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1 | | -#Finding the largest power of 2 (most significant bit in binary form), which is less than or |
| 1 | +#Finding the largest power of 2 (most significant bit in binary form), which is less than or |
2 | 2 | equal to the given number N. |
3 | 3 |
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4 | | -Let’s say binary form of a N is {1111}base2 which is equal to 15. |
| 4 | +Let’s say binary form of a N is {1111}base2 which is equal to 15. |
5 | 5 | 15 = 2^4-1, where 4 is the number of bits in N. |
6 | 6 |
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7 | 7 | This property can be used to find the largest power of 2 less than or equal to N. How? |
8 | 8 | If we somehow, change all the bits which are at right side of the most significant bit of N to 1, then the number will become x + (x-1) = 2 * x -1 , where x is the required answer. |
9 | 9 | Example: |
10 | | -Let’s say N = 21 = {10101}, here most significant bit is the 4th one. (counting from 0th digit) and so the answer should be 16. |
| 10 | +Let’s say N = 21 = {10101}, here most significant bit is the 4th one. (counting from 0th digit) and so the answer should be 16. |
11 | 11 | So lets change all the right side bits of the most significant bit to 1. Now the number changes to |
12 | | -{11111} = 31 = 2 * 16 -1 = Y (let’s say). |
13 | | -Now the required answer is (Y+1)>>1 or (Y+1)/2. |
| 12 | +{11111} = 31 = 2 * 16 -1 = Y (let’s say). |
| 13 | +Now the required answer is (Y+1)>>1 or (Y+1)/2. |
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