Why is the floating point number of 0.1 larger than 0.1, but when added 10 times, it is smaller than 1.0 [Part 2]

Part 1

This is a continuation of this article.

Why is the floating point number of 0.1 larger than 0.1, but it is smaller than 1.0 when added 10 times [Part 1]

(Copy from the first part) When 0.1 is represented by a floating point number

The sign, exponent, and mantissa when 0.1 is expressed as a floating point number are as follows.

Code= 0
index= 01111111011
mantissa= 1001100110011001100110011001100110011001100110011010

The exponent is 1019 in decimal. The floating-point exponent has 1023 added to it, so subtracting this gives -4.

(Copy from the first part) How much is added to the result when 0.1 is added

The table below shows the calculation results when 0.1 is added to the odd rows excluding the header, and the differences between the even rows.

(Copy from the first part) Procedure for adding floating point numbers

Addition of positive floating point numbers is done by the following procedure.

① Match the smaller index to the larger index (2) Adjust by reducing the mantissa as the index is increased. ③ Add the mantissa ④ If a carry occurs, add 1 to the index and reduce the mantissa by that amount. ⑤ Round the overflowing digits of the mantissa (even rounding)

(Copy from the first part) Protective girder and sticky bit

Guard digits are the area to store the overflowing digits. The Sticky bit is a boolean value that indicates whether or not the number below the protection digit contains a number greater than or equal to 1.

In the first part, it was mentioned that one protective digit is sufficient for adding positive numbers, but two digits are required when expanding to negative numbers. The following uses two protection digits and a sticky bit. The two digits (2 bits) of the protection digit are named Guard bit and Round bit from the top.

Calculation process when 0.1 is added 10 times

Check the calculation process when adding 0.1 step by step.

Expression used this time

This time, I will explain using the following figure.

0.1 + 0.1 calculation process

Check the calculation process of 0.1 + 0.1.

① Match the smaller index to the larger index (2) Adjust by reducing the mantissa as the index is increased. Since the indexes are the same, nothing is done.

③ Add the mantissa It is the result of adding

④ If a carry occurs, add 1 to the index and reduce the mantissa by that amount. Since there is a carry, add 1 to the exponent and divide the mantissa by 2. Since it is a binary number, dividing by 2 shifts the digit by one to the right (bit shift).

⑤ Round the overflowing digits of the mantissa (even rounding) The only overflowing bit is 0, so truncate it.

Above,

index= -3 (Decimal number)
mantissa= 1001100110011001100110011001100110011001100110011010 (Binary number)

Was obtained. If you convert this to a decimal number,

0.200000000000000011102230246251565404236316680908203125

is.

Calculation process of 0.2 (cumulative sum) + 0.1

Add 0.1 to the value obtained above.

① Match the smaller index to the larger index (2) Adjust by reducing the mantissa as the index is increased. The exponent of 0.2 is 1 larger, so add 1 to the exponent of 0.1 and shift the mantissa by 1 to the right.

③ Add the mantissa It is the result of adding up.

④ If a carry occurs, add 1 to the index and reduce the mantissa by that amount. Since there is a carry, add 1 to the exponent and shift the mantissa to the right 1.

⑤ Round the overflowing digits of the mantissa (even rounding) The overflowing digit is 10, which is exactly half, so round it in even numbers. This time, the least significant bit of the mantissa is 1, so round it up.

Above,

index= -2 (Decimal number)
mantissa= 11001100110011001100110011001100110011001100110100 (Binary number)

Was obtained. If you convert this to a decimal number,

0.3000000000000000444089209850062616169452667236328125

is.

In 0.2 (cumulative sum) + 0.1, 0.1 is originally larger than 0.1 in floating point number, and the added result is rounded up further, so print You will see results that are not what you expected when you did.

print(0.1 + 0.1 + 0.1)
# => 0.30000000000000004

Calculation process of 0.3 (cumulative sum) + 0.1

Add 0.1 to the value obtained above.

① Match the smaller index to the larger index (2) Adjust by reducing the mantissa as the index is increased. Since the exponent of 0.3 is 2 larger, add 2 to the exponent of 0.1 and shift the mantissa by 2 to the right.

