A new safe too keep all the richest you will gain from becoming a Clojure developer (hopefully). The safe has a 3 combination lock to protect your new found wealth, but just how safe is the safe?
Create a new Clojure projectλ︎
Pracitcalli Clojure CLI Config provides the
:project/create alias to create projects using deps-new project.
Designing the combination lockλ︎
Lets consider how we would create such a combination lock in Clojure.
- The combination is managed by three tumbler wheels
- Each tumbler wheel has the same range of numbers on then, 0 to 9
Each tumbler wheel could have all the numbers it contains within a Collection in Clojure. The simplest approach would be to put the numbers 0 to 9 into a Vector (an array-like collection).
As the numbers on the tumbler wheel are just a range between 0 and 9, then rather than type out all the numbers we can use the
range function to generate all the numbers for us.
When we give the range function one argument, it will create all the whole numbers from 0 to the number before that of the argument. In the following example, we give
range the argument of 10 and we receive the numbers from 0 to 9.
You can also give
range two arguments, such as '(range 5 15)'.
Be careful not to call the
rangefunction by itself, or it will try and generate an infinite range of numbers (until your computer memory is all used up).
Create all the Combinationsλ︎
Complete the following code (replacing the ,,,) to generate all the possible combinations of the lock
Instead of showing all the possible combinations, count all the combinations and return the total number of combinations
Take the code from the combinations and wrap it in the count function
To make our lock harder to break into, we should only allow the combinations where each tumbler wheel has a different number. So you cannot have combinations like 1-1-1, 1-2-2, 1-2-1, etc.
How many combinations does that give us?
Complete the following code to create a 3-tumbler wheel combination lock, where none of the numbers are the same
Hint: Beware not to enter (range) without an argument as Clojure may try and evaluate infinity
Suggested solution to the completed 3-lock challenges.
;; a 3 combination padlock ;; model the combinations (for [tumbler-1 (range 10) tumbler-2 (range 10) tumbler-3 (range 10)] [tumbler-1 tumbler-2 tumbler-3]) ;; now count the possible combinations (count (for [tumbler-1 (range 10) tumbler-2 (range 10) tumbler-3 (range 10)] [tumbler-1 tumbler-2 tumbler-3])) (count (for [tumbler-1 (range 10) tumbler-2 (range 10) tumbler-3 (range 10) :when (or (= tumbler-1 tumbler-2) (= tumbler-2 tumbler-3) (= tumbler-3 tumbler-1))] [tumbler-1 tumbler-2 tumbler-3])) ;; lets look at the combinations again, we can see that there is always at least 2 matching values. This is probably the opposite of what we want in real life. (for [tumbler-1 (range 10) tumbler-2 (range 10) tumbler-3 (range 10) :when (or (= tumbler-1 tumbler-2) (= tumbler-2 tumbler-3) (= tumbler-3 tumbler-1))] [tumbler-1 tumbler-2 tumbler-3])