Roman Gräf
Die Bitbreite einer Vorzeichenlosen Binärzahl lässt sich wie folgt errechnen:
\[b_2(X)=\lceil\frac{log(X+1)}{log(2)}\rceil\]
Für \(11758070_{10}\) Coronafälle ergibt sich eine Bitbreite von 24.
Die benötigten Stellen für Basis-7-Zahlen errechnet sich durch \(b_7(X)=\lceil\frac{X+1}{log(7)}\rceil\). Dementsprechend benötigt man für die Zahl \(11758070_{10}\) 9 Stellen.
| \(X\) | \(\lfloor\frac{X}{7}\rfloor\) | \(X\) mod \(7\) |
|---|---|---|
| \(11758070\) | \(\lfloor 11758070/7\rfloor=1679724\) | \(2\) |
| \(1679724\) | \(\lfloor 1679724/7\rfloor=239960\) | \(4\) |
| \(239960\) | \(\lfloor 239960/7\rfloor=34280\) | \(0\) |
| \(34280\) | \(\lfloor 34280/7\rfloor=4897\) | \(1\) |
| \(4897\) | \(\lfloor 4897/7\rfloor=699\) | \(4\) |
| \(699\) | \(\lfloor 699/7\rfloor=99\) | \(6\) |
| \(99\) | \(\lfloor 99/7\rfloor=14\) | \(1\) |
| \(14\) | \(\lfloor 14/7\rfloor=2\) | \(0\) |
| \(2\) | \(\lfloor 2/7\rfloor=0\) | \(2\) |
\(11758070_{10}=201641042_{7}\)
| \(X\) | \(\lfloor\frac{X}{7}\rfloor\) | \(X\) mod \(7\) |
|---|---|---|
| \(374827\) | \(\lfloor 374827/7\rfloor=53546\) | \(5\) |
| \(53546\) | \(\lfloor 53546/7\rfloor=7649\) | \(3\) |
| \(7649\) | \(\lfloor 7649/7\rfloor=1092\) | \(5\) |
| \(1092\) | \(\lfloor 1092/7\rfloor=156\) | \(0\) |
| \(156\) | \(\lfloor 156/7\rfloor=22\) | \(2\) |
| \(22\) | \(\lfloor 22/7\rfloor=3\) | \(1\) |
| \(3\) | \(\lfloor 3/7\rfloor=0\) | \(3\) |
\(374827_{10}=3120535_{7}\)
| \(X\) | \(\lfloor\frac{X}{7}\rfloor\) | \(X\) mod \(7\) |
|---|---|---|
| \(212164\) | \(\lfloor 212164/7\rfloor=30309\) | \(1\) |
| \(30309\) | \(\lfloor 30309/7\rfloor=4329\) | \(6\) |
| \(4329\) | \(\lfloor 4329/7\rfloor=618\) | \(3\) |
| \(618\) | \(\lfloor 618/7\rfloor=88\) | \(2\) |
| \(88\) | \(\lfloor 88/7\rfloor=12\) | \(4\) |
| \(12\) | \(\lfloor 12/7\rfloor=1\) | \(5\) |
| \(1\) | \(\lfloor 1/7\rfloor=0\) | \(1\) |
\(212164_{10}=1542361_{7}\)
| \(2\) | \(0\) | \(1\) | \(6\) | \(4\) | \(1\) | \(0\) | \(4\) | \(2\) | |
| \(+\) | \(3\) | \(1\) | \(2\) | \(0\) | \(5\) | \(3\) | \(5\) | ||
| Übertrag | \(1\) | \(1\) | \(1\) | ||||||
| \(=\) | \(2\) | \(0\) | \(5\) | \(0\) | \(6\) | \(1\) | \(6\) | \(1\) | \(0\) |
\(201641042_7+3120535_7=205061610_7\)
| \(2\) | \(0\) | \(5\) | \(0\) | \(6\) | \(1\) | \(6\) | \(1\) | \(0\) | |
| \(-\) | \(1\) | \(5\) | \(4\) | \(2\) | \(3\) | \(6\) | \(1\) | ||
| Übertrag | \(1\) | \(1\) | \(1\) | \(1\) | |||||
| \(=\) | \(2\) | \(0\) | \(3\) | \(2\) | \(1\) | \(6\) | \(2\) | \(1\) | \(6\) |
\(205061610_7-1542361_7=203216216_7\)
| \(a_i\) | \(X\) | \(X\times 7+a_i\) |
|---|---|---|
| \(2\) | \(0\) | \(2\) |
| \(0\) | \(2\) | \(14\) |
| \(3\) | \(14\) | \(101\) |
| \(2\) | \(101\) | \(709\) |
| \(1\) | \(709\) | \(4964\) |
| \(6\) | \(4964\) | \(34754\) |
| \(2\) | \(34754\) | \(243280\) |
| \(1\) | \(243280\) | \(1702961\) |
| \(6\) | \(1702961\) | \(11920733\) |
Nach einer Woche gibt es \(203216216_{7}=11920733_{10}\) aktive Fälle.
