Each player takes turns, removing 1 or 2 bananas from a basket of 50 bananas. The player who empties the basket wins.
What are the weights that should be used for the distances, and what should the matrix size be? Should the matrix change every time someone makes a move? Should player 1's moves be horizontal and player 2's moves vertical?
Thanks for the read
I'm not sure why you'd specifically want to use dynamic programming and/or Manhattan distance for this puzzle. This is a game for which you can find a fixed solution.
If you go first and there are 3 bananas, no matter what you play, I win. You pick one, I pick two, and vice versa. If there are six bananas, the same logic allows me to reduce the game to the 3 banana case. In fact, for any 3n bananas, I can reduce the game to 3(n-1) bananas. If the number of bananas isn't a multiple of three, then you can make it a multiple of three (By removing either one or two bananas), and ensure victory.
For k bananas, you always remove k % 3. If k % 3 == 0, you've lost unless your opponent makes a mistake, so play whatever you like. That's it.
I agree with #pdpi but if you insist of solving this problem with dynamic programing, then you should do something like this :
f(left_in_the_basket, mine)
if left_in_the_basket is 2 && turn is mine:
return 1
if left_in_the_basket is 1 && turn is mine:
return 1
if left_in_the_basket is 2 && turn is not mine:
return -1
if left_in_the_basket is 1 && turn is not mine:
return -1
return max (f(left_in_the_basket - 1, not mine), f(left_in_the_basket - 2, not mine))
Related
I've been practicing some probabilities for last few weeks around 10/week usually did the job, however as the topics got harder I've started struggling and now I'm completely stuck. I've been looking online and i did find similar examples but nothing to touch my case in particular. I will continue looking for an answer so even if u can't answer my specific cases links to online literature will be appreciated.
answers are welcome, however i would rather be more interested in explanation of how a problem works.
urn contains m white and k black balls. two players are drawing the balls one after another without putting them back into the urn. the winner is the first person to draw white ball. what is the probability that second player will be the winner? (k=4, m=4)
women tend to vote with probability of a, men tend to do the same with probability of b. the probability c tells us that if we take a couple one of them will not go voting. what are the chances that at least one of them will vote? (a=0.49, b=0.61, c=0.75)
we are sending a message of n bytes. to obtain higher chance of sending entire message without ruining it we use k different wires. What is the probability of sending an entire message through one of the wires without ruining it, if the probability of ruining any of the bytes at any of the wires is p. (p=0.06, n=7, k=6)
the basketball game finals play to N wins. after m + n games the result is m : n. what is the probability of winning the finals for the lagging(the team that's behind) team if it is known that the leading team wins each match with a probability of p? (m=3 n=2 N=5 p=0.36)
sorry for my English and any help will be appreciated
Just A
So Given m = k and both = 4. And each player takes a ball out per turn. Given the worst case scenario 4 good balls will be picked (K) there will be 2 turns of both players picking the black ball then the first player will be guaranteed the white ball next turn. Therefore you need to work out the probability for the first two turns.
Turn 1
- P1 turn 1 Chance for black= 4/8
- P2 turn 2 Change for white = 4/7
Therefore turn 1 chance = (4/8)*(4/7). = 28.57%
Turn 2
- P1 turn 1 Chance for black= 2/6
- P2 turn 2 Change for white = 4/5
Therefore turn 2 chance = (4/6)*(4/5). = 26.664% but to get to turn 2 is a 28.57% chance
Therefore its the first probability add the second probability given the first case occurs. So 28.57% + (26.44%*28.57%)= 36.12%
The third case is a always lose. Top tip then be the first player!
Turn 3 Instance win for player 2 therefore a loss. = 0%
It is ultimately a game of nim with certain modification.
Rules are as follows :
There are two players A and B.
The game is played with two piles of matches. Initially, the first pile contains N matches and the second one contains M matches.
The players alternate turns; A plays first.
On each turn, the current player must choose one pile and remove a positive number of matches (not exceeding the current number of matches on that pile) from it.
It is only allowed to remove X matches from a pile if the number of matches in the other pile divides X.
The player that takes the last match from any pile wins.
Both players play optimally.
My take :
let us say we have two piles 2 7. we have 3 cases to reduce the second pile to wiz : 2 1 , 2 3 , 2 5 If A is playing optimally he/she will go for 2 3 so that the only chance left for B is to do 2 1 and then A can go for 0 1 and win the game. The crust of the solution being that if A or B ever encounters any situation where it can directly lose in the next step then it will try their best to avoid it and use the situation to their advantage by just leaving at a state 1 step before that losing stage .
But this approach fails for some unknown test cases , is there any better approach to find the winner , or any other test case which defies this logic.
This is a classic dynamic programming problem. First, find a recurrence relationship that describes an outcome of a game in terms of smaller games. Your parameters here are X and Y, where X is the number of matches in one stack and Y in the other. How do I reduce this problem?
Well, suppose it is my turn with X matches, and suppose that Y is divisible by the numbers a1, a2, and a3, while x is divisible by b1, b2, b3. Then, I have six possible turns. The problem reduces to solving for (X-a1, Y) (X-a2, Y) (X-a3,Y), (X,Y-b1), (X,Y-b2), (X, Y-b3). Once these six smaller games are solved, if one of them is a winning game for me, then I make the corresponding move and win the game.
There is one more parameter, which is whose turn it is. This doubles the size of solvable problems.
