New answers tagged sequence
2
APL (Dyalog Unicode), 46 bytes
{∧/0<⍵:⊃⊃⍸A⍷↑{⍵((+/↑),⊢)⍣{</+/¨A⍵}⍵⍴1}¨⍳⊃A←⌽⍵}
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Anonymous prefix function, taking a vector of integers.
How?
∧/0<⍵: ⍝ If the given sequence contains any non-positive integers, then return nothing. Otherwise...
A←⌽⍵ ⍝ Set A to be the reverse of the given sequence
{...}¨⍳⊃ ⍝ For each n in 1 to last ...
2
Perl 5 -p, 20 bytes
/,/;$_=$`*$'&&$'**$`
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1
Javascript, 24 bytes - recursive version, 33 bytes - usage of Binet's formula, 46 bytes - continuous approximation with the "phi" itself
Usage of "Binet's formula" and a bitwise "OR" operator to round the result. Originally in the "Binet's formula" you have to subtract the so-called "smaller phi to the n-th power&...
1
///, 196 bytes
/h/ 3//g/ 2//f/ 1//e/aa//d/
a//c/ //b/ea//a/c /bc1bc
b1gfbdac1hhfecdaf 4 6 4fe dc1 5f0f0 5facdac15g0f5e d 7g1h5h5g1 7a d8g8 56 70 56g8 8c
9h6 84f26f26 84h6 9
bg52b dac462 462ec
b 924b daf716f716e
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0
Brachylog, 28 bytes
^₂g{∧{Ċ≥₁>ᵛ1&≜↔;?^ᵐ+}}ᶠ⁽o∋↙?
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How it works
^₂g{∧{Ċ≥₁>ᵛ1&≜↔;?^ᵐ+}}ᶠ⁽o∋↙?
^₂g 7 -> [49]
{∧{ }}ᶠ⁽ find first 49 outputs whose …
Ċ≥₁>ᵛ1 inputs are of form [X,Y],
≥₁ are ordered
>ᵛ1 ...
3
Charcoal, 57 bytes
Ｆ³⊞υ﹪ι²Ｆ”)⧴→↨w﹪f”«ＵＭυ⁺κ§υ⊖λ⊞υ⁰Ｆ›ⅉ⁰≔⪫✂υＩι±Ｉι¹ ι⟦⁺× ⊘⁻²⁴Ｌιι
Try it online! Link is to verbose version of code. Explanation:
Ｆ³⊞υ﹪ι²
Start by preparing a list 0 1 0. (Charcoal's indexing is cyclic, so rather than letting it wrap I need to provide a safety margin.)
Ｆ”)⧴→↨w﹪f”«
Loop over the compressed string 111113222667 which represents ...
3
APL (Dyalog Unicode), 81 75 bytes
⌊{⌽q⊃⍨⊃⍋|-/¨q←⊃,/2,/¨p/⍨B∊¨p←2(+/⍤↑,⊢,¯2-/⍤↑⊢)⍣{∧/B≤|¯2↑1⌽⍵}¨⍺,¨(-,⊢)⍳B←⍵}⌈
-6 bytes thanks to @Adám
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Explanation
This performs a similar algorithm as @notjagan's answer. For each C between -B and B inclusive, it produces the Lucas sequence containing A,C (where A and B are the minimum and maximum of the two ...
4
05AB1E, 26 bytes
13LεDÝc•6hö¢ðU•RNèF¦¨]»¦.c
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Explanation:
13L # Push a list in the range [1,13]
ε # Map each value `y` to:
D # Duplicate the value `y`
Ý # Pop and push a list in the range [0,`y`]
c # Take the binomial coefficient of `y` with each value in this list
...
2
Ruby, 91 bytes
66551112000090.digits.map{|i|n=0;$*.map!{|j|n+n=j}<<1;puts ($*[i..~i]*' ').center 23if i<9}
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Iterates over the digits of 66551112000090 (from the right), one digit for each row of the tree. Each digit, with the exception of the 9, specifies the number of elements to be omitted from the beginning and end of the ...
6
Python 2, 118 bytes
n=1
exec"a=n/6+n/10*4+(n==6);print' '.join(str(((2**n+1)**n>>n*k)%2**n)for k in range(a,n+1%n-a)).center(23);n+=1;"*13
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Expresses binomial coefficients inline like in my tip here with binom(n,k)=((2**n+1)**n>>n*k)%2**n. The number of entries cut off on each side for the n'th row (one-indexed) is ...
6
J, 67 bytes
(1785951#:~13#7)(0(,~' '#~12-<.@-:@#)@":-@[}.}.)&>1;2}.<@(!{:)\i.14
The pascal's triangle part is adapted from @ephemient's answer here.
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How it works
<@(!{:)\i.14
The first 14 rows of the triangle.
1;2}.
Drop the first two rows, prepend 1.
(1785951#:~13#7)
0 0 0 0 0 2 1 1 1 5 5 6 6 stored in base 7.
":-...
