Part 3: A suitcase has a combination lock with four wheels. Each wheel has ten d

Question

Part 3: A suitcase has a combination lock with four wheels. Each wheel has ten digits, 0 through 9 How many different combinations are there? Assume that a thief can try one combination every second, and that the thief will get lucky after trying about half the combinations, how long will it take the thief to open the briefcase?

If each wheel of the briefcase lock has only eight digits: 0, 1, 2, 3, 4, 5, 6, 7; how many different combinations are there? How long will it take our thief to open the lock?

Part 4: Suppose that a lock like the one described above has only two numbers, 0 and 1, on the wheels, and there are four wheels. How many combinations are there? What happens if an additional wheel is added? How many combinations, and how long to "crack?"

In general, what does adding another wheel do to the number of combinations?

Part 5: Explain what, if anything, the previous two exercises have to do with encryption and what, if anything, you learned from them.

Explanation / Answer

Part 3:

We are given that the briefcase has 4 wheels. And each wheel has 10 digits. Since the digits in the combinationn can repeat, for example, we can have a combination as 9999 or 8998 etc. So the maximum number of possible combinations would be:

!0 (first wheel) x 10 (second wheel) x 10 (third wheel) x 10 (fourth wheel) = 10000 combinations.

As our thief can try one combination in 1 second, and also the thief gets the right combination after trying half the total combinations(i.e., 5000 combinantions), it will take him 5000 seconds or around 1 hour and 22 minutes.

Again, Now that the wheel has 8 digits, the maximum number of possible combinations become:

8 x 8 x 8 x 8 = 4096 combinations

Which will take the theif 4096 seconds (around 1 hour and 8 seconds) to try all the combinations.

If the thief gets lucky again after trying half the combinations only (means 2048), it will take him 2048 seconds or 34 minutes and 8 seconds (approx).

Part 4:

Now each wheel has only 2 digits and there are 4 wheels, so possible number of combinations become:

2 x 2 x 2 x 2 = 16 combinations

When one more wheel is added, the number of combinations become,

2 x 2 x 2 x 2 x 2 = 32 combinations

This will take our thief 32 seconds to open the case if he tries all the combinations, or 16 seconds if he gets lucky again.

In general, adding another wheel inctreases the prevvious number of combinations by the times there are digits on the wheel.

Part 5:

The above two exercises are a great example of encryption. Let me explain how.

Suppose there is an encryption scheme where an alphabet in a message can be replaced by only a single other alaphabet to encode it. Like A can be replaced by P. This makes it easier for a theif to crack the code. As he woudl have to try only 25 combinations to crack the code for alphabet A (15 if he gets lucky).

Even if there are 4 A's in the message, the thief would have to make effort just to crack the first A in the message, i.e., omly 25 combinations. Consider these alphabets like the digits on the wheel

But, (please Note here), if we use an encoding scheme where each A can be replaced by any alphabet in the English language, then it becomes almost impossible to crack. For example, First A was replaced by p, seconds was replaced by H, third was replaced by T and so on. So if there are 4 A's in the message, there would be:

25 x 25 x 25 x 25 = 390,625 combinations.

Effective! Isn't it?

Hope this helps...

BEST WISHES!

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