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Challenges

# Length of a Sumac Sequence

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Heavily based on this closed challenge from SE.

# Description

A Sumac sequence starts with two non-zero integers $t_1$ and $t_2.$

The next term, $t_3 = t_1 - t_2$

More generally, $t_n = t_{n-2} - t_{n-1}$

The sequence ends when $t_n ≤ 0$. All values in the sequence must be positive.

# Challenge

Given two integers $t_1$ and $t_2$, compute the Sumac sequence, and output it's length.

If there is a negative number in the input, remove everything after it, and compute the length.

You may take the input in any way (Array, two numbers, etc.)

# Test Cases

(Sequence is included for clarification)

[t1,t2]   Sequence          n
------------------------------
[120,71]  [120,71,49,22,27] 5
[101,42]  [101,42,59]       3
[500,499] [500,499,1,498]   4
[387,1]   [387,1,386]       3
[3,-128]  [3]               1
[-2,3]    []                0
[3,2]     [3,2,1,1]         4


# Scoring

This is code-golf. Shortest answer in each language wins.

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# Raku, 19 bytes

{(|@_,*-*...0>*)-1}


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{                 }  # Anonymous code block
...         # Create a sequence
|@_                # Starting with the input
,*-*            # With each element being the difference between the previous two
0>*      # Until it is negative
(             )-1   # And return the length of the list minus one

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# Python 3, 49 47 44 35 bytes

a=lambda x,y:+(x>=0)and(1+a(y,x-y))


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-3 bytes thanks to @Hakerh400‭

-9 bytes thanks to @Jo King‭

# Python 2, 50 bytes

x,y=input()
a=0
while x>0:x,y,a=y,x-y,a+1
print(a)


An alternate solution I also came up with that I swear I should be able to make shorter, but can't think of it currently.

Try it online!

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# C (gcc), 32 bytes

f(a,b){return a>0?f(b,a-b)+1:0;}


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# Japt-N, 9 bytes

¨T©ÒßVVnU


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¨T©ÒßVVnU     :Implicit input of integers U & V
¨             :U is >=
T            :  0
Ò          :Negate the bitwise NOT of
ß         :Recursive call with arguments
V        :  U=V
VnU     :  V=U-V
:Implicit output of final result as an integer
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# JavaScript (Node.js), 27 25 bytes

(-2 thanks to JoKing)

f=(a,b)=>a<0?0:1+f(b,a-b)


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# Scala, 83 bytes

x=>x.init#::Stream.iterate(x)(x=>x++(x,x.tail).zipped.map(_-_))indexWhere(_.last<1)


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Takes input as a list of two numbers. The fact that the length can be 0 makes this a bit longer.

x =>  //List(t1, t2) a.k.a. the second partial sequence
x.init  //List(t1) a.k.a. the first partial sequence
#::     //prepended to
Stream  //an infinite stream of partial sequences
.iterate(x)  //starting with the second partial sequence
(x =>        //to which this function is repeatedly applied:
x ++         //Concatenate the previous sequence to the next values
(x, x.tail).zipped  //Zip the previous sequence and its tail
.map(_-_))          //And subtract each i-1th element from the ith element
indexWhere   //Find the index of the sequence where
(_.last<1)   //The last element is not positive

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# JavaScript (Node.js), 31 bytes

f=(a,b)=>a<0?0:b<0?1:1+f(b,a-b)


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# BQN, 15 bytesSBCS

{𝕨(1+⊢𝕊-)⟜𝕩⍟>0}


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A block function that takes $t_1$ as the left argument 𝕨 and $t_2$ as the right argument 𝕩. In BQN, 𝕊 indicates recursion on the current block function. Inside the function the idea is just to return 0 if 𝕨≤0 and otherwise advance the sequence by one, recurse, and add one to the resulting sequence length.

{𝕨(1+⊢𝕊-)⟜𝕩⍟>0}
0   # Default result of 0
𝕨         ⍟>    # If 𝕨>0:
(     )⟜𝕩      # Use 𝕩 as right arg (left is still 𝕨)
⊢ -         # Right argument ⊢ and difference -
𝕊          # Recurse


The unrelated 0s for sequence length and minimum 𝕨 are combined by making 0 the right argument and using Repeat (⍟). With 0 repetitions, Repeat will return the right argument 0, and with 1 repetition, it will call the operand function, which immediately replaces 0 with 𝕩 using After (𝕩 is a number, so it acts as a constant function).

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# Jelly, 9 bytes

Rȧ_@RÐ¿@L


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Takes $t_1$ on the left and $t_2$ on the right. R}ȧ_@RÐ¿L works given the arguments in the opposite order.

R            Ascending range from 1 to t_1 (truthy if positive),
ȧ           and:
Ð¿      collecting intermediate results, loop while
R        positive
@     on the reversed arguments:
_          subtract
@         the first from the second.
L    Return the length of the conjunction.


¿ and family, given a dyad, replace the right argument with the previous left argument on each successive iteration. This is incredibly bothersome for some tasks, but perfect for working with this exact kind of sequence.

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