Jamie's Blog

Problem description, from Project Euler

Find the sum of all the primes below N.

Solution

Brute-forced it again, but it still runs in less than 30 seconds (at least on my MacBook Pro). This one was simple using the Sieve of Eratosthenes from problem 7.

The sieve, collapsed because it’s the same as from problem 7:

; a
; http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
;1. Create a list of consecutive integers from two to n: (2, 3, 4, ..., n),
;2. Initially, let p equal 2, the first prime number,
;3. While enumerating all multiples of p starting from p2, strike them off from the original list,
;4. Find the first number remaining on the list after p (it's the next prime); let p equal this number,
;5. Repeat steps 3 and 4 until p2 is greater than n.
;6. All the remaining numbers in the list are prime.
; returns a lazy-sequence of the multiples of n

(defn sieve-of-eratosthenes
    [n]
    ; step 1 Create a list of consecutive integers from two to n: (2, 3, 4, ..., n),

    (defn sieve [rng p]
        ;(print "range: " rng "\n")
        (
            ; this lambda used to perform the actual recursion
            (fn [new-rng]
                ; step #5 Repeat steps 3 and 4 until p^2 is greater than n.
                (if (<= (* p p) n)
                    ; recurse
                    (sieve new-rng
                        ; step # 4 Find the first number remaining on the list after p
                        ; (it's the next prime); let p equal this number,
                        (first (filter #(> % p) new-rng))
                        )
                    ; end recursion otherwise
                    new-rng
                    )
                )
            ; this is an argument to the lambda above
            ; step #3 While enumerating all multiples of p starting from p2,
            ; strike them off from the original list,
            (filter
                #(or
                    (not= ; divisible by p
                        (rem % p)
                        0
                        )
                    ; greater than p^2
                    (< % (* p p))
                    )
                    rng
                )
            )
        )

    (sieve
        (range 2 n)
        2 ; step #2 Initially, let p equal 2, the first prime number
        )
    )

The new code:

(defn problem10 [max-nbr]
    (reduce + (sieve-of-eratosthenes max-nbr)) ; brute!
    )

Problem description, from Project Euler

Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Solution

I decided to store the data in a text file. Not only would it save me from embedding the number in the source, but also give me an opportunity to try some simple file handling in Clojure. There is nothing super notable here, it’s another brute-force attempt.

; use to reduce a list into a string
(defn str-join [delim l]
    (reduce (fn [res cur] (str res delim cur)) l)
    )
; use to parse an integer out of a string
(defn parseInt [n]
    (Integer/parseInt n)
    )

(defn problem08 [filename digit-count]
    ; used to read the file and remove any newlines
    (defn get-normalized-file-contents [filename]
        (str-join "" (remove #(= % \newline) (slurp filename)))
        )

    ; used to obtain the maximum product
    (defn find-numbers [haystack max-number]
        ; compute the product of the first 5 numbers in haystack
        (def tmp (reduce *
            (map parseInt (map str (take digit-count haystack)))
            ))

        ; are there more digits to try? if so, recurse
        (if (>= (count haystack) digit-count)
            ; work with the rest of the list
            (find-numbers
                (rest haystack)
                ; pass to this function the greater of the current
                ; maximum and the product of the first 5 digits in
                ; the current list
                (if (> tmp max-number) tmp max-number)
                )
            ; otherwise, return what we have determines to be the max number
            max-number
            )
        )

    (find-numbers
        (get-normalized-file-contents filename)
        -99
        )
    )

(problem08 "/pathtofile/data.txt" 5)

Problem description, from Project Euler

There exists exactly one Pythagorean triplet for which a + b + c = N. (N is known.)
Find the product abc.

Solution

This was a quick once since I just brute-forced it. I think there are some optimizations possible related to the idea that a^2 + b^2 < c^2.

