Problem Solving Heuristics

Stage 22 of 23 Strand 4 of 4 4 lessons

4 illustrated lessons, each teaching the why before the how.

Start with Working Backwards from the Goal in the app →
Revise Problem Solving Heuristics with flashcards →

Jump to a lesson

Working Backwards from the Goal

Ask what would produce the answer you want.

Starting at the goal and asking what would produce it turns an open search into a short chain of steps

Read the full lesson →

jugs of 3 and 5 units, get exactly 4start at the goalwhat would leave 4 in the large jug?

Instead of pouring at random, ask which state would give you the answer straight away.

to have 4 in the large jugso you need to be able to remove 1fill it and remove 1because 3 − 2 = 1the small jug holding 2 removes 1

Four is 5 − 1, so the question becomes how to take exactly one unit out of a full jug.

so get 2 into the small jug firstin the large jugfill 5, fill 3 from it, 2 are leftnow read it forwardempty 3, move the 2 across

That state is reachable, so the chain is complete and can be read forward as the solution.

the moveforwardtry, and see where it goesbackwardask what would produce it

Working backward narrows the search, because only some states could produce the goal.

Now you

the goalstep backwardwhat would produce itsomething you already have

Why undo the operations in reverse order?

the goalstep backwardwhat would produce itsomething you already have

A number is doubled, then 7 is added, giving 23. What was it?

Solving a Simpler Problem First

Shrink it, find the method, then scale up.

Shrinking a problem until it can be solved by hand often exposes the method the full problem needs

Read the full lesson →

how many diagonals has a polygon with 20 sides?

A 20-sided polygon has far too many diagonals to draw and count reliably.

diagonals4 sides25 sides56 sides9

Draw the small cases instead. Four, five and six sides can all be counted by hand.

each corner joins to all but 2 neighbors and itselfonce from each endn(n − 3) counts each diagonal twice

The small cases show the reason: each corner has n − 3 diagonals, and each is counted from both ends, so n(n − 3)/2.

n = 6: 6 × 3 ÷ 2 = 9checked on a small case, then used on the large onen = 20: 20 × 17 ÷ 2 = 170

Check the rule against a case you counted, then use it on the case you could not.

Now you

the hard caseshrink ita small case you can dothe method, then the hard case again

What is the point of solving a smaller version first?

the hard caseshrink ita small case you can dothe method, then the hard case again

You find a rule from three small cases. What must you still do?

From a Special Case to Every Case

A pattern noticed, then a pattern earned.

Checking special cases tests a result, and only a general argument proves it

Read the full lesson →

1 + 3 = 4a square again1 + 3 + 5 = 9and again1 + 3 + 5 + 7 = 16

Add the odd numbers in order and the totals keep coming out as squares.

conjecture: the first n odd numbers add to
the cases are evidence, not proofstill only a conjecture

That is a conjecture. The cases make it worth believing, but they do not prove it.

1357

An n by n square is built from L-shaped layers, and each layer holds an odd number of dots.

layer k adds 2k − 1 dotsand this covers every n at once
so n layers give

That argument covers every n, so the pattern is now proved for all of them.

Now you

cases that fitfrom the casesa conjecturefrom an argumenta proof

What does a special case of a proved result tell you?

cases that fitfrom the casesa conjecturefrom an argumenta proof

You have checked a claim for n = 1 to 10. What do you have?

Finding the Flaw in an Argument

A false ending means one step was not earned.

A false conclusion means some step is unjustified, and reading an argument means finding which one

Read the full lesson →

let a = bmultiply both sides by a
subtract b² from both sides

Every line so far follows from the one above, and each move is stated plainly.

(a + b)(a − b) = b(a − b)divide both sides by a − ba + b = b2b = b, so 2 = 1

The conclusion is false, so some step is unjustified, and only one of them is.

a = b, so a − b = 0the step that is not alloweddividing by a − b is dividing by 0

Since a = b, the factor a − b is 0, and dividing by 0 is the one move not allowed.

worth checkingdivideby something zerosquare rootsign lostassumethe conclusion used

A few moves account for most flaws, so check those first when a conclusion is absurd.

Now you

a = b(a + b)(a − b) = b(a − b)divide by a − ba + b = b2 = 1

An argument uses the statement it is trying to prove. What is that called?

a = b(a + b)(a − b) = b(a − b)divide by a − ba + b = b2 = 1

From x² = 9 an argument concludes x = 3. What is missing?

Continue your journey in the app — save your progress