一个波状号码是一个号码，其中其数字之间交替向上和向下像下面的编号：461902或708143，或甚至1010101，但不是123，因为2 <3。

写一个程序或功能，如果一个数是它返回一个truthy值起伏，否则一个falsy值。最短的代码获胜。

注意：一位数字是有效输入，但不视为udulant，因此isUndulant对于n <10 ，返回false。

J，45岁

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

样品使用：

   *./(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

我敢肯定，有一种更好的方法可以扭曲Insert /来一次完成更多的工作，但是我已经几个月没有J了，我需要重新开始。

Ruby，72 70个字符

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]}

用法和测试用例：

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

一位数字为false：

p u[5]       # <= false

连续的相同数字也返回false：

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

J，30个字节

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

与其他J答案不同的方法。

   * / 0 <（＃，] * {。* 1 _1 $〜＃）2-/ \ ai“：461902
1个
   * / 0 <（＃，] * {。* 1 _1 $〜＃）2-/ \ ai“：708143
1个
   * / 0 <（＃，] * {。* 1 _1 $〜＃）2-/ \ ai“：1010101
1个
   * / 0 <（＃，] * {。* 1 _1 $〜＃）2-/ \ ai“：123
0
   * / 0 <（＃，] * {。* 1 _1 $〜＃）（} .-} :) ai“：5
0

如果5个被认为是多余的，将短3个字符。

（pdf）eTeX，129个字符

\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

编译时pdfetex filename.tex 1324?提供pdf输出。TeX主要是一种排版语言，而输出到stdout大约需要20个字符。同样，对一位数字的奇怪要求（错误而不是真实）要花我26个字符。

Haskell，88 77 73 65个字符

z=tail>>=zipWith compare
q[]=0>1
q s=all(/=EQ)$s++z s
u=q.z.show

这需要常用的语言编译（或-X标志）NoMonomorphismRestriction。如果您不同意，我们必须添加4个字符并定义z：

z s=zipWith compare s$tail s

贤者，83 76字节

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

有了使用JBernardo的cmp（* [..]）的想法。在Sage中，uniq(...)是的别名list(set(...))。

编辑：只是注意到对于x <10，uniq(cmp(...)) == []不在上[[1],[-1]]。如果将x作为字符串而不是整数输入，我可以再输入4个字符！

Python：101 100个字符

缩小之前：

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:])))))

缩小后：

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:]))))

Python，134129个字符

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])

取消高尔夫：

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)

JavaScript，88个字符

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}

本质上，将数字转换为字符串并比较相邻的字符，将每个字符的期望值翻转。

K，41个字节

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

例如

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

CoffeeScript， 98 67 53个字节

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

测试：

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

未压缩：

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

J，44 39 36 31字节

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

用法与以前相同。

我没有注意到我的上一次编辑使得完全不需要使用0的不等式。:-)

先前的答案（+说明）：

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

用法：

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

答案分为四个部分：

1. u=.".;' ',.": 这将数字读为字符串":，将其拆分为以空格' ',.开头的字符列表，将其缝合在一起;，将其转换回数字".，然后存储结果，u=.这基本上使461902变成了4 6 1 9 0 2用J处理

2. */2~:/\2<:/\ 这对存储在u中的值进行运算。它获取每对字符，并检查左字符是否小于或等于右字符，2<:/\因此4 6 1 9 0 2变为1 0 1 01。然后获取此结果，并检查每对数字的不等式2~:/\所以1 0 1 0 1变成1 1 11。最后，将它们全部相乘得到0或1。*/这时如果不是两件事情，我们可以返回答案：当问题要求为0；和相等的数字被视为与“小于”相同，因此461900返回1而不是0。Bummer。走吧...

3. (1<#u) 这将检查u中存储的项目数#u是否大于1，如果它只是一个数字，则返回false。

4. (0=+/2=/\u) 这将获取存储在u中的每对数字并检查是否相等2=/\u。然后，它汇总答案并检查其是否为0。

然后，将数字2、3和4的结果相乘，以使数字满足问题中指定的要求（希望）为1。

Haskell，82个字节

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

在线尝试！

Python，119108字节

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]

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]))

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.

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.

Perl/re, 139 bytes

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

GolfScript, 48 bytes

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

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

JavaScript, 66 65 62 60 bytes

Takes input as a string, returns true for undulant numbers, an empty string (falsey) for single digit numbers and false otherwise.

([s,...a])=>a+a&&a.every(x=>eval(s+"<>"[++y%2]+x,s=x),y=s<a)

Try it

Run the Snippet below to test 0-9 and 25 random numbers <10,000,000.

f=
([s,...a])=>a+a&&a.every(x=>eval(s+"<>"[++y%2]+x,s=x),y=s<a)
tests=new Set([...Array(10).keys()])
while(tests.add(Math.random()*1e7|0).size<35);
o.innerText=[...tests].map(x=>(x=x+``).padStart(7)+` = `+JSON.stringify(f(x))).join`\n`

<pre id=o></pre>

Explanation

A few fun little tricks in this one so I think it warrants a rare explanation to a JS solution from me.

