(In reply to
re(2): please read by Ken Haley)
I ran the following program to try random selections of 4 cards, and in each case print on the first line the four numbers, in order, and the computed Z value. On the second line for each are the ak and Z values followed by the ak and Z values that the decoder, A, would deduce. In those instances where these differ from what B intended, the second line is also marked with an X.
Program:
match$ = "1234,1243,1324,1342,1423,1432,2134,2143,2314,2341,2413,2431,"
match$ = match$ + "3124,3142,3214,3241,3412,3421,4123,4132,4213,4231,4312,4321"
CLS
DO
REDIM taken(100)
FOR i = 1 TO 4
DO
n = INT(RND(1) * 100 + 1)
IF taken(n) = 0 THEN cd(i) = n: orig(i) = n: taken(n) = 1: EXIT DO
LOOP
NEXT
j = 0
FOR i = 1 TO 100
IF taken(i) THEN
j = j + 1
cd(j) = i
END IF
NEXT
seq$ = ""
FOR i = 1 TO 4
FOR j = 1 TO 4
IF orig(i) = cd(j) THEN seq$ = seq$ + LTRIM$(STR$(j))
NEXT
NEXT
Z = (INSTR(match$, seq$) + 4) / 5
FOR i = 1 TO 4
PRINT orig(i);
NEXT
PRINT Z
cd(5) = cd(1) + 100
FOR i = 1 TO 4
IF cd(i + 1) - cd(i) > 24 THEN ak = cd(i): EXIT FOR
NEXT
newNumb = ak + Z
s = 1
FOR r = 1 TO 5
IF newNumb < cd(s) THEN
recd(r) = newNumb
newNumb = 99999
ELSE
recd(r) = cd(s)
s = s + 1
END IF
NEXT
recd(6) = recd(1) + 100: recd(7) = recd(2) + 100
FOR i = 2 TO 6
diff = recd(i + 1) - recd(i - 1)
IF diff >= 25 THEN ak2 = recd(i - 1): Z2 = recd(i) - ak2: EXIT FOR
NEXT
PRINT ak; Z, ak2; Z2;
IF ak = ak2 AND Z = Z2 THEN PRINT : ELSE PRINT " X"
REM
LOOP
Results:
71 54 58 29 22
29 22 29 22
31 78 2 77 11
2 11 2 11
82 71 5 42 23
5 23 5 23
87 80 38 97 15
38 15 38 15
88 6 95 37 14
6 14 6 14
53 77 6 60 11
6 11 6 11
47 30 63 65 7
65 7 30 17 X
27 28 83 59 2
28 2 28 2
99 92 23 70 23
23 23 23 23
99 25 54 11 22
25 22 11 14 X
100 68 2 58 23
2 23 2 23
11 80 29 5 12
29 12 11 18 X
30 39 31 95 3
39 3 39 3
98 41 28 17 24
41 24 28 13 X
17 65 42 72 3
17 3 17 3
33 64 21 19 18
33 18 21 12 X
59 9 46 91 13
9 13 9 13
27 79 38 29 6
38 6 38 6
92 64 63 43 24
64 24 63 1 X
10 57 70 92 1
10 1 10 1
84 3 55 92 13
3 13 3 13
44 68 51 52 5
68 5 68 5
47 36 41 27 22
47 22 41 6 X
To take one of the X'ed examples, if the audience choices were
C D E F
47 30 63 65
Rearranged this is
30 47 63 65
D C E F for Z=7
30 47 63 65 130 shows ak=65
z+ak = 72
A then sees 30 47 63 65 72
and appends 130 and 147:
30 47 63 65 72 130 147
The difference 63-30 shows up immediately as >= 25, so A perceives ak = 30 and Z=17 for ak+Z=47, and so reconstructs
30 47 63 65 72
E B F C D
Is this mistaken decoding? No. Let's start out with
C D E F
65 72 30 63
30 63 65 72
E F C D Z=17
30 63 65 72 130 ak=30; Z+ak = 47
Present to A: 30 47 63 65 72; just as in the first problem where the audience chosen numbers were different. Both 47 30 63 65 and 65 72 30 63 resulted in the same coded value, 30 47 63 65 72. It was inevitable that different audience-chosen sets of 4 cards, in order, must share encoded sets of cards as there are only 75,287,520 of the latter and 94,109,400 of the former.
Out of the 23 computer-generated random sets of 4 cards, in 7 cases, the result of the decoding disagreed with the set being encoded.
Did Finland really use just 100 cards in the statement of the problem?
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Posted by Charlie
on 2007-05-13 03:09:48 |
re(3): please read
