Got a strategy for Last Biscuit?

2656 /www/nrich/html/content/id/2656/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> <a href="http://nrich.maths.org/public/viewer.php?obj_id=1186&part=index">Last Biscuit</a> is a game for two players. Below, you can play it against an expert - the computer. You'll need to come up with a good strategy to stand a chance! Each player can take biscuits in one of two ways: <ol><li>By taking any number they like from just one stack or</li> <li>By taking the same amount from both stacks.</li> </ol> <div>The winner is the person who takes the last biscuit/s. Think carefully!</div> <div></div><div> Look for patterns in your winning moves and use them to find the winning strategy.</div> <div>When you've found some winning moves, explain why they work.</div> Can you describe the winning strategy if the stacks start by containing 100 and 50 biscuits? <a href="/content/id/2656/LastBiscuit.swf">Click for Full Screen Version</a> <mdo:flash height="400" width="550"><param name="movie" value="/content/id/2656/LastBiscuit.swf" ></param><param name="flashplayerversion" value="7" ></param><param name="height" value="400" ></param><param name="width" value="550" ></param></mdo:flash> <a href="http://www.nrich.maths.org/public/viewer.php?obj_id=1186&part=index&refpage=titlesearch.php"></a> </mdoxml> <mdoxml version="1.0"> We received the most complete solution from Neil Poole : Stuff I've worked out: At the end of your turn, you must not leave: 1a) either 1, 2, 3, 4 or 5 biscuits in either stack 1b) 6 in one stack if there is 11 or more in the other 1c) 7 in one stack if there is 5 or more in the other 1d) 8 in one stack if there is 14 or more in the other 1e) 10 in one stack if there is 7 or more in the other 1f) 13 in one stack if there is 9 or more in the other 2) stacks with consecutive numbers of biscuits (eg. 4 in one and 5 in the other) OR stacks with the same number of biccies in each 3a) stacks with 2 difference from 4 in one, 6 in the other upwards. 3b) stacks with 3 difference from 5 in one, 8 in the other upwards. 3c) stacks with 4 difference from 7 in one, 11 in the other upwards 3d) stacks with 5 difference from 9 in one, 14 in the other upwards All the above are true except if you leave "winning combos" (see below). Winning combos (if you make them): 4) 1 in one stack, 2 in the other 5) 3 in one stack, 5 in the other (then opponent must leave either 1), 2) or 3)) 6) 4 in one stack, 7 in the other 7) 6 in one stack, 10 in the other 8) 8 in one stack, 13 in the other If the opponent does 1), 2) or 3) then you will be able to make 4), 5), 6), 7) or 8). Keep doing winning combinations until you can take the last biccy(-ies)!!!!!!! Other winning combos include 9 & 15, 11 & 18, 12 & 20, etc. There are an infinite amount of winning combinations. <div>To find one, find the lowest whole positive number that hasn't been used yet (that is one of the numbers of the winning combo) and add it to the lowest positive whole difference between the two nos in the combos not yet used to find the other number in the combo.</div> <div>For example: The lowest integer not yet used in any of our combos is 14, as 13 has been used in 8 & 13, 12 has been used in 12 & 20, etc.</div> <div>The difference between 11 & 18 is 7.</div> <div>The difference between 12 & 20 is 8.</div> <div>9 isn't the difference in any of our combos so far so we add 9 to 14 to get 23.</div> So our new winning combo is 14 & 23. Combos with the same difference as lower winning combos are "losing combos". Any number in a winning combo with a number higher than its pair (eg. 14 & 27 coz 27 is higher than 23) is a losing combo. There's probably an easier way to work all that out!!!!!!!!! More findings out: I think it's only ever possible to make one winning combo (if you can). I think that if you start, and the starting amount of biccies isn't a winning combo, then you will be able to make a winning combo. If the starting amount of biccies is a winning combo and you start, against the "expert computer" you've lost. If the expert computer starts, and the starting combo is a winning combo, you've won (if you're good enough). If the expert computer starts, and the starting combo isn't a winning combo, you've lost. If the computer hasn't just, and it isn't the first go of the game, you can always make a winning combo, and vice-versa. This is great Neil. Many thanks for such a full solution. Mike from the NRICH team has also been working on this problem and has added: I'm now convinced there's some interesting maths in this. It looked like w1(i) and w2(i) [the biscuits lefts in the ith winning position] were approximately linearly related to i, so I charted them in Excel and read off the regression line: Using a few 10s of results this gave: w1(i) is approximately 2.618i - 0.5026 w2(i) is approximately 1.618i - 0.5026 Interesting, since 1.618 is very close to the golden ratio phi, and 2.618 is (phi+1) or phi^2. That -0.5026 looks like it's caused by rounding down to an integer value. So, with phi = (\sqrt(5)+1)/2, I tried for 0 < = i < = 10000: w1(i) = floor(phi*i) w2(i) = floor((phi+1)*i) Perfect