Let's decrease the monotony

Question

...but hey, no need to be strict.

Given a non-empty array of strictly positive integers, determine if it is:

1. Monotone strictly decreasing. This means that each entry is strictly less than the previous one.
2. Monotone non-increasing, but not strictly decreasing. This means that each entry is less than or equal to the preceding, and the array does not fall in the above category.
3. None of the above.

Note the following corner cases:

• An array with a single number is monotone strictly decreasing (vacuously so).
• An array with the same number repeated is monotone non-increasing, but not strictly decreasing.

Rules

You may provide a program or a function

Input can be taken in any reasonable format: array, list, string with numbers separated by spaces, ...

You can choose any three consistent outputs for the three categories respectively. For example, outputs can be numbers 0, 1, 2; or strings 1 1, 1 0, empty string.

Shortest code in bytes wins

Test cases

Monotone strictly decreasing:

7 5 4 3 1
42 41
5

Monotone non-increasing, but not strictly decreasing:

27 19 19 10 3
6 4 2 2 2
9 9 9 9

None of the above:

1 2 3 2
10 9 8 7 12
4 6 4 4 2

Answer 1

Perl 6, 17 bytes

{[>](@_)+[>=] @_}

• Monotone strictly decreasing: 2
• Monotone non-increasing: 1
• Other: 0

Expanded:

{            # bare block lambda with implicit parameter list ｢@_｣
  [>]( @_ )  # reduce using ｢&infix:« > »｣
  +
  [>=] @_    # reduce using ｢&infix:« >= »｣
}

Answer 2

MATL, 10, 7 bytes

0hdX>ZS

Try it online! or verify all test cases!

3 bytes saved, thanks to @LuisMendo!

The outputs are

• Strictly decreasing: -1

• Non-increasing: 0

• Other: 1

Explanation:

0           % Push a '0'
 h          % Join the 0 to the end of the input array.
  d         % Get consecutive differences
   X>       % Get the largest difference
     ZS     % Get its sign
            % Implicitly print it

Answer 3

Jelly, 10 9 5 bytes

-Method found by DrMcMoylex, go give some credit!

;0IṠṀ

TryItOnline! or run all tests

Returns :-1=monotone strictly decreasing; 0=monotone non-increasing; 1=other.

How?

;0IṠṀ - Main link: list
;0    - concatenate with a zero
  I   - incremental differences
   Ṡ  - sign
    Ṁ - maximum

Answer 4

Mathematica, 22 bytes

Sign@*Max@*Differences

Unnamed function taking a list of numbers as input. Returns -1 if the list is strictly decreasing, 0 if it's nonincreasing but not strictly decreasing, and 1 if it's neither.

Pretty simple algorithm: take the differences of consecutive pairs, take the largest one, and take the sign of that largest one.

(I feel like there must exist some language in which this algorithm is 3 bytes....)

Regarding an array with a single entry: Differences yields an empty list; Max of an empty list gives -∞ (!); and Sign[-∞] evaluates to -1 (!!). So it actually works on this corner case. Gotta love Mathematica sometimes. (Indeed, the function also correctly labels an empty list as strictly decreasing.)

Answer 5

Haskell, 40 38 37 bytes

foldl min GT.(zipWith compare<*>tail)

Returns

• GT for Monotone strictly decreasing
• EQ for Monotone non-increasing
• LT else

compare compares two numbers and returns GT (EQ, LT) if the first number is greater than (equal to, less than) the second number. zipWith compare<*>tail compares neighbor elements. foldl min GT reduces the list of the comparison results with the min function starting with GT (note: LT < EQ < GT).

Edit: @xnor found 2 3 bytes. Thanks!

Answer 6

Common Lisp, 43 40 bytes

(defun f(x)`(,(apply'> x),(apply'>= x)))

This takes input as a Lisp list, and returns (T T), (NIL T) and (NIL NIL) to distinguish the 3 categories. Here it is running on the provided test cases:

CL-USER> (mapcar #'f '((7 5 4 3 1)
                       (42 41)
                       (5)
                       (27 19 19 10 3)
                       (6 4 2 2 2)
                       (9 9 9 9)
                       (1 2 3 2)
                       (10 9 8 7 12)
                       (4 6 4 4 2)))
((T T) (T T) (T T) (NIL T) (NIL T) (NIL T) (NIL NIL) (NIL NIL) (NIL NIL))

Answer 7

Python 2, 30 bytes

lambda l:max(map(cmp,l[1:],l))

-1 for strictly decreasing, 0 for weakly decreasing, +1 for non-decreasing

Using cmp to compare consecutive elements, and takes the maximum. This is done by removing the first element of one copy of the list, then mapping cmp . For example, l=[2,2,1] gives

l[1:]  2   1   None
l      2   2   1
cmp    0  -1   -1

which has max 0 because an equality exists.

