Answer:
The summary including its given problem is outlined in the following section on the interpretation.
Explanation:
That's not entirely feasible, since at least n similarities have to be made to order n quantities. Find the finest representation where the numbers of 1 to 10 have already been arranged.
⇒ 1 2 3 4 5 6 7 8 9 10
Let's say that we identify one figure as the key then compared it towards the numbers across the left. Whether the correct number is greater, therefore, left number, are doing nothing to switch the location elsewhere.
Because although the numbers have already been categorized 2 has always been compared to 1 which would be perfect, 3 becomes especially in comparison to 2 and so much more. This should essentially take 9 moves, or nearly O(n) moves.
If we switch that little bit already
⇒ 1 3 2 4 5 6 7 8 9 10
3 Is contrasted with 1. 2 will indeed be matched against 3 as well as 2. Since 2 has indeed been exchanged, it must, therefore, be matched with 1 as there might be a case whereby each number z exchanged is greater than the number Y as well as the quantity X < Y.
Only one adjustment expanded the steps which culminated in n+1.
Answer:
In print view, you're unable to make changes to your report.
Answer and Explanation:
The bootstarp loader is also know as bootstrapping a bootstrap loader is usually located in EPROM (erasable programmable read only memory). It is a non volatile memory. The booststrap loader is automatically exicuted by the processor when we turn on the computer.Non volatile memory is that memory in which retain their content even when the power is swithed off.
Answer:
3rd_player, player3 player_score
Answer:
1. 2588672 bits
2. 4308992 bits
3. The larger the data size of the cache, the larger the area of memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.
Explanation:
1. Number of bits in the first cache
Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))
total bits = 2^15 (1+14+(32*2^1)) = 2588672 bits
2. Number of bits in the Cache with 16 word blocks
Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))
total bits = 2^13(1 +13+(32*2^4)) = 4308992 bits
3. Caches are used to help achieve good performance with slow main memories. However, due to architectural limitations of cache, larger data size of cache are not as effective than the smaller data size. A larger cache will have a lower miss rate and a higher delay. The larger the data size of the cache, the larger the area of memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.