; flattens a nested list
(defn flatten [x] (let [s? #(instance? clojure.lang.Sequential %)] (filter (complement s?) (tree-seq s? seq x))))
; returns the result of the pythagorean equation
(defn pythagorean-triplet-c [a b]
    (. java.lang.Math
        (sqrt
            (+ (* a a)
                (* b b)
                )
            )
        )
    )

; returns the c term of a pythagorean triple
; nil if natural term does not exist
(defn pythagorean-triplet-c-natural [a b]
    (def result (pythagorean-triplet-c a b))
    (if (zero? (- result (int result)))
        (int result)
        nil
        )
    )
(defn problem09 [sum]
    (def res
        (map
            (fn [a]
                (map
                    (fn [b]
                        (def c (pythagorean-triplet-c-natural a b))
                        (if (and (not (nil? c))
                                (= sum (+ a b c))
                                )
                            (* a b c)
                            nil
                            )
                        )
                        (range 100 sum)
                    )
                )
            (range 100 sum)
            )
        )
    (first (distinct (filter #(not (nil? %)) (flatten res))))
    )

Problem Description, from Project Euler

What is the n-th prime number?

Problem Solution

This was simple to solve using the Sieve created for the solution to problem #3. To make the program variable, we start with finding all primes less than 1000. If that is not enough to solve the problem, we double this “maximum” value, and search for all primes less than 2000. Rinse, repeat.

The old stuff:

; http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
(defn multiples
    [n]
    (lazy-cat
        [1 n]
        (map
            (fn [a] (+ a n))
            (rest (multiples n))
            )
        )
    )

(defn sieve-of-eratosthenes
    [n]
    ; step 1 Create a list of consecutive integers from two to n: (2, 3, 4, ..., n),
    (defn sieve [rng p]
        ;(print "range: " rng "\n")
        (
            ; this lambda used to perform the actual recursion
            (fn [new-rng]
                ; step #5 Repeat steps 3 and 4 until p^2 is greater than n.
                (if (<= (* p p) n)
                    ; recurse
                    (sieve new-rng
                        ; step # 4 Find the first number remaining on the list after p
                        ; (it's the next prime); let p equal this number,
                        (first (filter #(> % p) new-rng))
                        )
                    ; end recursion otherwise
                    new-rng
                    )
                )
            ; this is an argument to the lambda above
            ; step #3 While enumerating all multiples of p starting from p2,
            ; strike them off from the original list,
            (filter
                #(or
                    (not= ; divisible by p
                        (rem % p)
                        0
                        )
                    ; greater than p^2
                    (< % (* p p))
                    )
                    rng
                )
            )
        )

    (sieve
        (range 2 n)
        2 ; step #2 Initially, let p equal 2, the first prime number
        )
    )

The new stuff:

(defn problem07 [nth]
    (defn try-find-nth-prime [primes max-number]
        (if (< (count primes) nth)
            ; if the list of primes is not long enough,
            ; double the maximum search number
            (try-find-nth-prime
                (sieve-of-eratosthenes (* max-number 2))
                (* max-number 2)
                )
            ; otherwise, take the nth member of the list
            (last (take nth primes))
            )            

        )
    (try-find-nth-prime (sieve-of-eratosthenes 1000) 1000)
    )

Problem Description, from Project Euler

Find the largest palindrome made from the product of two 3-digit numbers

Solution

I couldn’t think of many optimizations for this problem, other than avoiding duplicates when calculating the products of numbers. That’s why I just brute-forced it. The only (barely) interesting thing I did here was to write a function which recognized palindromes. Thinking about the problem now, I suppose we can predict whether the first and last digits of a product will be the same by looking at the hundreds and tens digits of the factors.

; flattens a nested list
(defn flatten [x] (let [s? #(instance? clojure.lang.Sequential %)] (filter (complement s?) (tree-seq s? seq x))))

(defn palindrome? [s]
    (if (or
        ; all 1 letter strings or
        ; nils are palindromes
        (<= (count s) 1)
        (nil? s)
        )
        true
        (and
            (= (first s) (last s))
            (palindrome? (take (- (count s) 2) (rest s)))
            )
        )
    )
(defn problem04 []
    (def x (range 100 1000))
    ; sort, then take the last
    (last (sort
        ; filter to only palindromes
        (filter
            #(palindrome? (str %))
            ; distinct products of all 3-digit numbers
            (distinct (flatten
                (map
                    (fn [n]
                        (map (fn [a] (* a n)) x))
                    x
                    )
                ))
            )
        ))
    )

Problem description, from Project Euler

What is the largest prime factor of the number N?