3
JavaScript (ES8), 137 bytes
_=>[a=[1],..."000021115566"].map((v,n)=>(s=(a=[i=1,...a.map(v=>v+~~a[i++])]).slice(v,n+2+!n-v).join` `).padStart(23+s.length>>1)).join`
`
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1
GAS x86-64 for Linux (Machine code), 76 68 bytes
Loop-free, because I thought it would be a fun idea. GCC 10.1.
Note/warning: for this to work, you have to turn off DEP and PIE (ASLR). See compilation instructions below.
Look, ma! No loops! Byte count is for the assembled opcodes of func + e + f. Works on 32 bit input. Sure, you'll eventually run out of ...
1
Actually, 16 bytes
"1,"◙01W;a+;◙',◙
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0
Python 2, 59 51 55 33 bytes
a=0
b=1
while 1:a,b=b,a+b;print a
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Thanks for @pppery for saving a whole 22 bytes!
0
Perl 5 -alp, 148 bytes
$_=9;while(!$o||--$F[0]){$o=0;if(++$_ ne reverse){for$i(0..(@a=/./g)){map{@b=@a;@b[$i,$_]=@b[$_,$i];$o+=($t=1*join'',@b)eq reverse$t}$i..$#a}}}say$o
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Output is on two lines. First line is the order; second line is the number.
1
Raku, 45 bytes
{|$_,{|keys abs([-] @_[++$,$++])∖@_}...^!*}
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Explanation
{ } # Anonymous codeblock
... # Generating a sequence
|$_, # Starting with the input
{ } # Where we ...
1
Wolfram Language (Mathematica), 50 bytes
#//.l:_[___,b_,a_,___]:>l<>d/;FreeQ[l,d=Abs[b-a]]&
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Takes a sequence with any head. If the sequence is already complete, returns it unchanged. Otherwise, returns the completed sequence in a StringJoin.
<> (StringJoin) is short, flattens Lists, does not operate on integer (non-string) ...
2
Python 3.8, 109 bytes
f=lambda l,n,a=0,b=0:f(l[1:],m,a+1-j,b+j)if l and(m:=n-((j:='i|'.index(l[0]))+1)*2)>=0 else[l!=''and m<0,a,b]
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2
Jelly, 14 bytes
I'm assuming, for now, that the "boolean" part of the output may be given as a Truthy/Falsey value.
Og©2ḤÄ’<a®‘ċⱮ3
A dyadic Link accepting the trees (a list of characters) on the left and the axe durability (an integer) on the right which yields a list of three integers, [is_broken, small_trees, big_trees] (Note that non-zero ...
1
Retina 0.8.2, 55 bytes
\d+
$*
(11)+1?(?<-1>(i)|(?<-1>(\|)))*($)?.*
$#4 $#3 $#2
Try it online! Takes input as [durability][trees] without a separator and outputs [complete] [big] [small]. Explanation:
\d+
$*
Convert the durability to unary.
(11)+1?
Capture half the durability as $#1.
(?<-1>(i)|(?<-1>(\|)))*
Count the trees as they ...
4
Python 2, 103 \$\cdots\$ 82 80 bytes
Saved 4 6 9 11 bytes thanks to ovs!!!
f=lambda s,n:n<sum(6&ord(t)%6for t in s)and f('<'+s[:-1],n)or map(s.count,"<i|")
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Inputs a string of trees \$s\$ (as is and |s) and an axe durability \$n\$.
Outputs a list of [axe broken, small trees, large trees] where axe broken is truthy if ...
4
R, 63 bytes
using characters exactly as described in challenge (or only 54 bytes using 0,1 to represent small & big trees, and outputting 0,1 to represent FALSE/TRUE axe breaking).
function(a,t)list(sum(c<-2+2*(t<"i"))>a,table(t[cumsum(c)<=a]))
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4
J, 37 34 33 bytes
(](-:;+/@#:@])(>:2*+/\)#])' i'i.]
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-1 byte thanks to Bubbler
Converts small to 1, big to 2. Now create a filter by doubling and scan summing, and apply filter to find entries less than or equal to the left input. Take just those entries, convert to binary, and sum to get the <num big>, <num small> part of the ...
3
Keg, 32 bytes
0:&®s?⑷¦i2⑹|\|4©s⑨®s±™⑸¿⅀0=⑻©s(.
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Takes input as trees, durability and outputs big, small and whether or not the axe breaks.
Somehow, Keg beat pyth.
7
JavaScript (Node.js), 52 bytes
Takes input as (b)(n), where b is a Buffer (using the characters described in the challenge) and n is the durability of the axe.
Returns a Boolean value and 2 integers as [ broken, [ small, big ]].
b=>n=>[b.some(c=>(n-=6&c+1)<0||!++a[c%3],a=[0,0]),a]
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How?
Given an ASCII code c, we use 6 & (c +...
1
Python 3, 56 55 bytes
Saved a byte thanks to Surculose Sputum!!!
lambda l:l.sort()!=len({b/a for a,b in zip(l,l[1:])})<2
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1
R, 38 bytes
function(x)!sd(diff(y<-sort(x))/y[-1])
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sort x
calculate difference between successive elements
divide by value of each element (except first) = fold-difference
check if results are all the same (standard deviation = zero)
Fails for negative fold-differences between elements (which seems to be a true geometric sequence, ...
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