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The problem i have goes as follows (simplified):
I have a board, represented as a matrix of n x m squares (n might equal m)
In it, there are p game pieces
Each game piece has a pre-defined speed, which is how many steps it can take in it's turn
Pieces can't overlap
There are three types of cells: those which don't require extra movements to be crossed (you loose 0 extra speed when going through), those which require 1 extra movement to be crossed and some which you simply can't get through (like a wall)
So, given a game piece in a certain [i,j] position in my game board, i want to find out:
a) All the places it can move to, with it's speed
b) The path to a certain [k,l] position in the board
Having a) solved, b) is almost trivial.
Currently the algorithm i'm using goes as follows, assuming a language where arrays of size n go from 0 to n-1:
Create a sqaure matrix of speed*2+1 size which represents the cost of moving as if all cells had no extra cost to be crossed (the piece is on the position [speed, speed])
Create another square matrix of speed*2+1 size which has the extra costs of each cell (those which can't be crossed because either it's a wall or there is another piece in it has a value of infinite)(the piece is on the position [speed, speed])
Create another square matrix of speed*2+1 size which is the sum of the former two(the piece is on the position [speed, speed])
Correct the latter matrix making sure the value of each cell is: the minimal cost of all the adjacent cells + 1 + the extra cost of the cell. If it isn't, i correct it and start with the matrix all over again.
An example:
P are pieces, W are walls, E are empty cells which require no extra movement, X are cells which require 1 extra movement to be crossed.
X,E,X,X,X
X,X,X,X,X
W,E,E,E,W
W,E,X,E,W
E,P,P,P,P
The first matrix:
2,2,2,2,2
2,1,1,1,2
2,1,0,1,2
2,1,1,1,2
2,2,2,2,2
The second matrix:
1,0,1,inf,1
1,1,1,1,1
inf,0,0,0,inf
inf,0,1,0,inf
0,inf,inf,inf,inf
The sum:
3,2,3,3,3
3,2,2,2,3
inf,1,0,1,inf
inf,1,2,1,inf
inf,inf,inf,inf,inf
Since [0,0] is not 2+1+1, i correct it:
The sum:
4,2,3,3,3
3,2,2,2,3
inf,1,0,1,inf
inf,1,2,1,inf
inf,inf,inf,inf,inf
Since [0,1] is not 2+1+0, i correct it:
The sum:
4,3,3,3,3
3,2,2,2,3
inf,1,0,1,inf
inf,1,2,1,inf
inf,inf,inf,inf,inf
Since [0,2] is not 2+1+1, i correct it:
The sum:
4,2,4,3,3
3,2,2,2,3
inf,1,0,1,inf
inf,1,2,1,inf
inf,inf,inf,inf,inf
Which one is the correct answer?
What I want to know is if this problem has a name I can search it by (couldn't find anything) or if anybody can tell me how to solve the point a).
Note that I want the optimal solution, so I went with a dynamic programming algorithm. Might random walkers be better? AFAIK, this solution is not failing (yet), but I have no proof of correctness for it, and I want to be sure it works.
A-star is a standard algorithm to determine shortest path give obstacles on a 2d board and cost per square of moving. You can also use it to test if a specific move is valid, but to actually generate all valid moves I would simply start ay the start position, move in each direction by one square mark which squares are valid and then repeat from each of your new places making sure not to visit the same square again. It will be a recursive algorithm calling itself at most 4 times on each call and will generate you valid moves efficiently. If there are constraints like how many squares you can move at once with different costs just pass the running total of how far you've come for each square.
I'm currently trying to solve an algorithm problem from last year's Polish Collegiate Championships which reads as follows:
The Lord Mayor of Bytetown plans to locate a number of radar speed
cameras in the city. There are n intersections in Bytetown numbered
from 1 to n, and n-1 two way street segments. Each of these street
segments stretches between two intersections. The street network
allows getting from each intersection to any other.
The speed cameras are to be located at the intersections (maximum one
per intersection), wherein The Lord Mayor wants to maximise the number
of speed cameras. However, in order not to aggravate Byteland
motorists too much, he decided that on every route running across
Bytetown roads that does not pass through any intersection twice there
can be maximum k speed cameras (including those on endpoints of the
route). Your task is to write a program which will determine where the
speed cameras should be located.
