Is there a bump?

Question

Given a list of positive integers determine if there is an element that is either greater than its two neighbors or less than its two neighbors (a "bump"). To be clear a bump can never be the first or last item of the list because they only have one neighbor.

Your program should output one of two consistent values each corresponding to either a list with no bumps or a list with bumps. What the values are is unimportant you may choose them yourself.

This is code-golf so answers will be scored in bytes with fewer bytes being better.

Test cases

[] -> False
[1] -> False
[1,2] -> False
[1,2,1] -> True
[1,2,2] -> False
[1,2,3] -> False
[1,2,2,1] -> False
[1,2,2,3] -> False
[1,2,1,2] -> True
[1,3,2] -> True
[3,1,2] -> True
[2,2,2] -> False

Answer 1

Jelly, 5 bytes

IṠIỊẠ

Returns 0 if there's a bump, 1 if not.

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How it works

IṠIỊẠ  Main link. Argument: A (integer array)

I      Increments; take all forward differences of A.
 Ṡ     Take the signs.
       The signs indicate whether the array is increasing (1), decreasing (-1), or
       constant at the corresponding point. A 1 followed by a -1 indicates a local
       maximum, a -1 followed by a 1 a local minimum.
  I    Increments; take the forward differences again.
       Note that 1 - (-1) = 2 and (-1) - 1 = -2. All other seven combinations of
       signs map to -1, 0, or 1.
   Ị   Insignificant; map each difference d to (-1 ≤ d ≤ 1).
    Ạ  All; return 1 if all differences are insignificant, 0 if not.

Answer 2

JavaScript (ES6), 38 bytes

Returns a boolean.

a=>a.some(x=n=>x*(x=a<n|-(a>(a=n)))<0)

Test cases

let f =

a=>a.some(x=n=>x*(x=a<n|-(a>(a=n)))<0)

console.log(f([]))              // False
console.log(f([1]))             // False
console.log(f([1,2]))           // False
console.log(f([1,2,1]))         // True
console.log(f([1,2,2]))         // False
console.log(f([1,2,3]))         // False
console.log(f([1,2,2,1]))       // False
console.log(f([1,2,2,3]))       // False
console.log(f([1,2,1,2]))       // True
console.log(f([1,3,2]))         // True
console.log(f([2,0,1]))         // True
console.log(f([2,2,2]))         // False
console.log(f([-4,100000,89]))  // True

How?

We use a to store the previous value of n. We set x to 1 if a < n, -1 if a > n or 0 if a = n. And we test whether old_x * x < 0, which is only possible if (old_x = 1 and x = -1) or (old_x = -1 and x = 1).

Because x is initialized to the anonymous callback function of some(), it is coerced to NaN during the first iteration, which makes the test falsy.

Answer 3

Haskell, 42 bytes

any(<0).f(*).f(-)
f a b=zipWith a b$tail b

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Explanation

First we have the function f that takes a binary function and a list and applies the binary function to every adjacent pair in the list.

Then our main function applies f(-) to the input list. This calculates the difference list. We then apply f(*) to the list to multiply every adjacent pair. Lastly we ask if any pair is less than zero.

A number in the end list can only be negative if it is the product of a negative and positive number from the difference list. Thus in order to produce a negative entry (and then return true) the original list must switch from increasing to decreasing or vice versa, that is it must have a bump.

Answer 4

Python 2, 43 bytes

f=lambda x,y,z,*t:0>(y-x)*(z-y)or f(y,z,*t)

Returns True if there's a bump, errors if there isn't. (allowed by default)

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Answer 5

Octave with Image Package, 34 32 bytes

2 bytes saved thanks to @StewieGriffin!

@(x)0||prod(im2col(diff(x),2))<0

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Explanation

Computes consecutive differences, arranges them in sliding blocks of length 2, obtains the product of each block, and tests if any such product is negative.

Answer 6

Haskell, 33 bytes

f(p:r@(c:n:_))=(c-p)*(c-n)>0||f r

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True if there's a bump, errors if there's not.