The shorter list is automatically extended with None, which is less than all numbers and so harmless. This phantom element also insulates against taking the min of an empty list when the input has length 1.

Answer 8

Brachylog, 7 bytes

>,1|>=,

Try it online!

This prints 1 for strictly decreasing, 0 for non-increasing and false. otherwise.

Explanation

  (?)>              Input is a strictly decreasing list
      ,1(.)         Output = 1
|                 Or
  (?)>=             Input is a non-increasing list
       ,(.)         Output is a free variable; gets automatically labeled as an integer at
                      the end of execution. Since its domain is [-inf, +inf], the first
                      value it takes is 0
                  Or
                    No other possibility, thus this main predicate is false.

Other 7 bytes solutions

>=!>,;1           Returns 0 for strictly decreasing, false. for non-increasing, 1 otherwise.

>=!>,1|           Returns 1 for strictly decreasing, false. for non-increasing, 0 otherwise.

Answer 9

R, 44 bytes

d=diff(scan());c(any(!d)&all(d<=0),all(d<0))

Reads input from stdin and prints the following depending on the input:

Output:

[1] FALSE TRUE: Monotone non-increasing

[1] TRUE FALSE: Monotone strictly decreasing

[1] FALSE FALSE: None of the above

Answer 10

JavaScript (ES6), 51 bytes

a=>a.some((e,i)=>e>a[i-1])+a.some((e,i)=>e>=a[i-1])

Returns 0 for strict decreasing, 1 for non-increasing, 2 otherwise.

Answer 11

05AB1E, 5 8 bytes

Bug fixed by Emigna, thanks! It uses the same method as DrMcMoylex's.

®¸ì¥Z0.S

®¸ì   Implicitly take input and appends -1 to it
¥     Yield deltas
 Z    Take the largest delta
  0.S Take its sign and implicitly display it

Try it online!

Output is:

-1 if strictly decreasing sequence
 0 if non-strictly decreasing sequence
 1 otherwise

Answer 12

Ruby, 37 bytes

->l{[d=l==l.sort.reverse,d&&l|[]==l]}

Output:[true,true], [true,false] or [false,false]

Answer 13

Haskell, 34 bytes

maximum.(zipWith compare<*>(1/0:))

Try it online!

Outputs LT,EQ,GT for strictly monotonic, non-increasing, and otherwise, respectively. Thanks to @xnor for pointing out that my previous version failed on singleton lists and suggesting this alternative.

Takes the maximum of the sighn of the differences between adjacent pairs, prepending with inf. (zipWith compare<*>(1/0:) applied to l computes compare l inf:l elementwise).

This is competitive with the existing Haskell answers and ekes out ahead by being pointfree and anonymous.

Answer 14

Mathematica, 15 11 bytes

##>0|##>=0&

This is a variadic function, taking all input integers as separate arguments.

• Strictly decreasing: True | True
• Non-increasing: False | True
• Neither: False | False

Note that | is not Or but Alternatives, which is part of pattern matching syntax, which explains why these expressions don't get evaluated to True, True, False, respectively.

The code itself is mostly an application of this tip. For example ##>0 is Greater[##, 0] but then ## expands to all the input values so we get something like Greater[5, 3, 2, 0], which itself means 5>3>2>0.

Answer 15

Racket, 44 bytes

(λ(x)(map(λ(p)(apply p`(,@x 0)))`(,>,>=)))

Invoked:

(map (λ(x)(map(λ(p)(apply p`(,@x 0)))`(,>,>=)))
 '((7 5 4 3 1)
   (42 41)
   (5)
   (27 19 19 10 3)
   (6 4 2 2 2)
   (9 9 9 9)
   (1 2 3 2)
   (10 9 8 7 12)
   (4 6 4 4 2)))

Result:

'((#t #t)
 (#t #t)
 (#t #t)
 (#f #t)
 (#f #t)
 (#f #t)
 (#f #f)
 (#f #f)
 (#f #f))

Answer 16

C++14, 85 bytes

int f(int x){return 3;}int f(int x,int y,auto...p){return((x>=y)+2*(x>y))&f(y,p...);}

Returns 3 (0b11) for strict decreasing, 1 (0b01) for non-increasing and 0 otherwise.

Ungolfed:

int f(int x) {return 3;}
int f(int x,int y,auto...p){
  return ((x>=y)+2*(x>y)) & f(y,p...);
}

I thought this was a perfect problem for C++17's folding expressions:

int g(auto...x){return(x>...)+(x>=...);}

Unfortunately it does not chain the relational operators but does

((x1>x2)>x3)>x4)...

which is not was wanted.

