in a suit. Group them by length and honor count:
Holdings(length,points) = { set of holdings with this length and honor count }
So
Holdings(3,7) = {AK2, AK3,...,AKT,AQJ}
and let
h(length,points) = |Holdings(length,points)|
Now list all shapes that match your shape condition (spades=5):
5-8-0-0
5-7-1-0
5-7-0-1
...
5-0-0-8
Note that the collection of all possible hand shapes has size 560, so this list is not huge.
For each shape, list the ways you can get the total honor points you are looking for by listing the honor points per suit. For example,
Shape Points per suit
5-4-4-0 10-3-0-0
5-4-4-0 10-2-1-0
5-4-4-0 10-1-2-0
5-4-4-0 10-0-3-0
5-4-4-0 9-4-0-0
...
Using our sets Holdings(length,points), we can compute the number of ways to get each of these rows.
For example, for the row 5-4-4-0 10-3-0-0, you'd have:
h(5,10)*h(4,3)*h(4,0)*h(0,0)
So, pick one of these rows at random, with relative probability based on the count, and then, for each suit, choose a holding at random from the correct Holdings() set.
Obviously, the wider the range of hand shapes and points, the more rows you will need to pre-compute. A little more code, you can still do this with some cards pre-determined - if you know where the ace of spades or west's whole hand or whatever.
[*] In theory, you can solve these conditional probability issues for smart stacking with multiple hands, but the solution to the problem would make it effective only for extremely rare types of deals. That's because the number of rows in the factory's table is roughly the product of the number of rows for stacking one hand times the number of rows for stacking the other hand. Also, the h() table has to be keyed on the number of ways of dividing the n cards amongst hand 1, hand 2, and other hands, which changes the number of values from roughly 2^13 to 3^13 possible values, which is about two orders of magnitude bigger.
Since the numbers are quite small here, you could just take the heuristic approach: Randomly deal your cards, evaluate the constraints and just deal again if they are not met.
Depending on how fast your computer is, it might be enough to do this:
Repeat:
do a random deal
Until the board meets all the constraints
As with all performance questions, the thing to do is try it and see!
edit I tried it and saw:
done 1000000 hands in 12914 ms, 4424 ok
This is without giving any thought to optimisation - and it produces 342 hands per second meeting your criteria of "North has 5 spades and 13-16 honour points". I don't know the details of your application but it seems to me that this might be enough.
I would go for this flow, which I think does not affect the randomness (other than by pruning solutions that do not meet constraints):
List in your program all possible combinations of "valued" cards whose total Honour points count is between 13 and 16. Then pick randomly one of these combinations, removing the cards from a fresh deck.
Count how many spades you already have among the valued cards, and pick randomly among the remaining spades of the deck until you meet the count.
Now pick from the deck as much non-spades, non-valued cards as you need to complete the hand.
Finally pick the other hands among the remaining cards.
You can write a program that generates the combinations of my first point, or simply hardcode them while accounting for color symmetries to reduce the number of lines of code :)
Since you want to practise bidding, I guess you will likely be having various forms of constraints (and not just 1S opening, as I guess for this current problem) coming up in the future. Trying to come up with the optimal hand generation tailored to the constraints could be a huge time sink and not really worth the effort.
I would suggest you use rejection sampling: Generate a random deal (without any constraints) and test if it satisfies your constraints.
In order to make this feasible, I suggest you concentrate on making the random deal generation (without any constraints) as fast as you can.
To do this, map each hand to a 12byte integer (the total number of bridge hands fits in 12 bytes). Generating a random 12 byte integer can be done in just 3, 4 byte random number calls, of course since the number of hands is not exactly fitting in 12 bytes, you might have a bit of processing to do here, but I expect it won't be too much.
Richard Pavlicek has an excellent page (with algorithms) to map a deal to a number and back.
See here: http://www.rpbridge.net/7z68.htm
I would also suggest you look at the existing bridge hand dealing software (like Deal 3.1, which is freely available) too. Deal 3.1 also supports doing double dummy analysis. Perhaps you could make it work for you without having to roll one of your own.
Hope that helps.