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Then the first player can never be beaten if he plays optimally.
Now we look at some variations of the Nim game.
Misere (Misere) Neem
Misere is a French word. The player who picks up the last stone in the Mijera Nimgame loses. Mijera nim ao xorsum>0
Winning Position. But the first player has to change the strategy a bit. In the last move, without picking up all the stones, only one stone is left, the second player is then forced to pick up the last stone. But if there is exactly 1 stone in each pile then xorsum will not work. Then you have to see whether the number of stacks is even or odd. If there are odd number stacks and each stack has 1 stone then the current player cannot be defeated.
Prime Power Game
A number is n
is given A player may divide n by a power of any prime number on his move. If the number becomes 1 then the current player wins.
Note that any number n can be written as a product of some prime number. For example if n=56700 we can write n=(2×2)×(3×3×3×3)×(5×5)×(7)=22×34×52×71. Now suppose you have a pile of 4 stones, and the set of numbers of stones {2,4,2,1}
. Now that it’s turned into a Nim game, you can pick up one or more stones from any pile!
Nimble
n at nimble game
There are cells denoted by 0 to n−1. Each room contains one or more coins.
Only one coin can be removed from a room per move and placed in a room on the left. The one who can’t move will lose. When all the stones reach the 0th house, no more moves will be possible.
Note that room 2 has 3 stones and each coin can be moved exactly 2 times. Again it is possible to move each coin in cell number 1 1 time and each coin in cell number 4 4 times.
We are i
We can think of each coin in the -th house as a pile of stones of size i since it is possible to move the coin exactly i times. So in the above example the set of stacks of stones would be {1,1,2,2,2,4,4,4,4,4,4}
. Size of 6 stacks is 4 because cell number 4 has 6 coins. Now the game has become a normal nimgame!
The Matrix Game
A n×m
Size grid is given. Each room has some stones. A player can pick some stones from one or more columns of any row. At least one stone must be picked up in each move. Whoever picks up the last stone wins.
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This game can also be easily converted into a normal Nim game. The total number of stones in each row should be thought of as a stack. After extracting the xorsum, the work is done.
insect neem
Pokar neem is a type of bogus neem. In Bogus Neem, stones can not only be removed, but stones can also be added. The rule of poker neem is that one or more stones can be added to any pile. If the initial size of a stack is m
Then the size cannot exceed m even after adding stones. And so that the game does not run indefinitely, any player k
Can’t add more than 12 stones.
In this game if the first player is in the winning state (xorsum>0
) then when the 2nd player adds some stones, the 1st player removes the stones and brings the game back to this state! So xorsum is the same as common neem
You can tell who will win the game.
In the next episode we will look at some more variations and learn about Grandi numbers. Now try to solve these problems in LightOZ. Some more variations of Neem can be found here.
Hand in Graph Theory 14 – Strongly Connected Components
A strongly connected component or SCC of a directed graph is a component that has a path from every node to every other node. In the figure below, each strongly connected component of a graph is shown with a different color.
Using the concept of depth first search finishing time we get O(V+E)
We can isolate the strongly connected components of a graph. Before reading this article, you must have an understanding of topological sorting and the discovery and finishing time of depth first search.
See the graph below:
Many people try to find out SCC in a wrong way at first. That is to treat all nodes that can be reached by DFS from any node as a component. But it is very easy to understand that it will not work, in the above graph if you run DFS from 1, all the nodes can be visited, but if you can go from 1 to 4, there is no way to go from 4 to 1, so they are not part of the same connected component. The problem with this method is that DFS moves out of the connected component and into another component. To solve this problem we will run DFS a bit smarter.
Two nodes u,v will be in the same SCC if and only if there is a path from u to v and also a path from v to u.
First we run DFS from 1 and write the discovery time and finishing time of all nodes. By visiting the nodes in the order 1,2,3,4,5,6,7 we get the starting/finishing time as shown below:
Now for ease of understanding let’s consider all nodes of strongly connected components as one big node:
Note that after ‘decomposing’ the graph in this way, the graph can no longer contain cycles, i.e. the graph becomes a DAG or directed acyclic graph. Now the large nodes can easily be arranged in a topological order, the order will be A,B,C.
Now notice that if a node D1 can be moved to another node D2 in DAG then D1 must be before D2 in the topological order. Again, we already know that the one that is first in the topological order has a longer finishing time because all other nodes have to return to that node after finishing their work.
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