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Difference between revisions of "2018 AIME I Problems/Problem 14"

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(→‎Solution: add more complete solution with DP table)
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==Solution==
 
==Solution==
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(incomplete, someone help format this)  
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This is easily solved by recursion/dynamic programming. To simplify the problem somewhat, let us imagine the seven vertices on a line <math>E \leftrightarrow P_4 \leftrightarrow P_5 \leftrightarrow S \leftrightarrow P_1 \leftrightarrow P_2 \leftrightarrow P_3 \leftrightarrow E</math>. We can count the number of left/right (L/R) paths of length <math>\le 11</math> that start at <math>S</math> and end at either <math>P_4</math> or <math>P_3</math>.
  
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Make a table showing how many ways there are to get to each vertex in a certain amount of jumps. <math>P_2, P_1, S, P_5</math> all equal the sum of their adjacent elements in the previous jump. <math>P_3</math> equals the previous <math>P_2</math>. <math>P_4</math> equals the previous <math>P_5</math>. Each <math>E</math> equals the previous adjacent element in the previous jump.
+
We can visualize the paths using the common grid counting method by starting at the origin <math>(0,0)</math>, so that a right (R) move corresponds to moving 1 in the positive <math>x</math> direction, and a left (L) move corresponds to moving 1 in the positive <math>y</math> direction. Because we don't want to move more than 2 units left or more than 3 units right, our path must not cross the lines <math>y = x+2</math> or <math>y = x-3</math>. Letting <math>p(x,y)</math> be the number of such paths from <math>(0,0)</math> to <math>(x,y)</math> under these constraints, we have the following base cases:
  
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Jump & E & P_3 & P_2 & P_1 & S & P_5 & P_4 & E
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<math>p(x,0) = \begin{cases}
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1 & x \le 3 \\
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0 & x > 3
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\end{cases}
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\qquad p(0,y) = \begin{cases}
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1 & y \le 2 \\
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0 & y > 3
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\end{cases}
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</math>
  
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0 0 0 0 0 1 0 0 0 \\
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and recursive step
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1 0 0 0 1 0 1 0 0 \\
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<math>p(x,y) = p(x-1,y) + p(x,y-1)</math> for <math>x,y \ge 1</math>.
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2 0 0 1 0 2 0 1 0 \\
 
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3 0 1 0 3 0 3 0 1 \\
 
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4 1 0 4 0 6 0 3 1 \\
 
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5 1 4 0 10 0 9 0 4 \\
 
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6 5 0 14 0 19 0 9 4 \\
 
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7 5 14 0 33 0 28 0 13 \\
 
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8 19 0 47 0 61 0 28 13 \\
 
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9 19 47 0 108 0 89 0 41 \\
 
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10 66 0 155 0 197 0 89 41 \\
 
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11 66 155 0 352 0 286 0 130 \\
 
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12 221 - - - - - - 130
 
  
 +
The filled in grid will look something like this, where the lower-left <math>1</math> corresponds to the origin:
  
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The number of ways to jump to the ends within 12 jumps is <math>221 + 130 = \boxed{351}</math>.
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<math>
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\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline
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0 & 0 & 0 & 0 & \textbf{89} & & & \\ \hline
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0 & 0 & 0 & \textbf{28} & 89 & & & \\ \hline
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0 & 0 & \textbf{9} & 28 & 61 & 108 & 155 & \textbf{155} \\ \hline
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0 & \textbf{3} & 9 & 19 & 33 & 47 & \textbf{47} & 0 \\ \hline
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\textbf{1} & 3 & 6 & 10 & 14 & \textbf{14} & 0 & 0 \\ \hline
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1 & 2 & 3 & 4 & \textbf{4} & 0 & 0 & 0 \\ \hline
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1 & 1 & 1 & \textbf{1} & 0 & 0 & 0 & 0 \\ \hline
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\end{tabular}
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</math>
  
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The bolded numbers on the top diagonal represent the number of paths from <math>S</math> to <math>P_4</math> in 2, 4, 6, 8, 10 moves, and the numbers on the bottom diagonal represent the number of paths from <math>S</math> to <math>P_3</math> in 3, 5, 7, 9, 11 moves. We don't care about the blank entries or entries above the line <math>x+y = 11</math>. The total number of ways is <math>1+3+9+28+89+1+4+14+47+155 = \boxed{351}</math>.
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 +
-scrabbler94
  
 
{{AIME box|year=2018|n=I|num-b=13|num-a=15}}
 
{{AIME box|year=2018|n=I|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:56, 9 March 2018


Let $SP_1P_2P_3EP_4P_5$ be a heptagon. A frog starts jumping at vertex $S$. From any vertex of the heptagon except $E$, the frog may jump to either of the two adjacent vertices. When it reaches vertex $E$, the frog stops and stays there. Find the number of distinct sequences of jumps of no more than $12$ jumps that end at $E$.

Solution

This is easily solved by recursion/dynamic programming. To simplify the problem somewhat, let us imagine the seven vertices on a line $E \leftrightarrow P_4 \leftrightarrow P_5 \leftrightarrow S \leftrightarrow P_1 \leftrightarrow P_2 \leftrightarrow P_3 \leftrightarrow E$. We can count the number of left/right (L/R) paths of length $\le 11$ that start at $S$ and end at either $P_4$ or $P_3$.

We can visualize the paths using the common grid counting method by starting at the origin $(0,0)$, so that a right (R) move corresponds to moving 1 in the positive $x$ direction, and a left (L) move corresponds to moving 1 in the positive $y$ direction. Because we don't want to move more than 2 units left or more than 3 units right, our path must not cross the lines $y = x+2$ or $y = x-3$. Letting $p(x,y)$ be the number of such paths from $(0,0)$ to $(x,y)$ under these constraints, we have the following base cases:

$p(x,0) = \begin{cases} 1 & x \le 3 \\ 0 & x > 3 \end{cases} \qquad p(0,y) = \begin{cases} 1 & y \le 2 \\ 0 & y > 3 \end{cases}$

and recursive step $p(x,y) = p(x-1,y) + p(x,y-1)$ for $x,y \ge 1$.

The filled in grid will look something like this, where the lower-left $1$ corresponds to the origin:

$\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline 0 & 0 & 0 & 0 & \textbf{89} & & & \\ \hline 0 & 0 & 0 & \textbf{28} & 89 & & & \\ \hline 0 & 0 & \textbf{9} & 28 & 61 & 108 & 155 & \textbf{155} \\ \hline 0 & \textbf{3} & 9 & 19 & 33 & 47 & \textbf{47} & 0 \\ \hline \textbf{1} & 3 & 6 & 10 & 14 & \textbf{14} & 0 & 0 \\ \hline 1 & 2 & 3 & 4 & \textbf{4} & 0 & 0 & 0 \\ \hline 1 & 1 & 1 & \textbf{1} & 0 & 0 & 0 & 0 \\ \hline \end{tabular}$

The bolded numbers on the top diagonal represent the number of paths from $S$ to $P_4$ in 2, 4, 6, 8, 10 moves, and the numbers on the bottom diagonal represent the number of paths from $S$ to $P_3$ in 3, 5, 7, 9, 11 moves. We don't care about the blank entries or entries above the line $x+y = 11$. The total number of ways is $1+3+9+28+89+1+4+14+47+155 = \boxed{351}$.

-scrabbler94

2018 AIME I (Problems • Answer Key • Resources)
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