I am new to combination and permutation related algorithms. Does anybody have any thoughts on how to program to solve this classical problem? There are 3 boxes(A,B,C) and 10 balls(1,2,3,...,10), we want to put all balls into the boxes. The result should be {Box A: ball 1; Box B: ball 2,3,4; Box C: ball 5 6 7 8 9 10}, {Box A: ball 1 2; Box B: ball 3 4 5; Box C: 7 8 9 10}, .... I want to get all combinations (not the number of different combinations).
Furthermore, what if there is a constraint that each box contains at most 4 balls?
Thank you.
You can put the first ball in any of three boxes, so you have three variants.
There are three variants for the second ball, three for the third and so on.
They are independent, so you have 3^10 variants, and each variant has 1:1 mapping with a number in range 0..3^10-1.
Consider number in ternary number system, so k-th ternary digit of number tells us what box (a=0,b=1,c=2) k-th ball belongs to.
Example for 3 balls:
Number 14 = 112 ternary, so first ball in C, second and third in B
For case of limited box size simple approach is recursive generation - arguments of recursion are list of available balls and current combination (list of boxes with balls and vacant places).
Related
I am trying to solve this apparently beginner level problem. But not even able to come up with a brute force approach.
Here is the problem statement:
Johnny has some difficulty memorizing the small prime numbers. So, his computer science teacher has asked him to play with the following puzzle game frequently.
The puzzle is a 3x3 board consisting of numbers from 1 to 9. The objective of the puzzle is to swap the tiles until the following final state is reached:
1 2 3
4 5 6
7 8 9
At each step, Johnny may swap two adjacent tiles if their sum is a prime number. Two tiles are considered adjacent if they have a common edge.
Help Johnny to find the shortest number of steps needed to reach the goal state.
Input
The first line contains t, the number of test cases (about 50). Then t test cases follow. Each test case consists of a 3x3 table describing a puzzle which Johnny would like to solve.
The input data for successive test cases is separated by a blank line.
Output
For each test case print a single line containing the shortest number of steps needed to solve the corresponding puzzle. If there is no way to reach the final state, print the number -1.
Example
Input:
2
7 3 2
4 1 5
6 8 9
9 8 5
2 4 1
3 7 6
Output:
6
-1
Here is a start on Python for finding the distance to ALL boards that are reachable from the start.
start = '123456789'
needed_moves = {start: 0}
work_queue = [start]
while (len(work_queue)):
board = work_queue.pop(0)
for next_board in next_move(board):
if next_board not in needed_moves:
needed_moves[next_board] = needed_moves[board] + 1
work_queue.append(next_board)
This is a standard pattern for a breadth first search. Just supply a next_move function and some glue for the tests cases and you're good.
For a depth first search, just switch the queue for a stack. Which really means do work_stack.pop() to pop the last element rather than work_queue.pop(0) to get the first one.
This question already has answers here:
combinations totaling to sum
(5 answers)
Closed 5 years ago.
The exact problem i am referring to and the number of distributions for the problem is computed here. I am interested in knowing those distributions explicitly.
For example, there are 5 balls and 3 boxes: one distribution is 2 balls in box 1, 2 in box 2, 1 in box 3 referred to as, say 221. Now i want to list all such possible distributions: -
212
131
104
.
.
.
One way is that i run the matlab command: perms([0,0,0,0,0,1,1,1]). This essentially generates all permutations of 5 balls and 2 sticks. but there is massive over-counting as the command perms does not recognize identical objects.
Very simple ... sort of.
function alloc(balls, boxes):
if boxes = 1
return [balls]
else
for n in range 0:balls
return alloc(balls-n, boxes-1)
That's the basic recursion logic: pick each possible quantity of balls, then recur on the remaining balls and one box fewer.
The list-gluing methods will be language-dependent; I leave them as an exercise for the student.
