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\begin_body

\begin_layout Standard
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\backslash
exerc1[10]
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How would Avogadro's number and Planck's constant (3) be represented in base 100,
 excess 50,
 four-digit floating point notation?
 (This would be the representatioon used by 
\family typewriter
MIX
\family default
,
 as in (4),
 when the byte size is 100.)
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answer
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hfil 
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Avogadro's number
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quad
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hfil 
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Planck's constant
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quad
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\backslash
hfil
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mixbox{
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byte{+}
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byte{62}
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byte{60}
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byte{22}
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byte{14}
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byte{0}}
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hfil
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mixbox{
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byte{+}
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byte{37}
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byte{66}
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byte{26}
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byte{10}
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byte{0}}
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hfil
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\backslash
rexerc4[16]
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Assume that 
\begin_inset Formula $b=10$
\end_inset

,
 
\begin_inset Formula $p=8$
\end_inset

.
 What result does Algorithm A give for 
\begin_inset Formula $(50,+.98765432)\oplus(49,+.33333333)$
\end_inset

?
 For 
\begin_inset Formula $(53,-.99987654)\oplus(54,+.10000000)$
\end_inset

?
 For 
\begin_inset Formula $(45,-.50000001)\oplus(54,+.10000000)$
\end_inset

?
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\begin_layout Standard
\begin_inset Formula 
\begin{align*}
(50,+.98765432)\oplus(49,+.33333333) & \to(50,+1.020987653)\to(51,+.10209877)\\
(53,-.99987654)\oplus(54,+.10000000) & \to(54,.000012346)\to(50,+.12346000)\\
(45,-.50000001)\oplus(54,+.10000000) & \to(54,+.09999999949999999)\to(53,+.99999999)
\end{align*}

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\backslash
rexerc5[24]
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Let us day that 
\begin_inset Formula $x\sim y$
\end_inset

 (with respect to a given radix 
\begin_inset Formula $b$
\end_inset

) if 
\begin_inset Formula $x$
\end_inset

 and 
\begin_inset Formula $y$
\end_inset

 are real numbers satisfying the following conditions:
\begin_inset Formula 
\begin{gather*}
\lfloor x/b\rfloor=\lfloor y/b\rfloor;\\
x\bmod b=0\iff y\bmod b=0;\\
0<x\bmod b<\tfrac{1}{2}b\iff0<y\bmod b<\tfrac{1}{2}b;\\
x\bmod b=\tfrac{1}{2}b\iff y\bmod b=\tfrac{1}{2}b;\\
\tfrac{1}{2}b<x\bmod b<b\iff\tfrac{1}{2}b<y\bmod b<b.
\end{gather*}

\end_inset

Prove that if 
\begin_inset Formula $f_{v}$
\end_inset

 is replaced by 
\begin_inset Formula $b^{-p-2}F_{v}$
\end_inset

 between steps A5 and A6 of Algorithm A,
 where 
\begin_inset Formula $F_{v}\sim b^{p+2}f_{v}$
\end_inset

,
 the result of that algorithm will be unchanged.
 (If 
\begin_inset Formula $F_{v}$
\end_inset

 is an integer and 
\begin_inset Formula $b$
\end_inset

 is even,
 this operation essentially truncates 
\begin_inset Formula $f_{v}$
\end_inset

 to 
\begin_inset Formula $p+2$
\end_inset

 places while remembering whether any nonzero digits have been dropped,
 thereby minimizing the length of register that is needed for the addition in step A6.)
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answer 
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Note that the conditions are equivalent to saying that 
\begin_inset Formula $\lfloor2x/b\rfloor=\lfloor2y/b\rfloor$
\end_inset

 and that 
\begin_inset Formula $2x\bmod b=0\iff2y\bmod b=0$
\end_inset

.
 This would mean that 
\begin_inset Formula $\lfloor2F_{v}/b\rfloor=\lfloor2b^{p+1}(b^{-p-2}F_{v})\rfloor=\lfloor2b^{p+1}f_{v}\rfloor$
\end_inset

