(x=scan())%in%(2*4^(n=0:x)-2^n-1)
One challenge on code golf is to find the shortest possible code to identify whether or not an integer belongs to the binary cyclops numbers which binary expansion is 0, 101, 11011, 1110111, 111101111, &tc. The n-th such number being
this leads to the above solution in R (26 bits). The same length as the C solution [which I do not get]
f(n){n=~n==(n^=-~n)*~n/2;}
And with shorter versions in many esoteric languages I had never heard of, like the 8 bits Brachylog code
ḃD↔Dḍ×ᵐ≠
or the 7 bits Jelly
B¬ŒḂ⁼SƊ
As a side remark, since this was not the purpose of the game, the R code is most inefficient in creating a set of size (x+1), with most terms being Inf.
Xi'an's Og 
September 1, 2019 at 9:38 am
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March 28, 2019 at 9:08 pm
This is the closest I can get to the logic of the C solution in R. Hoping this survives the forum software:
cyclops = function(n) {m=bitwXor(n,n+1);n==(m+2)*floor(m/2)}
If n=(2^k-1)*(2^(k+1)+1), then using the binary representation of integers, a bitwise exclusive OR comparison between n and n+1 gives us m=2^(k+1)-1. So m+2 gives the larger factor and floor(m/2) the smaller one. If we can reconstruct n from its factors then it is a cyclops number.
You can shave off a few characters, avoiding brackets and the floor function with:
cyclops = function(n) {m=bitwXor(n,n+1);2*n+2==m*m+m}
The C code is extremely compact because bit-wise operators are built in (we need the bitwXor function), you can rely on integer arithmetic (R coerces to double), and you can overwrite the input in the middle of an expression (you can’t do this in a functional language).
March 28, 2019 at 10:24 pm
Thank you Martyn! Makes more sense to me than the C solution. But can you gain a few bytes by not using a function and scan() instead?
March 29, 2019 at 11:25 am
Here’s a version with scan().
m=bitwXor(n<-scan(),n+1);n==(m+2)*(m-1)/2
This is vectorized so you can give as many numbers as you want.
March 29, 2019 at 12:26 pm
42 bytes, just a wee bit above the R solution of 26 bytes!
March 28, 2019 at 7:19 am
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