write a function findlongestword that takes an array of words and returns the length of the longest one.

This function can be used to modify a two-level dictionary by providing the first-level key (`key1`), the second-level key (`key2`), and the value (`value`) that should be stored at `dict2[key1][key2]`.

To write the function `update_dict2(dict2, key1, key2, value)`, we can follow these steps:
1.
```python
def update_dict2(dict2, key1, key2, value):
   if key1 in dict2:
       dict2[key1][key2] = value
   else:
       dict2[key1] = {key2: value}
```

2.

The function `update_dict2` takes four parameters: `dict2`, `key1`, `key2`, and `value`.

Inside the function, we first check if `key1` exists in `dict2` using the `in` operator. If `key1` is already a key in `dict2`, we update the value of `dict2[key1][key2]` to the provided `value`.

If `key1` is not a key in `dict2`, we create a new dictionary with `key2` as the key and `value` as the value, and assign it to `dict2[key1]`. This ensures that the structure of `dict2` is maintained, with `key1` as the first-level key and `key2` as the second-level key.

This function modifies `dict2` in-place, meaning the changes are made directly to the original dictionary.
```python
dict2 = {"a": {"x": 1, "y": 2}, "b": {"x": 3, "y": 4}}
key1 = "b"
key2 = "y"
value = 5

update_dict2(dict2, key1, key2, value)

print(dict2)
```

Output:
```python
{"a": {"x": 1, "y": 2}, "b": {"x": 3, "y": 5}}
```

In this example, we have an existing dictionary `dict2` with two first-level keys: "a" and "b". The second-level keys for both "a" and "b" are "x" and "y". We want to update the value of `dict2["b"]["y"]` to 5. After calling the `update_dict2` function, we can see that the value has been successfully updated in the dictionary.

This function can be used to modify a two-level dictionary by providing the first-level key (`key1`), the second-level key (`key2`), and the value (`value`) that should be stored at `dict2[key1][key2]`.

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The complete question is,

CSc 120: Update a 2-level dictionary Definitions A two-level dictionary is analogous to a list-of-lists (aka "2D-list), except that it involves dictionaries rather than lists: . Alist-of-lists L is a list where each element L[ someidx) is itself a list; and LIaomeidx) anotheridx] gives us a value stored in L . Analogously, a two-level dictionary D is a dictionary-of-dictionaries: D[ somekey] is itself a dictionary, and DI somekey]Ianotherkey] gives us a value stored in the two-level dictionary D. In this example, somekey is called the first-leve key and anotherkey is called the second-level key. In the examples below, DD is assumed to be the following two-level dictionary: bbb' Cstring4, ddd string5 ece 'string6, 'Estring'. ccc· { .aaa . ; .string 8., .bbb. ; .string9. } Thus, we have: DD bbb' cec' is the value string4', while DD ccbbb' is string9 Expected Behavior Write a function update dict2(dict2, keyl, key2, value), where diet2 is a two-level dictionary: key1 is the first-level key: key2 is the second-level key: and value is a value to be stored at dict2[key1key21. This function should return a dictionary obtained by updating dict2 such that in the resulting dictionary, which we refer to as newdictz, the following holds newdict2[ key1keyz- value. Examples DD is the two-level dictionary shown above (under Definitions) 1. update dict2 (DD, 'aaa,'cce,12) return value: 2. update_dict2 (DD, 'aaa', 9g,string17) return value: 3. update dict2(DD, 'ggg', 'aaa, 'string17") return value:

In a packet-switched network, without message segmentation, the time to move the entire 5 Mbit message from the source host to the first packet switch is determined by the link capacity of 10 Mbps. It takes 5 milliseconds to transmit the message.

When the message is segmented into 400 packets, each 20 kilobits long, the time to move the first packet from the source host to the first switch is still 5 milliseconds. The second packet will be fully received at the first switch when the first packet is being sent from the first switch to the second switch, resulting in no additional delay.

(a) Without message segmentation, the time to move the entire 5 Mbit message from the source host to the first packet switch can be calculated using the formula: Time = Message Size / Link Capacity. Here, the message size is 5 Mbits and the link capacity is 10 Mbps. Thus, the time to move the message is 5 milliseconds.

Since each switch in store-and-forward packet switching requires the complete packet to be received before forwarding, the total time to move the message from the source host to the destination host would also be 5 milliseconds, assuming there are no propagation, queuing, and processing delays.

(b) When the message is segmented into 400 packets, each 20 kilobits long, the time to move the first packet from the source host to the first switch remains the same as before, 5 milliseconds. This is because the link capacity and the packet size remain unchanged.

As for the second packet, it will be fully received at the first switch when the first packet is being sent from the first switch to the second switch. Since the link capacity is greater than the packet size, there is no additional delay for the second packet. This is because the transmission of the first packet does not affect the reception of subsequent packets as long as the link capacity is sufficient to handle the packet size.

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a) The memory capacity is 64 bytes.

b) The total number of blocks in memory is 32.

c) The number of cache lines is 256.

In this scenario, the word size is given as 2.1 words per byte, meaning each word occupies 1 byte. The cache size is specified as 512 bytes. Therefore, the memory capacity can be calculated by multiplying the word size and the cache size, resulting in 2.1 words/byte * 512 bytes = 1075.2 words. Since the word size is given as an integer, the memory capacity is rounded down to the nearest integer, resulting in a capacity of 64 bytes.

To determine the total number of blocks in memory, we divide the memory capacity by the word size: 64 bytes / 2.1 words/byte = 30.47619 words. Again, since the word size is given as an integer, the number of blocks is rounded down to the nearest integer, resulting in 32 blocks.

The number of cache lines can be calculated by dividing the cache size by the word size: 512 bytes / 2.1 words/byte = 243.80952 words. Once again, rounding down to the nearest integer, we find that the number of cache lines is 256.

In summary, the memory capacity is 64 bytes, the total number of blocks in memory is 32, and the number of cache lines is 256.

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To solve the voting problem and work with an Excel file in MATLAB, a function can be created that takes the name of the Excel file as an argument. It will read the voting results from Sheet1 and display them, and write the results on a new sheet within the same Excel file.

To address the task, a MATLAB function needs to be implemented. The function should accept the name of the Excel file as an input parameter. Using MATLAB's built-in functions for Excel file manipulation, the function can read the voting results from Sheet1 and display them on the screen.

Furthermore, the function should create a new sheet within the same Excel file to write the results. This can be achieved by using appropriate functions for creating worksheets and writing data to them in MATLAB.

By combining these steps, the function will successfully solve the voting problem, read the voting results from the Excel file, display them on the screen, and write the results to a new sheet within the same file.

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