Input
The first line of input contains two integers n and k (1 <= n, k <=
1000000): the number of intersections in Bytetown and maximum number
of speed cameras which can be set up on an individual route. The lines
that follow describe Bytetown street network: the i-th line contains
two integers a_i and b_i (1 <= a_i, b_i <= n), meaning that there is a
two-way street segment which joins two intersections numbered a_i and
b_i.
Output
The first output line should produce m: the number describing the
maximum number of speed cameras, that can be set up in Byteland. The
second line should produce a sequence of m numbers describing the
intersections where the speed cameras should be constructed. Should
there be many solutions, your program may output any one of them.
Example
For the following input data:
5 2
1 3
2 3
3 4
4 5
one of the correct results is:
3 1 2 4
So judging by how many teams solved it, I'm guessing it can't be too hard but still, I got stuck almost immediately with no idea as to how to move on. Since we know that "on every route running across Bytetown roads that does not pass through any intersection twice there can be maximum k speed camera", I guess we first have to somehow dissect the graphs into components being possible routes around the town. This alone seems like a really hard thing to do cause supposing there's an intersection with four motorways coming out of it, it already creates three possible directions for every enter point, thus making 12 routes. Not to mention how the situation complicates when there's more such four-handed intersections.
Maybe I'm approaching the task from the wrong angle? Could you please help?
It seems greedy works here
while k >= 2
mark all leaves of the tree and remove them
k = k - 2;
if ( k == 1 )
mark any 1 of remaining vertices
Easy problem and known algorithm:
I have a big array with 100 members. First X members are 0, and the rest are 1. Find X.
I am solving it by a binary search: Check member 50, if it is 0 - check member 75, etc, until I find adjacent 0 and 1.
I am looking for an optimized algorithm for the same problem in 2-dimensions:
I have 2-dimensional array 100*100. Those members that are on rows 0-X AND on columns 0-Y are 0, and the rest are 1. How to find Y and X?
Edit : The optimal solution consists in two simple binary search.
I'm very sorry for the long and convoluted post I did below. What the problem fundamentally consists in is to find a point in a space that contains 100*100 elements. The best you can do is to divide at each step this space in two. You can do it in a convoluted way (the one I did in the rest of the post) But if you realize that a binary search on the X axis still divides the research space in two at each step, (the same goes for the Y axis) then you understand that it's optimal.
I still let the thing I did, and I'm sorry that I made some peremptory affirmations in it.
If you're looking for a simple algorithm (though not optimal) just run the binary search twice as suggested.
However, if you want an optimal algorithm, you can look for the boundary on X and on Y at the same time. (You have to note that the two algorithm have same asymptotical complexity, but the optimal algorithm will still be faster)
In all the following graphics, the point (0, 0) is in the bottom left corner.
Basically when you choose a point and get the result, you cut your space in two parts. When you think about it that is actually the biggest amount of information you can extract from this.
If you choose the point (the black cross) and the result is 1 (red lines), this means that the point you're looking for can not be in the gray space (thus must be in the remaining white area)
On the other hand, if the value is 0 (blue lines), this means that the point you're looking for can not be in the gray area (thus must be in the remaining white area)
So, if you get one 0 result and one 1 result, this is what you'll get :
The point you're looking for is either in rectangle 1, 2 or 3. You just need to check the two corners of rectangle 3 to know which of the 3 rectangle is the good one.
So the algorithm is the following :
Note where are the bottom left and top right corner of the rectangle you're working with.
Do a binary search along the diagonal of the rectangle until you've stumbled at least once on a 1 result and once a 0 result.
Check the 2 other corners of the rectangle 3 (you'll necessary already know the values of the two corners on the diagonal) It is possible to check only one corner to know the right rectangle (but you'll have to check the two corners if the right rectangle is the rectangle 3)
Determine if the point you're looking for is in rectangle 1, 2 or 3
Repeat by reducing the problem to the good rectangle until the final rectangle is reduced to a point : it's the value you're looking for
Edit : if you want the supremum optimality, you'd not the when you choose the point (50, 50), you do not cut the space in equal part. One is three time bigger than the other. Ideally, you'll choose a point that cuts the space in two equal regions (area-wise)
You should compute once at the beginning the value of factor = (1.0 - 1.0/sqrt(2.0)). Then when you want to cut bewteen values a and b, choose the cutting point as a + factor*(b-a). When you cut the initial 100x100 rectangle at the point (100*factor, 100*factor) the two regions will have an area (100*100)/2, thus the convergence will be quicker.
