Mountain range numbers

Question

Mountain range number

A number is a mountain range number if the inequalities satisfied by their consecutive digits alternate. In a way, looking at the number's digits should exhibit a /\/\/\... or a \/\/\/... pattern.

More formally, if our number n has \$k\$ digits

$$n = d_1d_2d_3\cdots d_k$$

then n is a mountain range number if

$$\begin{cases}d_1 > d_2 \\ d_2 < d_3 \\ d_3 > d_4 \\ \cdots \end{cases} \vee \begin{cases}d_1 < d_2 \\ d_2 > d_3 \\ d_3 < d_4 \\ \cdots \end{cases}$$

Your task

Given an integer with 3 or more digits, output a Truthy value if the number is a mountain range number or Falsy otherwise.

Input

A positive integer n with 3 or more digits, in any reasonable format, e.g.

• Integer
• String
• List of digits

Test cases

Truthy

1324 -> Truthy
9191 -> Truthy
12121 -> Truthy
121212 -> Truthy
1212121 -> Truthy
19898 -> Truthy

Falsy

(Added another Falsy test case as per the comments, some answers might not cover the 4422 test case)

123 -> Falsy
321 -> Falsy
4103 -> Falsy
2232 -> Falsy
1919199 -> Falsy
4422 -> Falsy

This is code-golf so shortest answer in bytes wins! Standard loopholes are forbidden.

Answer 1

R, 44 40 bytes

_{crossed out 44 is still regular 44}

-1 byte thanks to Giuseppe.

function(x,d=diff)all(d(sign(d(x)))^2>3)

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Computes the differences of the signs of the differences of the input. These must all be equal to 2 or -2, i.e. the square must equal 4; checking that the square is >3 is sufficient.

If two consecutive digits are equal, there will be a 0 in the signs of differences, leading to a difference of signs of differences equal to 1 or -1. If three consecutive digits are in ascending or descending order, then the corresponding differences will be of the same sign, leading to a difference of signs of differences equal to 0. If neither of these occurs, the number is a mountain range number.

---

Old version (included as it might be golfable):

R, 44 43 bytes

-1 byte thanks to Giuseppe.

function(x)all(s<-sign(diff(x)),rle(s)$l<2)

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Computes the signs of the differences of consecutive digits. Then verifies that

• none of the signs are 0s (would correspond to 2 equal consecutive digits);
• the runs of the signs are all equal to 1, i.e. no 2 consecutive signs are equal.

Answer 2

05AB1E, 6 bytes

¥ü*0‹P

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With truthy and falsy reversed, this would be 5 bytes:

¥ü*dZ

TIO

Answer 3

JavaScript (ES6),  35  33 bytes

a=>!a.some(p=v=>a*(a=p-(p=v))>=0)

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Commented

a =>                // a[] = input list of digits,
                    //       re-used to store the last difference 
  !a.some(          //
    p =             // initialize p to a non-numeric value
    v =>            // for each v in a[]:
      a * (         //   multiply a by
        a =         //     the new value of a defined as
          p -       //       the difference between p and
          (p = v)   //       the new value of p, which is v
      )             //
      >= 0          //   the test fails if this is non-negative
  )                 // end of some()

Answer 4

Japt -!, 7 bytes

Takes input as a digit array.

äÎä* dÄ

Try it

Answer 5

Haskell, 57 55 47 44 42 bytes

all(<0).z(*).z(-)
z f(x:s)=zipWith(f)s$x:s

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Takes input as a list of digits.

• -2 by swapping the order of s and x:s

• -8 by using a different helper function

• -3 by using partial application and pointfree code

• -2 by excluding f= from the submission (which I didn't realize was allowed :P)

xnor improved my answer using >>=.

Answer 6

Python, 47 bytes

f=lambda a,b,*l:l==()or(a-b)*(b-l[0])*f(b,*l)<0

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Takes input splatted like f(1,2,3,4). Same idea as my second Haskell answer.

Answer 7

Jelly, 7 6 bytes

A benchmarking solution.

A monadic link taking as input the list of digits

I×Ɲ<0Ạ

You can try it online or verify all test cases.

