# Inplace (Fixed space) M x N size matrix transpose | Updated

About four months of gap (missing GFG), a new post. Given an M x N matrix, transpose the matrix without auxiliary memory.It is easy to transpose matrix using an auxiliary array. If the matrix is symmetric in size, we can transpose the matrix inplace by mirroring the 2D array across it’s diagonal (try yourself). How to transpose an arbitrary size matrix inplace? See the following matrix,

```a b c       a d g j
d e f  ==>  b e h k
g h i       c f i l
j k l```

As per 2D numbering in C/C++, corresponding location mapping looks like,

```Org element New
0     a     0
1     b     4
2     c     8
3     d     1
4     e     5
5     f     9
6     g     2
7     h     6
8     i    10
9     j     3
10    k     7
11    l    11```

Note that the first and last elements stay in their original location. We can easily see the transformation forms few permutation cycles.

• 1->4->5->9->3->1  – Total 5 elements form the cycle
• 2->8->10->7->6->2 – Another 5 elements form the cycle
• 0  – Self cycle
• 11 – Self cycle

From the above example, we can easily devise an algorithm to move the elements along these cycles. How can we generate permutation cycles? Number of elements in both the matrices are constant, given by N = R * C, where R is row count and C is column count. An element at location ol (old location in R x C matrix), moved to nl (new location in C x R matrix). We need to establish relation between ol, nl, R and C. Assume ol = A[or][oc]. In C/C++ we can calculate the element address as,

`ol = or x C + oc (ignore base reference for simplicity)`

It is to be moved to new location nl in the transposed matrix, say nl = A[nr][nc], or in C/C++ terms

`nl = nr x R + nc (R - column count, C is row count as the matrix is transposed)`

Observe, nr = oc and nc = or, so replacing these for nl,

`nl = oc x R + or -----> [eq 1]`

after solving for relation between ol and nl, we get

```ol     = or x C     + oc
ol x R = or x C x R + oc x R
= or x N     + oc x R    (from the fact R * C = N)
= or x N     + (nl - or) --- from [eq 1]
= or x (N-1) + nl```

OR,

`nl = ol x R - or x (N-1)`

Note that the values of nl and ol never go beyond N-1, so considering modulo division on both the sides by (N-1), we get the following based on properties of congruence,

```nl mod (N-1) = (ol x R - or x (N-1)) mod (N-1)
= (ol x R) mod (N-1) - or x (N-1) mod(N-1)
= ol x R mod (N-1), since second term evaluates to zero
nl = (ol x R) mod (N-1), since nl is always less than N-1```

A curious reader might have observed the significance of above relation. Every location is scaled by a factor of R (row size). It is obvious from the matrix that every location is displaced by scaled factor of R. The actual multiplier depends on congruence class of (N-1), i.e. the multiplier can be both -ve and +ve value of the congruent class.Hence every location transformation is simple modulo division. These modulo divisions form cyclic permutations. We need some book keeping information to keep track of already moved elements. Here is code for inplace matrix transformation,

`// Program for in-place matrix transpose`
`#include <stdio.h>`
`#include <iostream>`
`#include <bitset>`
`#define HASH_SIZE 128`
`using` `namespace` `std;`
`// A utility function to print a 2D array of size nr x nc and base address A`
`void` `Print2DArray(``int` `*A, ``int` `nr, ``int` `nc)`
`{`
`    ``for``(``int` `r = 0; r < nr; r++)`
`    ``{`
`        ``for``(``int` `c = 0; c < nc; c++)`
`            ``printf``(``"%4d"``, *(A + r*nc + c));`
`        ``printf``(``"\n"``);`
`    ``}`
`    ``printf``(``"\n\n"``);`
`}`
`// Non-square matrix transpose of matrix of size r x c and base address A`
`void` `MatrixInplaceTranspose(``int` `*A, ``int` `r, ``int` `c)`
`{`
`    ``int` `size = r*c - 1;`
`    ``int` `t; ``// holds element to be replaced, eventually becomes next element to move`
`    ``int` `next; ``// location of 't' to be moved`
`    ``int` `cycleBegin; ``// holds start of cycle`
`    ``int` `i; ``// iterator`
`    ``bitset<HASH_SIZE> b; ``// hash to mark moved elements`
`    ``b.reset();`
`    ``b[0] = b[size] = 1;`
`    ``i = 1; ``// Note that A[0] and A[size-1] won't move`
`    ``while` `(i < size)`
`    ``{`
`        ``cycleBegin = i;`
`        ``t = A[i];`
`        ``do`
`        ``{`
`            ``// Input matrix [r x c]`
`            ``// Output matrix 1`
`            ``// i_new = (i*r)%(N-1)`
`            ``next = (i*r)%size;`
`            ``swap(A[next], t);`
`            ``b[i] = 1;`
`            ``i = next;`
`        ``}`
`        ``while` `(i != cycleBegin);`
`        ``// Get Next Move (what about querying random location?)`
`        ``for` `(i = 1; i < size && b[i]; i++)`
`            ``;`
`        ``cout << endl;`
`    ``}`
`}`
`// Driver program to test above function`
`int` `main(``void``)`
`{`
`    ``int` `r = 5, c = 6;`
`    ``int` `size = r*c;`
`    ``int` `*A = ``new` `int``[size];`
`    ``for``(``int` `i = 0; i < size; i++)`
`        ``A[i] = i+1;`
`    ``Print2DArray(A, r, c);`
`    ``MatrixInplaceTranspose(A, r, c);`
`    ``Print2DArray(A, c, r);`
`    ``delete``[] A;`
`    ``return` `0;`
`}`

