Sure: I'm saying figuring out the formula from a handful of examples is
tricky. But I can do that, and I already know Clojure has conditionals. I'm
trying to learn something about how the logic engine works.

lvh

On Thu, Oct 31, 2019 at 5:23 PM Satyam Ramawat <satyamrama...@gmail.com>
wrote:

> Laurens,
>
> Have you tried checking up to what extent it goes?
>
>
>
> On Thu, Oct 31, 2019 at 9:44 PM Laurens Van Houtven <_...@lvh.io> wrote:
>
>> I always need all of them. Figuring out how many that are combinatorially
>> is surprisingly tricky!
>>
>> On Thu, Oct 31, 2019 at 11:33 AM Satyam Ramawat <satyamrama...@gmail.com>
>> wrote:
>>
>>> I mean you can limit the answer according to need, by running into loop
>>> until end users value satisfy, will can be either 5,20 or N. so it will
>>> throw output according to run of a loop.
>>>
>>> On Thu, 31 Oct 2019 at 4:28 PM, Laurens Van Houtven <_...@lvh.io> wrote:
>>>
>>>> Satyam: you're suggesting limiting the answers asked from run, or
>>>> something else?
>>>>
>>>> On Thu, Oct 31, 2019 at 11:24 AM Satyam Ramawat <
>>>> satyamrama...@gmail.com> wrote:
>>>>
>>>>> On a very simple way, you can restrict by using if statement by
>>>>> crossing checking the count/length of the list,
>>>>> https://clojuredocs.org/clojure.core/if
>>>>>
>>>>> On Tuesday, October 29, 2019 at 11:06:54 PM UTC, Laurens Van Houtven
>>>>> wrote:
>>>>>>
>>>>>> Hi,
>>>>>>
>>>>>> I'm trying to divide a list into 3 lists such that (= my-list (concat
>>>>>> a b c)). I guess you could say I'm writing concato :-)
>>>>>>
>>>>>> (l/run n
>>>>>>   [a b c]
>>>>>>   (l/fresh [A B] ;; uppercase are internal accumulators
>>>>>>     (l/appendo a b A)
>>>>>>     (l/appendo A c B)
>>>>>>     (l/== B '(p q r s))))
>>>>>>
>>>>>> ... running this program with small n (let's say 5) works fine. Ask
>>>>>> for more answers than there are (e.g. 20) and it ostensibly runs forever.
>>>>>>
>>>>>> 1. Why?
>>>>>> 2. How do I fix that? (I've tried reordering the goals.)
>>>>>>
>>>>>> thanks
>>>>>> lvh
>>>>>>
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