I Ching divination - Books - Famous Chinese Shopping City

March 8, 2014


		1 Introduction 2 Methods 3 See also

Famous Chinese Shopping City Books

I Ching divination

Wikipedia

Among the many forms of Divination|divination is a method using the I Ching or Book of Changes. The book is structured as an 8x8 matrix of sixty-four hexagrams representing the states and the dynamic relationships of the eight elements, each represented by a trigram. Throughout China's region of cultural influence (including Korea, Japan and Vietnam), scholars have added comments and interpretation to this work, one of the most important in ancient Chinese culture; it has also attracted the interest of many thinkers in the West. See the I Ching main article for historical and philosophical information.

The process of consulting the book as an oracle involves determining the hexagram by a method of random generation and then reading the text associated with that hexagram, and is a form of bibliomancy.

Each line of a hexagram determined with these methods is either stable ("young") or changing ("old"); thus, there are four possibilities for each line, corresponding to the cycle of change from yin to yang and back again:

old yin (yin changing into yang), which has the number 6 and symbol <strike>---</strike>x<strike>---</strike>

young yang (unchanging yang), which has the number 7 and symbol <strike>--------</strike>

young yin (unchanging yin), which has the number 8 and symbol <strike>---</strike>  <strike>---</strike>

old yang (yang changing into yin), which has the number 9 and symbol <strike>---o---</strike>

Once a hexagram is determined, each line has been determined as either changing (old) or unchanging (young). Since each changing line is seen as being in the process of becoming its opposite, a new hexagram can be formed by transposing each changing yin line with a yang line, and vice versa. Thus, further insight into the process of change is gained by reading the text of this new hexagram and studying it as the result of the current change.

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Methods

Several of the methods use a randomising agent to determine each line of the hexagram. These methods produce a number which corresponds to the numbers of changing or unchanging lines discussed above, and thus determines each line of the hexagram.

Cracks on turtle shell

The turtle shell oracle is probably the earliest record of fortune telling.
The diviner would apply heat to a piece of a turtle shell (sometimes with a hot poker), and interpret the resulting cracks.
The cracks were sometimes annotated with inscriptions, the oldest Chinese writings that have been discovered.
This oracle predated the earliest versions of the Zhou Yi (dated from about 1100 BC) by hundreds of years.

A variant on this method was to use ox shoulder bones. When thick material was to be cracked, the underside was thinned by carving with a knife.

Yarrow stalks

The yarrow stalk method of divination was the next major oracular method after the turtle shell. It was comparatively quick and easy to perform. A yarrow stalk is a piece of dried stem from the yarrow plant, approximately 15-18 inches in length.

However, the yarrow divination is not a truly randomized method, since it is statistical bias|statistically biased toward certain answers. While it is unlikely that the ancient Chinese knew the mathematics|mathematical justification for why this was so, some ancient practitioners were probably aware of the bias through the empiricism|empirical evidence of many repeated divinations.

The yarrow stalk method is performed as follows:

use fifty dried stalks of the yarrow plant and a large clear table space

set aside one stalk to represent unity, using forty-nine stalks for the remainder of the ritual

for each of the six lines of the hexagram (starting from the bottom and moving up)

* divide and count the stalks three times as follows

** gather the stalks into the left hand

** split them randomly into two bundles with the left thumb

** place the two bundles separately, as left and right piles, onto the table

** take one stalk from the right side pile, hold it between the little finger and ring finger of the left hand

** pick up the left side pile in the left hand

** count the stalks from the pile into separate piles of four, until four or fewer remain

** hold this remainder between the ring and middle finger of the left hand

** pick up the right side pile in the left hand

** count the stalks from the pile into separate piles of four, until four or fewer remain

** hold this remainder between the index and middle finger of the left hand

** set aside all the stalks held between fingers of the left hand

** count the number of piles of four stalks

** if this is not the third iteration, gather all the piles of four together to repeat the dividing and counting process

* after the third iteration, the number of piles of four stalks will be six, seven, eight or nine

* determine the current line of the hexagram from this number: 6 = old yin, 7 = young yang, 8 = young yin, 9 = old yang.

once six lines have been determined the hexagram is formed

Using this method, the probabilities of each type of line are as follows:

old yin: 1 in 16 (0.0625)

young yang: 5 in 16 (0.3125)

young yin: 7 in 16 (0.4375)

old yang: 3 in 16 (0.1875)

By way of explanation:

You start with 50, then subtract 1 to become 49. Then divide into two piles, P and 49-P, on the table and then keep 1 on the left hand

If P mod 4 is 1, 49 - 1 - P would be 3 ) all results into 4
If P mod 4 is 2, 49 - 1 - P would be 2 )
If P mod 4 is 3, 49 - 1 - P would be 1 )

and hence the number will become 49 -1 - 4 = 44 with a probability of 3/4.