Output:

```   1   2   3   4   5   6
7   8   9  10  11  12
13  14  15  16  17  18
19  20  21  22  23  24
25  26  27  28  29  30

1   7  13  19  25
2   8  14  20  26
3   9  15  21  27
4  10  16  22  28
5  11  17  23  29
6  12  18  24  30```

In given array of elements like [a1b2c3d4e5f6g7h8i9j1k2l3m4]. Convert it to [abcdefghijklm1234567891234]. The program should run inplace. What we need is an inplace transpose. Given below is code.

`#include <stdio.h>`
`#include <iostream>`
`#include <bitset>`
`#define HASH_SIZE 128`
`using` `namespace` `std;`
`typedef` `char` `data_t;`
`void` `Print2DArray(``char` `A[], ``int` `nr, ``int` `nc) {`
`   ``int` `size = nr*nc;`
`   ``for``(``int` `i = 0; i < size; i++)`
`      ``printf``(``"%4c"``, *(A + i));`
`   ``printf``(``"\n"``);`
`}`
`void` `MatrixTransposeInplaceArrangement(data_t A[], ``int` `r, ``int` `c) {`
`   ``int` `size = r*c - 1;`
`   ``data_t t; ``// holds element to be replaced, eventually becomes next element to move`
`   ``int` `next; ``// location of 't' to be moved`
`   ``int` `cycleBegin; ``// holds start of cycle`
`   ``int` `i; ``// iterator`
`   ``bitset<HASH_SIZE> b; ``// hash to mark moved elements`
`   ``b.reset();`
`   ``b[0] = b[size] = 1;`
`   ``i = 1; ``// Note that A[0] and A[size-1] won't move`
`   ``while``( i < size ) {`
`      ``cycleBegin = i;`
`      ``t = A[i];`
`      ``do` `{`
`         ``// Input matrix [r x c]`
`         ``// Output matrix 1`
`         ``// i_new = (i*r)%size`
`         ``next = (i*r)%size;`
`         ``swap(A[next], t);`
`         ``b[i] = 1;`
`         ``i = next;`
`      ``} ``while``( i != cycleBegin );`
`      ``// Get Next Move (what about querying random location?)`
`      ``for``(i = 1; i < size && b[i]; i++)`
`         ``;`
`      ``cout << endl;`
`   ``}`
`}`
`void` `Fill(data_t buf[], ``int` `size) {`
`   ``// Fill abcd ...`
`   ``for``(``int` `i = 0; i < size; i++)`
`   ``buf[i] = ``'a'``+i;`
`   ``// Fill 0123 ...`
`   ``buf += size;`
`   ``for``(``int` `i = 0; i < size; i++)`
`      ``buf[i] = ``'0'``+i;`
`}`
`void` `TestCase_01(``void``) {`
`   ``int` `r = 2, c = 10;`
`   ``int` `size = r*c;`
`   ``data_t *A = ``new` `data_t[size];`
`   ``Fill(A, c);`
`   ``Print2DArray(A, r, c), cout << endl;`
`   ``MatrixTransposeInplaceArrangement(A, r, c);`
`   ``Print2DArray(A, c, r), cout << endl;`
`   ``delete``[] A;`
`}`
`int` `main() {`
`   ``TestCase_01();`
`   ``return` `0;`
`}`

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rakesh