,
 and that 
\begin_inset Formula $2b^{p+1}f_{v}\in\mathbb{Z}\iff2b^{p+1}(2b^{-p-2}F_{v})\in\mathbb{Z}$
\end_inset

.
 If 
\begin_inset Formula $2b^{p+1}f_{v}\in\mathbb{Z}$
\end_inset

,
 then 
\begin_inset Formula $b^{-p-2}F_{v}=f_{v}$
\end_inset

 and nothing changes.
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\begin_layout Standard
Now let's assume that 
\begin_inset Formula $2b^{p+1}f_{v}\notin\mathbb{Z}$
\end_inset

,
 and let's denote with 
\begin_inset Formula $^{\prime}$
\end_inset

 the values obtained by Algorithm A when substituting 
\begin_inset Formula $f_{v}$
\end_inset

 by
\begin_inset Formula $b^{-p-2}F_{v}$
\end_inset

.
 Because 
\begin_inset Formula $b^{p}f_{u}\in\mathbb{Z}$
\end_inset

,
 we have 
\begin_inset Formula $b^{p+2}f_{w}\sim b^{p+2}f'_{w}$
\end_inset

 and 
\begin_inset Formula $2b^{p+1}f_{w},2b^{p+1}f'_{w}\notin\mathbb{Z}$
\end_inset

.
 This means that,
 in step N1,
 
\begin_inset Formula $|f|\geq1\iff|f'|\geq1$
\end_inset

 and 
\begin_inset Formula $f\neq0$
\end_inset

,
 and N4,
 if run,
 preserves these conditions.
 It also means that step N3 runs at most once,
 as the opposite would imply that 
\begin_inset Formula $|bf'_{w}|<1/b$
\end_inset

,
 which happens if and only if 
\begin_inset Formula $|bf_{w}|<1/b$
\end_inset

,
 if and only if 
\begin_inset Formula $|f_{w}|<1/b^{2}$
\end_inset

,
 but 
\begin_inset Formula $f_{u}\geq.1$
\end_inset

 because inputs are normalized,
 so this would imply 
\begin_inset Formula $f_{v}>.09$
\end_inset

 and 
\begin_inset Formula $e_{u}-e_{v}\leq1$
\end_inset

,
 so 
\begin_inset Formula $f_{w}$
\end_inset

 would have at most 
\begin_inset Formula $p+1$
\end_inset

 digits and 
\begin_inset Formula $2b^{p+1}f_{w}\in\mathbb{Z}\#$
\end_inset

.
 Therefore N2 produces the same results for 
\begin_inset Formula $f$
\end_inset

 and 
\begin_inset Formula $f'$
\end_inset

 every time it runs and,
 when we reach step N5,
 
\begin_inset Formula $b^{p+1}f\sim b^{p+1}f'$
\end_inset

 and 
\begin_inset Formula $2b^{p}f,2b^{p}f'\notin\mathbb{Z}$
\end_inset

,
 so step N5 produces the same result in both cases and the rest of the algorithm runs over the same state and produces the same result.
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\backslash
rexerc11[M20]
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\end_inset

Give an example of normalized,
 excess 50,
 eight-digit floating decimal numbers 
\begin_inset Formula $u$
\end_inset

 and 
\begin_inset Formula $v$
\end_inset

 for which rounding overflow occurs in multiplication.
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answer 
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\end_inset

An illustrative example would be
\begin_inset Formula 
\[
(50,.99999990)\otimes(50,.10000001).
\]

\end_inset

The result would be 
\begin_inset Formula $(50,.0999999999999990)$
\end_inset

,
 which becomes 
\begin_inset Formula $(50,.10000000)$
\end_inset

 after shifting left in N3 and then right in N4 after N5 rounds up to 
\begin_inset Formula $(49,1.)$
\end_inset

.
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