Compute the number of ways the frog can move from A to B.

Question

A frog is travelling from point A $(0,0)$ to point B $(5,6)$ but each step can only be $1$ unit up or $1$ unit to the right. Also, the frog refuses to move three steps in the same direction consecutively. Compute the number of ways the frog can move from A to B.

Is this a $3$D DP problem? If we remove the "three steps in the same direction consecutively" condition then we can define $f(i,j)$ to be the number of paths from $(0,0)$ to $(i,j)$ then from recurrence/DP we have

$$f(i,j) = f(i-1,j)+f(i,j-1)$$ where we will go one step back for bottom and result in $f(i,j-1)$ and one step left and result in $f(i-1,j)$. Then for no of steps we need to add two options. How do we modify it by adding these condition? Is there specifically a Dynamic programming approach? It will be useful for me to solve these problems under a similar concept.

This is a math question. I would be very happy if you can explain me without code. Using a tree and a recurrence relation. For simplicity you can also reduce the dimension from $(5,6)$ to any smaller number.

Answer 1

The bivariate generating function for the words on the alphabet $\{ L, U \}$ not containing the subwords $LLL$, $UUU$ is

$$ g(x,y) = \frac{1}{1-x-y + \frac{x^3}{1+x+x^2} + \frac{y^3}{1+y+y^2} } $$

This is deduced with the Jackson-Goulden cluster method. The two bad words don't have any overlap with each other, and do have (all possible) overlaps with themselves. So we have the diagonal system (using $w$ to denote the weight function)

$$\begin{aligned} w(C[LLL]) &= -w(LLL) - (w(L)+w(LL))w(C[LLL]) \\ w(C[UUU]) &= -w(UUU) - (w(U)+w(UU))w(C[UUU]) \\ \end{aligned}$$

i.e.

$$\begin{aligned} w(C[LLL]) &= -x^3 - (x+x^2)w(C[LLL]) \\ w(C[UUU]) &= -y^3 - (y+y^2)w(C[UUU]) \\ \end{aligned}$$

Then

$$ g(x,y) = \frac{1}{1-w(\text{all alphabets})-w[LLL]-w[UUU]} \\ = \frac{1}{1-x-y + \frac{x^3}{1+x+x^2} + \frac{y^3}{1+y+y^2} }. $$

From this rational generating function a recurrence can be acquired. But it is kind of cumbersome. We can massage it to

$$ g(x,y) = \frac{1}{1 - \frac{x+x^2}{1+x+x^2} - \frac{y+y^2}{1+y+y^2} } = \frac{(1+x+x^2)(1+y+y^2)}{(1+x+x^2)(1+y+y^2) - x(1+x)(1+y+y^2) - y(1+y)(1+x+x^2) } = \frac{(1+x+x^2)(1+y+y^2)}{1-xy(1+x)(1+y) } \\ = (1+x+x^2)(1+y+y^2) \sum_{j=0}^\infty (xy(1+x)(1+y))^j \\ = (1+x+x^2)(1+y+y^2) \sum_{j=0}^\infty \sum_{a=0}^j \sum_{b=0}^j \binom{j}{a} \binom{j}{b} x^{j+a}y^{j+b} \\ $$

And then extract the terms $x^5y^6$. The answer: $113$.

Answer 2

An elementary (if tedious and error prone) method.

Note: I am including this at the OP's request. In practice, I would automate the search...the pencil and paper method is a bit too error prone to be relied on. That said, it is fast and not especially difficult. Took me well under $10$ minutes by hand (which won't mean much if an error is detected).

We need to make $6$ up moves and $5$ right moves. The only constraint on the order is that we can't make three consecutive moves of the same type. So, arrange the $6$ up moves in a line and consider the seven gaps between them. The $5$ right moves must go in those gaps, no more than two of them can go in a single gap. Label the gaps $g_i$ from left to right (so $g_1,g_7$ are on the ends).

We consider the ways to choose the empty gaps, consistent with the constraint.