Solution

In previous problems, I was using a list of primes I got somewhere else. I figured it would be relevant to implement a prime-finding algorithm. I decided to go with simplicity, and implement the Sieve of Eratosthenes. It’s basically the algorithm you would come up with yourself if you wanted to figure out how to find all the prime numbers within a certain range.

Say we want to find all the prime numbers between 1 and 100. Start with the first prime number 2. Remove from the list all multiples of 2. Work with the next prime number 3. Remove from the list all multiples of 3. Continue until the square number you pick is greater than the last number in the list. This is because no number after that will have a multiple in the list.

For the part where we factor primes, I used the algorithm I came up with for problem 5, even though it is inelegant and has a kludgy fix.

Overall, I was happy with my solution. I think I am beginning to think more “functionally.”

; flattens a nested list
(defn flatten [x] (let [s? #(instance? clojure.lang.Sequential %)] (filter (complement s?) (tree-seq s? seq x))))

; http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
(defn sieve-of-eratosthenes
    [n]
    (defn sieve-of-eratosthenes-helper
        [n]
        ; step 1 Create a list of consecutive integers from two to n: (2, 3, 4, ..., n),

        (defn sieve [rng p]
            (
                ; this lambda used to perform the actual recursion
                (fn [new-rng]
                    ; step #5 Repeat steps 3 and 4 until p^2 is greater than n.
                    (if (<= (* p p) n)
                        ; recurse
                        (sieve new-rng
                            ; step # 4 Find the first number remaining on the list after p
                            ; (it's the next prime); let p equal this number,
                            (first (filter #(> % p) new-rng))
                            )
                        ; end recursion otherwise
                        new-rng
                        )
                    )
                ; this is an argument to the lambda above
                ; step #3 While enumerating all multiples of p starting from p2,
                ; strike them off from the original list,
                (filter
                    #(or
                        (not= ; divisible by p
                            (rem % p)
                            0
                            )
                        ; greater than p^2
                        (< % (* p p))
                        )
                        rng
                    )
                )
            )

        (sieve
            (range 2 n)
            2 ; step #2 Initially, let p equal 2, the first prime number
            )
        )

    (sieve-of-eratosthenes-helper n)
    )

(defn problem03 [n]
    ; begin by generating primes below the sqrt of the number.
    ; I'm just guessing this will be good enough. I started
    ; with a guess of 1000, which was woefully small
    (def prime-cache (sieve-of-eratosthenes (. java.lang.Math (sqrt n))))
    (defn prime?
        [n]
        (not (nil? (some #(= n %) prime-cache)))
        )
    ; prime-factorization algorithm from solving previous
    ; project Euler problems such as #5
    (defn prime-factorization
        [n]
        (defn pf-helper
            []
            (def first-divisible
                (first (filter
                    #(zero?
                        (rem n %)
                        )
                    (rest prime-cache))
                    )
                )

            (def result (/ n first-divisible))

            (def new-result
                (if (not (prime? result))
                    (prime-factorization result)
                    [result]
                    )
                )

            ; i don't know why I need to set this again.
            ; the recursion seems to mutate a value that
            ; should not be in its scope
            (def first-divisible
                (first (filter
                    #(zero?
                        (rem n %)
                        )
                    (rest prime-cache))
                    )
                )

            (if (nil? new-result)
                nil
                (if (nil? first-divisible)
                    [new-result]
                    (flatten [new-result first-divisible])
                    )
                )
            )

        (if (prime? n)
            [n]
            (pf-helper)
            )
        )

    ; factor!
    (prime-factorization n)

    )

Problem description, from Project Euler

Find the difference between the sum of the squares of the first N natural numbers and the square of the sum.