Da \(1000000_{10}\) eine Bitbreite von 20 hat und wir zwei Zahlen haben, brauchen wir \(20+2=22\) Paritätsbits, sowie \(2*20=40\) Datenbits falls wir Genesungen und Neuerkrankungen seperat übertragen. Eine Übertragung von Fallzahlen bis zu \(1000000000_{10}\) braucht eine Bitbreite von \(30\). Dementsprechend ist eine Übertragung aller aktiven Fälle ohne Parität (\(30\) Bits) effizienter als eine Übertragung der Genesungen und Neuerkrankungen mit Parität (\(62\) Bits).
| P | |||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | |
| 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | X |
Das durchgestrichene 1 Bit muss eine null sein, da Längs und Querparität für diese Zelle falsch sind. Dementsprechend liegen \(100932_{10}\) Genesungen vor.
| Ziffer | \(x_0\) | \(x_1\) | \(x_2\) | \(x_3\) | \(a\) | \(b\) | \(c\) | \(d\) | \(e\) | \(f\) | \(g_1\) | \(g_2\) | \(h\) | \(i\) | \(j\) | \(k\) | \(l\) | \(m\) | \(n\) | \(dp\) | \(dk\) |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| 2 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 3 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 4 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 5 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 6 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 7 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 8 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 9 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
\(a=(x_0\lor x_1\lor x_2\lor \neg x_3)\land (x_0\lor \neg x_1\lor x_2\lor x_3)\)
\(b=(x_0\lor \neg x_1\lor x_2\lor \neg x_3)\land (x_0\lor \neg x_1\lor \neg x_2\lor x_3)\)
\(c=(x_0\lor x_1\lor \neg x_2\lor x_3)\)
\(d=(x_0\lor x_1\lor x_2\lor \neg x_3)\land (x_0\lor \neg x_1\lor x_2\lor x_3)\land (x_0\lor \neg x_1\lor \neg x_2\lor \neg x_3)\)
\(e=(\neg x_0\land \neg x_1\land \neg x_2\land \neg x_3)\lor (\neg x_0\land \neg x_1\land x_2\land \neg x_3)\lor\) \((\neg x_0\land x_1\land x_2\land \neg x_3)\lor (x_0\land \neg x_1\land \neg x_2\land \neg x_3)\)
\(f=(x_0\lor x_1\lor x_2\lor \neg x_3)\land (x_0\lor x_1\lor \neg x_2\lor x_3)\land (x_0\lor x_1\lor \neg x_2\lor \neg x_3)\land\) \((x_0\lor \neg x_1\lor \neg x_2\lor \neg x_3)\)
\(g_1=(x_0\lor x_1\lor x_2\lor x_3)\land (x_0\lor x_1\lor x_2\lor \neg x_3)\land (x_0\lor \neg x_1\lor \neg x_2\lor \neg x_3)\)
\(g_2=(x_0\lor x_1\lor x_2\lor x_3)\land (x_0\lor x_1\lor x_2\lor \neg x_3)\land (x_0\lor \neg x_1\lor \neg x_2\lor \neg x_3)\)
\(h=0\)
\(i=0\)
\(j=(\neg x_0\land \neg x_1\land \neg x_2\land \neg x_3)\lor (\neg x_0\land \neg x_1\land \neg x_2\land x_3)\)
\(k=(\neg x_0\land \neg x_1\land \neg x_2\land \neg x_3)\)
