Mixing
Extract and Divide
Clash Royale CLAN TAG#URR8PPP
up vote
7
down vote
favorite
Challenge
For a given positive integer $n$:
- Repeat the following until $n < 10$ (until $n$ contains one digit).
- Extract the last digit.
- If the extracted digit is even (including 0) multiply the rest of the integer by $2$ and add $1$ ( $2n+1$ ). Then go back to
step 1else move tostep 4. - Divide the rest of the integer with the extracted digit (integer / digit) and add the remainder (integer % digit), that is your new $n$.
Note: In case n < 10 at the very beginning obviously you don't move to step 2. Skip everything and print $[n, 0]$.
Input
n: Positive integer.
Output
[o, r]: o: the last n, r: number of repeats.
Step-by-step example
For input n = 61407:
Repeat 1: 6140_7 -> 6140 / 7 = 877, 6140 % 7 = 1 => 877 + 1 = 878
Repeat 2: 87_8 -> 2 * 87 + 1 = 175 (even digit)
Repeat 3: 17_5 -> 17 / 5 = 3, 17 % 5 = 2 => 3 + 2 = 5 ( < 10, we are done )
The output is [5, 3] (last n is 5 and 3 steps repeats).
Rules
There are really no rules or restrictions just assume n > 0. You can either print or return the output.
code-golf number arithmetic division
asked Sep 1 at 15:26
DimChtz
601111
 |Â
show 1 more comment
up vote
7
down vote
favorite
Challenge
For a given positive integer $n$:
- Repeat the following until $n < 10$ (until $n$ contains one digit).
- Extract the last digit.
- If the extracted digit is even (including 0) multiply the rest of the integer by $2$ and add $1$ ( $2n+1$ ). Then go back to
step 1else move tostep 4. - Divide the rest of the integer with the extracted digit (integer / digit) and add the remainder (integer % digit), that is your new $n$.
Note: In case n < 10 at the very beginning obviously you don't move to step 2. Skip everything and print $[n, 0]$.
Input
n: Positive integer.
Output
[o, r]: o: the last n, r: number of repeats.
Step-by-step example
For input n = 61407:
Repeat 1: 6140_7 -> 6140 / 7 = 877, 6140 % 7 = 1 => 877 + 1 = 878
Repeat 2: 87_8 -> 2 * 87 + 1 = 175 (even digit)
Repeat 3: 17_5 -> 17 / 5 = 3, 17 % 5 = 2 => 3 + 2 = 5 ( < 10, we are done )
The output is [5, 3] (last n is 5 and 3 steps repeats).
Rules
There are really no rules or restrictions just assume n > 0. You can either print or return the output.
code-golf number arithmetic division
asked Sep 1 at 15:26
DimChtz
601111
I'm pretty sure that most answers from here can be trivially modified to answer this question.
– Rushabh Mehta
Sep 1 at 16:00
@Arnauld No, the output needs to be two numbers (returned or printed). I guess you could convert the decimal number to string, replace.with whitespace and print it.
– DimChtz
Sep 1 at 16:06
1
Do we enter step 2 if n<10?
– AlexRacer
Sep 1 at 16:55
@Arnauld That is not the same for n=2 and n=4.
– AlexRacer
Sep 1 at 17:00
1
Yes. It's much clearer now. :)
– Arnauld
Sep 1 at 23:42
 |Â
show 1 more comment
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Challenge
For a given positive integer $n$:
- Repeat the following until $n < 10$ (until $n$ contains one digit).
- Extract the last digit.
- If the extracted digit is even (including 0) multiply the rest of the integer by $2$ and add $1$ ( $2n+1$ ). Then go back to
step 1else move tostep 4. - Divide the rest of the integer with the extracted digit (integer / digit) and add the remainder (integer % digit), that is your new $n$.
Note: In case n < 10 at the very beginning obviously you don't move to step 2. Skip everything and print $[n, 0]$.
Input
n: Positive integer.
Output
[o, r]: o: the last n, r: number of repeats.
Step-by-step example
For input n = 61407:
Repeat 1: 6140_7 -> 6140 / 7 = 877, 6140 % 7 = 1 => 877 + 1 = 878
Repeat 2: 87_8 -> 2 * 87 + 1 = 175 (even digit)
Repeat 3: 17_5 -> 17 / 5 = 3, 17 % 5 = 2 => 3 + 2 = 5 ( < 10, we are done )
The output is [5, 3] (last n is 5 and 3 steps repeats).
Rules
There are really no rules or restrictions just assume n > 0. You can either print or return the output.
code-golf number arithmetic division
asked Sep 1 at 15:26
DimChtz
601111
Challenge
For a given positive integer $n$:
- Repeat the following until $n < 10$ (until $n$ contains one digit).
- Extract the last digit.
- If the extracted digit is even (including 0) multiply the rest of the integer by $2$ and add $1$ ( $2n+1$ ). Then go back to
step 1else move tostep 4. - Divide the rest of the integer with the extracted digit (integer / digit) and add the remainder (integer % digit), that is your new $n$.
Note: In case n < 10 at the very beginning obviously you don't move to step 2. Skip everything and print $[n, 0]$.
Input
n: Positive integer.
Output
[o, r]: o: the last n, r: number of repeats.
Step-by-step example
For input n = 61407:
Repeat 1: 6140_7 -> 6140 / 7 = 877, 6140 % 7 = 1 => 877 + 1 = 878
Repeat 2: 87_8 -> 2 * 87 + 1 = 175 (even digit)
Repeat 3: 17_5 -> 17 / 5 = 3, 17 % 5 = 2 => 3 + 2 = 5 ( < 10, we are done )
The output is [5, 3] (last n is 5 and 3 steps repeats).
Rules
There are really no rules or restrictions just assume n > 0. You can either print or return the output.
code-golf number arithmetic division
asked Sep 1 at 15:26
DimChtz
601111
edited Sep 2 at 12:12
asked Sep 1 at 15:26
DimChtz
601111
asked Sep 1 at 15:26
DimChtz
601111
asked Sep 1 at 15:26
DimChtz
601111
601111
I'm pretty sure that most answers from here can be trivially modified to answer this question.
– Rushabh Mehta
Sep 1 at 16:00
@Arnauld No, the output needs to be two numbers (returned or printed). I guess you could convert the decimal number to string, replace.with whitespace and print it.
– DimChtz
Sep 1 at 16:06
1
Do we enter step 2 if n<10?
– AlexRacer
Sep 1 at 16:55
@Arnauld That is not the same for n=2 and n=4.
– AlexRacer
Sep 1 at 17:00
1
Yes. It's much clearer now. :)
– Arnauld
Sep 1 at 23:42
 |Â
show 1 more comment
I'm pretty sure that most answers from here can be trivially modified to answer this question.
– Rushabh Mehta
Sep 1 at 16:00
@Arnauld No, the output needs to be two numbers (returned or printed). I guess you could convert the decimal number to string, replace.with whitespace and print it.
– DimChtz
Sep 1 at 16:06
1
Do we enter step 2 if n<10?
– AlexRacer
Sep 1 at 16:55
@Arnauld That is not the same for n=2 and n=4.
– AlexRacer
Sep 1 at 17:00
1
Yes. It's much clearer now. :)
– Arnauld
Sep 1 at 23:42
I'm pretty sure that most answers from here can be trivially modified to answer this question.
– Rushabh Mehta
Sep 1 at 16:00
I'm pretty sure that most answers from here can be trivially modified to answer this question.
– Rushabh Mehta
Sep 1 at 16:00
@Arnauld No, the output needs to be two numbers (returned or printed). I guess you could convert the decimal number to string, replace
. with whitespace and print it.– DimChtz
Sep 1 at 16:06
@Arnauld No, the output needs to be two numbers (returned or printed). I guess you could convert the decimal number to string, replace
. with whitespace and print it.– DimChtz
Sep 1 at 16:06
1
1
Do we enter step 2 if n<10?
– AlexRacer
Sep 1 at 16:55
Do we enter step 2 if n<10?
– AlexRacer
Sep 1 at 16:55
@Arnauld That is not the same for n=2 and n=4.
– AlexRacer
Sep 1 at 17:00
@Arnauld That is not the same for n=2 and n=4.
– AlexRacer
Sep 1 at 17:00
1
1
Yes. It's much clearer now. :)
– Arnauld
Sep 1 at 23:42
Yes. It's much clearer now. :)
– Arnauld
Sep 1 at 23:42
 |Â
show 1 more comment
17 Answers
17
active
oldest
votes
up vote
6
down vote
R, 93 bytes
function(n,r=0,e=n%%10,d=(n-e)/10)"if"(n<10,c(n,r),"if"(e%%2,f(d%/%e+d%%e,r+1),f(d*2+1,r+1)))
Try it online!
answered Sep 1 at 19:28
duckmayr
43115
2
Save some bytes by getting rid of second "if", reversingn<10condition and using%/%in d definition : TIO
– Kirill L.