Answer 7

R, 48 bytes

function(x)any(apply(embed(diff(x),2),1,prod)<0)

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How it works step-by-step using c(1,4,1,4) as example:

> x=c(1,4,1,4)
> diff(x)
[1]  3 -3  3
> embed(diff(x),2)
     [,1] [,2]
[1,]   -3    3
[2,]    3   -3
> apply(embed(diff(x),2),1,prod)
[1] -9 -9
> any(apply(embed(diff(x),2),1,prod)<0)
[1] TRUE

As a bonus, here is a solution of similar length and concept using package zoo:

function(x)any(zoo::rollapply(diff(x),2,prod)<0)

Answer 8

Perl 6, 39 bytes

{so~(.[1..*]Zcmp$_)~~/'re L'|'s M'/}

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$_ is the list argument to this anonymous function. .[1..*] is the same list, but with the first element dropped. Zcmp zips the two lists together with the cmp operator, resulting in a list of Order values. For example, for an input list 1, 2, 2, 2, 1 this would result in the list More, Same, Same, Less.

Now we just need to know whether that list contains two adjacent elements More, Less or Less, More. The trick I used is to convert the list to a space-delimited string with ~, then test whether it contains either substring re L or s M. (The first one can't be just e L because Same also ends with an "e".)

The smart match operator returns either a Match object (if the match succeeded) or Nil (if it didn't), so so converts whatever it is into a boolean value.

Answer 9

Ruby, 55 46 bytes

->a{a.each_cons(3).any?{|x,y,z|(y-x)*(y-z)>0}}

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A lambda accepting an array and returning boolean.

-9 bytes: Replace (x<y&&y>z)||(x>y&&y<z) with (y-x)*(y-z)>0 (thanks to GolfWolf)

->a{
  a.each_cons(3)              # Take each consecutive triplet
    .any?{ |x,y,z|            # Destructure to x, y, z
      (y-x)*(y-z) > 0         # Check if y is a bump
    }
}

Answer 10

Wolfram Language (Mathematica), 40 bytes

MatchQ[{___,x_,y_,z_,___}/;x<y>z||x>y<z]

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Answer 11

R, 58 56 bytes

function(x)any(abs(diff(sign(diff(c(NA,x)))))>1,na.rm=T)

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Saved 2 bytes thanks to Giuseppe

Answer 12

J, 16 15 bytes

-1 byte thanks to FrownyFrog

1 e.0>2*/\2-/\]

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Original: 16 bytes

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

2-/\] - differences of each adjacent items

2*/\ - products of each adjacent items

[:<./ - the minimum

0> - is negative?

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Answer 13

Java 8, 108 104 101 86 84 79 72 bytes

a->{int i=a.length,p=0;for(;i-->1;)i|=p*(p=a[i]-a[i-1])>>-1;return-i>1;}

-2 bytes thanks to @OlivierGrégoire.
-13 bytes thanks to @Nevay.

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Answer 14

PostgreSQL 173 bytes

SELECT DISTINCT ON(a)a,x>j and x>k OR x<least(j,k)FROM(SELECT a,x,lag(x,1,x)OVER(w)j,lead(x,1,x)OVER(w)k FROM d WINDOW w AS(PARTITION BY rn ORDER BY xn))d ORDER BY 1,2 DESC;
     a     | c 
-----------+---
 {1}       | f
 {1,2}     | f
 {1,2,1}   | t
 {1,2,1,2} | t
 {1,2,2}   | f
 {1,2,2,1} | f
 {1,2,2,3} | f
 {1,2,3}   | f
 {1,3,2}   | t
 {2,2,2}   | f
 {3,1,2}   | t
(11 rows)

Answer 15

Japt, 11 bytes

-5 bytes thanks to @ETHproductions

ä- mÌäa d>1

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This uses Dennis's algorithm

Answer 16

Japt, 9 bytes

ä- ä* d<0

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A mashup of Oliver's answer with the approach used by several other answers.

Answer 17

Attache, 39 bytes

Any&:&{_*~?Sum[__]}@Slices&2@Sign@Delta

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Pretty happy with how this turned out.

Explanation

This is a composition of four functions:

Delta
Sign
Slices&2
Any&:&{_*~?Sum[__]}

Delta gets the differences between elements. =

Then, Sign is applied to each difference, giving us an array of 1s, 0s, and -1s. =

Then, Slices&2 gives all slices of length two from the array, giving all pairs of differences.

Finally, Any&:&{_*~?Sum[__]} is equivalent to, for input x:

Any[&{_*~?Sum[__]}, x]
Any[[el] -> { el[0] and not (el[0] + el[1] = 0) }, x]

This searches for elements which sum to zero but are not zero. If any such pair of elements exist, then there is a bump.

