We review the book "The Responsible Entrepreneur" written by Carol Sanford. 

She introduces us to the entrepreneur who is responsible - a special breed whochallenge and refine cultural assumptions, laws and regulations and even the process of governance. As Kevin Jones, founder of SoCap said, this book is about being courageous and in fact he dares the readers to read and implement the bolduseful advice. 

Let us take a look at what Carol means by Responsible Entrepreneurs and what theyachieve. She says these entrepreneurs are required to do and think far beyond what is usually required of business leaders. The Responsible Entrepreneurs offers us this'blueprint'. By understanding the archetype  most aligned with their goals,entrepreneurs will learn how to grow their business and even change the game. 

In this book she includes the following: 

Some extraordinary people have changed the game for the better. 

How modern archetypes are altering the future. 
Why new measures of accomplishment are based on the success of the whole. 

She says such entrepreneurs have the courage to take on what they don't yet know how to do and the dedication to build the capability to do it. These entrepreneursare driven by the realization that the society and the planet need something big from them and that, if they don't rise to the challenge, the work may not get done. 

Entrepreneuralism is about personal agency and the development of the will. While entrepreneurs are inventing new products, industries, sources of capital and models of enterprise, the responsible entrepreneurs are doing that and more because they want to create the right platform and support an ecosystem.  They run successful business while deeply caring about the innovation that can not only help them secure their business but also transform the world. 

They enjoy developing the acumen needed to work on all parts of the business. 

They maintain their own motivation, getting stimulus, training and innovation when they need it. They hold a positive attitude. 

They are willing to take on big challenges that stretch them beyond what theycurrently know how to do and to ride the roller coaster of needing to continually rise to the occassion. 

They tend to focus the game changing aspirations in one of four distinct domains 

Industries - where the work is to disrupt and replace the automatic patterns 

Social systems - where the work is to move upstream to the causes of social problems and address them at their source. 

Cultural paradigms - where the enhance the belief systems into something moreholistic and embracing 

Foundational agreements - where the work is to renew and vitalize the deeper intention behind the governing documents, such as a corporate charter. 


From these four domains she draws parallels to four archetypes 

Warrior - The warrior protects the values of a community. In the business world, this work takes place within the industry. 

Clown - The clown pokes fun at collective self-centeredness and unconsciousness,opening space for humility and heartfelt appreciation of others. 
Hunter - The hunter perpetuates life by strengthening the mutual exchange between the tribe and the natural world. 

Headman - The headman awakens the individuals to their potential and inspiresthem to work with others in order to contribute to something larger thanthemselves. 


The four archetypes are all necessary to the healthy functioning of the society. Infact if any one of them is missing, the society becomes vulnerable. 


To the entreprenueur, these archetypes translate into four unique roles as follows: 

The warrior who changes an industry is the realization entrepreneur. He is driven by the vision of an improved reality. 

The clown who changes social systems is the reconnection entrepreneur. She reveals the gap in our cognition regarding the impact of existing social systems 

The hunter who changes cultural paradigms is the reciprocity entrepreneur. She tries to strike a balance between what is taken and what is given. 

The headman or headwoman who changes the connection to foundation agreements is the regenerative entrepreneur.  She seeks to reveal and evolve the inherent potential of founding agreements that create the accepted structures within which the society operates. 


The four roles do different work. For each of them the author takes examples and the principles or pillars that they operate on. 

A realization entrepreneur transforms an industry by renewing its purpose andvalues. The warrior archetype embodies this. Her four pillars are: 


Perfecting an industry: The first thing that a realization entrepreneur transforms is her goal – from making just a better product to one that satisfies the customers and the success of the industry stakeholders. She anticipates what has not been realized or wanted. 


Integrity beyond reproach: A realization entrepreneur must also upgrade thesources from which the business actions originate. By being transparent, she harbors the trust and respect that is needed to fulfil the role of industry transformer. 


Principled Precision: The work of a realization requires rigor and precision.Precision comes from being true to the nature of the work. 