If P mod 4 is 4, 49 - 1 - P would be 4 ) results into 8

and hence the number becomes 49 - 1 - 8 = 40, with a probability of 1/4.

As hereafter the pile is divisible by 4, from the second iteration (which repeat in the third iteration) the probability is quite different, as follows (using example of 44):

From 44 it divides into two piles X and 44-X on the table with 1 on the left hand

If X mod 4 is 2, 44 - 1 - X would be 1 ) all reseults into a pile of 3
If X mod 4 is 1, 44 - 1 - X would be 2 )

and hence the number will become 44 - 1 - 3 = 40 with a probability of 2/4.

If X mod 4 is 4, 44 - 1 - X would be 3 ) all reseults into a pile of 7
If X mod 4 is 3, 44 - 1 - X would be 4 )

and hence the number becomes 44 - 1 - 7 = 36, with a probability of 2/4.

Altogether that would generate the following probability tree:

50 - 1 -> 49

-> 44 (3/4) -> 40 (2/4) -> 36 (2/4) i.e. as 9 at probability 3*2*2/64 = 12/64
-> 32 (2/4) i.e. as 8 at probability 3*2*2/64 = 12/64
-> 36 (2/4) -> 32 (2/4) i.e. as 8 at probability 3*2*2/64 = 12/64
-> 28 (2/4) i.e. as 7 at probability 3*2*2/64 = 12/64
-> 40 (1/4) -> 36 (2/4) -> 32 (2/4) i.e. as 8 at probability 1*2*2/64 = 4/64
-> 28 (2/4) i.e. as 7 at probability 1*2*2/64 = 4/64
-> 32 (2/4) -> 28 (2/4) i.e. as 7 at probability 1*2*2/64 = 4/64
-> 24 (2/4) i.e. as 6 at probability 1*2*2/64 = 4/64

Collecting all these branches one arrives at:

Probability(6) should be ( 4)/64 = 4/64 = 1/16 for old yin
Probability(9) should be (12 )/64 = 12/64 = 3/16 for old yang
Probability(7) should be (12+ 4+ 4)/64 = 20/64 = 5/16 for young yang
Probability(8) should be (12+12+ 4)/64 = 28/64 = 7/16 for young yin

This is the same from the probability above. The correct probability has been used also in the marble, bean, dice and two or four coin methods below. This probability is significantly different from that of the three-coin method, because the required amount of accuracy occupies four binary bits of information, so three coins is one bit short. In terms of chances-out-of-sixteen, the three-coin method yields 2,2,6,6 instead of 1,3,5,7 for old-yin, old-yang, young-yang, young-yin respectively.

Coins

Three-coin method

The three coin method came into currency over a thousand years later. The quickest, easiest, and most popular method by far, it has largely supplanted the yarrow stalks. However, it is significant that the probabilities of this method differ from the yarrow stalks.

Using this method, the probabilities of each type of line are as follows:

old yang: 1 in 8 (0.125)

old yin: 1 in 8 (0.125)

young yang: 3 in 8 (0.375)

young yin: 3 in 8 (0.375)

While there is one method for tossing three coins (once for each line in the hexagram), there are several ways of checking the results.

= How the coins are tossed =

use three coins with distinct "head" and "tail" sides

for each of the six lines of the hexagram, beginning with the first (bottom) line and ending with the sixth (top) line:

toss all three coins

write down the resulting line

once six lines have been determined, the hexagram is formed

= How the line is determined from the coin toss =
The numerical method:

assign the value 3 to each "head" result, and 2 to each "tail" result

total all the coin values

the total will be six, seven, eight or nine

determine the current line of the hexagram from this number: 6 = old yin, 7 = young yang, 8 = young yin, 9 = old yang.

