Typically, a CLEP exam is taken by students who

Show the array that results from the following sequence of key insertions using a hashing system under the given conditions: 5,

sergejj [24]

Answer:

a) Linear probing is one of the hashing technique in which we insert values into the hash table indexes based on hash value.

Hash value of key can be calculated as :

H(key) = key % size ;

Here H(key) is the index where the value of key is stored in hash table.

----------

Given,

Keys to be inserted are : 5 , 205, 406,5205, 8205 ,307

and size of the array : 100.

First key to be inserted is : 5

So, H(5) = 5%100 = 5,

So, key 5 is inserted at 5th index of hash table.

-----

Next key to inserted is : 205

So, H(205) = 205%100 = 5 (here collision happens)

Recompute hash value as follows:

H(key) =(key+i) % size here i= 1,2,3...

So, H(205) =( 205+1)%100 = 206%100 = 6

So, key 205 is inserted in the 6th index of the hash table.

----------

Next Key to be inserted : 406

H(406) = 406%100 = 6(collision occurs)

H(406) =(406+1) %100 = 407%100 = 7

So, the value 406 is inserted in 7the index of the hash table.

-----------------

Next key : 5205

H(5205) = 5205%100 = 5(collision)

So, H(5205) = (5205+1)%100 = 6( again collision)

So, H(5205) = 5205+2)%100 = 7(again collision)

So, H(5205) = (5205+3)%100 = 8 ( no collision)

So, value 5205 is inserted at 8th index of the hash table.

-------------

Similarly 8205 is inserted at 9th index of the hash table because , we have collisions from 5th to 8th indexes.

-------------

Next key value is : 307

H(307) = 307%100 = 7(collision)

So, (307+3)%100 = 310%100 = 10(no collision)

So, 307 is inserted at 10th index of the hash table.

So, hash table will look like this:

Key index

5 5

205 6

406 7

5205 8

8205 9

307 10

b) Quadratic probing:

Quadratic probing is also similar to linear probing but the difference is in collision resolution. In linear probing in case of collision we use : H(key) = (key+i)%size but here we use H(key) =( key+i^2)%size.

Applying Quadratic probing on above keys:.

First key to be inserted : 5.

5 will go to index 5 of the hash table.

-------

Next key = 205 .

H(205) = 205%100 = 5(collision)

So. H(key)= (205+1^2)%100 = 6(no collision)

So, 205 is inserted into 6th index of the hash table.

--------

Next key to be inserted 406:

So, 406 %100 = 6 (collision)

(406+1^2)%100 = 7(no collision)

So, 406 is moved to 7th index of the hash table.

----------

Next key is : 5205

So, 5205%100 = 5 (collision)

So, (5205+1^2)%100 = 6 ( again collision)

So, (5205+2^2)%100 = 9 ( no collision)

So, 5205 inserted into 9th index of hash table.

-----------

Next key is 8205:

Here collision happens at 5the , 6the , 9th indexes,

So H(8205) = (8205+4^2)%100 = 8221%100 = 21

So, value 8205 is inserted in 21st index of hash table.

--------

Next value is 307.

Here there is collision at 7the index.

So, H(307) = (307+1^2)%100 = 308%100= 8.

So, 307 is inserted at 8the index of the hash table.

Key Index

5 5

205 6

406 7

5205 9

8205 21

307 8

3 0

3 years ago

Write a function in python that computes and returns the sum of the digits for any integer that is between 1 and 999, inclusive.

DanielleElmas [232]

Answer:

def sum_digits(number):

total = 0

if 1 <= number <= 999:

while number > 0:

r = int (number % 10)

total +=r

number /= 10

else:

return -1

return total

print(sum_digits(658))

Explanation:

Write a function named sum_digits that takes one parameter, number

Check if the number is between 1 and 999. If it is, create a while loop that iterates until number is greater than 0. Get the last digit of the number using mudulo and add it to the total. Then, divide the number by 10 to move to the next digit. This process will continue until the number is equal to 0.

If the number is not in the given range, return -1, indicating invalid range.

Call the function and print the result

3 0

3 years ago

Keith would like to compare data he has collected from research. The data includes the electrical output

liberstina [14]

Answer: my test says line graph soooo….

Explanation: if its what the test says

4 0

2 years ago

How do I persuade my parents to allow me to have social media? I will be 15 in a week or so, so I am old enough. I'm pulling my

Flauer [41]

Answer:

my explanation is above my comment :)

Explanation:

6 0

3 years ago

Write a procedure named Str_find that searches for the first matching occurrence of a source string inside a target string and r

kirill115 [55]

Answer: Provided in the explanation section

Explanation:

Str_find PROTO, pTarget:PTR BYTE, pSource:PTR BYTE

.data

target BYTE "01ABAAAAAABABCC45ABC9012",0

source BYTE "AAABA",0

str1 BYTE "Source string found at position ",0

str2 BYTE " in Target string (counting from zero).",0Ah,0Ah,0Dh,0

str3 BYTE "Unable to find Source string in Target string.",0Ah,0Ah,0Dh,0

stop DWORD ?

lenTarget DWORD ?

lenSource DWORD ?

position DWORD ?

.code

main PROC

INVOKE Str_find,ADDR target, ADDR source

mov position,eax

jz wasfound ; ZF=1 indicates string found

mov edx,OFFSET str3 ; string not found

call WriteString

jmp quit

wasfound: ; display message

mov edx,OFFSET str1

call WriteString

mov eax,position ; write position value

call WriteDec

mov edx,OFFSET str2

call WriteString

quit:

exit

main ENDP

;--------------------------------------------------------

Str_find PROC, pTarget:PTR BYTE, ;PTR to Target string

pSource:PTR BYTE ;PTR to Source string

;

; Searches for the first matching occurrence of a source

; string inside a target string.

; Receives: pointer to the source string and a pointer

; to the target string.

; Returns: If a match is found, ZF=1 and EAX points to

; the offset of the match in the target string.

; IF ZF=0, no match was found.

;--------------------------------------------------------

INVOKE Str_length,pTarget ; get length of target

mov lenTarget,eax

INVOKE Str_length,pSource ; get length of source

mov lenSource,eax

mov edi,OFFSET target ; point to target

mov esi,OFFSET source ; point to source

; Compute place in target to stop search

mov eax,edi ; stop = (offset target)

add eax,lenTarget ; + (length of target)

sub eax,lenSource ; - (length of source)

inc eax ; + 1

mov stop,eax ; save the stopping position

; Compare source string to current target

cld

mov ecx,lenSource ; length of source string

L1:

pushad

repe cmpsb ; compare all bytes

popad

je found ; if found, exit now

inc edi ; move to next target position

cmp edi,stop ; has EDI reached stop position?

jae notfound ; yes: exit

jmp L1 ; not: continue loop

notfound: ; string not found

or eax,1 ; ZF=0 indicates failure

jmp done

found: ; string found

mov eax,edi ; compute position in target of find

sub eax,pTarget

cmp eax,eax ; ZF=1 indicates success

done:

ret

Str_find ENDP

END main

cheers i hoped this helped !!

6 0

3 years ago