# Why all problems are not in P complexity?

İf Somebody wants to play perfect chess.I consider for finding solution to this problem,possible minimum data compression method which describes with 10 bit maximum possible data is in the range of 2^10.Then with some formula which can take input as next move in the chess,There should be some formula that can change all equations which are in the chess solving algorithm.Therefore ,if we assume in the chess.There should be 10^120 possible games.With approximately 400 bit with if conditions we can calculate all possible game's best strategy and with same thinking way all math problems can be solved with P complexity.Why there are differences with complexity classes?Short answer:İn karnaugh map some variation cannot be described shortly. I understood my mistake vote to delete.

• Your question is really hard to read, but I think you may not realize that there are time complexity and space complexity of algorithms. If you somehow manage that you algorithm uses only polynomial space, it does not mean that it will use polynomial time. The second thing you may be missing is that time complexity analysis is not done on a high level, where something might seem polynomial. About the if-statements that you mention: How many of them will there be? – Sandro Lovnički Nov 10 '18 at 23:59
• İf there are 5 if statements which are nested and when first if statement decided other four if statenents are changed then if fourth if statement decided other three if statements are changed. – etfwrtfrew etghtrth Nov 11 '18 at 0:16
• I am afraid that I am seeing this question as an intentional or unintentional attempt to puzzle people with fuzzy and gray logic. I cannot see OP's intention to clarify the question. – John L. Nov 11 '18 at 3:09

Computational complexity considers problems with inputs and looks at how the cost of the computation increases as the input gets longer. For example, you could generalize chess from an $$8\times 8$$ board to an arbitrary $$n\times n$$ board and ask how the time taken to compute which side can force checkmate varies as a function of $$n$$. Certain versions of this problem (it depends on exactly what rules you take) have been proven to require an amount of time that is exponential in $$n$$, which implies that they can't be solved in only a polynomial amount of time.