Sep 1 at 20:24
add a comment |Â
up vote
4
down vote
Python 2, 78 77 76 73 bytes
Recursive version. Saved 3 bytes thanks to Jonathan Frech.
f=lambda n,r=0:n>9and f(n%2*sum(divmod(n/10,n%10))or n/10*2+1,-~r)or[n,r]
Try it online!
74 bytes
Full program.
n,i=input(),0
while n>9:i+=1;r,R=n/10,n%10;n=[r-~r,r/R+r%R][R&1]
print n,i
Try it online!
answered Sep 1 at 16:20
Mr. Xcoder
30.3k758193
1
Since2divides10, isn%10%2not equivalent ton%2?
– Jonathan Frech
Sep 1 at 23:47
1
Great point by @JonathanFrech that will get you to 73 bytes. I think the following form would also work, and is also 73 bytes:def g(n,r=0):u,v=n/10,n%10;return u and g([u-~u,u/v+u%v][v&1],r+1)or(n,r)
– mathmandan
Sep 2 at 21:16
Somehow Jonathan's message did not appear in my inbox... Indeed, you're right, editing ASAP.
– Mr. Xcoder
Sep 2 at 21:17
add a comment |Â
up vote
3
down vote
JavaScript (ES6), 62 59 bytes
f=(n,r=0)=>(x=n/10|0)?f((n%=10)&1?x/n+x%n|0:x-~x,r+1):[n,r]
Try it online!
Commented
f = (n, r = 0) => // n = input, r = number of iterations
(x = n / 10 | 0) ? // x = floor(n / 10); if x is not equal to 0:
f( // do a recursive call:
(n %= 10) & 1 ? // n = last digit; if odd:
x / n + x % n | 0 // use quotient + remainder of x / n
: // else:
x - ~x, // use x - (-x - 1) = x + x + 1 = 2x + 1
r + 1 // increment the number of iterations
) // end of recursive call
: // else:
[n, r] // return [ final_result, iterations ]
answered Sep 1 at 15:58
Arnauld
63.6k580268
add a comment |Â
up vote
3
down vote
05AB1E (legacy), 19 bytes
[DgD–#ÃÂćDÈi·>ë‰O]?
Explanation:
[ Start infinite loop
DgD Get the length
–# If it's 1, print the loop index and break.
ÃÂćD Extract: "hello" -> "hell", "o"
Èi if even,
·> Double and increment
ë else,
‰O Divmod, and sum the result (div and mod).
] End if statement & infinite loop
? Print
Try it online!
answered Sep 1 at 16:51
Okx
12k27100
add a comment |Â
up vote
3
down vote
Julia 1.0, 81 80 79 bytes
function d(n,c=0)l,r=n÷10,n%10;n>9 ? d(r%2<1 ? 2l+1 : l÷r+l%r,c+1) : [n,c]end
Try it online!
- -1 (thanks Mr. Xcoder)
answered Sep 1 at 19:09
Rustem B.
513
Whis is my first golfcode :)
– Rustem B.
Sep 1 at 19:11
Welcome to PPCG!
– Giuseppe
Sep 1 at 19:45
@Giuseppe thanks
– Rustem B.
Sep 1 at 19:49
1
Save a byte using<1rather than==0.
– Mr. Xcoder
Sep 1 at 20:50
add a comment |Â
up vote
2
down vote
Ruby, 61 bytes
f=->n,r=0n>9?(a=n%10;n/=10;f[a%2<1?n-~n:n/a+n%a,r+1]):[n,r]
Try it online!
answered Sep 1 at 18:10
Kirill L.
2,4061116
add a comment |Â
up vote
2
down vote
Retina, 79 bytes
(d+)[02468]$
;$.(_2*$1*
(d+)(.)
$2*_;$1*
}`(_+);(1)*(_*)
;$.($3$#2*
^;*
$.&;
Try it online! Explanation:
(d+)[02468]$
;$.(_2*$1*
If the last digit is even, multiply the rest of the integer by 2 and add 1. This always results in an odd number, so we don't need to repeat this.
(d+)(.)
$2*_;$1*
Split off the last digit and convert to unary for the divmod.
}`(_+);(1)*(_*)
;$.($3$#2*
Divide the rest of the integer by the last digit and add the remainder. Then repeat the whole program until there's only one digit left.
^;*
$.&;
Count the number of operations performed.
answered Sep 1 at 20:50
Neil
75.1k744170
add a comment |Â
up vote
2
down vote
Perl 6, 61 bytes
tail kv $_,0!!$0*2+1...!/(.+)(.)/: 2
Try it online!
Returns (r, o).
answered Sep 1 at 22:01
nwellnhof
3,503714
add a comment |Â
up vote
2
down vote
Brachylog, 43 bytes
Feels messy and bad but here it is anyway
;0hl1&
Try it online!
answered Sep 2 at 9:10
Kroppeb
90628
add a comment |Â
up vote
2
down vote
Swift 4, 112 104 100 93 bytes
var n=Int(readLine()!)!,r=0,p=n%10;while n>9n=[n/10*2+1,n/10/p+n/10%p][n%2];r+=1;print(n,r)
Try it online!
Prints n r
- -7 (Thanks to Mr. Xcoder)
answered Sep 2 at 7:52
paper1111
1214
Very nice first answer (+1)!var n=Int(readLine()!)!,r=0,p=n%10;while n>9n=[n/10*2+1,n/10/p+n/10%p][n%2];r+=1;print(n,r)should save you 7 bytes (golfing the conditional).
– Mr. Xcoder
Sep 3 at 12:24
@Mr.Xcoder Thanks!
– paper1111
Sep 4 at 12:28
add a comment |Â
up vote
1
down vote
Jelly, 19 bytes
dd/Sɗ:Ḥ‘ɗḂ?ƬâµṖṖṪ,LƊ
Try it online!
answered Sep 1 at 21:08
Erik the Outgolfer
29.4k42698
add a comment |Â
up vote
1
down vote
Common Lisp, 145 bytes
A function which takes the n provided and returns the list (n r). Lisp syntax is fun.
(defun f(n &optional(r 0))(if(< n 10)`(,n,r)(let((h(floor n 10))(l(mod n 10)))(if(evenp l)(f(1+(* 2 h))(1+ r))(f(+(mod h l)(floor h l))(1+ r)))))
answered Sep 2 at 6:50
theemacsshibe
214
1
Welcome to the site =D
– Luis felipe De jesus Munoz
Sep 2 at 12:48
Thankyou very much, Luis.
– theemacsshibe
Sep 3 at 8:46
add a comment |Â
up vote
1
down vote
Haskell, 78 bytes
f n|n<10=(n,0)|(d,m)<-divMod n 10=(1+)<$>f(last$2*d+1:[div d m+mod d m|odd n])
Try it online!
answered Sep 2 at 20:48
nimi
29.9k31881
add a comment |Â
up vote
1
down vote
VBA (Excel), 145?, 142, 139, 123 bytes
Condensed Form:
Sub t(n)
While n>10
e=Right(n, 1)
n=Left(n,Len(n)-1)
If e Mod 2=0Then n=n*2+1 Else n=Int(n/e)+n Mod e
c=c+1
wend
Debug.?n;c
End Sub
Expanded (with comments)
Sub test(n) 'For a given positive integer n
'Repeat the following until n<10
While n > 10
'Extract the last digit
e = Right(n, 1)
'Remove the last digit
n = Left(n, Len(n) - 1)
'If the extracted digit is even...
If e Mod 2 = 0 Then
'Multiply the rest of the integer by 2 and add 1
n = n * 2 + 1
Else
'Divide the rest of the integer with the extracted digit and add the remainder
n = Int(n / e) + n Mod e
End If
'Keep count
c = c + 1
'End the Loop
Wend
'Give the output
Debug.Print (n ; c)
End Sub
(Do excuse me, as this is my first post. I will happily edit! Right now, it looks like to test it you open the Immediate Window and type 't n' where n is your number to test :D)
answered Sep 2 at 21:27
seadoggie01
1316
add a comment |Â
up vote
1
down vote
8086 machine code, 35 bytes
00000000 31 c9 31 d2 bb 0a 00 f7 f3 85 c0 74 14 f6 c2 01 |1.1........t....|
00000010 75 05 d1 e0 40 e2 eb 89 d3 88 f2 f7 f3 01 d0 e2 |u...@...........|
00000020 e1 f7 d9 |...|
00000023
Input: AX = n
Output: DX = o, CX = r
Assembled from:
xor cx, cx
next: xor dx, dx
mov bx, 10
div bx
test ax, ax
jz done
test dl, 1
jnz odd
shl ax, 1
inc ax
loop next
odd: mov bx, dx
mov dl, dh
div bx
add ax, dx
loop next
done: neg cx
answered Sep 1 at 17:00
user5434231
1,556512
add a comment |Â
up vote
1
down vote
Red, 131 bytes
func[n][s: 0 while[n > 9][r: do form take/last t: form n t: do t
n: either r % 2 = 1[t / r +(t % r)][2 * t + 1]s: s + 1]print[n s]]
Try it online!
