Algorithms Q&A - Recent questions and answers in Informed Search

Answered: Stationary Distribution for this conditional probability table

Sun, 26 Apr 2026 13:16:22 +0000

Hello Professor,

Let the stationary distribution be:

pi = (s, c, r)

where:

s = probability of Sun
c = probability of Cloudy
r = probability of Rain

For a stationary distribution, the probabilities should stay the same after one transition. So we need:

pi = pi P

Using the table, the equations are:

s = 0.6s + 0.4c + 0.1r

c = 0.3s + 0.2c + 0.4r

r = 0.1s + 0.4c + 0.5r

Also, since these are probabilities:

s + c + r = 1

From the first equation:

s = 0.6s + 0.4c + 0.1r

0.4s = 0.4c + 0.1r

4s = 4c + r

r = 4s - 4c

From the second equation:

c = 0.3s + 0.2c + 0.4r

0.8c = 0.3s + 0.4r

Substitute r = 4s - 4c:

0.8c = 0.3s + 0.4(4s - 4c)

0.8c = 0.3s + 1.6s - 1.6c

2.4c = 1.9s

c = 19s / 24

Now use r = 4s - 4c:

r = 4s - 4(19s / 24)

r = 4s - 19s / 6

r = 24s / 6 - 19s / 6

r = 5s / 6

Now use the total probability equation:

s + c + r = 1

s + 19s / 24 + 5s / 6 = 1

s + 19s / 24 + 20s / 24 = 1

63s / 24 = 1

s = 24 / 63

s = 8 / 21

Now find c:

c = 19s / 24

c = 19 / 63

Now find r:

r = 5s / 6

r = 20 / 63

So the stationary distribution is:

Sun = 8 / 21

Cloudy = 19 / 63

Rain = 20 / 63

In decimal form:

Sun = 0.381

Cloudy = 0.302

Rain = 0.317

Final answer:

pi = (8 / 21, 19 / 63, 20 / 63)

This means that in the long run, the weather is sunny about 38.1% of the time, cloudy about 30.2% of the time, and rainy about 31.7% of the time

Answered: Admissible Heuristic for Chutes and Ladders

Thu, 16 Apr 2026 10:39:31 +0000

An admissible heuristic can be built simply by ignoring chutes and keeping ladders.

Let h(i) be the minimum number of dice throws needed to reach square 100 from square i in this relaxed game. Since removing chutes makes the game easier, the true cost in the real game is never smaller than this value, so h(i) is admissible.

Steps:

1. States 1...100

2. For each square s, for each die result d (1...6):

- move to t = s + d if t <= 100

- if t is the foot of a ladder, replace it by the ladder top,

- do not apply chutes

3. This gives a directed graph where every edge has cost 1 throw.

4. Run BFS backward from square 100 or BFS forward from each node in order to get the shortest path length to 100.

5. Set that distance as h(s)

------------------------------

Why admissible? The relaxed game has all real moves except the harmful effect of chutes, therefore, its shortest path cost is a lower bound on the true shortest path cost.

Solve this grid world.

Tue, 06 May 2025 15:13:38 +0000

Solve the final V* values for the given grid world. All given values are exit states. Fill in all missing cells. Don’t use value iteration. Show all your work/logic/steps, etc.

Assume: Gamma = 0.9. Noise Model: [0.8, 0.1, 0.1], that is, 80% chance of going in the intended direction and 10% in each of perpendicular ones.

Using pen and paper is preferable.

Weird N Puzzle

Tue, 04 Mar 2025 16:03:53 +0000

You are given a weirdly shaped n puzzle (sliding puzzle).

The best way to imagine the shape is that think of 10 x 10 n puzzle (a rather large one), and that some tiles in the center are just fixed - they don't move. (See Slack for an example).

For this problem, what is a good heuristic, from an A Star algorithm perspective? Clearly Manhattan Distance still works, but is that the best?