Answer 17

Python 2, 61 74 bytes

+13 bytes for the single number input

x=map(str,input())
print[2,eval(">".join(x))+eval(">=".join(x))][len(x)>1]

Requires input in bracket list form like [3,2,1]. Returns 2 for strict decreasing, 1 for non-increasing and 0 otherwise.

Old solution:

print eval(">".join(x))+eval(">=".join(x))

Answer 18

Python 3, 81 52 bytes (Thanks to FryAmTheEggMan)

e=sorted
lambda a:(a==e(a)[::-1])+(e({*a})[::-1]==a)

Try it online !

Answer 19

Haskell, 36 bytes

f l=[scanl1(min.(+x))l==l|x<-[0,-1]]

(+x) is because haskell mis-interprets (-x) as a value instead of a section. I wonder if the whole expression can be profitably made pointfree.

Answer 20

Befunge, 50 bytes

&: >~1+#^_v>:0`|
1\:^  @.$$<-@.2_
-: ^    >& ^   >

Try it online!

Accepts input as a sequence of int separated by spaces, and returns 0 if strictly decreasing, 1 if non-strictly decreasing, 2 otherwise.

Since reading befunge is kind of impossible if you don't know the language, this is the algorithm in pseudocode:

push(input())

while( getchar()!=EOF ){
  push(input())
  subtract()
  duplicate()
  if(pop()>0){
    subtract() //we are doing a-(a-b), so b is now on top
    duplicate()
  }
  else{
    if(pop()==0){
      push(1) //the sequence is not strictly decreasing
      swap()
      duplicate()
    }
    else{
      push(2) //the sequence has just increased
      output(pop)
      return
    }
  }
}
pop()
pop()
output(pop())

*in befunge memory is a stack which starts with an infinite amount of 0 on it. pop(), push(x), input() and output(x) are self-explainatory, the other pseudofunctions i used work like this:

function duplicate(){
  a=pop()
  push(a)
  push(a)
}

function subtract(){
  a=pop()
  b=pop()
  push(b-a)
}

function swap(){
  a=pop()
  b=pop()
  push(a)
  push(b)
}

Funge!

---

Previous version, just 41 bytes but invalid since it requires a 0 to terminate the input sequence (or using an interpreter like this)

&:  >&:|>:0`|
1\v@.$_<-@.2_
- >:v  >^   >

Try it online!

Answer 21

J, 14 bytes

Monadic verb taking the list on the right, returning 1 for strictly decreasing, 0 for weakly decreasing, and _1 otherwise.

*@([:<./2-/\])

Takes the sign * of the minimum <./ of consecutive differences 2-/\ of the list. J doesn't swap the order of the differences when taking them so e.g. the sequence is strictly decreasing if these are all positive. Notably, <./ returns positive infinity on zero-element lists.

In use at the REPL:

   *@([:<./2-/\]) 3
1
   *@([:<./2-/\]) 3 2
1
   *@([:<./2-/\]) 3 2 2
0
   *@([:<./2-/\]) 3 2 2 3
_1

Answer 22

C, 68 67 Bytes

A function f, which is passed an array of ints (l) preceded by its length (n, also an int). Returns 3 if monotone strictly decreasing, 1 if monotone non-increasing, but not strictly decreasing, 0 otherwise.

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

Un-golfed slightly for readability:

int f_semiungolfed(int n, int* l) {
    return (n < 2) ? 3 : ((l[0] > l[1]) * 2 | l[0] >= l[1]) & f(n - 1, l + 1);
}

Rearranged and commented to show logic:

int f_ungolfed(int n, int* l) {
    int case1 = 0, case2 = 0, recursion = 0;
    if (n < 2) { // Analogous to the ternary conditional I used - n < 2 means we have a single-element/empty list
        return 3; // Handles the vacuous-truth scenario for single lists
    } else {
        case1 = l[0] > l[1]; // The first case - are the two numbers in the list strictly decreasing? (case1 is 1 if so)
        case2 = l[0] >= l[1]; // The second case - are the two numbers strictly non-increasing (case2 is 1 if so)
        recursion = f_ungolfed(n - 1, l + 1); // Recursively call ourselves on the "rest" of the list (that is, everything other than the first element). Consider that comparison is transitive, and that we already have a vacuous-truth scenario covered.
        case1 *= 2; // Shift case1's value over to the left by one bit by multiplying by 2. If case1 was 1 (0b01), it's now 2 (0b10) - otherwise it's still 0 (0b00)
        return (case1 | case2) & recursion; 
        // The bitwise OR operator (|) will combine any 1-bits from case1's value (either 0b10 or 0b00) or case2's value (either 0b01 or 0b00) into either 3, 2, 1, or 0 (0b11, 0b10, 0b01, or 0b00 respectively).
        // The bitwise AND operator (&) will combine only matching 1-bits from (case1|case2) and the return value of the recursive call - if recursion = 0b11 and case1|case2 = 0b01, then the return value will be 0b01.
    }
}

