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

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For vertices <math>S, P_1, P_2, P_5,</math> the number of ways to get there after <math>n</math> jumps is the sum of the number of ways to get to the adjacent vertices after <math>n-1</math> jumps. For vertices <math>P_3,</math> and <math>P_4,</math> the number of ways to get there after <math>n</math> jumps is he number of ways to get to <math>P_2</math> and <math>P_5</math> after <math>n-1</math> jumps.
 
For vertices <math>S, P_1, P_2, P_5,</math> the number of ways to get there after <math>n</math> jumps is the sum of the number of ways to get to the adjacent vertices after <math>n-1</math> jumps. For vertices <math>P_3,</math> and <math>P_4,</math> the number of ways to get there after <math>n</math> jumps is he number of ways to get to <math>P_2</math> and <math>P_5</math> after <math>n-1</math> jumps.
  
−
<math> \begin{tabular}{l|*{7}{c}}
+
<math> \begin{tabular}{|l|c|c|c|c|c|c|c|}
−
Jump \#           & S  & 1  & 2  & 3  & E           & 4  & 5  \\ \hline
+
\hline
−
1                 & 0   & 1   & 0   & 0   & \textbf{0}  & 0   & 1   \\ \hline
+
Jump \# & \(S\)   & \(P_1\) & \(P_2\) & \(P_3\) & \(E\)        & \(P_4\) & \(P_5\) \\ \hline
−
2                 & 2   & 0   & 1   & 0   & \textbf{0}  & 1   & 0   \\ \hline
+
1       & 0       & 1       & 0       & 0       & \textbf{0}  & 0       & 1       \\ \hline
−
3                 & 0   & 3   & 0   & 1   & \textbf{1}  & 0   & 3   \\ \hline
+
2       & 2       & 0       & 1       & 0       & \textbf{0}  & 1       & 0       \\ \hline
−
4                 & 6   & 0   & 4   & 0   & \textbf{1}  & 3   & 0   \\ \hline
+
3       & 0       & 3       & 0       & 1       & \textbf{1}  & 0       & 3       \\ \hline
−
5                 & 0   & 10 & 0   & 4   & \textbf{3}  & 0   & 9   \\ \hline
+
4       & 6       & 0       & 4       & 0       & \textbf{1}  & 3       & 0       \\ \hline
−
6                 & 19 & 0   & 14 & 0   & \textbf{4}  & 9   & 0   \\ \hline
+
5       & 0       & 10     & 0       & 4       & \textbf{3}  & 0       & 9       \\ \hline
−
7                 & 0   & 33 & 0   & 14 & \textbf{9}  & 0   & 28 \\ \hline
+
6       & 19     & 0       & 14     & 0       & \textbf{4}  & 9       & 0       \\ \hline
−
8                 & 61 & 0   & 47 & 0   & \textbf{14}  & 28 & 0   \\ \hline
+
7       & 0       & 33     & 0       & 14     & \textbf{9}  & 0       & 28     \\ \hline
−
9                 & 0   & 108 & 0   & 47 & \textbf{28}  & 0   & 89 \\ \hline
+
8       & 61     & 0       & 47     & 0       & \textbf{14}  & 28     & 0       \\ \hline
−
10                 & 197 & 0   & 155 & 0   & \textbf{47}  & 89 & 0   \\ \hline
+
9       & 0       & 108     & 0       & 47     & \textbf{28}  & 0       & 89     \\ \hline
−
11                 & 0   & 352 & 0   & 155 & \textbf{89}  & 0   & 286 \\ \hline
+
10     & 197     & 0       & 155     & 0       & \textbf{47}  & 89     & 0       \\ \hline
−
12                 & 638 & 0   & 507 & 0   & \textbf{155} & 286 & 0   \\ \hline
+
11     & 0       & 352     & 0       & 155     & \textbf{89}  & 0       & 286     \\ \hline
 +
12     & 638     & 0       & 507     & 0       & \textbf{155} & 286     & 0       \\ \hline
 
\end{tabular} </math>
 
\end{tabular} </math>
  

Latest revision as of 19:45, 15 October 2024

Problem

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 1

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:

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

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 $1+3+9+28+89+1+4+14+47+155 = \boxed{351}$.