I have all the (unsorted) cards

Tue, 04 Mar 2025 15:50:53 +0000

You are given 52 playing cards, organized in 4 columns and 13 rows. However, these are all Aces, in 4 different colors: red, blue, green and yellow. These are all jumbled up. You are supposed to organize them such that each column has one color of Aces only. You can decide which color you want to go in which column. The only move that you can move is to swap/exchange two cards that horizontally or vertically adjacent. That is, just vertical or horizontal movement, like the n-puzzle, and no diagonal moves.

Analyze the problem, and present an algorithm for this. Present your heuristic. No need to explain how A-Star works itself. Describe without using any pseudocode.

Answered: Solve this GridWorld

Mon, 06 May 2024 20:21:15 +0000

10    10    10    10    10
10    a       b       a    10
10    b       c       b    10
10    a       b       a    10
10    10    10    10    10

a = 0.5*(0.8*10 + 0.1*b + 0.1*10) = 0.5*(9+0.1*b)
b = 0.5*(0.8*10 + 0.1*a + 0.1*b) = 0.5*(8 + 0.1*a + 0.1*b)
c = 0.5*(0.8*b + 0.1*b + 0.1*b) = 0.5*(b)
On solving, a = 4.72 b = 4.459 c = 2.23

Answered: Admissible Heuristic for n-Queens Problem

Sat, 04 May 2024 22:55:17 +0000

To calculate the heuristic that considers both vertical, horizontal, and diagonal attacks, we need to sum up the counts of attacking pairs for each of these categories. Let's denote these counts as follows:

V: Number of pairs of queens attacking each other vertically.
H: Number of pairs of queens attacking each other horizontally.
D: Number of pairs of queens attacking each other diagonally.

The final admissible heuristic would then be the sum of these counts:

Heuristic=V+H+D

This heuristic is admissible because it provides a lower bound on the number of moves required to reach the goal state. It's consistent because moving a queen to a different row in the same column, to a different column, or diagonally will either decrease the count of vertical, horizontal, or diagonal attacks, or leave it unchanged.

Answered: Give an example of a task environment that is..

Sat, 04 May 2024 22:49:19 +0000

Examples for given combinations:

Episodic and Discrete:
- Example: Playing a game of tic-tac-toe.
- Explanation: In tic-tac-toe, each game is discrete and episodic. The game consists of a finite number of discrete states (the possible board configurations), and each episode (game) has a clear starting point (an empty board) and ending point (win, lose, or draw).
Continuous and Episodic:
- Example: Skiing down a mountain.
- Explanation: Skiing down a mountain can be considered a continuous and episodic task. Each skiing session is an episode, with a clear starting point (the top of the mountain) and ending point (reaching the bottom or stopping at a certain point). The action space (movements on the slopes) is continuous, as there are infinite possible movements and trajectories.
Continuous, Non-Deterministic, and Non-Episodic:
- Example: Autonomous driving in a city.
- Explanation: Autonomous driving in a city can be a continuous, non-deterministic, and non-episodic task. The environment is continuous because the car's movement and surroundings form a continuous space (e.g., streets, traffic flow). It's non-deterministic because external factors such as weather conditions, road conditions, and the behavior of other drivers introduce uncertainty into the system. It's non-episodic because there's no clear division between episodes – the driving task continues indefinitely without clear start or end points, especially in urban environments with continuous traffic flow.

Answered: A* algorithm for fruit sorting

Fri, 22 Dec 2023 18:39:24 +0000

By graph-representing the problem, generating a cost function and heuristic function, and creating successors to travel across the state space, the A* approach may be used to efficiently arrange a set of fruits in a 3x10 array. Using the A* method, we may reduce the number of movements required to reach the goal state, where the fruits are arranged in ascending order of size.

This method ensures that the algorithm explores the state space wisely by taking into account both the estimated cost (heuristic) and the actual cost (moves). We can re-create the optimal path by going back and using the solution's final steps to physically arrange the fruits in the desired sequence.

