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Volgvan
3 years ago
11

Suppose you have a string matching algorithm that can take in (linear)strings S and T and determine if S is a substring (contigu

ous) of T. However,you want to use it in the situation where S is a linear string but T is a circularstring, so it has no beginning or ending position. You could break T at eachcharacter and solve the linear matching problem|T|times, but that wouldbe very inefficient. Show how to solve the problem by only one use of thestring matching algorithm.
Computers and Technology
2 answers:
Zielflug [23.3K]3 years ago
8 0

Answer:

no seishsssssssssssssssssssss

telo118 [61]3 years ago
8 0
I have no clue how this app works
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In the paragraph group, click the Dialog box Launcher. ,and select the Alignment drop - down menu to set your justified text. You can also use the keyboard shortcut , Ctrl + J to justify your text

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2 years ago
A Chain of dry-cleaning outlets wants to improve its operations by using data from devices at individual locations to make real-
Svetllana [295]

The technology would the business combination with its current cloud operations to make this possible are ERP and loT.

<h3>What is the ERP?</h3>

Enterprise resource planning (ERP) refers to a shape of a software program application that companies use to manipulate each day industrial organization sports activities in conjunction with accounting, procurement, task management, chance manage and compliance, and supply chain operations.

  1. Given the organization has a cloud setup, we need a company solution (ERP system) which needs to be to be had to all stores and loT to build up facts from devices and deliver it over to the ERP system.
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7 0
2 years ago
Panes created using the vertical split bar scroll together horizontally. true or false.
Nataly_w [17]
I think the statement given above is false. Panes created using the vertical split bar do not scroll together horizontally. <span>The two </span>panes scroll together vertically, but scroll<span> independently </span>horizontally<span>. Hope this answers the question. Have a nice day.</span>
4 0
3 years ago
Weak Induction
slega [8]

Answer:

Following are the answer to this question:

Explanation:

In option 1:

The value of n is= 7, which is (base case)

\to 3^7

when n=k for the true condition:

\to 3^k

when n=k+1 it tests the value:

\to 3^{(k+1)}= 3^k,3\\\to < (k!) 3 \ substituting \ equation \\\to

since k>6  hence the value is KH>3 hence proved.

In option 2:

when:

for n=1:(base case)

\log(1!)

0<=0 \\ condition is true

when the above statement holds value n=1

when n=k

\log(k!)

when n=k+1

\log(k+1)!=\log(k!)+\log(k+1)\\

             

\because k \log k      [\therefore KH>K \Rightarrow  \log(KH>\loK)]

In option 3:

when n=1:

A_1 \cup B=A_1 \cup B\\\\

when n=k

\to (A_1\cap A_2 \cap.....A_k) \cup B\\=(A_1\cup B) \cap(A_2\cup B_2)....(A_k \capB).....(a)\\\to n= k+1\\ \to (A_1\cap A_2 \cap.....A_{kH}) \cup B= (A_1\cup B)\\\\\to  [(A_1\cap A_2 \cap.....A_{k}) \cup B]\cap (A_{KH}\cup B)\\\\\to  [(A_1\cup B) \cap (A_2 \cup B) \cap (A_3\cup B).....(A_k\cup B)\cap (A_{k+1} \cup B)\\\\  \ \ \ \ \ \ substituting \ equation \ a \\\\

hence n=k+1 is true.

7 0
2 years ago
Based on the code you created in this Unit, propose a way to re-use most of the code (with different information in the variable
lisov135 [29]

Answer:

Is this a question or an answer

7 0
3 years ago
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