(Solution by scrabbler94)

Solution 2

Let $E_n$ denotes the number of sequences with length $n$ that ends at $E$. Define similarly for the other vertices. We seek for a recursive formula for $E_n$. \begin{align*} E_n&=P_{3_{n-1}}+P_{4_{n-1}} \\ &=P_{2_{n-2}}+P_{5_{n-2}} \\ &=P_{1_{n-3}}+P_{3_{n-3}}+S_{n-3}+P_{4_{n-3}} \\ &=(P_{3_{n-3}}+P_{4_{n-3}})+S_{n-3}+P_{1_{n-3}} \\ &=E_{n-2}+S_{n-3}+P_{1_{n-3}} \\ &=E_{n-2}+P_{1_{n-4}}+P_{5_{n-4}}+S_{n-4}+P_{2_{n-4}} \\ &=E_{n-2}+(S_{n-4}+P_{1_{n-4}})+P_{5_{n-4}}+P_{2_{n-4}} \\ &=E_{n-2}+(E_{n-1}-E_{n-3})+S_{n-5}+P_{4_{n-5}}+P_{1_{n-5}}+P_{3_{n-5}} \\ &=E_{n-2}+(E_{n-1}-E_{n-3})+(S_{n-5}+P_{1_{n-5}})+(P_{4_{n-5}}+P_{3_{n-5}}) \\ &=E_{n-2}+(E_{n-1}-E_{n-3})+(E_{n-2}-E_{n-4})+E_{n-4} \\ &=E_{n-1}+2E_{n-2}-E_{n-3} \\ \end{align*} Computing a few terms we have $E_0=0$, $E_1=0$, $E_2=0$, $E_3=1$, and $E_4=1$.

Using the formula yields $E_5=3$, $E_6=4$, $E_7=9$, $E_8=14$, $E_9=28$, $E_{10}=47$, $E_{11}=89$, and $E_{12}=155$.

Finally adding yields $\sum_{k=0}^{12}E_k=\boxed{351}$.

~ Nafer

Solution 3

For vertices $S, P_1, P_2, P_5,$ the number of ways to get there after $n$ jumps is the sum of the number of ways to get to the adjacent vertices after $n-1$ jumps. For vertices $P_3,$ and $P_4,$ the number of ways to get there after $n$ jumps is he number of ways to get to $P_2$ and $P_5$ after $n-1$ jumps.

$\begin{tabular}{|l|c|c|c|c|c|c|c|} \hline Jump \# & \(S\) & \(P_1\) & \(P_2\) & \(P_3\) & \(E\) & \(P_4\) & \(P_5\) \\ \hline 1 & 0 & 1 & 0 & 0 & \textbf{0} & 0 & 1 \\ \hline 2 & 2 & 0 & 1 & 0 & \textbf{0} & 1 & 0 \\ \hline 3 & 0 & 3 & 0 & 1 & \textbf{1} & 0 & 3 \\ \hline 4 & 6 & 0 & 4 & 0 & \textbf{1} & 3 & 0 \\ \hline 5 & 0 & 10 & 0 & 4 & \textbf{3} & 0 & 9 \\ \hline 6 & 19 & 0 & 14 & 0 & \textbf{4} & 9 & 0 \\ \hline 7 & 0 & 33 & 0 & 14 & \textbf{9} & 0 & 28 \\ \hline 8 & 61 & 0 & 47 & 0 & \textbf{14} & 28 & 0 \\ \hline 9 & 0 & 108 & 0 & 47 & \textbf{28} & 0 & 89 \\ \hline 10 & 197 & 0 & 155 & 0 & \textbf{47} & 89 & 0 \\ \hline 11 & 0 & 352 & 0 & 155 & \textbf{89} & 0 & 286 \\ \hline 12 & 638 & 0 & 507 & 0 & \textbf{155} & 286 & 0 \\ \hline \end{tabular}$

$0+0+1+1+3+4+9+14+28+47+89+155=\boxed{351}.$

Video Solution

https://www.youtube.com/watch?v=uWNExJc3hok

2018 AIME I (Problems • Answer Key • Resources)
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