Run your binary search twice. First determine X by running binary search on the last row and then determine Y by running binary search on last column.
Simple solution: go first in X-direction and then in Y-direction.
Check (0,50); If it is 0, check (0,75); until You find adjacent 0 and 1. Then go to Y direction from there.
Second solution:
Check member (50,50). If it is 1, check (25,25), until You find 0. Continue, until You find adjacent (X,X) and (X+1,X+1) that are 0 and 1. Then test (X,X+1) and (X+1,X). Neither or one of them will be 1. If neither, You are finished. If only one, say for example (X+1,X), then You know that the box's size is between (X+1,X) and (100,X). Use binary search to find box's height.
EDIT: As Chris pointed out, it seems that the simple approach is faster.
Second solution (modified):
Check member (50,50). If it is 1, check (25,25), until You find 0. Continue, until You find adjacent (X,X) and (X+1,X+1) that are 0 and 1. Then test (X,X+1). If it is 1, then do binary search on line (X,X+1)...(X,100). Else do binary search on line (X,X)...(100,X).
Even then I am probably beating a dead horse here. If it will be faster, then by neglible amount. This is just for theoretical fun. :)
EDIT 2 As Fezvez and Chris put it, binary search divides the search space in two most efficiently; My approach divides the area to 1/4 and 3/4 pieces. Fezvez pointed out that this could be remedied by calculating the dividing factor beforehand (but that would be extra calculation). In modified version of my algorithm I choose the direction where to go (X or Y direction), which effectively also divides the search space in two, and then conduct binary search. To conclude, this shows that this approach will always be a bit slower. (and more complicated to implement.)
Thank You, Igor Oks, for interesting question. :)
Use binary search on both dimensions and the 1D case:
Start with j=50. Now the 1-D array obtained by varying i is of the desired form - so find X from 1D case.
If X = 100 (i.e. no ones), then make j=75 (middle of the range in j dimension) and repeat.
If X < 100, then you have found it. All that is left is to fix i=X and find Y from the 1D case.
I'm trying to find the optimal solution for a little puzzle game called Twiddle (an applet with the game can be found here). The game has a 3x3 matrix with the number from 1 to 9. The goal is to bring the numbers in the correct order using the minimum amount of moves. In each move you can rotate a 2x2 square either clockwise or counterclockwise.
I.e. if you have this state
6 3 9
8 7 5
1 2 4
and you rotate the upper left 2x2 square clockwise you get
8 6 9
7 3 5
1 2 4
I'm using a A* search to find the optimal solution. My f() is simply the number of rotations needed. My heuristic function already leads to the optimal solution (if I modify it, see the notice a t the end) but I don't think it's the best one you can find. My current heuristic takes each corner, looks at the number at the corner and calculates the manhatten distance to the position this number will have in the solved state (which gives me the number of rotation needed to bring the number to this postion) and sums all these values. I.e. You take the above example:
6 3 9
8 7 5
1 2 4
and this end state
1 2 3
4 5 6
7 8 9
then the heuristic does the following
6 is currently at index 0 and should by at index 5: 3 rotations needed
9 is currently at index 2 and should by at index 8: 2 rotations needed
1 is currently at index 6 and should by at index 0: 2 rotations needed
4 is currently at index 8 and should by at index 3: 3 rotations needed
h = 3 + 2 + 2 + 3 = 10
Additionally, if h is 0, but the state is not completely ordered, than h = 1.
But there is the problem, that you rotate 4 elements at once. So there a rare cases where you can do two (ore more) of theses estimated rotations in one move. This means theses heuristic overestimates the distance to the solution.
My current workaround is, to simply excluded one of the corners from the calculation which solves this problem at least for my test-cases. I've done no research if really solves the problem or if this heuristic still overestimates in some edge-cases.
So my question is: What is the best heuristic you can come up with?