Solution

This one actually seemed to be easier than the others. No filters required!

    (defn problem06
        [upper]
        (defn sq [n] (* n n))
        (def l (range 1 (+ upper 1)))
        (-
            (sq (reduce + l))
            (reduce + (map sq l))
            )
        )

The biggest hangup I had was that when I originally read the problem, I thought I’d need a power function, and then spent at least 15 minutes determining there was no Clojure core power function. I found some alternatives here:

     ; for integers (cuz it's faster)
     (. (. java.math.BigInteger (valueOf 2)) (pow 5))
     ; for doubles
     (. java.lang.Math (pow 2 -3))

Problem description, from Project Euler

What is the smallest number that is evenly divisible by all of the numbers from 1 to N?

Solution

The math

Basically, we are finding the least common multiple (LCM) of the numbers (grade school math?). This can be done by generating the prime factors of each of the numbers. Once the factorization is complete, determine the greatest number of times a prime appears in any number. For example, the prime factors of 16 are (2,2,2,2) and of 20 are (2,2,5). 2 appears at most 4 times in numbers from 1-20, so we stick with (2,2,2,2). Repeat for all prime factors of the numbers 1 through 20.

An alternative way to think about this, is to list all the greatest powers of primes below 20, and find the product of these.
2^4 * 3^2 * 5 * 7 * 11 * 13 * 17 * 19. Doing it this way in the program would be quicker, but I decided to factor each number.

My first attempt

This solution is UGLY! For one, I hardcoded a list of primes. I do not yet know how to generate a list of primes quickly for arbitrary N, although I believe that will be required for future Euler problems (and Google will tell me how to do so).

Also, the structure of the program doesn’t seem truly functional to me. I may amend this post with a revision later. I did learn a number of great things in this including:

• How to flatten a sequence using tree-seq
• Working with hash maps

The biggest headache was the prime-factorization function, which I probably could’ve written in another language in 15-30 minutes. Instead, it took me maybe 6 hours to get the algorithm right, and I had to use a weird hack. I think I have some misunderstanding about the scope of a def call. I’m sure the topic will become illuminated as I proceed.

(defn problem05
    [upper]

    (defn flatten [x] (let [s? #(instance? clojure.lang.Sequential %)] (filter (complement s?) (tree-seq s? seq x)))) 

