The document outlines a template for implementing a simple transposition cipher for an electrical engineering assignment, providing guidelines and an example of the encryption process. It includes specifications for input and output formats, resource limits, and sample code in C to read a matrix and perform operations on it, along with instructions for filling out a function. The document also offers contact details for help and a reminder to seek clarification on any questions via Piazza.

1. For any help regarding Electrical Engineering Assignment Help
visit :- https://www.programminghomeworkhelp.com/
Email :- support@programminghomeworkhelp.com
or call us at :- +1 678 648 4277
programminghomeworkhelp.com

2. Problem
Averysimpletranspositioncipherencrypt(S )canbedescribedbythefollowingrules:
1. Ifthelengthof Sis1 or2, thenencrypt(S )is S.
2. IfSisastringof N characterss1s2s3. . . sNandk = IN /2j, then
enc(S) = encrypt(sk sk−1 ... s2s1)+ encrypt(sN sN−1 ... sk+1)
where+ indicatesstringconcatenation.
Forexample,encrypt("Ok") = "Ok" andencrypt("12345678") ="34127856".
Writeaprogramto implementthiscipher,givenanarbitrarytextfileinputupto 16 MB insize.Start withthe
templateprogramfoundatprovided in the file loop.data.zip as a basis for your program. Inthisprogram,youwill
seeamostlycompletefunctionto readafileinto adynamicallyallocated stringasrequiredforthisproblem.
programminghomeworkhelp.com
size t getstr( char **str, FILE *input ) {
size t chars to read = BLOCK SIZE; size t length = 0;
II ...snipped... see template file size t chars = 0;
while( ( chars = fread( *str + length, 1, chars to read, input ) ) ) { II you fill this out
}
II ...snipped... see template file return length;
}
Read through the code carefully, make sure you understand it, and complete the inner part of the while loop.
Look up realloc andthe <string.h> header. If you have any questions about the provided code or don’tknow
why something is structured the way it is, pleaseaskabout it on Piazza.
Youwill alsoseeanemptyfunction“encrypt”,whichyoushouldfillout.
void encrypt( char *string, size t length ) {
II you fill this out
}
Resource Limits
Forthisproblemyouareallotted3secondsof runtimeandupto 32MB of RAM.
Input Format
Lines 1. ..: The whole file (can be any number of lines) should be read in asa string.

3. 21
aeyrleT sttf!enn aod
Test
early
and often!
Output Format
Line1: One integer:thetotalnumberof charactersinthestring Lines
2. . . :The encipheredstring.
Output Explanation
Here’seachcharacterinthestringaswearesupposedto readit in, separatedwith ‘.’ sowecanseethe newlines
andspaces:
.T.e.s.t.n.e.a.r.l.y.n.a.n.d. .o.f.t.e.n.!.
The stringisfirstsplitinhalfandtheneachhalfreversed,andthefunctioncalledrecursively;youcan seethe
recursiongoingonhere:
.y.l.r.a.e.n.t.s.e.T. .!.n.e.t..f.o. .d.n.a.n.
.n.a.n.d. .o.
.e.a.r.l.y.
.a.e. .y.l.r
I
.T.e.s.t.n.
.e.T. .n.t.s.
I
.f.t.e.n.!.
.t.f. .!.n.e
I
.y. .r.l. .n. .s.t. .!. .e.n. I
.n.a.n. .o. .d.
I
.n. .n.a. .o. .d. .
This diagrammakesit lookabit morecomplicatedthanit actuallyis.Youcanseethatthesampleis correct by
reading off the leaves of the tree from left to right–it’s the enciphered string we want.
.a.e.y.r.l.e.T.n.s.t.t.f.!.e.n.n.n.a.o.d. .
programminghomeworkhelp.com

4. Solution
/*PROG: matrix2
LANG: C
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct Matrix_s {
size_t R, C;
int *index;
} Matrix;
Matrix*
allocate_matrix( size_t R, siz
e_t C ) {
Matrix *matrix
= malloc( sizeof( Matrix ) );
matrix->R = R;
matrix->C = C;
matrix->index = malloc( R *
C * sizeof( int ) );
return matrix;
}
programminghomeworkhelp.com