More readable:
f: func [ n ] [
s: 0
while [ n > 9 ] [
r: do form take/last t: form n
t: do t
n: either r % 2 = 1
[ t / r + (t % r) ]
[ 2 * t + 1 ]
s: s + 1
]
print [ n s ]
]
answered Sep 4 at 9:13
Galen Ivanov
4,7571829
add a comment |Â
up vote
-2
down vote
C++, 65 bytes
Assume that n, r and o variables are already defined above :)
r=0;doint m=n/10,d=n%10;n=n&1?m/d+m%d:m*2+1;++r;while(n>9);o=n;
p.s. Sorry for such design of post, I have not figured out how to do it well yet...
answered Sep 1 at 22:42
Jin X
11
1
Hello and welcome to PPCG. As it stands, your answer is not conforming to our default I/O, which essentially calls for either full programs or functions. Take a look at other C++ submissions to get a feel for how to construct these wrappers. Also, please do not be discouraged by your one downvote -- not knowing our rule set on your first post is fine and should not be punished. They may remove it if you fix your answer!
– Jonathan Frech
Sep 1 at 23:37
You may also find TIO a useful tool to both test your submission and auto-generate your post.
– Jonathan Frech
Sep 1 at 23:41
Your answer in its current form is invalid. Please either edit it to conform to our default I/O or delete it.
– Jonathan Frech
Sep 5 at 13:24
add a comment |Â
17 Answers
17
active
oldest
votes
17 Answers
17
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
R, 93 bytes
function(n,r=0,e=n%%10,d=(n-e)/10)"if"(n<10,c(n,r),"if"(e%%2,f(d%/%e+d%%e,r+1),f(d*2+1,r+1)))
Try it online!
answered Sep 1 at 19:28
duckmayr
43115
2
Save some bytes by getting rid of second "if", reversingn<10condition and using%/%in d definition : TIO
– Kirill L.
Sep 1 at 20:24
add a comment |Â
up vote
6
down vote
R, 93 bytes
function(n,r=0,e=n%%10,d=(n-e)/10)"if"(n<10,c(n,r),"if"(e%%2,f(d%/%e+d%%e,r+1),f(d*2+1,r+1)))
Try it online!
answered Sep 1 at 19:28
duckmayr
43115
2
Save some bytes by getting rid of second "if", reversingn<10condition and using%/%in d definition : TIO
– Kirill L.
Sep 1 at 20:24
add a comment |Â
up vote
6
down vote
up vote
6
down vote
R, 93 bytes
function(n,r=0,e=n%%10,d=(n-e)/10)"if"(n<10,c(n,r),"if"(e%%2,f(d%/%e+d%%e,r+1),f(d*2+1,r+1)))
Try it online!
answered Sep 1 at 19:28
duckmayr
43115
R, 93 bytes
function(n,r=0,e=n%%10,d=(n-e)/10)"if"(n<10,c(n,r),"if"(e%%2,f(d%/%e+d%%e,r+1),f(d*2+1,r+1)))
Try it online!
answered Sep 1 at 19:28
duckmayr
43115
answered Sep 1 at 19:28
duckmayr
43115
answered Sep 1 at 19:28
duckmayr
43115
answered Sep 1 at 19:28
duckmayr
43115
43115
2
Save some bytes by getting rid of second "if", reversingn<10condition and using%/%in d definition : TIO
– Kirill L.
Sep 1 at 20:24
add a comment |Â
2
Save some bytes by getting rid of second "if", reversingn<10condition and using%/%in d definition : TIO
– Kirill L.
Sep 1 at 20:24
2
2
Save some bytes by getting rid of second "if", reversing
n<10 condition and using %/% in d definition : TIO– Kirill L.
Sep 1 at 20:24
Save some bytes by getting rid of second "if", reversing
n<10 condition and using %/% in d definition : TIO– Kirill L.
Sep 1 at 20:24
add a comment |Â
up vote
4
down vote
Python 2, 78 77 76 73 bytes
Recursive version. Saved 3 bytes thanks to Jonathan Frech.
f=lambda n,r=0:n>9and f(n%2*sum(divmod(n/10,n%10))or n/10*2+1,-~r)or[n,r]
Try it online!
74 bytes
Full program.
n,i=input(),0
while n>9:i+=1;r,R=n/10,n%10;n=[r-~r,r/R+r%R][R&1]
print n,i
Try it online!
answered Sep 1 at 16:20
Mr. Xcoder
30.3k758193
1
Since2divides10, isn%10%2not equivalent ton%2?
– Jonathan Frech
Sep 1 at 23:47
1
Great point by @JonathanFrech that will get you to 73 bytes. I think the following form would also work, and is also 73 bytes:def g(n,r=0):u,v=n/10,n%10;return u and g([u-~u,u/v+u%v][v&1],r+1)or(n,r)
– mathmandan
Sep 2 at 21:16
Somehow Jonathan's message did not appear in my inbox... Indeed, you're right, editing ASAP.
– Mr. Xcoder
Sep 2 at 21:17
add a comment |Â
up vote
4
down vote
Python 2, 78 77 76 73 bytes
Recursive version. Saved 3 bytes thanks to Jonathan Frech.
f=lambda n,r=0:n>9and f(n%2*sum(divmod(n/10,n%10))or n/10*2+1,-~r)or[n,r]
Try it online!
74 bytes
Full program.
n,i=input(),0
while n>9:i+=1;r,R=n/10,n%10;n=[r-~r,r/R+r%R][R&1]
print n,i
Try it online!
answered Sep 1 at 16:20
Mr. Xcoder
30.3k758193
1
Since2divides10, isn%10%2not equivalent ton%2?
– Jonathan Frech
Sep 1 at 23:47
1
Great point by @JonathanFrech that will get you to 73 bytes. I think the following form would also work, and is also 73 bytes:def g(n,r=0):u,v=n/10,n%10;return u and g([u-~u,u/v+u%v][v&1],r+1)or(n,r)
– mathmandan
Sep 2 at 21:16
Somehow Jonathan's message did not appear in my inbox... Indeed, you're right, editing ASAP.
– Mr. Xcoder
Sep 2 at 21:17
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Python 2, 78 77 76 73 bytes
Recursive version. Saved 3 bytes thanks to Jonathan Frech.
f=lambda n,r=0:n>9and f(n%2*sum(divmod(n/10,n%10))or n/10*2+1,-~r)or[n,r]
Try it online!
74 bytes
Full program.
n,i=input(),0
while n>9:i+=1;r,R=n/10,n%10;n=[r-~r,r/R+r%R][R&1]
print n,i
Try it online!
answered Sep 1 at 16:20
Mr. Xcoder
30.3k758193
Python 2, 78 77 76 73 bytes
Recursive version. Saved 3 bytes thanks to Jonathan Frech.
f=lambda n,r=0:n>9and f(n%2*sum(divmod(n/10,n%10))or n/10*2+1,-~r)or[n,r]
Try it online!
74 bytes
Full program.
n,i=input(),0
while n>9:i+=1;r,R=n/10,n%10;n=[r-~r,r/R+r%R][R&1]
print n,i
Try it online!
answered Sep 1 at 16:20
Mr. Xcoder
30.3k758193
edited Sep 2 at 21:19
answered Sep 1 at 16:20
Mr. Xcoder
30.3k758193
answered Sep 1 at 16:20
Mr. Xcoder
30.3k758193
answered Sep 1 at 16:20
Mr. Xcoder
30.3k758193
30.3k758193
1
Since2divides10, isn%10%2not equivalent ton%2?
– Jonathan Frech
Sep 1 at 23:47
1
Great point by @JonathanFrech that will get you to 73 bytes. I think the following form would also work, and is also 73 bytes:def g(n,r=0):u,v=n/10,n%10;return u and g([u-~u,u/v+u%v][v&1],r+1)or(n,r)
– mathmandan
Sep 2 at 21:16
Somehow Jonathan's message did not appear in my inbox... Indeed, you're right, editing ASAP.
– Mr. Xcoder
Sep 2 at 21:17
add a comment |Â
1
Since2divides10, isn%10%2not equivalent ton%2?
– Jonathan Frech
Sep 1 at 23:47
1
Great point by @JonathanFrech that will get you to 73 bytes. I think the following form would also work, and is also 73 bytes:def g(n,r=0):u,v=n/10,n%10;return u and g([u-~u,u/v+u%v][v&1],r+1)or(n,r)
– mathmandan
Sep 2 at 21:16
Somehow Jonathan's message did not appear in my inbox... Indeed, you're right, editing ASAP.
