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Challenges Fibonacci without consecutive digits

Pyth, 25 23 bytes @f-1aM.:jT;2u+Gs>2GyQU2 Try it online! First, we generate the Fibonacci numbers: u+Gs>2GyQU2 U2 range(2), the list [0, 1] u Apply the follow...

posted 9mo ago by isaacg‭ · edited 9mo ago by isaacg‭

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#2: Post edited by

isaacg‭ · 2024-03-01T17:12:57Z (9 months ago)
Found a shorter way to generate fibonacci numbers.

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~~# [Pyth], 25 bytes~~
~~@f-1aM.:jT;2hM.u,eNsNyQU2~~
~~[Try it online!][TIO-lt8wf7zj]~~
First, we generate the Fibonacci numbers:
hM.u,eNsNyQU2
~~U2 range(2), the list [0, 1]~~
~~.u Apply the following function iteratively,~~
saving all results.
,eNsN [end(N), sum(N)], where N is the previous list
~~yQ Twice the input number of repetitions~~
hM Take the first entry of each pair.
I think there might be a shorter way to do this part, perhaps by using `u` rather than `.u`?
Next, we check for consecutive digits.
@f-1aM.:jT;2
f Filter on the following function
jT; Take the base ten representation of the number.
; accesses T's value in the external scope.
.: 2 Take all pairs of digits.
aM Map pairs to their absolute difference.
-1 Check if 1 is not among the differences.
@ Index the filter result at the input position.
[Pyth]: https://github.com/isaacg1/pyth
~~[TIO-lt8wf7zj]: https://tio.run/##K6gsyfj/3yFN1zDRV88qK8TaKMNXr1Qn1a/YrzIw1Oj/fyMDAA "Pyth – Try It Online"~~

# [Pyth], <s>25</s> 23 bytes
@f-1aM.:jT;2u+Gs>2GyQU2
[Try it online!][TIO-lt8wtkyl]
First, we generate the Fibonacci numbers:
u+Gs>2GyQU2
U2 range(2), the list [0, 1]
u Apply the following function iteratively,
returning the final result.
+G Add to the end of the current list
s>2G The sum of the last two elements of the current list
yQ Repeat for twice the input number of repetitions
Next, we check for consecutive digits.
@f-1aM.:jT;2
f Filter on the following function
jT; Take the base ten representation of the number.
; accesses T's value in the external scope.
.: 2 Take all pairs of digits.
aM Map pairs to their absolute difference.
-1 Check if 1 is not among the differences.
@ Index the filter result at the input position.
[Pyth]: https://github.com/isaacg1/pyth
[TIO-lt8wtkyl]: https://tio.run/##K6gsyfj/3yFN1zDRV88qK8TaqFTbvdjOyL0yMNTo/38jAwA "Pyth – Try It Online"

#1: Initial revision by

isaacg‭ · 2024-03-01T17:05:46Z (9 months ago)

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# [Pyth], 25 bytes

    @f-1aM.:jT;2hM.u,eNsNyQU2

[Try it online!][TIO-lt8wf7zj]

First, we generate the Fibonacci numbers:

    hM.u,eNsNyQU2
               U2    range(2), the list [0, 1]
      .u             Apply the following function iteratively,
                     saving all results.
        ,eNsN        [end(N), sum(N)], where N is the previous list
             yQ      Twice the input number of repetitions
    hM               Take the first entry of each pair.

I think there might be a shorter way to do this part, perhaps by using `u` rather than `.u`?

Next, we check for consecutive digits.

    @f-1aM.:jT;2
     f              Filter on the following function
            jT;     Take the base ten representation of the number.
                    ; accesses T's value in the external scope.
          .:   2    Take all pairs of digits.
        aM          Map pairs to their absolute difference.
      -1            Check if 1 is not among the differences.
    @               Index the filter result at the input position.

[Pyth]: https://github.com/isaacg1/pyth
[TIO-lt8wf7zj]: https://tio.run/##K6gsyfj/3yFN1zDRV88qK8TaKMNXr1Qn1a/YrzIw1Oj/fyMDAA "Pyth – Try It Online"

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