Undulant numbers

Question

An undulant number is a number where its digits go up and down like the following number: 461902 or 708143, or even 1010101, but not 123, because 2 < 3. write function which returns true if a number is undulant, false otherwise. The shortest function wins.

NOTE: For completeness, single digit numbers are accepted by the function and the function would return false (I imagine this case as non-undulant number), thus isUndulant returns false for n < 10.

Answer 1

J, 45

*./(n>9),(}:(=-)}.)(}:*@-}.)n#:~10$~>.10^.n=.

Sample use:

   *./(n>9),(}:(=-)}.)(}:*@-}.)n#:~10$~>.10^.n=. 461902
1
   *./(n>9),(}:(=-)}.)(}:*@-}.)n#:~10$~>.10^.n=. 708143
1
   *./(n>9),(}:(=-)}.)(}:*@-}.)n#:~10$~>.10^.n=. 1010101
1
   *./(n>9),(}:(=-)}.)(}:*@-}.)n#:~10$~>.10^.n=. 123
0
   *./(n>9),(}:(=-)}.)(}:*@-}.)n#:~10$~>.10^.n=. 5
0

I'm pretty sure there's a finer way of twisting Insert / to do more of the work in a go, but I've been J-less for months, I need to get back to it.

Answer 2

Ruby, 72 70 characters

Q=10;k=->n,v{(n%Q-n/Q%Q)*v<0?k[n/Q,-v]:n<Q};u=->n{n>9&&k[n,-1]|k[n,1]}

Usage and testcases:

p u[10101]   # <= true
p u[708143]  # <= true
p u[2421]    # <= false
p u[1231]    # <= false
p u[873]     # <= false

Single digits yield false:

p u[5]       # <= false

Consecutive identical digits also return false:

p u[66]      # <= false
p u[1221]    # <= false

Answer 3

Haskell, 88 characters

q[]=9<3
q[e]=e/=EQ
q(a:b:s)=a/=EQ&&a/=b&&q(b:s)
w s=zipWith compare s$tail s
u=q.w.show

Answer 4

Python: 101 100 characters

Before minification:

undulate = (lambda n: n > 9
            and all(cmp(*digits) == (i % 2) * 2 - 1
                    for i, digits
                    in enumerate(zip(min(`n`,`n`[1:]), 
                                     max(`n`,`n`[1:])))))

After minification:

a=lambda b:b>9and all(cmp(*c)==d%2*2-1 for d,c in enumerate(zip(min(`b`,`b`[1:]),max(`b`,`b`[1:]))))

Answer 5

Sage, 76 83char

f=lambda x:uniq(cmp(*`x`[i-2:i][::(-1)^i])for i in[2..len(`x`)])in[[1],[-1]]

Got the idea to use cmp(*[..]) from JBernardo. In Sage, uniq(...) is an alias for list(set(...)).

Edit: just noticed that for x < 10, uniq(cmp(...)) == [], which isn't on [[1],[-1]]. If x were input as a string, instead of an integer, I could get another 4 characters out!

Answer 6

J, 30

*/0<(#,]*{.*1 _1$~#)2-/\a.i.":

A different approach than the other J answers.

   */0<(#,]*{.*1 _1$~#)2-/\a.i.":461902
1
   */0<(#,]*{.*1 _1$~#)2-/\a.i.":708143
1
   */0<(#,]*{.*1 _1$~#)2-/\a.i.":1010101
1
   */0<(#,]*{.*1 _1$~#)2-/\a.i.":123
0
   */0<(#,]*{.*1 _1$~#)(}.-}:)a.i.":5
0

Would be 3 characters shorter if 5 were considered undulant.

Answer 7

(pdf)eTeX, 129 chars

\def\a#1#2{\if#2?\ifx\r\s\def\s{1}\else
True\end\fi\fi\edef\t{\pdfstrcmp{#2}{#1}}\ifx\s\t
False\end\fi\let\s\t\a#2}\expandafter\a

Compiling with pdfetex filename.tex 1324? gives a pdf output. TeX is primarily a typesetting language, and outputting instead to stdout would take around 20 more chars. Also the strange requirement for one-digit numbers (false rather than true) takes me 26 chars.

Answer 8

Python, 134 129 chars

def f(x):d=[cmp(*i)for i in zip(`x`,`x`[1:])]if x>9 else[0];n=d[0]>0;return all(i<0 for i in d[n::2])&all(i>0 for i in d[n<1::2])

Ungolfed:

def f(x):
    if x>9:
        d = [cmp(*i)for i in zip(`x`,`x`[1:])] #difference of x[i] and x[i+1]
    else:
        d = [0]       #trick to return False if x<10 using less chars
    n = d[0]>0        #First digit is -1 or 1?
    neg = d[n::2]     #negative numbers if x is Undulant
    pos = d[not n::2] #positive numbers if x is Undulant

    #check if all negs are -1 and all pos are 1 and return value
    return all(i<0 for i in neg) and all(i>0 for i in pos)

Answer 9

JavaScript, 88 chars

function _(i){i+='';c=i[0];f=i[a=x=1];for(g=f<c;d=i[x++];c=d)g^=a&=g?d<c:d>c;return!f^a}

In essence, turn the number into a string and compare adjacent characters, flipping the expectation for each.

