def square(number):
if number < 1 or number > 64:
raise ValueError('square must be between 1 and 64')
return 1 << number - 1
def total():
return (1 << 64) - 1
Instead of using math for calculation, you can set a bit in the correct position for the number of grains on a square.
To understand how this works, consider just two squares that are represented in binary bits as 00.
You use the left-shift operator to set 1 at the position needed to make the correct decimal value.
- To set the one grain on Square One you shift
1for0positions to the left. So, ifnumberis1for square One, you subtractnumberby1to get0, which will not move it any positions to the left. The result is binary01, which is decimal1. - To set the two grains on Square Two you shift
1for1position to the left. So, ifnumberis2for square Two, you subtractnumberby1to get1, which will move it1position to the left. The result is binary10, which is decimal2.
| Square | Shift Left By | Binary Value | Decimal Value |
|---|---|---|---|
| 1 | 0 | 0001 | 1 |
| 2 | 1 | 0010 | 2 |
| 3 | 2 | 0100 | 4 |
| 4 | 3 | 1000 | 8 |
For total we want all of the 64 bits set to 1 to get the sum of grains on all sixty-four squares.
The easy way to do this is to set the 65th bit to 1 and then subtract 1.
To go back to our two-square example, if we can grow to three squares, then we can shift 1 two positions to the left for binary 100,
which is decimal 4.
By subtracting 1 we get 3, which is the total number of grains on the two squares.
| Square | Binary Value | Decimal Value |
|---|---|---|
| 3 | 0100 | 4 |
| Square | Sum Binary Value | Sum Decimal Value |
|---|---|---|
| 1 | 0001 | 1 |
| 2 | 0011 | 3 |