match on all pairs! If phi is involved there must be some Fibonacci sequences around. In fact there are lots of them. The winning pairs are the even pairs of consecutive numbers in a union of Fibonacci sequences. If F(1,2) is the Fibonacci sequence that starts 1,2,3,5,8... and F(1,3) starts 1,3,4,7,11,.... you can use F(1,3) to fill in the gaps left by F(1,2). F(2,4) fills in the gaps left by the previous 2, and so on. This all makes a certain amount of sense since the ratio of successive Fibonacci numbers is phi - and that was the gradient of the regression line. It's still all just conjecture though - I haven't been able to prove any of it so far. Assuming it's all true, it's possible to write a slick algorithm that works for very large numbers of biscuits - essentially by using the inverse linear relationship to find the nearest winning pairs that can be reached from any position. Unfortunately, the calculation is still a bit much to do in one's head - but it avoids the need to tabulate all pairs up to the one that works. I tried that and so far as I can tell it works fine. See <a href="http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibrab.html#PhiPlot">http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibrab.html#PhiPlot</a> This page is all relevant to the winning pairs in Last Biscuit. It's talking about the same function I needed to calculate them - floor(i*phi) - defining this as the spectrum of phi. There's an interesting section on how the spectra of 2 numbers A and B can partition the integers into 2 sets iff 1/A + 1/B = 1. The smaller and larger numbers of the winning pairs form such a partition with A = phi and B = phi^2. 1/phi + 1/phi^2 = 1 since phi + 1 = phi^2. <a href="http://www.numericana.com/answer/games.htm">http://www.numericana.com/answer/games.htm</a> goes into some detail linking this version, other nim like games and the grundy numbers. After that background, the following paper might be more understandable <a href="http://www.math.temple.edu/%7Exysun/wythoff/wythoff_2.pdf%20">http://www.math.temple.edu/~xysun/wythoff/wythoff_2.pdf</a> I think it contains a proof that last biscuit winning pairs are a wythoff sequence - but it ain't easy! </mdoxml> <mdoxml version="1.0"> <h3>Why do this problem?</h3> Strategy games are always good for developing mathematical thinking. This game is interesting because although the 'overall' strategy is difficult, students can usefully analyse particular cases. This will require clear recording of results and careful analysis of the logical possibilities. They will then be able to put their strategy into action by beating the computer, which never makes a mistake in its moves, although this will still require clear application of the procedure of the strategy. <h3>Possible approach</h3> <div>This game is difficult and most students will need to play it (and lose!) several times to start to get a feel for how to win. Note there there is no 'obvious overall strategy', so students should play the game for a while and then come back together to share ideas. Suggest that students consider the cases for low numbers of biscuits first (less than 10 of each type).</div> <div>It might become apparent that certain configurations are known to be a 'win' for person to play next. When one of these configurations is found it could be shared with the group.</div> <div>Using known winning configurations might allow students to develop winning configurations for larger numbers of biscuits.</div> <div>Throughout, clarity of thinking, analysis of the game position and clear recording of results should be encouraged.</div> <div>Once several winning positions have been found, can any patterns be found? Could students make any conjectures as to the form of winning configurations?</div> <h3>Key questions</h3> <ul> <li>What is the smallest 'certain lose' position?</li> <li>How could we prevent the computer from putting us into this position?</li> </ul> <h3>Possible extension</h3> <div>This structure allows a rich analysis which very able students might enjoy. Various questions of proof are as follows:</div> <ul> <li>Can you prove that (1, 2) is the only winning configuration with a difference of 1 in the total number of biscuits?</li> <li>How many winning configurations differ by 2, 3 or n biscuits?</li> </ul> <div>You might also like to play <a href="http://nrich.maths.org/public/viewer.php?obj_id=402&part=">Nim</a> , which is a version of this game with multiple counters.</div> <h3>Possible support</h3> <div>Focus to start with on numbers of biscuits less than 5 of each type. Can a strategy be devised to win for these individual cases?</div> <div>Students could play in pairs if the computer is winning too often. They could also play with counters instead of the interactivity.