The Mod Stack This was a stack I devised to enable me to play with with memorized deck routines. It is not meant as a replacement for a memorized deck. However, it works very well well if you just occasionally want to try somethin out ! with a few minutes work, you can compute the card at any position, or the the position of any card. The advantae of it over a memorized memorized deck is that it is easy to learn and you don"t have to mentally maintain it, the disadvantae is that it is not as fast as a memorized deck. I really wanted a mathematical stack that would transform #$ numbers into #$ different numbers and do it very simply. simply. Then by doin some simple math math in my head I could compute which card was at which position. position. The key was to keep the math simple. simple. %oe M. Turner pointed me to &'ichard"s Stack(, devised by 'ichard )hrich, which I played with. The problem there is that that it re*uired oin from decimal to octal, and while that"s that"s not too difficult it took too lon to do the calculations. calculations. I wanted somethin very simple. simple. However, many simple computations came up with a stack that was in a clearly oranized stack ! if all the colors alternate reularly, it"s not very random lookin, for e+ample. In the stack I am presentin here, the touhest math math is multiplyin or dividin dividin by four. If you can do that, you"ll have no trouble with the math. The key idea of 'ichard"s Stack was reversin the numbers. I used this idea and came came up with a stack that I find very simple to compute, yet will put the cards in a pretty mi+ed up order. irst, I have to assin a value to each card. This turned out to be very important. important. The value aain had to be easy to compute, but the simplest ways turned out to leave the deck lookin ordered, so I didn"t want to use them. In the end, multiplication by four turned out to be the best solution, combined with the -HaSe order to ive each suit a value ! -/0 H/1 S/$ /2 Take each card value and multiply it by four. Then add the suit value listed above. So if it"s a club, add nothin. 3dd one for hearts, two for spades, and three for diamonds. diamonds. That"s the card value. So for an e+ample, let"s look look at the # of iamonds. The value is $2 ! # times times 4 is $0, plus 2 for the diamond. To reverse this step to o from a number value to a card, just divide the number by four. The remainder determines the suit ! no remainder is a club, 1 is a diamond, $ spades, and three diamonds, just as before. before. So take $2, divide by 4 and you you have # remainder 2. Three is diamonds, so it"s a five of diamonds.
The only math is multiplyin and dividin by four, and that"s as hard as it ets. There is an e+ception, thouh, and that"s the kins. I ave the kins values 1, $, 2, and #$ in -HaSe order ! 5in of -lubs is 1, etc. Havin that e+ception for the kins kept the ne+t step much simpler. So now each card has a value. The ne+t step is to morph that value to place the card in a random lookin position in the deck. To do this, just reverse the diits. So the five of diamonds has a value of $2. 'eversin the diits tells us that the card is in location 2$. or another simple e+ample, let"s look at the seven of spades. Seven times four is $6, plus $ for the spades, ives a value of 20. 'eversin the diits in 20 ives us 02 ! so the card is in location 02, the third card in the stack. Similarly, the kin of spades with a value of three 7or 02, just imaine a zero before the three8 is the 20 th card in the stack. 9ou will *uickly realize that this only works for half the deck. The other half will ive you a card position above #0 ! for e+ample, the : of Hearts has a value of : times 4 / 2; plus 1 for bein a heart / 2<=reversin the diits ives us <2. I wanted to subtract #0 which would be simple, but that takes us to a location already used. So for this stack, subtract 4#. This is M)-H easier than it sounds as you can use a trick. >icture the two diits in <2. ?ow lower the first one by # and raise the second one by #. or the numbers in this stack, that will always work. I picture it as movin # from the first number to the second. So <2 becomes $6. < <@# $
2 2A# 6
More e+amples=let"s look at 2:. 'everse the diits to et :2. That"s hiher than #0, so we take five from the : and move it to the 2, makin it 46. So the card with value 2: is the 46th card in the stack.72: divided by 4 is :, reminder 2, so the card is the nine of diamonds.8 Bet"s do one more e+ample like that. 2# reverses to #2. This is hiher than fifty, so move five from the first diit to the second. #@# is zero, and 2A# is 6, so this maps to the 6th location in the stack. So let"s take the seven of diamonds. That has a value of 21 7< times 4 / $6 plus three for the diamond is 21.8 1 is less than five, so that is in deck position 12. The only e+ceptions are stack numbers #0,#1,and #$. Since they are hiher than #0, this method won"t work on those numbers. #1 and #$ ive the same value as 1 and $, and #0 ends up bein #, also a repeat. So I just leave those alone. This means the CS, C, and 5 just map to themselves. My stack thus ends in several court cards in a row, but that"s not abnormal enouh to be noticed.
