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I am doing a problem CCHESS, here is the link ( http://www.spoj.pl/problems/CCHESS/ ) to the problem.
The question is as follows:
In the country of Rome, Chess is a royal game. For evey move the players had to give some bucks to the Emperor Jurg. The LGMs or Little Green Men, are very good player of chess. But as the chess is a expensive game, thats why it is royal, they asked you to help them find the minimum bucks which they had to pay for moving their Knight from one position to another. Any number of steps can be used to reach the destination.
Constraints:
The chess has a dimension of 8X8, and the index of left bottom cell (0, 0).
Knight move only in a standard way, i.e. 2 row and 1 col or 1 row and 2 col.
If in a step Knight move from (a, b) to (c, d), then LGM had to pay a*c + b*d bucks to Emperor Jurg.
0 ≤ a, b, c, d ≤ 7
Input
There are 100-150 test cases. Each test case is composed of four space separeated integers.The first two numbers, a, b, are the starting position of the Knight and the next two, c, d, are the destination of the Knight. Read upto End Of File.
Output
For each test case, print the minimum amount of bucks they had to pay in separate line. If its impossible to reach the destination then print -1.
Example
Input:
2 5 5 2
4 7 3 2
1 2 3 4
Output:
42
78
18
Explanation for test case #1:
2 5 5 2
For moving Knight from
(2, 5) to (5, 2)
in minimum cost, one of the path is
(2, 5) -> (3, 3) ->(5, 2)
Bucks paid:
(2, 5) = 0
(2, 5) -> (3, 3) = 0 + (2*3 + 5*3) = 21
(3, 3) -> (5, 2) = 21 + (3*5 + 3*2) = 42
To infinity and beyond...
I have done this problem using brute force, i.e. checking recursively all the possible paths but i think i am missing somewhere to find a direct approach, because numerous submissions are of 0.00 where as my recursive approach got accepted in 0.3s .
Any help would be appreciated.
Construct a graph G=(V,E) where
V is the set of coordinates in the grid {v=(x,y)}
E is the set of edges between vertices
Assign weights on the edges where weight is (v1.x * v2.x + v1.y*v2.y)
Use Dijkstra's algorithm to find the shortest path (1 source - 1 destination)
source = (a,b) and destination = (c,d)
If there is no path report -1.
The number of vertices are limited to (8*8) = 64
The number of edges are limited to 64 * (8) = 512
as the knight can move to at most 8 other coordinates from one place.
Try A* algorithm, with heuristic = manhattan_distance/3 .
Related
onsider a graph of five vertices whose vertices are labelled 1 to 5. The only edges present in the graph are one each from 1 to 2, 1 to 3, 1 to 4 and 1 to 5. Let the time taken to travel from 1 to 2, 3, 4 and 5 be 5, 5, 1 and 1 units respectively. Also assume that if it takes time t to travel from vertex a to b, then it takes the same time to travel from b to a.We wish to select a walk from vertex 1 to some other vertex, back to vertex 1 and so on till each vertex (except vertex 1 - the source) is visited exactly once.
Let the initial instant be t = 0. Let the times of visit of vertices 2, 3, 4 and 5 from t = 0 be t2, t3, t4 and t5.We wish to minimise the sum of t2, t3, t4 and t5.
Find out the minimum possible sum of the given times.
i am unable to understand the question itself
This is my understanding of the question:
If you visit nodes 2, 3 ,4, and 5 in that order, the times will be:
t2 = 5 (going from 1 to 2),
t3 = 15 (5 to get to 2, 5 to return from 2 back to 1, and 5 to go from 1 to 3),
t4 = 21 (15 + 5 + 1),
t5 = 23 (21 + 1 + 1).
The sum is 64. You can get a better time with a different order, your task is to find the best (minimum) sum.
To minimise the total time of visiting, we select the path which takes minimum time first because this time piles up on all other vertices. After we come back to 1,we then select the path with second minimum time and we repeat this process each time after we come back to 1.
Let the times in sorted order be a,b,c and d for vertices v1,v2,v3 and v4. Then vertex v1 is visited at time t = a.We come back to 1 at t = 2a.We then reach the second vertex at t = 2a + b and come back to 1 at t = 2a + 2b. Continuing similarly, vertex v3 is visited at t = 2a + 2b + c and v4 is visited at t = 2a + 2b + 2c + d. So, the minimum value of the sum of times would be 7a + 5b + 3c + d.
