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Print the first n numbers of the fibonacci sequence in one expression

So, I've been chatting a bit with Python lately, and I'm trying to find a way to output the nth number of a fibonacci sequence in a single expression. This is the code that I have written so far:

(lambda f: f if f<2 else (f-1)+(f-2))(n)
# n == 1 -> 1
# n == 2 -> 1
# n == 3 -> 3
# n == 4 -> 5
# n == 5 -> 7
....

However, as I noted above, this simply outputs a set of odd numbers. I am confused why this is happening, because if I have to rewrite this as a named lambda function, it will look something like this:

f = lambda n: n if n<2 else f(f-1)+f(f-2)
# f(1) -> 1
# f(2) -> 1
# f(3) -> 2
# f(4) -> 3
...
# f(10) -> 55
...

, Lambda Calculus, , , , Python . Y combinator -, , -.

, , , , - . , - :

[(lambda f: f if f<2 else (f-1)+(f-2))(n) for n in range(10)]

x .

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JavaScript, # C, .

EDIT: , :

[(lambda f: (lambda x: f(lambda v: x(x)(v)))(lambda x: f(lambda v: x(x)(v))))(lambda f:(lambda n: n if n<2 else f(n-1)+f(n-2)))(y) for y in range(10)]

, , , , , - .

+6
5

:

(lambda f: (4 << f * (3 + f)) // ((4 << 2 * f) - (2 << f) - 1) & ((2 << f) - 1))(n)

:

0, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, ...

1, . .

@lehiester F71, 308061521170130 308061521170129 .

+2

, :

>>> g = lambda f: f if f < 2 else g(f-1)+g(f-2)
>>> [g(n) for n in range(10)]
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
+3

, , , - . , .

fib = (lambda n: (lambda fib: fib(fib, [], n, None))(lambda fib, arr, i, _: arr if i == 0 else fib(fib, arr, i-1, arr.append(1) if len(arr) < 2 else arr.append(arr[-1]+arr[-2]))))

, . (lambda fib: fib(fib, [], n, None)) , . - . .

lambda fib, arr, i, _: arr if i == 0 else fib(fib, arr, i-1, arr.append(1) if len(arr) < 2 else arr.append(arr[-1]+arr[-2]))) . fib , arr - , , , fib n . , i 0. - : , arr.

, . N=500 .

, (, , , - ):

(lambda n: ((lambda fib: fib(fib,n+1))(lambda fib, i: (1 if i <= 2 else fib(fib,i-2) + fib(fib,i-1)))))(N)

, @cdlane:

%timeit [0, 1] + [(4<<n*(3+n)) // ((4<<2*n)-(2<<n)-1) & ((2<<n)-1) for n in range(N)][1:]
10 loops, best of 3: 88.3 ms per loop

, @lehiester:

%timeit [int(round((lambda n: ((1+5**0.5)**n-(1-5**0.5)**n)/(2**n*5**0.5))(x))) for x in range(N)]
1000 loops, best of 3: 1.49 ms per loop

:

%timeit (lambda n: (lambda fib: fib(fib, [], n, None))(lambda fib, arr, i, _: arr if i == 0 else fib(fib, arr, i-1, arr.append(1) if len(arr) < 2 else arr.append(arr[-1]+arr[-2]))))(N)
1000 loops, best of 3: 434 us per loop

, :

%timeit (lambda n: (lambda arr, fib_supp: [arr] +  [fib_supp(arr) for i in xrange(n)])([], (lambda arr: arr.append(1) if len(arr) < 2 else arr.append(arr[-1]+arr[-2])))[0])(N)
1000 loops, best of 3: 346 us per loop

, . , setitem . , , . , , , :

%timeit (lambda n: (lambda arr, fib_supp: any(fib_supp(i, arr) for i in xrange(2,n)) or arr)([1] * n, (lambda i, arr: arr.__setitem__(i,(arr[i-1]+arr[i-2])))))(N)
1000 loops, best of 3: 385 us per loop
+3

- , - - Python, reflexive, .