Overall, the A* algorithm provides a logical and effective solution to the problem at hand, maximizing the arrangement of fruits in the array while taking constraints and allowable motions into mind.

Answered: What is the code level difference between graph search and tree search?

Thu, 14 Dec 2023 02:57:34 +0000

In tree search, the algorithm does not keep track of visited states. It explores nodes without checking whether a similar state has been encountered before.
This can lead to revisiting the same state multiple times, which might result in inefficiency and redundant computations.
Tree search is more straightforward to implement but can be less efficient in terms of time and space.

In graph search, the algorithm keeps track of visited states to avoid revisiting them. This is typically done using a set or some data structure to store the explored states.
This ensures that the algorithm does not waste time revisiting the same states, improving both time and space efficiency.
The additional bookkeeping for visited states makes the implementation slightly more complex than tree search.

Answered: Solve the V* values for this grid world MDP - 3 x 4

Tue, 02 May 2023 21:29:57 +0000

Here I have formed solution by using policy iteration approach. I am assuming it will move to the right always. Only one action per state. We are given a discount factor (γ) of 0.9 and noise of 0.8, 0.1, 0.1. The first column has terminal states with a value of -1, and the last column has terminal states with a value of 1. There are no living rewards.

a - go to the right always, fixed policy.

We have the following grid world:

-1	0	0	1
-1	0	0	1
-1	0	0	1

It can be shown as :

-1	A	B	1
-1	C	D	1
-1	E	F	1

Policy Iteration:

A = 0.9* (0.8 * B + 0.1 * A + 0.1 * C)

B = 0.9* (0.8 * 1 + 0.1 * B + 0.1 * D )

C = 0.9* (0.8 * D + 0.1 * A + 0.1 * E )

D = 0.9* (0.8 * 1 + 0.1 * B + 0.1 * F )

E = 0.9* (0.8 * F + 0.1 * C + 0.1 * E )

F = 0.9* (0.8 * 1 + 0.1 * D + 0.1 * F )

--------------------------------------------------

A = 0.72B + 0.09A + 0.09C

B = 0.72 + 0.09B + 0.09D

C = 0.72D + 0.09A + 0.09E

D = 0.72 + 0.09B + 0.09F

E = 0.72F + 0.09C + 0.09E

F = 0.72 + 0.09D + 0.09F

-------------------------------------------------

0.72B + 0.09A + 0.09C - A = 0

0.72 + 0.09B + 0.09D - B = 0

0.72D + 0.09A + 0.09E - C = 0

0.72 + 0.09B + 0.09F - D = 0

0.72F + 0.09C + 0.09E - E = 0

0.72 + 0.09D + 0.09F - F = 0

------------------------------------------------

Now we have :

0.72B - 0.91A + 0.09C = 0

0.72 - 0.91B + 0.09D = 0

0.72D + 0.09A + 0.09E - C = 0

0.72 + 0.09B + 0.09F - D = 0

0.72F + 0.09C - 0.91E = 0

0.72 + 0.09D - 0.91F = 0

We can see that :

0.72 - 0.91F + 0.09D = 0

0.72 - 0.91B + 0.09D = 0

It means that B = F

---------------------------------

Now we have :

0.72B - 0.91A + 0.09C = 0

0.72 - 0.91B + 0.09D = 0

0.72D + 0.09A + 0.09E - C = 0

0.72 + 0.09B + 0.09B - D = 0

0.72B + 0.09C - 0.91E = 0

We can see that :

- 0.09C + 0.91E = 0.91A - 0.09C

It means that : A = E

----------------------------------

Now we have :

0.72B - 0.91A + 0.09C = 0

0.72 - 0.91B + 0.09D = 0

0.72D + 0.18A - C = 0

0.72 + 0.18B - D = 0

0.72B + 0.09C - 0.91A = 0

We can see that

0.91B - 0.09D = -0.18B + D

1.09B = 1.09D

B = D = F

----------------------------------

Now we have :