Test cases (courtesy IDEOne):

{7, 5, 4, 3, 1}: 3
{42, 41}: 3
{5}: 3
{27, 19, 19, 10, 3}: 1
{6, 4, 2, 2, 2}: 1
{9, 9, 9, 9}: 1
{1, 2, 3, 2}: 0
{10, 9, 8, 7, 12}: 0
{4, 6, 4, 4, 2}: 0

Answer 23

Retina, 41 bytes

\d+
$*
A`\b(1+) 1\1
S`\b$
\b(1+) \1\b.*|$

Try it online! (The first line enables a linefeed-separated test suite.)

• Strictly decreasing: 2
• Non-increasing: 3
• Neither: 1

Explanation

\d+
$*

Converts the input unary.

A`\b(1+) 1\1

The regex here matches an increasing pair of consecutive numbers. If this is the case, the input can clearly not be non-increasing. The A denotes it as an "anti-grep" stage which means that the input line is discarded and replaced with the empty string if the regex matches.

S`\b$

This is a split stage which is used to append a linefeed to the input only if the input wasn't discarded. So we've got two possible outcomes so far: non-increasing inputs get a linefeed at the end and others are still empty.

\b(1+) \1\b.*|$

Finally, we count the number of matches of this regex. The regex either matches to identical numbers (and then everything to the end of the string to avoid multiple matches of this kind for inputs like 1 1 1 1), or the "end of the input". Let's go through the three types of inputs:

• Strictly decreasing: the first part of the regex can't match because all values are unique, but the $ matches. Now $ isn't exactly "the end of the string". It can also match in front of a trailing linefeed. So we'll actually get two matches from it, one at the end of the input, and one after the linefeed we inserted.
• Non-increasing: now the first part of the regex also provides a match, and we end up with three matches.
• Neither: remember that we took care to turn the input into an empty string, so now $ matches only once.

Answer 24

Vim, 69 bytes

V}JYu:sor!n
V}JP:%s/^\(.*\)\n\1/m\r\1
:%s/\vm\n.*<(.+) \1.*/n
:g/\d/d

Try it online!

This is vanilla vim (TIO supports V (vim), which is backwards compatible). Takes input as a list of integers, one per line. Outputs m for strictly monotonic, n for non-increasing, and empty string otherwise.

V}              " select everything
  J             " space-join it all into one line
   Y            " yank that
    u           " undo back to one per line
     :sor!n     " reverse sort (the whole file, by line) numerically
V}J             " space join again
   P            " paste the previously yanked original list
                " now we have the list on line 1 and a sorted copy on line 2

Next we have two regexes.

:%s/^\(.*\)\n\1/m\r\1

Matches two copies of the same line and replaces the first with "m".

:%s/\vm\n.*<(\d+) \1.*/n

Matches that "m" followed by a line containing <(.+) \1, i.e. a word boundary followed by <foo> <foo>, and if found replaces everything with n. The \v at the start says "all characters are special" so we don't have to escape any of <()+.

There's a bit of careful cleverness here: we don't need a word boundary after the \1 because the first number can't be a prefix of the second in a reverse sorted list. Furthermore, even though (.+) can match several numbers (as opposed to (\d+) which would match only one for an extra byte), this won't result in any false positives for the same reason: we can't have a repeated string of multiple numbers in a reverse sorted list unless they're all the same, in which case we want to match them anyway.

The state after all this is one of:

monotonic     non-increasing      other
---------     --------------      -----
m             n                   3 1 2
3 2 1                             3 2 1

So, to collapse these states, we just delete any line which still has a digit: :g/\d/d, resulting in the desired output.

I do wonder whether the regexes can be golfed some (maybe avoiding some cross-line stuff?) but the corner cases can be a bit tricky.

