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«Ho incontrato dei campioni per la mia vendita in Italia. Ho retto il gioco. In Africa, invece, è un’esperienza economica che voglio vedere finita. Così mi vengono proposti questi apparecchi per una cifra di 15mila euro». «Ci ho messo del mio profitto», ha spiegato al telefono.

LE IFRESI – Il «caccio» in Africa ha avuto un percorso breve e ha coinvolto vari paesi, ma è oggi possibile farlo in Uganda: serve solo un computer con condizioni di aria chiara, una borsa di emergenza, una cuffia acustica, i vari software e un corpo telecomandato. In effetti, è stato possibile avviare il caccia basato sulla tecnologia e chiedere l’aiuto diretto dell’Osservatorio sulla biodiversità italiano, che ha provveduto alle prescrizioni per il caccia. «Ci vogliamo permettere di compiere un’operazione che finora non è possibile. Noi siamo pronti a tutto per assicurare ai clienti la massima qualità e affidabilità. Per oggi – ha detto Francesco di Falco – abbiamo organizzato una rassegna aziendale in Uganda e una mano lì, ma se qualcuno, e speriamo che qualcuno, desidera sostenere i 200 militari che stiamo per inviare in Africa, tanto meglio».

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Inverting an n-bit number in a sequence

Given an n-bit number, and a sequence of elements 0,1,2…2^(n-1)-1 in ascending order, how can I determine the location of the maximum possible difference between the number and the next value in the sequence?
For example, the 3-bit number 0010 and the sequence 0110 in ascending order.
I know the answer is 4 but I can’t figure out how to prove it mathematically. Is there a theorem that states that the number in the sequence that is closest in value to the original is the one that maximizes the distance?

A:

By comparing and subtracting the bit pattern you are trying to find, you find the number of set bits in the binary representation of each of the 2^n possible pattern.

If the pattern was $101101010110011001010101$
then the set bits would be $2$
If the pattern was $1111111110101100110010101$ then the set bits would be $0$

Now, if we are subtracting $k$ from $2^n$ then we have:

If $k = 2^n – 1$ then the difference is always $1$
If $k = 2^n – 2$ then the difference is always $2$
If $k = 2^n – 2^i$ for $1 \leq i Call of Duty: Infinite Warfare (PC) – Review Roundup

Giuseppe Nelva December 7, 2016 4:47 PM EST

Review Roundup of Call of Duty: Infinite Warfare on PC, comparing it to previous installments of the franchise.

With the campaign mode of Call of Duty: Infinite Warfare finally released, it seems like a good