0.72B - 0.91A + 0.09C = 0

0.72 - 0.91B + 0.09D = 0

0.72B + 0.18A - C = 0

0.72 + 0.18B - B = 0

We can see that:

- 0.91A + 0.09C = 0.18A - C

A = C = E

So it can be shown as :

0.72B - 0.82A = 0

0.72 - 0.82B = 0

--------------------------------------------------------

B = D = F = 0.72/0.82 ≈ 0.88

A = C = E = ( 0.72 * B ) / 0.82 = (0.72 * 0.878) / 0.82 ≈ 0.77

So it can be shown as :

-1	0.77	0.88	1
-1	0.77	0.88	1
-1	0.77	0.88	1

Answered: Admissible Heuristic for n-puzzle if multiple tiles can be slided at once

Tue, 07 May 2019 13:24:05 +0000

for any tile, it can swap with the empty tile，and each swap counts for max(|y1-y2|,|x1-x2|) moves, and if they are in the same row or same col, this move will count for 0. every time, if the empty tile is not in the right place(the tile the empty tile should be in goal position), then swap it with the number should be in this tile in the goal position(to make this number matched). If the empty tile is in goal position, then swap it with the farthest mismatched number, do until to the goal position. And heuristic is the sum of moves.

Answered: 8 puzzle that can be solved or not?

Tue, 30 Apr 2019 23:14:59 +0000

I can delete some states with the potential rule in parity order.

Answered: Is Harmonic Mean of admissible heuristics an admissible heuristic?

Sat, 13 Apr 2019 02:41:13 +0000

Hi Professor Arora,

I think it is. Firstly, when we talk about Harmonic Mean, we have assumed that all heuristic functions are not equal to 0, otherwise they can't be denominators. Then, after we have this assumption, Harmonic Mean is obviously admissible.

Proof:

For all h_i in {h₁, ... , h_n}, h_i <= C^*, where C^* is the fact cost (this is the definition of admissibility).
Since h_i <= C^*, we have 1 / h_i >= 1 / C^* for all h_i not equal to 0.
By 2, we get 1 / h₁ + ... + 1 / h_n >= n / C^*.
Multiply C^* / (1 / h₁ + ... + 1 / h_n) on both sizes, we have C^* >= n / (1 / h₁ + ... + 1 / h_n), i.e. H <= C^*.

Q.E.D.

Answered: Converting ternary constraint into binary constraints

Sat, 09 Feb 2019 16:36:43 +0000

We can use a new data structure.

For example, make an auxiliary variable t = [x, z-x].

t.1 = x

t.2 = y

sum(t) = z

Answered: Manhattan Distances, but Linear Interference...

Wed, 06 Feb 2019 17:47:04 +0000

For every liner conflict, it will take at least 2 more moves, like choose one of them to move down and up. We can add 2 for heuristic as some kinds of penalty.

There seems to be a lot of good solutions to this problem.

http://idm-lab.org/bib/abstracts/papers/aaai07b.pdf

Answered: In the context of informed search, what is the difference between complete, optimal and optimally efficient algorithm?

Tue, 20 Mar 2018 03:48:56 +0000

The complete means the algorithm addresses all possible inputs, optimal means algorithm will return the best solution, optimally efficient means no other methods will cost less.

A* is complete because heuristic is always less than the real and finally it will lead to the optimal solution. The prove is same with why A* is optimal.

A* is optimal g(n)+h(n)<= real cost, g(final)+h(0)=final cost, so the algorithm will travel some possible nodes which costs less than real final cost. But the final solution must be optimal.

A* is optimally efficient: if we design h, method doesn't matter.

Answered: N-Queens Count all conflicts in O(n) time

Sat, 03 Mar 2018 03:13:26 +0000

We can count line by line to determine how many queens in a line.

For example, we take the horizontal direction as row by row from top to bottom.

In each row, we count how many queens in this row and get the number of conflicts= ((number of queens-1)+1)*(number of queens-1)/2.

There are almost n line in each direction and there is 4 direction.

So cost=4*n=O(n).