So, the answer is 32.
so can we start the walk with vertex 5 =1+1=2
vertex 4 = 1+1=2
vertex 3=5+5=10
vertex 2= 5
total=2+2+10+5=19
Note that your total never changes, no matter the order in which you visit the nodes, so you cannot minimize it. You may want to clarify this with the teacher, but the problem definition states "time instance," not duration.
Think of it as starting a timer at time t-1 Then, for your path above (t-1 = 0): t-5(vertex-5) = 1, t-4 = 3, t-2 = 9, t-3 = 19. The sum is 32. Much better than 64, so you are on the right track!
Dijkstra's is used to find the shortest path between nodes, and it doesn't apply here. This problem is much easier.
Question: You are given the following inputs:
3
0 0 1
3 1 1
6 0 9
The first line is the number of points on the graph.
The rest of the lines contain the points on the graph, and their reward. For example:
0 0 1 would mean at point (0,0) [which is the starting point] you are given a reward of 1.
3 1 1 would mean at point (3,1) you are given a reward of 1.
6 0 9 would mean at point (6, 0) you are given a reward of 9.
Going from point a, to point b costs 1.
Therefore if you go from (0,0) -> (3,1) -> (6,0) your reward is 11-2 (cost of traversing 2 nodes) * sqrt(10).
Goal: Determine the maximum amount of rewards you can make (the total amount of reward you collect - the cost) based on the provided inputs.
How would I go about solving this? It seems like dynamic programming is the way to go, but I am not sure where to start.
I am searching for a method to create, in a fast way a random matrix A with the follwing properties:
A = transpose(A)
A(i,i) = 0 for all i
A(i,j) >= 0 for all i, j
sum(A) =~ degree; the sum of rows are randomly distributed by a distribution I want to specify (here =~ means approximate equality).
The distribution degree comes from a matrix orig, specifically degree=sum(orig), thus I know that matrices with this distribution exist.
For example: orig=[0 12 7 5; 12 0 1 9; 7 1 0 3; 5 9 3 0]
orig =
0 12 7 5
12 0 1 9
7 1 0 3
5 9 3 0
sum(orig)=[24 22 11 17];
Now one possible matrix A=[0 11 5 8, 11 0 4 7, 5 4 0 2, 8 7 2 0] is
A =
0 11 5 8
11 0 4 7
5 4 0 2
8 7 2 0
with sum(A)=[24 22 11 17].
I am trying this for quite some time, but unfortunatly my two ideas didn't work:
version 1:
I switch Nswitch times two random elements: A(k1,k3)--; A(k1,k4)++; A(k2,k3)++; A(k2,k4)--; (the transposed elements aswell).
Unfortunatly, Nswitch = log(E)*E (with E=sum(sum(nn))) in order that the Matrices are very uncorrelated. As my E > 5.000.000, this is not feasible (in particular, as I need at least 10 of such matrices).
version 2:
I create the matrix according to the distribution from scratch. The idea is, to fill every row i with degree(i) numbers, based on the distribution of degree:
nn=orig;
nnR=zeros(size(nn));
for i=1:length(nn)
degree=sum(nn);
howmany=degree(i);
degree(i)=0;
full=rld_cumsum(degree,1:length(degree));
rr=randi(length(full),[1,howmany]);
ff=full(rr);
xx=i*ones([1,length(ff)]);
nnR = nnR + accumarray([xx(:),ff(:)],1,size(nnR));
end
A=nnR;
However, while sum(A')=degree, sum(A) systematically deviates from degree, and I am not able to find the reason for that.
Small deviations from degree are fine of course, but there seem to be systmatical deviations in particulat of the matrices contain in some places large numbers.
I would be very happy if somebody could either show me a fast method for version1, or a reason for the systematic deviation of the distribution in version 2, or a method to create such matrices in a different way. Thank you!
Edit:
This is the problem in matsmath's proposed solution:
Imagine you have the matrix:
orig =
0 12 3 1
12 0 1 9
3 1 0 3
1 9 3 0
with r(i)=[16 22 7 13].
Step 1: r(1)=16, my random integer partition is p(i)=[0 7 3 6].
Step 2: Check that all p(i)<=r(i), which is the case.
