Continuation to the Introduction, and some worked examples

Now we will see how to complete the details of the identification of the 'hidden' card.

Recall from the Introduction our illustrative example 17, 38, 62, 116 and 120, where A knows ' i ' to be 3 and ' s ' to be 2, and where P - on receipt of the 4 cards - knows ' s ' to be '2'. Recall that renumbered value is 113 (= (- s ) (mod 5)).

• 113 = 5. 22 + (5 - 2)
• so A needs to signal ' 22 ' to P
• in base factorial 22 is 3.3!+2.2!+0.1!
• so A needs signal 22 with the triple (3, 2, 0)

Agreement on how that may be done . Consider this quadruple of the 4 passed c 's: (38, 62, 17, 120), and read the entries from left to right:

• the 1st entry is '38', and of the following three, 1 of them is less than 38
• the 2nd entry is '62', and of the following two, 1 of them is less than 62
• the 3rd entry is '17', and of the following one, none of them ( 0 ) is less than 17

Now, simply make the understanding that passing the four cards 17 (= ), 38 (= ), 62 (= ) and 120 (= ) in the left-to-right order (38, 62, 17, 120) would signal the number 1.3! + 1.2! + 0.1! (=8).

But, wishing to signal the number 22 (= 3 .3!+ 2 .2!+ 0 .1!) from our example above, we should signal by passing in the left-to-right order (120, 62, 17, 38) since:

• the 1st entry is '120', and of the following three, 3 of them are less than 120
• the 2nd entry is '62', and of the following two, 2 of them are less than 62
• the 3rd entry is '17', and of the following one, none of them ( 0 ) is less than 17

The point is simply that for any quadruple (A, B, C, D) of distinct numbers:

• the number of numbers from [B, C, D] that are less than A is 0, 1, 2 or 3 (gives ' ')
• the number of numbers from [C, D] that are less than B is 0, 1 or 2 (gives ' ')
• the number of numbers from [D] that are less than C is 0 or 1 (gives ' ')

P then knows that the hidden has shifted value 113 (= 5.(the decoded 22 )) + (5 - s )) , and its actual value satisfies the vital congruence:

(mod 5)

P then knows that cannot be 17, 38 nor 62 (they are less than the shifted value, and so must be and ), and so must be , allowing P know that i is 3. Adding 3 to 113 gives the hidden card 116.

A telegraphic worked example with n = 4