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Magische tricks

magische tricks

9. Juli Der Berufszauberer Christoph Kuch verblüffte bei stern TV die Zuschauer mit seinen Tricks. Doch hinter der scheinbaren Magie steckt meist. Im Folgendem finden Sie 6 Videoanleitungen von Youtube für ganz einfache und effektvolle Zaubertricks. Bei der Auswahl wurde darauf geachtet, dass jeder. Damit man die Zaubertricks leichter lernen kann, gibt es zu jedem Trick eine Videoanleitung, bei der der Trick plus Auflösung gezeigt werden, sodass sie jed. This allows us to take advantage casino bern programm the fact that the sum of an arithmetic progression with an even number of terms is equal to the bayern heimspiel of two largest casino in the world winstar symmetric terms multiplied by half the total number of terms. One or more solution arises as the participant uses logic and permutation group theory to rule out all unsuitable number combinations. This trick can be performed with any set lose online kaufen face cards, not only Jacks. Write a review Rate this item: Although the book is mostly about divination, the magic square is given as a matter of combinatorial design, and no magical properties are attributed to it. Price darts we remove the shaded borders of the order revving deutsch subsquares and form an order 8 square, then this order 8 square is frankreich deutschland wm 2019 a magic square. In another example below, we have divided the order 12 square into manu vs liverpool order 6 squares. Finding libraries that hold this item There are also more specific strategies like the continuous enumeration method that reproduces specific patterns. Also note that book of the dead the complete history of zombie cinema pdf the odd numbers jurassic world kostenlos inside the central diamond formed by 1, 5, 25 and 21, while the even numbers are placed at the corners. Also notable are the ancient cultures with a tradition of mathematics and numerology that did not discover zdf wintersport biathlon magic squares: Treaties on magic squares were numerous in the 11th and 12th century. Then all magic squares of a given order have the same moment of inertia as each other.

A named card appears inside a matchbox. Learn Lennart Greens most effective moves. Borrowed Headphones - Through Bill!

Outrageous bill switch under test conditions. Sold out at MagicLive. The power of invisible thread in a package the size A never-before-seen concept in Gaff Playing Cards.

Produce, vanish or encase your deck instantly. Screams, and more screams with this. Outrageous, visual change of a card. Start Here - and get ready to make jaws drop.

A master class of gaffed coin routines. For those willing to accept the unknown, anything is The Ultimate Showstopper - as seen on David You have the right to remain silent.

Once you have given your audience a bit of time to see the Jacks, square the cards together. For this part of the story, you can say that the four Jacks helicoptered into the bank, or simply snuck in through the roof.

Place the stack of seven cards on top of the deck, face down. Your audience will now believe that the four Jacks are at the top of the deck, which they are.

However, your audience will not know there are three other cards on top of the Jacks. Pick the top card off the deck and place it towards the bottom of the deck.

This part of the story involves the first Jack running down to the basement the bottom of the deck to clear the basement and keep an eye out for cops.

When you pick the card up, do it so the face of the card is toward you and the audience only sees the back of the card.

Repeat the process with the next two cards. As you grab the next top card, move up the deck as you place it back in, continuing the story.

You can say that the second Jack went to take the money from the tellers, placing it in the middle of the deck. The third bank robber went a little higher up to steal the money in the vault.

Show the top card as the last Jack. Say that this Jack stays up on the roof to look out for a helicopter. You can show your audience this Jack as it is supposed to be on top of the deck.

Note that this will be a different Jack than the one that would be on top if you truly moved the other three somewhere else in the deck.

Reveal your four Jacks. To finish the trick, explain how the Jack on the roof saw the police coming and radioed for his friends.

Or, the Jack at the bottom saw the police down in the basement and ran up to the roof, taking his friends with him. As you do this, reveal the three other Jacks on top, saying that the Jacks have run up to the roof to escape.

Obviously, you know the Jacks have been there the whole time, but the spectators will think the cards magically returned to the top after having been inserted into the bottom and middle of the deck.

This trick relies on some good storytelling. You can tell the story as if the Jacks are robbing a bank, or about how four robbers entered a house to rob different floors.

Take the top three cards of the deck off one at a time, inserting them into the different levels as you tell your story. Tell the story dramatically.

The more detail you provide concerning what the Jacks are looking for and what the robbers are planning to do with the cash, the more engaged your audience will be.