Full dress inspection: This is about readiness to engage in a wider field of action.  This is a sign of maturity if the company is well prepared and able to be at the top of the game and everyone participates in it. 


A reconnection entrepreneur goes beyond the products and services by finding theunderlying causes rather than the symptoms. He goes both upstream and into the
future to see how to bring about a systemic change and to make them accessible andintelligible to us. 


His four pillars are : 


Evoking conscience: He cultivates the virtue of relentless caring because that willgive him the tenacity to work in this arena. 


Relinquishing attachment: Learning to see and relinquish attachments requiresdisciplined practice. Attachments can build walls and run the show, so we must bediligent in controlling it. 


Evolving potential: He tries to understand how the system he wants to affect behaves when it is healthy. 


Destabilizing thinking to invite reflections: He breaks peoples unconscious thoughtpatterns and attachments. 


The reciprocity entrepreneur is an expansionist and a fosterer. He will illuminate the larger whole within which a business is embedded. He will bring into a group whatever is outside it that will improve its health and continued evolution 


His Pillars are  

1) Wholeness: this  is the overarching goal for this entrepreneur 

2) Significance: He has to step up one or two levels of system to meet theexpectations of the shareholders 

3) Destiny : He knows the evolution of the group and that’s what he improves 

4) Camaraderie : He uses this to foster the group for ambitious change effort 


Lastly, a regenerative entrepreneur acts as  a headman or headwoman. She revitalizes the people and is focused on the core purpose of governance that provides structure stability and opportunity to all. 

Her pillars are: 

1) Transformation : This is the overarching goal for this role. 

2) Accomplishment: She enables the accomplishment of her downstreamcustomers 

3) Impossible dream: She dares to think the impossible in order to change the world 

4) Dialogue:  She works on change through skillful engagement rather than top down directives. 


With these four different roles, you target the four different domains that makes youa Responsible Entrepreneur 

node* BuildTree(int values[], int count, int start, int stop)
{
if (start > count) return NULL;
node* current = (node*) malloc(sizeof node);
memset((void*) current, 0, sizeof node);
current->value = values[start];
if (start >= stop) return current;
int left = start + 1;
int right = stop;
for (int k = start+1; k < stop; k++)
if (values[k] > current->value){
right = k;
break;
}
current->left = BuildTree(values, count, left, right-1);
current->right = BuildTree(values, count, right, stop);
return current;
}


int main(int argc, char* argv[])
{
int values[255] = {0};
int count = 0;

readElementsInOrder((int*)values,&count);
for (int i = 0; i < count; i++)
{
printf("%d", values[i]);
}
printf("\r\nOutput:\r\n");

node* root = NULL;
root = BuildTree(values, count, 0, count-1);
printPreOrder(root);

printf("\r\nwith both children\r\n");
node* victim = getNode(root, 7); // with both children
treeDelete(root, victim);
printPreOrder(root);
printf("\r\nwith left child\r\n");
victim = getNode(root, 8); // with left child
treeDelete(root, victim);
printPreOrder(root);
printf("\r\nwith no child\r\n");
victim = getNode(root, 2); // with no child
treeDelete(root, victim);
printPreOrder(root);
printf("\r\nwith right child\r\n");
victim = getNode(root, 4); // with right child
treeDelete(root, victim);
printPreOrder(root);
printf("finish");
}

