New answers tagged integer
2
ARM T32 machine code, 34 bytes
b510 0001 d00c 2001 2204 3204 2302 3302
3001 001c 4364 1b14 d4f7 07a4 41a1 d1f6
bd10
Following the AAPCS, this takes a 0-based index in r0 and returns the 0-based entry at that index in r0.
Assembly:
.section .text
.syntax unified
.global sequence
.thumb_func
sequence:
push {r4, lr} @ Save r4 and lr to the stack
...
0
Charcoal, 10 bytes
Ｉ⌊Φθ⁼↔ι⌈↔θ
Try it online! Link is to verbose version of code. Takes input as a list of two (actually arbitrary many) integers (which need not be distinct). Explanation: Outputs the element with the greatest absolute value, or the less of two elements tied for absolute value. Explanation:
θ Input list
Φ Filter where
...
0
Jelly, 15 bytes
Ḥ½ḶoƊ€ḊŻĖS€ḂFTḣ
A monadic Link accepting \$i\$ which yields a list of the first \$i\$ terms.
Try it online!
How?
Ḥ½ḶoƊ€ḊŻĖS€ḂFTḣ - Link: positive integer, i
Ḥ - double -> 2i
€ - for each (x in [1..2i]):
Ɗ - last three links as a monad, f(x):
½ - square root (x) -> r
Ḷ ...
1
Charcoal, 10 bytes
ＩＥＮ‽Ｘ²¦¹²⁸
Try it online! Link is to verbose version of code. Explanation:
Ｎ Input integer
Ｅ Map over implicit range
Ｘ²¦¹²⁸ 2¹²⁸
‽ Random integer (0-indexed)
Ｉ Cast to string
Implicitly print
0
05AB1E, 19 bytes
Prints an infinite list of the 1-based sequence.
∞ε©tLε®+Éy®≠›‹]˜ƶ0K
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∞ε ] # for n in [1, 2, ...]:
© # store n in the register
tL # range from 1 to sqrt(n)
ε ] # for k in this range:
®+É # is n+k odd?
y®≠› ...
1
Jelly, 18 bytes
²‘½€RḂ¬ÐeḊŻ$€FT1;ḣ
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A monadic link taking an integer \$n\$ and returning the first \$n\$ terms of the sequence.
4
Python 3, 67 bytes
Prints the 0-based sequence forever.
d=n=k=0
while 1:k+=1;k**=k*k<=n;n+=k<2;k+n&1<(k>1%n)!=print(d);d+=1
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Commented:
d=n=k=0
while 1:
k+=1 # increment k
k**=k*k<=n # k=k**1=k if k*k<=n, else k=k**0=1
n+=k<2 # increment n if k is equal to 1
...
2
COW, 183 bytes
oomMMMmoOMMMmoOoomMMMmoOMMMmOomOoMMMmoOmoOmoOMMMMMMmoOMMMmOomOoMMMmoOmoOmoOMMMMMMmoOMMMmOoMOOmOomOomOomOomOomOoMOOMOomoOmoOMOOMoOmoOmoOMOOMoOmoOmoOMOoMMMOOOmooOOOmooOOOmooMMMmoomoOOOM
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Returns the smaller integer.
It increases and decreases both, the first to reach 0 (with precedence to the decreasing one) is the smaller.
[0]: n ...
0
Excel, 62 bytes
=MID(CONCAT((RANDARRAY(A1*128)>0.5)*1),SEQUENCE(A1,,,128),128)
Link to Spreadsheet
Excel doesn't handle 128 bit integers. What this returns is technically n strings of 128 random 1s and 0s, which could be interpreted as n 128 bit binary integers. If Excel did handle 128 bit integers, then answer would be =RANDARRAY(A1,,,2^128-1,1) which ...
2
Japt, 5 bytes
Outputs true for the first number being higher or false for the second.
²+V>²
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4
Vyxal, 5 bytes
²?Ḣ+⇧
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Takes input as a list and returns [1,0] for first being bigger, [0,1] for second being bigger.