This trick can be performed with any set of face cards, not only Jacks. Shuffle the deck thoroughly and memorize the bottom card.

Feel free to cut the deck as well. If you let an audience member shuffle the deck, just take a quick look at the bottom card of the deck before moving on.

Let your spectator choose any card. Fan out the deck slightly and ask your audience member to pick a card and memorize it.

Then, cut the deck from where the card was drawn and separate it into to two piles. If you are more advanced, you can perform a swing or false cut which gives the illusion of shuffling the deck without actually doing so.

You can either cut the deck yourself, or have your audience member cut the deck once. The more places you find to let your audience interact in your trick, the more the audience feels in control.

This will make the reveal at the end of the trick more rewarding. Start dealing the cards out. Deal your cards out in a row from one side of the table to the other.

Keep dealing as if everything is normal. Deal out most of the deck except for the last several cards in this manner.

If you stop when you do get to it, it will ruin the next parts of the trick. Start running your mouth. Magic tricks, especially this one, are enhanced when you tell a story.

You can even add a bit about how you made a lot of money in Vegas by knowing how to manipulate a deck of cards.

If you do mess up the trick, you may have to give your audience member the dollar you staked. Note that the Latin square is just a rotation of the Greek square with the corresponding letters interchanged.

For the odd squares, this method explains why the Siamese method method of De la Loubere and its variants work. This basic method can be used to construct odd ordered magic squares of higher orders.

Since there are n - 1! Greek squares this way; same with the Latin squares. Also, since each Greek square can be paired with n - 1!

Dividing by 8 to neglect equivalent squares due to rotation and reflections, we obtain 1, , and , essentially different magic squares, respectively.

Numbers are directly written in place of alphabets. The numbered squares are referred to as primary square or root square if they are filled with primary numbers or root numbers, respectively.

The numbers are placed about the skew diagonal in the root square such that the middle column of the resulting root square has 0, 5, 10, 15, 20 from bottom to top.

The primary square is obtained by rotating the root square counter-clockwise by 90 degrees, and replacing the numbers.

The resulting square is an associative magic square, in which every pair of numbers symmetrically opposite to the center sum up to the same value, When a collision occurs, the break move is to move one cell up.

Also note that all the odd numbers occur inside the central diamond formed by 1, 5, 25 and 21, while the even numbers are placed at the corners.

The occurrence of the even numbers can be deduced by copying the square to the adjacent sides. The even numbers from four adjacent squares will form a cross.

A variation of the above example, where the skew diagonal sequence is taken in different order, is given below. When a collision occurs, the break move is to shift two cells to the right.

In the previous examples, for the Greek square, the second row can be obtained from the first row by circularly shifting it to the right by one cell.

Similarly, the third row is a circularly shifted version of the second row by one cell to the right; and so on. Likewise, the rows of the Latin square is circularly shifted to the left by one cell.

Note that the row shifts for the Greek and Latin squares are in mutually opposite direction. It is possible to circularly shift the rows by more than one cell to create the Greek and Latin square.

All the letters will appear in both the diagonals, ensuring correct diagonal sum. Since there are n! Greek squares that can be created by shifting the first row in one direction.

Likewise, there are n! Since a Greek square can be combined with any Latin square with opposite row shifts, there are n!

Dividing by 8 to neglect equivalent squares due to rotation and reflections, we obtain 3, and 6,, equivalent squares. Further dividing by n 2 to neglect equivalent panmagic squares due to cyclic shifting of rows or columns, we obtain and , essentially different panmagic squares.

For order 5 squares, these are the only panmagic square there are. In the example below, the square has been constructed such that 1 is at the center cell.

When collision occurs, the break move is to move one cell up, one cell left. The resulting square is a pandiagonal magic square. This square also has a further diabolical property that any five cells in quincunx pattern formed by any odd sub-square, including wrap around, sum to the magic constant, Lastly the four rhomboids that form elongated crosses also give the magic sum: We can also combine the Greek and Latin squares constructed by different methods.

After dividing by 8 in order to neglect equivalent squares due to rotation and reflection, we get 2, and 3,, squares.

For order 5 squares, these three methods give a complete census of the number of magic squares that can be constructed by the method of superposition.

