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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.
Task definition:
I have a matrix of natural numbers. The task is to find path from the top-left corner of matrix to bottom-right corner of matrix and dial maximum score.
Rules of navigation: if you are located in [i][j] you can move:
a) to [i][j-1], [i][j+1], [i+1][j] cells and dial zero points
b) to [i+1][j+1] and dial matrix[i][j] points
Little example:
Assume you have score 50and matrix
0 3 5 3 2
4 7 2 5 2
4 3 5 2 5
Assume you are in [1][1] cell (matrix[1][1] = 7). You can navigate to:
a) [1][0] cell with 50 score
b) [1][2] cell with 50 score
c) [2][1] cell with 50 score
d) [2][2] cell with 57 score
What a problem:
I solve this task in very slow way...
I try to implement in with help of recursion. It's easy if you just want to find maximum score. Something like
public int loop(int i, int j) {
int left = loop(i, j-1);
int top = loop(i-1, j);
int diagonal = loop(i-1,j-1) + matrix[i-1][j-1];
return maximum(left, top, diagonal);
}
BUT, I want to find a path with maximum score! And it's very time/memory consuming.
Why it's time/memory consuming:
And there is one problem: I need store path-collection and pass it as a parameter to the loop method. But loop method forks on each iteration and I have to copy path-collection thee times an iteration. Otherwise, each of loop forks will modify common path-collection and finally I will have in it all possible paths. I mean if between left, top & diagonal the biggest is left that we must not to include paths linked with top and diagonal.
Question:
How to solve it in right way?
EDIT:
Actually there is no need to find the full path. It only need to find point's in which you dial a score (in which you make a diagonal moves)
You don't need dynamic programming nor brute force for this!
To see why, let's analyze the rules:
You can move in direction j freely (left & right), so there's no reason to be careful about that direction - you can move into the optimial horizontal postion whenever you want.
Once you increase i (down) there's no way back (though you can increase i without gaining points). Each increase of i should net the maximal amount of points.
You gain points by leaving a cell, but you can only ever leave a row once.
That means you can subdivide this problem and do not need dynamic programming: you can move to the optimal j location, then take one diagonal step; repeat until done.
The optimal i step is moving from a non-last cell in a row with the highest value. You can't move from the last cell because there's no diagonal move possible - so if your matrix has only one column (or row for that matter) you'll never gain points. You can't lose points because the values are natural numbers (but if negative numbers were allowed, you can still skip a row).
In more detail, the optimal path is then found by...
Does the matrix have just one column or row? Move right repeatedly without gaining points then end the program. You can't do much here.
find the maximum value in the current row, ignoring the last value.
generate 'j' moves towards a maximally-valued cell, then move diagonally.
If you're not on the last row, go back to step 2.
You're on the last row and cannot gain more points; just generate moves towards the bottom-right corner to finish your path.
That's it!
Note that there may be multiple maximal paths, your problem specification doesn't guarantee a unique solution.
EDIT: If you don't need the actual path, but just the numbers you scored, the algorithm is much easier - remove or disregard the last row and last column, then for each i (row) return the maximum value in that row.
EDIT:
I misread the question to being just moving down and to the right (ie: j could only change to j or j+1.) so this answer is wrong.
You can use dynamic programming to solve this problem. Greedy doesn't exactly work because you can only travel "down and to the right".
The naive dynamic programming solution would essentially "work backwards" in a literal sense and start from the bottom-right and compute max score when starting at that cell.
Starting from the right-left, and from bottom-up, you can compute the best score you can get from that score simply. You do this for the m x n matrix, then you start from the top left and choose the direction that has the max score.
I need some advice. I'm developing a game similar to Flow Free wherein the gameboard is composed of a grid and colored dots, and the user has to connect the same colored dots together without overlapping other lines, and using up ALL the free spaces in the board.
My question is about level-creation. I wish to make the levels generated randomly (and should at least be able to solve itself so that it can give players hints) and I am in a stump as to what algorithm to use. Any suggestions?
