CODEWITHSUNDEEP
pythonnumpybinary
user1205371
user1205371

Reputation:

Largest power of 2 no larger than a given value

Looking for something along the lines of:

t = 1
get_floor_bin(t)
#1

t = 2
get_floor_bin(t)
#2

t = 3
get_floor_bin(t)
#2

t = 7
get_floor_bin(t)
#4

t = 15
get_floor_bin(t)
#8

t = 16
get_floor_bin(t)
#16

My current approach involves creating a list of binary numbers, and searching for each individual closest floor number within the list, but I was wondering if there are more clever approaches.

Thanks!

Upvotes: 3

Views: 230

Answers (3)

holdenweb
holdenweb

Reputation: 37013

Is this what you need? Your explanation isn't terribly easy to understand.

for i in 1, 2, 3, 7, 15, 16:
    print 1 << (i.bit_length() - 1)

This gives:

1
2
2
4
8
16

Upvotes: 18

isedev
isedev

Reputation: 19601

The expression 1 << int(math.floor(math.log(x,2))) will give you the correct result:

import math
for x in (1,2,3,7,15,16):
    print x, 1 << int(math.floor(math.log(x,2)))
1 1
2 2
3 2
7 4
15 8
16 16

Upvotes: 2

zhangxaochen
zhangxaochen

Reputation: 34017

Not sure if this way is fast enough:

In [394]: def largestPower(n):
     ...:     return int('1'+'0'*(len(bin(n))-3), 2)

In [395]: largestPower(15)
Out[395]: 8

In [396]: largestPower(1)
Out[396]: 1

In [397]: largestPower(7)
Out[397]: 4

Upvotes: 0

Related Questions