Pascal-Like Triangles and Fibonacci-Like Sequences
The international conference “The 20th Japan Conference on Discrete and Computational Geometry, Graphs, and Games (JCDCG^3 2017)”
August 29th- September 1st, 2017 (Tokyo University of Science, Tokyo, Japan )
http://www.jcdcgg.u-tokai.ac.jp
up a card from a box. The game starts with player A1. When a player receives the box, he or she draws out a card, and these cards are not returned to the box. The first player to draw a red card loses the game, and the game ends. This game is mathematically the same as a Russian roulette game.
Example. 4 players play the game, and let f(n,m) be the probability that the first player loses in the game. We arrange f(n,m) vertically for n=1,2,3,.. horizontally for m=1,2,3,...,n.
ps(k-1)round s rounds s rounds … … Θ1 h th round h rounds … The case that the first player picks up a red card in (k − 1)ps + h th round … … … Θ̍ Θ2 s rounds s rounds … … … Θ̍ Θ2 s rounds s rounds
⎢ ⎥ ⎦ ⎥ cycles for the first Player’s picking up red cards, The number of combination of red cards is n−(k−1) ps−h C m−1 . k=1 n−m+sp−h+1 sp ⎢ ⎣ ⎢ ⎥ ⎦ ⎥ ∑ h=1 s ∑
of red cards when the first player loses the game. U(n,m) = n−(k−1)ps−h C m−1 k=1 n−m+sp−h+1 sp ⎢ ⎣ ⎢ ⎥ ⎦ ⎥ ∑ h=1 s ∑ F(n,m) = U(n,m) n C m The probability is
the sequence by b4,1(n). When p = 2 and s = 2, we denote the sequence by b2,2(n). Then b4,1(n+1)=b2,2(n) for n=1,2,3 (mod 4). Please see the relation between these sequences.
p). The sequence produced by diagonals sum of game of p-persons who pick up 2 cards in each turn is very similar to the sequence for 2p persons who pick up 1 card in each turn.
player B picks up s2 cards in each turn. Player A loses the game when he collects g1 red cards, and Player B loses the game when he collects g2 red cards. Note that a player loses the game when he picks up one card in the previous game.
cards m = 3. s1=2,s2=3, A’s goal : g1=3, B’s goal : g2=2, " " # # # " " # # # D D D D D T T Player A cannot collect three red cards, and Player B cannot collect two red cards. D
A loses the game is We are going to present an outline of the proof. U(n, m, s1, s2, g1, g2) = s1 X h=1 ( min(b n m+g1 h+s1 s1 c,b n+s1+s2 h s1+s2 c,b n m+s1+s2+g1+g2 1 s1+s2 c) X k=d g1+s1 h s1 e min(g2 1,m g1,(k 1)s2) X v=max(0,m n+(k 1)(s1+s2) g1+h ((k 1)s2 Cv ⇥ h+(k 1)s1 1 Cg1 1 ⇥ n (k 1)(s1+s2) h Cm v g1 )
1 Cg1 1 ⇥ n (k 1)(s1+s2) h Cm v g1 min(b n m + g1 h s1 c + 1, b n h s1 + s2 c + 1, b n m + g1 + g2 1 s1 + s2 c + 1) min(g2 1, m g1, (k 1)s2) v = max(0, m n + (k 1)(s1 + s2) g1 + h k = d g1 h s1 e + 1
collects g1 red cards. The number of combinations is Since there are (k-1)s2 places, we have A non-negative integer v should be less than g2. Otherwise B loses the game. Since player A collect g1 cards, player B collect v cards. From the remaining m-g1 cards.
we have Now we have all the inequalities for k and v, and hence U(n, m, s1, s2, g1, g2) = s1 X h=1 ( min(b n m+g1 h s1 c+1,b n h s1+s2 c+1,b n m+g1+g2 1 s1+s2 c+1) X k=d g1 h s1 e+1 min(g2 1,m g1,(k 1)s2) X v=max(0,m n+(k 1)(s1+s2) g1+h ((k 1)s2 Cv ⇥ h+(k 1)s1 1 Cg1 1 ⇥ n (k 1)(s1+s2) h Cm v g1 ) g1 h m + n + s1 s1 k g1 + g2 h m + n + s1 + s2 1 s1 + s2 k
(n, m, s1, s2, g1, g2) = U (n 1, m 1, s1, s2, g1, g2) + U (n 1, m, s1, s2, g1, g2) m g1 + g2 m g1 + g2 In this case there is no draw for the game. We omit the proof here.
in certain part of the triangle. 6 O N 6 O N 6 O N 6 O N 6 O N 6 O N 0 0 0 0 0 0 2 6 6 2 3 9 9 3 0 0 0 0 0 0 0 0 0 1 0 1 1 0 1 2 1 0 1 3 3 1 0 3 10 12 6 1 0 6 22 31 21 7 1 In the triangle on the right side, red numbers are calculated by this formula. U(n,m) - ( U(n-1,m)+U(n-1,m-1) ). {s1 ,s2 }={2,3} {g1 ,g2 }={2,4}
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