how to cast integer into string ?

I am trying to solve this question. 2^(15) = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26. What is the sum of the digits of the number 2^(1000)? This is how I solved in python. sum([int(x) for x in str(pow(2,1000))]) For lisp, I don't know how to implement that algorithm ... This is my sketch on how to slove ... (loop for x across "12345" summing (- (char-int x) 48)) Still I need to cast (expt 2 1000) into string from integer ... How can I do that ?

On Dec 7, 12:13=A0pm, thurahlain...@gmail.com wrote: > I am trying to solve this question. > > 2^(15) =3D 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 =3D 26. > What is the sum of the digits of the number 2^(1000)? > > This is how I solved in python. > sum([int(x) for x in str(pow(2,1000))]) > > For lisp, I don't know how to implement that algorithm ... > This is my sketch on how to slove ... > > (loop for x across "12345" summing (- (char-int x) 48)) > > Still I need to cast (expt 2 1000) into string from integer ... > How can I do that ? One way: (prin1-to-string (expt 2 1000))

On Dec 7, 8:41 pm, thurahlain...@gmail.com wrote: > > One way: (prin1-to-string (expt 2 1000)) > > (loop for x across (prin1-to-string (expt 2 1000)) summing (- (char- > int x) 48)) > > It works ! Thanks ... It would be slightly better to use CHAR-CODE. But you can also use DIGIT-CHAR-P. CL-USER 4 > (digit-char-p #\3) 3 CL-USER 5 > (reduce #'+ (write-to-string (expt 2 1000)) :key #'digit- char-p) 1366

On Dec 7, 11:13=A0am, thurahlain...@gmail.com wrote: > I am trying to solve this question. > > 2^(15) =3D 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 =3D 26. > What is the sum of the digits of the number 2^(1000)? > > This is how I solved in python. > sum([int(x) for x in str(pow(2,1000))]) > > For lisp, I don't know how to implement that algorithm ... > This is my sketch on how to slove ... > > (loop for x across "12345" summing (- (char-int x) 48)) As an alternative, (defun integer-digit-list (n &optional (radix 10)) "Returns a list of the digits (base radix) of an integer. Least digit is in position 0." (check-type n integer) (if (zerop n) (list 0) (labels ((digit-list (n) (multiple-value-bind (upper digit) (floor n radix) (cons digit (when (> upper 0) (digit-list upper)))))) (digit-list (abs n))))) CL-USER> (reduce '+ (integer-digit-list (expt 2 1000))) 1366 CL-USER> (reduce '+ (integer-digit-list (expt 2 1000) 16)) 1 CL-USER> Wade

On Sun, 07 Dec 2008 10:13:40 -0800, thurahlaing06 wrote: > I am trying to solve this question. > > 2^(15) = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26. What > is the sum of the digits of the number 2^(1000)? > > This is how I solved in python. > sum([int(x) for x in str(pow(2,1000))]) > > For lisp, I don't know how to implement that algorithm ... This is my > sketch on how to slove ... > > (loop for x across "12345" summing (- (char-int x) 48)) > > Still I need to cast (expt 2 1000) into string from integer ... How can > I do that ? Others have already provided solutions. I wanted to add mine because it is important to realize that you do not need the intermediate list of digits, and you can proceed from either end (+ is commutative). So (defun sum-digits (i) (declare ((integer 0) i)) ;; we want a nonnegative integer (let ((sum 0)) (tagbody top (multiple-value-bind (quotient remainder) (truncate i 10) (incf sum remainder) (when (plusp quotient) (setf i quotient) (go top)))) sum)) (sum-digits (expt 2 1000)) Some consider tagbody ugly - it is their loss :-) Tamas