Answer 18

Octave, 33 bytes

@(x)0||abs(diff(sign(diff(x))))>1

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Explanation:

@(x)                           % Anonymous function taking x as input
                  diff(x)       % Takes the difference between consecutive elements
             sign(diff(x))      % The sign of the differences
        diff(sign(diff(x)))     % The difference between the signs
    abs(diff(sign(diff(x)))>1   % Check if the absolute value is 2
@(x)abs(diff(sign(diff(x)))>1   % Output as matrices that are treated truthy or falsy

Answer 19

MATL, 8 bytes

dZSd|1>a

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Answer 20

Husk, 7 bytes

V<0Ẋ*Ẋ-

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Explanation

V<0Ẋ*Ẋ-  Implicit input, say [2,5,5,1,4,5,3]
     Ẋ-  Consecutive differences: [3,0,-4,3,1,-2]
   Ẋ*    Consecutive products: [0,0,-12,3,-2]
V<0      Is any of them negative? Return 1-based index: 3

Answer 21

05AB1E, 7 bytes

¥ü‚P0‹Z

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Explanation

¥         # calculate delta's
 ü‚       # pair each element with the next element
   P      # product of each pair
    0‹    # check each if less than 0
      Z   # max

Answer 22

Brachylog, 10 bytes

s₃.¬≤₁∧¬≥₁

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Succeeds (true.) if there is a bump, and fails (false.) if there is no bump.

Explanation

This is fairly readable already:

s₃.           There is a substring of the input…
  .¬≤₁        …which is not non-decreasing…
      ∧       …and…
       ¬≥₁    …which is not non-increasing

Answer 23

Brachylog, 10 bytes

s₃s₂ᶠ-ᵐ×<0

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Not nearly as neat and elegant as @Fatalize's existing 10 byte answer, but it works!

s₃   % There exists a substring of three elements [I,J,K] in the array such that

s₂ᶠ  % When it's split into pairs [[I,J],[J,K]]

-ᵐ   % And each difference is taken [I-J, J-K]

×    % And those differences are multiplied (I-J)*(J-K)
     % (At a bump, one of those will be negative and the other positive. 
     % At other places, both differences will be positive, or both negative, 
     %  or one of them 0 - ultimately resulting in a non-negative product.)

<0   % The product is negative

Answer 24

APL (Dyalog), 15 19 20 bytes

2 bytes saved thanks to @EriktheOutgolfer

{2>≢⍵:0⋄2∊|2-/×2-/⍵}

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1 for bump, 0 for no bump.

Answer 25

Python 2, 60 bytes

lambda l:any(p>c<n or p<c>n for p,c,n in zip(l,l[1:],l[2:]))

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Pretty much the same thing, thought it would be shorter though...

Python 2, 63 bytes

f=lambda l:l[3:]and(l[0]>l[1]<l[2]or l[0]<l[1]>l[2]or f(l[1:]))

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Answer 26

Julia 0.6, 57 56 bytes

l->any(p>c<n||p<c>n for(p,c,n)=zip(l,l[2:end],l[3:end]))

Basically just totallyhuman's python answer. -1 byte from user71546

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Julia 0.6, 39 bytes

f(x,y,z,a...)=x>y<z||x<y>z||f(y,z,a...)

Lispy recursion style, aka Dennis's python answer. Returns true when a bump exists, otherwise throws an error. This should maybe be 42 bytes since you have to splat it when calling. Eg for a=[1,2,1] you call as f(a...). f(a)=f(a...) would remove that need, but is longer. I need to get better a recursion, and I don't really like writing code that throws an error.

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Answer 27

Husk, 9 bytes

▼mεẊ-Ẋo±-

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Uses Dennis's algorithm.

Answer 28

Perl 5, 54 + 2 (-pa) = 56 bytes

map$\|=$F[$_-1]!=(sort{$a-$b}@F[$_-2..$_])[1],2..$#F}{

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Answer 29

Pyt, 11 7 bytes

₋ʁ*0<Ʃ±

Outputs 1 if there is a bump, 0 otherwise

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Port of Wheat Wizard's Haskell answer

---

Old way (11 bytes):

₋±₋Å1≤ĐŁ↔Ʃ=

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Returns False if there is a bump, True otherwise

Port of Dennis' Jelly answer

Answer 30

Windows Batch, 126 120 114 bytes

I was pretty excited about my first use of the shift command.

This script generally works for numbers within 9 digits(no matter positive or negative.)