An alternative is to count the "tails":

3 tails = old yin

2 tails = young yang

1 tail = young yin

0 tails = old yang

Another alternative is this simple mnemonic based on the dynamics of a group of three people. If they are all boys, for example, the masculine prevails. But, if there is one girl with two boys, the feminine prevails! So:

all tails = old yin

one tail = young yin

one head = young yang

all heads = old yang

Two-coin method

Some purists contend that there is a problem with the three-coin method because its probabilities differ from the more ancient yarrow-stalk method. Others would argue that the yarrow stalk method was flawed, and the three coins method was actually superior. In fact, over the centuries there have even been other methods used for consulting the oracle.

If want an easier and faster way of consulting the oracle with a method that has the same probabilities as the yarrow stalk method, here's a method using two coins (with two tosses per line):

first toss of the two coins: if both are "heads," use a value of 2; otherwise, value is 3

second toss: a "head" has a value of 2, a "tail" a value of 3. Add the two values from this toss and the value from the first toss.

the sum of the three vales will be 6 (old yin), 7 (young yang), 8 (young yin), or 9 (old yang). This provides the first (bottom) line of the hexagram.

Repeat the process for each remaining line.

Four coins
If you're comfortable with binary, four coins can be very quick and easy, and like 2 coins matches the probablities of the yarrow-stalk method. Here's a table showing the different combinations of four coin throws and their binary sum and corresponding line (six lines making a full changing hexagram starting at the bottom). To calculate the binary sum of a four coin throw, place the coins in a line, then add up all the heads using 8 for the left-most coin, then 4, 2 and 1 for a head in the right-most position. The full explantaion relating it to the yarrow stalk method is at OrganicDesign:I Ching / Divination.
<table cellspacing=0 cellpadding=0><tr>
<td valign=top style=text-align:center>
<table border=1 borderwidth=1 cellspacing=0 cellpadding=4 style=text-align:center>
<tr><td>Sum<td>Coins<td>Line
<tr><td>0<td>T T T T<td><strike>---</strike>x<strike>---</strike>
<tr><td>1<td>T T T H<td><strike>---o---</strike>
<tr><td>2<td>T T H T<td><strike>---o---</strike>
<tr><td>3<td>T T H H<td><strike>---o---</strike>
</table>
<td width=50> 
<td valign=top style=text-align:center>
<table border=1 borderwidth=1 cellspacing=0 cellpadding=4 style=text-align:center>
<tr><td>Sum<td>Coins<td>Line
<tr><td>4<td>T H T T<td><strike>-------</strike>
<tr><td>5<td>T H T H<td><strike>-------</strike>
<tr><td>6<td>T H H T<td><strike>-------</strike>
<tr><td>7<td>T H H H<td><strike>-------</strike>
</table>
<td width=50> 
<td valign=top style=text-align:center>
<table border=1 borderwidth=1 cellspacing=0 cellpadding=4 style=text-align:center>
<tr><td>Sum<td>Coins<td>Line
<tr><td>8<td>H T T T<td><strike>-------</strike>
<tr><td>9<td>H T T H<td><strike>---</strike> <strike>---</strike>
<tr><td>10<td>H T H T<td><strike>---</strike> <strike>---</strike>
<tr><td>11<td>H T H H<td><strike>---</strike> <strike>---</strike>
</table>
<td width=50> 
<td valign=top style=text-align:center>
<table border=1 borderwidth=1 cellspacing=0 cellpadding=4 style=text-align:center>
<tr><td>Sum<td>Coins<td>Line
<tr><td>12<td>H H T T<td><strike>---</strike> <strike>---</strike>
<tr><td>13<td>H H T H<td><strike>---</strike> <strike>---</strike>
<tr><td>14<td>H H H T<td><strike>---</strike> <strike>---</strike>
<tr><td>15<td>H H H H<td><strike>---</strike> <strike>---</strike>
</table>
</table>

Dice

Using coins will quickly reveal some problems: while shaking the coins in cupped hands, it's hard to know whether they are truly being tumbled; when flipping the coins, they tend to bounce and scatter. It's much easier to use a die as a coin-equivalent: if an odd number of pips shows, it counts as "heads"; if an even number of pips shows, as "tails." Obviously, the 50/50 probability is preserved -- and rolling dice turns out to be easier and quicker than flipping coins. Thus the three-coin method will use three dice.

Dice can also be used for the two-coin method. It is best to use two pairs of dice, each pair having its own color -- e.g., a pair of blue dice and a pair of white dice, such as are commonly found in backgammon sets. One pair can then be designated the "first toss" in the two-coin method, and the other the "second toss." One roll of four dice will then determine a line, with probabilities matching the yarrow-stalk method.