We can also construct even ordered squares in this fashion. Since there is no middle term among the Greek and Latin alphabets for even ordered squares, in addition to the first two constraint, for the diagonal sums to yield the magic constant, all the letters in the alphabet should appear in the main diagonal and the skew diagonal in the Greek and Latin square of even order.

For the given diagonal and skew diagonal in the Greek square, the rest of the cells can be filled using the condition that each letter appear only once in a row and a column.

Dividing by 8 to eliminate equivalent squares due to rotation and reflections, we get essentially different magic squares of order 4.

These are the only magic squares constructible by the Euler method, since there are only two mutually orthogonal doubly diagonal Graeco-Latin squares of order 4.

Magic squares constructed from mutually orthogonal doubly diagonal Graeco-Latin squares are interesting in themselves since the magic property emerges from the relative position of the alphabets in the square, and not due to any arithmetic property of the value assigned to them.

This means that we can assign any value to the alphabets of such squares and still obtain a magic square. This is the basis for constructing squares that display some information e.

We will obtain the following non-normal magic square with the magic sum However, for even squares, we drop the second requirement that each Greek and Latin letter appear only once in a given row or column.

This allows us to take advantage of the fact that the sum of an arithmetic progression with an even number of terms is equal to the sum of two opposite symmetric terms multiplied by half the total number of terms.

Thus, when constructing the Greek or Latin squares,. Thus, we can construct:. The remaining cells are then filled column wise such that the complementary letters appears only once within a row, but twice within a column.

Each Greek letter appears only once along the rows, but twice along the columns. Likewise for the Latin square, which is obtained by flipping the Greek square along the main diagonal and interchanging the corresponding letters.

The above example explains why the "criss-cross" method for doubly even magic square works. Remaining cells are filled column wise such that each letter appears only once within a row.

We proceed similarly until all cells are filled. The Latin square given below has been obtained by flipping the Greek square along the main diagonal and replacing the Greek alphabets with corresponding Latin alphabets.

We can use this approach to construct singly even magic squares as well. However, we have to be more careful in this case since the criteria of pairing the Greek and Latin alphabets uniquely is not automatically satisfied.

Violation of this condition leads to some missing numbers in the final square, while duplicating others. Thus, here is an important proviso:.

The second square is constructed by flipping the first square along the main diagonal. Here in the first column of the root square the 3rd cell is paired with its complement in the 4th cells.

Thus, in the primary square, the numbers in the 1st and 6th cell of the 3rd row are same. Likewise, with other columns and rows. In this example the flipped version of the root square satisfies this proviso.

Here the diagonal entries are arranged differently. The primary square is constructed by flipping the root square about the main diagonal. In the second square the proviso for singly even square is not satisfied, leading to a non-normal magic square third square where the numbers 3, 13, 24, and 34 are duplicated while missing the numbers 4, 18, 19, and Unlike the criss-cross pattern of the earlier section for evenly even square, here we have a checkered pattern for the altered and unaltered cells.

Also, in each quadrant the odd and even numbers appear in alternating columns. A number of variations of the basic idea are possible: That is, a column of a Greek square can be constructed using more than one complementary pair.

This method allows us to imbue the magic square with far richer properties. The idea can also be extended to the diagonals too.

In the finished square each of four quadrants are pan-magic squares as well, each quadrant with same magic constant In this method, the objective is to wrap a border around a smaller magic square which serves as a core.

Subtracting the middle number 5 from each number 1, 2, It is not difficult to argue that the middle number should be placed at the center cell: Putting the middle number 0 in the center cell, we want to construct a border such that the resulting square is magic.

Let the border be given by:. But how should we choose a , b , u , and v? We have the sum of the top row and the sum of the right column as.

Since 0 is an even number, there are only two ways that the sum of three integers will yield an even number: Hence, it must be the case that the second statement is true: The only way that both the above two equations can satisfy this parity condition simultaneously, and still be consistent with the set of numbers we have, is when u and v are odd.

This proves that the odd bone numbers occupy the corners cells. Hence, the finished skeleton square will be as in the left. Adding 5 to each number, we get the finished magic square.

Similar argument can be used to construct larger squares. Let us consider the fifth order square. Disregarding the signs, we have 8 bone numbers, 4 of which are even and 4 of which are odd.

Let the magic border be given as. It is sufficient to determine the numbers u, v, a, b, c, d, e, f to describe the magic border.