\(l=0\)
\(m=0\)
\(n=0\)
\(dp=0\)
\(dk=0\)
\(G=\overline{\overline{45}(4+5)}\stackrel{\text{T12}}{=}\overline{(\overline{4}+\overline{5})(4+5)}\stackrel{\text{T8}}{=}\overline{\overline{4}4+\overline{4}5+\overline{5}4+\overline{5}5}\) \(G\stackrel{\text{T5}}{=}\overline{\overline{4}5+\overline{5}4}\stackrel{\text{T12}}{=}\overline{\overline{4}5}\cdot\overline{\overline{5}4}\stackrel{\text{T12}}{=}(\overline{4}+5)(\overline{5}+4)\stackrel{\text{T8}}{=}(\overline{4}+5)\overline{5}+(\overline{4}+5)4\) \(G\stackrel{\text{T8}}{=}\overline{4}\cdot\overline{5}+\overline{5}5+\overline{4}4+45\stackrel{\text{T5}}{=}\overline{4}\cdot\overline{5}+45\) \(F=0\overline{4}(125+\overline{3}5+\overline{3}\cdot\overline{5}+123)+\overline{0}12(\overline{3}4\overline{5}+\overline{3}4\overline{0}+\overline{3}45)+1\overline{2}3G\) \(\stackrel{\text{T5}}{=}0\overline{4}(125+\overline{3}+123)+\overline{0}12(\overline{3}4\overline{5}+\overline{3}4\overline{0}+\overline{3}45)+1\overline{2}3G\) \(\stackrel{\text{T8}}{=}0\overline{4}125+0\overline{4}\cdot\overline{3}+0\overline{4}123+\overline{0}12\overline{3}4{5}+\overline{0}12\overline{3}4\overline{0}+\overline{0}12\overline{3}4\overline{5}+1\overline{2}3G\) \(F=\overline{4}125+0\overline{4}\cdot\overline{3}+0\overline{4}123+\overline{0}12\overline{3}4\overline{5}+\overline{0}12\overline{3}4\overline{0}+\overline{0}12\overline{3}4\overline{5}+1\overline{2}3(\overline{4}\cdot\overline{5}+45)\) \(\stackrel{\text{T8}}{=}\overline{4}125+0\overline{4}\cdot\overline{3}+0\overline{4}123+\overline{0}12\overline{3}4\overline{5}+\overline{0}12\overline{3}4\overline{0}+\overline{0}12\overline{3}4\overline{5}+1\overline{2}3\overline{4}\cdot\overline{5}+1\overline{2}345\) \(\stackrel{\text{T6}}{=}12\overline{4}5+0\overline{3}\cdot\overline{4}+0123\overline{4}+\overline{0}12\overline{3}4\overline{5}+\overline{0}12\overline{3}4\overline{5}+\overline{0}\cdot\overline{0}12\overline{3}4+1\overline{2}3\overline{4}\cdot\overline{5}+1\overline{2}345\) \(\stackrel{\text{T3}}{=}12\overline{4}5+0\overline{3}\cdot\overline{4}+0123\overline{4}+\overline{0}12\overline{3}4\overline{5}+\overline{0}12\overline{3}4\overline{5}+\overline{0}12\overline{3}4+1\overline{2}3\overline{4}\cdot\overline{5}+1\overline{2}345\) \(\stackrel{\text{T8}}{=}12\overline{4}5+0\overline{3}\cdot\overline{4}+0123\overline{4}+\overline{0}12\overline{3}4(\overline{5}+\overline{5}+ONE)+1\overline{2}3\overline{4}\cdot\overline{5}+1\overline{2}345\) \(\stackrel{\text{T2}}{=}12\overline{4}5+0\overline{3}\cdot\overline{4}+0123\overline{4}+\overline{0}12\overline{3}4+1\overline{2}3\overline{4}\cdot\overline{5}+1\overline{2}345\)