(Disclaimer: This is for a university project, so this is a bit of homework. But I'm free to use any resource if can come up with, so it's okay to ask you guys. Also I will credit Stackoverflow for helping me ;) )
Simplicity is often most effective. Consider the nine digits (in the rows-first order) as forming a single integer. The solution is represented by the smallest possible integer i(g) = 123456789. Hence I suggest the following heuristic h(s) = i(s) - i(g). For your example, h(s) = 639875124 - 123456789.
You can get an admissible (i.e., not overestimating) heuristic from your approach by taking all numbers into account, and dividing by 4 and rounding up to the next integer.
To improve the heuristic, you could look at pairs of numbers. If e.g. in the top left the numbers 1 and 2 are swapped, you need at least 3 rotations to fix them both up, which is a better value than 1+1 from considering them separately. In the end, you still need to divide by 4. You can pair up numbers arbitrarily, or even try all pairs and find the best division into pairs.
All elements should be taken into account when calculating distance, not just corner elements. Imagine that all corner elements 1, 3, 7, 9 are at their home, but all other are not.
It could be argued that those elements that are neighbors in the final state should tend to become closer during each step, so neighboring distance can also be part of heuristic, but probably with weaker influence than distance of elements to their final state.
This is a followup to my earlier question about deciding if a hand is ready.
Knowledge of mahjong rules would be excellent, but a poker- or romme-based background is also sufficient to understand this question.
In Mahjong 14 tiles (tiles are like
cards in Poker) are arranged to 4 sets
and a pair. A straight ("123") always
uses exactly 3 tiles, not more and not
less. A set of the same kind ("111")
consists of exactly 3 tiles, too. This
leads to a sum of 3 * 4 + 2 = 14
tiles.
There are various exceptions like Kan
or Thirteen Orphans that are not
relevant here. Colors and value ranges
(1-9) are also not important for the
algorithm.
A hand consists of 13 tiles, every time it's our turn we get to pick a new tile and have to discard any tile so we stay on 13 tiles - except if we can win using the newly picked tile.
A hand that can be arranged to form 4 sets and a pair is "ready". A hand that requires only 1 tile to be exchanged is said to be "tenpai", or "1 from ready". Any other hand has a shanten-number which expresses how many tiles need to be exchanged to be in tenpai. So a hand with a shanten number of 1 needs 1 tile to be tenpai (and 2 tiles to be ready, accordingly). A hand with a shanten number of 5 needs 5 tiles to be tenpai and so on.
I'm trying to calculate the shanten number of a hand. After googling around for hours and reading multiple articles and papers on this topic, this seems to be an unsolved problem (except for the brute force approach). The closest algorithm I could find relied on chance, i.e. it was not able to detect the correct shanten number 100% of the time.
Rules
I'll explain a bit on the actual rules (simplified) and then my idea how to tackle this task. In mahjong, there are 4 colors, 3 normal ones like in card games (ace, heart, ...) that are called "man", "pin" and "sou". These colors run from 1 to 9 each and can be used to form straights as well as groups of the same kind. The forth color is called "honors" and can be used for groups of the same kind only, but not for straights. The seven honors will be called "E, S, W, N, R, G, B".
Let's look at an example of a tenpai hand: 2p, 3p, 3p, 3p, 3p, 4p, 5m, 5m, 5m, W, W, W, E. Next we pick an E. This is a complete mahjong hand (ready) and consists of a 2-4 pin street (remember, pins can be used for straights), a 3 pin triple, a 5 man triple, a W triple and an E pair.
Changing our original hand slightly to 2p, 2p, 3p, 3p, 3p, 4p, 5m, 5m, 5m, W, W, W, E, we got a hand in 1-shanten, i.e. it requires an additional tile to be tenpai. In this case, exchanging a 2p for an 3p brings us back to tenpai so by drawing a 3p and an E we win.
1p, 1p, 5p, 5p, 9p, 9p, E, E, E, S, S, W, W is a hand in 2-shanten. There is 1 completed triplet and 5 pairs. We need one pair in the end, so once we pick one of 1p, 5p, 9p, S or W we need to discard one of the other pairs. Example: We pick a 1 pin and discard an W. The hand is in 1-shanten now and looks like this: 1p, 1p, 1p, 5p, 5p, 9p, 9p, E, E, E, S, S, W. Next, we wait for either an 5p, 9p or S. Assuming we pick a 5p and discard the leftover W, we get this: 1p, 1p, 1p, 5p, 5p, 5p, 9p, 9p, E, E, E, S, S. This hand is in tenpai in can complete on either a 9 pin or an S.