#codingquestions
Tree operations
        typedef struct _node{
  int value;
  node* left;
  node* right;
 }node;
 node* treeMinimum(node* root)
 {
  node* current = root;
  while(current->left)
   current = current->left;
  return current;
 }
 node* treeParent(node* root, node* z)
 {
  node* parent = NULL;
  node* current = root;
  while(current){
   if(current == z) return parent;
   if (current->value < z->value){
    parent = current;
    current = current->right;
   }else{
    parent = current;
    current = current->left;
      }
  }
  // z was not found in tree
  if (parent != NULL
      && parent->left != z
      && parent->right != z)
   parent = NULL;
  return parent;
 }
 node* treeSuccessor(node*root, node* z)
 {
  if (z->right)
   return treeMinimum(z->right);
  node* y = treeParent(root, z);
  while (y && z == y->right)
  {
   z = y;
   y = treeParent(root, y);
  }
  return y;
 }
 node* treeDelete(node*root, node* z)
 {
  node* y = NULL;
  node* x = NULL;
  node* px = NULL;
  node* py = NULL;
  if (z->left == NULL || z->right == NULL)
   y = z;
  else
   y = treeSuccessor(root, z);
  if (y->left != NULL)
   x = y->left;
  else
   x = y->right;
  //if (x != NULL)
  // px = py;
  py = treeParent(root, y);
  if (py == NULL)
   root = x;
      // root->value = x->value;
      // root->left = x->left;
      // root->right = x->right;
      // delete x;
  else if (y == py->left)
      py->left = x;
  else
   py->right = x;
  if (y != z)
   z->value = y->value;
  return y;
 }
 bool exists(node* root, int val)
 {
  if (root == NULL) return false;
  node* current = root;
  while(current)
  {
   if (current->value == val) return true;
   if (current->value < val) current = current->right;
   else current = current->left;
  }
  return false;
 }

 void readElementsInOrder(int* pValue, int* pCount)
 {
  char buffer[255] = {'5',' ', '3', ' ','2', ' ', '4', ' ', '7', ' ', '6',' ', '8'};
  sscanf(buffer, "%s");
  char seps[] = " ";
  char* token = NULL;
  //int values[255];
  int* values = pValue;
  token = strtok(buffer, seps);
  int i = 0;
  while (token != NULL)
  {
   values[i] = atoi(token);
   token = strtok(NULL, seps);
   i++;
  }
  *pCount = i;
 }
 for (int i = 0; i < count; i++)
 {
  printf("%d", values[i]);
 }

5324768

[IMO Shortlist 2013, C3] 

A crazy physicist discovered a new kind of particle which he called an imon. Some pairs of imons in the lab can be entangled, and each imon can participate in many entanglement relations. The physicist has found a way to perform the following two kinds of operations with these particles, one operation at a time. 

(i) If some imon is entangled with an odd number of other imons in the lab, then the physicist can destroy it. 

(ii) At any moment, he may double the whole family of imons in the lab by creating a copy I’ of each imon I. During this procedure, the two copies I’ and J’ become entangled if and only if the original imons I and J are entangled, and each copy I’ becomes entangled with its original imon I; no other entanglements occur or disappear at this moment.  

Show that after a finite number of operations, he can ensure that no pair of particles is entangled. 

The imons and their cloning are better visualized with a  data structure that represents the entanglements and the imons. We therefore use a graph with vertices as imons and edges as entanglements. Further, in order to differentiate every imon, we color them each differently. From graph coloring we know that a graph with n vertices can be colored with n colors so that every two connected vertices have different colors. 

Now if we could reduce these colors to just one it would mean a graph with no  edges – our desired goal. 

In other words, our strategy is to assume a graph G that admits proper coloring for n colors so that no two connected vertices have different colors and then show that a sequence of operations can result in a graph which admits coloring of n-1.  

To show this we do the following: 

We repeatedly apply operation 1 to any appropriate vertices  until there are no more. This  reduces the number of vertices and results in a graph with all even edges. Since this is a subset, it will still honor a proper coloring in n colors. We fix the colorings. 

Now we apply the second operation to the graph. This doubles the vertices but now we can still color them differently using n colors because for the new vertices around an original vertex of color k,  we color them with  k+1 mod n. The original vertices have different colors. Their clones also have different colors from the original now. Furthermore between the clones the coloring will be different because n > 1.  And finally all the vertices have now od d number of edges. Therefore those vertices that have the color n can now be destroyed leaving the graph with coloring in n – 1 colors. 