Note: image shows the objective of Flow Free, and it is the same objective of what I am developing.
Thanks for your help. :)
Consider solving your problem with a pair of simpler, more manageable algorithms: one algorithm that reliably creates simple, pre-solved boards and another that rearranges flows to make simple boards more complex.
The first part, building a simple pre-solved board, is trivial (if you want it to be) if you're using n flows on an nxn grid:
For each flow...
Place the head dot at the top of the first open column.
Place the tail dot at the bottom of that column.
Alternatively, you could provide your own hand-made starter boards to pass to the second part. The only goal of this stage is to get a valid board built, even if it's just trivial or predetermined, so it's worth keeping it simple.
The second part, rearranging the flows, involves looping over each flow, seeing which one can work with its neighboring flow to grow and shrink:
For some number of iterations...
Choose a random flow f.
If f is at the minimum length (say 3 squares long), skip to the next iteration because we can't shrink f right now.
If the head dot of f is next to a dot from another flow g (if more than one g to choose from, pick one at random)...
Move f's head dot one square along its flow (i.e., walk it one square towards the tail). f is now one square shorter and there's an empty square. (The puzzle is now unsolved.)
Move the neighboring dot from g into the empty square vacated by f. Now there's an empty square where g's dot moved from.
Fill in that empty spot with flow from g. Now g is one square longer than it was at the beginning of this iteration. (The puzzle is back to being solved as well.)
Repeat the previous step for f's tail dot.
The approach as it stands is limited (dots will always be neighbors) but it's easy to expand upon:
Add a step to loop through the body of flow f, looking for trickier ways to swap space with other flows...
Add a step that prevents a dot from moving to an old location...
Add any other ideas that you come up with.
The overall solution here is probably less than the ideal one that you're aiming for, but now you have two simple algorithms that you can flesh out further to serve the role of one large, all-encompassing algorithm. In the end, I think this approach is manageable, not cryptic, and easy to tweek, and, if nothing else, a good place to start.
Update: I coded a proof-of-concept based on the steps above. Starting with the first 5x5 grid below, the process produced the subsequent 5 different boards. Some are interesting, some are not, but they're always valid with one known solution.
Starting Point
5 Random Results (sorry for the misaligned screenshots)
And a random 8x8 for good measure. The starting point was the same simple columns approach as above.
Updated answer: I implemented a new generator using the idea of "dual puzzles". This allows much sparser and higher quality puzzles than any previous method I know of. The code is on github. I'll try to write more details about how it works, but here is an example puzzle:
Old answer:
I have implemented the following algorithm in my numberlink solver and generator. In enforces the rule, that a path can never touch itself, which is normal in most 'hardcore' numberlink apps and puzzles
First the board is tiled with 2x1 dominos in a simple, deterministic way.
If this is not possible (on an odd area paper), the bottom right corner is
left as a singleton.
Then the dominos are randomly shuffled by rotating random pairs of neighbours.
This is is not done in the case of width or height equal to 1.
Now, in the case of an odd area paper, the bottom right corner is attached to
one of its neighbour dominos. This will always be possible.
Finally, we can start finding random paths through the dominos, combining them
as we pass through. Special care is taken not to connect 'neighbour flows'
which would create puzzles that 'double back on themselves'.
Before the puzzle is printed we 'compact' the range of colours used, as much as possible.
The puzzle is printed by replacing all positions that aren't flow-heads with a .
My numberlink format uses ascii characters instead of numbers. Here is an example:
$ bin/numberlink --generate=35x20
Warning: Including non-standard characters in puzzle
35 20
....bcd.......efg...i......i......j
.kka........l....hm.n....n.o.......
.b...q..q...l..r.....h.....t..uvvu.
....w.....d.e..xx....m.yy..t.......
..z.w.A....A....r.s....BB.....p....
.D.........E.F..F.G...H.........IC.
.z.D...JKL.......g....G..N.j.......
P...a....L.QQ.RR...N....s.....S.T..
U........K......V...............T..