² # [x^2, y^2]
?Ḣ # [y]
+ # [x^2+y, y^2]
⇧ # Grade up
2
Jelly, 4 bytes
Ḥc2Ụ
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Alternative grade-up solution. Uses (2n)C2 = n(2n-1) as the grade key function. This function maps 0, 1, -1, 2, -2, 3, -3, ... to 0, 1, 3, 6, 10, 15, 21, ..., i.e. the triangular numbers.
This allows a shorter APL solution:
APL(Dyalog Unicode), 5 bytes SBCS
⍋2!+⍨
Try it on APLgolf!
1
Jelly, 5 4 bytes
²+ḊỤ
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Terrible port of xnor's Python solution, also stealing from hyper-neutrino's Jelly solution ovs's output format ([1, 2] if y is greater, [2, 1] if x is greater). Takes input as a list [x, y].
² [x^2, y^2]
+ plus
Ḋ [y]:
²+Ḋ [x^2 + y, y^2].
Ụ Grade up.
2
Vyxal, 7 bytes
ȧ4*?±-⇧
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Oh yes, jelly port. Very fun.
⟨0|1⟩ for second integer being yes, ⟨1|0⟩ for first integer being yes.
Explained
ȧ4*?±-⇧
ȧ4* # 4 * abs(input) [vectorised] # call this x
?± # sign_of(input) [vectorised - <0 = -1, =0 = 0, >0 = 1] # call this y
- # x - y [vectorised element-wise] # call this z
⇧ # ...
1
Scala, 37 bytes
_.maxBy(x=>if(x>0)x*2-1 else x.abs*2)
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Very simple, simply takes a list containing the two number, maps each to its index in the sequence, and finds the max according to that.
1
C (gcc), 20 bytes
f(a,b){a=a*a>b*b-b;}
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Port of xnor's Python answer.
1
APL (Dyalog Unicode), 6 bytes
A port of hyper-neutrino's Jelly answer. Outputs are swapped.
⍋×-4×|
Try it online! g gets the "higher" number according to the output.
11
Python, 20 bytes
lambda x,y:x*x>y*y-y
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Checks whether \$x^2>y^2-y \$.
Outputs True/False for whether x comes after y in the listing. Note that we're not asked handle the case x==y.
To see that this works, note that each of the x*x value in the table is greater than all the y*y-y values to the left of it, but none to the right of it.
...
6
J, 8 bytes
>&(-3^|)
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Or verbose: (x - 3 ^ abs(x)) > (y - 3 ^ abs(y)).
This maps the inputs to the following sequence to compare them in:
0 1 _1 2 _2 3 _3 4 _4 5 _5
_1 _2 _4 _7 _11 _24 _30 _77 _85 _238 _248
If sorted input as output is allowed, then 7 byte /:*.@%: is another fun way to sort /: the input list: *. ...
1
Wolfram Language (Mathematica), 19 bytes
Most@*SortBy[-# #&]
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-23 bytes from @att
3
Python 3, 29 26 bytes
Returns True if a is a at a higher index than b, False otherwise.
lambda a,b:a*a+(a<0)/2>b*b
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1
Python 2, 30 bytes
lambda x,y:(x*x,x<0)>(y*y,y<0)
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This is one of those irritating ones that can't be shortened with a loop because the loop is only over 2 items.
The abs(x) -> x*x trick was taken from ovs' answer.
6
Desmos, 21 bytes
f(x,y)=\{xx-x<yy,0\}
(Newline is required)
Edit: Just noticed that xnor's answer is very similar to mines, but I found this independently.
Function is \$f(x,y)\$, and it returns \$0\$ if \$x\$ has a higher index, otherwise returns \$1\$.
This seems to work, but I just randomly stumbled upon it, so I can't explain how it works.
Try It On ...
3
Jelly, 5 bytes
²Ḥ_ṠỤ
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-1 byte thanks to ovs
-1 byte thanks to Unrelated String
Returns [1, 2] if the first integer comes first and [2, 1] for the second integer.