Step 3:
My random matrix starts looks like
A =
0 7 3 6
7 0 . .
3 . 0 .
6 . . 0
with the new row sum vector rnew=[r(2)-p(2),...,r(n)-p(n)]=[15 4 7]
Second iteration (here the problem occures):
Step 1: rnew(1)=15, my random integer partition is p(i)=[0 A B]: rnew(1)=15=A+B.
Step 2: Check that all p(i)<=rnew(i), which gives A<=4, B<=7. So A+B<=11, but A+B has to be 15. contradiction :-/
Edit2:
This is the code representing (to the best of my knowledge) the solution posted by David Eisenstat:
orig=[0 12 3 1; 12 0 1 9; 3 1 0 3; 1 9 3 0];
w=[2.2406 4.6334 0.8174 1.6902];
xfull=zeros(4);
for ii=1:1000
rndmat=[poissrnd(w(1),1,4); poissrnd(w(2),1,4); poissrnd(w(3),1,4); poissrnd(w(4),1,4)];
kkk=rndmat.*(ones(4)-eye(4)); % remove diagonal
hhh=sum(sum(orig))/sum(sum(kkk))*kkk; % normalisation
xfull=xfull+hhh;
end
xf=xfull/ii;
disp(sum(orig)); % gives [16 22 7 13]
disp(sum(xf)); % gives [14.8337 9.6171 18.0627 15.4865] (obvious systematic problem)
disp(sum(xf')) % gives [13.5230 28.8452 4.9635 10.6683] (which is also systematically different from [16, 22, 7, 13]
Since it's enough to approximately preserve the degree sequence, let me propose a random distribution where each entry above the diagonal is chosen according to a Poisson distribution. My intuition is that we want to find weights w_i such that the i,j entry for i != j has mean w_i*w_j (all of the diagonal entries are zero). This gives us a nonlinear system of equations:
for all i, (sum_{j != i} w_i*w_j) = d_i,
where d_i is the degree of i. Equivalently,
for all i, w_i * (sum_j w_j) - w_i^2 = d_i.
The latter can be solved by applying Newton's method as described below from a starting solution of w_i = d_i / sqrt(sum_j d_j).
Once we have the w_is, we can sample repeatedly using poissrnd to generate samples of multiple Poisson distributions at once.
(If I have time, I'll try implementing this in numpy.)
The Jacobian matrix of the equation system for a 4 by 4 problem is
(w_2 + w_3 + w_4) w_1 w_1 w_1
w_2 (w_1 + w_3 + w_4) w_2 w_2
w_3 w_3 (w_1 + w_2 + w_4) w_3
w_4 w_4 w_4 (w_1 + w_2 + w_3).
In general, let A be a diagonal matrix where A_{i,i} = sum_j w_j - 2*w_i. Let u = [w_1, ..., w_n]' and v = [1, ..., 1]'. The Jacobian can be written J = A + u*v'. The inverse is given by the Sherman--Morrison formula
A^-1*u*v'*A^-1
J^-1 = (A + u*v')^-1 = A^-1 - -------------- .
1 + v'*A^-1*u
For the Newton step, we need to compute J^-1*y for some given y. This can be done straightforwardly in time O(n) using the above equation. I'll add more detail when I get the chance.
First approach (based on version2)
Let your row sum vector given by the matrix orig [r(1),r(2),...,r(n)].
Step 1. Take a random integer partition of the integer r(1) into exactly n-1 parts, say p(2), p(3), ..., p(n)
Step 2. Check if p(i)<=r(i) for all i=2...n. If not, go to Step 1.
Step 3. Fill out your random matrix first row and colum by the entries 0, p(2), ... , p(n), and consider the new row sum vector [r(2)-p(2),...,r(n)-p(n)].
Repeat these steps with a matrix of order n-1.
The point is, that you randomize one row at a time, and reduce the problem to searching for a matrix of size one less.
As pointed out by OP in the comment, this naive algorithm fails. The reason is that the matrices in question have a further necessary condition on their entries as follows:
FACT:
If A is an orig matrix with row sums [r(1), r(2), ..., r(n)] then necessarily for every i=1..n it holds that r(i)<=-r(i)+sum(r(j),j=1..n).
That is, any row sum, say the ith, r(i), is necessarily at most as big as the sum of the other row sums (not including r(i)).
