10
Pyth, \$(ω^ω, 6)\$
>_PE_P
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Compares the reversed prime factorizations of the two inputs.
Pyth, \$(ε_0, 17)\$
L_SmylP#U_dPb>yEy
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L def y(b):
Pb prime factors of b
m map over d:
_d -d
U range [-d, …, -1]
P# ...
9
///, 80 bytes
/!1/1!//!/S'SS'.'//1/S.B'S.B|B'SS.B|S|.SS'B|S'SS|B.SxBxSSxBxS|B..S//B/\\//S/\//!
Takes input in unary as 1s appended to the program. Produces output in unary as .s.
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The program sends the ! to the end of the input, then makes several more substitutions. After completing the substitutions that exist in the initial program, each 1 ...
9
K (ngn/k), 43 41 bytes
{{+/'y*/:(u*|+\|+/x*u)-x*++\u:=#x}\x:\:x}
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Brutally golfed by ngn.
-2 bytes because it is allowed to output starting with A itself.
input: matrix A with size n * n
{{...}\x:\:x} scan over the list of n copies of A
{...} inner fn: x is running matrix, y is A
u:=#x u: identity matrix of size n
x*++...
9
Haskell, \$(\omega, 3)\$
This solution is dead simple, really can't be beat in Haskell.
(<)
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Haskell, \$(\omega+n,19+2\left\lfloor\log_{10}(n+1)\right\rfloor)\$
Here's a scheme to get \$\omega + n\$ for any \$n\in\mathbb{N}\$. Just replace the 9s below with one more than \$n\$. This will make the first \$n\$ numbers bigger than all other ...
7
R, 113 109 103 96 95 bytes
Or R>=4.1, 88 bytes by replacing the word function with \.
-4 bytes and another -7 thanks to @Dominic van Essen.
function(A,B=A)for(i in 2:nrow(A))B=print((diag(sum(b<-diag(B))-cumsum(b))-B*upper.tri(B))%*%A)
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Straightforward approach.
The terms on the diagonal are calculated as cumsum-sum.
Solution shorter ...
7
JavaScript (V8), 21 bytes
b=>g=n=>n&&1+g(n/b|0)
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7
Python 2, 28 bytes
f=lambda x,b:x and-~f(x/b,b)
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thanks @Kevin Cruijssen for -1 byte by using python 2 instead of python 3
7
Pari/GP, 29 bytes
n->!sumdiv(n,d,Str(d)>Str(n))
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Using the hint.
7
05AB1E, 5 bytes
Using the hint, suggested by Kevin Cruijssen
ѧ{θQ
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Cast the divisors Ñ to string §, sort {, is the last one θ equal to the input Q?
Or 9 bytes without the hint, -1 byte thanks to Kevin Cruijssen:
ÑÂøεS∍‹}P
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Ñ # divisors of the input
Âø # zip with its reverse
ε } # map over the divisor ...
6
Jelly, 13 12 bytes
Ø.,æ*ÆḊN,ÆṭƲ
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Takes input as [v, u] on the left, and a on the right. Outputs as [β, α]
-1 byte thanks to ovs!
Uses xnor's formula that
$$\alpha = \operatorname{tr} \left( \begin{bmatrix} 0 & 1 \\ v & u \end{bmatrix}^a \right) \\
\beta = -(-v)^a = - \det \left( \begin{bmatrix} 0 & 1 \\ v & u \end{bmatrix}^a \...
6
Python 2, 182 bytes
def E(a,b):
if b:x,y,d=E(b,a%b);return y,x-a/b*y,d
return 1,0,a
def f(a,b,c,d):x,y,g=E(a,b);z,w,h=E(c,d);j=E(g,h)[2];return(b/g,-a/g,0,0),(0,0,d/h,-c/h),(-x*h/j,-y*h/j,z*g/j,w*g/j)
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Yes, it is possible to solve it without fancy built-ins. And this gives relatively good results for a non-LLL one.
How it works
I derived a ...
6
Python 3 + numpy, 117, 99, 97, 91, 89(@Dingus), 88 bytes
from numpy import*
def f(M):
N=M;T=triu(M==M)
for _ in M:N=T@N[T&T.T,print(N)]*M-T*N@M
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Old version
Old version
Old version
Old version
Old version
Detailed explanation
(refers to the latest and greatest versison (88 bytes))
We start with a sizeable upfront investment to create ...
6
APL (Dyalog Extended), 2 bytes (SBCS)
Anonymous tacit infix function taking \$b\$ as left argument and \$n\$ as right argument.
≢⊤
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≢ tally the digits of
⊤ the anti-base (i.e. the representation in the given base)
Note that this is wasteful in that it actually does the base conversion.
APL (Dyalog Unicode), 4 bytes
Anonymous tacit infix ...