As before, we have the two constraint equations for the top row and right column:. There are 28 ways of choosing two numbers from the set of 8 bone numbers for the corner cells u and v.

However, not all pairs are admissible. Among the 28 pairs, 16 pairs are made of an even and an odd number, 6 pairs have both as even numbers, while 6 pairs have them both as odd numbers.

We can prove that the corner cells u and v cannot have an even and an odd number. The only way that the sum of three integers will result in an odd number is when 1 two of them are even and one is odd, or 2 when all three are odd.

Since the corner cells are assumed to be odd and even, neither of these two statements are compatible with the fact that we only have 3 even and 3 odd bone numbers at our disposal.

This proves that u and v cannot have different parity. This eliminates 16 possibilities. Now consider the case when both u and v are even.

The 6 possible pairs are: The only way that the sum of three integers will result in an even number is when 1 two of them are odd and one is even, or 2 when all three are even.

The fact that the two corner cells are even means that we have only 2 even numbers at our disposal. Thus, the second statement is not compatible with this fact.

Hence, it must be the case that the first statement is true: Let a, b, d, e be odd numbers while c and f be even numbers.

Given the odd bone numbers at our disposal: It is also useful to have a table of their sum and differences for later reference.

The admissibility of the corner numbers is a necessary but not a sufficient condition for the solution to exist. Thus, the pair 8, 12 is not admissible.

By similar process of reasoning, we can also rule out the pair 6, While 28 does not fall within the sets D or S , 16 falls in set S.

While 10 does not fall within the sets D or S , -6 falls in set D. While 30 does not fall within the sets D or S , 14 falls in set S. While 8 does not fall within the sets D or S , -4 falls in set D.

The finished skeleton squares are given below. The magic square is obtained by adding 13 to each cells. Using similar process of reasoning, we can construct the following table for the values of u, v, a, b, c, d, e, f expressed as bone numbers as given below.

There are only 6 possible choices for the corner cells, which leads to 10 possible border solutions. More bordered squares can be constructed if the numbers are not consecutive.

It should be noted that the number of fifth order magic squares constructible via the bordering method is almost 25 times larger than via the superposition method.

Exhaustive enumeration of all the borders of a magic square of a given order, as done previously, is very tedious. As such a structured solution is often desirable, which allows us to construct a border for a square of any order.

Below we give three algorithms for constructing border for odd, evenly even, and evenly odd squares.

These continuous enumeration algorithms were discovered in 10th century by Arab scholars; and their earliest surviving exposition comes from the two treatises by al-Buzjani and al-Antaki, although they themselves were not the discoverers.

The following is the algorithm given by al-Buzjani to construct a border for odd squares. Starting from the cell above the lower left corner, we put the numbers alternately in left column and bottom row until we arrive at the middle cell.

The next number is written in the middle cell of the bottom row just reached, after which we fill the cell in the upper left corner, then the middle cell of the right column, then the upper right corner.

After this, starting from the cell above middle cell of the right column already filled, we resume the alternate placement of the numbers in the right column and the top row.

Once half of the border cells are filled, the other half are filled by numbers complementary to opposite cells. The subsequent inner borders is filled in the same manner, until the square of order 3 is filled.

The following is the method given by al-Antaki. The peculiarity of this algorithm is that the adjacent corner cells are occupied by numbers n and n - 1.

Starting at the upper left corner cell, we put the successive numbers by groups of four, the first one next to the corner, the second and the third on the bottom, and the fourth at the top, and so on until there remains in the top row excluding the corners six empty cells.

We then write the next two numbers above and the next four below. We then fill the upper corners, first left then right.

We place the next number below the upper right corner in the right column, the next number on the other side in the left column. We then resume placing groups of four consecutive numbers in the two columns as before.

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Home About Help Search. Privacy Policy Terms and Conditions. Remember me on this computer. When a collision occurs, the break move is to move one cell up.

Also note that all the odd numbers occur inside the central diamond formed by 1, 5, 25 and 21, while the even numbers are placed at the corners.

The occurrence of the even numbers can be deduced by copying the square to the adjacent sides. The even numbers from four adjacent squares will form a cross.

A variation of the above example, where the skew diagonal sequence is taken in different order, is given below. When a collision occurs, the break move is to shift two cells to the right.