In light of this, a revised algorithm is possible. Note that in Step 2b. we check if the new row sum vector has the property discussed above.
Step 1. Take a random integer partition of the integer r(1) into exactly n-1 parts, say p(2), p(3), ..., p(n)
Step 2a. Check if p(i)<=r(i) for all i=2...n. If not, go to Step 1.
Step 2b. Check if r(i)-p(i)<=-r(i)+p(i)+sum(r(j)-p(j),j=2..n) for all i=2..n. If not, go to Step 1.
Step 3. Fill out your random matrix first row and colum by the entries 0, p(2), ... , p(n), and consider the new row sum vector [r(2)-p(2),...,r(n)-p(n)].
Second approach (based on version1)
I am not sure if this approach gives you random matrices, but it certainly gives you different matrices.
The idea here is to change some parts of your orig matrix locally, in a way which maintains all of its properties.
You should look for a random 2x2 submatrix below the main diagonal which contains strictly positive entries, like [[a,b],[c,d]] and perturbe its contents by a random value r to [[a+r,b-r],[c-r,d+r]]. You make the same change above the main diagonal too, to keep your new matrix symmetric. Here the point is that the changes within the entries "cancel" each other out.
Of course, r should be chosen in a way such that b-r>=0 and c-r>=0.
You can pursue this idea to modify larger submatrices too. For example, you might choose 3 random row coordinates r1, r2, r2 and 3 random column coordinates c1, c2, and c3 and then make changes in your orig matrix at the 9 positions (ri,cj) as follows: you change your 3x3 submatrix [[a b c],[d e f], [g h i]] to [[a-r b+r c] [d+r e f-r], [g h-r i+r]]. You do the same at the transposed places. Again, the random value r must be chosen in a way so that a-r>=0 and f-r>=0 and h-r>=0. Moreover, c1 and r1, and c3 and r3 must be distinct as you can't change the 0 entries in the main diagonal of the matrix orig.
You can repeat such things over and over again, say 100 times, until you find something which looks random. Note that this idea uses the fact that you have existing knowledge of a solution, this is the matrix orig, while the first approach does not use such knowledge at all.
You start with a stack of n sticks. Each player removes 2 or 3 sticks each turn. The player who removes the last stick wins. If their is only one stick left, the game is a draw.
I need to determine who will win a game of n stacks in general, expressed as a function of n. However, unless we start the game with 2, 3 or 5 sticks, it is always possible to direct the game so that it ends in a draw. I drew the game tree for 9 sticks and more and it is always possible to prevent a loss by making choices that lead to one stick remaining. How can I write a winning rule for this given problem?
However, unless we start the game with 2, 3 or 5 sticks, it is always possible to direct the game so that it ends in a draw.
I don't think this is true.
For example, suppose we start with 10 sticks. If you remove x sticks, I will always remove 5-x sticks. This will mean that after one turn each there are 5 sticks left, and after two turns each I have won.
The same will apply for any multiple of 5.
Now consider other possible values modulo 5.
Here's the rule:
Losing Position: n = 5k
Draw: n = 5k+1 or n = 5k+4
Winning Position: n = 5k+2 or n = 5k+3
You can observe the pattern by building a table like the one shown below:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 .....
L D W W D L D W W D L D W W D L D
Steps for Building the table:
Observe that you lose if n = 0, mark it as L.
When n = 1, it is draw, mark it as D.
When n = 2, you can only draw 2 sticks. So, your opponent happens to face n=0 which is losing, so you win, mark it as W.
For n = 3, you can take 2 or 3 sticks, so, your opponent can end up at 1 or 0 sticks. 1 is a draw and 0 is loss for him and therefore win for us. So, we will chose win, mark it as W.
For n=4, similarily, opponent can end up at 2 or 1. So, we can draw or lose. We will choose draw, mark it as D.
For n = 5, we can either make our opponent end up at 2 or 3. Both of them are win for him. So, we lose. Mark it as L.
Basically, to determine the state(L, W or D) of a number n, we need to look at states of already computed states n-2 and n-3.
Observe that this pattern (LDWWD) repeats after that.
We all know that the minimum number of moves required to solve the classical towers of hanoi problem is 2n-1. Now, let us assume that some of the discs have same size. What would be the minimum number of moves to solve the problem in that case.