The key is to find all possible moves, and recur for each of them, keeping a storage of already solved games for efficiency.
The base case needs to be figured out naturally.
I was doing some coding challenges and a problem came up that said roughly this:
"Two players each taking turns starting with player one. There are N
sticks given, each player takes 1, 2 or 3 sticks on their turn, the
player to take the last stick loses, the goal is to find an algorithm
that lets player one win with certainty (not always possible, player two is supposed to take turns that will ensure victory) and output 1, 2 or 3 as
the starting amount of sticks taken or 0 if it's impossible to win.
Input is N. Example: Input:2 Output:1"
I tried to think about it but all I came up with is that it would take checking every possible outcome because of all the possibilities that could be chained together if N is big. I also thought that if the last stick is to be taken by player 2 so as to not lose, that is N-1 is taken by player 1 (whether by taking N-1 only or N-1, N-2 or N-1, N-2, N-3) leaving N to player 2, that is the only way to ensure victory.
It turned out that the solution was (N-1) mod 4, but I can't understand why that is the case.
So my question is how do you approach a problem like that and why is the solution a modulo? Also is there a way to spot modulo problems like these? Other coders did it fairly quickly so I suppose practice makes perfect, but I have no idea where to start from.
It is modulo 4 because if one player has the advantage, he can keep the same advantage by taking 3 sticks if the first player took 1, 2 if the first player took 2, and 1 stick if the first player took 3. The other player simply doesn't have any control anymore.
Take the problem backwards :
You don't have to care about a big N, you just need to analyze what the situation looks like when only 4 sticks or less are left.
Who will win when there are 1, 2, 3 or 4 sticks left?
Who will win when there are 4n+1, 4n+2, 4n+3 or 4n+4 sticks left?
Given D discs, P poles, and the initial starting positions for the disks, and the required final destination for the poles, how can we write a generic solution to the problem?
For example,
Given D=6 and P=4, and the initial starting position looks like this:
5 1
6 2 4 3
Where the number represents the disk's radius, the poles are numbered 1-4 left-right, and we want to stack all the disks on pole 1.
How do we choose the next move?
The solution is (worked out by hand):
3 1
4 3
4 1
2 1
3 1
(format: <from-pole> <to-pole>)
The first move is obvious, move the "4" on top of the "5" because that its required position in the final solution.
Next, we probably want to move the next largest number, which would be the "3". But first we have to unbury it, which means we should move the "1" next. But how do we decide where to place it?
That's as far as I've gotten. I could write a recursive algorithm that tries all possible places, but I'm not sure if that's optimal.
We can't.
More precisely, as http://en.wikipedia.org/wiki/Tower_of_Hanoi#Four_pegs_and_beyond says, for 4+ pegs, proving what the optimal solution is is an open problem. There is a known very good algorithm, that is widely believed to be optimal, for the simple case where the pile of disks is on one peg and you want to transfer the whole pile to another. However we do not have an algorithm, or even a known heuristic, for an arbitrary starting position.
If we did have a proposed algorithm, then the open problem would presumably be much easier.
Well I have been through many sites teaching on how to solve it, but was wondering how to create it. I am not interested much in the coding aspects of it, but wanted to know more on the algorithms behind it. For example, when the grid is generated with 10 mines or so, I would use any random function to distribute itself across the grid, but then again how do I set the numbers associated to it and decide which box to be opened? I couldn't frame any generic algorithm on how would I go about doing that.
Perhaps something in the lines of :
grid = [n,m] // initialize all cells to 0
for k = 1 to number_of_mines
get random mine_x and mine_y where grid(mine_x, mine_y) is not a mine
for x = -1 to 1
for y = -1 to 1
if x = 0 and y = 0 then
grid[mine_x, mine_y] = -number_of_mines // negative value = mine
else
increment grid[mine_x + x, mine_y + y] by 1
That's pretty much it...
** EDIT **
Because this algorithm could lead into creating a board with some mines grouped too much together, or worse very dispersed (thus boring to solve), you can then add extra validation when generating mine_x and mine_y number. For example, to ensure that at least 3 neighboring cells are not mines, or even perhaps favor limiting the number of mines that are too far from each other, etc.
** UPDATE **
I've taken the liberty of playing a little with JS bin here came up with a functional Minesweeper game demo. This is simply to demonstrate the algorithm described in this answer. I did not optimize the randomness of the generated mine position, therefore some games could be impossible or too easy. Also, there are no validation as to how many mines there are in the grid, so you can actually create a 2 by 2 grid with 1000 mines.... but that will only lead to an infinite loop :) Enjoy!
You just seed the mines and after that, you traverse every cell and count the neighbouring mines.
Or you set every counter to 0 and with each seeded mine, you increment all neighbouring cells counters.
If you want to place m mines on N squares, and you have access to a random number generator, you just walk through the squares remaining and for each square compute (# mines remaining)/(# squares remaining) and place a mine if your random number is equal to or below that value.
Now, if you want to label every square with the number of adjacent mines, you can just do it directly:
count(x,y) = sum(
for i = -1 to 1
for j = -1 to 1
1 if (x+i,y+j) contains a mine
0 otherwise
)
or if you prefer you can start off with an array of zeros and increment each by one in the 3x3 square that has a mine in the center. (It doesn't hurt to number the squares with mines.)
This produces a purely random and correctly annotated minesweeper game. Some random games may not be fun games, however; selecting random-but-fun games is a much more challenging task.