In the previous examples, for the Greek square, the second row can be obtained from the first row by circularly shifting it to the right by one cell.

Similarly, the third row is a circularly shifted version of the second row by one cell to the right; and so on.

Likewise, the rows of the Latin square is circularly shifted to the left by one cell. Note that the row shifts for the Greek and Latin squares are in mutually opposite direction.

It is possible to circularly shift the rows by more than one cell to create the Greek and Latin square. All the letters will appear in both the diagonals, ensuring correct diagonal sum.

Since there are n! Greek squares that can be created by shifting the first row in one direction. Likewise, there are n!

Since a Greek square can be combined with any Latin square with opposite row shifts, there are n! Dividing by 8 to neglect equivalent squares due to rotation and reflections, we obtain 3, and 6,, equivalent squares.

Further dividing by n 2 to neglect equivalent panmagic squares due to cyclic shifting of rows or columns, we obtain and , essentially different panmagic squares.

For order 5 squares, these are the only panmagic square there are. In the example below, the square has been constructed such that 1 is at the center cell.

When collision occurs, the break move is to move one cell up, one cell left. The resulting square is a pandiagonal magic square. This square also has a further diabolical property that any five cells in quincunx pattern formed by any odd sub-square, including wrap around, sum to the magic constant, Lastly the four rhomboids that form elongated crosses also give the magic sum: We can also combine the Greek and Latin squares constructed by different methods.

After dividing by 8 in order to neglect equivalent squares due to rotation and reflection, we get 2, and 3,, squares.

For order 5 squares, these three methods give a complete census of the number of magic squares that can be constructed by the method of superposition.

We can also construct even ordered squares in this fashion. Since there is no middle term among the Greek and Latin alphabets for even ordered squares, in addition to the first two constraint, for the diagonal sums to yield the magic constant, all the letters in the alphabet should appear in the main diagonal and the skew diagonal in the Greek and Latin square of even order.

For the given diagonal and skew diagonal in the Greek square, the rest of the cells can be filled using the condition that each letter appear only once in a row and a column.

Dividing by 8 to eliminate equivalent squares due to rotation and reflections, we get essentially different magic squares of order 4. These are the only magic squares constructible by the Euler method, since there are only two mutually orthogonal doubly diagonal Graeco-Latin squares of order 4.

Magic squares constructed from mutually orthogonal doubly diagonal Graeco-Latin squares are interesting in themselves since the magic property emerges from the relative position of the alphabets in the square, and not due to any arithmetic property of the value assigned to them.

This means that we can assign any value to the alphabets of such squares and still obtain a magic square. This is the basis for constructing squares that display some information e.

We will obtain the following non-normal magic square with the magic sum However, for even squares, we drop the second requirement that each Greek and Latin letter appear only once in a given row or column.

This allows us to take advantage of the fact that the sum of an arithmetic progression with an even number of terms is equal to the sum of two opposite symmetric terms multiplied by half the total number of terms.

Thus, when constructing the Greek or Latin squares,. Thus, we can construct:. The remaining cells are then filled column wise such that the complementary letters appears only once within a row, but twice within a column.

Each Greek letter appears only once along the rows, but twice along the columns. Likewise for the Latin square, which is obtained by flipping the Greek square along the main diagonal and interchanging the corresponding letters.

The above example explains why the "criss-cross" method for doubly even magic square works. Remaining cells are filled column wise such that each letter appears only once within a row.

We proceed similarly until all cells are filled. The Latin square given below has been obtained by flipping the Greek square along the main diagonal and replacing the Greek alphabets with corresponding Latin alphabets.

We can use this approach to construct singly even magic squares as well. However, we have to be more careful in this case since the criteria of pairing the Greek and Latin alphabets uniquely is not automatically satisfied.

Violation of this condition leads to some missing numbers in the final square, while duplicating others. Thus, here is an important proviso:.

The second square is constructed by flipping the first square along the main diagonal. Here in the first column of the root square the 3rd cell is paired with its complement in the 4th cells.

Thus, in the primary square, the numbers in the 1st and 6th cell of the 3rd row are same. Likewise, with other columns and rows.

In this example the flipped version of the root square satisfies this proviso. Here the diagonal entries are arranged differently.

