- 10,716
- 7,324
The second one isn't a solution as it violates a requirement.What we get are the two solutions
[tex]
n_p = 5, n_c = 7, n_b = 100-5-7 = 88\\
n_p = 0, n_c = 20, n_b = 100-20 = 80
[/tex]
The second one isn't a solution as it violates a requirement.What we get are the two solutions
[tex]
n_p = 5, n_c = 7, n_b = 100-5-7 = 88\\
n_p = 0, n_c = 20, n_b = 100-20 = 80
[/tex]
Right, I thought it violated the spirit of the excercise but didn't read carefully enough! I've edited my solution.The second one isn't a solution as it violates a requirement.
I will solve it with the correction.23. John wants to buy sweets for his party. He has $100 to spend. A box Belgian pralines costs $7, chips cost $3, and for 50 cents he gets a chocolate bar. What does he have to buy, if he wants at least one of each and spend the whole $100 to buy exactly 100 pieces in total?
But Matlab won't get you the Nobel prize for finding the GUT - pattern recognition might!Technically the fastest way is to copy-paste the table into Matlab and have it sum the numbers for you as this is what I started with and nobody has answered yet ... This saves all the time spent on noticing the pattern.
Theorists use Matlab a lot.But Matlab won't get you the Nobel prize for finding the GUT - pattern recognition might!
As not everyone will be familiar with the names: A dime is 10 cent, a quarter is 25 cent.In a tin are 10 dimes and 5 quarters. Against a stake one can shake out 3 coins, which one can keep.
26. In a tin are 10 dimes and 5 quarters. Against a stake one can shake out 3 coins, which one can keep.
What stake is the offer worth?
The rules are not clear to me. Should the sum of all your cards be 15 or the sum of a subset of your cards? If the former, then you have at best drawn when your opponent forces you above 15. It is easy as player one to force your opponent above 15 and as player two to block all possible wins (as there will always be at most one single card that brings player one to exactly 15). If the latter, there is an obvious winning strategy (P1:6, P2:9, P1: 8, P2: 1/7, P1: 7/1 and win 8+7 or 6+8+1)28. Nine cards with numbers one to nine are open in the middle of the table. Two players draw alternatively a card. Winner is who first has a hand that totals fifteen. Is there a winning strategy for the first player, for the second, or will optimal strategies always end with a draw? Why?
The dictionary plus Google translate both said 'alternative'.The rules are not clear to me. Should the sum of all your cards be 15 or the sum of a subset of your cards? If the former, then you have at best drawn when your opponent forces you above 15. It is easy as player one to force your opponent above 15 and as player two to block all possible wins (as there will always be at most one single card that brings player one to exactly 15). If the latter, there is an obvious winning strategy (P1:6, P2:9, P1: 8, P2: 1/7, P1: 7/1 and win 8+7 or 6+8+1)
I also think the word you are looking for is ”alternatingly”?
I disagree. In particular@lpetrich's solution is correct.
is not correct. Under your rules this would bring him to 17 for a certain loss or draw and so taking 8 would not be a good choice unless it is to avoid a loss. Also the assertion that 1,2,3,4,5 would sum to 15 is irrelevant because what matters is what comes before, not only what is about to come.Player #1 still has an opportunity for a two-move win, and plays one of the remaining two cards that will make it possible. I select 8 for definiteness.
I disagree. It is not tic-tac toe. To consider it tic-tac-toe you must first show that the only combination that can win is with three cards, but both 2, 3, 4, and indeed 5 cards (as in 12345) are possible wins under the rules presented.Imagine the cards are arranged as a magic square:
4 9 2
3 5 7
8 1 6
Now it is tic-tac-toe.
Wiktionary says:The dictionary plus Google translate both said 'alternative'.
Edit: Just as an example. The following (given your board configuration) is a win for x:alternatively
adverb
1. in an alternative way.
-------
alternatingly
adverb
1. In an alternating manner; taking turns.
To show this you still need to show that all 4- and 5-combos that win are unattainable. This is not trivial. Yes, you can show that there will be no 3-combo that wins, but you must show that it is impossible to devise a strategy that wins with 4 or 5 cards (which you, per my argument, cannot do) without winning the 3-in-a-row of tic-tac-toe. I honestly think my argument is much simpler. There are also bad tic-tac-toe moves that will not lose you this game, which to me suggests that the strategy applicable to this game is much easier. Also, the first move of the second player can be forced, unlike in tic-tac-toe.It is tic-tac-toe since I assumed an optimal strategy from both players. Of course you can lose deliberately. You just cannot win.
Sorry, but I don't see it. I see too many caveats with that argument for it to hold water even informally. (If it was that easy then it should be possible to at least make it plausible to me that it is viable.) It is also certainly not the argument presented in #138 and I cannot see that argument being correct either.The tic-tac-toe argument is easy to see and sufficient in this frame.
Tic-tac-toe is not necessarily decided in three moves. Also, tic-tac-toe strategy places less constraints on the second player than this game does, for example for the first move. For example, if you pick 9 then I have to pick 6 in this game or you win. In tic-tac-toe I can pick 5. The only optimal response to me picking 5 in this game is picking 6 to win. In tic-tac-toe it is to force a draw, which can be done with several different moves.A player can always chose the tic-tac-toe strategy ##S## of the square in post #140 above and there is nothing the other one can do about it. If you are first to draw, I will choose my card according to ##S##. If I start, then you will have to follow ##S##, for otherwise I will win. There will be no more than 3 moves, because it will be decided in 3. Only possibility is, that there is a combination of 15 with 3 cards which isn't a row, column or diagonal of the square. But if so, I will follow ##S## and will win since you opened up the possibility of e.g. a row in 3 moves. There is no way you can beat ##S##.
The best word for this use in British English is alternately.P.S.: You were right with alternating. I made a mistake. Likely due todaynighttime.