A similar distribution to yarrow stalks is possibly using two dice, 1 eight-sided (1d8), and 1 twenty-sided
(1d20). Roll both of them at once per line.

If the 1d20 is an even number
then
if the 1d8 = 1 -X- moving yin (1/16 pobability)
if the 1d8 = 2 - 8 - - yin (7/16 pobability)
If the 1d20 is an odd number:
then
if the 1d8 = 1 - 5 --- yang (5/16 pobability)
if the 1d8 = 6 - 8 -0- moving yang (3/16 pobability)

Marbles or beads (method of sixteen)

This method is a recent innovation, designed to be quick like the coin method, while giving the same probabilities as the yarrow stalk method.

use sixteen marbles of four different colours but the same size, distributed as follows

* 1 marble of a colour representing old yin (such as blue)

* 5 marbles of a colour representing young yang (such as white)

* 7 marbles of a colour representing young yin (such as black)

* 3 marbles of a colour representing old yang (such as red)

place all the marbles in a bag or other opaque container

for each of the six lines of the hexagram

* shake all sixteen marbles together in the container to "shuffle" them

* draw out one marble

* the marble drawn determines the current line of the hexagram

* replace the marble in the container

once six lines have been determined, the hexagram is formed

A good source of marbles is a (secondhand) Chinese checkers set: 6 colors, 10 marbles each.

Using this method, the probabilities of each type of line are the same as the distribution of the colours, as follows:

old yin: 1 in 16 (0.0625)

young yang: 5 in 16 (0.3125)

young yin: 7 in 16 (0.4375)

old yang: 3 in 16 (0.1875)

An improvement on this method uses 16 beads of four different colors but with the same size and shape (i.e., indistinguishable by touch), strung beads being much more portable than marbles. You take the string and, without looking, grab a bead a random. The comments above apply to this method as well.

Rice grains

For this method, either rice grains, or small seeds are used.

One picks up a few seeds between the middle finger and thumb. Carefully and respectfully place them on a clean sheet of paper. Repeat this process six times, keeping each cluster of seeds in a separate pile --- each pile represents one line. One then counts the number of seeds in each cluster, starting with the first pile, which is the base line. If there is an even number of seeds, then the line is yin, otherwise the line is yang --- except if there is one seed, in which case one redoes that line.

One then asks the question again, and picks up one more cluster of seeds. Count the number of seeds you have, then keep subtracting six, until you have six seeds or less. This gives you the number of the line that specifically represent your situation. It is not a moving Line. If you do not understand your answer, you may rephrase the question, and ask it a second time.

Calligraphy brush strokes

Calendric systems

There is a component of Taoist thought which is concerned with numerological/cosmological systems. This has also been applied to the I Ching as well. The noted Chinese Neo-Confucian philosopher Shao Yung (1011-1077 CE) is the one who has done the most work in popularizing this concept and in developing/publishing oracular systems based on them. This is the most sophisticated usage of I Ching oracular systems.

The most readily accessible of these methods (the easiest to learn to do, and also to use) is called the Plum Blossom Oracle. In fact, however, there are several variants of this method. One method uses the number of brushstrokes used in writing the question along with the date and time of the inquiry. Another method simply uses the date and time without an actual question. There are other variants as well, including not using date and time at all. The resulting numbers are used to select the trigrams (in either the Early Heaven or the Later Heaven sequence), which then identify the hexagram of the answer. It is also possible to find Plum Blossom Oracle computer programs to more easily and efficiently do the calculations.

The most accurate of these calendric methods is also the most complex. This is called the Ho Map Lo Map Rational Number method (and has been published in Sherrill and Chu's "Astrology of I Ching"). It uses a very complicated series of operations with a series of tables to generate series of predictions which are entirely calendar-based.

The method set out in "Astrology of I Ching" has been reported to contain an error, leading to improper hexagrams sometimes being generated. However, the system can never produce the "missing" trigrams Li and Tui as a representation of the earthly force at a particular moment in time, since they are both assigned odd numeric values when the Later Heaven cycle of trigrams is superimposed on the so-called Magic Square of Three:

4....9....2

3....5....7

8....1....6

The earthly numbers are all even and thus the system is not flawed even though—being a composite method involving several layers - it is far from being seamless.

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