IceM
(The December 31 Game) Two players alternately name dates. On each move, a player can increase the month or the day of the month, but not both. The starting position is January 1, and the player who names December 31 wins. According to the rules, the first player can start by naming some day in January after the first, or the first of some month after January. For example (Jan. 10, Mar. 10, Mar. 15, Apr. 15, Apr. 25, Nov. 25, Nov. 30, Dec. 30, Dec. 31) is an instance of the game won by the first player. Derive a winning strategy for the first player. (Hint: use strong induction to describe the "winning dates".)
I am aware that the pattern is DAYS - MONTH = 19 and if you hold onto the 19th date (IE: if you move it from January 1st to January 20th on the first turn you will forever win).
I'm more or less interested in seeing how someone did the strong induction proof. So, I basically am looking for wording.
I.E:
My attempt at wording my proof by strong induction (criticism is welcomed!):
The strategy holds for the first player when the month is November and the day is the 30th, as the second player will either have to advance the month or the day leaving the first player the ability to swoop in and take the option he or she has left to win the game (reach december 31st). This proves the basis step.
Suppose we start with many dates d and many months m. If the first player takes the 20th of january, then regardless of what the second player does, whether it be to advance the month or the date, the first player wins.
Otherwise, if the first player takes a different m or d for some m or d. The econd player will need to take a m or d such that the difference between the days and months is not 19.
The remainder of the game is equivalent to d-m and the second player always has to have the difference between the days and months of the spot he or she moves to not equal to 19.
The induction hypothesis tells us then that the first player wins this game.
Answer
If the first player can name Nov 30, she wins.
Which means: If the first player can name Oct 29, she wins. Because whatever date the other player names, the first player can then name either Dec 31 (winning) or Nov 30 (and be guaranteed to win).
Which means, by the same reasoning, that if the first player can name Sep 28, she wins.
Work your way back and you should be able to determine that player 1 can name a winning date on her first turn.
EDIT:
I think I can tell that you understand, but the wording is a bit confusing. I would try to word it like a normal (strong) induction, i.e. prove it for n=1, then show that, if it is true for n=1 up to n=k-1, then it is also true for n=k. i.e.:
theorem: that saying (month,day) = (13-n, 32-n) wins.
Proof for n=1: n=1 means (12,31) = December 31, obviously wins.
Assume true for n=1 up to n=k-1
Then for n=k: player 1 says (13-k, 32-k). If player 2 adds to the month i.e. says (13-m, 32-k), where m<k, then player 1 says (13-m, 32-m), which wins (by our assumption, because 1 <= m <= k-1). Similarly if player 2 adds to the day by saying (13-k, 32-m), player 1 says (13-m, 32-m) which also wins.
So we have proven the theorem by strong induction: if (13-n, 32-n) wins for n = 1 to k-1, we have shown it also wins for n=k.
So armed with the theorem, we can find a winning strategy: that player 1 can jump to a winning date on his or her first turn; and at any subsequent turn can always jump to a winning date.
Try to rewrite that in your own words.
If the first player can name Nov 30, she wins.
Which means: If the first player can name Oct 29, she wins. Because whatever date the other player names, the first player can then name either Dec 31 (winning) or Nov 30 (and be guaranteed to win).
Which means, by the same reasoning, that if the first player can name Sep 28, she wins.
Work your way back and you should be able to determine that player 1 can name a winning date on her first turn.
EDIT:
I think I can tell that you understand, but the wording is a bit confusing. I would try to word it like a normal (strong) induction, i.e. prove it for n=1, then show that, if it is true for n=1 up to n=k-1, then it is also true for n=k. i.e.:
theorem: that saying (month,day) = (13-n, 32-n) wins.
Proof for n=1: n=1 means (12,31) = December 31, obviously wins.
Assume true for n=1 up to n=k-1
Then for n=k: player 1 says (13-k, 32-k). If player 2 adds to the month i.e. says (13-m, 32-k), where m<k, then player 1 says (13-m, 32-m), which wins (by our assumption, because 1 <= m <= k-1). Similarly if player 2 adds to the day by saying (13-k, 32-m), player 1 says (13-m, 32-m) which also wins.
So we have proven the theorem by strong induction: if (13-n, 32-n) wins for n = 1 to k-1, we have shown it also wins for n=k.
So armed with the theorem, we can find a winning strategy: that player 1 can jump to a winning date on his or her first turn; and at any subsequent turn can always jump to a winning date.
Try to rewrite that in your own words.
Birthday Party games(moms 31st bday)?
Q. moms 31st and we need fun and games. there will be family and kids and plz no video game answers.
thx and plz help**************!!!!!!!!
thx and plz help**************!!!!!!!!
Answer
A fun game is always twister. We play it at thanksgiving and its always a hoot. We also set up volleyball in the yard and do that. (we got the set for like 20 bucks at walmart, but i dont know if they have it anymore) also we team up and do pictionary and games like gestures and catch phrase. We have people ranging from 6-65 and everyone has fun. Hope u have a great day!
A fun game is always twister. We play it at thanksgiving and its always a hoot. We also set up volleyball in the yard and do that. (we got the set for like 20 bucks at walmart, but i dont know if they have it anymore) also we team up and do pictionary and games like gestures and catch phrase. We have people ranging from 6-65 and everyone has fun. Hope u have a great day!
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