In article <8abe9b94-c4ed-4df4-8f82-750f1b40c33d@x8g2000yqk.googlegroups.com>, Wade <wade.humeniuk@gmail.com> wrote: > On Dec 7, 11:13�am, thurahlain...@gmail.com wrote: > > I am trying to solve this question. > > > > 2^(15) = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26. > > What is the sum of the digits of the number 2^(1000)? > > > > This is how I solved in python. > > sum([int(x) for x in str(pow(2,1000))]) > > > > For lisp, I don't know how to implement that algorithm ... > > This is my sketch on how to slove ... > > > > (loop for x across "12345" summing (- (char-int x) 48)) > > As an alternative, > > (defun integer-digit-list (n &optional (radix 10)) > "Returns a list of the digits (base radix) of an integer. Least > digit > is in position 0." > (check-type n integer) > (if (zerop n) > (list 0) > (labels > ((digit-list (n) > (multiple-value-bind (upper digit) > (floor n radix) > (cons digit (when (> upper 0) (digit-list upper)))))) > (digit-list (abs n))))) > > CL-USER> (reduce '+ (integer-digit-list (expt 2 1000))) > 1366 > CL-USER> (reduce '+ (integer-digit-list (expt 2 1000) 16)) > 1 > CL-USER> > > Wade If put in a library it might be worth to have recursion replaced with iteration (-> stack size limit). -- http://lispm.dyndns.org/

On Dec 7, 3:15=A0pm, Rainer Joswig <jos...@lisp.de> wrote: > In article > <8abe9b94-c4ed-4df4-8f82-750f1b40c...@x8g2000yqk.googlegroups.com>, > > > > =A0Wade <wade.humen...@gmail.com> wrote: > > On Dec 7, 11:13=A0am, thurahlain...@gmail.com wrote: > > > I am trying to solve this question. > > > > 2^(15) =3D 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 =3D 2= 6. > > > What is the sum of the digits of the number 2^(1000)? > > > > This is how I solved in python. > > > sum([int(x) for x in str(pow(2,1000))]) > > > > For lisp, I don't know how to implement that algorithm ... > > > This is my sketch on how to slove ... > > > > (loop for x across "12345" summing (- (char-int x) 48)) > > > As an alternative, > > > (defun integer-digit-list (n &optional (radix 10)) > > =A0 "Returns a list of the digits (base radix) of an integer. =A0Least > > digit > > =A0 =A0is in position 0." > > =A0 (check-type n integer) > > =A0 (if (zerop n) > > =A0 =A0 =A0 (list 0) > > =A0 =A0 =A0 (labels > > =A0 =A0 =A0((digit-list (n) > > =A0 =A0 =A0 =A0 (multiple-value-bind (upper digit) > > =A0 =A0 =A0 =A0 =A0 =A0 (floor n radix) > > =A0 =A0 =A0 =A0 =A0 (cons digit (when (> upper 0) =A0(digit-list upper)= ))))) > > =A0 =A0(digit-list (abs n))))) > > > CL-USER> (reduce '+ (integer-digit-list (expt 2 1000))) > > 1366 > > CL-USER> (reduce '+ (integer-digit-list (expt 2 1000) 16)) > > 1 > > CL-USER> > > > Wade > > If put in a library it might be worth to have recursion > replaced with iteration (-> stack size limit). > Here is a iterative solution. (defun integer-digit-list (n &optional (radix 10)) "Returns a list of the digits base radix of an integer. Least digit is in position 0." (check-type n integer) (setf n (abs n)) (loop collect (multiple-value-bind (upper digit) (floor n radix) (setf n upper) digit) while (> n 0))) Wade