²Ḥ_ṠỤ Main Link; accepts a list
² x ** 2
Ḥ 2 * (x ** 2)
_Ṡ 2 * (x ** 2) - sign(x)
Ụ Grade up
0 => 0
1 => 1
-1 => 3
2 => 7
-2 => 9
3 =&...
5
Husk, 7 bytes
F>m€Θݱ
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Returns 1 if the second is larger than the first, 0 otherwise
1
CJam, 20 19 bytes
ri{L{2mr+}128*2bN}*
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CJam, 27 24 bytes (Old method)
ri{8,{G*YG#(mr\m<}%:+N}*
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Edit: had a better idea using the "repeat N times" operator. Saved 3 characters.
Edit 2: tried a completely new algorithm. Saved 4 characters.
Edit 3: went for the string manipulation side of things. Saved 1 character.
2
JavaScript (Browser), 99 bytes
n=>crypto.getRandomValues(a=new BigUint64Array(n*2)).reduce((r,v,i)=>i&1?[v<<64n|a[i^1],...r]:r,[])
This feature is introduced eailer this month by this post. And it currently requires a recent version of Firefox / Chromium Nightly build to run.
It is a sad story that words crypto, getRandomValues, ...
2
Wolfram Language (Mathematica), 25 24 bytes
–1 byte thanks to Bubbler!
RandomInteger[4^64-1,#]&
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I wrote this and even I can't find anything interesting to say about it
2
PowerShell, 57 bytes 41 bytes
@(1..$n|%{(Random 1.0)*("2*"*128+1|iex)})
Explanation: 1..$n enumerates integers from 1 to $n, % is an alias to ForEach-Object, Random decimal between 0 and 1.0, multiply by 2^128 through this clever obfuscation. @() encapsulates the result in an array, save 3 bytes if a actual array isn't needed.
Try it online!
...
2
Scala, 49 bytes
def f(n:Int)=Seq.fill(n)(BigInt(128,util.Random))
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2
R, 24 22 bytes
Edit: -1 byte thanks to pajonk
runif(scan())*4^64%/%1
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Output is an integer uniformly selected from the range 0...2^128-1.
R does not natively support 128-bit inteters, so the data type of the output is a floating-point representation (click here for a TIO link that dispays all the digits, instead of displaying in scientific ...
3
Julia 1.0, 18 bytes
!n=rand(UInt128,n)
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sorry, boring useful built-in
2
Raku, 17 bytes
{roll ^4**64: $_}
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^4**64 is the range of numbers [0, 2¹²⁸). The roll method returns randomly selected numbers from that range, in the quantity of $_, the input argument.
1
Japt, 7 bytes
ÇMq2pIÑ
Ç // Create the range [0..U) where U is the input, then map each value to
Mq // a random non-negative integer smaller than
2 // 2
p // to the power of
IÑ // 64 * 2.
Try it here.
1
Perl 5, 51 bytes
sub{join('',map chr rand 256,1..16*pop)=~/.{16}/sg}
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Creates n random 128 bit integers. Or n strings of 16 random 8 bit chars. How to print the array elements as decimal numbers can be seen by clicking the "Try it online" link.
2
C (gcc), 68 42 bytes
f(n,a)char*a;{for(n*=16;n--;)a[n]=rand();}
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takes an 0 array and the number requested as input
the array I taken as *char to be manipulated byte by byte so that we set every 8bit to rand() which is RAND_MAX e.g. At least 32767 and thus guaranteed to affect every bit
numbers printed as decimal thanks to this good answer ...
3
Factor, 33 31 bytes
[ [ 128 2^ random ] replicate ]
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replicate Create a sequence given a number of elements and a generating quotation.
128 2^ 2128
random Return a uniformly random number (via Factor's default Mersenne Twister PRNG) from 0 to whatever number is on top of the data stack minus one.
2
Ruby, 29 Bytes
->n{n.times.map{rand 2**128}}
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3
Befunge-98 (PyFunge), 57 37 bytes
&'wa+0> #;\1-:9j;+1?;#*2\_$.1-:!#@_1j
Try it online! Note that the input has to have a trailing space because of a bug in the interpreter.
