Arraypath with various step values
@pbejse wrote:.... I guess "random" means any order of the four given distance [ 15 18 21 24 ] but no repeats Yes? ....
I hope not, because that would directly contradict Message 3. So I'll be interested in what the intent really is -- it would have been better to have it all spelled out in detail from the beginning.
Kent Cooper, AIA 
@Kent1Cooper wrote:it would have been better to have it all spelled out in detail from the beginning.
Yes, and to quote @devitg
- For better understanding, and maybe get further help, please upload such sample.dwg -
🙂
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From the month of October til today , our friend @devitg posted that same reply, an estimated number of 10 times .
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@Anonymous wrote:
All circles are moved considering the new distance.
That explains why you wanted Arraypath.
But anyway, the "move" can still be incorporated in the code if needed.
Hi,
using this routine there is no control on the last step/distance
Do you think could be possible to set the function to try several times to create the series of circles until the last distance is within the two value?
example: (taking a look on the image above) if the values were 31 and 35, the first array would be good, the second would be erase and do again.
These two values and the times the routine should try to do the operation can be fixed within the code (it is not necessary to enter them every time in the dialog box).
I know it's not always possibile have "a correct sequence of random distances" to get a good/valid value for the last distance, for this reason I decided to set a maximum of attempts. If no one of these attempts is valid, the routine will draw the last "sequence of random distances".
It could be very useful if in case of "no solution/not found a correct sequence of random distances" the last two circles changed color.
Thank you for any help!
@Anonymous wrote:
..... The counter go to negative value, so the cycle goes on indefinitely.
I was trying resetting attempts (variable itr-max) ... with these two alternatives (the syntax of the IF loop condition is still not clear to me
(if (itr-max<0) (setq itr-max 200)) (setq itr-max (1- itr-max) ;; lower the counter(if (< itr-max 0) ;;very strange for me this syntax (setq itr-max 200)) (setq itr-max (1- itr-max) ;; lower the counterIs it the right point to edit?
Why syntax are wrong?
The second one is closer. AutoLisp functions always start with the function name, followed by the arguments the function should use. But if you don't want it to go below 0, it should check whether its less than 1, not whether it's less than 0 -- you don't want to wait until it's already a negative value before doing something to avoid a negative value. Or if it's always stepping 1 at a time, you could just check whether it's equal to 0. And you have one right parenthesis in the wrong place, and one missing. Try this:
(if
(= itr-max 0) ; the test expression
(setq itr-max 200) ; the 'then' expression if the test returns anything other than nil
;; [I moved the other right parenthesis later]
(setq itr-max (1- itr-max)) ; the 'else' expression if the test returns nil
;; [I added the right parenthesis to close the (setq) function]
) ; close the (if) function after both 'then' and 'else' expressions
Kent Cooper, AIA
@Anonymous wrote:
.... Does LISP syntax would always the then and else istructions? Could I do a IF loop without the else instructions?
....
(if
(= itr-max 0)
(setq itr-max 200) then instruction((setq itr-max (1- itr-max) ;; lower the counter else instructions
dlt-lst2 nil))) ;; reset length list
but got a syntax error on load...
Yes, you can do an (if) function with no 'else' expression, if you don't want it to do anything when the 'test' expression returns nil. See >Help for (if)<, in which the [elseexpr] in the arguments list is in brackets, meaning it's optional.
Your blue parentheses are the problem there [and you need to add at least one semicolon to comment out the words "then instruction" at the end of that line]. The (setq) function without further wraparound is the 'else' expression. If you added those parentheses because of the two variables in the (setq) function, they're not needed -- you can set any number of variables within one (setq) function.
EDIT: But come to think of it.... Without studying dozens of earlier messages to know what you're doing with this, I wonder whether you don't want the length list set to nil along with jumping up the counter, rather than along with lowering the counter. Should it be this instead?
(if
(= itr-max 0) ; test -- when counting down reaches 0,
(setq ; then
itr-max 200 ; jump counter up
dlt-lst2 nil ;; reset length list
) ; setq
(setq itr-max (1- itr-max)) ;; lower the counter else instructions
) ; if
Kent Cooper, AIA