– Mr. Xcoder
Sep 2 at 21:17
1
1
Since
2 divides 10, is n%10%2 not equivalent to n%2?– Jonathan Frech
Sep 1 at 23:47
Since
2 divides 10, is n%10%2 not equivalent to n%2?– Jonathan Frech
Sep 1 at 23:47
1
1
Great point by @JonathanFrech that will get you to 73 bytes. I think the following form would also work, and is also 73 bytes:
def g(n,r=0):u,v=n/10,n%10;return u and g([u-~u,u/v+u%v][v&1],r+1)or(n,r)– mathmandan
Sep 2 at 21:16
Great point by @JonathanFrech that will get you to 73 bytes. I think the following form would also work, and is also 73 bytes:
def g(n,r=0):u,v=n/10,n%10;return u and g([u-~u,u/v+u%v][v&1],r+1)or(n,r)– mathmandan
Sep 2 at 21:16
Somehow Jonathan's message did not appear in my inbox... Indeed, you're right, editing ASAP.
– Mr. Xcoder
Sep 2 at 21:17
Somehow Jonathan's message did not appear in my inbox... Indeed, you're right, editing ASAP.
– Mr. Xcoder
Sep 2 at 21:17
add a comment |Â
up vote
3
down vote
JavaScript (ES6), 62 59 bytes
f=(n,r=0)=>(x=n/10|0)?f((n%=10)&1?x/n+x%n|0:x-~x,r+1):[n,r]
Try it online!
Commented
f = (n, r = 0) => // n = input, r = number of iterations
(x = n / 10 | 0) ? // x = floor(n / 10); if x is not equal to 0:
f( // do a recursive call:
(n %= 10) & 1 ? // n = last digit; if odd:
x / n + x % n | 0 // use quotient + remainder of x / n
: // else:
x - ~x, // use x - (-x - 1) = x + x + 1 = 2x + 1
r + 1 // increment the number of iterations
) // end of recursive call
: // else:
[n, r] // return [ final_result, iterations ]
answered Sep 1 at 15:58
Arnauld
63.6k580268
add a comment |Â
up vote
3
down vote
JavaScript (ES6), 62 59 bytes
f=(n,r=0)=>(x=n/10|0)?f((n%=10)&1?x/n+x%n|0:x-~x,r+1):[n,r]
Try it online!
Commented
f = (n, r = 0) => // n = input, r = number of iterations
(x = n / 10 | 0) ? // x = floor(n / 10); if x is not equal to 0:
f( // do a recursive call:
(n %= 10) & 1 ? // n = last digit; if odd:
x / n + x % n | 0 // use quotient + remainder of x / n
: // else:
x - ~x, // use x - (-x - 1) = x + x + 1 = 2x + 1
r + 1 // increment the number of iterations
) // end of recursive call
: // else:
[n, r] // return [ final_result, iterations ]
answered Sep 1 at 15:58
Arnauld
63.6k580268
add a comment |Â
up vote
3
down vote
up vote
3
down vote
JavaScript (ES6), 62 59 bytes
f=(n,r=0)=>(x=n/10|0)?f((n%=10)&1?x/n+x%n|0:x-~x,r+1):[n,r]
Try it online!
Commented
f = (n, r = 0) => // n = input, r = number of iterations
(x = n / 10 | 0) ? // x = floor(n / 10); if x is not equal to 0:
f( // do a recursive call:
(n %= 10) & 1 ? // n = last digit; if odd:
x / n + x % n | 0 // use quotient + remainder of x / n
: // else:
x - ~x, // use x - (-x - 1) = x + x + 1 = 2x + 1
r + 1 // increment the number of iterations
) // end of recursive call
: // else:
[n, r] // return [ final_result, iterations ]
answered Sep 1 at 15:58
Arnauld
63.6k580268
JavaScript (ES6), 62 59 bytes
f=(n,r=0)=>(x=n/10|0)?f((n%=10)&1?x/n+x%n|0:x-~x,r+1):[n,r]
Try it online!
Commented
f = (n, r = 0) => // n = input, r = number of iterations
(x = n / 10 | 0) ? // x = floor(n / 10); if x is not equal to 0:
f( // do a recursive call:
(n %= 10) & 1 ? // n = last digit; if odd:
x / n + x % n | 0 // use quotient + remainder of x / n
: // else:
x - ~x, // use x - (-x - 1) = x + x + 1 = 2x + 1
r + 1 // increment the number of iterations
) // end of recursive call
: // else:
[n, r] // return [ final_result, iterations ]
answered Sep 1 at 15:58
Arnauld
63.6k580268
edited Sep 1 at 16:45
answered Sep 1 at 15:58
Arnauld
63.6k580268
answered Sep 1 at 15:58
Arnauld
63.6k580268
answered Sep 1 at 15:58
Arnauld
63.6k580268
63.6k580268
add a comment |Â
add a comment |Â
up vote
3
down vote
05AB1E (legacy), 19 bytes
[DgD–#ÃÂćDÈi·>ë‰O]?
Explanation:
[ Start infinite loop
DgD Get the length
–# If it's 1, print the loop index and break.
ÃÂćD Extract: "hello" -> "hell", "o"
Èi if even,
·> Double and increment
ë else,
‰O Divmod, and sum the result (div and mod).
] End if statement & infinite loop
? Print
Try it online!
answered Sep 1 at 16:51
Okx
12k27100
add a comment |Â
up vote
3
down vote
05AB1E (legacy), 19 bytes
[DgD–#ÃÂćDÈi·>ë‰O]?
Explanation:
[ Start infinite loop
DgD Get the length
–# If it's 1, print the loop index and break.
ÃÂćD Extract: "hello" -> "hell", "o"
Èi if even,
·> Double and increment
ë else,
‰O Divmod, and sum the result (div and mod).
] End if statement & infinite loop
? Print
Try it online!
answered Sep 1 at 16:51
Okx
12k27100
add a comment |Â
up vote
3
down vote
up vote
3
down vote
05AB1E (legacy), 19 bytes
[DgD–#ÃÂćDÈi·>ë‰O]?
Explanation:
[ Start infinite loop
DgD Get the length
–# If it's 1, print the loop index and break.
ÃÂćD Extract: "hello" -> "hell", "o"
Èi if even,
·> Double and increment
ë else,
‰O Divmod, and sum the result (div and mod).
] End if statement & infinite loop
? Print
Try it online!
answered Sep 1 at 16:51
Okx
12k27100
05AB1E (legacy), 19 bytes
[DgD–#ÃÂćDÈi·>ë‰O]?
Explanation:
[ Start infinite loop
DgD Get the length
–# If it's 1, print the loop index and break.
ÃÂćD Extract: "hello" -> "hell", "o"
Èi if even,
·> Double and increment
ë else,
‰O Divmod, and sum the result (div and mod).
] End if statement & infinite loop
? Print
Try it online!
answered Sep 1 at 16:51
Okx
12k27100
edited Sep 1 at 17:37
answered Sep 1 at 16:51
Okx
12k27100
answered Sep 1 at 16:51
Okx
12k27100
answered Sep 1 at 16:51
Okx
12k27100
12k27100
add a comment |Â
add a comment |Â
up vote
3
down vote
Julia 1.0, 81 80 79 bytes
function d(n,c=0)l,r=n÷10,n%10;n>9 ? d(r%2<1 ? 2l+1 : l÷r+l%r,c+1) : [n,c]end
Try it online!
- -1 (thanks Mr. Xcoder)
answered Sep 1 at 19:09
Rustem B.
513
Whis is my first golfcode :)
– Rustem B.
Sep 1 at 19:11
Welcome to PPCG!
– Giuseppe
Sep 1 at 19:45
@Giuseppe thanks
– Rustem B.
Sep 1 at 19:49
1
Save a byte using<1rather than==0.
– Mr. Xcoder
Sep 1 at 20:50
add a comment |Â
up vote
3
down vote
Julia 1.0, 81 80 79 bytes
function d(n,c=0)l,r=n÷10,n%10;n>9 ? d(r%2<1 ? 2l+1 : l÷r+l%r,c+1) : [n,c]end
Try it online!
- -1 (thanks Mr. Xcoder)
answered Sep 1 at 19:09
Rustem B.
513
Whis is my first golfcode :)
– Rustem B.
Sep 1 at 19:11
Welcome to PPCG!
– Giuseppe
Sep 1 at 19:45
@Giuseppe thanks
– Rustem B.
Sep 1 at 19:49
1
Save a byte using<1rather than==0.
– Mr. Xcoder
Sep 1 at 20:50
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Julia 1.0, 81 80 79 bytes
function d(n,c=0)l,r=n÷10,n%10;n>9 ? d(r%2<1 ? 2l+1 : l÷r+l%r,c+1) : [n,c]end
Try it online!
- -1 (thanks Mr. Xcoder)
answered Sep 1 at 19:09
Rustem B.
513
Julia 1.0, 81 80 79 bytes
function d(n,c=0)l,r=n÷10,n%10;n>9 ? d(r%2<1 ? 2l+1 : l÷r+l%r,c+1) : [n,c]end
Try it online!
