CODEWITHSUNDEEP
python
Ben Longo
Ben Longo

Reputation: 171

Fastest way to find nearest triangle number?

In python I need a function that takes an integer and returns the absolute value of n minus the nearest triangle number to n. The way I'm doing it now is by generating a list of all the triangle numbers up to n. Then using:

def closest(myList, myNumber):
    c = min(myList, key=lambda x:abs(x-myNumber))

To find the closest triangle number.

The end result list should look like:

[0, 0, 1 , 0 , 1 ,1 , 0 , 1, 2 , 1, 0 , 1, 2, 2, 1, 0, etc.]

If you have another method of generating the same result that's faster go for it.

Upvotes: 2

Views: 3458

Answers (5)

Varun Narisetty
Varun Narisetty

Reputation: 155

In my solution I have used the fact that let K = sqrt(2*N) if N is a triangular number then it will be Kth triangular number We know Kth triangular number will be K*(K+1)/2

Here is my solution in Python

from math import sqrt
from math import floor
def close_triangle(n):
m = floor(sqrt(2*n))
t1 = m*(m+1)/2       #next triangular number to n
m = m-1
t2 = m*(m+1)/2       #previous triangular number to n
diff1 = t1-n
diff2 = n - t2
if diff1 < diff2 :
  return t1
else:
  return t2
print "Nearest trianguler number";
print close_triangle(13)

Upvotes: 2

CT Zhu
CT Zhu

Reputation: 54380

A little bit of math will bring a more analytical solution:

from math import sqrt
def close_triangle2(n):
    m=int(0.5*(-1+sqrt(1+8*n))) #solve it for the explicit formula
    tan0=(m*m+m)/2              #the closest one is either this
    tan1=(m*m+3*m+2)/2          #or this
    if (n-tan0)>(tan1-n):
        return tan1-n
    else:
        return n-tan0

It will become slightly faster than the loop version from @perreal when the number is large:

In [87]:

%timeit close_triangle(111111111)
100000 loops, best of 3: 5 µs per loop
In [86]:

%timeit close_triangle2(111111111)
100000 loops, best of 3: 4.13 µs per loop

If you wonders:

In [94]:

close_triangle2((30**10 * (30**10+1)) / 2 + 100)
Out[94]:
100L
In [95]:

close_triangle((30**10 * (30**10+1)) / 2 + 100)
Out[95]:
100L

In [102]:

%timeit close_triangle((30**10 * (30**10+1)) / 2 + 100)
10000 loops, best of 3: 17.9 µs per loop
In [103]:

%timeit close_triangle2((30**10 * (30**10+1)) / 2 + 100)
100000 loops, best of 3: 12 µs per loop

Upvotes: 4

aruisdante
aruisdante

Reputation: 9105

You can leverage the triangle-root property to quickly calculate the nearest triangle number to a number, and from there easily calculate your value

from math import sqrt

def triangle_number(tnumber):
    '''
    returns the tnumberth triangle number (I.E 3 = 6)
    '''
    return (tnumber*(tnumber+1))/2

def triangle_root(number):
    '''
    returns the triangle root of a number (I.E 6 = 3)
    '''
    return (sqrt(8*number+1)-1)/2

def nearest_root(number):
    '''
    Calculates the nearest whole triangle root to a number (I.E. 5 = 3)
    '''
    t_root = triangle_root(number)
    return round(t_root)

def find_delta(number):
    '''
    Given a number, returns abs(n-nearest_triangle)
    '''
    return abs(number - triangle_number(nearest_root(number)))

Of course you could condense this into a single function for efficiency, I just spaced it out for clarity.

Upvotes: 2

linpingta
linpingta

Reputation: 2620

I have one solution that doesn't need calculate every triangle number first, but I am not sure whether it's best.

As formula of triange number is

    A(n) = n*(n-1)/2       (n is integer)

For input x, it should be in middle of A(n) and A(n+1)

    A(n) <= x <= A(n+1)

For A(n) we have:

    n * (n - 1) / 2 <= x
    (n - 1)^2 < n * (n-1) <= 2*x
    n < sqrt(2*x) + 1

For A(n+1) we have:

    (n + 1)^2 > (n + 1) * n >= 2*x
    n > sqrt(2*x) - 1

Now, if we have input number x, we could rapidly find n that makes x in [A(n),A(n+1)], then we only need to calculate distance of [abs(A(n) - x),abs(A(n+1) - x)] and make smallar one as real output.

Upvotes: 0

perreal
perreal

Reputation: 98088

May not be the best way but should be faster than your method:

def triangle_around(n):
    k = int((2*n)**0.5)
    p = (k**2 + k) / 2 
    while p < n:
        k, p = k + 1, p + k + 1 
    return (p - k, p)

def close_triangle(n):
    t = triangle_around(n)
    return min(t[1] - n, n - t[0])

print close_triangle(8)    # gives 2

Upvotes: 1

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