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# Iterative fac = fn|n| y = 1 for i in 1..n y = y*i end return y end # Recursive fac = |n| 1 if n==0 else n*fac(n-1) # Functional fac = |n| (1..n).reduce(1,|x,y| x*y) # As a dynamic system Fac = |[n,y]| [n+1,y*(n+1)] fac = |n| (Fac^n)([0,1])[1] fac = |n| Fac.orbit([0,1]).skip(n)()[1] # Tail-recursive function call(f,n,y) x = f(n,y) while x: Function x = x() end return x end Fac = |n,y| y if n==0 else || Fac(n-1,y*n) fac = |n| call(Fac,n,1) # Recursive, by Y combinator Y = |F| (|x| x(x))(|x| F(|n| x(x)(n))) fac = Y(|f||n| 1 if n==0 else n*f(n-1)) # Corecursive fac = |n| fn*|| y=1; k=1 while true yield y y, k = y*k, k+1 end end.skip(n)() # By counting all permutations inside of the # space of all mappings from 0..n-1 to 0..n-1 fac = |n| (list(n)^n).count(|t| (0..n-1).all(|x| x in t)) # By generating all permutations and counting them use itertools: permutations fac = |n| permutations(n).count() # Print them to the terminal (0..9).map(|n| [n,fac(n)]).each(print)
# Generate all permutations of a list function permutations(a) if len(a)<=1 return [copy(a)] else b = [] x = a[..0] for p in permutations(a[1..]) for i in 0..len(a)-1 b.push(p[..i-1]+x+p[i..]) end end return b end end # Return an interator instead of a list function permutations(a) return fn*|| if len(a)<=1 yield copy(a) else x = a[..0] for p in permutations(a[1..]) for i in 0..len(a)-1 yield p[..i-1]+x+p[i..] end end end end end for n in 0..4 permutations(list(n)).each(print) end # Generate all permutations of an ordered # list in lexicographical order function permutations(a) if len(a)==0 return [[]] else b = [] for i in 0..len(a)-1 for x in permutations(a[..i-1]+a[i+1..]) b.push([a[i]]+x) end end return b end end # Return an iterator instead of a list function permutations(a) return fn*|| if len(a)==0 yield [] else for i in 0..len(a)-1 for x in permutations(a[..i-1]+a[i+1..]) yield [a[i]]+x end end end end end # Take any iterable perm = |a| permutations(list(a))
# Generate all k-permutations of an ordered # list in lexicographical order perm = |a,k=len(a)| [[]] if k==0 else list([a[i]]+x for i in len(a) for x in perm(a[..i-1]+a[i+1..],k-1))
# as a list prod = |*a| [[]] if a==[] else list( [x]+t for x in a[0] for t in prod(*a[1..])) # as a set prod = |*a| {[]} if a==[] else set( [x]+t for x in a[0] for t in prod(*a[1..])) # as an iterator prod = |*a| iter([[]]) if a==[] else ( [x]+t for x in a[0] for t in prod(*a[1..])) # as a list prod = |*a| a.reduce([[]],|x,y| (x*list(y)).map(|[t,s]| t+[s]))
There is a recursive formula for the Bernoulli numbers:
This formula can be implemented directly, taking advantage of exact rational numbers in combination with long integers.
# memoizing fixed point combinator use functional: fix # rational numbers use math.rational: rat # binomial coefficient use math.cf: bc B = fix({},|B,n| 1 if n==0 else 1-rat(1,n+1)*(0..n-1).sum(|k| B(k)*bc(n+1,k))) for n in 0..8 print([n,B(n)]) end # [0, 1] # [1, 1/2] # [2, 1/6] # [3, 0] # [4, -1/30] # [5, 0] # [6, 1/42] # [7, 0] # [8, -1/30] print(B(60)) # -1215233140483755572040304994079820246041491/56786730
Alternatively:
# factorial, # Eulerian numbers of the first order, # Stirling numbers of the second kind use math.cf: fac, euler1, stirling2 B = |n| rat( n*(0..n-1).sum(|k| (-1)^k*euler1(n-1,k)), 2^n*(2^n-1) ) B = |n| (0..n).sum(|k| rat((-1)^k*fac(k)*stirling2(n+1,k+1),k+1)) B = |n| rat(1,2) if n==1 else (0..n).sum(|k| rat((-1)^k*fac(k)*stirling2(n,k),k+1)) B = |n| rat(n,2^(n+1)-2)*(0..n-1).sum(|k| rat(fac(k)*stirling2(n,k+1),(-2)^k))