</div> </mdoxml> <mdoxml version="1.0"> Try working backwards. Fibonacci sequences might help. Play <a href="http://www.nrich.maths.org/public/viewer.php?obj_id=1269&part=index&refpage=monthindex.php"> GOT IT</a> and try to find a strategy for winning at this game. </mdoxml> <mdoxml version="1.0"> Description Last Biscuit is a game that challenges you to use a systematic approach to deduce a winning strategy. Can you find a strategy that will beat the computer? Solution: You can beat your opponent if you leave the following number of biscuits: 1, 2 3, 5 4, 7 6, 10 8, 13 9, 15 11, 18 12, 20 14, 23 16, 26 17, 28 19, 31 21, 34 22, 36 24, 39 25, 41 27, 44 29, 47 30, 49 32, 52 33, 54 35, 57 37, 60 38, 62 40, 65 42, 68 43, 70 45, 73 46, 75 48, 78 50, 81 I'm now convinced there's some interesting maths in this. It looked like w1(i) and w2(i) [the biscuits lefts in the ith winning position] were approximately linearly related to i, so I charted them in Excel and read off the regression line: Using a few 10s of results this gave: w1(i) is approximately 2.618i - 0.5026 w2(i) is approximately 1.618i - 0.5026 Interesting, since 1.618 is very close to the golden ratio phi, and 2.618 is (phi+1) or phi^2. That -0.5026 looks like it's caused by rounding down to an integer value. So, with phi = (\sqrt(5)+1)/2, I tried for 0 < = i < = 10000: w1(i) = floor(phi*i) w2(i) = floor((phi+1)*i) Perfect match on all pairs! If phi is involved there must be some Fibonacci sequences around. In fact there are lots of them. The winning pairs are the even pairs of consecutive numbers in a union of Fibonacci sequences. If F(1,2) is the Fibonacci sequence that starts 1,2,3,5,8... and F(1,3) starts 1,3,4,7,11,.... you can use F(1,3) to fill in the gaps left by F(1,2). F(2,4) fills in the gaps left by the previous 2, and so on. This all makes a certain amount of sense since the ratio of successive Fibonacci numbers is phi - and that was the gradient of the regression line. It's still all just conjecture though - I haven't been able to prove any of it so far. Assuming it's all true, it's possible to write a slick algorithm that works for very large numbers of biscuits - essentially by using the inverse linear relationship to find the nearest winning pairs that can be reached from any position. Unfortunately, the calculation is still a bit much to do in one's head - but it avoids the need to tabulate all pairs up to the one that works. I tried that and so far as I can tell it works fine. An Excel spreadsheet so you can see what I mean is saved in common > site content. See http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibrab.html#PhiPlot This page is all relevant to the winning pairs in Last Biscuit. It's talking about the same function I needed to calculate them - floor(i*phi) - defining this as the spectrum of phi. There's an interesting section on how the spectra of 2 numbers A and B can partition the integers into 2 sets iff 1/A + 1/B = 1. The smaller and larger numbers of the winning pairs form such a partition with A = phi and B = phi^2. 1/phi + 1/phi^2 = 1 since phi + 1 = phi^2. http://home.att.net/~numericana/answer/games.htm goes into some detail linking this version other nim like games and the grundy numbers. After that background, the following paper might be more understandable http://www.math.temple.edu/~xysun/wythoff/wythoff_2.pdf I think it contains a proof that last biscuit winning pairs are a wythoff sequence - but it ain't easy! You can beat your opponent if you leave the following number of biscuits: 1, 2 3, 5 4, 7 6, 10 8, 13 9, 15 11, 18 12, 20 14, 23 16, 26 17, 28 19, 31 21, 34 22, 36 24, 39 25, 41 27, 44 29, 47 30, 49 32, 52 33, 54 35, 57 37, 60 38, 62 40, 65 42, 68 43, 70 45, 73 46, 75 48, 78 50, 81 A strategy to find winning moves is to note that $(0,0)$ is a winning pair and then find the complete set of winning pairs by induction. If $(A_n,B_n),A_n \leq B_n$ is the last move discovered, then $A_{n+1}$ is the minimum positive integer that does not appear in a winning pair yet, and $B_{n+1} = A_{n+1} + 1 + (B_n - A_n)$. $(A_{n+1},B_{n+1})$ must be a new winning move because it is impossible to generate a lower winning move from it in one turn. Neither $A_{n+1}$ nor $B_{n+1}$ appear in a lower pair,so taking from a single stack fails. Also, no lower winning pair has a difference of $A_{n+1}-B_{n+1}$,so taking from both stacks will fail. Plotting $A_{n}$ against $n$ and $B_{n}$ against $n$ reveals near linear relationships, which turn out to be represented exactly by the equations:- \begin{eqnarray} A_n&=& \mbox{floor}(n\phi)\\ B_n &=& \mbox{floor}(n\phi^2) \end{eqnarray} where the floor function rounds values to the next lowest integer, and $\phi$ is the golden section. A good way to discover these relationships is to plot them in Excel and infer them from the regression line equations. For a proof, see <a href="http://www.math.temple.edu/~xysun/wythoff/wythoff_2.pdf">ref [10] in this paper by Xinyu Sun</a> Even when you know the winning pairs, it is not trivial to calculate the next best move from a given starting position, since the above formulae must be inverted (the floor function complicates matters) and applied with some care. The winning move from a (500,1000) start will take you to (500, 309). </mdoxml> 2 5 0 0 1 1 0 Got a strategy for Last Biscuit? Can you beat the computer in the challenging strategy game? 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