Find the position of rightmost set bit of decimal in ruby
Ruby program for Find the position of rightmost set bit of decimal. Here more information.
# Ruby Program for
# Find position of rightmost set bit
class BitPosition
# Finding the right most active bit position
def rightmostActiveBit(n)
if (n <= 0)
return
end
# Example
# ———————
# n = 320
# n = (00101000000) Binary
# -n = (11011000000) (2s)
# Calculate log2
# ——————————————
# Formula : log(n & -n) / log(2) + 1
# -----------------------------------
# log(320 & -320) / log(2) + 1)
# Here : log(320 & -320) = log(64) = 4.158883
# log(2) = 0.693147
# (4.158883 / 0.693147) + 1 = (7) position
# ————————————————————————————————————————
# Calculate rightmost active bits
result = (Math.log(n & -n) / Math.log(2) + 1).to_i
print(" Number : ", n ," Result : ", result, "\n")
end
end
def main()
task = BitPosition.new()
# Test Cases
# 320 = Binary(101000000)
task.rightmostActiveBit(320)
# (1000) = Binary(1111101000)
task.rightmostActiveBit(1000)
# (153) = Binary(10011001)
task.rightmostActiveBit(153)
# (354) = Binary(101100010)
task.rightmostActiveBit(354)
# 160 = Binary(10100000)
task.rightmostActiveBit(160)
end
main()
Output
Number : 320 Result : 7
Number : 1000 Result : 4
Number : 153 Result : 1
Number : 354 Result : 2
Number : 160 Result : 6
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.
New Comment