The bad selections are those that contain (at least) one of $(g_2,g_3), (g_3,g_4), (g_4,g_5), (g_5,g_6)$

Case I: exactly two empty gaps. We've seen that there are exactly $4$ bad choices, and $\binom 72-4=\boxed {17}$

Case II: exactly three empty gaps. the bad choices are now:

$$(g_1,g_2,g_3), (g_2,g_3,g_4), (g_2,g_3, g_5), (g_2,g_3,g_6), (g_2,g_3,g_7)$$ $$(g_1,g_3,g_4),(g_3,g_4,g_5),(g_3,g_4,g_6),(g_3,g_4,g_7)$$ $$(g_1,g_4,g_5),(g_2,g_4,g_5), (g_4,g_5,g_6),(g_4,g_5,g_7)$$ $$(g_1,g_5,g_6),(g_2,g_5,g_6),(g_3,g_5,g_6), (g_5,g_6,g_7)$$

Thus there are $17$ bad choices, so $\binom 73-17=18$ good ones. For each good choice, we need to put two right moves in one non-empty gap and one each in the other three, so there are $4$ ways to finish. Thus, there are $4\times 18=\boxed {72}$ .

Case III. exactly four empty gaps. In this case, since there are only three occupied gaps, it is easier to count the good choices. Looking at the list of excluded pairs (of empty gaps) we see that we must occupy at least one of $g_2,g_3$ and at least one of $g_5,g_6$ and then we only have one more free choice. Not hard to conclude that the good selections are: $$(g_1,g_3,g_5), (g_2,g_3,g_5),(g_3,g_5,g_6),(g_2,g_4,g_5)$$ $$(g_2,g_4,g_6),(g_3,g_4,g_5),(g_3,g_4,g_6),(g_3,g_5,g_7)$$

Given a good choice we need to place two of the right moves in two of the occupied gaps and one right move in the other, so there are three choices for each good choice. Thus there are $8\times 3=\boxed {24}$.

Finally: in total there are $17+72+24=\boxed {113}$ possible solutions.

Sanity Check for case III: it is easy to see that $(g_a,g_b,g_c)$ is a good selection if and only if $(g_{8-c},g_{8-b},g_{8-a})$ is and the above list satisfies that.

Answer 3

Alternative approach, which is based on partitioning an integer.

To partition $~5~$ into terms all less than 3:

• P5a: 2 - 2 - 1

• P5b: 2 - 1 - 1 - 1

• P5c: 1 - 1 - 1 - 1 - 1

To partition $~6~$ into terms all less than 3:

• P6a: 2 - 2 - 2

• P6b: 2 - 2 - 1 - 1

• P6c: 2 - 1 - 1 - 1 - 1

• P6d: 1 - 1 - 1 - 1 - 1 - 1

When combining a P5x with a P6y, if they both have the same number of terms, then either can go first. Otherwise, the one that has one extra term must go first.

Now, it is simply case work:

---

$\underline{\text{P5a : P6a}}$

$$2 \times 3 = 6.$$

The first factor represents that either P5a or P6a can have the first move. The second factor is the number of distinct ways that the P5a = 2 - 2 - 1 can distribute its moves.

---

$\underline{\text{P5a : P6b}}$

$$1 \times \binom{3}{1} \times \binom{4}{2} = 18.$$

---

$\underline{\text{P5b : P6a}}$

$$1 \times 4 = 4.$$

---

$\underline{\text{P5b : P6b}}$

$$2 \times 4 \times 6 = 48.$$

---

$\underline{\text{P5b : P6c}}$

$$1 \times 4 \times 5 = 20.$$

---

$\underline{\text{P5c : P6b}}$

$$1 \times 6 = 6.$$

---

$\underline{\text{P5c : P6c}}$

$$2 \times 5 = 10.$$

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$\underline{\text{P5c : P6d}}$

$$1 = 1.$$

---

$\underline{\text{Final Computation}}$

$$6 + 18 + 4 + 48 + 20 + 6 + 10 + 1 = 113.$$

Answer 4

This is not a 3D DP problem (in an algorithm sense) given the constraint that the frog can't move $k$ steps in the same direction consecutively ($k=3$ in this case), because the state of the frog is finite. That said, we can use another dimension to track its state:

1. It has moved 1 unit up consecutively;
2. It has moved 2 unit up consecutively;
3. It has moved 1 unit to the right consecutively;
4. It has moved 2 unit to the right consecutively.

