New answers tagged arithmetic
1
Pyth, 8 bytes
._eS.+aZ
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Literal translation of James's MATL answer.
0
Branch, 38 bytes
^\4^%/;c1^>/0c45^?[.0]anc2^%/49c105^?.
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Explanation
Implicit - the initial node's value is set to the first argument
^\4 Create a parent and sibling and set the right child to 4
% Modulo
/; Go to the left child and copy the mod result
c1 Go to the right child (automatically ...
2
Vim, 69 bytes
V}JYu:sor!n
V}JP:%s/^\(.*\)\n\1/m\r\1
:%s/\vm\n.*<(.+) \1.*/n
:g/\d/d
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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 ...
0
Python 3, 113 bytes
def f(L):
z=0
for i in range(0,len(L)-1):
if L[i]-L[i+1]<0:
return 2
if L[i]-L[i+1]==0:
z=1
return z
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Could be shorter if there were a sign function. This outputs 0 for strictly decreasing, 1 for weakly decreasing and 2 for none of the above.
0
Branch, 2 bytes
+#
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Explanation
+ adds the values of the children of the current node. If any nodes are missing, they are usually created with a value of 0; however, binary operations will instead read from STDIN (this usually gets 0 if the input is exhausted, but if the input was empty to begin with, feof doesn't ...
2
Yggdrasil, 4 bytes
+__$
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Yggdrasil automatically substitutes 2 characters for different values when turning the source code in a binary tree. % represents a null byte and _ represents a None value. Yggdrasil then traverses the tree and replaces Nones with sequential command line arguments (i.e. the first encountered, depth-first, is the first ...
1
Japt, 20 bytes
No big shenanigans, just recursively calculates the result.
§2?1:ßU-ßUÉ)+ßU-ßU-2
§2 // If the input is less than or equal to 2
?1 // return 1,
: // otherwise
ß // recursively recall with
U-ßUÉ // input minus recursive recall input minus one
)+ ...
1
Husk, 17 bytes
!¡λṁ!¹m≠→L¹↑_2)ḋ3
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Feels too long. I'll post something with fix soon.
1
V (vim), 93 bytes
:s/1\n/a
D@"i1 <esc>
qqYplllA]<esc>0i[<esc>:s/\(\d\+\) \(\d\+\) /\1+\2,/g
C<c-r>=<c-r>"
<Esc><c-o>V}J0i <esc>@qq@qdd:s/ i1/1
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Special casing 1 for <c-o> was a bit annoying, but the rest plays out smoothly.
Possible byte saves can be in the large regex, and maybe removing ...
2
Branch, 98 bytes
\^//C//70/105/122Z/zc\/66/117/z/za1O[/ob3^%Vc/v?[./]b5^%Wc\w?[./]a/vbw^*/0bo^?[#0]a10.o}O/;b101^-]
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Explanation
The first part basically loads "Fizz" and "Buzz" along the left branches of two trees, so it looks like:
0
/ \
0 0
/| / \
F 0 B ...
5
Jelly, 12 bytes
:g/æl/;g/}ɗ/
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Takes input as [[list of numerators], [list of denominators]]. +2 bytes to take input as a list of [numerator, denominator] pairs
How it works
:g/æl/;g/}ɗ/ - Main link. Takes a pair of lists [N, D] on the left
/ - Columnwise reduce by:
g - GCD
: - Divide, reducing the fractions ...
0
Duocentehexaquinquagesimal, 6 bytes
5Q¬$††
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0
Duocentehexaquinquagesimal, 5 bytes
3Σª;â
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0
Python 3, 56 bytes
def f(n):
k=1
for i in range(2,n+1):
k=k*i
return k
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2
Scratch, 62 bytes
when gf clicked
ask()and wait
say(round((answer)/([sqrt v]of(2
2
Vyxal, d, 5 bytes
5vτ½⌈
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A port of the Jelly answer which is a port of short husk answer.