- -1 (thanks Mr. Xcoder)
answered Sep 1 at 19:09
Rustem B.
513
edited Sep 1 at 21:33
answered Sep 1 at 19:09
Rustem B.
513
answered Sep 1 at 19:09
Rustem B.
513
answered Sep 1 at 19:09
Rustem B.
513
513
Whis is my first golfcode :)
– Rustem B.
Sep 1 at 19:11
Welcome to PPCG!
– Giuseppe
Sep 1 at 19:45
@Giuseppe thanks
– Rustem B.
Sep 1 at 19:49
1
Save a byte using<1rather than==0.
– Mr. Xcoder
Sep 1 at 20:50
add a comment |Â
Whis is my first golfcode :)
– Rustem B.
Sep 1 at 19:11
Welcome to PPCG!
– Giuseppe
Sep 1 at 19:45
@Giuseppe thanks
– Rustem B.
Sep 1 at 19:49
1
Save a byte using<1rather than==0.
– Mr. Xcoder
Sep 1 at 20:50
Whis is my first golfcode :)
– Rustem B.
Sep 1 at 19:11
Whis is my first golfcode :)
– Rustem B.
Sep 1 at 19:11
Welcome to PPCG!
– Giuseppe
Sep 1 at 19:45
Welcome to PPCG!
– Giuseppe
Sep 1 at 19:45
@Giuseppe thanks
– Rustem B.
Sep 1 at 19:49
@Giuseppe thanks
– Rustem B.
Sep 1 at 19:49
1
1
Save a byte using
<1 rather than ==0.– Mr. Xcoder
Sep 1 at 20:50
Save a byte using
<1 rather than ==0.– Mr. Xcoder
Sep 1 at 20:50
add a comment |Â
up vote
2
down vote
Ruby, 61 bytes
f=->n,r=0n>9?(a=n%10;n/=10;f[a%2<1?n-~n:n/a+n%a,r+1]):[n,r]
Try it online!
answered Sep 1 at 18:10
Kirill L.
2,4061116
add a comment |Â
up vote
2
down vote
Ruby, 61 bytes
f=->n,r=0n>9?(a=n%10;n/=10;f[a%2<1?n-~n:n/a+n%a,r+1]):[n,r]
Try it online!
answered Sep 1 at 18:10
Kirill L.
2,4061116
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Ruby, 61 bytes
f=->n,r=0n>9?(a=n%10;n/=10;f[a%2<1?n-~n:n/a+n%a,r+1]):[n,r]
Try it online!
answered Sep 1 at 18:10
Kirill L.
2,4061116
Ruby, 61 bytes
f=->n,r=0n>9?(a=n%10;n/=10;f[a%2<1?n-~n:n/a+n%a,r+1]):[n,r]
Try it online!
answered Sep 1 at 18:10
Kirill L.
2,4061116
edited Sep 1 at 18:21
answered Sep 1 at 18:10
Kirill L.
2,4061116
answered Sep 1 at 18:10
Kirill L.
2,4061116
answered Sep 1 at 18:10
Kirill L.
2,4061116
2,4061116
add a comment |Â
add a comment |Â
up vote
2
down vote
Retina, 79 bytes
(d+)[02468]$
;$.(_2*$1*
(d+)(.)
$2*_;$1*
}`(_+);(1)*(_*)
;$.($3$#2*
^;*
$.&;
Try it online! Explanation:
(d+)[02468]$
;$.(_2*$1*
If the last digit is even, multiply the rest of the integer by 2 and add 1. This always results in an odd number, so we don't need to repeat this.
(d+)(.)
$2*_;$1*
Split off the last digit and convert to unary for the divmod.
}`(_+);(1)*(_*)
;$.($3$#2*
Divide the rest of the integer by the last digit and add the remainder. Then repeat the whole program until there's only one digit left.
^;*
$.&;
Count the number of operations performed.
answered Sep 1 at 20:50
Neil
75.1k744170
add a comment |Â
up vote
2
down vote
Retina, 79 bytes
(d+)[02468]$
;$.(_2*$1*
(d+)(.)
$2*_;$1*
}`(_+);(1)*(_*)
;$.($3$#2*
^;*
$.&;
Try it online! Explanation:
(d+)[02468]$
;$.(_2*$1*
If the last digit is even, multiply the rest of the integer by 2 and add 1. This always results in an odd number, so we don't need to repeat this.
(d+)(.)
$2*_;$1*
Split off the last digit and convert to unary for the divmod.
}`(_+);(1)*(_*)
;$.($3$#2*
Divide the rest of the integer by the last digit and add the remainder. Then repeat the whole program until there's only one digit left.
^;*
$.&;
Count the number of operations performed.
answered Sep 1 at 20:50
Neil
75.1k744170
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Retina, 79 bytes
(d+)[02468]$
;$.(_2*$1*
(d+)(.)
$2*_;$1*
}`(_+);(1)*(_*)
;$.($3$#2*
^;*
$.&;
Try it online! Explanation:
(d+)[02468]$
;$.(_2*$1*
If the last digit is even, multiply the rest of the integer by 2 and add 1. This always results in an odd number, so we don't need to repeat this.
(d+)(.)
$2*_;$1*
Split off the last digit and convert to unary for the divmod.
}`(_+);(1)*(_*)
;$.($3$#2*
Divide the rest of the integer by the last digit and add the remainder. Then repeat the whole program until there's only one digit left.
^;*
$.&;
Count the number of operations performed.
answered Sep 1 at 20:50
Neil
75.1k744170
Retina, 79 bytes
(d+)[02468]$
;$.(_2*$1*
(d+)(.)
$2*_;$1*
}`(_+);(1)*(_*)
;$.($3$#2*
^;*
$.&;
Try it online! Explanation:
(d+)[02468]$
;$.(_2*$1*
If the last digit is even, multiply the rest of the integer by 2 and add 1. This always results in an odd number, so we don't need to repeat this.
(d+)(.)
$2*_;$1*
Split off the last digit and convert to unary for the divmod.
}`(_+);(1)*(_*)
;$.($3$#2*
Divide the rest of the integer by the last digit and add the remainder. Then repeat the whole program until there's only one digit left.
^;*
$.&;
Count the number of operations performed.
answered Sep 1 at 20:50
Neil
75.1k744170
answered Sep 1 at 20:50
Neil
75.1k744170
answered Sep 1 at 20:50
Neil
75.1k744170
answered Sep 1 at 20:50
Neil
75.1k744170
75.1k744170
add a comment |Â
add a comment |Â
up vote
2
down vote
Perl 6, 61 bytes
tail kv $_,0!!$0*2+1...!/(.+)(.)/: 2
Try it online!
Returns (r, o).
answered Sep 1 at 22:01
nwellnhof
3,503714
add a comment |Â
up vote
2
down vote
Perl 6, 61 bytes
tail kv $_,0!!$0*2+1...!/(.+)(.)/: 2
Try it online!
Returns (r, o).
answered Sep 1 at 22:01
nwellnhof
3,503714
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Perl 6, 61 bytes
tail kv $_,0!!$0*2+1...!/(.+)(.)/: 2
Try it online!
Returns (r, o).
answered Sep 1 at 22:01
nwellnhof
3,503714
Perl 6, 61 bytes
tail kv $_,0!!$0*2+1...!/(.+)(.)/: 2
Try it online!
Returns (r, o).
answered Sep 1 at 22:01
nwellnhof
3,503714
edited Sep 1 at 22:44
answered Sep 1 at 22:01
nwellnhof
3,503714
answered Sep 1 at 22:01
nwellnhof
3,503714
answered Sep 1 at 22:01
nwellnhof
3,503714
3,503714
add a comment |Â
add a comment |Â
up vote
2
down vote
Brachylog, 43 bytes
Feels messy and bad but here it is anyway
;0hl1&
Try it online!
answered Sep 2 at 9:10
Kroppeb
90628
add a comment |Â
up vote
2
down vote
Brachylog, 43 bytes
Feels messy and bad but here it is anyway
;0hl1&
Try it online!
answered Sep 2 at 9:10
Kroppeb
90628
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Brachylog, 43 bytes
Feels messy and bad but here it is anyway
;0hl1&
Try it online!
answered Sep 2 at 9:10
Kroppeb
90628
Brachylog, 43 bytes
Feels messy and bad but here it is anyway
;0hl1&
Try it online!
answered Sep 2 at 9:10
Kroppeb
90628
answered Sep 2 at 9:10
Kroppeb
90628
answered Sep 2 at 9:10
Kroppeb
90628
answered Sep 2 at 9:10
Kroppeb
90628
90628
add a comment |Â
add a comment |Â
up vote
2
down vote
Swift 4, 112 104 100 93 bytes
var n=Int(readLine()!)!,r=0,p=n%10;while n>9n=[n/10*2+1,n/10/p+n/10%p][n%2];r+=1;print(n,r)
Try it online!