The part generating a random 128-bit number bit by bit is explained below:
'wa+0> #;\1-:9j;+1?;#*2\_\.@
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'wa+ push loop counter, ...
12
x86-64 machine code, 10 bytes
Fortunately, there is a built-in RNG. Requires the RDRAND instruction.
00000000: c1e1 020f c7f0 abe2 fac3 ..........
Disassembled:
shl ecx, 2
label:
rdrand eax
stosd
loop label
ret
Outputs to a pre-allocated buffer in rdi. n is passed in ecx (arrays with 2^30 elements and more are not supported).
Try it ...
5
JavaScript (Node.js), 72 70 bytes
Saved 2 bytes thanks to @RedwolfPrograms
The helper function g builds a 128-bit integer, one bit at a time. This seems a bit long...
n=>Array(n).fill(128n).map(g=k=>k--&&BigInt(Math.random()<.5)|2n*g(k))
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1
Jelly, 7 bytes
Ø⁷2*X’)
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A monadic link taking a single argument and returning a list of random 128-bit unsigned integers of that length.
Explanation
) | For each value from 1..n
Ø⁷ | 128
2* | 2 ** that
X | Random Int from 1 to that
’ | Subtract 1
If the output can be from 1 to \$2^{128}\$ rather than 0 to \$2^{128} ...
2
Haskell, 64 bytes
k=length
f l|sum l==0=l|h:t<-l=f t#(f.init)l
a#b|k a>k b=a|1>0=b
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f return longest sublist by checking if list sum is 0 or choosing the best result coming from recursively inspecting tail and init
a# b used by f to choose longest result
2
Python 2
54 53 bytes: Try it online!
lambda n:map(randrange,[4**64]*n)
from random import*
69 bytes: Try it online!
lambda n:[int(os.urandom(i).encode('hex'),i)for i in[16]*n]
import os
Python 3
54 bytes: Try it online!
lambda n:[*map(randbits,[128]*n)]
from secrets import*
57 56 bytes: Try it online!
lambda n:[*map(randrange,[4**64]*n)]
from random ...
5
Vyxal, 7 bytes
ƛ₇ƛ₀℅;B
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ƛ # Map...
₇ƛ ; # Map 128 to...
₀℅ # Random choice from digits of 10
B # Convert to base 10
Or if it's allowed not to complete within the lifetime of the universe, 5 bytes:
ƛ₇Eʁ℅
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ƛ # Map...
₇E # 2 ** 128
ʁ # Range (0...x)
℅ # Choose a random item
It's trying to ...
1
Stax, 7 bytes
é╟φQRl:
Run and debug it
1
Charcoal, 75 bytes
≔⁻²⪪024420¹ε≔⁻²⪪343101¹δＮθＦＮＦθ«Ｊ×κ§ε⊖ι×κ§δ⊖ιＦ⊕κ«ＧＨV→^←².Ｍ§ε⊕ι§δ⊕ι»»≔ＬＫＡζ⎚Ｉζ
Try it online! Link is to verbose version of code. Takes input as size and sections. Explanation: Far too long because it doesn't try to use a formula, but guarantees 100% correct answers.
≔⁻²⪪024420¹ε≔⁻²⪪343101¹δ
Create displacement vectors for each of the six ...
0
JavaScript (Node.js), 38 36 bytes
n=>k=>n*n*k+(k<6&&n*4+2-k-(n>1&k>4))
-2 bytes thanks to Shaggy
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4
Python 2, 38 bytes
lambda n,k:k*n*n+(4*n-k+2-k/5%n)*(k<6)
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Not sure about its correctness. But it at least passed all testcases.
-2 bytes by dingledooper
-5 bytes by xnor
0
Haskell, 88 bytes
f=head.filter((0==).sum).concat.iterate(concat.zipWith map[tail,init].replicate 2).(:[])
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Top 50 recent answers are included