Explained
5vτ½⌈
5vτ # convert each digit of the input to base 5
½ # halve each item in that list (halving vectorises all the way down)
⌈ # ceiling each item in that list
# -d deep sums the list and implicitly outputs
3
Charcoal, 13 bytes
ＩΣ⭆Ｓ⭆↨Ｉι⁵Ｌ↨λ³
Try it online! Link is to verbose version of code. Explanation:
Ｓ Convert input to a string
⭆ Map over characters and join
ι Current character
Ｉ Cast to integer
↨ ⁵ Convert to base 5
⭆ Map over base 5 digits and join
λ Current ...
2
Retina 0.8.2, 20 15 bytes
.
$*1,
1{5}|11?
Try it online! Link includes test cases. Explanation:
.
$*1,
Convert each digit to unary separately.
1{5}|11?
Count the number of 5s, 2s and 1s needed to make each digit.
4
Python 2, 37 bytes
f=lambda n:n and n/5%2-n%5/-2+f(n/10)
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Uses a formula rather than a lookup table for each digit. n/5%2 counts the five-cent coin, and subtracting -n%5/-2 is equivalent to adding (n%5+1)/2 for the one- and two-cent coins.
n%10 0 1 2 3 4 5 6 7 8 9
-----------------------------
n/5%2 0 0 0 0 0 1 1 1 1 1
0-n%5/-2 0 1 ...
4
Wolfram Language (Mathematica), 39 bytes
⌈.4#⌉-⌊#/5⌋&@*IntegerDigits/*Tr
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2
Grok, 19 bytes
:Yp:Y%zp/Yp1%-I,Wzq
This outputs remainder,division. Additionally, inputs must be passed through STDIN; it can't be piped from another command.
Explanation:
:Yp # Takes the first integer from STDIN and duplicates it.
:Y # Takes the second integer from STDIN and duplicates it to the register.
%z ...
3
Python 2, 48 bytes
f=lambda x:x and int("0112212233"[x%10])+f(x/10)
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49 bytes in Python 3 because you'd need // for floor division.
14
Husk, 11 9 8 bytes
Edit: -1 byte thanks to caird coinheringaahing
ṁo⌈½ṁB5d
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d # get the digits
ṁB5 # convert them all to base-5
# (this gives a 1 for each 5-denomination coin needed,
# as well as the leftover for each digit.
# We'll need 2 more coins for those with leftover 3 or 4,
# ...
5
05AB1E, 11 bytes
S<•δ¬Èº•sèO
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Same approach as my Jelly answer.
How it works
S<•δ¬Èº•sèO - Program. Input: n
S - Cast n to digits
< - Decrement
•δ¬Èº• - Compressed integer: 1122122330
sè - Using n's digits, index into the digits of 1122122330
O - Sum
Kudos to Kevin Cruijssen's excellent ...
8
R, 54 51 47 45 bytes
Edit: converted to console input instead of a full function to try not to fall behind Robin Ryder's answer
d=utf8ToInt(scan(,''))-48;sum(d>0,d>5,d%%5>2)
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14
JavaScript (ES7), 36 35 bytes
Similar to other answers. Using a Black Magic formula instead of a lookup table.
f=n=>n&&(n%10)**29%3571%4+f(n/10|0)
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Here is a script that looks for \$(p,m)\$ pairs such that \$(n^p\bmod m)\bmod 4=a_n\$ for all \$n\in[0..9]\$.
It's worth noting that this code takes IEEE-754 precision errors into ...
3
C (gcc), 39 bytes
f(n){n=n?f(n/10)+""[n%10]:0;}
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JavaScript (Node.js), 37 bytes
f=n=>n&&+"0112212233"[n%10]+f(n/10|0)
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10
Haskell, 55…41 40 bytes
-1 byte thanks to xnor, for using a string instead of the hard-coded list.
a=0:tail[i+read[j]|i<-a,j<-"0112212233"]
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a is the infinite sequence.
How?
It's not hard to find the recursive formula
$$
a(n)=a\left(\left\lfloor\frac{n}{10}\right\rfloor\right)+a(n \operatorname{mod} 10)
$$
with the base cases ...
