General Report on Tunny


23d Page 82



BR 3407 0450/24/2 WD 23/2 COL 7

T 6020

1P2/L R 3106 A 1553 Φ 27.8 ST 1624
K1 K2
02 08 E 1631
02 20 E 1625
11 09 A 1633
16 07 A 1629
18 08 D 1629
19 31 C 1625
37 21 E 1637
40 28 B 1655    B 102  3.7Φ
40 28 C 1655

SET K1 K2  40 28

SPAN IN 1000'S

0271
0245

0268
0249

0301
0214

0280
0237

0268
0247

0261
0257

SPAN 4000-END IN 500'S

0139
0119

0129
0128

0131
0128

0130
0129

SPAN 01-4500

1259
1064    2B  195 B 98   4.1Φ

SPANNING 01-4500

SET K1 K2  40 28  GOOD
    The items in the first line are message number, time and date sent, wheel day and Colossus number.

    T 6020 is the text length, measured as a check, as soon as the tape is on Colossus.

    1P2/L is typewriterese for 1+2/ Χ2 x; this run is chosen because the chit (not preserved) was so marked [23E].

    R 3106 is the number of places looked at, i.e. the number of places where Χ2 = x.

    At random the expected number (A) of these when 1+2 = . is
½ × 3106 = 1553.

    Φ (typewriterese for σ) 27.8 is the standard deviation of 1+2 = . viz. ½ = ½ . This is of course an application of the formula quoted in 21 (b), that if random proportional frequency in a normal distribution is p, the standard deviation is . A table of ½ and ¼ is provided at each Colossus.

    ST 1624 is the set total, i.e. Colossus is set so as not to display or print any smaller score. Because this is a two-wheel run, ST is taken as A + 2½σ.

    The best score is 3.7σ, not even "good" [23C(a)] but worth spanning [23F(o)]. In each pair of span scores the upper is 1+2 = ., the lower 1+2 = x : this makes it easy to see where a slide occurs, evidently between 4000 and 5000, for 5000 - 6000 shows almost no bulge and 4000 - 5000 only a small bulge. 4000 - 5000 is therefore spanned in 500's and the bulge of 1+2 = . is seen to cease at about the 4500th letter: it is therefore believed that there is a message slide here and the subsequent runs are done spanning 1 - 4500: the sigma-age is now 4.1 instead of 3.7; the setting is therefore "good".

    Here the operator makes the mistake of neglecting a 4-letter count for ΔD1, ΔD2 [23E(h)] this is easily reconstructed and would read
        .   .    592
        .  x    498
        x x    667
        x  .    567

Because xx is so strong, this would have suggested the run 3+4x/1x2x which would in fact yield a score of 7.9σ.

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