To avoid drawing this text in length even more, you can read up on more example at wikipedia or using one of the various search results at google. All of them are a bit more technical though, so I hope the above description suffices.
Algorithm
As stated, I'd like to calculate the shanten number of a hand. My idea was to split the tiles into 4 groups according to their color. Next, all tiles are sorted into sets within their respective groups to we end up with either triplets, pairs or single tiles in the honor group or, additionally, streights in the 3 normal groups. Completed sets are ignored. Pairs are counted, the final number is decremented (we need 1 pair in the end). Single tiles are added to this number. Finally, we divide the number by 2 (since every time we pick a good tile that brings us closer to tenpai, we can get rid of another unwanted tile).
However, I can not prove that this algorithm is correct, and I also have trouble incorporating straights for difficult groups that contain many tiles in a close range. Every kind of idea is appreciated. I'm developing in .NET, but pseudo code or any readable language is welcome, too.
I've thought about this problem a bit more. To see the final results, skip over to the last section.
First idea: Brute Force Approach
First of all, I wrote a brute force approach. It was able to identify 3-shanten within a minute, but it was not very reliable (sometimes too a lot longer, and enumerating the whole space is impossible even for just 3-shanten).
Improvement of Brute Force Approach
One thing that came to mind was to add some intelligence to the brute force approach. The naive way is to add any of the remaining tiles, see if it produced Mahjong, and if not try the next recursively until it was found. Assuming there are about 30 different tiles left and the maximum depth is 6 (I'm not sure if a 7+-shanten hand is even possible [Edit: according to the formula developed later, the maximum possible shanten number is (13-1)*2/3 = 8]), we get (13*30)^6 possibilities, which is large (10^15 range).
However, there is no need to put every leftover tile in every position in your hand. Since every color has to be complete in itself, we can add tiles to the respective color groups and note down if the group is complete in itself. Details like having exactly 1 pair overall are not difficult to add. This way, there are max around (13*9)^6 possibilities, that is around 10^12 and more feasible.
A better solution: Modification of the existing Mahjong Checker
My next idea was to use the code I wrote early to test for Mahjong and modify it in two ways:
don't stop when an invalid hand is found but note down a missing tile
if there are multiple possible ways to use a tile, try out all of them
This should be the optimal idea, and with some heuristic added it should be the optimal algorithm. However, I found it quite difficult to implement - it is definitely possible though. I'd prefer an easier to write and maintain solution first.
An advanced approach using domain knowledge
Talking to a more experienced player, it appears there are some laws that can be used. For instance, a set of 3 tiles does never need to be broken up, as that would never decrease the shanten number. It may, however, be used in different ways (say, either for a 111 or a 123 combination).
Enumerate all possible 3-set and create a new simulation for each of them. Remove the 3-set. Now create all 2-set in the resulting hand and simulate for every tile that improves them to a 3-set. At the same time, simulate for any of the 1-sets being removed. Keep doing this until all 3- and 2-sets are gone. There should be a 1-set (that is, a single tile) be left in the end.
Learnings from implementation and final algorithm
I implemented the above algorithm. For easier understanding I wrote it down in pseudocode:
Remove completed 3-sets
If removed, return (i.e. do not simulate NOT taking the 3-set later)
Remove 2-set by looping through discarding any other tile (this creates a number of branches in the simulation)
If removed, return (same as earlier)
Use the number of left-over single tiles to calculate the shanten number
By the way, this is actually very similar to the approach I take when calculating the number myself, and obviously never to yields too high a number.
This works very well for almost all cases. However, I found that sometimes the earlier assumption ("removing already completed 3-sets is NEVER a bad idea") is wrong. Counter-example: 23566M 25667P 159S. The important part is the 25667. By removing a 567 3-set we end up with a left-over 6 tile, leading to 5-shanten. It would be better to use two of the single tiles to form 56x and 67x, leading to 4-shanten overall.
