COMP5001 Artificial Intelligence Week 3 — Heuristics and Logic

Dr Sakib Anwar
University of Winchester
2026/27

Remember Week 1?

Every person on the island is a knight or a knave. Knights always tell the truth. Knaves always lie.

In Week 1 you solved this puzzle. What are A and B? How did you work it out? Groups · 5 min

Answer: A is a knave and B is a knight.
One way: suppose A is a knight. Then “we are both knaves” is true, so A is a knave. Impossible! So A is a knave, the sentence is false, and B must be a knight.

There are only four possibilities: knight or knave, for A and for B. Later today a computer checks all four, and finds the same answer. It is called model checking.

Where Week 2 left off

explored by BFS (in order)path

Last week BFS explored 29 squares to find the 17-step path. When you solved this maze yourself in Week 1, did you look at every square?

No. You could see where B was, so you headed towards it.

Uninformed search

DFS and BFS. They know nothing about where the goal is, so they try squares in a fixed order.

Informed search

Uses extra knowledge about the problem, such as a guess at how far each square is from the goal, to look in the right direction first.

A heuristic is a guess

The Week 1 maze. C and D are two squares to compare. The grey numbers (last click) are each square’s Manhattan distance to B: its h.

Ignore the walls. Which square is closer to B: C or D?

D: 0 rows and 6 columns away, so 6. C: 4 rows and 8 columns away, so 12.

Can you really get from D to B in 6 steps?

No: the wall forces a detour of 10 steps. A heuristic is a guess, and it can be wrong. But it is very quick to work out.

Heuristic

h(n): a quick estimate of the cost from square n to the goal. In a maze, the Manhattan distance: rows apart + columns apart, ignoring walls.

Greedy best-first search

7 = h, the guess of steps to Bfrontier: purple number = its hexplored (in order)path

Rule: always take the square in the frontier with the smallest h (the small number in its corner).

Steps 2–3. Up and right have the same h. On a tie, greedy takes the square added first: up.
Steps 9–10. Blocked by the wall at the top, it goes down: h rises to 7, then 8. The squares still waiting have h = 11 or 12, so these still win.
Step 18. B, after only 18 squares. BFS needed 29.
Path: 17 steps, the shortest.

Greedy was faster than BFS and still found the shortest path. Is it always this good?

Where greedy goes wrong

7 = h, the guess of steps to Bfrontier: purple number = its hexplored (in order)path

A new maze. Which way would you go from A? Which way will greedy go?

Step 2. Right has h = 11; down has h = 13. Greedy goes right.
Step 9. The corner: h = 4. B looks close, but there is a wall in the way.
Steps 10–13. The way turns back to the left and h rises to 8. Down (13) still looks worse, so greedy keeps going.
Step 21. B, after 21 squares.
Path: 20 steps.

Key idea

The shortest path goes down and along the bottom: 16 steps. Greedy only looks at h, the guess to go. It never counts the steps it has already taken.

A*: steps so far + guess to go

route to Xroute to Y

X and Y are both waiting in the frontier. Which should the search take next?

	X	Y
`h(n)`: guess to B	7	11
`g(n)`: steps from A so far	11	5
`f(n)` = `g(n)` + `h(n)`	18	16

Greedy looks only at h, so it takes X.

Counting the steps already taken, Y is the better bet.

A* search

Take the frontier square with the smallest f(n) = g(n) + h(n): the cost so far plus the guessed cost to go.

A* step by step

7 = h, the guess of steps to Bfrontier: purple number = its fexplored: big number = fpath

Rule: take the square with the smallest f = g + h.

Steps 1–9. Along the top route f stays at 12: each step adds 1 to g and takes 1 off h.
Steps 10–11. Round the corner f jumps to 14, so A* also tries the square below A (f = 14).
Steps 12–27. One more square on top (f = 16), then the top route reaches 18. The bottom route stays at 16, so A* follows it to B.
Path: 16 steps, the shortest.

`maze4.txt`	Squares explored	Path
BFS	32	16
Greedy	21	20
A*	27	16

Press Play. The big number in each explored square is now f, not the step number. The corner number is still h. Steps 1 to 9: f = 12 all the way along the top route. It looks perfect. Step 10: the route turns back left; g goes up and h goes up, so f jumps by 2. A* immediately looks at the other option, the square below A (f = 14). After that it compares the two routes: the top route's next square has f = 18, the bottom route's squares all have f = 16. Watch the bottom corridor fill up. The table: A* explored more squares than greedy, but fewer than BFS, and of the two informed searches it is the only one that guarantees the shortest path (next slide says when). Ties: A* takes the square nearer B first. Students' own versions may explore a slightly different number of squares.

A good guess never overestimates

Two ways to measure Salisbury to Winchester: along the road, or in a straight line.

Can the straight line between two towns ever be longer than the road between them?

No. So straight-line distance never overestimates. Nor does Manhattan distance in a maze: walls only make the real route longer.