Prints n r
- -7 (Thanks to Mr. Xcoder)
answered Sep 2 at 7:52
paper1111
1214
Very nice first answer (+1)!var n=Int(readLine()!)!,r=0,p=n%10;while n>9n=[n/10*2+1,n/10/p+n/10%p][n%2];r+=1;print(n,r)should save you 7 bytes (golfing the conditional).
– Mr. Xcoder
Sep 3 at 12:24
@Mr.Xcoder Thanks!
– paper1111
Sep 4 at 12:28
add a comment |Â
up vote
2
down vote
Swift 4, 112 104 100 93 bytes
var n=Int(readLine()!)!,r=0,p=n%10;while n>9n=[n/10*2+1,n/10/p+n/10%p][n%2];r+=1;print(n,r)
Try it online!
Prints n r
- -7 (Thanks to Mr. Xcoder)
answered Sep 2 at 7:52
paper1111
1214
Very nice first answer (+1)!var n=Int(readLine()!)!,r=0,p=n%10;while n>9n=[n/10*2+1,n/10/p+n/10%p][n%2];r+=1;print(n,r)should save you 7 bytes (golfing the conditional).
– Mr. Xcoder
Sep 3 at 12:24
@Mr.Xcoder Thanks!
– paper1111
Sep 4 at 12:28
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Swift 4, 112 104 100 93 bytes
var n=Int(readLine()!)!,r=0,p=n%10;while n>9n=[n/10*2+1,n/10/p+n/10%p][n%2];r+=1;print(n,r)
Try it online!
Prints n r
- -7 (Thanks to Mr. Xcoder)
answered Sep 2 at 7:52
paper1111
1214
Swift 4, 112 104 100 93 bytes
var n=Int(readLine()!)!,r=0,p=n%10;while n>9n=[n/10*2+1,n/10/p+n/10%p][n%2];r+=1;print(n,r)
Try it online!
Prints n r
- -7 (Thanks to Mr. Xcoder)
answered Sep 2 at 7:52
paper1111
1214
edited Sep 4 at 12:27
answered Sep 2 at 7:52
paper1111
1214
answered Sep 2 at 7:52
paper1111
1214
answered Sep 2 at 7:52
paper1111
1214
1214
Very nice first answer (+1)!var n=Int(readLine()!)!,r=0,p=n%10;while n>9n=[n/10*2+1,n/10/p+n/10%p][n%2];r+=1;print(n,r)should save you 7 bytes (golfing the conditional).
– Mr. Xcoder
Sep 3 at 12:24
@Mr.Xcoder Thanks!
– paper1111
Sep 4 at 12:28
add a comment |Â
Very nice first answer (+1)!var n=Int(readLine()!)!,r=0,p=n%10;while n>9n=[n/10*2+1,n/10/p+n/10%p][n%2];r+=1;print(n,r)should save you 7 bytes (golfing the conditional).
– Mr. Xcoder
Sep 3 at 12:24
@Mr.Xcoder Thanks!
– paper1111
Sep 4 at 12:28
Very nice first answer (+1)!
var n=Int(readLine()!)!,r=0,p=n%10;while n>9n=[n/10*2+1,n/10/p+n/10%p][n%2];r+=1;print(n,r) should save you 7 bytes (golfing the conditional).– Mr. Xcoder
Sep 3 at 12:24
Very nice first answer (+1)!
var n=Int(readLine()!)!,r=0,p=n%10;while n>9n=[n/10*2+1,n/10/p+n/10%p][n%2];r+=1;print(n,r) should save you 7 bytes (golfing the conditional).– Mr. Xcoder
Sep 3 at 12:24
@Mr.Xcoder Thanks!
– paper1111
Sep 4 at 12:28
@Mr.Xcoder Thanks!
– paper1111
Sep 4 at 12:28
add a comment |Â
up vote
1
down vote
Jelly, 19 bytes
dd/Sɗ:Ḥ‘ɗḂ?ƬâµṖṖṪ,LƊ
Try it online!
answered Sep 1 at 21:08
Erik the Outgolfer
29.4k42698
add a comment |Â
up vote
1
down vote
Jelly, 19 bytes
dd/Sɗ:Ḥ‘ɗḂ?ƬâµṖṖṪ,LƊ
Try it online!
answered Sep 1 at 21:08
Erik the Outgolfer
29.4k42698
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Jelly, 19 bytes
dd/Sɗ:Ḥ‘ɗḂ?ƬâµṖṖṪ,LƊ
Try it online!
answered Sep 1 at 21:08
Erik the Outgolfer
29.4k42698
Jelly, 19 bytes
dd/Sɗ:Ḥ‘ɗḂ?ƬâµṖṖṪ,LƊ
Try it online!
answered Sep 1 at 21:08
Erik the Outgolfer
29.4k42698
answered Sep 1 at 21:08
Erik the Outgolfer
29.4k42698
answered Sep 1 at 21:08
Erik the Outgolfer
29.4k42698
answered Sep 1 at 21:08
Erik the Outgolfer
29.4k42698
29.4k42698
add a comment |Â
add a comment |Â
up vote
1
down vote
Common Lisp, 145 bytes
A function which takes the n provided and returns the list (n r). Lisp syntax is fun.
(defun f(n &optional(r 0))(if(< n 10)`(,n,r)(let((h(floor n 10))(l(mod n 10)))(if(evenp l)(f(1+(* 2 h))(1+ r))(f(+(mod h l)(floor h l))(1+ r)))))
answered Sep 2 at 6:50
theemacsshibe
214
1
Welcome to the site =D
– Luis felipe De jesus Munoz
Sep 2 at 12:48
Thankyou very much, Luis.
– theemacsshibe
Sep 3 at 8:46
add a comment |Â
up vote
1
down vote
Common Lisp, 145 bytes
A function which takes the n provided and returns the list (n r). Lisp syntax is fun.
(defun f(n &optional(r 0))(if(< n 10)`(,n,r)(let((h(floor n 10))(l(mod n 10)))(if(evenp l)(f(1+(* 2 h))(1+ r))(f(+(mod h l)(floor h l))(1+ r)))))
answered Sep 2 at 6:50
theemacsshibe
214
1
Welcome to the site =D
– Luis felipe De jesus Munoz
Sep 2 at 12:48
Thankyou very much, Luis.
– theemacsshibe
Sep 3 at 8:46
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Common Lisp, 145 bytes
A function which takes the n provided and returns the list (n r). Lisp syntax is fun.
(defun f(n &optional(r 0))(if(< n 10)`(,n,r)(let((h(floor n 10))(l(mod n 10)))(if(evenp l)(f(1+(* 2 h))(1+ r))(f(+(mod h l)(floor h l))(1+ r)))))
answered Sep 2 at 6:50
theemacsshibe
214
Common Lisp, 145 bytes
A function which takes the n provided and returns the list (n r). Lisp syntax is fun.
(defun f(n &optional(r 0))(if(< n 10)`(,n,r)(let((h(floor n 10))(l(mod n 10)))(if(evenp l)(f(1+(* 2 h))(1+ r))(f(+(mod h l)(floor h l))(1+ r)))))
answered Sep 2 at 6:50
theemacsshibe
214
edited Sep 2 at 6:56
answered Sep 2 at 6:50
theemacsshibe
214
answered Sep 2 at 6:50
theemacsshibe
214
answered Sep 2 at 6:50
theemacsshibe
214
214
1
Welcome to the site =D
– Luis felipe De jesus Munoz
Sep 2 at 12:48
Thankyou very much, Luis.
– theemacsshibe
Sep 3 at 8:46
add a comment |Â
1
Welcome to the site =D
– Luis felipe De jesus Munoz
Sep 2 at 12:48
Thankyou very much, Luis.
– theemacsshibe
Sep 3 at 8:46
1
1
Welcome to the site =D
– Luis felipe De jesus Munoz
Sep 2 at 12:48
Welcome to the site =D
– Luis felipe De jesus Munoz
Sep 2 at 12:48
Thankyou very much, Luis.
– theemacsshibe
Sep 3 at 8:46
Thankyou very much, Luis.