WW...X.......Z0..M.................
1....X...23..Z0..........M....44...
5.......Y..Y....6.........C.......p
5...P...2..3..6..VH.......O.S..99.I
........E.!!......o...."....O..$$.%
.U..&&..J.\\.(.)......8...*.......+
..1.......,..-...(/:.."...;;.%+....
..c<<.==........)./..8>>.*.?......#
.[..[....]........:..........?..^..
..._.._.f...,......-.`..`.7.^......
{{......].....|....|....7.......#..
And here I run it through my solver (same seed):
$ bin/numberlink --generate=35x20 | bin/numberlink --tubes
Found a solution!
┌──┐bcd───┐┌──efg┌─┐i──────i┌─────j
│kka│└───┐││l┌─┘│hm│n────n┌o│┌────┐
│b──┘q──q│││l│┌r└┐│└─h┌──┐│t││uvvu│
└──┐w┌───┘d└e││xx│└──m│yy││t││└──┘│
┌─z│w│A────A┌┘└─r│s───┘BB││┌┘└p┌─┐│
│D┐└┐│┌────E│F──F│G──┐H┐┌┘││┌──┘IC│
└z└D│││JKL┌─┘┌──┐g┌─┐└G││N│j│┌─┐└┐│
P──┐a││││L│QQ│RR└┐│N└──┘s││┌┘│S│T││
U─┐│┌┘││└K└─┐└─┐V││└─────┘││┌┘││T││
WW│││X││┌──┐│Z0││M│┌──────┘││┌┘└┐││
1┐│││X│││23││Z0│└┐││┌────M┌┘││44│││
5│││└┐││Y││Y│┌─┘6││││┌───┐C┌┘│┌─┘│p
5││└P│││2┘└3││6─┘VH│││┌─┐│O┘S┘│99└I
┌┘│┌─┘││E┐!!│└───┐o┘│││"│└─┐O─┘$$┌%
│U┘│&&│└J│\\│(┐)┐└──┘│8││┌*└┐┌───┘+
└─1└─┐└──┘,┐│-└┐│(/:┌┘"┘││;;│%+───┘
┌─c<<│==┌─┐││└┐│)│/││8>>│*┌?│┌───┐#
│[──[└─┐│]││└┐│└─┘:┘│└──┘┌┘┌┘?┌─^││
└─┐_──_│f││└,│└────-│`──`│7┘^─┘┌─┘│
{{└────┘]┘└──┘|────|└───7└─────┘#─┘
I've tested replacing step (4) with a function that iteratively, randomly merges two neighboring paths. However it game much denser puzzles, and I already think the above is nearly too dense to be difficult.
Here is a list of problems I've generated of different size: https://github.com/thomasahle/numberlink/blob/master/puzzles/inputs3
The most straightforward way to create such a level is to find a way to solve it. This way, you can basically generate any random starting configuration and determine if it is a valid level by trying to have it solved. This will generate the most diverse levels.
And even if you find a way to generate the levels some other way, you'll still want to apply this solving algorithm to prove that the generated level is any good ;)
Brute-force enumerating
If the board has a size of NxN cells, and there are also N colours available, brute-force enumerating all possible configurations (regardless of wether they form actual paths between start and end nodes) would take:
N^2 cells total
2N cells already occupied with start and end nodes
N^2 - 2N cells for which the color has yet to be determined
N colours available.
N^(N^2 - 2N) possible combinations.
So,
For N=5, this means 5^15 = 30517578125 combinations.
For N=6, this means 6^24 = 4738381338321616896 combinations.
In other words, the number of possible combinations is pretty high to start with, but also grows ridiculously fast once you start making the board larger.
Constraining the number of cells per color
Obviously, we should try to reduce the number of configurations as much as possible. One way of doing that is to consider the minimum distance ("dMin") between each color's start and end cell - we know that there should at least be this much cells with that color. Calculating the minimum distance can be done with a simple flood fill or Dijkstra's algorithm.