I         Take the forward differences
  Ɲ       and for each pair,
 ×        multiply them together.
   <0     Check if those are below 0.
      Ạ   Check if this array of booleans only contains Truthy values.

-1 byte thanks to @79037662

Answer 8

Bash, 147 144 bytes

M(){
a=${1:0:1}
d=x
i=1
while [ $i -lt ${#1} ]
do
b=${1:$i:1}
case $d$((a-b)) in
[ux]-*)d=d;;*0|u*|d-*)return 1;;*)d=u;;esac
a=$b
let i++
done
}

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I seem to like trying shell submissions, and learned some bash-isms in golfing this one.

$((a-b)) is equivalent to $(( $a - $b )) -- apparently you don't need the $ inside a $(( )) construct.

There is a ++ operator, works in $(( )) and in let

Subtracting letters is accepted, strangely. One of my samples in the TIO reads "xy", and apparently $((a-b)) evaluates a to x, and then variable x to an empty string and the empty string as numeric zero, and comparable for b and y. If I set x and y in the environment, those values are used.

Edit: -3 bytes by not putting whitespace after ;;, thanks to S.S.Anne

Answer 9

Haskell, 37 bytes

all(<0).g(*).g(-)
g=(=<<tail).zipWith

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Takes the zipWith-based answer of 79037662 and generalizes out the pattern of

g(?) = \s->zipWith(?)(tail s)s

that applies the operator (?) to pairs of adjacent elements. This is shortened to the pointfree g=(=<<tail).zipWith.

We first apply g(-) to the input to take differences of consecutive elements, then g(*) to take products of those consecutive differences. Then, we check that these products are all negative, which means that consecutive differences must be opposite in sign.

---

Haskell, 40 bytes

f(a:b:t)=t==[]||(a-b)*(b-t!!0)<0&&f(b:t)

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The idea is a bit clearer to see in the slightly less-golfed form:

42 bytes

f(a:b:c:t)=(a-b)*(b-c)<0&&f(b:c:t)
f _=1>0

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We check that the first three digits (a,b,c) have the a->b steps and b->c steps going opposite directions by checking that the differences a-b and b-c have opposite signs, that is, their product is negative. Then we recurse to the list without its first element until the list has fewer than 3 elements, where it's vacuously true.

An alternative to check suffixes directly turned out longer:

43 bytes

f l=and[(a-b)*(b-c)<0|a:b:c:t<-scanr(:)[]l]

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Answer 10

Python 2, 65 58 bytes

lambda A:all((x-y)*(y-z)<0for x,y,z in zip(A,A[1:],A[2:]))

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Answer 11

Ruby -nl, 57 41 bytes

Replaces each character in the input string with the cmp comparison (<=> in Ruby) between it and the next character $'[0] (if there is no next character, remove the character instead). Then, check if the resulting string consists entirely of alternating 1 and -1.

gsub(/./){$&<=>$'[0]}
p~/^1?(-11)*(-1)?$/

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Old Solution, 57 bytes

Check for duplicate consecutive numbers first by checking if the input string matches /(.)\1/ and inverting it. If no such pairs are found, replace each character with true or false based on whether its cmp style comparisons (<=>) to the character before it $`[-1] and after it $'[0] are not equal. (If there is no character before or after it, the <=> returns nil, which is definitely not equal to whatever the other character comparison returns.) Finally, it checks if the result does not contain an f (meaning no falses were returned).

p ! ~/(.)\1/&&gsub(/./){($`[-1]<=>$&)!=($&<=>$'[0])}!~/f/

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Answer 12

Excel (Insider build ver. 1912), 122 Bytes

A1 'Input
B1 =SEQUENCE(LEN(A1))
C1 =MID(A1,B1#,1)
D1 =SIGN(IF(NOT(B1#-1),C1-C2,C1#-INDEX(C1#,B1#-1)))
E1 =(SUM(D1#)=D1*ISODD(LEN(A1)))*PRODUCT(D1#) 'Output

Returns ±1 (truthy) or 0 (falsy)

Explanation (can add more detail if people are interested)