@TerryA, , x Fibonacci .

, :

[int(round((lambda n: ((1+5**0.5)**n-(1-5**0.5)**n)/(2**n*5**0.5))(x)))
 for x in range(10)]

, x , , float x=600 , , - @cdlane, x=71, .. x in range(72).

EDIT: @cdlane , x . , , .

[0, 1] + [(4<<n*(3+n)) // ((4<<2*n)-(2<<n)-1) & ((2<<n)-1)
          for n in range(10)][1:]
+1

- Python

-, , , , python, lambdas

U = lambda f: f (f)

Y = U (lambda h: lambda f: f (lambda x: h (h) (f) (x)))

loop = Y (lambda recur: lambda acc: lambda a: lambda b: lambda n:
  acc if n == 0 else
    recur (acc + [a]) (a + b) (a) (n - 1))

fibonacci = loop ([]) (0) (1)

print (fibonacci (10))
# [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]

, lambdas U, Y, loop fibonacci - ,

# in Y, replace U with its definition
Y = (lambda f: f (f)) (lambda h: lambda f: f (lambda x: h (h) (f) (x)))
# in loop, replace Y with its definition
loop = (lambda f: f (f)) (lambda h: lambda f: f (lambda x: h (h) (f) (x))) (lambda recur: lambda acc: lambda a: lambda b: lambda n:
  acc if n == 0 else
    recur (acc + [a]) (a + b) (a) (n - 1))
# in fibonacci, replace loop with its definition
fibonacci = (lambda f: f (f)) (lambda h: lambda f: f (lambda x: h (h) (f) (x))) (lambda recur: lambda acc: lambda a: lambda b: lambda n:
  acc if n == 0 else
    recur (acc + [a]) (a + b) (a) (n - 1)) ([]) (0) (1)

fibonacci , - print...

# in print, replace fibonacci with its definition
print ((lambda f: f (f)) (lambda h: lambda f: f (lambda x: h (h) (f) (x))) (lambda recur: lambda acc: lambda a: lambda b: lambda n:
  acc if n == 0 else
    recur (acc + [a]) (a + b) (a) (n - 1)) ([]) (0) (1) (10))
# [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]

,

( Python), -

,

JavaScript. JS , , , ( ) - ,

const U = f =>
  f (f)

const Y =
  U (h => f => f (x => h (h) (f) (x)))

const loop = Y (recur => acc => a => b => n =>
  n === 0
    ? acc
    : recur (acc.concat ([a])) (a + b) (a) (n - 1))

const fibonacci =
  loop ([]) (0) (1)

console.log (fibonacci (10))
// [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
Hide result
// in Y, replace U with its definition
Y = (f => f (f)) (h => f => f (x => h (h) (f) (x)))
// in loop, replace Y with its definition
loop = (f => f (f)) (h => f => f (x => h (h) (f) (x))) (recur => acc => a => b => n =>
  n === 0
    ? acc
    : recur (acc.concat ([a])) (a + b) (a) (n - 1))
// in fibonacci, replace loop with its definition
fibonacci = (f => f (f)) (h => f => f (x => h (h) (f) (x))) (recur => acc => a => b => n =>
  n === 0
    ? acc
    : recur (acc.concat ([a])) (a + b) (a) (n - 1)) ([]) (0) (1)

fibonacci , - console.log...

, ,

console.time ('fibonacci (500)')
console.log ((f => f (f)) (h => f => f (x => h (h) (f) (x))) (recur => acc => a => b => n =>
  n === 0
    ? acc
    : recur (acc.concat ([a])) (a + b) (a) (n - 1)) ([]) (0) (1) (500))
console.timeEnd ('fibonacci (500)')
// [ 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ... 490 more items ]
// fibonacci (500): 3 ms
Hide result

. Y-combinator . , . , ^ _ ^

+1

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