– theemacsshibe
Sep 3 at 8:46
add a comment |Â
up vote
1
down vote
Haskell, 78 bytes
f n|n<10=(n,0)|(d,m)<-divMod n 10=(1+)<$>f(last$2*d+1:[div d m+mod d m|odd n])
Try it online!
answered Sep 2 at 20:48
nimi
29.9k31881
add a comment |Â
up vote
1
down vote
Haskell, 78 bytes
f n|n<10=(n,0)|(d,m)<-divMod n 10=(1+)<$>f(last$2*d+1:[div d m+mod d m|odd n])
Try it online!
answered Sep 2 at 20:48
nimi
29.9k31881
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Haskell, 78 bytes
f n|n<10=(n,0)|(d,m)<-divMod n 10=(1+)<$>f(last$2*d+1:[div d m+mod d m|odd n])
Try it online!
answered Sep 2 at 20:48
nimi
29.9k31881
Haskell, 78 bytes
f n|n<10=(n,0)|(d,m)<-divMod n 10=(1+)<$>f(last$2*d+1:[div d m+mod d m|odd n])
Try it online!
answered Sep 2 at 20:48
nimi
29.9k31881
answered Sep 2 at 20:48
nimi
29.9k31881
answered Sep 2 at 20:48
nimi
29.9k31881
answered Sep 2 at 20:48
nimi
29.9k31881
29.9k31881
add a comment |Â
add a comment |Â
up vote
1
down vote
VBA (Excel), 145?, 142, 139, 123 bytes
Condensed Form:
Sub t(n)
While n>10
e=Right(n, 1)
n=Left(n,Len(n)-1)
If e Mod 2=0Then n=n*2+1 Else n=Int(n/e)+n Mod e
c=c+1
wend
Debug.?n;c
End Sub
Expanded (with comments)
Sub test(n) 'For a given positive integer n
'Repeat the following until n<10
While n > 10
'Extract the last digit
e = Right(n, 1)
'Remove the last digit
n = Left(n, Len(n) - 1)
'If the extracted digit is even...
If e Mod 2 = 0 Then
'Multiply the rest of the integer by 2 and add 1
n = n * 2 + 1
Else
'Divide the rest of the integer with the extracted digit and add the remainder
n = Int(n / e) + n Mod e
End If
'Keep count
c = c + 1
'End the Loop
Wend
'Give the output
Debug.Print (n ; c)
End Sub
(Do excuse me, as this is my first post. I will happily edit! Right now, it looks like to test it you open the Immediate Window and type 't n' where n is your number to test :D)
answered Sep 2 at 21:27
seadoggie01
1316
add a comment |Â
up vote
1
down vote
VBA (Excel), 145?, 142, 139, 123 bytes
Condensed Form:
Sub t(n)
While n>10
e=Right(n, 1)
n=Left(n,Len(n)-1)
If e Mod 2=0Then n=n*2+1 Else n=Int(n/e)+n Mod e
c=c+1
wend
Debug.?n;c
End Sub
Expanded (with comments)
Sub test(n) 'For a given positive integer n
'Repeat the following until n<10
While n > 10
'Extract the last digit
e = Right(n, 1)
'Remove the last digit
n = Left(n, Len(n) - 1)
'If the extracted digit is even...
If e Mod 2 = 0 Then
'Multiply the rest of the integer by 2 and add 1
n = n * 2 + 1
Else
'Divide the rest of the integer with the extracted digit and add the remainder
n = Int(n / e) + n Mod e
End If
'Keep count
c = c + 1
'End the Loop
Wend
'Give the output
Debug.Print (n ; c)
End Sub
(Do excuse me, as this is my first post. I will happily edit! Right now, it looks like to test it you open the Immediate Window and type 't n' where n is your number to test :D)
answered Sep 2 at 21:27
seadoggie01
1316
add a comment |Â
up vote
1
down vote
up vote
1
down vote
VBA (Excel), 145?, 142, 139, 123 bytes
Condensed Form:
Sub t(n)
While n>10
e=Right(n, 1)
n=Left(n,Len(n)-1)
If e Mod 2=0Then n=n*2+1 Else n=Int(n/e)+n Mod e
c=c+1
wend
Debug.?n;c
End Sub
Expanded (with comments)
Sub test(n) 'For a given positive integer n
'Repeat the following until n<10
While n > 10
'Extract the last digit
e = Right(n, 1)
'Remove the last digit
n = Left(n, Len(n) - 1)
'If the extracted digit is even...
If e Mod 2 = 0 Then
'Multiply the rest of the integer by 2 and add 1
n = n * 2 + 1
Else
'Divide the rest of the integer with the extracted digit and add the remainder
n = Int(n / e) + n Mod e
End If
'Keep count
c = c + 1
'End the Loop
Wend
'Give the output
Debug.Print (n ; c)
End Sub
(Do excuse me, as this is my first post. I will happily edit! Right now, it looks like to test it you open the Immediate Window and type 't n' where n is your number to test :D)
answered Sep 2 at 21:27
seadoggie01
1316
VBA (Excel), 145?, 142, 139, 123 bytes
Condensed Form:
Sub t(n)
While n>10
e=Right(n, 1)
n=Left(n,Len(n)-1)
If e Mod 2=0Then n=n*2+1 Else n=Int(n/e)+n Mod e
c=c+1
wend
Debug.?n;c
End Sub
Expanded (with comments)
Sub test(n) 'For a given positive integer n
'Repeat the following until n<10
While n > 10
'Extract the last digit
e = Right(n, 1)
'Remove the last digit
n = Left(n, Len(n) - 1)
'If the extracted digit is even...
If e Mod 2 = 0 Then
'Multiply the rest of the integer by 2 and add 1
n = n * 2 + 1
Else
'Divide the rest of the integer with the extracted digit and add the remainder
n = Int(n / e) + n Mod e
End If
'Keep count
c = c + 1
'End the Loop
Wend
'Give the output
Debug.Print (n ; c)
End Sub
(Do excuse me, as this is my first post. I will happily edit! Right now, it looks like to test it you open the Immediate Window and type 't n' where n is your number to test :D)
answered Sep 2 at 21:27
seadoggie01
1316
edited Sep 2 at 23:42
answered Sep 2 at 21:27
seadoggie01
1316
answered Sep 2 at 21:27
seadoggie01
1316
answered Sep 2 at 21:27
seadoggie01
1316
1316
add a comment |Â
add a comment |Â
up vote
1
down vote
8086 machine code, 35 bytes
00000000 31 c9 31 d2 bb 0a 00 f7 f3 85 c0 74 14 f6 c2 01 |1.1........t....|
00000010 75 05 d1 e0 40 e2 eb 89 d3 88 f2 f7 f3 01 d0 e2 |u...@...........|
00000020 e1 f7 d9 |...|
00000023
Input: AX = n
Output: DX = o, CX = r
Assembled from:
xor cx, cx
next: xor dx, dx
mov bx, 10
div bx
test ax, ax
jz done
test dl, 1
jnz odd
shl ax, 1
inc ax
loop next
odd: mov bx, dx
mov dl, dh
div bx
add ax, dx
loop next
done: neg cx
answered Sep 1 at 17:00
user5434231
1,556512
add a comment |Â
up vote
1
down vote
8086 machine code, 35 bytes
00000000 31 c9 31 d2 bb 0a 00 f7 f3 85 c0 74 14 f6 c2 01 |1.1........t....|
00000010 75 05 d1 e0 40 e2 eb 89 d3 88 f2 f7 f3 01 d0 e2 |u...@...........|
00000020 e1 f7 d9 |...|
00000023
Input: AX = n
Output: DX = o, CX = r
Assembled from:
xor cx, cx
next: xor dx, dx
mov bx, 10
div bx
test ax, ax
jz done
test dl, 1
jnz odd
shl ax, 1
inc ax
loop next
odd: mov bx, dx
mov dl, dh
div bx
add ax, dx
loop next
done: neg cx
answered Sep 1 at 17:00
user5434231
1,556512
add a comment |Â
up vote
1
down vote
up vote
1
down vote
8086 machine code, 35 bytes
00000000 31 c9 31 d2 bb 0a 00 f7 f3 85 c0 74 14 f6 c2 01 |1.1........t....|
00000010 75 05 d1 e0 40 e2 eb 89 d3 88 f2 f7 f3 01 d0 e2 |u...@...........|
00000020 e1 f7 d9 |...|
00000023
Input: AX = n
Output: DX = o, CX = r
Assembled from:
xor cx, cx
next: xor dx, dx
mov bx, 10
div bx
test ax, ax
jz done
test dl, 1
jnz odd
shl ax, 1
inc ax
loop next
odd: mov bx, dx
mov dl, dh
div bx
add ax, dx
loop next
done: neg cx
answered Sep 1 at 17:00
user5434231
1,556512
8086 machine code, 35 bytes
00000000 31 c9 31 d2 bb 0a 00 f7 f3 85 c0 74 14 f6 c2 01 |1.1........t....|
00000010 75 05 d1 e0 40 e2 eb 89 d3 88 f2 f7 f3 01 d0 e2 |u...@...........|
00000020 e1 f7 d9 |...|
00000023
Input: AX = n
Output: DX = o, CX = r
Assembled from:
xor cx, cx
next: xor dx, dx
mov bx, 10
div bx
test ax, ax
jz done
test dl, 1
jnz odd
shl ax, 1
inc ax
loop next
odd: mov bx, dx
mov dl, dh
div bx
add ax, dx
loop next
done: neg cx
answered Sep 1 at 17:00
user5434231
1,556512
edited Sep 3 at 3:24
answered Sep 1 at 17:00
user5434231
1,556512
answered Sep 1 at 17:00
user5434231
1,556512
answered Sep 1 at 17:00
user5434231
1,556512
1,556512
add a comment |Â
add a comment |Â
up vote
1
down vote
Red, 131 bytes
func[n][s: 0 while[n > 9][r: do form take/last t: form n t: do t
n: either r % 2 = 1[t / r +(t % r)][2 * t + 1]s: s + 1]print[n s]]
Try it online!