B1 =SEQUENCE(LEN(A1)) ' Generates a spill array from 1 to the length of the input
C1 =MID(A1,B1#,1) ' Splits characters into rows. Using each value in the spill array B1#
                  ' as a charcter index
D1 =SIGN(IF(NOT(B1#-1), ' Choose different value on the first cell
           C1-C2, ' Use the opposite of the first difference between digits
           C1#-INDEX(C1#,B1#-1))) ' get the difference between each digit and the previous
E1 =(SUM(D1#)=D1*ISODD(LEN(A1))) ' Sum the digit differences, if the 
                                 ' input length is even check if 0, else check if equal to
                                 ' thefirst row of the differences
       *PRODUCT(D1#))            ' ensure there aren't any repeated digits

Tests

Answer 13

Jelly, (5?) 6 bytes

5 if we may invert the truthy/falsey output (strip the trailing ¬).

IṠIỊẸ¬

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Answer 14

J, 15 bytes

[:*/0>2*/\2-/\]

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-7 bytes thanks to RGS's technique

Answer 15

Charcoal, 29 27 bytes

ＵＭθ⁻ι§θ⊕κＵＭθ×ι§θ⊕κ›⁰⌈…θ⁻Ｌθ²

Try it online! Link is to verbose version of code. Takes input as a list of digits and outputs as a Charcoal boolean (- for a mountain range number, otherwise no output). Explanation:

ＵＭθ⁻ι§θ⊕κ

Take consecutive differences (cyclic, so includes difference between last and first digit).

ＵＭθ×ι§θ⊕κ

Take consecutive products (again, cyclic).

›⁰⌈…θ⁻Ｌθ²

All results bar the last two must be negative.

Answer 16

APL+WIN, 17 15 bytes

2 bytes saved thanks to Jo King

Prompts for list of digits:

×/1=0>×2×/-2-/⎕

Try it online! Courtesy of Dyalog Classic

Answer 17

Python 3, ^{101 \$\cdots\$ 103} 94 bytes

Added 13 bytes to fix error kindly pointed out by @ChasBrown.
Saved 9 bytes thanks to @ChasBrown!!!

def f(l):x=[a<b for a,b in zip(l[1:],l)];return all(a!=b for a,b in zip(x[1:]+l[1:],x[:-1]+l))

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Answer 18

Burlesque, 22 bytes

XX2COqcm^m2COPD{0.<}al

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XX      # Explode into digits
2CO     # 2-grams ("abc"->{"ab" "bc"})
qcm^m   # Compare each via UFO operator
2CO     # 2-grams
PD      # Product
{0.<}al # All less than 0

Answer 19

K (ngn/k), 14 bytes

&/0>2_*':-':$:

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$: as string

-': subtract (as ascii codes) each prior; implicit 0 before first

*': multiply by each prior; implicit 1 before first

2_ drop first 2 elements

&/0> all negative?

Answer 20

Brachylog, 9 bytes

¬{s₃.o↙Ḋ}

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Takes a list of digits as input.

Explanation

¬{      }       It is impossible…
  s₃            …to find a subsequence of 3 elements…
    .o↙Ḋ        …which is already ordered

Slight subtility: o↙Ḋ is used to check whether the digits are increasing or decreasing. By default, o (which is the same as o₀) is for increasing order, and o₁ is for decreasing order. By using o↙Ḋ (Ḋ being an integer between 0 and 9), we check that the whole predicate is impossible for o₀, or o₁, or o₂, …, o₉. o₂ to o₉ are not implemented and thus will fail, which doesn’t impact the program as a whole.

If true. is an acceptable falsy value, and false. an acceptable truthy value (which I don’t think it should be), then you should be able to remove these 3 bytes: ¬{…}.

Answer 21

Java (JDK), 95 83 bytes

p->{int i=0,j=1;for(;p.length>-~++i;)j=(p[i-1]-p[i])*(p[i]-p[i+1])<0?j:0;return j;}

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Thanks to all in the comments for improvements - especially bit-shifting which I never would have thought of!!

Answer 22

C (gcc), 59 bytes

d;m(int*s){for(d=*s/s[1];s[1]&&s[1]/ *s-d;d^=1)s++;s=s[1];}

Takes as input a wide string of digits and returns zero if that number is a mountain range number.

-12 bytes thanks to ceilingcat!

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