When we repeatedly reduce the colors with the above strategy, we are left with vertices and no edges as required. 
#codingexercise
Print count of headers listed in a file

#include "stdafx.h"
#include "stdio.h"
#include "string.h"

static int Hash (const char * pSrc, size_t len);

int main(int argc, char* argv[])
{
static const char filename[] = "foo.txt";

static const int MAX_COUNT = 255;

static int frequency[MAX_COUNT] = {0};

FILE* fd = fopen(filename, "r");

if ( fd != NULL )
{
char line[MAX_COUNT + 1];

while ( fgets(line, sizeof line, fd) != NULL )
{
char* p = strchr(line, ':');

if (p != NULL)
{
frequency[Hash(_strlwr(line), (size_t)(p-line))%MAX_COUNT] += 1;
}
}

fclose(fd);

char header[MAX_COUNT + 1] = {0};

printf( "Enter the header:\n " );

scanf("%255s", header);

header[MAX_COUNT] = '\0';

printf( "%s occurs %d times.\n", header, frequency[Hash(_strlwr(header), strlen(_strlwr(header)))%MAX_COUNT] );
}

return 0;
}

static int Hash (const char * pSrc, size_t len)
{
  size_t res = 0;

  while (len--) res = (res << 1) ^ *pSrc++;

  return (int)res;
}


If we want to print all headers by sorted order, we could do something like this:
#dirty

typedef struct _headers {

char header[MAX_COUNT+1];

int hash;

int frequency;

} headers;

int comp (const header * elem1, const header * elem2)

{

    int f = elem1->frequency;

    int s = elem2->frequency;

    if (f > s) return  1;

    if (f < s) return -1;

    return 0;

}

int main(int argc, char* argv[])

{

    headers* x  = (header*) malloc(MAX_COUNT * sizeof(headers));

    memset((void*)x, 0, sizeof headers);

    

    qsort (x, sizeof(x)/sizeof(*x), sizeof(*x), comp);

    for (int i = 0 ; i < MAX_COUNT ; i++){

        headers[i] = malloc(sizeof(struct headers));

        // populate header and frequency

    }

    qsort (x, sizeof(x)/sizeof(*x), sizeof(*x), comp);

    // print headers encountered in sorted order

    for (int i = 0 ; i < MAX_COUNT ; i++){

        if (x[i]){

           printf("%s",x[i]->header);

        }

    }     

    return 0;

}

[Bulgaria 2005]

For positive integers t, a, b, a (t, a, b)-game is a two player game defined by the following rules. Initially, the number t is written on a blackboard. In his first move, the first player replaces t with either t – a or t – b. Then, the second player subtracts either a or b from this number, and writes the result on the blackboard, erasing the old number. After this, the first player once again erases either a or b from the number written on the blackboard, and so on. The player who first reaches a negative number loses the game. Prove that there exist infinitely many values of t for which the first player has a winning strategy for all pairs (a, b) with (a + b) = 2005.

Note that the players subtract a or b from the given number so if they were to alternate the number would keep reducing by 2005 each time. Its easy for any player to know whether the number substracted was a or b because the previous number and the new number are both on the board one after the other. The winning player has to ensure that the number reduces by 2005 by alternating a or b as per the previous move of the opponent.

In other words, if the winning player has a strategy for x then he has a winning strategy for x + 2005k
Since k can be a multiple of 2005, there are infinitely many numbers possible for this player with the above winning strategy.

The players reduce the numbers by a or b so ultimately there is a value v1 or v2 possible before the number becomes negative. Every other number on the board is therefore a multiple of v1 + a , v1 + b, v1 + a+ b or v2 +a , v2 + b and v2 +a + b.

If it is v2, we call it a favorable position and the first player wins for v2. we say that the strategy works for him.

If it is v1 then the first player. he will win only if v1 +a + b = v2 and v1+a and v1+b are not favorable but that means v1 + a - b and v1 + b - a are favorable but that means v1 + a and v1 + b are favorable which is a contradiction.