4
Husk, 23 bytes
L◄Lfo=⁰Σ↑o≤⁰LṖƒ(+İ€m*10
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Extremely slow past 11.
Explanation
L◄Lfo=⁰Σ↑o≤⁰LṖƒ(+İ€m*10
ƒ( create an infinite list using:
İ€ currency denomination builtin: [1,1/2,1/5,...200]
+ plus
m*10 the input mapped to *10
this gives [1,1/...
7
R, 64 52 50 45 bytes
sum(c(1,2,1:3)[.6*utf8ToInt(scan(,""))-28.4])
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Same strategy as Delfad0r's Haskell answer, which is nicely explained.
First, scan(,"") reads in input as a string. Then, utf8ToInt(...)-48 takes a string of digits and converts it to a vector of integer digits. This works out shorter than taking input as ...
6
Jelly, 7 bytes
Db5FHĊS
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Steals Ports Dominic Van Essen's Husk answer, be sure to upvote that!
How it works
Db5FHĊS - Main link. Takes an integer n on the left
D - Convert to digits
b5 - Convert each digit to base 5
F - Flatten
H - Halve each
Ċ - Ceiling of each
S - SUm
Jelly, 11 bytes
Dị“FȮŀO’D¤S
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0
M4, 49 bytes
define(f,`ifelse($1,0,1,`eval($1*f(decr($1)))')')
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Usage
f(n)dnl where n is an integer.
3
Haskell, 34 bytes
maximum.(zipWith compare<*>(1/0:))
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Outputs LT,EQ,GT for strictly monotonic, non-increasing, and otherwise, respectively. Thanks to @xnor for pointing out that my previous version failed on singleton lists and suggesting this alternative.
Takes the maximum of the sighn of the differences between adjacent pairs, ...
2
Wolfram Language (Mathematica), 57 56 bytes
MaximalBy[Range@#-1,#>9&&1+#0@@Times@@@RealDigits@#&,1]&
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Returns the value wrapped in a list.
Since MaximalBy chooses the maximum based on Mathematica's internal ordering, choosing a value for the base case (when n≤9) isn't necessary.
1
C (gcc), 95 94 bytes
Saved a byte thanks to ceilingcat!!!
a;b;c;d;r;f(n){for(b=0;n--;r=c>=b?b=c,n:r)for(c=0,d=n;a=d,d=d>9;++c)for(;a;a/=10)d*=a%10;d=r;}
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0
Python 3, 230 bytes
def f(n):
s="1"
while 2>1:
if int(s)%n==0:
return s
if "0" in s:
for j in range(0,len(s)):
if s[j]=="0":
q=j
t=s[:q]
for k in range(q,len(s)):
t=t+str(1-int(s[k]))
s=t
else:
s=str(10**len(s))
3
Haskell, 82 bytes
f n=snd$minimum$((,)=<<p)<$>[0..n-1]
p n|n>9=p(product$read.pure<$>show n)-1
p n=0
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Golfing Delfad0r's solution by using the decorate-sort-undecorate idiom to find the maximizing value.
It's easier to see how it works in this slightly longer version.
83 bytes
f n=snd$minimum[(-p i,i)|i<-[0..n-1]]
p n|...
6
Python 2, 70 bytes
lambda n:max(range(n),key=g)
g=lambda x:x<10or-~g(eval('*'.join(`x`)))
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The helper function g recursively computes multiplicative persistence, and the main function in the top line finds value that maximizes g among the half-open range from 0 to n. It works out that max chooses the earlier element in case of a tie for ...
3
Retina, 60 bytes
.+
*
L$`.
$.`
+/\d.+/_~(`.
$&$*
)`^
.+¶#$$.(
.
#
D`#+
¶*$
¶
Try it online! Link includes test cases. Explanation:
.+
*
L$`.
$.`
List the numbers up to n.
+/\d.+/_
Repeat while there are numbers of at least two digits...
~(`.
$&$*
)`^
.+¶#$$.(
... convert the numbers into Retina expressions that calculate their digital product ...
