Euler Project 37

The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.

Find the sum of the only eleven primes that are both truncatable from left to right and right to left.

NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.

My Solution

(defun primep (n)

(cond ((= 2 n) t)

((evenp n) 'nil)

((= 1 n) 'nil)

(t (let ((limit (floor (sqrt n))))

(do ((i 3 (incf i 2)))

((> i limit) t)

(if (= 0 (mod n i)) (return-from primep 'nil)))))))

(defun enumerate-interval (low high)

(loop for i from low to high

collect i))

(defun primes(n)

(remove-if #'(lambda(x) (null x)) (loop for i from 3 to n by 2

collect (if (and (primep i) (not (null i))) i))))

(defun leftCycle(num)

(let ((numberstr (format nil "~a" num))

(IsLeftCycle T))

(loop for i from 0 to (- (length numberstr) 1)

do(if (not (primep (parse-integer numberstr :start i :end (length numberstr))))

(progn (setq IsLeftCycle nil)

(return IsLeftCycle))))

IsLeftCycle))

(defun rightCycle(num)

(let ((numberstr (format nil "~a" num))

(IsRightCycle T))

(loop for i from 1 to (length numberstr)

do (if (not (primep (parse-integer numberstr :start 0 :end (+ i))))

(progn (setq IsRightCycle nil)

(return IsRightCycle))))

IsRightCycle))

(defun Euler37(n)

(loop for i from 21 to n

when (and (rightCycle i)

(leftCycle i))

sum i))