Two jugs riddle - Systematic Problem Solving

Submitted by Atanu Chaudhuri on Fri, 09/10/2020 - 00:55

two jugs riddle

A combination of intuitive and systematic approach to problem solving would give you best results

The riddle

You have two jugs. Capacity of one is exactly 5 litre of water and of the other 3 litre of water. You also have a tank full of all the water that you may need to fill or empty the jugs.

You may pour or fill any number of times but finally you have to have exactly 4 litre of water in the 5 litre jug. No more no less.

Part I: Can you do it?

Part II: If you can, how many ways can you do it without repeating any move?

Total recommended time: 15 minutes.

Solution of two jug riddle: Stage 1: First level abstraction and specifying all possibilities of an outcome exhaustively

As you won't readily have two such jugs, you will,

imagine in your mind the objects of jugs as well as,
the actions of filling and emptying water.

The process being simple, you won't have any difficulty in deciding the outcome of an action that you have taken—all in your mind. This is first level abstraction.

For example, just fill the 3 litre jug fully from the tank and empty it into the 5 litre jug. You would be able to visualize the OUTCOME of these two actions precisely as,

3 litre jug empty and 3 litre water in the 5 litre jug.

Have we covered all aspects of the outcome?

No. We have surely missed one important aspect.

Solution Stage 2: Precise and exhaustive specification of the outcome

We have missed the 2 litre of empty volume in the 5 litre jug. Add this to make the specification of outcome precise and exhaustive,

3 litre empty volume in the 3 litre jug,
3 litre of water in the 5 litre jug, and also,
2 litre of empty volume yet to be filled in the 5 litre jug.

You have now used two techniques of systematic problem solving,

First level abstraction of the objects and the actions on the objects,
Exhaustive listing out all the possible outcomes of specific actions.

Solution Stage 3: Second level abstraction to CONVERT OUTCOME of specific actions into EQUATIONS with NUMBERS

It is not difficult to translate the outcome of specific actions into addition or subtraction of two simple digits.

For example, if you fill the 3 litre jug from fully filled 5 litre jug, the outcomes or volume status of the two jugs would be represented by,

5 Litre jug: $5 - 3 = 2$: 2 litre water remains with 3 litre volume empty, and,

3 litre jug: $0 + 3 = 3$: 3 litre jug is fully filled.

Solution Stage 4: Specifying the goal and the last action to reach the goal

You will now specify the goal precisely as,

5 litre jug has 4 litre water, as 3 litre jug cannot hold 4 litre of water.

How can you reach this goal by the LAST of a series of actions?

This is the crucial question that will lead you quickly to the solution.

Without much thinking and using the second level abstraction of expressing jug volume status as a simple equation, you can write the first way to reach 4 as,

$3 + 1 = 4$.

Translating back into filling and pouring actions, the equation means,

Action: Pouring 3 litre of water from 3 litre jug into 5 litre jug containing 1 litre of water,

Or, Pouring 1 litre of water from 3 litre jug into 5 litre jug containing 3 litre of water.

Working backwards, you have to solve now the next problem of forming 1 litre in 5 litre jug with 3 litre in 3 litre jug or 3 litre in 5 litre jug and 1 litre in 3 litre jug. Essentially the problem boils down now,

To get exactly 1 litre in any of the two jugs.

How would you get 1 as the result of simple additions or subtractions of the jug capacities 5 and 3?

Abstraction technique is applied again.

As you have already started using the addition or subtraction of two numbers to get the water content in a jug, you would quickly remember that 3 plus 3 is 6 and then subtracting 5 would give you 1, that is,

$(3 + 3) - 5 = 1$,

Or, $3 + 3 = 6$, followed by, $6 - 5 = 1$.

Two equations.

This is,

Expressing actions and outcomes in abstracted form of two arithmetic equations—easy to discover and express.

You can sense now that you have reached a point very near to the solution.

Translating the two equations to filling and emptying actions and jug volume status, you would describe these two actions as,

Step 1. Fill 3 litre jug.

Step 2. Empty 3 litre jug into 5 litre jug.

Step 3. Fill 3 litre jug again.

Step 4. Fill 2 litre empty volume in 5 litre jug carefully from the fully filled 3 litre jug.

1 litre water remains in 3 litre jug and 5 litre jug is fully filled.

You have to move the 1 litre into 5 litre jug,

Step 5. Empty 5 litre jug into the tank.

Step 6. Pour 1 litre water from 3 litre jug into empty 5 litre jug thus emptying the 3 litre jug.

It is dead easy now. Only two steps are left,

Step 7. Fill the empty 3 litre jug.

Step 8. Empty it into the 5 litre jug resulting, $1+ 3 = 4$.