More readable:
f: func [ n ] [
s: 0
while [ n > 9 ] [
r: do form take/last t: form n
t: do t
n: either r % 2 = 1
[ t / r + (t % r) ]
[ 2 * t + 1 ]
s: s + 1
]
print [ n s ]
]
answered Sep 4 at 9:13
Galen Ivanov
4,7571829
add a comment |Â
up vote
1
down vote
Red, 131 bytes
func[n][s: 0 while[n > 9][r: do form take/last t: form n t: do t
n: either r % 2 = 1[t / r +(t % r)][2 * t + 1]s: s + 1]print[n s]]
Try it online!
More readable:
f: func [ n ] [
s: 0
while [ n > 9 ] [
r: do form take/last t: form n
t: do t
n: either r % 2 = 1
[ t / r + (t % r) ]
[ 2 * t + 1 ]
s: s + 1
]
print [ n s ]
]
answered Sep 4 at 9:13
Galen Ivanov
4,7571829
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Red, 131 bytes
func[n][s: 0 while[n > 9][r: do form take/last t: form n t: do t
n: either r % 2 = 1[t / r +(t % r)][2 * t + 1]s: s + 1]print[n s]]
Try it online!
More readable:
f: func [ n ] [
s: 0
while [ n > 9 ] [
r: do form take/last t: form n
t: do t
n: either r % 2 = 1
[ t / r + (t % r) ]
[ 2 * t + 1 ]
s: s + 1
]
print [ n s ]
]
answered Sep 4 at 9:13
Galen Ivanov
4,7571829
Red, 131 bytes
func[n][s: 0 while[n > 9][r: do form take/last t: form n t: do t
n: either r % 2 = 1[t / r +(t % r)][2 * t + 1]s: s + 1]print[n s]]
Try it online!
More readable:
f: func [ n ] [
s: 0
while [ n > 9 ] [
r: do form take/last t: form n
t: do t
n: either r % 2 = 1
[ t / r + (t % r) ]
[ 2 * t + 1 ]
s: s + 1
]
print [ n s ]
]
answered Sep 4 at 9:13
Galen Ivanov
4,7571829
answered Sep 4 at 9:13
Galen Ivanov
4,7571829
answered Sep 4 at 9:13
Galen Ivanov
4,7571829
answered Sep 4 at 9:13
Galen Ivanov
4,7571829
4,7571829
add a comment |Â
add a comment |Â
up vote
-2
down vote
C++, 65 bytes
Assume that n, r and o variables are already defined above :)
r=0;doint m=n/10,d=n%10;n=n&1?m/d+m%d:m*2+1;++r;while(n>9);o=n;
p.s. Sorry for such design of post, I have not figured out how to do it well yet...
answered Sep 1 at 22:42
Jin X
11
1
Hello and welcome to PPCG. As it stands, your answer is not conforming to our default I/O, which essentially calls for either full programs or functions. Take a look at other C++ submissions to get a feel for how to construct these wrappers. Also, please do not be discouraged by your one downvote -- not knowing our rule set on your first post is fine and should not be punished. They may remove it if you fix your answer!
– Jonathan Frech
Sep 1 at 23:37
You may also find TIO a useful tool to both test your submission and auto-generate your post.
– Jonathan Frech
Sep 1 at 23:41
Your answer in its current form is invalid. Please either edit it to conform to our default I/O or delete it.
– Jonathan Frech
Sep 5 at 13:24
add a comment |Â
up vote
-2
down vote
C++, 65 bytes
Assume that n, r and o variables are already defined above :)
r=0;doint m=n/10,d=n%10;n=n&1?m/d+m%d:m*2+1;++r;while(n>9);o=n;
p.s. Sorry for such design of post, I have not figured out how to do it well yet...
answered Sep 1 at 22:42
Jin X
11
1
Hello and welcome to PPCG. As it stands, your answer is not conforming to our default I/O, which essentially calls for either full programs or functions. Take a look at other C++ submissions to get a feel for how to construct these wrappers. Also, please do not be discouraged by your one downvote -- not knowing our rule set on your first post is fine and should not be punished. They may remove it if you fix your answer!
– Jonathan Frech
Sep 1 at 23:37
You may also find TIO a useful tool to both test your submission and auto-generate your post.
– Jonathan Frech
Sep 1 at 23:41
Your answer in its current form is invalid. Please either edit it to conform to our default I/O or delete it.
– Jonathan Frech
Sep 5 at 13:24
add a comment |Â
up vote
-2
down vote
up vote
-2
down vote
C++, 65 bytes
Assume that n, r and o variables are already defined above :)
r=0;doint m=n/10,d=n%10;n=n&1?m/d+m%d:m*2+1;++r;while(n>9);o=n;
p.s. Sorry for such design of post, I have not figured out how to do it well yet...
answered Sep 1 at 22:42
Jin X
11
C++, 65 bytes
Assume that n, r and o variables are already defined above :)
r=0;doint m=n/10,d=n%10;n=n&1?m/d+m%d:m*2+1;++r;while(n>9);o=n;
p.s. Sorry for such design of post, I have not figured out how to do it well yet...
answered Sep 1 at 22:42
Jin X
11
edited Sep 1 at 23:01
answered Sep 1 at 22:42
Jin X
11
answered Sep 1 at 22:42
Jin X
11
answered Sep 1 at 22:42
Jin X
11
11
1
Hello and welcome to PPCG. As it stands, your answer is not conforming to our default I/O, which essentially calls for either full programs or functions. Take a look at other C++ submissions to get a feel for how to construct these wrappers. Also, please do not be discouraged by your one downvote -- not knowing our rule set on your first post is fine and should not be punished. They may remove it if you fix your answer!
– Jonathan Frech
Sep 1 at 23:37
You may also find TIO a useful tool to both test your submission and auto-generate your post.
– Jonathan Frech
Sep 1 at 23:41
Your answer in its current form is invalid. Please either edit it to conform to our default I/O or delete it.
– Jonathan Frech
Sep 5 at 13:24
add a comment |Â
1
Hello and welcome to PPCG. As it stands, your answer is not conforming to our default I/O, which essentially calls for either full programs or functions. Take a look at other C++ submissions to get a feel for how to construct these wrappers. Also, please do not be discouraged by your one downvote -- not knowing our rule set on your first post is fine and should not be punished. They may remove it if you fix your answer!
– Jonathan Frech
Sep 1 at 23:37
You may also find TIO a useful tool to both test your submission and auto-generate your post.
– Jonathan Frech
Sep 1 at 23:41
Your answer in its current form is invalid. Please either edit it to conform to our default I/O or delete it.
– Jonathan Frech
Sep 5 at 13:24
1
1
Hello and welcome to PPCG. As it stands, your answer is not conforming to our default I/O, which essentially calls for either full programs or functions. Take a look at other C++ submissions to get a feel for how to construct these wrappers. Also, please do not be discouraged by your one downvote -- not knowing our rule set on your first post is fine and should not be punished. They may remove it if you fix your answer!
– Jonathan Frech
Sep 1 at 23:37
Hello and welcome to PPCG. As it stands, your answer is not conforming to our default I/O, which essentially calls for either full programs or functions. Take a look at other C++ submissions to get a feel for how to construct these wrappers. Also, please do not be discouraged by your one downvote -- not knowing our rule set on your first post is fine and should not be punished. They may remove it if you fix your answer!
– Jonathan Frech
Sep 1 at 23:37
You may also find TIO a useful tool to both test your submission and auto-generate your post.
– Jonathan Frech
Sep 1 at 23:41
You may also find TIO a useful tool to both test your submission and auto-generate your post.
– Jonathan Frech
Sep 1 at 23:41
Your answer in its current form is invalid. Please either edit it to conform to our default I/O or delete it.
– Jonathan Frech
Sep 5 at 13:24
Your answer in its current form is invalid. Please either edit it to conform to our default I/O or delete it.
– Jonathan Frech
Sep 5 at 13:24
add a comment |Â
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I'm pretty sure that most answers from here can be trivially modified to answer this question.
– Rushabh Mehta
Sep 1 at 16:00
@Arnauld No, the output needs to be two numbers (returned or printed). I guess you could convert the decimal number to string, replace
.with whitespace and print it.– DimChtz
Sep 1 at 16:06
1
Do we enter step 2 if n<10?
– AlexRacer
Sep 1 at 16:55
@Arnauld That is not the same for n=2 and n=4.
– AlexRacer
Sep 1 at 17:00
1
Yes. It's much clearer now. :)
– Arnauld
Sep 1 at 23:42