C6 BGR (Bulgaria)
On a 999x 999 board a limp rook can move in the following way: From any square it can move
to any of its adjacent squares, i.e. a square having a common side with it, and every move
must be a turn, i.e. the directions of any two consecutive moves must be perpendicular. A non-
intersecting route of the limp rook consists of a sequence of pairwise di fferent squares that the
limp rook can visit in that order by an admissible sequence of moves. Such a non-intersecting
route is called cyclic, if the limp rook can, after reaching the last square of the route, move
directly to the fi rst square of the route and start over.
How many squares does the longest possible cyclic, non-intersecting route of a limp rook
visit?

Let us color the cells with four colors A, B, C and D in the following way
for (i,j) equivalent to (0,0) mod 2 use A
for (i,j) equivalent to (0,1) mod 2 use B
for (i,j) equivalent to (1,0) mod 2 use C
for (i,j) equivalent to (1,1) mod 2 use D

From an A-cell, the rook has to visit a B-cell or a C-cell. In the first case, the order of the colors of the cells visited is given by A, B, D, C, A, B, D, C ... and in the second case the order is given by A, D, B, A, C, D, B ... Since the route is closed it must contain the same number of cells of each color.
There are only 499^2 A-cells.It can be shown that the rook cannot visit all A cells on its route and hence the maximum possible number of cells in a route is 4( 499 ^2 -1). The four squares deducted from the total happen to be the four corners of the board.

Assume that the route passes through every single A cell. Color the A cells in black and white in a chessboard manner. A cells that are two cells apart are different colors. Since the number of A cells is odd, the rook cannot always alternate between black and white A cells. Let the two A cells of the same color which are four rook steps apart be (a,b) and (a+2, b+2) respectively.

There is only one path to take between these two squares. Let this path be (a,b), (a,b+1), (a+1,b+1), (a+1,b+2) and (a+2,b+2) we can say (a,b+1) is B otherwise we can interchange the rows and columns to be such that it is B.
Now let us look at the A cell vertically above which is (a,b+2) The only way the rook passes through it is via (a-1,b+2), (a,b+2) and(a,b+3) in this order. Hence to connect these two parts of the path, there must be  a path connecting the cell (a,b+3) and (a,b) and also a path connecting (a+2,b+2) and (a-1,b+2)
But these are opposite vertices of a quadrilateral and paths are outside so they must interesect. But an intersection is only possible if a cell is visited twice and this is a contradiction.
Hence the number of cells must be at most 4. (499^2 -1)

[IMO Shortlist 2009, C1] 

Consider 2009 cards, each having one gold side and one black side, lying on parallel on a long table. Initially all cards show their gold sides. Two players, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of 50 consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins. 

a) Does the game necessarily end? 

b) Does there exist a winning strategy for the starting player? 

If the players start from left, their set of flipped cards don't overlap and they can take turns to reach 2000. During each turn 50 more cards are converted to black starting from the left. After that the player turning the remaining nine cards wins. Since 50 * 40 = 2000, the player starting first has a wining strategy and wins. This method and outcome assumes that none of the cards flipped once already are flipped again. Even if the non-starting player choose to play from the right or overlaps the cards he plays with the ones he played earlier, the starting player has pushed the leftmost possible card towards the right and reduced the window to play in and since he is left with turning the nine cards wherever they may occur, he can win. 

