New answers tagged arithmetic
1
AWK, 19 bytes
1+($0=int($0/2^.5))
The 1+(...) is solely for the purpose of evaluating 0 or 1. Otherwise the expression evaluates to false and doesn't print anything.
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1
Ruby, 20 bytes
->n{(n/2**0.5).to_i}
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1
Haskell, 32 bytes
h n=last[x|x<-[0..n],2*x^2<=n^2]
This submission works for arbitrary precision integers, but is very slow
because it checks every number \$x\$ below \$n\$ for whether it satisfies the equation
\$ 2x^2\leq n^2\$, then takes the last one. Thus the runtime is exponential in the bit length of the input number.
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Haskell, ...
1
Pari/GP, 17 bytes
Another arbitrary-precision answer.
n->sqrtint(n^2\2)
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0
GolfScript, 17 bytes
~3/{~@@>{}{;}if}%
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Explanation
~ # Evaluate the input
3/ # Split the input into chunks of 3
{ }% # Map every item in the input
~@@> # Is the 1st item < second item?
{}{;}if # If false, discard the current item
0
GolfScript, 3 bytes
Input is to be taken as two separate lines. In GolfScript, a preceding 0 is allowed for numbers.
n/0
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Explanation
# Two inputs taken with newlines
n/ # Split the input (separated by newlines) with newlines
0 # Add a 0 to prevent a preceding 0
# Implicit concatenation
GolfScript, 3 bytes
~+)
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...
2
naz, 56 bytes
6a8m2x1v1x1f1r3x1v2e0m1o0x1x2f2r3x1v3e0m1o0x1x3f1a1o0x1f
Takes two natural numbers as input, separated by a space.
Explanation (with 0x commands removed)
6a8m2x1v # Set variable 1 equal to 48 ("0")
1x1f1r3x1v2e0m1o # Function 1
# Read the first byte of input, removing it from the input string
# Jump ...
0
R, 19 bytes
function(i)i%/%2^.5
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1
Charcoal, 46 bytes
≔⁰θ≔⁰ηＦ↨÷ＸＮ²¦²¦⁴«≔⁺×θ⁴ιθ≦⊗η¿›θ⊗η«≧⁻⊕⊗ηθ≦⊕η»»Ｉη
Try it online! Link is to verbose version of code. Performs an arbitrary-precision floored integer square root of n²/2 using the binary square root algorithm as demonstrated e.g. by Dr. Math. Explanation:
≔⁰θ≔⁰η
Set the accumulator and result to zero.
Ｆ↨÷ＸＮ²¦²¦⁴«
Loop over the base 4 ...
2
Burlesque, 8 bytes
@2r@|/R_
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@2 # Push 2.0
r@ # Sqrt it
|/ # Cast input to number, divide input by 2
R_ # Round to nearest
3
C (gcc), Precision limited by built-in types, 42 36 bytes
__int128 f(__int128 n){n/=sqrtl(2);}
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Floor for the most part but the last output is ceiling.
Uses GCC's __int128 type: shorter in text length than unsigned long, can represent every value in unsigned long, and determined to not be a builtin type. Stay tuned for 6-8 weeks to get ...
1
cQuents, 11 bytes
#|1:A_/2^.5
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Explanation
#|1 output the first term
: mode: sequence
each term equals:
A input
_/ //
2 2
^ **
.5 .5
10
8087 FPU machine code, 11 bytes
Unassembled listing:
D9 E8 FLD1 ; load a 1 constant (need to make a 2)
D8 C0 FADD ST, ST(0) ; ST = 1+1 = 2
D9 FA FSQRT ; ST = SQRT(2)
DE F9 FDIVP ST(1), ST ; ST = N / ST
DF 1F FISTP QWORD PTR [BX] ; *BX = ROUND(ST)
C3 RET ; return ...
3
C (gcc), 23 \$\cdots\$ 53 50 bytes
typedef unsigned long long L;L f(L x){x/=sqrt(2);}
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Saved 6 bytes thanks to a'_'!!!
Added 38 bytes to fix type error kindly pointed out by S.S. Anne.
Saved 3 bytes thanks to rtpax!!!
2
JavaScript (Node.js) arbitrary-precision, 62 58 bytes
Thanks to Arnauld saving 4 bytes
(n,v=n*n/2n,m=x=>x-(y=v/x+x>>1n)>>1n?m(y):y)=>v<2n?v:m(1n)
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This is sqrt(n*n/2) after golfing the iterative Newton method sqrt() from https://stackoverflow.com/a/53684036.
2
PowerShell, 67 bytes
param([uint64]$n)($n/[math]::Sqrt(2)).ToString("G17")-replace'\..*'
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.NET (and thus, by extension, PowerShell) doesn't have a BigDecimal, so we're limited to Double or Decimal. However, the [math]::Sqrt() function only works on Double, so there we're stuck. So far, so standard. We then specify precision with G17, which ...
