solution to 2.32. One of the best exercises in this section.

diff --git a/src/sicp/ex2_32.clj b/src/sicp/ex2_32.clj
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+(ns sicp.ex2_32
+  (:use [clojure.test]))
+
+(defn subsets [s]
+  (if (empty? s)
+    (list ())
+    (let [r (subsets (rest s))]
+      (concat r (map #(cons (first s) %) r)))))
+
+(comment
+"
+Let s be '(1 2 3).
+
+1. (subset '(1 2 3)) => (subset '(2 3)) => (subset '(3)) => (subset '())
+2. At this point subset returns.
+3. r becomes (). value of s is '(3).
+4. concat () with result of (map #(cons 3 %) '()) => (() (3))
+5. returns to previous call, where s is '(2 3).
+6. (map #(cons 2 %) '(() (3))) => ((2) (2 3))
+7. concat the above with previous r => (() (3) (2) (2 3))
+8. return to the previous call, where s is '(1 2 3)
+9. (map #(cons 1 %) '(() (3) (2) (2 3))) => ((1) (1 3) (1 2) (1 2 3))
+10. Append the above with previous value of r => (() (3) (2) (2 3) (1) (1 3) (1 2) (1 2 3))
+"
+)
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