So far you can take a card, compute a value and transform that value into a stack location. It would be nice to take a stack location, compute the value, and transform the value back to a card. This can be done by reversin the steps. Start with an easy one ! location 12. 'everse the diits, and it"s card value 21. ?ow divide 4 into 21. 4 times < is $6, with a remainder of three. The remainder maps to the -haSe card values 7-,H,S, @D 0,1,$,28 so this is a < of iamonds. If you want to know what card is in location $6, just reverse the diits to et 6$. 6$ is hiher than #0, so move five from the 6 to the $ ! 6@# is 2, and $A# is <, so 6$ becomes 2<. ?ow divide 2< by 4. 4 oes evenly into 2; : times, so this is : remainder 1. This is a : of Hearts.
iurin Eut the ?e+t -ard in the Stack >ierre Fmmanuel fiured this out and it"s pretty cool. If you fiure out a card in the stack you can always fiure out the ne+t card by oin from the card number to the stack number, addin one to that, and oin back to a card value. Gut it"s actually *uite a bit easier. Say you"re lookin at the five of clubs. That card value is $0 7five times four, plus nothin for the clubs, and you have $0.8 'everse the numbers and it"s in stack position $. So what"s in stack position 2 If you know the card value, add ten to it ! and that is the value of the card in the ne+t position. So for our card value of $0, the ne+t card will have the value 20 ! that is 4 times < remainder $, so Seven of Spades. So once you can et from a card to a card value, just addin ten will always tell you the ne+t card in the stack 3lmost. If the value you end up with is over #0, this won"t work. Gut >ierre found a pretty cool trick for the rest. In brief, instead of addin ten, take the second diit and add five to it. If that is larer than ten, then add the two diits toether. That sounds complicated, but it really is *uite easy. or card values 40, 41, 4$, 42, and 44 take the second diit plus # ! #, ;, <, 6, :. or card values 4#, 4;, 4<, and 46, take the second diit and subtract 4 ! 1, $, 2, 4. In one sentence, if the second diit is below five, then add five. If it"s above five, subtract four. 3lso, note that the last several cards are set, so if you"re at value 4:, the ne+t three are #0, #1, and #$ so don"t follow the steps above
3 table with all the values follows. If you want to use this stack, use this table to enerate a practice deck. Jet a sharpie marker, and on the top of each card write the card value. En the bottom, write the stack location. I usually do this by takin a new deck and makin 12 piles of each suit so each card is in -HS in a pile by itself. Then stack those piles up as shown in the riht column so that they are in order 75-, 5H, 5S, 3-, 3H, etc.8 Then on the top of each card, write the card number 701, 0$, 02, etc. !always write both diits8 ?ow redo your practice deck so that it"s in the stack order as shown in the left side 7$S, #-,
Stack K
'eversed
Mod #0
3dd #
-ard Lalue
-ard K
-ard Lalue
01
10
10
10
$S
1
5-
0$
$0
$0
$0
#-
$
5H
02
20
20
20
2
5S
04
40
40
40
10-
4
3-
0#
#0
0
#
3H
#
3H
0;
;0
10
1#
2
;
3S
0<
<0
$0
$#
;H
<
3
06
60
20
2#
6
6
$-
0:
:0
40
4#
%H
:
$H
10
01
1
1
5-
10
$S
11
11
11
11
$
11
$
1$
$1
$1
$1
#H
1$
2-
12
21
21
21
<
12
2H
14
41
41
41
10H
14
2S
1#
#1
1
;
3S
1#
2
1;
;1
11
1;
4-
1;
4-
1<
<1
$1
$;
;S
1<
4H
16
61
21
2;
:-
16
4S
1:
:1
41
4;
%S
1:
4
$0
0$
$
$
5H
$0
#-
$1
1$
1$
1$
2-
$1
#H
$$
$$
$$
$$
#S
$$
#S
$2
2$
2$
2$
6-
$2
#
$4
4$
4$
4$
10S
$4
;-
$#
#$
$
<
3
$#
;H
$;
;$
1$
1<
4H
$;
;S
$<
<$
$$
$<
;
$<
;
$6
6$
2$
2<
:H
$6
<-
$:
:$
4$
4<
%
$:
20
02
2
2
5S
20
21
12
12
12
2H
21
<
2$
$2
$2
$2
#
2$
6-
22
22
22
22
6H
22
6H
24
42
42
42
10
24
6S
2#
#2
2
6
$-
2#
6
2;
;2
12
16
4S
2;
:-
2<
<2
$2
$6
<-
2<
:H
26
62
22
26
:S
26
:S
2:
:2
42
46
C-
2:
:
40
04
4
4
3-
40
10-
41
14
14
14
2S
41
10H
4$
$4
$4
$4
;-
4$
10S
42
24
24
24
6S
42
10
44
44
44
44
%-
44
%-
4#
#4
4
:
$H
4#
%H
4;
;4
14
1:
4
4;
%S
4<
<4
$4
$:
4<
%
46
64
24
2:
:
46
C-
4:
:4
44
4:
CH
4:
CH
#0
#0
CS
#0
CS
#1
#1
C
#1
C
#$
#$
5
#$
5
3ppendi+ 3fter sharin this stack with the Maic -af it was pointed out to me that it shared a lot in common with the Gart Hardin -ard System. It does indeed use a lot of the same tricks to et to a similar place. The difference seems to be that we started with a different card order, and Gart adds either nothin, 12, $; or 2: to it based on suit. I multiply the value by four and add the suit. Most of the rest of our stacks seem to just line riht up. That bein said, I had never heard of his system when I devised mine, so that was *uite a surprise