4
dc, 5 bytes
d*2/v
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Takes input and leaves output on the stack.
dc automatically uses arbitrary-precision integers, and supports a precision of 0 decimal places by default, thus automatically "rounding". So taking the square-root of 2 will yield 1. Instead, this solution squares the input, by duplicating it and * multiplying both the items ...
2
Whitespace, 122 103 bytes
[S S T T N
_Push_-1][S S S N
_Push_0][S N
S _Dupe_0][T N
T T _Read_STDIN_as_integer][T T T _Retrieve_input][S N
S _Dupe_input][N
T S T N
_If_0_Jump_to_Label_ZERO][N
S S N
_Create_Label_LOOP][S N
T _Swap_top_two][S S S T N
_Push_1][T S S S _Add][S N
T _Swap_top_two][S N
S _Dupe_input][S N
S _Dupe_input][T ...
3
PHP, 17 bytes
<?=$argn/2**.5|0;
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Uses @Niphram's truncate method (which in PHP also has the ability to convert the float to an int)
I know it's trendy to say PHP is to be hated, but I kinda came to like its oddities, and it gives me a chance to add an original answer
EDIT: saved 4 bytes using <?= php tag (no need to echo)
EDIT2: ...
3
wx, 3 bytes
It's W, with just one instruction added: square root. Turns out that this is very useful! (P.S. the built-in was added before the challenge.)
2Q/
Explanation
2Q % Find the square root of 2
a / % Divide the input by it
% If one operand is an integer,
% the program will automatically
% try to trunctuate to an integer
2
CJam, 9 bytes
CJam has mQ, but unfortunately it trunctuates to an integer ... Another port of Lyxal's answer.
q~2 .5#/i
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Explanation
q~ e# Take input & evaluate
2 e# Take 2 to the power of ...
.5# e# ... 0.5 (equal to square root)
/ e# Divide the input by it
i e# Convert to integer
3
Pyth, 6 bytes
The division auto-casts the number to a decimal!? (In seriousness, is there a square root function in Pyth?)
/Q@2 2
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Explanation
@2 2 to the power of
2 1/2 (effectively calculates math.sqrt(2))
/Q Divide the (evaluated) input by that number
2
TI-BASIC, 5 bytes
int(Ans√(2⁻¹
Built-ins are great.
Input is a number in Ans.
Output is what is specified in the challenge.
Explanation:
√(2⁻¹ ;get the square root of 1/2
Ans ;get the input (Ans)
;implicit multiplication
int( ;truncate
;implicit print of Ans
Note: TI-BASIC is a tokenized ...
6
Japt, 3 bytes
z2q
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z is the floor division method and q is the nth-root method, defaulting to square root when it's not passed an argument.
12
JavaScript (ES6), 12 bytes
i=>i/2**.5|0
Uses a binary or to truncate the result
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6
Jelly, 15 bytes
³²:2_²:Ẹ¡:2+µƬṪ
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An arbitrary precision Jelly answer that uses the Newton-Raphson method to find the correct answer. Uses only integer arithmetic operations so the intermediate values are all Python big ints rather than getting cast as floats which would lose precision. The integer result equates to the floor of what would be ...
8
Java 8, 18 bytes
n->n/=Math.sqrt(2)
Limited to a maximum of \$9{,}223{,}372{,}036{,}854{,}775{,}807\$ (signed 64-bit integer).
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Explanation:
n-> // Method with long as both parameter and return-type
n/= // Divide the input by:
Math.sqrt(2) // The square-root of 2
// The `/=` sets the divided ...
14
Scratch 3.0, 7 blocks/62 bytes
Try it online Scratch!
As SB Syntax:
when gf clicked
ask()and wait
say(round((answer)/([sqrt v]of(2
It's always fun to usual visual languages! At least I have built-ins this time.
2
MathGolf, 4 bytes
2√/i
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Explanation:
2√ # Take the square-root of 2
/ # Divide the (implicit) input-integer by this
i # Cast it to an integer, truncating any decimal values
# (after which the entire stack joined together is output implicitly as result)
3
APL (Dyalog Extended), 5 bytesSBCS
Full program. Prompts stdin for zero or more numbers.
⌈⎕÷√2
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⌈ ceiling of
⎕ console input
÷ divided by
√ the square root of
2 two
14
Python 3, 19 17 bytes
A different python answer
lambda x:x//2**.5
-2 bytes thanks to @Mukundan
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2
Haskell, 20 bytes
f n=round$n/(sqrt 2)
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6
Mathematica, 17 14 13 bytes / 12 7 characters
⌊#/√2⌋&
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-3 bytes because Mathematica accepts the char √, which I copied from this MathGolf answer.
-1 byte, -5 characters, as per @Mark S. suggestion, by using ⌊⌋.