Admissible heuristic

Never overestimates the true cost to the goal. A* with an admissible heuristic always finds the shortest path.

What if h guessed too high? Say h(Y) were 30 on the last slide.

Then f(Y) = 5 + 30 = 35, worse than the top route. A* would go the long way and return the 20-step path.

The road map from last week. A sat-nav doing A* on real roads uses straight-line distance as h. A heuristic that never guesses too high cannot make the best route look worse than it is, so A* never skips it. The overestimate example: if Y wrongly looked 30 steps from B, A* would reach B along the top route (f = 20) before it ever took Y (f = 35). CS50 also mentions "consistent" heuristics. Students do not need this; Manhattan and straight-line distance are both. If someone asks what other heuristics there are: the notes' section on admissible heuristics has a table (diagonal moves, road maps, the 8-puzzle, h = 0, an inflated guess) and the "relax the problem" rule for inventing one: drop a rule, and the easier problem's exact cost is your guess. This ends the search half. Seminar sheet, Activities 2 and 3: greedy and A* by hand on a small map, and which heuristics never overestimate. Short pause, then logic.

What is logic?

Which of these can be true or false? Whole class

Sentence	True or false?
Winchester is in Hampshire.	true
2 + 2 = 5.	false
Close the door!	neither: an order
Is it raining?	neither: a question
It is raining.	true or false, depending on the day

Statement

A sentence that is either true or false. Orders and questions are not statements.

Logic

A precise way to write statements, with rules for working out what follows from them. It is the language of symbolic AI from Week 1.

Today we write statements in logic, then let a computer work out what follows from them.

Knowledge-based agents

What we know

If it didn’t rain, Harry visited Hagrid today.
Harry visited Hagrid or Dumbledore today, but not both.
Harry visited Dumbledore today.
Harry did not visit Hagrid today. (from 2 and 3)
It rained today. (from 1 and 4)

Did it rain today? None of the three sentences says so. Whole class

Harry visited Dumbledore, and not both, so he did not visit Hagrid.

If it hadn’t rained, he would have visited Hagrid. He didn’t, so it rained.

Knowledge-based agent

An agent that stores what it knows as sentences (statements written in a logic), and reasons with them to reach new conclusions.

Inference

Working out new sentences from old ones, as we just did for sentences 4 and 5.

Writing sentences with symbols

Give each statement a short name:

It rained today. → rain
Harry visited Hagrid today. → hagrid
Harry visited Dumbledore today. → dumbledore

Propositional symbol

A short name for one statement. In any world, it is either true or false. Textbooks often use single letters: P, Q, R.

Sentence 1 says: “If it didn’t rain, Harry visited Hagrid today.” Which of our statements are inside it?

Two: rain (with a “not”) and hagrid, joined by “if … then”.

Propositional logic

Symbols, joined by five words: not, and, or, if … then and if and only if. The next two slides give each word a symbol.

Joining sentences: not, and, or

not: “It did not rain.” → ¬rain
and: “Harry visited Hagrid and Dumbledore.” → hagrid ∧ dumbledore
or: “Harry visited Hagrid or Dumbledore.” → hagrid ∨ dumbledore

Read ¬ as “not”, ∧ as “and”, ∨ as “or”.

“Or” means at least one

P ∨ Q is true when P is true, or Q is true, or both. To say “but not both”, add ¬(P ∧ Q).

When is each sentence true? Fill in the table. Pairs · 3 min

P	Q	¬P	P ∧ Q	P ∨ Q
false	false	true	false	false
false	true	true	false	true
true	false	false	false	true
true	true	false	true	true

P and Q stand for any two sentences. Each row is one possible world.

How do we write sentence 2, “Harry visited Hagrid or Dumbledore, but not both”?

(hagrid ∨ dumbledore) ∧ ¬(hagrid ∧ dumbledore)

If … then, and if and only if

if … then: “If it is raining, then I am indoors.” → rain → indoors

Read rain → indoors as “rain implies indoors”.

if and only if: “I am indoors if, and only if, it is raining.” → rain ↔ indoors

Key idea

P → Q is false only when P is true and Q is false. P ↔ Q is true when P and Q are the same.

I promised: “If it is raining, I am indoors.” In which worlds did I break my promise? Whole class

`rain`	`indoors`	rain → indoors	rain ↔ indoors
false	false	true	true
false	true	true	false
true	false	false: broken!	false
true	true	true	true

When it is not raining, the promise says nothing about where I am, so it is not broken: the sentence counts as true.

Models: every possible world

World 1
rain true
hagrid true

World 2
rain true
hagrid false

World 3
rain false
hagrid true

World 4
rain false
hagrid false

3 symbols: 8 worlds. n symbols: 2ⁿ worlds.
Each new symbol doubles the number.

Two symbols, rain and hagrid. How many different worlds are possible? Whole class

Model

One possible world: a true or false value for every symbol.

Knowledge base (KB)

The sentences an agent knows are true.

Suppose the KB is one sentence, rain ∨ hagrid. Which worlds does it rule out?