Example, let us assume that there are three discs. In the classical problem, the minimum number of moves required would be 7. Now, let us assume that the size of disc 2 and disc 3 is same. In that case, the minimum number of moves required would be:
Move disc 1 from a to b.
Move disc 2 from a to c.
Move disc 3 from a to c.
Move disc 1 from b to c.
which is 4 moves. Now, given the total number of discs n and the sets of discs which have same size, find the minimum number of moves to solve the problem. This is a challenge by a friend, so pointers towards solution are welcome. Thanks.
Let's consider a tower of size n. The top disk has to be moved 2n-1 times, the second disk 2n-2 times, and so on, until the bottom disk has to be moved just once, for a total of 2n-1 moves. Moving each disk takes exactly one turn.
1 moved 8 times
111 moved 4 times
11111 moved 2 times
1111111 moved 1 time => 8 + 4 + 2 + 1 == 15
Now if x disks have the same size, those have to be in consecutive layers, and you would always move them towards the same target stack, so you could just as well collapse those to just one disk, requiring x turns to be moved. You could consider those multi-disks to be x times as 'heavy', or 'thick', if you like.
1
111 1 moved 8 times
111 collapse 222 moved 4 times, taking 2 turns each
11111 -----------> 11111 moved 2 times
1111111 3333333 moved 1 time, taking 3 turns
1111111 => 8 + 4*2 + 2 + 1*3 == 21
1111111
Now just sum those up and you have your answer.
Here's some Python code, using the above example: Assuming you already have a list of the 'collapsed' disks, with disks[i] being the weight of the collapsed disk in the ith layer, you can just do this:
disks = [1, 2, 1, 3] # weight of collapsed disks, top to bottom
print sum(d * 2**i for i, d in enumerate(reversed(disks)))
If instead you have a list of the sizes of the disks, like on the left side, you could use this algorithm:
disks = [1, 3, 3, 5, 7, 7, 7] # size of disks, top to bottom
last, t, s = disks[-1], 1, 0
for d in reversed(disks):
if d < last: t, last = t*2, d
s = s + t
print s
Output, in both cases, is 21, the required number of turns.
It completely depends on the distribution of the discs that are the same size. If you have n=7 discs and they are all the same size then the answer is 7 (or n). And, of course the standard problem is answered by 2n-1.
As tobias_k suggested, you can group same size discs. So now look at the problem as moving groups of discs. To move a certain number of groups, you have to know the size of each group
examples
1
n=7 //disc sizes (1,2,3,3,4,5,5)
g=5 //group sizes (1,1,2,1,2)
//group index (1,2,3,4,5)
number of moves = sum( g-size * 2^( g-count - g-index ) )
in this case
moves = 1*2^4 + 1*2^3 + 2*2^2 + 1*2^1 + 2*2^0
= 16 + 8 + 8 + 2 + 2
= 36
2
n=7 //disc sizes (1,1,1,1,1,1,1)
g=1 //group sizes (7)
//group index (1)
number of moves = sum( g-size * 2^( g-count - g-index ) )
in this case
moves = 7*2^0
= 7
3
n=7 //disc sizes (1,2,3,4,5,6,7)
g=7 //group sizes (1,1,1,1,1,1,1)
//group index (1,2,3,4,5,6,7)
number of moves = sum( g-size * 2^( g-count - g-index ) )
in this case
moves = 1*2^6 + 1*2^5 + 1*2^4 + 1*2^3 + 1*2^2 + 1*2^1 + 1*2^0
= 64 + 32 + 16 + 8 + 4 + 2 + 1
= 127
Interesting note about the last example, and the standard hanoi problem: sum(2n-1) = 2n - 1
I wrote a Github gist in C for this problem. I am attaching a link to it, may be useful to somebody, I hope.
Modified tower of Hanoi problem with one or more disks of the same size
There are n types of disks. For each type, all disks are identical. In array arr, I am taking the number of disks of each type. A, B and C are pegs or towers.
Method swap(int, int), partition(int, int) and qSort(int, int) are part of my implementation of the quicksort algorithm.
Method toh(char, char, char, int, int) is the Tower of Hanoi solution.
How it is working: Imagine we compress all the disks of the same size into one disk. Now we have a problem which has a general solution to the Tower of Hanoi. Now each time a disk moves, we add the total movement which is equal to the total number of that type of disk.