    ; cheated by using a list of primes. could be replaced by implementation of sieve of eratosthenes from problem 7
    ; http://www.math.utah.edu/~pa/math/primelist.html
    (def prime-cache [1 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 631 641 643 647 653 659 661 673 677 683 691 701 709 719 727 733 739 743 751 757 761 769 773 787 797 809 811 821 823 827 829 839 853 857 859 863 877 881 883 887 907 911 919 929 937 941 947 953 967 971 977 983 991 997 1009 1013 1019 1021 1031 1033 1039 1049 1051 1061 1063 1069 1087 1091 1093 1097 1103 1109 1117 1123 1129 1151 1153 1163 1171 1181 1187 1193 1201 1213 1217 1223 1229 1231 1237 1249 1259 1277 1279 1283 1289 1291 1297 1301 1303 1307 1319 1321 1327 1361 1367 1373 1381 1399 1409 1423 1427 1429 1433 1439 1447 1451 1453 1459 1471 1481 1483 1487 1489 1493 1499 1511 1523 1531 1543 1549 1553 1559 1567 1571 1579 1583 1597 1601 1607 1609 1613 1619 1621 1627 1637 1657 1663 1667 1669 1693 1697 1699 1709 1721 1723 1733 1741 1747 1753 1759 1777 1783 1787 1789 1801 1811 1823 1831 1847 1861 1867 1871 1873 1877 1879 1889 1901 1907 1913 1931 1933 1949 1951 1973 1979 1987 1993 1997 1999 2003 2011 2017 2027 2029 2039 2053 2063 2069 2081 2083 2087 2089 2099 2111 2113 2129 2131 2137 2141 2143 2153 2161 2179 2203 2207 2213 2221 2237 2239 2243 2251 2267 2269 2273 2281 2287 2293 2297 2309 2311 2333 2339 2341 2347 2351 2357 2371 2377 2381 2383 2389 2393 2399 2411 2417 2423 2437 2441 2447 2459 2467 2473 2477 2503 2521 2531 2539 2543 2549 2551 2557 2579 2591 2593 2609 2617 2621 2633 2647 2657 2659 2663 2671 2677 2683 2687 2689 2693 2699 2707 2711 2713 2719 2729 2731 2741 2749 2753 2767 2777 2789 2791 2797 2801 2803 2819 2833 2837 2843 2851 2857 2861 2879 2887 2897 2903 2909 2917 2927 2939 2953 2957 2963 2969 2971 2999 3001 3011 3019 3023 3037 3041 3049 3061 3067 3079 3083 3089 3109 3119 3121 3137 3163 3167 3169 3181 3187 3191 3203 3209 3217 3221 3229 3251 3253 3257 3259 3271 3299 3301 3307 3313 3319 3323 3329 3331 3343 3347 3359 3361 3371 3373 3389 3391 3407 3413 3433 3449 3457 3461 3463 3467 3469 3491 3499 3511 3517 3527 3529 3533 3539 3541 3547 3557 3559 3571 3581 3583 3593 3607 3613 3617 3623 3631 3637 3643 3659 3671 3673 3677 3691 3697 3701 3709 3719 3727 3733 3739 3761 3767 3769 3779 3793 3797 3803 3821 3823 3833 3847 3851 3853 3863 3877 3881 3889 3907 3911 3917 3919 3923 3929 3931 3943 3947 3967 3989 4001 4003 4007 4013 4019 4021 4027 4049])

    ; return primes below a certain number
    (defn primes
        [max-prime]
        ; we should actually have code to add to
        ; the prime cache, but it's not necessary
        ; to solve the problem
        (filter #(< % max-prime) prime-cache)
        )
    ; determines whether or not a number is a prime number
    (defn prime?
        [n]
        (not (nil? (some #(= n %) prime-cache)))
        )
    (defn prime-factorization
        [n]
        (defn pf-helper
            []
            (def first-divisible
                (first (filter
                    #(zero?
                        (rem n %)
                        )
                    (rest (primes n)))
                    )
                )
            ;(print "first-divisible: " first-divisible "\n")

            (def result (/ n first-divisible))
            ;(print "result: " result "\n")

            (def new-result
                (if (not (prime? result))
                    (prime-factorization result)
                    [result]
                    )
                )
            ;(print "new-result:" new-result "\n")
            ; i don't know why I need to set this again.
            ; the recursion seems to mutate a value that
            ; should not be in its scope
            (def first-divisible
                (first (filter
                    #(zero?
                        (rem n %)
                        )
                    (rest (primes n)))
                    )
                )
            (if (nil? new-result)
                nil
                (if (nil? first-divisible)
                    [new-result]
                    (flatten [new-result first-divisible])
                    )
                )
            )

        (if (prime? n)
            [n]
            (pf-helper)
            )
        )

    ; will return a hash where each list member is
    ; a key and the values are the count of the occurences
    ; of the member in the list
    (defn cnt-hash
        [haystack]
        (if (nil? haystack)
            {}
            ; increment the value in the hash
            (
                ; tail recursion?
                (def store (cnt-hash (rest haystack)))
                (def needle (first haystack))
                (assoc store needle
                    (if (contains? store needle)
                        (inc (get store needle))
                        1
                        )
                    )
                )
            )
        )
    ; looks at each list and returns the maximum
    ; number of occurences of each element in each
    ; list
    (defn list-max-occurences
        [l]
        (defn helper-reduce
            [h1 h2]
            (merge-with
                ; merges and returns the larger value
                ; if there is a key conflict
                (fn [a b] (if (> a b) a b))
                h1
                h2
                )
            )
        (reduce
            helper-reduce
            (map cnt-hash l)
            )
        )

