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

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Let <math>SP_1P_2P_3EP_4P_5</math> be a heptagon. A frog starts jumping at vertex <math>S</math>. From any vertex of the heptagon except <math>E</math>, the frog may jump to either of the two adjacent vertices. When it reaches vertex <math>E</math>, the frog stops and stays there. Find the number of distinct sequences of jumps of no more than <math>12</math> jumps that end at <math>E</math>.
 
Let <math>SP_1P_2P_3EP_4P_5</math> be a heptagon. A frog starts jumping at vertex <math>S</math>. From any vertex of the heptagon except <math>E</math>, the frog may jump to either of the two adjacent vertices. When it reaches vertex <math>E</math>, the frog stops and stays there. Find the number of distinct sequences of jumps of no more than <math>12</math> jumps that end at <math>E</math>.
  
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==Solution==
+
==Solution 1==
 
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>.
 
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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(Solution by scrabbler94)
 
(Solution by scrabbler94)
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== Solution 2 ==
 
== Solution 2 ==
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Let <math>E_n</math> denotes the number of sequences with length <math>n</math> that ends at <math>E</math>. Define similarly for the other vertices. We seek for a recursive formula for <math>E_n</math>.
 
Let <math>E_n</math> denotes the number of sequences with length <math>n</math> that ends at <math>E</math>. Define similarly for the other vertices. We seek for a recursive formula for <math>E_n</math>.
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
E_n&=P_{3_{n-1}}+P_{5_{n-1}} \\
+
E_n&=P_{3_{n-1}}+P_{4_{n-1}} \\
−
&=P_{2_{n-2}}+P_{4_{n-2}} \\
+
&=P_{2_{n-2}}+P_{5_{n-2}} \\
−
&=P_{1_{n-3}}+P_{3_{n-3}}+S_{n-3}+P_{5_{n-3}} \\
+
&=P_{1_{n-3}}+P_{3_{n-3}}+S_{n-3}+P_{4_{n-3}} \\
−
&=(P_{3_{n-3}}+P_{5_{n-3}})+S_{n-3}+P_{1_{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}+S_{n-3}+P_{1_{n-3}} \\
−
&=E_{n-2}+P_{1_{n-4}}+P_{4_{n-4}}+S_{n-4}+P_{2_{n-4}} \\
+
&=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_{4_{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_{5_{n-5}}+P_{1_{n-5}}+P_{3_{n-5}} \\
+
&=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_{5_{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-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} \\
 
&=E_{n-1}+2E_{n-2}-E_{n-3} \\
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~ Nafer
 
~ Nafer
 +
 +
== Solution 3 ==
 +
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|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} </math>
 +
 +
<math>0+0+1+1+3+4+9+14+28+47+89+155=\boxed{351}.</math>
  
 
==Video Solution==
 
==Video Solution==
 
https://www.youtube.com/watch?v=uWNExJc3hok
 
https://www.youtube.com/watch?v=uWNExJc3hok
−
 
−
==Solution 3 (One page solution)==
 
−
 
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[[Image:Aaaaaaaaaa.jpg|700px]]
 
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(Solution by mathleticguyyy)
 
  
 
{{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}}
  
−
==Video Solution==
+
[[Category:Intermediate Combinatorics Problems]]
−
To see a comprehensive video covering this, visit :
 
−
https://www.youtube.com/watch?v=uWNExJc3hok
 

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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