:L
@if %3.==. exit/b
@if %2 gtr %1 if %3 lss %2 goto:e
@if %2 lss %1 if %3 gtr %2 goto:e
@shift
@goto:L
:e
@echo T

Answer 31

Wolfram Language (Mathematica), 37 36 bytes

FreeQ[(d=Differences)@Sign@d@#,-2|2]&

Gives the opposite of the test case answers (False and True reversed). Prepend a ! to switch to the normal form.

OR

Abs@(d=Differences)@Sign@d@#~FreeQ~2&

Also reversed output, so replace FreeQ with MatchQ for normal form.

Explanation: Take the sign of the differences of the sequence. Iff the resulting sequence includes {1,-1} or {-1,1} there is a bump. The absolute value the differences of {1,-1} or {-1,1} is 2 in either case.

Shave off another byte by squaring the final list instead of taking the absolute value:

FreeQ[(d=Differences)@Sign@d@#^2,4]&

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Answer 32

Perl, 35 bytes

Includes +3 for -p

bump.pl:

#!/usr/bin/perl -p
s%\S+ %$a*1*($a=$&-$')%eg;$_=/-/

Run as:

bump.pl <<< "3 1 2"

Answer 33

C (gcc), 101 97 bytes

r,i;f(a,n)int*a;{for(i=n|1;--i;a[i]-=a[i-1]);for(r=0;--n>0;r=a[n]<0&a[n-1]>0|a[n]>0&a[n-1]<0|r);}

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Differences the list, then looks for two neighboring differences that are not zero and have opposite sign.

Answer 34

C (gcc), 78 bytes

Variant: loop

r,i;f(a,n)int*a;{for(r=n>2,i=0;i<n-2;)r=(a[i++]-a[i])*(a[i]-a[i+1])<0&&r;i=r;}

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Variant: recursive

f(a,n)int*a;{n=(n<3)?0:((a[--n]-a[--n])*(a[n]-a[n-1])<0&&((n<2)?1:f(a,n+1)));}

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The criteria is that a[i]-a[i+1] and a[i+1]-a[i+2] are non-zero and have opposite sign. i=r is just a way to return the value (described here).

P. S. Additional two test cases may reveal bugs: { 1, 2, 1, 2, 2 } and { 2, 2, 1, 2, 1 }

Answer 35

APL (Dyalog Classic), 15 bytes

0∨.>2×/2-/⊃,⊃,⊢

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Answer 36

Retina 0.8.2, 43 bytes

.+
$*
1`(1+)1¶\1¶1\1|¶(1+)(?<!\2¶\2)¶(?!\2)

Try it online! Takes input on separate lines and outputs 0 or 1. Explanation:

.+
$*

Convert to unary.

1`            |

Count at most one matching regex.

  (1+)1¶\1¶1\1

Match a number that is surrounded by greater numbers on both sides.

               ¶(1+)(?<!\2¶\2)¶(?!\2)

Match a number that is not surrounded by greater or equal numbers on either side, i.e. both are less.

Answer 37

Python, 91 bytes

def f(a):return[i for i,j in enumerate(a[1:-1])if(j>max(a[i+2],a[i])or j<min(a[i+2],a[i]))]

An empty array is a falsey value, while a populated array is truthy

Answer 38

Perl 6, 50 47 bytes

{?grep {0>(.[1]-.[2])*[-] .[^2]},.rotor(3=>-2)}

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Answer 39

GFORTH 71 Bytes

: B ROT 2DUP = -1 = IF DROP ROT = -1 = IF  ELSE ." TRUE " THEN THEN ;

Output:

1 2 1 B TRUE  ok
19 4 19 B TRUE  ok
5 4 6 B  ok

Answer 40

Elm, 102 bytes

f a b=case b of
 c::d::e->a c d::f a(d::e)
 _->[]
a g=case g of
 []->1<0
 i::k->i<0||a k
a<<f(*)<<f(-)

Explanation

This works very similar to my Haskell answer. Except Elm is missing all of the functions that do stuff so I had to make every thing from the ground up. You can test it by running the following here

import Html exposing (text)
f a b=case b of
 c::d::e->a c d::f a(d::e)
 _->[]
a g=case g of
 []->1<0
 i::k->i<0||a k
g=a<<f(*)<<f(-)
s x=case x of
 True->"True"
 False->"False"
main=g[1,1,1,1,1,0,1]|>s|>text