In article <58423cde-97e0-4f13-ab4d-8d74140068b6@l42g2000yqe.googlegroups.com>, Wade <wade.humeniuk@gmail.com> wrote: > On Dec 7, 3:15�pm, Rainer Joswig <jos...@lisp.de> wrote: > > In article > > <8abe9b94-c4ed-4df4-8f82-750f1b40c...@x8g2000yqk.googlegroups.com>, > > > > > > > > �Wade <wade.humen...@gmail.com> wrote: > > > On Dec 7, 11:13�am, thurahlain...@gmail.com wrote: > > > > I am trying to solve this question. > > > > > > 2^(15) = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26. > > > > What is the sum of the digits of the number 2^(1000)? > > > > > > This is how I solved in python. > > > > sum([int(x) for x in str(pow(2,1000))]) > > > > > > For lisp, I don't know how to implement that algorithm ... > > > > This is my sketch on how to slove ... > > > > > > (loop for x across "12345" summing (- (char-int x) 48)) > > > > > As an alternative, > > > > > (defun integer-digit-list (n &optional (radix 10)) > > > � "Returns a list of the digits (base radix) of an integer. �Least > > > digit > > > � �is in position 0." > > > � (check-type n integer) > > > � (if (zerop n) > > > � � � (list 0) > > > � � � (labels > > > � � �((digit-list (n) > > > � � � � (multiple-value-bind (upper digit) > > > � � � � � � (floor n radix) > > > � � � � � (cons digit (when (> upper 0) �(digit-list upper)))))) > > > � �(digit-list (abs n))))) > > > > > CL-USER> (reduce '+ (integer-digit-list (expt 2 1000))) > > > 1366 > > > CL-USER> (reduce '+ (integer-digit-list (expt 2 1000) 16)) > > > 1 > > > CL-USER> > > > > > Wade > > > > If put in a library it might be worth to have recursion > > replaced with iteration (-> stack size limit). > > > > Here is a iterative solution. > > (defun integer-digit-list (n &optional (radix 10)) > "Returns a list of the digits base radix of an integer. Least digit > is in position 0." > (check-type n integer) > (setf n (abs n)) > (loop collect > (multiple-value-bind (upper digit) > (floor n radix) > (setf n upper) > digit) > while (> n 0))) > > Wade Yep. or using SETF instead of MULTIPLE-VALUE-BIND: (defun integer-digit-list (n &optional (radix 10)) "Returns a list of the digits base radix of an integer. Least digit is in position 0." (check-type n integer) (setf n (abs n)) (loop with digit do (setf (values n digit) (floor n radix)) collect digit while (plusp n))) -- http://lispm.dyndns.org/

Tamas K Papp <tkpapp@gmail.com> wrote: +--------------- | Others have already provided solutions. I wanted to add mine because it | is important to realize that you do not need the intermediate list of | digits, and you can proceed from either end (+ is commutative). So | | (defun sum-digits (i) | (declare ((integer 0) i)) ;; we want a nonnegative integer | (let ((sum 0)) | (tagbody | top | (multiple-value-bind (quotient remainder) (truncate i 10) | (incf sum remainder) | (when (plusp quotient) | (setf i quotient) | (go top)))) | sum)) | | (sum-digits (expt 2 1000)) | | Some consider tagbody ugly - it is their loss :-) +--------------- I cheerfully use TAGBODY when it's appropriate, but I don't consider the above to be such a case, especially given that there's a considerably simpler way *without* it, viz.: (defun sum-digits (i) (assert (and (integerp i) (>= i 0))) (loop while (plusp i) summing (multiple-value-bind (quotient remainder) (truncate i 10) (setf i quotient) remainder))) -Rob p.s. Note that I changed your DECLARE to an ASSERT, since DECLARE is a promise *by* the programmer, not a check *of* the programmer/user! I suppose I could have written (ASSERT (TYPEP I '(INTEGER 0 *))), but the TYPEP is likely to be slower. ----- Rob Warnock <rpw3@rpw3.org> 627 26th Avenue <URL:http://rpw3.org/> San Mateo, CA 94403 (650)572-2607

* (Rob Warnock) <Z7-dnfnvaPE4NqHUnZ2dnUVZ_sbinZ2d@speakeasy.net> : Wrote on Sun, 07 Dec 2008 23:07:49 -0600: | I cheerfully use TAGBODY when it's appropriate, but I don't consider | the above to be such a case, especially given that there's a considerably | simpler way *without* it, viz.: | | (defun sum-digits (i) | (assert (and (integerp i) (>= i 0))) | (loop while (plusp i) | summing (multiple-value-bind (quotient remainder) | (truncate i 10) | (setf i quotient) | remainder))) | | | p.s. Note that I changed your DECLARE to an ASSERT, since DECLARE is | a promise *by* the programmer, not a check *of* the programmer/user! | I suppose I could have written (ASSERT (TYPEP I '(INTEGER 0 *))), | but the TYPEP is likely to be slower. Since we're picking nits anyway :-} you could use (check-type i (integer 0)) instead of the assert -- Madhu

thurahlaing06@gmail.com writes: >> One way: (prin1-to-string (expt 2 1000)) > (loop for x across (prin1-to-string (expt 2 1000)) summing (- (char- > int x) 48)) > > It works ! Thanks ... No it doesn't work here (EBCDIC system). Use: (loop :for x :across (prin1-to-string (expt 2 1000)) :summing (digit-char-p x)) Yes, DIGIT-CHAR-P returns what you want, despite of its name. -- __Pascal Bourguignon__