(N.B. Note that this entire next section only discusses the number of cells, but does not say anything about their locations)
In your example, this means (not counting the start and end cells)
dMin(orange) = 1
dMin(red) = 1
dMin(green) = 5
dMin(yellow) = 3
dMin(blue) = 5
This means that, of the 15 cells for which the color has yet to be determined, there have to be at least 1 orange, 1 red, 5 green, 3 yellow and 5 blue cells, also making a total of 15 cells.
For this particular example this means that connecting each color's start and end cell by (one of) the shortest paths fills the entire board - i.e. after filling the board with the shortest paths no uncoloured cells remain. (This should be considered "luck", not every starting configuration of the board will cause this to happen).
Usually, after this step, we have a number of cells that can be freely coloured, let's call this number U. For N=5,
U = 15 - (dMin(orange) + dMin(red) + dMin(green) + dMin(yellow) + dMin(blue))
Because these cells can take any colour, we can also determine the maximum number of cells that can have a particular colour:
dMax(orange) = dMin(orange) + U
dMax(red) = dMin(red) + U
dMax(green) = dMin(green) + U
dMax(yellow) = dMin(yellow) + U
dMax(blue) = dMin(blue) + U
(In this particular example, U=0, so the minimum number of cells per colour is also the maximum).
Path-finding using the distance constraints
If we were to brute force enumerate all possible combinations using these color constraints, we would have a lot less combinations to worry about. More specifically, in this particular example we would have:
15! / (1! * 1! * 5! * 3! * 5!)
= 1307674368000 / 86400
= 15135120 combinations left, about a factor 2000 less.
However, this still doesn't give us the actual paths. so a better idea would be to a backtracking search, where we process each colour in turn and attempt to find all paths that:
doesn't cross an already coloured cell
Is not shorter than dMin(colour) and not longer than dMax(colour).
The second criteria will reduce the number of paths reported per colour, which causes the total number of paths to be tried to be greatly reduced (due to the combinatorial effect).
In pseudo-code:
function SolveLevel(initialBoard of size NxN)
{
foreach(colour on initialBoard)
{
Find startCell(colour) and endCell(colour)
minDistance(colour) = Length(ShortestPath(initialBoard, startCell(colour), endCell(colour)))
}
//Determine the number of uncoloured cells remaining after all shortest paths have been applied.
U = N^(N^2 - 2N) - (Sum of all minDistances)
firstColour = GetFirstColour(initialBoard)
ExplorePathsForColour(
initialBoard,
firstColour,
startCell(firstColour),
endCell(firstColour),
minDistance(firstColour),
U)
}
}
function ExplorePathsForColour(board, colour, startCell, endCell, minDistance, nrOfUncolouredCells)
{
maxDistance = minDistance + nrOfUncolouredCells
paths = FindAllPaths(board, colour, startCell, endCell, minDistance, maxDistance)
foreach(path in paths)
{
//Render all cells in 'path' on a copy of the board
boardCopy = Copy(board)
boardCopy = ApplyPath(boardCopy, path)
uRemaining = nrOfUncolouredCells - (Length(path) - minDistance)
//Recursively explore all paths for the next colour.
nextColour = NextColour(board, colour)
if(nextColour exists)
{
ExplorePathsForColour(
boardCopy,
nextColour,
startCell(nextColour),
endCell(nextColour),
minDistance(nextColour),
uRemaining)
}
else
{
//No more colours remaining to draw
if(uRemaining == 0)
{
//No more uncoloured cells remaining
Report boardCopy as a result
}
}
}
}
FindAllPaths
This only leaves FindAllPaths(board, colour, startCell, endCell, minDistance, maxDistance) to be implemented. The tricky thing here is that we're not searching for the shortest paths, but for any paths that fall in the range determined by minDistance and maxDistance. Hence, we can't just use Dijkstra's or A*, because they will only record the shortest path to each cell, not any possible detours.