③ Add the mantissa It is the result of adding up.

④ If a carry occurs, add 1 to the index and reduce the mantissa by that amount. No carry up, so do nothing

⑤ Round the overflowing digits of the mantissa (even rounding) The overflowing digit is 10, which is exactly half, so round it in even numbers. This time, the least significant bit of the mantissa is 0, so it is truncated.

Above,

index= -2 (Decimal number)
mantissa= 1001100110011001100110011001100110011001100110011010 (Binary number)

Was obtained. If you convert this to a decimal number,

0.40000000000000002220446049250313080847263336181640625

is.

Calculation process of 0.4 (cumulative sum) + 0.1

Add 0.1 to the value obtained above.

① Match the smaller index to the larger index (2) Adjust by reducing the mantissa as the index is increased. Since the exponent of 0.4 is 2 larger, add 2 to the exponent of 0.1 and shift the mantissa by 2 to the right.

③ Add the mantissa It is the result of adding up. All values in the mantissa are now 0.

④ If a carry occurs, add 1 to the index and reduce the mantissa by that amount. Since there is a carry, add 1 to the exponent and shift the mantissa by 1 to the right.

⑤ Round the overflowing digits of the mantissa (even rounding) The overflowing digit is 010, so it will be truncated.

Above,

index= -1 (Decimal number)
mantissa= 0000000000000000000000000000000000000000000000000000 (Binary number)

Was obtained. Converting this to a decimal number,

0.5

is. It was a coincidence that it became perfect.

Calculation process of 0.5 (cumulative sum) + 0.1

Add 0.1 to the value obtained above.

① Match the smaller index to the larger index (2) Adjust by reducing the mantissa as the index is increased. Since the exponent of 0.5 is 3 larger, add 3 to the exponent of 0.1 and shift the mantissa by 3 to the right.

③ Add the mantissa It is the result of adding up.

④ If a carry occurs, add 1 to the index and reduce the mantissa by that amount. No carry up, so do nothing

⑤ Round the overflowing digits of the mantissa (even rounding) The overflowing digit is 010, so it will be truncated.

Above,

index= -1 (Decimal number)
mantissa= 11001100110011001100110011001100110011001100110011 (Binary number)

Was obtained. Converting this to a decimal number,

0.59999999999999997779553950749686919152736663818359375

is.

Calculation process of 0.6 (cumulative sum) + 0.1, 0.7 (cumulative sum) + 0.1, 0.8 (cumulative sum) + 0.1

Since 0.6 to 0.8 are almost the same as 0.5 (cumulative sum) + 0.1, they are omitted. Values less than 1 are added in this range due to value truncation.

0.9 (cumulative sum) + 0.1 calculation process

This is almost the same as 0.5 (cumulative sum) + 0.1, but let's see what the result looks like.

① Match the smaller index to the larger index (2) Adjust by reducing the mantissa as the index is increased. Since the exponent of 0.9 is 3 larger, add 3 to the exponent of 0.1 and shift the mantissa by 3 to the right.

③ Add the mantissa It is the result of adding up.

④ If a carry occurs, add 1 to the index and reduce the mantissa by that amount. No carry up, so do nothing

⑤ Round the overflowing digits of the mantissa (even rounding) The overflowing digit is 010, so it will be truncated.

Above,

index= -1 (Decimal number)
mantissa= 1111111111111111111111111111111111111111111111111111 (Binary number)

Was obtained. Converting this to a decimal number,

0.99999999999999988897769753748434595763683319091796875

is.

Up to this point, we have confirmed the operation when 0.1 is added 10 times.

Summary

The answer is, why is 0.1 slightly larger than 0.1 when converted to a floating point number, but less than 1.0 when added 10 times? That's because when you add two floating point numbers, you round the numbers that don't fit in the mantissa.

Reference material

-About floating point arithmetic

• 3.6 Floating Point(PDF)
• Why you need a Guard bit, in addition to the Round and Sticky bits(PDF) -Floating Point Rounding Direction and Properties (1) -decimal --- Arithmetic operations on decimal fixed and floating point numbers

Tools used, etc.

-Floating Point Internal Representation Simulator -Bin, octa, decimal, hexadecimal reciprocal conversion tool