prepend,append-Sequence

Question

Task

The prepend,append-Sequence is defined recursively, like this

• a(1) = 1
• a(n) = a(n-1).n , if n is even
• a(n) = n.a(n-1) , if n is odd

where the . represents an integer concatenation.

So the first few terms are: 1,12,312,3124,53124,531246,7531246,... This is A053064.

Your task is, given an integer a> 0 to return n, such that the nth element in the prepend,append-Sequence is equal to a and if no such n exists return 0, a negative number or error out etc.

Rules

• Input can be taken as an integer, string, list of characters/digits etc.
• Output can be printed to STDOUT or returned (integer, string etc. is fine)
• On invalid input & in the case no such n exists your program may do anything but return a positive integer (eg. loop forever, return 0 etc.)
• You may choose to use 0-indexing, but then the output in case no n exists cannot be 0

Test cases

1 -> 1
12 -> 2
21 -> 0
123 -> 0
312 -> 3
213 -> 0
211917151311975312468101214161820 -> 21
2119171513119753102468101214161820 -> 0
333129272523211917151311975312468101214161820222426283031 -> 0
999795939189878583817977757371696765636159575553514947454341393735333129272523211917151311975312468101214161820222426283032343638404244464850525456586062646668707274767880828486889092949698100 -> 100

• Mr. Xcoder: More formal: a(n-1)*(int(log(n))+1)+n and n*(int(log(n))+1)+a(n-1)?
• Wheat Wizard: @Mr.Xcoder I would call that less formal :P
• Mr. Xcoder: @JonathanAllan That is already in the question for ~10 minutes.
• user41805: I suggest allowing errors for invalid inputs.
• Mr. Xcoder: I suggest allowing undefined behaviour for invalid inputs.

Answer 1

JavaScript (ES6), 40 bytes

Takes input as a string. Throws a recursion error if no index is found.

f=(n,s=k='1')=>n==s?k:f(n,++k&1?k+s:s+k)

Demo

f=(n,s=k='1')=>n==s?k:f(n,++k&1?k+s:s+k)
console.log(f('1')) // 1
console.log(f('12')) // 2
console.log(f('312')) // 3
console.log(f('211917151311975312468101214161820')) // 21
console.log(f('999795939189878583817977757371696765636159575553514947454341393735333129272523211917151311975312468101214161820222426283032343638404244464850525456586062646668707274767880828486889092949698100')) // 100

• Rick Hitchcock: I think you can save a byte with this: f=(n,s=k='1')=>n-s?f(n,++k&1?k+s:s+k):k
• Arnauld: @RickHitchcock Unfortunately, that would force Number comparisons and introduce false positives on large inputs caused by the loss of precision.
• Rick Hitchcock: Gotcha. It works on the test cases but was unsure how it would handle other situations.

Answer 2

C# (.NET Core), (削除) 83, 80, 60 (削除ここまで) 59 bytes

n=>{int i=0;for(var t="";t!=n;)t=++i%2<1?t+i:i+t;return i;}

Try it online!

Takes the input as a string into a lambda function. 1-indexed. Returns the index of the value for truthy, or infitnitely loops for a "falsey"

Answer 3

Python 2, 63 bytes

-1 byte thanks to @EriktheOutgolfer.

f=lambda x,i='1',j=2:i!=`x`and f(x,[i+`j`,`j`+i][j%2],j+1)or~-j

Try it online!

Python 2, 64 bytes

-18 bytes thanks to @officialaimm, because I didn't notice erroring out was allowed!

x,i,j=input(),'1',1
while i!=x:j+=1;i=[i+`j`,`j`+i][j%2]
print j

Try it online!

Python 2, 82 bytes (does not loop forever)

This one returns 0 for invalid inputs.

def f(n,t="",i=1):
 while len(t)<len(n):t=[t+`i`,`i`+t][i%2];i+=1
 print(n==t)*~-i

Try it online!