Answer 25

Axiom, 114 bytes

m(c:List(INT)):INT==(i:=r:=1;repeat(~index?(i+1,c)=>break;c.i<c.(i+1)=>return 0;if c.i=c.(i+1)then r:=2;i:=i+1);r)

Ungolfed

-- if [a,b,..n] decrescente ritorna 1
--          non crescente   ritorna 2
--          altrimenti      ritorna 0  
m(c:List(INT)):INT==
   i:=r:=1
   repeat
      ~index?(i+1,c)=>break 
      c.i<c.(i+1)   =>return 0
      if c.i=c.(i+1) then r:=2
      i:=i+1
   r

Results

(x) -> m([3,1])=1, m([1,1])=2, m([])=1, m([1])=1, m([1,3])=0
   (x)  [1= 1,2= 2,1= 1,1= 1,0= 0] 

Answer 26

APL, 16 bytes

(a≡a[⍒a])×1+a≡∪a

Note: enter one element array as eg a←1⍴3 otherwise: a←4 3 2 1

Interpreting output:

2 Monotone strictly decreasing
1 Monotone non-increasing, but not strictly decreasing
0 None of the above

Idea: test for monotonicity by comparing original to sorted array, check for non-increasing by comparing to array with removed duplications.

(And I think it can be improved...)

Answer 27

LabVIEW, 12 nodes, 18 wires ==> 48 bytes by convention

No functions hidden in the other case frames, just a single wire across.

Answer 28

Ceylon, 86 bytes

Object m(Integer+l)=>let(c=l.paired.map(([x,y])=>x<=>y))[if(!smaller in c)equal in c];

The function takes the input as its parameters, and returns a tuple of zero or one booleans – [false] for Monotone strictly decreasing, [true] for Monotone non-increasing, but not strictly decreasing, and [] for None of the above.

It can be used like this:

shared void run() {
    print("Monotone strictly decreasing:");
    print(m(7, 5, 4, 3, 1));
    print(m(42, 41));
    print(m(5));

    print("Monotone non-increasing, but not strictly decreasing:");
    print(m(27, 19, 19, 10, 3));
    print(m(6, 4, 2, 2, 2));
    print(m(9, 9, 9, 9));

    print("None of the above:");
    print(m(1, 2, 3, 2));
    print(m(10, 9, 8, 7, 12));
    print(m(4, 6, 4, 4, 2));
}

Output:

Monotone strictly decreasing:
[false]
[false]
[false]
Monotone non-increasing, but not strictly decreasing:
[true]
[true]
[true]
None of the above:
[]
[]
[]

An ungolfed and commented version:

// Let's decrease the monotony! 
//
// Question:  http://codegolf.stackexchange.com/q/101036/2338
// My Answer: http://codegolf.stackexchange.com/a/101309/2338


// Define a function which takes a non-empty list `l` of Integers)
// (which arrive as a tuple) and returns an Object (actually
// a `[Boolean*]`, but that is longer.) 
Object m(Integer+ l) =>
        // the let-clause declares a variable c, which is created by taking
        // pairs of consecutive elements in the input, and then comparing
        // them. This results in lists like those (for the example inputs):
        // { larger, larger, larger, larger }
        // { larger }
        // {}
        // { larger, equal, larger, larger }
        // { larger, larger, equal, equal }
        // { equal, equal, equal }
        // { smaller, smaller, larger }
        // { larger, larger, larger, smaller }
        // { smaller, larger, equal, larger }  
        let (c = l.paired.map( ([x,y]) => x<=>y) )
            // now we analyze c ...
            // If it contains `smaller`, we have an non-decreasing sequence.
            // We return `[]` in this case (an empty tuple).
            // Otherwise we check whether `equal` is in the list, returning
            // `[true]` (a non-strictly decreasing sequence) if so,
            // and `[false]` (a strictly decreasing sequence) otherwise.
            [if(!smaller in c) equal in c];

Answer 29

Clojure, 34 bytes

#(if(apply > %)1(if(apply >= %)2))

Very straight-forward, returns 1 for if strictly decreasing, 2 if non-increasin and nil otherwise.

Also tried avoiding apply with macro's ~@ but it is just longer at 43 chars (this results in [1 2 nil]):

(defmacro f[& i]`(if(> ~@i)1(if(>= ~@i)2)))

[(f 7 5 4 3 1)
 (f 27 19 19 10 3)
 (f 1 2 3 2)]

Answer 30

Pip, 8 bytes

O$>g$>=g

Full program. Takes the input list as command-line arguments. Outputs 11 for strictly decreasing, 01 for non-increasing, 00 for neither.

Try it online!

Explanation

This approach works because Pip's comparison operators, like Python's, chain together: 4>3>2 is true, rather than being (4>3)>2 (false) as in C. And the same behavior holds when the comparison operators are modified with the $ fold meta-operator.

          g is list of command-line args (implicit)
 $>g      Fold g on >
O         Output without newline
    $>=g  Fold g on >=
          Print (implicit)