You can use unique() to get rid of identical rows generated by perms():
A = unique(perms([0,0,0,0,0,1,1,1]), 'rows');
% `A` will contain all combinations, not permutations, of [0,0,0,0,0,1,1,1]
I have a row with numbers 1:n. I'm looking to add a second row also with the numbers 1:n but these should be in a random order while satisfying the following:
No positions have the same number in both rows
No combination of numbers occurs twice
For example, in the following
Row 1: 1 2 3 4 5 6 7 ...
Row 2: 3 6 15 8 13 12 7 ...
the number 7 occurs at the same position in both rows 1 and 2 (namely position 7; thereby not satisfying rule 1)
while in the following
Row 1: 1 2 3 4 5 6 7 ...
Row 2: 3 7 15 8 13 12 2 ...
the combination of 2+7 appears twice (in positions 2 and 7; thereby not satisfying rule 2).
It would perhaps be possible – but unnecessarily time-consuming – to do this by hand (at least up until a reasonable number), but there must be quite an elegant solution for this in MATLAB.
This problem is called a derangment of a permutation.
Use the function randperm, in order to find a random permutation of your data.
x = [1 2 3 4 5 6 7];
y = randperm(x);
Then, you can check that the sequence is legal. If not, do it again and again..
You have a probability of about 0.3 each time to succeed, which means that you need roughly 10/3 times to try until you find it.
Therefore you will find the answer really quickly.
Alternatively, you can use this algorithm to create a random derangment.
Edit
If you want to have only cycles of size > 2, this is a generalization of the problem.
In it is written that the probability
in that case is smaller, but big enough to find it in a fixed amount of steps. So the same approach is still valid.
This is fairly straightforward. Create a random permutation of the nodes, but interpret the list as follows: Interpret it as a random walk around the nodes, and if node 'b' appears after node 'a', it means that node 'b' appears below node 'a' in the lists:
So if your initial random permutation is
3 2 5 1 4
Then the walk in this case is 3 -> 2 -> 5 -> 1 -> 4 and you creates the rows as follows:
Row 1: 1 2 3 4 5
Row 2: 4 5 2 3 1
This random walk will satisfy both conditions.
But do you wish to allow more than one cycle in your network? I know you don't want two people to have each other's hat. But what about 7 people, where 3 of them have each other's hats and the other 4 have each other's hats? Is this acceptable and/or desirable?
Andrey has already pointed you to randperm and the rejection-sampling-like approach. After generating a permutation p, an easy way to check whether it has fixed point is any(p==1:n). An easy way to check whether it contains cycles of length 2 is any(p(p)==1:n).
So this gets permutations p of 1:n fulfilling your requirements:
p=[];
while (isempty(p))
p=randperm(n);
if any(p==1:n), p=[];
elseif any(p(p)==1:n), p=[];
end
end
Surrounding this with a for loop and for each counting the iterations of the while loop, it seems that one needs to generate on average 4.5 permutations for every "valid" one (and 6.2 if cycles of length three are not allowed, either). Very interesting.
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.
i got this task last week but can't find a good algorithm to solve the problem. So here is the description:
You can build a stable tower with building cubes by not putting bigger cubes to smaller ones and if you don't put harder cubes into lighter ones. Make a programme which gives you the highest possible tower with n number of cubes.
Input:
In the first row of in.txt there is the number of cubes n (1 =< n =<1000). the following n line consisting two positive integer, a cube's sidelength and weight (both of them are not higher than 2000) there are no similar cubes which sidelength and wieght is the same
Output:
you have to write the highest possible stable tower's m number into out.txt. into the second row you have to write in the ordinal number m of the tower from bottom to top. by the height of the tower we mean the amount of all of the cubes's sidelength (not the number of cubes). if there are more than one solution, you can give whatever you want
example for input and output:
input:
5
10 3
20 5
15 6
15 1
10 2
and the output:
3
2 1 5
here are limits on the program. Time limit: 0.2 sec. Memory limit: 16 Mb
I hope you can help me to solve this. thx for all help
The relationship "block A can be placed on top of block B" defines a partial order on the blocks. You can use Kahn's algorithm (aka "topological sort") to turn this into a total order, which you can then traverse in depth order to find the longest path.