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andrew-mc [135]
2 years ago
9

Why does binary addition have a rule for 1 + 1 + 1 if only two numbers are being added together?

Computers and Technology
1 answer:
dalvyx [7]2 years ago
5 0

Answer:

See below

Explanation:

Because sometimes you have to 'carry' a 1 over to the ext column when adding two binary numbers

Example :

 1 1 1

<u>+1 1 1 </u>    <==== starting in the first R column add  1 + 1  to get 0 and carry 1

                then the next column you will add   1 + 1 + 1  = 1   and carry 1 again

                    then 1 + 1 + 1= 1  and carry 1 again (to column 4)  to get

1   1  1 0

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Answer:

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3 years ago
Define the acronym WAECUP and its components
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Assuming that the components are the same as what's included in the acronym, that should be your answer.

Hope this helps!

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3 years ago
Suppose you have two arrays of ints, arr1 and arr2, each containing ints that are sorted in ascending order. Write a static meth
telo118 [61]
Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.

public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
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if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
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So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.

A quick explanation:

We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.

The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.


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3 years ago
What's the main idea??
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My computer wont let me download this pdf

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