To fix, we simple have to remove the wrong optimization, leading to this code:
Remove completed 3-sets
Remove 2-set by looping through discarding any other tile
Use the number of left-over single tiles to calculate the shanten number
I believe this always accurately finds the smallest shanten number, but I don't know how to prove that. The time taken is in a "reasonable" range (on my machine 10 seconds max, usually 0 seconds).
The final point is calculating the shanten out of the number of left-over single tiles. First of all, it is obvious that the number is in the form 3*n+1 (because we started out with 14 tiles and always subtracted 3 tiles).
If there is 1 tile left, we're shanten already (we're just waiting for the final pair). With 4 tiles left, we have to discard 2 of them to form a 3-set, leaving us with a single tile again. This leads to 2 additional discards. With 7 tiles, we have 2 times 2 discards, adding 4. And so on.
This leads to the simple formula shanten_added = (number_of_singles - 1) * (2/3).
The described algorithm works well and passed all my tests, so I'm assuming it is correct. As stated, I can't prove it though.
Since the algorithm removes the most likely tiles combinations first, it kind of has a built-in optimization. Adding a simple check if (current_depth > best_shanten) then return; it does very well even for high shanten numbers.
My best guess would be an A* inspired approach. You need to find some heuristic which never overestimates the shanten number and use it to search the brute-force tree only in the regions where it is possible to get into a ready state quickly enough.
Correct algorithm sample: syanten.cpp
Recursive cut forms from hand in order: sets, pairs, incomplete forms, - and count it. In all variations. And result is minimal Shanten value of all variants:
Shanten = Min(Shanten, 8 - * 2 - - )
C# sample (rewrited from c++) can be found here (in Russian).
I've done a little bit of thinking and came up with a slightly different formula than mafu's. First of all, consider a hand (a very terrible hand):
1s 4s 6s 1m 5m 8m 9m 9m 7p 8p West East North
By using mafu's algorithm all we can do is cast out a pair (9m,9m). Then we are left with 11 singles. Now if we apply mafu's formula we get (11-1)*2/3 which is not an integer and therefore cannot be a shanten number. This is where I came up with this:
N = ( (S + 1) / 3 ) - 1
N stands for shanten number and S for score sum.
What is score? It's a number of tiles you need to make an incomplete set complete. For example, if you have (4,5) in your hand you need either 3 or 6 to make it a complete 3-set, that is, only one tile. So this incomplete pair gets score 1. Accordingly, (1,1) needs only 1 to become a 3-set. Any single tile obviously needs 2 tiles to become a 3-set and gets score 2. Any complete set of course get score 0. Note that we ignore the possibility of singles becoming pairs. Now if we try to find all of the incomplete sets in the above hand we get:
(4s,6s) (8m,9m) (7p,8p) 1s 1m 5m 9m West East North
Then we count the sum of its scores = 1*3+2*7 = 17.
Now if we apply this number to the formula above we get (17+1)/3 - 1 = 5 which means this hand is 5-shanten. It's somewhat more complicated than Alexey's and I don't have a proof but so far it seems to work for me. Note that such a hand could be parsed in the other way. For example:
(4s,6s) (9m,9m) (7p,8p) 1s 1m 5m 8m West East North
However, it still gets score sum 17 and 5-shanten according to formula. I also can't proof this and this is a little bit more complicated than Alexey's formula but also introduces scores that could be applied(?) to something else.
Take a look here: ShantenNumberCalculator. Calculate shanten really fast. And some related stuff (in japanese, but with code examples) http://cmj3.web.fc2.com
The essence of the algorithm: cut out all pairs, sets and unfinished forms in ALL possible ways, and thereby find the minimum value of the number of shanten.
The maximum value of the shanten for an ordinary hand: 8.
That is, as it were, we have the beginnings for 4 sets and one pair, but only one tile from each (total 13 - 5 = 8).
Accordingly, a pair will reduce the number of shantens by one, two (isolated from the rest) neighboring tiles (preset) will decrease the number of shantens by one,
a complete set (3 identical or 3 consecutive tiles) will reduce the number of shantens by 2, since two suitable tiles came to an isolated tile.
Shanten = 8 - Sets * 2 - Pairs - Presets
Determining whether your hand is already in tenpai sounds like a multi-knapsack problem. Greedy algorithms won't work - as Dialecticus pointed out, you'll need to consider the entire problem space.