4
Scratch, 454 bytes
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This one was fun to make! It takes 13 seconds to process the first 100,000 numbers. Alternatively, 48 blocks.
when gf clicked
set[H v]to(
set[M v]to(
set[N v]to(-1
ask()and wait
repeat(answer
change(N)by(1
set[C v]to(
delete all of[B v
repeat(length of(N
change[C v]by(1
add(letter(C)of(N))to[B v
end
set[P v]to(
repeat until&...
4
Befunge-98 (FBBI), 107 105 108 bytes
+3 for a bugfix with the tiebreaks.
Definitely some golfing left to do...
< v:-1_;# $ <;v#:&
e:p1 9;j`N' <^;#1p4
;\0:<v; +1:$<^
+\1\v>:9`! #^_\1
>$$ ^ @.N'<
^#::<X'*%ap55/a_
Try it online! The code contains two null bytes, indicated by N above
How?
slightly outdated
Product of the ...
2
R, 59 55 bytes
Edit: -2 bytes thanks to Giuseppe, quickly superseded by -4 bytes (a different way) thanks to Robin Ryder
n=scan();while(print(n)>9)n=prod(utf8ToInt(c(n,""))-48)
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A different method to Giuseppe's answer for the same number of bytes, here as a full recursive function instead of the (often-shorter) approach of taking ...
1
Befunge-98 (FBBI), 43 42 bytes
&;1\v>#;:.:9`!#@_;
>$$ ^
^#::< '*%ap25/a_
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The first line takes input, prints the values and runs until a value <=9 is reached. The third line computes the product of digits of an integer (26 bytes on its own).
4
JavaScript (Node.js), 68 bytes
g=n=>n>9&&-~g(eval([...n+''].join`*`));f=n=>n--?g(h=f(n))<g(n)?n:h:0
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5
Julia 0.7, 58 54 bytes
n->findmax(.!(0:n-1))[2]-1
!n=n>9?!prod(digits(n))+1:0
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4
Charcoal, 18 bytes
ＦＮ⊞υ∧›ι⁹⊕§υΠιＩ⌕υ⌈υ
Try it online! Link is to verbose version of code. Explanation:
ＦＮ
Loop over the range 0..n-1.
⊞υ
Push to the predefined list...
∧›ι⁹
... zero if the current index is less than 10, otherwise...
⊕§υΠι
... increment the previously calculated persistence of the digital product of the current index.
Ｉ⌕υ⌈υ
Print the ...
2
Excel, 59 bytes
=LET(y,COLUMN(A:O),SUM((0&MID(SEQUENCE(A1),y,1))*10^(y-1)))
Explanation
=LET(
y,COLUMN(A:O) 'y = [1..15] horz.; Excel is accurate to 15 places
SUM((0&MID(SEQUENCE(A1),y,1))*10^(y-1)) 'final result
SEQUENCE(A1) '[1..A1] vertical
(0&...
6
R, 94 91 87 bytes
Edit: thanks to Kirill L. for spotting 2 (!) bugs, and also saving 3 bytes, and -4 more bytes thanks to Robin Ryder
which.max(Map(f<-function(x)`if`(x>9,1+f(prod(utf8ToInt(c(x,""))-48)),0),1:scan()-1))-1
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Started as a rather unimaginative construction based around Giuseppe's 'Multiplicative persistence' ...
3
Haskell, 86 85 84 bytes
-1 byte thanks to xnor for removing the f 1=0 case.
f n|n>1,x<-f$n-1,p x>=p(n-1)=x
f n=n-1
p n|n>9=1+p(product$read.pure<$>show n)
p n=1
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6
Jelly, 10 bytes
ḶDP$ƬL$ÐṀḢ
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Jelly, 10 bytes
ḶDP$Ƭ€ẈMḢ’
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How they work
ḶDP$ƬL$ÐṀḢ - Main link. Takes n on the left
Ḷ - Unlength; [0, 1, 2, ..., n-1]
$ÐṀ - Return the maximal elements under the previous 2 links:
$Ƭ - Iterate the previous 2 links until reaching a fixed point, collecting all steps:
D ...
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