    (def occurences
        (list-max-occurences
            ; returns a list of lists of prime-factors
            ; of the numbers between 1 and upper
            (map prime-factorization (range 1 (+ 1 upper)))
            )
        )

    ; returns the result of multiplying each key with its value
    (def products
        (map
            (fn [k] (. (. java.math.BigInteger (valueOf k)) (pow (get occurences k)))) ;(. java.lang.Math (pow k (get occurences k)))); (* k (get occurences k)))
            (keys occurences)
            )
        )
    (reduce * products)
)

I’m still not using the most current version of Clojure, so I may be rewriting functions which already exist.

Problem 2 Description, from Project Euler

Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed four million.

Solutions

My first attempt

The trickiest thing for me was figuring out how to write the fibonacci function in an elegant way. I know that with functional languages, recursion is supposed to be super elegant, and my solution was anything but.

    (defn problem02
        []
        (defn fib
            [termCount]
            (defn fib-help
                [termCount, terms]
                (def rterms (reverse terms))
                (def lst (first rterms))
                (def nlst (first (rest rterms)))
                (if (< (count terms) termCount)
                    (fib-help
                        termCount
                        (reverse (cons (+ lst nlst) rterms))
                        )
                    terms
                    )
                )
            (fib-help termCount, `(1, 2))
            )
        (reduce + (filter
            #(and
                (< % 4000000)
                (even? %)
                )
            (fib 200)
            ))

    )
    (problem02)

A solution by another person

The [site I linked to in the previous article](http://grok-code.com/367/learning-clojure-with-project-euler/) as a much more elegant solution. I really had trouble figuring out how to do an elegant Fibonacci algorithm, although I knew that mine was inelegant!

   (defn e2 [limit]
     (let [fibs2 (lazy-cat [0 1]
                           (map + fibs2 (rest fibs2)))]
       (reduce + (filter #(zero? (mod % 2))
                         (take-while #(< % limit) fibs2)))))

I love the user of *lazy-cat*. I was trying to search for how to do this, but wasn’t sure what this was called. I only knew that the *iterate* function used something like this (I guess I could’ve checked it’s source). I find the use of map interesting, but confusing because it seems as if *fibs2* would return a list and that there should somehow be duplicate values in that list. I’m sure I will come to understand more later.

After PhillyETE, I wanted to try out some Clojure, so I decided to solve the first problem of Project Euler. I coded up my own solution using some of the references on the Clojure site.

Project Euler Problem Description, from Project Euler

Find the sum of all the multiples of 3 or 5 below N.
(N is 1000 is this Project Euler problem.)

Solutions

My first attempt

    (def nmbrs (range 0 1000))
    (def multiples
        (filter
            #(and
                (not= % 0)
                (or
                    (=
                        (rem % 3)
                        0
                        )
                    (=
                        (rem % 5)
                        0
                        )
                    )
                )
            nmbrs
            )
        )
    (reduce + multiples)

Another attempt by a different author

Afterwards, I came across a terser, more elegant solution on a site that seems to have had the same idea as I did, Learning Clojure with Project Euler. The site had a really elegant looking solution:

    ; http://grok-code.com/367/learning-clojure-with-project-euler/
    (defn problem01 [limit]
     (reduce + (filter #(or (zero? (mod % 3))
                            (zero? (mod % 5)))
                       (take (- limit 1) (iterate inc 1)))))
    (problem 1000)

My revised attempt

The use of take and iterate are really interesting. I haven’t encountered these “lazy” functions, and was surprised when I tried to run (iterate inc 1) in the REPL. I decided to modify my version with some of the things I learned from the version above, and make it more terse.

    (defn problem01
        [upper]
        (reduce +
            (filter
                #(or
                    (zero?(rem % 3))
                    (zero?(rem % 5))
                    )
                (range 1 upper)
                ))
        )

I’ll be working on some more of the problems in the coming days.

(PS. If you are wondering why I use rem instead of mod, it’s because I am using an older version of clojure with TextMate–I’ve yet to upgrade)