5
Husk, 4 bytes
Σṁdḣ
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ḣ makes a range from 1 to input
ṁd flatmap digits
Σ sum
5
Zsh, 20 bytes
My first Zsh answer, cobbled together using these very helpful posts.
<<<${#v=$[[##$2]$1]}
Try it online or run the test suite.
Full program taking \$n\$ and \$b\$ as arguments.
Commented
<<< # print
${# } # length in characters of
v= # set dummy variable v to
$[[##$2]...
5
Proton, 67 bytes
f=(n,k)=>k?f("".join(str(n.count(x))+x for x:sorted(set(n))),k-1):n
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Python 3, 78 bytes
f=lambda n,k:k and f("".join(str(n.count(x))+x for x in sorted({*n})),k-1)or n
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-7 bytes thanks to Mukundan314
5
Jelly, 8 bytes
ṢŒrUFVµ¡
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Credit to hyper-neutrino for the µ trick, and for telling me I was wrong :P
Full program. Takes n then k on the command line
How it works
ṢŒrUFVµ¡ - Main link. Takes n on the left, k on the right
µ - Last links as a monadic chain f(n):
Ṣ - Sort the digits of n
Œr - Run length encode
UF - ...
5
Wolfram Language (Mathematica), 55 bytes
ToString@#2<>#&@@@Sort@Tally@Characters@#<>""&~Nest~##&
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Input ["n", k].
5
JavaScript (Node.js), 40 bytes
x=>b=>b.flatMap(v=>x?x-(x=~-x/v|0)*v:[])
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5
Python 2, \$\phi_{\Omega^\omega}(0)\cdot\omega\$, 266 bytes
l=len
T=lambda i,s='(',d=1:T(i/2,s+['),','('][i%2],d+i*2%4-1)if d else(eval(s)[0],i)
L=lambda A,B:any(A==b or L(A,b)for b in B)or all(L(a,B)for a in A)and(l(A)-l(B),0)<(0,next((L(a,b)for a,b in zip(A,B)if a!=b),0))
def f(a,b):A,a=T(a);B,b=T(b);return(a,0)<(b,L(A,B))
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At least I ...
5
R, 45 40 39 38 bytes
Or R>=4.1, 31 bytes by replacing the word function with \.
function(n,k=1:n)any(c("",k[!n%%k])>n)
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Using the hint (proof in @alephalpha's answer).
Outputs FALSE for digit small number and TRUE otherwise.
Converts list of divisors k[!n%%k] to character by prepending empty string. Then compares the vector ...
5
Python 2, 44 bytes
lambda n:any(`i`>`n`>n%i<1for i in range(n))
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Outputs True/False negated. Looks for an i that's a divisor of n and is lexicographically smaller than n. The two conditions are combined into an inequality chain, connected by Python 2's willingness to compare strings and numbers (strings are bigger).
This chain also ...
4
Charcoal, 5 bytes
ＩＬ↨ＮＮ
Try it online! Link is to verbose version of code. Works for any b>1. Unfortunately Base doesn't accept implicit arguments, so I have to explicitly input the n and b. Explanation:
Ｎ Input `n` as an integer
Ｎ Input `b` as an integer
↨ Convert `n` to base `b` as an array
Ｌ Take the length
Ｉ Cast to ...
4
Japt, 4 bytes
sV l
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4
Retina 0.8.2, 35 bytes
\d+
$*
+`\b(1+) (\1)+1*
$1 $#2$*
Try it online! Link includes test cases. Takes input in the order b, n. Explanation:
\d+
$*
Convert b and n to unary.
+`\b(1+) (\1)+1*
$1 $#2$*
Repeatedly integer divide n by b until n is less than b. With each division, prefix a space to b.
Count the total number of spaces, including the ...
4
05AB1E, 2 bytes
вg
First input is the base \$b\$, second input is \$n\$.
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Using logarithm like most other answers is 4 bytes:
.n>ï
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Explanation:
# First (implicit) input-integer = `b`
# Second (implicit) input-integer = `n`
в # Convert n to a base-b list
g ...
4
Pyth, 4 3 bytes
ljF
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# implicitly output
l # len of
jF # convert first element in input to base second element of input
-1 byte thanks to Mr. Xcoder
4
Vyxal, 7 bytes
(søeRf∑
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( # k times...
s # Sort
øe # Run length encode
R # Reverse each
f # Flatten
∑ # Sum
4
Python 3, \$(\omega^2,28)\$
lambda a,b:(a&-a,a)<(b&-b,b)
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Old version
This orders by number of trailing zeros (of binary rep) first and then by the number itself.
4
Haskell, \$ \omega \cdot 2\$, 22 bytes
(.q).(<).q
q=odd>>=(,)
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Based off Grain Ghost's Haskell answers. The function q is pointfree for q x=(odd x,x). That is, we sort numbers by whether they're odd, then the number itself, which gives the evens followed by the odds. We can generalize this idea as:
Haskell, \$ \omega \cdot 4\$, 24 ...
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