You have got your first solution in 8 steps.

No random thinking, no trial and error because you have started from mathematically true GOAL equation $1 + 3 = 4$ and used Working Backwards technique to discover how to get 1 litre in the 5 litre jug.

The key techniques you have used are:

Abstraction to convert outcome of a series of filling and emptying actions as simple equations of the form of $a \pm b=c$, with all three of $a$, $b$ and $c$ less than or equal to maximum volume 5.
Awareness at all stages of fill and empty volume of both the jugs using Exhaustive specifying technique.
Specifying the FINAL GOAL state and last action to reach the goal state as one addition of two digits resulting in 4, that is, $1 + 3 = 4$.
Working backwards technique to form the new penultimate (1 before the final) goal as, "Form 1 litre in 5 litre jug".
Reverse abstraction to translate each equation of the form of $a \pm b=c$ into filling and emptying actions.

Systematic and not difficult at all.

The key idea of course has been to form the final goal in the form of, $1 + 3 = 4$.

You have just completed solving the first part of the puzzle. The second part still remains to be solved.

Solution of the second part: Finding all possible solutions: Stage 1: Using abstract Goal equation

The final goal has to be revised now to the answer of the crucial question,

How many combinations of getting 4 as a result of addition or subtraction of two digits each less than or equal to 5 are possible?

You will think mathematically and that's child's play.

Answer is, there are only three possible ways to get 4,

$1 + 3 = 4$: your first solution,
$5 - 1 = 4$, and,
$2 + 2 = 4$.

The first number in the left hand side of each equation represents the water content in 5 litre jug because 4 litre has to be formed finally in the 5 litre jug.

With this additional clarification in your mind, naturally you wonder whether $3 + 1 = 4$ could result in a solution.

Let us see.

You already know how to form 1 as a result of additions and subtractions of 3 and 5. It was,

$3 + 3 - 5 = 1$.

The 1 litre was formed in the 3 litre jug in just two steps.

If only you could somehow store this 1 litre elsewhere and fill the 3 litre jug, you would have a new solution.

But you don't have any storing place other than the two jugs. This is an insurmountable barrier.

That's why we have ignored the $3 + 1 = 4$ as an infeasible combination.

Exploring possible combination of $5 - 1 = 4$: Applying Working backwards technique

Translating $5 - 1 = 4$ into fill and pour actions, the last action means,

Pour exactly 1 litre from fully filled 5 litre jug into the 3 litre jug thus filling it fully.

Work backwards one step.

The 3 litre jug must have had then exactly 2 litre of water and 1 litre empty volume.

It is again very simple to get 2 as a result of an operation of 3 and 5,

$5 - 3 = 2$.

Translate back to get the second solution straightaway,

Step 1. Fill 5 litre jug.

Step 2. Fill 3 litre jug from 5 litre jug. 2 litre water remains in 5 litre jug.

Step 3. Empty 3 litre jug in the tank.

Step 4. Pour 2 litre of water from 5 litre jug to the 3 litre jug that has now 1 litre empty volume.

Step 5. Fill 5 litre jug.

Step 6. Fill carefully the 3 litre jug from 5 litre jug transferring exactly 1 litre to the 3 litre jug. 4 litres water remain in the 5 litre jug.

Got the second solution in only 6 steps, much quicker.

To know if any more solution is possible, you will have to analyze possible solutions using the remaining combination $2 + 2 = 4$.

Exploring possible solution for the combination $2 + 2 = 4$: Abstraction and Deductive Reasoning identifying insurmountable barrier

You already know how to get 2 litre of water in just two steps.

Now you are faced with the insurmountable barrier—you don't have a separate place where you can store the 2 litres of water safely and create another 2 litres of water using the two empty jugs freely.

Once you have formed 2 litre in any of the jugs, it won't again be possible to create another 2 litre as you don't have the jugs free to use in empty state any more.

So $2 + 2 = 4$ is again a combination that won't give you a solution.

You can say confidently that only two solutions to the puzzle are possible, and no more.

And now do an exercise.

Forget all that we have explained and just remember a possible final state as the arithmetic equation, $1+3=4$. In other words, just remember the key idea.

Now go ahead to solve the problem again.

End note

You may have reached one and even the two solutions without using the techniques and reasoning steps and all by yourself. If so, congratulations. Good work.

But saying confidently that these two are the only two solutions won't have been easy, isn't it?

Systematic problem solving techniques follow from common sense and nothing alien or completely new individually.

Together though it is a new way of thinking that guarantees solution and that too quicker than usual.

Our recommendation

Never give up intuitive problem solving.

In fact that should always be your first foray into the problem.

Back it up with a habit of analysis and systematic thinking.

This combination would give you the best results in solving any problem.

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