5. void destroy_matrix(
Matrix *matrix ) {
free( matrix->index );
free( matrix );
}
typedef enum {
REGULAR = 0,
TRANSPOSE = 1
} Transpose;
// Allowing reading a
matrix in as either
regular or transposed
Matrix*
read_matrix( FILE *inp
ut, Transpose orient ) {
size_t R, C;
fscanf( input, "%zu
%zu", &R, &C );
Matrix *matrix =
NULL;
programminghomeworkhelp.com

6. if( orient == REGULAR
) {
matrix =
allocate_matrix( R, C );
for( size_t r = 0; r <
matrix->R; ++r ) {
for( size_t c = 0; c <
matrix->C; ++c ) {
fscanf( input, "%d",
&matrix->index[c + r *
C] );
}
}
} else if( orient ==
TRANSPOSE ) {
matrix =
allocate_matrix( C, R );
for( size_t r = 0; r <
matrix->C; ++r ) {
for( size_t c = 0; c <
matrix->R; ++c ) {
fscanf( input, "%d",
&matrix->index[r + c *
R] );
}
}
programminghomeworkhelp.com

7. } else {
fprintf( stderr, "Error: unknown
orientation %d.n", orient );
exit( EXIT_FAILURE );
}
return matrix;
}
void print_matrix( FILE *output,
Matrix *matrix ) {
fprintf( output, "%zu %zun", matrix-
>R, matrix->C );
for( size_t r = 0; r < matrix->R; ++r )
{
for( size_t c = 0; c < matrix->C - 1;
++c ) {
fprintf( output, "%d ", matrix-
>index[c + r * matrix->C] );
}
fprintf( output, "%dn", matrix-
>index[matrix->C - 1 + r * matrix->C]
);
}
}
programminghomeworkhelp.com

8. Matrix* product_matrix(
Matrix *a, Matrix *b ) {
if( a->C != b->C ) {
printf( "Error: tried to
multiply
(%zux%zu)x(%zux%zu)n",
a->R, a->C, b->C, b->R );
exit( EXIT_FAILURE );
}
Matrix *prod =
allocate_matrix( a->R, b->R
);
size_t nRows = prod->R,
nCols = prod->C, nInner = a-
>C;
for( size_t r = 0; r < nRows;
++r ) {
for( size_t c = 0; c <
nCols; ++c ) {
prod->index[c + r *
nCols] = 0;
for( size_t i = 0; i <
nInner; ++i ) {
prod->index[c + r *
nCols] += a->index[i + r *
programminghomeworkhelp.com

9. >index[i + c * nInner];
}
}
}
return prod;
}
int main(void) {
FILE *fin
= fopen( "matrix2.in", "r" );
if( fin == NULL ) {
printf( "Error: could not
open matrix2.inn" );
exit( EXIT_FAILURE );
}
Matrix *a = read_matrix(
fin, REGULAR );
Matrix *b = read_matrix(
fin, TRANSPOSE );
fclose( fin );
Matrix *c = product_matrix(
programminghomeworkhelp.com

10. a, b );
FILE *output
= fopen( "matrix2.out", "w" );
if( output == NULL ) {
printf( "Error: could not
open matrix2.outn" );
exit( EXIT_FAILURE );
}
print_matrix( output, c );
fclose( output );
destroy_matrix( a );
destroy_matrix( b );
destroy_matrix( c );
return 0;
}
programminghomeworkhelp.com

11. Below is the output using the
test data:
matrix2: 1: OK [0.006
seconds] 2: OK [0.007
seconds] 3: OK [0.007
seconds] 4: OK [0.019
seconds] 5: OK [0.017
seconds] 6: OK [0.109
seconds] 7: OK [0.178
seconds] 8: OK [0.480
seconds] 9: OK [0.791
seconds]10: OK [1.236
seconds]11: OK [2.088
seconds]
programminghomeworkhelp.com