• 0xffcourse: Ninja'd :D 65 bytes
• Mr. Xcoder: @officialaimm Thanks a lot! I didn't notice erroring out / loop forever was allowed.
• Erik the Outgolfer: Save a byte with a lambda: f=lambda x,i='1',j=2:i!=`x`and f(x,[i+`j`,`j`+i][j%2],j+1)or~-j
• Mr. Xcoder: @EriktheOutgolfer Wait, it throws recursion error for everything, although I set sys.setrecursionlimit(). Can you provide a tio?
• Erik the Outgolfer: @Mr.Xcoder Does it throw an error for x=1? Or x=12? I thought it only threw such an error for at least x=151311975312468101214 or something.

Answer 4

Jelly, 12 bytes

Rs2ZU1¦ẎVμ€i

Try it online!

Explanation:

Rs2ZU1¦ẎVμ€i
 μ€ Eval this link for each (automatic [1..n] range)
R Range
 s2 Split in pieces of: 2
 Z Zip
 U1¦ Only keep index: 1 of: Vectorized reverse
 Ẏ Flatten 1-deep
 V Concatenate string versions and eval
 i Find index of y in x (y = implicit input)

Answer 5

05AB1E, 14 bytes

$vDNÌNFs}«})Ik

Try it online! or as a Test suite

Explanation

0-indexed.
Returns -1 if the input is not in the sequence.

$ # push 1 and input
 v # for each y,N (element, index) in input do:
 D # duplicate top of stack
 NÌ # push N+2
 NF } # N times do:
 s # swap the top 2 elements on the stack
 « # concatenate the top 2 elements on the stack
 }) # end loop and wrap in a list
 Ik # get the index of the input in this list

• Okx: Haha, this is basically my solution with the g removed and the append/prepend thing shortened. I'll delete my answer
• Emigna: @Okx: Oh yeah, I see you golfed yours down to almost exactly this only minutes after my post. Great minds ;)

Answer 6

R, 73 bytes

p=paste0;n=scan(,'');l='';while(l!=n){F=F+1;l="if"(F%%2,p(F,l),p(l,F))};F

Reads from stdin and returns the value of the index (implicitly printed). Infinite loops when the value isn't in the sequence. F is by default FALSE which is cast to 0 when used in arithmetic.

Try it online!

Answer 7

Haskell, (削除) 75 (削除ここまで) (削除) 71 (削除ここまで) 57 bytes

f n=[i|i<-[1..],(show=<<reverse[1,3..i]++[2,4..i])==n]!!0

Takes n as a string.

Try it online!

Answer 8

Husk, ^{(削除) 17 (削除ここまで)} 15 bytes

£moiṁsṁṠehGJC2N

Try it online!

1-indexed. Returns 0 if not in the sequence.

-2 bytes from Leo, GJ!

• Dominic van Essen: Nice, particularly the z*İ_ trick to reverse alternate elements. 15 bytes by using string input...
• Leo: String input doesn't really work because values are not lexicographically ordered: Try it online!
• Leo: There is a shorter way by scanning with J, though: Try it online!

Answer 9

Vyxal, 12 bytes

ɾɖ≬wJṘyRYvṅḟ

Try it Online! Uses 0-indexing, outputs -1 for invalid.

 ḟ # Find index of input in...
 ɖ # Scan
ɾ # range(1, input)
 ≬--- # by
 wJ # Concatenate
 Ṙ # Reverse
 yRY # Reverse every other element
 vṅ # Concatenate each list into a number

Answer 10

Mathematica, 135 bytes

s=t={};x=1;While[x<5!,{s~AppendTo~#&,s~PrependTo~#&}[[x~Mod~2+1]]@x;AppendTo[t,FromDigits@Flatten[IntegerDigits/@s]];x++];t~Position~#&

Answer 11

Jelly, (削除) 19 18 (削除ここまで) 15 bytes

+ḂḶṚm2;RḤ$ṁμ€Vi

A monadic link taking and returning integers.