()=>

We start, simply, with an anonymous function which takes the integer string as an argument when called.

[s,...a]

That argument is immediately destructured into 2 parameters: s being the first character in the string and a being an array containing the remaining characters (e.g. "461902" becomes s="4" and a=["6","1","9","0","2"]).

a+a&&

First, we concatenate a with itself, which casts both occurrences to strings. If the input is a single digit number then a will be empty and, therefore, become and empty string; an empty string plus an empty string is still an empty string and, because that's falsey in JS, we stop processing at the logical AND and output our empty string. In all other cases a+a will be truthy and so we continue on to the next part of the function.

a.every(x=>)

We'll be checking if every element x in a returns true when passed through a function.

y=s<a

This determines what our first comparison will be (< or >) and then we'll alternate from there. We check if the string s is less than the array a, which gets cast to a string in the process so, if s is less than the first character in a, y will be true or false if it's not.

s+"<>"[++y%2]+x

We build a string with the current value of s at the beginning and x at the end. In between, we index into the string "<>" by incrementing y, casting its initial boolean value to an integer, and modulo by 2, giving us 0 or 1.

eval()

Eval that string.

s=x

Finally, we pass a second argument to eval, which it ignores, and use it to set the value of s to the current value of x for the next iteration.

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.

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.

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)).

CJam, 30 bytes

CJam is newer than this challenge, so this does not compete for the green checkmark, but it's not a winner anyway (although I'm sure this can actually be golfed quite a bit).

l"_1=\+{_@-\}*;]"_8'*t+~{W>},!

Test it here.

How it works

Firstly, I'm doing some string manipulation (followed by eval) to save 5 bytes on duplicate code:

"..."_8'*t+~
"..."        "Push this string.":
     _       "Duplicate.";
      8'*t   "Replace the 8th character (the -) with *.";
          +~ "Concatenate the strings and evaluate.";

So in effect my code is

l_1=\+{_@-\}*;]_1=\+{_@*\}*;]{W>},!

First, here is how I deal with the weird special case of a single digit. I copy the digit at index 1 and prepend it to the number. We need to distinguish 3 cases:

• The first two digits are different, like 12..., then we get 212..., so the start is undulant, and won't affect whether the entire number is undulant.
• The first two digits are the same, like 11..., then we get 111.... Now the start is not undulant, but the number wasn't undulant anyway, so this won't affect the result either.
• If the number only has one digit, the digit at index 1 will be the first digit (because CJam's array indexing loops around the end), so this results in two identical digits, and the number is not undulant.

Now looking at the code in detail:

l_1=\+{_@-\}*;]_1=\+{_@*\}*;]{W>},!
l                                   "Read input.";
 _1=\+                              "Prepend second digit.";
      {_@-\}*                       "This fold gets the differences of consecutive elments.";
             ;]                     "Drop the final element and collect in an aray.";
               _1=\+                "Prepend second element.";
                    {_@*\}*         "This fold gets the products of consecutive elments.";
                           ;]       "Drop the final element and collect in an aray.";
                             {W>},  "Filter out non-negative numbers.";
                                  ! "Logical not.";

I'm sure there is a shorter way to actually check digits (of length greater 1) for whether they are undulant (in particular, without using two folds), but I couldn't find it yet.

Prolog 87 bytes

u(X) :- number_codes(X,C),f(C).
f([_,_]).
f([A,B,C|L]) :- (A<B,B>C;A>B,B<C),f([B,C|L]).

To run it, just save it as golf.pl, open a prolog interpreter (e.g. gprolog) in the same directory then do:

consult(golf).
u(101010).

It will give true if the number is undulant, otherwise just no.

Mathematica, 46 bytes

#!=Sort@#&&#!=Reverse@Sort@#&[IntegerDigits@n]

Examples (spaces are not required):

# != Sort@# && # != Reverse@Sort@# &[IntegerDigits@5]
# != Sort@# && # != Reverse@Sort@# &[IntegerDigits@123]
# != Sort@# && # != Reverse@Sort@# &[IntegerDigits@132]
# != Sort@# && # != Reverse@Sort@# &[IntegerDigits@321]

(*  out *)
False  False  True  False

Scala, 141 133 129 97 bytes

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.

Perl, 78 bytes

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

Q, 71 bytes

{$[x>9;any all a=#[;(1 -1;-1 1)](#)a:1_signum(-':){"I"$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

Julia 0.6, 62 bytes

f(x,a=sign.(diff(digits(x))))=x>9&&-a*a[1]==(-1).^(1:endof(a))

Takes in a number, returns true for Undulant, and false for not. Eg f(163) returns true.

f(x,a=sign.(diff(digits(x))))=x>9&&-a*a[1]==(-1).^(1:endof(a))
f(x,                        )                                   # function definition
    a=sign.(diff(digits(x)))                                    # default 2nd argument is array of differences of signs of digits
                              x>9&&                             # short circuiting and to catch cases under 10
                                   -a*a[1]                      # make the first element of a always -1
                                          ==(-1).^(1:endof(a))  # check that a is an array of alternating -1 and 1 of correct length

Try it online!