likesforests> "10^150 would be a better lower-bound of the complexity of finding "a solution" to chess, based on our new knowledge."
JG27Pyth> "Calculating all chess positions, all games, this is a SUFFICIENT condition for solving chess... but the suggestion is they represent NECESSARY conditions. Why?"
That's not what this number represents. 10^150 represents a lower-bound on the complexity of calculating a single winning solution to chess. It has nothing to do with calculating all chess positions or all games.
Step #1: Assume such a solution exists--or else it's even more complex. :)
Step #2: How long would a game between two perfect chess engines last? It's obviously more than 30 moves considering the state of modern opening theory. Games between 2300+ humans often last 40 moves. Games between 3000+ engines often last 100 moves. Games between perfect engines would probably last longer but we'll err on the low side (to come up with a lower-bound) and say 100 moves.
Step #3: How many moves do we have to examine for White? We'll assume by some magic (mathematical formula, efficient pruning, etc) you can select White's best move 100% perfectly and say 1 move.
Step #4: How many moves do we have to examine for Black? To prove you have a winning solution you need to examine all of Black's legal moves. DeGroot's did extensive analysis and discovered that, on average, a player has 30 legal moves.
So (1 x 30) ^ 100 = (30 ^ 2) ^ 50 = (900)^50 =~ (10^3)^50 = 10^150.
JG27Pyth> "I have designed a game which has which has more or less the same number of possible moves as ordinary chess.
This example is crude as you say, but we can apply the same steps to your game:
Step #1: OK
Step #2: 1 move.
Step #3: 1 move.
Step #4: N/A - Black has no part in your example.
So 1^1 = 1. The complexity of your game is 1.
If resources = beer, cheese doodles and hard rock music, then I can say with absolute certainty that the answer is "yes."