ReSy-Float (non-binary base) to decimal conversion

Question

Given the ReSy-Float number (exam WS17)

with base=4, 3 exponent bits and 2 mantissa bits (not counting hidden bit?), how can we convert it into a decimal value?

I tried the following approach, but it yields incorrect results:

edit: added original screenshot of exam

Answer 1

This is the Resy-float format where we do not have any hidden bit.

ß = 4³-1 div 2 = 31

Exponent E = 2.4² + 1.4¹ + 3.4⁰ - 31 = 8

Mantissa M = (2.4¹ + 3.4⁰).4^1-2 = 11.4^-1

The represented number x is +M.B^E = (11.4^-1).4⁸ = +11.4⁷= 180224

Comments

This would make things clear, but that would mean the exponent is 3 bits long and the mantissa is 2 bits long, right? In the exam however, we are given the following information:
http://prntscr.com/u1i2js (this is a screenshot)
This would mean that the mantissa is 3 bits long and the exponent is 2 bits long. Is that an error or am I missing something?

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So the subscript in screenshot says B,m,e . And for the floating point number, we have the subscript 4,2,3 . That means base=4, mantissa=2 and exponent=3 bits long.

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Oh, you're right. Somehow I mixed things up. Thank you very much.

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Ok one more small question: Doesn't it say in the screenshot, that this ReSyFloat-format does have a hidden bit ("zzgl. dem 'hidden bit'")?

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I guess there is confusion over the wordings. Let me try to explain.

The IEEE-754 standard has the concept of a hidden bit which is always 1, and therefore not explicitly included in the number format.

In the Resy-float number format, we do NOT have the concept of a hidden bit. Instead all bits are explicitly shown in the number format. This is what the screenshot says. That the "Hidden-bit" which is hidden in the IEEE-754 standard is explicitly shown in the Resy-float number format.

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Ah, now I really understand it! Thank you!!

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Where did the 4^(1-2) for the calculation for the mantissa come from?

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That is because in the floating point number format, the m digit mantissa is represented as a fixed point number with m-1 digits to the right of the decimal point. Therefore to get the mantissa value, one has to multiply the radix-B value of m digit mantissa by B^{-(m-1)}. In this case, B=4 and m=2. Thus 4^{-(2-1)} = 4^{1-2}.

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I see. Thank you!

Answer 2

You use 2^3-1 -1 for the offset \beta. Since it's base 4 there should be 4^....

Also note that \beta = (B^e-1) div 2 = (4³-1) div 2 = 31

Comments

Thank you, but even if I calculate the bias as 4^(3-1)-1 the result is still in the range of 10^16, which is a lot larger than the solution.

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I added the correct formula for \beta to the answer. Does it work with that?

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The result is still in the millions. But isn't (4^3-1)/2=8 ?

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No, the -1 is not in the exponent. I think you computed the mantissa wrong: 2.3_4 = 2.75. And 2.75 * 4^(39-31) = 180224

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Ah, ok. How did you calculate the mantissa? (I can't make sense of 2.3_4 at the moment)

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Mantissa is 23_4 (last two digits). Since we don't have a hidden bit it read as a fixed-point number 2.3_4. Converting to base 10 that is 2 + 3*1/4 = 2.75

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Ok, I got it now, thanks!