The objective of checkers is to get as many kings as possible and try to defeat the opponent. You want the opponent to have zero checkers and you have more than zero.
It is physically impossible to fold a piece of paper in half more than 8 times. However, assuming you could do it (though it would be easier to cut the pile so far in half and put one half on top of the other), then: After 1 fold the stack has 2 sheets After 2 folds the stack has 4 sheets After 3 folds the stack has 8 sheets After n folds the stack has 2^n sheets After 50 folds the stack will be 2⁵⁰ sheets thick As each sheet is 0.1mm, the stack will be: 2⁵⁰ × 0.1 mm = 112589990684262.4 mm thick = 112589990.6842624 km thick ≈ 1.126 × 10¹¹ m thick
Yes
Technically, this is impossible as two checkers will always lie in a row. However, how about like this: ................... . @ ............. ..@ @ ......... ..@ ... @ ..... ..@ @ @ @ . ................... (@ = checker) (. = table top, used to ensure picture stays as designed)
in Chinese Checkers, you can move any way you want. Yes, backwards too.
There is no limit, so up to 15. But you are unwise in most circumstances to stack more than about 5, because it does limit your options when you have large stacks.
no, you cant.
Chinese checkers is played by between 2-6 opponents, each of whom start with 10 men.
yes
The difference between checkers and Chinese checkers is that Chinese checkers are used with marbles.
Postfix expressions are the simplest to evaluate with a stack, for example: 2 3 4 + * 2 (stack: 2) 3 (2 3) 4 (2 3 4) + (2 12) * (14)
Jonathan Stack was born on June 2, 1957.
Jonathan Stack was born on June 2, 1957.
sumerian checkers are what they are called regular checkers
Delete is mostly commonly known as pop in stack. The last element inserted into the stack is removed from the stack. Here is an illustration:Consider the stack with the elements 1,2,3,4 inserted in order.1->2->3->4\topThe top of the stack will be pointing to 4. When pop operation is performed, 4 is removed from the stack and the top is made to point to 3. The stack then becomes:1->2->3\top
in dynamic stack we don't have to initialize the size of array while in static stack we have 2 initialize it ......
dos ene one no the answer