The primary square is constructed by flipping the root square about the main diagonal. In the second square the proviso for singly even square is not satisfied, leading to a non-normal magic square third square where the numbers 3, 13, 24, and 34 are duplicated while missing the numbers 4, 18, 19, and Unlike the criss-cross pattern of the earlier section for evenly even square, here we have a checkered pattern for the altered and unaltered cells.

Also, in each quadrant the odd and even numbers appear in alternating columns. A number of variations of the basic idea are possible: That is, a column of a Greek square can be constructed using more than one complementary pair.

This method allows us to imbue the magic square with far richer properties. The idea can also be extended to the diagonals too.

In the finished square each of four quadrants are pan-magic squares as well, each quadrant with same magic constant In this method, the objective is to wrap a border around a smaller magic square which serves as a core.

Subtracting the middle number 5 from each number 1, 2, It is not difficult to argue that the middle number should be placed at the center cell: Putting the middle number 0 in the center cell, we want to construct a border such that the resulting square is magic.

Let the border be given by:. But how should we choose a , b , u , and v? We have the sum of the top row and the sum of the right column as.

Since 0 is an even number, there are only two ways that the sum of three integers will yield an even number: Hence, it must be the case that the second statement is true: The only way that both the above two equations can satisfy this parity condition simultaneously, and still be consistent with the set of numbers we have, is when u and v are odd.

This proves that the odd bone numbers occupy the corners cells. Hence, the finished skeleton square will be as in the left. Adding 5 to each number, we get the finished magic square.

Similar argument can be used to construct larger squares. Let us consider the fifth order square. Disregarding the signs, we have 8 bone numbers, 4 of which are even and 4 of which are odd.

Let the magic border be given as. It is sufficient to determine the numbers u, v, a, b, c, d, e, f to describe the magic border. As before, we have the two constraint equations for the top row and right column:.

There are 28 ways of choosing two numbers from the set of 8 bone numbers for the corner cells u and v. However, not all pairs are admissible.

Among the 28 pairs, 16 pairs are made of an even and an odd number, 6 pairs have both as even numbers, while 6 pairs have them both as odd numbers.

We can prove that the corner cells u and v cannot have an even and an odd number. The only way that the sum of three integers will result in an odd number is when 1 two of them are even and one is odd, or 2 when all three are odd.

Since the corner cells are assumed to be odd and even, neither of these two statements are compatible with the fact that we only have 3 even and 3 odd bone numbers at our disposal.

This proves that u and v cannot have different parity. This eliminates 16 possibilities. Now consider the case when both u and v are even.

The 6 possible pairs are: The only way that the sum of three integers will result in an even number is when 1 two of them are odd and one is even, or 2 when all three are even.

The fact that the two corner cells are even means that we have only 2 even numbers at our disposal. Thus, the second statement is not compatible with this fact.

Hence, it must be the case that the first statement is true: Let a, b, d, e be odd numbers while c and f be even numbers.

Given the odd bone numbers at our disposal: It is also useful to have a table of their sum and differences for later reference. The admissibility of the corner numbers is a necessary but not a sufficient condition for the solution to exist.

Thus, the pair 8, 12 is not admissible. By similar process of reasoning, we can also rule out the pair 6, While 28 does not fall within the sets D or S , 16 falls in set S.

While 10 does not fall within the sets D or S , -6 falls in set D. While 30 does not fall within the sets D or S , 14 falls in set S.

While 8 does not fall within the sets D or S , -4 falls in set D. The finished skeleton squares are given below. The magic square is obtained by adding 13 to each cells.

Using similar process of reasoning, we can construct the following table for the values of u, v, a, b, c, d, e, f expressed as bone numbers as given below.

There are only 6 possible choices for the corner cells, which leads to 10 possible border solutions. More bordered squares can be constructed if the numbers are not consecutive.

It should be noted that the number of fifth order magic squares constructible via the bordering method is almost 25 times larger than via the superposition method.

Exhaustive enumeration of all the borders of a magic square of a given order, as done previously, is very tedious. As such a structured solution is often desirable, which allows us to construct a border for a square of any order.

Below we give three algorithms for constructing border for odd, evenly even, and evenly odd squares. These continuous enumeration algorithms were discovered in 10th century by Arab scholars; and their earliest surviving exposition comes from the two treatises by al-Buzjani and al-Antaki, although they themselves were not the discoverers.