Only world 4, where both are false. Knowledge rules worlds out.

Entailment: what follows

If “It is a Tuesday in January” is true, must “It is January” be true? What about the other way round? Whole class

Yes. Every world where it is a Tuesday in January is a world where it is January.

No. On a Wednesday in January, “It is January” is true, but “It is a Tuesday in January” is not.

Entailment

A entails B when B is true in every world where A is true: B follows from A. Written A ⊨ B.

Our question is always: KB ⊨ query? Is the query true in every world that fits what we know?

Model checking

Knowledge base

¬rain → hagrid
hagrid ∨ dumbledore
¬(hagrid ∧ dumbledore)
dumbledore

Query: rain. Did it rain?

Model checking

List every world. Keep the worlds where the KB is true. If the query is true in all of them, then KB ⊨ query.

Three symbols, so 8 worlds. Which ones fit everything we know?

`rain`	`hagrid`	`dumbledore`	Ruled out by
false	false	false	4
false	false	true	1: no rain, so Hagrid
false	true	false	4
false	true	true	3: not both
true	false	false	4
true	false	true	fits everything
true	true	false	4
true	true	true	3: not both

rain is true in every world that is left, so KB ⊨ rain. It rained.

Knights and knaves, solved

A said: “We are both knaves.” Write what we know as sentences. Whole class

Each person is a knight or a knave, not both. → (AKnight ∨ AKnave) ∧ ¬(AKnight ∧ AKnave), and the same for B
If A is a knight, what A said is true. → AKnight → (AKnave ∧ BKnave)
If A is a knave, what A said is false. → AKnave → ¬(AKnave ∧ BKnave)

Four symbols give 16 worlds. The first sentence leaves only the four on the right.

A is a	B is a	Ruled out?
knight	knight	yes: a knight said “both knaves”
knight	knave	yes: a knight said “both knaves”
knave	knight	no: it fits
knave	knave	yes: a knave said something true

One world is left, so the KB entails A is a knave and B is a knight: the answer you found in Week 1, now checked by a method a computer can follow.

In code: `harry.py`


from logic import *

rain = Symbol("rain")
hagrid = Symbol("hagrid")
dumbledore = Symbol("dumbledore")

knowledge = And(
    Implication(Not(rain), hagrid),
    Or(hagrid, dumbledore),
    Not(And(hagrid, dumbledore)),
    dumbledore
)

print(model_check(knowledge, rain))

logic.py gives Symbol, Not, And, Or, Implication, Biconditional and model_check.

Match the lines of knowledge to sentences 1–4 on the model-checking slide. What does the last line print?

Same four sentences, in the same order. And(…) joins them: the KB is true only when all four are.

It prints True: the knowledge base entails rain.

What model_check does, in plain steps:

for each world (every true/false choice of the symbols): if the KB is true in this world: if the query is false here: return False return True

Heuristics and logic in the real world

Sat-navs and games: A* with a good heuristic, such as straight-line distance.
Computer chips: model checking tries every state of a design, to show that a bug can never happen. Its inventors won the 2007 Turing Award.
Railways: logic is used to check that signals can never send two trains onto the same track.
Product options: logic works out which options can go together, for example when you build a car online.

Discuss

A railway signalling system has been checked by model checking: in every world that fits its knowledge base, no two trains share a track. Can we trust it completely? What could still go wrong? Groups · 10 min

The check is only as good as the knowledge base. A missing or wrong sentence, or something the symbols do not cover (a broken sensor), is outside the proof. People who know the railway must check the sentences, and the system still needs testing.

Thank you

Practical: week04-knowledge/logic in the module code (harry.py)
Next week: Knowledge and Planning. Faster ways to reason than checking every world, then planning with facts and actions
Reading: Russell & Norvig, sections 3.5–3.6 and 7.1–7.4; Poole & Mackworth, sections 3.6 and 5.1
Sakib.Anwar@winchester.ac.uk

Adapted from CS50’s Introduction to Artificial Intelligence with Python by Brian Yu and David J. Malan, Harvard University, licensed under CC BY-NC-SA 4.0.

COMP5001 Artificial Intelligence Week 3 — Heuristics and Logic

Remember Week 1?

Where Week 2 left off

Uninformed search

Informed search

A heuristic is a guess

Heuristic

Greedy best-first search

Where greedy goes wrong

Key idea

A*: steps so far + guess to go

A* search

A* step by step

A good guess never overestimates

Admissible heuristic

What is logic?

Statement

Logic

Knowledge-based agents

What we know

Knowledge-based agent

Inference

Writing sentences with symbols

Propositional symbol

Propositional logic

Joining sentences: not, and, or

“Or” means at least one

If … then, and if and only if

Key idea

Models: every possible world

Model

Knowledge base (KB)

Entailment: what follows

Entailment

Model checking

Knowledge base

Model checking

Knights and knaves, solved

In code: harry.py

Heuristics and logic in the real world

Discuss

Thank you

In code: `harry.py`