The recurrence relations can thus be written as $$ \begin{aligned} f(i, j, 1) &= f(i,j-1,3) + f(i,j-1,4),\\ f(i,j, 2) &= f(i - 1, j, 1),\\ f(i, j, 3) &= f(i-1, j, 1) + f(i -1, j, 2),\\ f(i, j, 4) &= f(i, j-1, 3) \end{aligned} $$ with the initial values $f(1, 0, 1) = 1$ and so on. The third variable of the function $f$ corresponds to one of the 4 scenarios above. The size of the DP array should be the product of $B$'s coordinates and the number of states. Our desired value would be $\sum_{l=1}^4{f(x,y,l)}$ for some $B=(x,y)$.

This could, however, become a 3D DP problem like you said if $k$ is also mutable. The recurrence pattern would be lengthy while more or less the same as $k=3$. But in your case, the result would be $113$.

Here's a validation:

def dp(x, y):
    dp = np.zeros((x + 1, y + 1, 5), dtype=int)
dp[1, 0, 0] = dp[2, 0, 1] = dp[0, 1, 2] = dp[0, 2, 3] = 1
for i in range(1, x + 1):
    for j in range(1, y + 1):
        dp[i, j, 0] = dp[i - 1, j, 2] + dp[i - 1, j, 3]
        dp[i, j, 1] = dp[i - 1, j, 0]
        dp[i, j, 2] = dp[i, j - 1, 0] + dp[i, j - 1, 1]
        dp[i, j, 3] = dp[i, j - 1, 2]

return sum(dp[-1, -1])


print(dp(5, 6))

It gives exactly $113$.

This can be generalized for any $(x, y, k)$.

Answer 5

I am adding another answer, which is inspired by lulu's answer. Excerpting from lulu's answer:

We need to make $6$ up moves and $5$ right moves. The only constraint on the order is that we can't make three consecutive moves of the same type. So, arrange the $6$ up moves in a line and consider the seven gaps between them. The $5$ right moves must go in those gaps, no more than two of them can go in a single gap.

Note:
You also have an added constraint. Ignoring the $~2~$ outer gaps, focus only on the $~5~$ inner gaps. You can not have two consecutive inner gaps, each of size $~0.~$ This is because you (also) can not have $~3~$ consecutive up moves.

Setting $~n = 5, k = 6,~$ the desired enumeration is the number of solutions to

• $x_1 + x_2 + \cdots + x_{k+1} = n.$

• $x_1, ~x_2, ~\cdots, ~x_{k+1} \in \{0,1,2\}.$

• For $~i \in \{2, ~\cdots, ~k-1\},~$ you can not have
  $x_i = 0 = x_{i+1}.$

The purpose of this answer is to extend lulu's analysis to provide a general closed form formula, in terms of $~n~$ and $~k,~$ for the desired enumeration.

If you ignore the third bullet point above, then you could simply follow the model in this answer.

---

Because of the third bullet point, a special strategy will be needed.

Assume that of the $~k - 1~$ inner variables, that are represented by $~x_2, ~x_3, ~\cdots, x_k,~$ that exactly $~r~$ of them are forced to equal $~0,~$ and that the other $~(k - 1 - r)~$ of these variables are forced to each be $~\geq 1.$

Then, the lower bound on $~r~$ is $~\max( ~0, ~k - 1 - n ~),~$ and the upper bound on $~r~$ is $~\displaystyle \left\lfloor \frac{(k-1) + 1}{2}\right\rfloor = \left\lfloor\frac{k}{2}\right\rfloor.$ The upper bound reflects that you must avoid having two consecutive inner variables that are both equal to $~0.$

Let $~r~$ be any element in the appropriate range, and let $~f(n,k,r)~$ denote the number of permitted ways of having exactly $~r~$ of the $~(k-1)~$ inner variables equal to $~0.~$

Then, let $~g(n,k,r)~$ denote the enumeration of the number of solutions to

• $x_1 + x_2 + \cdots + x_{k+1} = n.$

• $x_1, ~x_{k+1} \in \{0,1,2\}.$

• $x_2, ~\cdots, ~x_{r+1} = 0.$

• $x_{r+2}, ~\cdots, ~x_k \in \{1,2\}.$

Then, the overall computation will be:

$$\sum_{r ~\text{in range}} [ ~f(n,k,r) \times g(n,k,r) ~].$$

---

When the lower bound for $~r~$ is $~0,~$ then for $~r \in \{0,1\},~$ you have that $~f(n,k,r) = \displaystyle \binom{k-1}{r}.~$