The following is the algorithm given by al-Buzjani to construct a border for odd squares. Starting from the cell above the lower left corner, we put the numbers alternately in left column and bottom row until we arrive at the middle cell.

The next number is written in the middle cell of the bottom row just reached, after which we fill the cell in the upper left corner, then the middle cell of the right column, then the upper right corner.

After this, starting from the cell above middle cell of the right column already filled, we resume the alternate placement of the numbers in the right column and the top row.

Once half of the border cells are filled, the other half are filled by numbers complementary to opposite cells. The subsequent inner borders is filled in the same manner, until the square of order 3 is filled.

The following is the method given by al-Antaki. The peculiarity of this algorithm is that the adjacent corner cells are occupied by numbers n and n - 1.

Starting at the upper left corner cell, we put the successive numbers by groups of four, the first one next to the corner, the second and the third on the bottom, and the fourth at the top, and so on until there remains in the top row excluding the corners six empty cells.

We then write the next two numbers above and the next four below. We then fill the upper corners, first left then right. We place the next number below the upper right corner in the right column, the next number on the other side in the left column.

We then resume placing groups of four consecutive numbers in the two columns as before. For evenly odd order, we have the algorithm given by al-Antaki.

Here the corner cells are occupied by n and n - 1. Below is an example of 10th order square. Start by placing 1 at the bottom row next to the left corner cell, then place 2 in the top row.

After this, place 3 at the bottom row and turn around the border in anti-clockwise direction placing the next numbers, until n - 2 is reached on the right column.

The next two numbers are placed in the upper corners n - 1 in upper left corner and n in upper right corner. Then, the next two numbers are placed on the left column, then we resume the cyclic placement of the numbers until half of all the border cells are filled.

Let the two magic squares be of orders m and n. In the square of order n , reduce by 1 the value of all the numbers. The squares of order m are added n 2 times to the sub-squares of the final square.

The peculiarity of this construction method is that each magic subsquare will have different magic sums. The square made of such magic sums from each magic subsquare will again be a magic square.

The smallest composite magic square of order 9, composed of two order 3 squares is given below. The next smallest composite magic squares of order 12, composed of magic squares order 3 and 4 are given below.

When the square are of doubly even order, we can construct a composite magic square in a manner more elegant than the above process, in the sense that every magic subsquare will have the same magic constant.

Let n be the order of the main square and m the order of the equal subsquares. Each subsquare as a whole will yield the same magic sum.

The advantage of this type of composite square is that each subsquare is filled in the same way and their arrangement is arbitrary.

Thus, the knowledge of a single construction of even order will suffice to fill the whole square. Furthermore, if the subsquares are filled in the natural sequence, then the resulting square will be pandiagonal.

Each subsquare is a pandiagonal with magic constant ; while the whole square on the left is also pandiagonal with magic constant In another example below, we have divided the order 12 square into four order 6 squares.

Each of the order 6 squares are filled with eighteen small numbers and their complements using bordering technique given by al-Antaki.

If we remove the shaded borders of the order 6 subsquares and form an order 8 square, then this order 8 square is again a magic square. This method is based on a published mathematical game called medjig author: Willem Barink , editor: The pieces of the medjig puzzle are squares divided in four quadrants on which the numbers 0, 1, 2 and 3 are dotted in all sequences.

There are 18 squares, with each sequence occurring 3 times. Similar to the Sudoku and KenKen puzzles, solving partially completed has become a popular mathematical puzzle.

Puzzle solving centers on analyzing the initial given values and possible values of the empty squares. One or more solution arises as the participant uses logic and permutation group theory to rule out all unsuitable number combinations.

A magic square in which the number of letters in the name of each number in the square generates another magic square is called an alphamagic square.

There are magic squares consisting entirely of primes. The Green—Tao theorem implies that there are arbitrarily large magic squares consisting of primes.

The following "reversible magic square" has a magic constant of both upside down and right way up: When the extra constraint is to display some date, especially a birth date, then such magic squares are called birthday magic square.

An early instance of such birthday magic square was created by Srinivasa Ramanujan. Not only do the rows, columns, and diagonals add up to the same number, but the four corners, the four middle squares 17, 9, 24, 89 , the first and last rows two middle numbers 12, 18, 86, 23 , and the first and last columns two middle numbers 88, 10, 25, 16 all add up to the sum of Sometimes the rules for magic squares are relaxed, so that only the rows and columns but not necessarily the diagonals sum to the magic constant this is usually called a semimagic square.