For just one more byte (but 5 more characters) I can always round to the nearest integer with
Round[#/√2]&
5
05AB1E, 3 bytes
2t÷
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-1 byte thanks to @Grimmy
Yet another port of my Keg answer for the sake of completion.
Explained
2t÷
2t # Push the square root of two
÷ # Integer division
🍟🍅
Still no ketchup.
2
Python 3, 22 21 bytes
lambda x:int(x/2**.5)
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-1 byte thanks to @RGS. Thanks for reminding me that implicit decimals exist
Just a port of my Keg answer. Nothing special here.
2
Keg, 6 bytes
21½Ë/ℤ
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This defines the function f as:
Taking a single parameter, then
Calculating the square root of 2 by raising it to the power of 0.5, then
Dividing the parameter by root 2, then
Casting the result to an integer (truncating / flooring the result) and returning it.
The footer is to define the test cases in a nice way.
...
45
Regex (ECMAScript+(?*)), 1169 929 887 853 849 bytes
Regex was never designed to do mathematics. It has no concept of arithmetic. However, when input is taken in the form of bijective unary, as a sequence of identical characters in which the length represents a natural number, it is possible to do a wide range of operations, building up from the simple ...
0
Haskell, 48 bytes
f=fromEnum;g a b=toEnum(div((f a)+(f b))2)::Char
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0
Brainf*ck, 39 bytes
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,>,>,<[->+<]<[->>-<<]>,[->-<]>[-->+<]>.
The input/output are the ASCII values of characters. Using "aaz0" gives "%" as output, as it corresponds to the case
97 122
97 48
which is solved by moving 37 units of liquid, the % symbol.
Could shave a couple of bytes if the ...
0
Burlesque, 33 bytes
riXX<-J2ENj2en2?***{XX++}ms10.%z?
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ri # Read input as int
XX # Explode into digits
<- # Reverse
J2EN # Duplicate and take every 2nd starting from 1
j2en # Take the other half
2?* # Multiply each digit by 2
** # Remerge arrays
{XX++}ms # Sum the digits of ...
1
MathGolf, 4 bytes
‼/%]
Inputs in the order b a, output as a list containing two integers.
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Explanation:
‼ # Apply the following two commands to the stack separately:
/ # Division (which will be integer division if the given arguments are integers)
% # Modulo
# (both builtins will use the implicit input-integers for their two ...
1
Whitespace, 68 bytes
[S S S N
_Push_0][S N
S _Dupe_0][T N
T T _Read_STDIN_as_integer][T T T _Retrieve_input1][S N
S _Dupe_input1][S N
S _Dupe_input1][T N
T T _Read_STDIN_as_integer][T T T _Retrieve_input2][T S T S _Integer_divide][T N
S T _Print_as_integer][S S S T S T S N
_Push_10_newline][T N
S S _Print_as_character][S S S N
...
1
Perl 6 Raku, 12 11 bytes
(*-*-*+*)/2
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Edit: I realized that I could take the parameters in any order. Four parameters are: liquid in container 1, liquid in container 2, volume of container 1, volume of container 2.
0
Rogex, 15 bytes
701300001600200
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The actual language isn't on TIO yet, so a python 3.8 interpreter with the program in the code section and the python interpreter split in the header and footer is provided
Rogex is a newly developed language that I've recently discovered, so I though I'd give it a crack and write an answer.
Explained
...
0
W, 3 bytes
You know, we can't beat the built-ins.
1S/
Explanation
a1S % Stack : 1 a
/ % Divide.(1/a)
% Output.
0
PHP, 165 156 110 bytes
$o=[$j=$i=$s=$m=0];for(;$i<strlen($n=$argn);)$o[]=$n[$i++]=='<'?++$m:--$s;for(;$j<=$i;)echo$o[$j]-$s,$n[$j++];
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Ungolfed
<?php
$min=0;
$max=0;
$current=0;
$order = [0];
for ($i=0;$i<strlen($argn);$i++){
$compare=$argn[$i];
if ($compare=='<') {
$max++;
$current=$max;
} ...
0
W r, 3 bytes
/@m
Explanation
abab % Implicit ops
/ % Divide a by b. Stack: a b (a/b)
@ % Roll down to show 2 operands. Stack: (a/b) a b
m % Modulo the operands. Stack: (a/b) (a%b)
% Output the whole stack
```
0
Haskell, 95 bytes
s=show
('<':y)!(a:b)=s a++'<':y!b
(x:y)!z=s(last z)++x:y!init z
_!(y:_)=s y
f x=x![0..length x]
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1
Burlesque, 9 bytes
)**avavL[
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If I'm allowed to report the ceiling, can save 2 chars )**foL[ fo is defined as avpd which calculates the average and then ceilings it.
)** # Map ord(a)
av # Calculate average
av # Floor the resulting double
L[ # Int to Char
1
Ahead, 7 bytes
ii+2/o@
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Top 50 recent answers are included