One way of finding these paths would be to use a multi-dimensional array for the board, where
each cell is capable of storing multiple waypoints, and a waypoint is defined as the pair (previous waypoint, distance to origin). The previous waypoint is needed to be able to reconstruct the entire path once we've reached the destination, and the distance to origin
prevents us from exceeding the maxDistance.
Finding all paths can then be done by using a flood-fill like exploration from the startCell outwards, where for a given cell, each uncoloured or same-as-the-current-color-coloured neigbour is recursively explored (except the ones that form our current path to the origin) until we reach either the endCell or exceed the maxDistance.
An improvement on this strategy is that we don't explore from the startCell outwards to the endCell, but that we explore from both the startCell and endCell outwards in parallel, using Floor(maxDistance / 2) and Ceil(maxDistance / 2) as the respective maximum distances. For large values of maxDistance, this should reduce the number of explored cells from 2 * maxDistance^2 to maxDistance^2.
I think you'll want to do this in two steps. Step 1) find a set of non-intersecting paths that connect all your points, then 2) Grow/shift those paths to fill the entire board
My thoughts on Step 1 are to essentially perform Dijkstra like algorithm on all points simultaneously, growing together the paths. Similar to Dijkstra, I think you'll want to flood-fill out from each of your points, chosing which node to search next using some heuristic (My hunch says chosing points with the least degrees of freedom first, then by distance, might be a good one). Very differently from Dijkstra though I think we might be stuck with having to backtrack when we have multiple paths attempting to grow into the same node. (This could of course be fairly problematic on bigger maps, but might not be a big deal on small maps like the one you have above.)
You may also solve for some of the easier paths before you start the above algorithm, mainly to cut down on the number of backtracks needed. In specific, if you can make a trace between points along the edge of the board, you can guarantee that connecting those two points in that fashion would never interfere with other paths, so you can simply fill those in and take those guys out of the equation. You could then further iterate on this until all of these "quick and easy" paths are found by tracing along the borders of the board, or borders of existing paths. That algorithm would actually completely solve the above example board, but would undoubtedly fail elsewhere .. still, it would be very cheap to perform and would reduce your search time for the previous algorithm.
Alternatively
You could simply do a real Dijkstra's algorithm between each set of points, pathing out the closest points first (or trying them in some random orders a few times). This would probably work for a fair number of cases, and when it fails simply throw out the map and generate a new one.
Once you have Step 1 solved, Step 2 should be easier, though not necessarily trivial. To grow your paths, I think you'll want to grow your paths outward (so paths closest to walls first, growing towards the walls, then other inner paths outwards, etc.). To grow, I think you'll have two basic operations, flipping corners, and expanding into into adjacent pairs of empty squares.. that is to say, if you have a line like
.v<<.
v<...
v....
v....
First you'll want to flip the corners to fill in your edge spaces
v<<<.
v....
v....
v....
Then you'll want to expand into neighboring pairs of open space
v<<v.
v.^<.
v....
v....
v<<v.
>v^<.
v<...
v....
etc..
Note that what I've outlined wont guarantee a solution if one exists, but I think you should be able to find one most of the time if one exists, and then in the cases where the map has no solution, or the algorithm fails to find one, just throw out the map and try a different one :)
You have two choices:
Write a custom solver
Brute force it.
I used option (2) to generate Boggle type boards and it is VERY successful. If you go with Option (2), this is how you do it:
Tools needed:
Write a A* solver.
Write a random board creator
To solve:
Generate a random board consisting of only endpoints
while board is not solved:
get two endpoints closest to each other that are not yet solved
run A* to generate path
update board so next A* knows new board layout with new path marked as un-traversable.
At exit of loop, check success/fail (is whole board used/etc) and run again if needed
The A* on a 10x10 should run in hundredths of a second. You can probably solve 1k+ boards/second. So a 10 second run should get you several 'usable' boards.
Bonus points:
When generating levels for a IAP (in app purchase) level pack, remember to check for mirrors/rotations/reflections/etc so you don't have one board a copy of another (which is just lame).
Come up with a metric that will figure out if two boards are 'similar' and if so, ditch one of them.
