From today I'll be putting the solutions of SICP on the web (which I took some time to finish). This is done to motivate myself to finish the SICP faster.
Also the solution will be in COMMON LISP rather than SCHEME since I want to learn the LISP also.
Till now I have finished till excersise 1.16 of SICP of which I have only left excersise 1.5 and 1.14( I never understood what is nickel etc ?).
Till now I found the excersise 1.11 coolest. Well it taught me what can be achieved by recursion also can be passed through state varaiables as a function (aka procedure it takes time to get out of C/C++ mindset :) ).
Probably I already knew that but the way it explains I have never seen in any book.
You can review the solutions and add the comments:
Pascal Triangle: Excerise 1.12
(defun pascal (rowName columnName) ( if (or (= rowName 1) (= columnName rowName)) 1 (if (= columnName 1) 1 (+ (pascal (- rowName 1) (- columnName 1)) (pascal (- rowName 1) columnName)))))
Excersise: 1.11 (iterative process only as recursive is straight forward)
(defun f (counter limit result fn1 fn2 fn3) (if (= counter limit) result (f (+ counter 1) limit (+ fn1 (+ (* 2 fn2) (* 3 fn3))) (+ fn1 (+ (* 2 fn2) (* 3 fn3))) fn1 fn2)))
COMMON-LISP-USER> (setf res1 0)
COMMON-LISP-USER> (f 3 4 res1 3 2 1)
10
3 comments:
Hi Rahul,
check out my blog man.
Rahul .. this is cool. Good going!
As Vimal pointed out its O(n) rather than O(log n).
The correct solution is
(defun f (base counter limit product mult) (if (< (* 2 counter) limit) (f base (* 2 counter) limit (* product product) mult) (if (= counter limit) (* product
mult) (f base 1 (- limit counter) base (* product mult)))))
It was able to calculate as 9999 pow 9999 successfully.
(f 9999 1 9999 1 9999)
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