If the game is played from right to left, the player going first converts fifty cards from the right to black. The player going second can undo almost all of this except one by choosing just one more card to the left of this fifty and converting them back to gold again. The first card on the right is now black and left out as the game progresses towards the left.In a subsequent move, the second card from the right which is now gold is left out as the fifty cards to its left are flipped. Thus the game inches one by one to the left, leaving a trail of alternate black and gold. Since there are odd number of cards to begin with there are odd number of moves possible. After each odd number of moves, the player who doesn't start first always finds a gold card to make a move. Therefore this player always win and the starting player doesn't.
Note there is an official answer that there is no winning strategy for the starting player. This we have shown is not necessarily the case because it does not consider the starting player to play from the left.
The official answer however does prove that the game will end by saying that each card can have two states only and so they represent a binary notation of 2009 1s or 0s  and this number reduces on each move. If 0s represent black and leading 0s are allowed and because a black card is a final state in that it cannot be used to start a move, each move decreases the current number and therefore the game must end.
Since there are forty turns minimum, we can reduce the game to playing the forty cards without any consecutive cards to be flipped.
#codingexercise
Flip_gold_and_black_cards(ref Cards[] cards)
{
var index = get_first_gold_card_from_left(ref cards);
flip_next_fifty_cards_at(ref cards, index);
}
int get_first_gold_card_from_left(ref cards)
{
for (int I=0; I <cards.length; I++)
     if (cards[I].color == gold) return I;
return cards.length;
}
void flip_next_fifty_cards_at(ref cards, int start)
{
for (int I = start; I < cards.length; I++)
  cards[I]. color = cards[I].color == gold ? black:gold;
}
}
#BookReview
We review the book "The Responsible Entrepreneur" written by Carol Sanford.
She introduces us to the entrepreneur who is responsible - a special breed who challenge and refine cultural assumptions, laws and regulations and even the process of governance. As Kevin Jones, founder of SoCap said, this book is about being courageous and in fact he dares the readers to read and implement the bold useful advice.
Let us take a look at what Carol means by Responsible Entrepreneurs and what they achieve. She says these entrepreneurs are required to do and think far beyond what is usually required of business leaders. The Responsible Entrepreneurs offers us this 'blueprint'. By understanding the archetype  most aligned with their goals, entrepreneurs will learn how to grow their business and even change the game.
In this book she includes the following:
Some extraordinary people have changed the game for the better.
How modern archetypes are altering the future.
Why new measures of accomplishment are based on the success of the whole.

She says such entrepreneurs have the courage to take on what they don't yet know how to do and the dedication to build the capability to do it. These entrepreneurs are driven by the realization that the society and the planet need something big from them and that, if they don't rise to the challenge, the work may not get done.
Entrepreneuralism is about personal agency and the development of the will. While entrepreneurs are inventing new products, industries, sources of capital and models of enterprise, the responsible entrepreneurs are doing that and more because they want to create the right platform and support an ecosystem.  They run successful business while deeply caring about the innovation that can not only help them secure their business but also transform the world.
They enjoy developing the acumen needed to work on all parts of the business.
They maintain their own motivation, getting stimulus, training and innovation when they need it. They hold a positive attitude.
They are willing to take on big challenges that stretch them beyond what they currently know how to do and to ride the roller coaster of needing to continually rise to the occassion.
They tend to focus the game changing aspirations in one of four distinct domains
Industries - where the work is to disrupt and replace the automatic patterns
Social systems - where the work is to move upstream to the causes of social problems and address them at their source.
Cultural paradigms - where the enhance the belief systems into something more holistic and embracing
Foundational agreements - where the work is to renew and vitalize the deeper intention behind the governing documents, such as a corporate charter.

From these four domains she draws parallels to four archetypes
Warrior - The warrior protects the values of a community. In the business world, this work takes place within the industry.
Clown - The clown pokes fun at collective self-centeredness and unconsciousness, opening space for humility and heartfelt appreciation of others.
Hunter - The hunter perpetuates life by strengthening the mutual exchange between the tribe and the natural world.
Headman - The headman awakens the individuals to their potential and inspires them to work with others in order to contribute to something larger than themselves.

The four archetypes are all necessary to the healthy functioning of the society. In fact if any one of them is missing, the society becomes vulnerable.

To the entreprenueur, these archetypes translate into four unique roles as follows:
The warrior who changes an industry is the realization entrepreneur. He is driven by the vision of an improved reality.
The clown who changes social systems is the reconnection entrepreneur. She reveals the gap in our cognition regarding the impact of existing social systems
The hunter who changes cultural paradigms is the reciprocity entrepreneur. She tries to strike a balance between what is taken and what is given.
The headman or headwoman who changes the connection to foundation agreements is the regenerative entrepreneur.  She seeks to reveal and evolve the inherent potential of founding agreements that create the accepted structures within which the society operates.

The four roles do different work. For each of them the author takes examples and the principles or pillars that they operate on.