Answer 10

Python, 119 108

def u(x):l=[cmp(i,j)for i,j in zip(`x`,`x`[1:])];print x>9and all([i*j<0 for i,j in zip(l,l[1:])])and l!=[0]

Answer 11

Python 105 101 100 chars

c=lambda r,t:len(r)<2 or(cmp(*r[:2])==t and c(r[1:],-t))
u=lambda x:x>9and c(`x`,cmp(*`x`[:2])or 1)

Recursive solution. c(r,t) checks if first char of r is less (t==-1) or greater (t==1) of second char, and call opposite check on shortened string.

Answer 12

J, 44 39 36 31 characters

*/2(0<#@],0>*/\)*2-/\".;' ',.":

Usage as before.

I hadn't noticed that my last edit made the inequality with 0 check completely unnecessary. :-)

Previous answer (+ explanation):

(0=+/2=/\u)*(1<#u)**/2~:/\2<:/\u=.".;' ',.":

Usage:

    (0=+/2=/\u)*(1<#u)**/2~:/\2<:/\u=.".;' ',.":461902
1

The answer has four parts:

1. u=.".;' ',.": This reads in the number as a string ":, splits it into a list of characters preceded by spaces ' ',., stitches it back together ;, converts it back to numbers ". and then stores the result u=. This basically turns 461902 into 4 6 1 9 0 2 which I find easier to process in J.

2. */2~:/\2<:/\ This operates on the value stored in u. It takes each pair of characters and checks if the left one is less than or equal to the right one 2<:/\ so 4 6 1 9 0 2 becomes 1 0 1 0 1. It then takes the result of this and checks each pair of numbers for inequality 2~:/\ so 1 0 1 0 1 becomes 1 1 1 1. Finally it multiplies them all together to get either a 0 or a 1 */ At this point we could return the answer if it weren't for 2 things: a single digit returns 1 when the question requires a 0; and equal numbers are treated the same as 'less than' so 461900 returns 1 instead of 0. Bummer. On we go...

3. (1<#u) This checks if the number of items stored in u #u is greater than 1 and returns false if it's just a single digit number.

4. (0=+/2=/\u) This takes each pair of numbers stored in u and checks for equality 2=/\u. It then sums the answers and checks if it has 0.

The results of parts 2, 3 and 4 are then multiplied together to (hopefully) produce a 1 when the number meets the requirements specified in the question.

Answer 13

CoffeeScript, 98 67 53

(n)->0!in((n[i]>=c^(n[0]<n[1])+i)%2for c,i in n[1..])

tests:

[
    '01010101' # true
    '12345'    # false
    '1010101'  # true
    '887685'   # false
    '9120734'  # true
    '090909'   # true
]

uncompressed:

undulant = (n) ->
    direction = n[0] < n[1]
    return n.split('').every (cur, i) ->
        prev = arr[i-1] or 10 * direction
        +(prev >= cur) is (direction+i)%2

Answer 14

k (41 chars)

{(x>9)&~max(=). 1_'-':'1_'(<':;>':)@\:$x}

eg.

{(x>9)&~max(=). 1_'-':'1_'(<':;>':)@\:$x}1212130659
1b

Answer 15

Python, 155 chars

g=lambda a,b:all(x>y for x,y in zip(a,b))
u=lambda D:g(D[::2],D[1::2])&g(D[2::2],D[1::2])
def U(n):D=map(int,str(n));return(n>9)&(u(D)|u([-d for d in D]))

Answer 16

C++, 94 chars

bool u(int N){int K,P,Q,U=1,D=1;while(N>9)P=N%10,Q=(N/=10)%10,K=D,D=U&Q<P,U=K&Q>P;return U^D;}

same method as my Erlang awnser with a for loop rather than recursion.

Answer 17

Haskell, 82 characters

c=cycle[(<),(>)]
l!n=n>9&&and(zipWith3($)l(show n)$tail$show n)
u n=c!n||((>):c)!n

Answer 18

Golfscript, 48

Hoping to beat J, my first time using Golfscript. Didn't quite succeed.

[`..,(<\1>]zip{..$=\-1%.$=-}%(\{.@*0<*}/abs

Answer 19

Perl/re, 139

Doing everything in regex is kind of a bad idea.

/^(?:(.)(?{local$a=$1}))?(?:(?>((.)(?(?{$a lt$3})(?{local$a=$3})|(?!)))((.)(?(?{$a gt$5})(?{local$a=$5})|(?!))))*(?2)?)(?(?{pos>1})|(?!))$/

I'm using Perl 5.12 but I think this will work on Perl 5.10. Pretty sure 5.8 is out though.

for (qw(461902 708143 1010101 123 5)) {
    print "$_ is " . (/crazy regex goes here/ ? '' : 'not ') . "undulant\n";
}

461902 is undulant
708143 is undulant
1010101 is undulant
123 is not undulant
5 is not undulant

Answer 20

PowerShell, 88

Naïve and trivial. I will golf later.

filter u{-join([char[]]"$_"|%{if($n){[Math]::Sign($n-$_)+1}$n=$_})-notmatch'1|22|00|^$'}

My test cases.

