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CodingChallenge_Roomkey_Clark_Benham.clj

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+(ns room-key-coding-challenge.core)
+
+(def n 10001); the number prime which will be found
+
+(defn is-not-prime [number & [factors]] 
+           (boolean (some integer? (map #(/ number %) factors))));;tests if a number is not a prime, based on a vector of all potential prime factors
+
+(defn upper-bound [k] (Math/ceil (* k (+ (Math/log k)  (Math/log (Math/log k)) -1 (* 1.8  (Math/log (Math/log k)) (/  (Math/log k) ))))))	;;k*(ln(k)+ln(ln(k)-1+1.8*ln(ln(k))/ln(k))) formula for the maximium value of the Kth prime for k>13
+(def upper-value (int (upper-bound n)));;finds the maximium possible integer value of the nth prime
+
+(def biggest-factor (Math/ceil (Math/sqrt upper-value)));;to test if the  prime, see if division with a diffrent number will produce an integer, sqrt(number) is the biggest possible factor; sqrt(n)*sqrt(n)=n
+(def num-factors (Math/ceil (/ (- biggest-factor 5) 2)));;subtracting values as the first 5 primes are given; testing increments by 2 so needs half as many steps
+
+
+(def factors (reduce (fn [primes number] 
+     (if (is-not-prime number primes)
+   		primes
+   		(conj primes number)));;if a number is a primes it is appended to the vector of primes; this is where vector of potential factors is created
+   [2 3 5 7 11](take num-factors (iterate (partial + 2) 13))));;starts at 13 as the lower bound estimation for prime numbers is true for n>13; incrementing by 2 as no even number will be prime
+
+
+;;function returns the nth prime from a starting odd number, given the vector of all possible prime factors
+(defn nth-prime [start-prime prime-factors count] 
+  (loop [search start-prime
+         prime-factors prime-factors
+         count count]
+    (if (is-not-prime search prime-factors)
+      (recur (inc (inc search)) prime-factors count);;if is not prime increase the number being test by 2(even number won't be prime) and rerun
+      (if (> count 1) 
+        (recur (inc (inc search)) prime-factors (dec count)) 
+        search))));;if is prime decrease how many more primes need to be found 
+
+(nth-prime (last factors) factors (- n (count factors)));;finds the prime number 10001- current prime# "spots" away from the given input
+
+
+
+
+;;code below would be used for finding extremely large primes, as it takes a mathematical bound on the range
+;(defn lower-bound [k] (Math/floor (* k (+ (Math/log k) (Math/log (Math/log k)) -1))));;minmum value the nth prime could be n*(ln(n)+ln(ln(n))-1)
+;(def lower-value (int (lower-bound n)))
+;(defn next-prime [start-prime prime-factors] 
+;  (loop [search start-prime
+;         prime-factors prime-factors]
+;    (if (is-not-prime search prime-factors)
+;      (recur (inc search) prime-factors)
+;      search      )))
+;;(next-prime lower-value factors)