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 training code problems, and on this one I am having problems to solve it, can you give me some tips how to solve it please.
The problem is taken from here:
https://www.ieee.org/documents/IEEEXtreme2008_Competitition_book_2.pdf
Problem 12: Cynical Times.
The problem is something like this (but do refer to above link of the source problem, it has a diagram!):
Your task is to find the sequence of points on the map that the bomber is expected to travel such that it hits all vital links. A link from A to B is vital when its absence isolates completely A from B. In other words, the only way to go from A to B (or vice versa) is via that link.
Due to enemy counter-attack, the plane may have to retreat at any moment, so the plane should follow, at each moment, to the closest vital link possible, even if in the end the total distance grows larger.
Given all coordinates (the initial position of the plane and the nodes in the map) and the range R, you have to determine the sequence of positions in which the plane has to drop bombs.
This sequence should start (takeoff) and finish (landing) at the initial position. Except for the start and finish, all the other positions have to fall exactly in a segment of the map (i.e. it should correspond to a point in a non-hit vital link segment).
The coordinate system used will be UTM (Universal Transverse Mercator) northing and easting, which basically corresponds to a Euclidian perspective of the world (X=Easting; Y=Northing).
Input
Each input file will start with three floating point numbers indicating the X0 and Y0 coordinates of the airport and the range R. The second line contains an integer, N, indicating the number of nodes in the road network graph. Then, the next N (<10000) lines will each contain a pair of floating point numbers indicating the Xi and Yi coordinates (1 < i<=N). Notice that the index i becomes the identifier of each node. Finally, the last block starts with an integer M, indicating the number of links. Then the next M (<10000) lines will each have two integers, Ak and Bk (1 < Ak,Bk <=N; 0 < k < M) that correspond to the identifiers of the points that are linked together.
No two links will ever cross with each other.
Output
The program will print the sequence of coordinates (pairs of floating point numbers with exactly one decimal place), each one at a line, in the order that the plane should visit (starting and ending in the airport).
Sample input 1
102.3 553.9 0.2
14
342.2 832.5
596.2 638.5
479.7 991.3
720.4 874.8
744.3 1284.1
1294.6 924.2
1467.5 659.6
1802.6 659.6
1686.2 860.7
1548.6 1111.2
1834.4 1054.8
564.4 1442.8
850.1 1460.5
1294.6 1485.1
17
1 2
1 3
2 4
3 4
4 5
4 6
6 7
7 8
8 9
8 10
9 10
10 11
6 11
5 12
5 13
12 13
13 14
Sample output 1
102.3 553.9
720.4 874.8
850.1 1460.5
102.3 553.9
Pre-process the input first, so you identify the choke points. Algorithms like Floyd-Warshall would help you.
Model the problem as a Heuristic Search problem, you can compute a MST which covers all choke-points and take the sum of the costs of the edges as a heuristic.
As the commenters said, try to make concrete questions, either here or to the TA supervising your class.
Don't forget to mention where you got these hints.
The problem can be broken down into two parts.
1) Find the vital links.
These are nothing but the Bridges in the graph described. See the wiki page (linked to in the previous sentence), it mentions an algorithm by Tarjan to find the bridges.
2) Once you have the vital links, you need to find the smallest number of points which given the radius of the bomb, will cover the links. For this, for each link, you create a region around it, where dropping the bomb will destroy it. Now you form a graph of these regions (two regions are adjacent if they intersect). You probably need to find a minimum clique partition in this graph.
Haven't thought it through (especially part 2), but hope it helps.
And good luck in the contest!
I think Moron' is right about the first part, but on the second part...
The problem description does not tell anything about "smallest number of points". It tells that the plane flies to the closest vital link.
So, I think the part 2 will be much simpler:
Find the closest non-hit segment to the current location.
Travel to the closest point on the closest segment.
Bomb the current location (remove all segments intersecting a circle)
Repeat until there are no non-hit vital links left.
This straight-forward algorithm has a complexity of O(N*N), but this should be sufficient considering input constraints.