Answer 21

JavaScript, 112

function(n,d,l,c,f){while(l=n%10,n=n/10|0)d=n%10,c?c>0?d>=l?(f=0):(c=-c):d<=l?(f=0):(c=-c):(c=d-l,f=1);return f}

You only need to pass it one argument. I could probably golf this further with a for loop.

Answer 22

Erlang, 137 123 118 chars

u(N)->Q=N div 10,u(Q,N rem 10,Q>0,Q>0). u(0,_,D,U)->D or U;u(N,P,D,U)->Q=N rem 10,u(N div 10,Q,U and(Q<P),D and(Q>P)).

Answer 23

Q, 71

{$[x>9;any all a=#[;(1 -1;-1 1)](#)a:1_signum(-':){"I"$x}each -3!x;0b]}

sample usage

q){$[x>9;any all a=#[;(1 -1;-1 1)](#)a:1_signum(-':){"I"$x}each -3!x;0b]} 5
0b
q){$[x>9;any all a=#[;(1 -1;-1 1)](#)a:1_signum(-':){"I"$x}each -3!x;0b]} 10101
1b
q){$[x>9;any all a=#[;(1 -1;-1 1)](#)a:1_signum(-':){"I"$x}each -3!x;0b]} 01010
1b
q){$[x>9;any all a=#[;(1 -1;-1 1)](#)a:1_signum(-':){"I"$x}each -3!x;0b]} 134679
0b
q){$[x>9;any all a=#[;(1 -1;-1 1)](#)a:1_signum(-':){"I"$x}each -3!x;0b]} 123456
0b
q){$[x>9;any all a=#[;(1 -1;-1 1)](#)a:1_signum(-':){"I"$x}each -3!x;0b]} 132436
1b

Answer 24

perl, 78 chars

sub u{@_=split//,$_=shift;s/.(?=.)/($&cmp$_[$+[0]])+1/ge;chop;$#_&&!/00|1|22/}

Answer 25

Scala 141, 133, 129 97:

def u(n:Int):Boolean=n>9&&{
val a=n%10
val b=(n/10)%10
a!=b&&n<99||(a-b*b-(n/100)%10)<0&&u(n/10)}

With a = n % 10, b = (n/10) % 10, c = (n/100) % 10

if a > b and b < c or 
   a < b and b > c

Then a-b * b-c is either x*-y or -x*y with x and y as positive numbers, and the product is in both cases negative, but for -x*-y or x*y (a < b < c or a > b > c) the product is always positive.

The rest of the code is handling special cases: one digit, two digits, two identical digits.

Answer 26

def is_undulant(n):
   digits = [int(i) for i in list(str(n))]
   diffs = []
   for i in range(len(digits) - 1):
        diffs.append(cmp(digits[i], digits[i+1]))
   if len(diffs) == 1:
       if diffs[0] == 0:
           return False
   else:
       for i in range(len(diffs) - 1):
           if diffs[i] * diffs[i+1] != -1:
               return False
   return True

Answer 27

PHP, 151 chars

function i($n){$n=str_split($n);@$p=$n[1]==null?:($m=($n[0]>=$n[1])?-1:1)*10;foreach($n as$k=>$v)if(($k%2*2-1)*($v-$p)*@$m<=0&$p=$v)return!1;return!0;}

ungolfed(-ish):

function is_undulant($n){

      $n=str_split($n);
      @$p=$n[1]==null?:($m = ($n[0] >= $n[1])? -1 : 1)*10;
      foreach ($n as$k=>$v)
        if ( ($k%2 * 2 - 1) * ($v - $p) * @$m <= 0 & $p=$v ) return !1;

      return !0;
}

Answer 28

Different approach I think. 124 chars

def undulant(x):
    x = `x`
    return any((int(x[1])-int(x[0]))*-1**i*(int(x[i])-int(x[i-1]))>0 for i in range(2,len(x)))

Answer 29

Scala, 194 characters

def u(i:Int):Boolean={val x=i.toString;if(!x.matches("(\\d)\\1")){"(?<=.*)(\\d)(?=(\\d).*)".r.replaceAllIn(x,m=>if(m.group(1)<m.group(2))"<"else">").matches("((<>)+<?)|((><)+>?)\\d")}else false}

Hmm...regex seemed like a good idea at the time...

Answer 30

C# 147 142

bool U(int i){if(i<10)return 0>1;var s=i+"";var b=s[0]>s[1];for(int j=2;j<s.Length;j++){var x=s[j-1]>s[j];if(x==b)return 0>1;b=x;}return 1>0;}

Working sample: http://ideone.com/b8ubr