Try it online! (very slow - takes ~50s on TIO just to confirm that 3124 is at index 4)

For a much faster version use the previous 18 byter (only checks up to the length of the input, which is sufficient).

How?

+ḂḶṚm2;RḤ$ṁμ€Vi - Link: number, v
 μ€ - perform the monadic link to the left for €ach k in [1,2,3,...v]
 - (v can be big, lots of k values makes it slow!)
 Ḃ - modulo k by 2 = 1 if odd 0 if even
+ - add to k = k+isOdd(k)
 Ḷ - lowered range = [0,1,2,...,k+isOdd(k)]
 Ṛ - reverse = [k+isOdd(k),...,2,1,0])
 m2 - modulo slice by 2 = [k+isOdd(k),k+isOdd(k)-2,...,3,1]
 $ - last two links as a monad:
 R - range(k) = [1,2,3,...,k]
 Ḥ - double = [2,4,6,...,2k]
 ; - concatenate = [k+isOdd(k),k+isOdd(k)-2,...,3,1,2,4,6,...,2k]
 ṁ - mould like range(k) = [k+isOdd(k),k+isOdd(k)-2,...,3,1,2,4,6,...,k-isOdd(k)]
 - (this is a list of the integers to be concatenated for index k)
 V - evaluate as Jelly code (yields a list of the concatenated integers)
 i - first index of v in that (or 0 if not found)

• Okx: How long would it take to compute 211917151311975312468101214161820?
• Jonathan Allan: A long, long time :p
• Okx: Yes, but how long?
• Jonathan Allan: Well looks like it's order v squared where v is the input integer.
• Erik the Outgolfer: @JonathanAllan Technically you call that 8 :p

Answer 12

Swift 4, 92 bytes

This loops infinitely for invalid cases, so I didn't include them in the testing link.

func f(x:String){var i="1",j=1;while i != x{j+=1;i=[i+String(j),String(j)+i][j%2]};print(j)}

Test Suite.

Amusingly, it is longer with a closure:

var f:(String)->Int={var i="1",j=1;while i != 0ドル{j+=1;i=[i+String(j),String(j)+i][j%2]};return j}

Test Suite.

Answer 13

Haskell, (削除) 115 (削除ここまで) 85 bytes

s=read.(show=<<)
f 1=1
f x|odd x=s[x,f$x-1]
f x=s[f$x-1,x]
g x=[n|n<-[1..],x==f n]!!0

Try it online!

• Wheat Wizard: @BruceForte I actually managed to save 30 thanks to your suggestion.

Answer 14

Perl 5, 54 + 1 (-n) = 55 bytes

$a=++,ドル%2?,ドル.$a:$a.,ドルwhile length$a<length;say/$a/&&,ドル

Try it online!

Returns nothing if not found.

Answer 15

Japt, 17 bytes

@P=PiXv n X;\P}a1

Try it online! or check (valid) test cases

Takes input as a string or integer. On invalid input, continues "forever" looking for a solution (thus why I didn't include them in the test cases).

Explanation:

@P=PiXv n X;\P}a1 
 # Implicitly start with P = ""
@ }a1 # Repeat the function for each integer X > 0 until one returns true
 P=Pi X # Insert X into P at index:
 Xv # X is divisible by 2?
 n # Times -1
 ;\P # Return whether P now equals the input
 # Implicitly output the last value of X

The "insert at index" portion might be a bit confusing, so I'll add more detail here. In Japt, NvD is a function which returns 1 if N is divisible by D, and 0 otherwise. If D is not provided, it defaults to 2. Thus Xv here is equal to 1 if X is even, and 0 if X is odd. NnD is a function that returns D - N. If D is not provided it defaults to 0, effectively returning -N. In this program, that results in -1 if X is even, and still 0 if X is odd. Finally, in Japt indexing negative numbers count from the end of a string or array. Thus the segment PiXv n X evaluates to Pi0 X if X is odd, prepending X, but it evaluates to Pi-1 X if X is even, appending it instead.