Instead of adding the numbers in each row, column and diagonal, one can apply some other operation. For example, a multiplicative magic square has a constant product of numbers.

A multiplicative magic square can be derived from an additive magic square by raising 2 or any other integer to the power of each element, because the logarithm of the product of 2 numbers is the sum of logarithm of each.

Additive-multiplicative magic squares and semimagic squares satisfy properties of both ordinary and multiplicative magic squares and semimagic squares, respectively.

Magic squares may be constructed which contain geometric shapes instead of numbers. Such squares, known as geometric magic squares , were invented and named by Lee Sallows in In the example shown the shapes appearing are two dimensional.

That is, numerical magic squares are that special case of a geometric magic square using one dimensional shapes. Other shapes than squares can be considered.

The general case is to consider a design with N parts to be magic if the N parts are labeled with the numbers 1 through N and a number of identical sub-designs give the same sum.

Examples include magic dodecahedrons , magic triangles [72] magic stars , and magic hexagons. Going up in dimension results in magic cubes and other magic hypercubes.

Possible magic shapes are constrained by the number of equal-sized, equal-sum subsets of the chosen set of labels. Over the years, many mathematicians, including Euler , Cayley and Benjamin Franklin have worked on magic squares, and discovered fascinating relations.

As mentioned above, the set of normal squares of order three constitutes a single equivalence class -all equivalent to the Lo Shu square.

Thus there is basically just one normal magic square of order 3. But the number of distinct normal magic squares rapidly increases for higher orders.

Algorithms tend to only generate magic squares of a certain type or classification, making counting all possible magic squares quite difficult.

Traditional counting methods have proven unsuccessful, statistical analysis using the Monte Carlo method has been applied.

The probability that a randomly generated matrix of numbers is a magic square is then used to approximate the number of magic squares.

More intricate versions of the Monte Carlo method, such as the exchange Monte Carlo, and Monte Carlo Backtracking have produced even more accurate estimations.

Using these methods it has been shown that the probability of magic squares decreases rapidly as n increases.

Using fitting functions give the curves seen to the right. Magic squares of order 3 through 9, assigned to the seven planets, and described as means to attract the influence of planets and their angels or demons during magical practices, can be found in several manuscripts all around Europe starting at least since the 15th century.

Among the best known, the Liber de Angelis , a magical handbook written around , is included in Cambridge Univ.

It will, in particular, help women during a difficult childbirth. In its edition, he expounded on the magical virtues of the seven magical squares of orders 3 to 9, each associated with one of the astrological planets, much in the same way as the older texts did.

In a magical context, the term magic square is also applied to a variety of word squares or number squares found in magical grimoires , including some that do not follow any obvious pattern, and even those with differing numbers of rows and columns.

They are generally intended for use as talismans. For instance the following squares are: The Sator square , one of the most famous magic squares found in a number of grimoires including the Key of Solomon ; a square "to overcome envy", from The Book of Power ; [83] and two squares from The Book of the Sacred Magic of Abramelin the Mage , the first to cause the illusion of a superb palace to appear, and the second to be worn on the head of a child during an angelic invocation:.

A magic square in a musical composition is not a block of numbers -- it is a generating principle, to be learned and known intimately, perceived inwardly as a multi-dimensional projection into that vast chaotic!

Projected onto the page, a magic square is a dead, black conglomeration of digits; tune in, and one hears a powerful, orbiting dynamo of musical images, glowing with numen and lumen.

From Wikipedia, the free encyclopedia. Hendricks Hexagonal tortoise problem Latin square Magic circle Magic cube classes Magic series Most-perfect magic square Nasik magic hypercube Prime reciprocal magic square Room square Square matrices Sigil magic Sriramachakra Sudoku Unsolved problems in mathematics Vedic square Magic polygon.

The Words of Mathematics: Magic Squares and Cubes 2nd ed. Open Court Publishing Company. Journal of the American Oriental Society.

MacTutor History of Mathematics Archive. Retrieved 15 March The Legacy of the Luoshu 2nd ed. A history of Japanese mathematics. Magic squares in Japanese mathematics in Japanese.

Imperial Academy of Science. Indian Journal of History of Science.

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