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The Magic Cafe Forum Index » » Shuffled not Stirred » » Harding Method G2 better than Method G (0 Likes) Printer Friendly Version

glowball
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For this Method G2 I am using the SDHC pattern (with S=0 and D=1) because it at least keeps the meaningful visual hearts=2 and clubs=3 in the calculations, but most importantly the SDHC imaginary pattern spreads the suit colors in the physical deck.

This Method G2 gets rid of the ugly -1 calculation that was in Method G (thank you mod stack evanthx).

Note: Kings are to be treated as value zero not value thirteen.

In Method G2 there are two cards to memorize:
QC at position 6
KS at position 7

Method G2 Harding physical stack:
2H, 5S, 7H, 10S, QH, QC, KS, 8C, JD, KD, 2C, 5D, 7C, 10D, AH, 4S, 6H, 9S, JH, KH, 3S, 5H, 8S, 10H, AC, 4D, 6C, 9D, JC, KC, 3D, 5C, 8D, 10C, 2S, 4H, 7S, 9H, QS, AS, 3H, 6S, 8H, JS, 2D, 4C, 7D, 9C, QD, AD, 3C, 6D
End of method G2 physical Harding stack.

The 2 algorithms (card to number and number to card) are below:

Spectator says card name. The magician (or confederate) calculates the position (If Queen of Clubs then position is six, if King of Spades then position is seven otherwise do the normal calculations below):
a. Multiply 4 times the target card value.
b. Add the pip point/lobe count. Note Spades equals zero, diamonds equal=1, Hearts=2, Clubs=3.
c. Flip the two digits.
d. If greater than 52 do the -5+5 (same as subtracting 45).

Spectator says number, The magician (or confederate) calculates the card name (If number is 6 then card is Queen of clubs, if number is 7 then card is King of Spades otherwise do the normal calculations below):
1. Flip the two digit number.
2. If now greater than 50 (yes greater than 50) then -5+5 (subtr 45).
3. Divide by constant 4.
4. quotient is the card value, the remainder tells the suit (use SDHC note zero remainder means Spades).

Note that
remainder 0 means Spades,
remainder 1 means Diamonds,
remainder 2 means Hearts,
remainder 3 means Clubs.

Note if the spectator names a number less than 10 then treat it as though it had a leading zero such as 4 means 04. If they say 9 it means 09 and then you take it through the appropriate steps.
glowball
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Mnemonic memory trick for the two cards that must be memorized: The Battle ax Queen of Clubs (holding her three ball club) and her cousin the King of Spades (holding his shovel) are the 6th and 7th person in line somewhere.

Or maybe you just visualize American Gothic painting with the old woman holding a three knob club and the farmer man instead of a pitchfork is holding a shovel. And she's got the number six written above her head and he's got the number seven written above his head. If you picture this in your mind you'll never forget it.
glowball
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Note the reason for the "if greater than 50" (and not "if greater than 52" in the "number to card" calculation) is because of the craziness around the imaginary position 6 and 7 that need to be divided by 4 for the AH (position 15) and AC (position 25). Don't try to do anything extra or try to figure it out, just follow the algorithm.
ltrblst
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Is it possible in this stack to calculate directly the previous and next card?

In the original Harding stack you can.
glowball
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Yes. I think it would be pretty much the same way as the original Harding stack ie: you know the position of a card and you want to know what the next card is then you mentally add one and then mentally do the number to card algorithm. For the previous card you would subtract one from the position and then do the number to card algorithm.

If using the Harding the stack check out using my Midas card:

https://www.themagiccafe.com/forums/view......orum=205

My 79-year-old brain does not want to work that hard so I will just use the Aronson stack that I have memorized 15 some years ago!

Good luck!
ltrblst
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Quote:
On Aug 9, 2022, glowball wrote:
Yes. I think it would be pretty much the same way as the original Harding stack ie: you know the position of a card and you want to know what the next card is then you mentally add one and then mentally do the number to card algorithm. For the previous card you would subtract one from the position and then do the number to card algorithm.


Thank you for the answer, but actually it's much easier: using the Harding stack to calculate the next card you don't need to know the position, that would not be fast or practical.

For example using rhe regular Harding stack for most cards you just need to subtract 3 and advance the suit in CHSD order, e.g.

8H (12) 5S (13) 2D (14)...

There are a couple of exceptions but it's an unusual feature for an algorithmic stack.

That's why I love the Harding stack.

(that and transformation to and from NDO)
hcs
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Quote:
On Aug 10, 2022, ltrblst wrote:
Thank you for the answer, but actually it's much easier: using the Harding stack to calculate the next card you don't need to know the position, that would not be fast or practical.
For example using rhe regular Harding stack for most cards you just need to subtract 3 and advance the suit in CHSD order, e.g.
8H (12) 5S (13) 2D (14)...
There are a couple of exceptions but it's an unusual feature for an algorithmic stack.
That's why I love the Harding stack.
(that and transformation to and from NDO)
Only about 11 exceptions.
glowball
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I did not realize this, so thanks guys. The standard Harding stack is looking better and better, although I think G2 makes a better looking spread.
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