In the remainder of this section, it is assumed that $~r~$ is in range, and that $~r \geq 2.$

To visualize the computation of $~f(n,k,r),~$ temporarily assume that $~n~$ is large enough, and that $~k = 8, ~r = 3,~$ and consider the following tableau:

x2 x3=0 x4 x5=0 x6 x7=0 x8

In the above tableau, since $~k = 8,~$ the inner variables are represented by $~x_2, ~x_3, ~\cdots, ~x_8.~$ So, you have a tableau with $~7~$ positions. The $~3~$ variables set to $~0~$ create $~4~$ islands. Reading the size of these islands from left to right, you can let $~y_1, ~y_2, y_3, ~y_4~$ denote the size of these islands.

In the above tableau, since you are starting with $~7~$ positions, and since you have $~3~$ of these positions taken by the zero-variables of $~x_3, x_5, x_7,~$ you have that $~y_1 + y_2 + y_3 + y_4 = 7 - 3 = 4.$

Further, the placement of the $~3~$ zero variables will be satisfactory if and only if the variables $~y_2, y_3~$ are both greater than $~0.~$

Therefore, for $~r~$ in range, with $~r \geq 2, ~~f(n,k,r)~$ equals the enumeration of the number of solutions to

• $y_1 + \cdots + y_{r+1} = k - 1 - r.$

• $y_1, ~y_{r+1} \in \Bbb{Z_{\geq 0}}.$

• $y_2, ~\cdots, ~y_r \in \Bbb{Z_{\geq 1}}.$

By basic Stars and Bars theory, for $~r~$ in range and $~r \geq 2:$

$$f(n,k,r) = \binom{[k - 1 - r] - [r - 1] + r}{r} = \binom{k-r}{r}.$$

In the present problem, with $~n = 5, ~k = 6,~$ you have that $~r \in \{0,1,2,3\}.~$ Then, $~f(5,6,0) = 1, ~f(5,6,1) = 5,~$ and for $~r \geq 2,~$ $~f(5,6,r) = \displaystyle \binom{6-r}{r}.$

---

Assuming that $~r~$ is in range, to compute $~g(n,k,r)~$ the original problem of

• $x_1 + x_2 + \cdots + x_{k+1} = n.$

• $x_1, ~x_{k+1} \in \{0,1,2\}.$

• $x_2, ~\cdots, ~x_{r+1} = 0.$

• $x_{r+2}, ~\cdots, ~x_k \in \{1,2\}.$

should be adjusted by eliminating the variables forced to equal $~0,~$ and re-indexing the remaining variables. With this adjustment, the enumeration problem becomes

• $x_1 + x_2 + \cdots + x_{k+1-r} = n.$

• $x_1, ~x_{k+1-r} \in \{0,1,2\}.$

• $x_{2}, ~\cdots, ~x_{k-r} \in \{1,2\}.$

The lower bounds of the variables are changed to $~0~$ by further changing the enumeration to

• $x_1 + x_2 + \cdots + x_{k+1-r} = n - (k-r-1) = (n + 1 + r - k).$

• $x_1, ~x_{k+1-r} \in \{0,1,2\}.$

• $x_{2}, ~\cdots, ~x_{k-r} \in \{0,1\}.$

Then, to compute $~g(n,k,r)~$ you can use the methods given in this answer.

In the present problem, with $~n = 5, ~k = 6,~$ you have that $~g(5,6,r)~$ equals the number of solutions to:

• $x_1 + x_2 + \cdots + x_{7-r} = r.$

• $x_1, ~x_{7-r} \in \{0,1,2\}.$

• $x_{2}, ~\cdots, ~x_{6-r} \in \{0,1\}.$

So, $~g(5,6,0) = 1, ~g(5,6,1) = 6, ~g(5,6,2) = 12, ~g(5,6,3) = 10.$

Therefore, the final enumeration is

$$\sum_{r=0}^3 f(5,6,r) \times g(5,6,r)$$

$$= [ ~1 \times 1 ~] + [ ~5 \times 6 ~] + [ ~6 \times 12 ~] + [ ~1 \times ~10] = 113.$$