Social Engineering: Theory and Practice

Social Engineering: Theory and Practice
PO8 <bart@cs.uoregon.edu>
March 23, 1997

The Netrunner runner Prep card Social Engineering is an oft-used
card: it provides both an early-game speedup and a way of
dealing with particularly nasty ice in the late game.  It,
along with its early-game counterpart Inside Job, is an
important runner-deck tool.  However, Social Engineering
suffers somewhat from a bad reputation.  It is all too easy for
the runner to lose more bits than they can afford at a critical
time during the game.  And yet, if the Runner risks too few
bits, the chances of success are not very high.

The inventors of the branch of Mathematics known as Game Theory
wanted to answering these very sorts of economic questions.
Game Theory can tell us exactly how to play Social Engineering
optimally in a given situation.  In this article, I will
describe these optimal strategies for Social Engineering, and
give some suggestions about how to convert this theory into
practice.

Here's the rules for Social Engineering, as taken from
BitByte's (<adriant@greenie.com>) spoiler list
(http://home.pacific.net.sg/~bitbyte).

	Social Engineering: Prep (1) [1.0 U]
	Hide at least (2) from your pool in your hand; the Corp
	then guesses how many bits you hid. If the Corp guesses
	correctly, lose that many bits. Otherwise, choose a
	data fort and a piece of ICE on that fort.  Then make a
	run on that fort, during which you automatically pass
	that piece of ICE.

Assume a pool of n bits.  Then, let us choose to risk i bits
[2 <= i <= n] with probability proportional to 1/i.  We must
normalize, so we are chosing to risk i bits with probability
	p(i) = (1/i) / (1/2 + 1/3 + ... + 1/n)
Thus, when n = 5, we have probabilities
	i	p(i)
	2	0.39
	3	0.26
	4	0.19
	5	0.15
Now imagine that the corp chooses any pure strategy (i.e., it
always guesses some particular number of bits).  Its expected
return on this strategy (i.e., the number of bits it expects to
cost the runner due to guessing correctly) is identical.
	i	e(i)
	2	0.78
	3	0.78
	4	0.78
	5 	0.78
In other words, no matter what pure strategy the corp picks, it
will cost the runner 1.78 bits, on average, to play social
engineering.  It can be proven from this that any combination of
pure corp strategies will achieve the same result, and thus
this strategy is optimal for the runner (although the proof
is outside the scope of this paper).

Unfortunately, the above "optimal" strategy is based on a
fairly unrealistic assumption.  Presumably the runner
is playing this card because they will receive some utility
from a successful run, not just to avoid losing bits.
Unfortunately, there is no clear "value of being able to pass a
selected piece of ice" in the game.  Let us arbitrarily set
this value at 3 bits, just to see how it affects our
calculations.  (Why 3 bits?  Inside Job, which costs 2 bits,
will let us pass the 1st piece of ice for free.  Let us assume
that the value of passing an arbitrary piece of ice is half
again as much.  We will generalize later in any case.)

We will now use a different strategy, namely; we will risk i
bits with probability 1/i+3.  Normalizing gives us
	p(i) = (1/(i+3)) / (1/5 + 1/6 + ... + 1/(n+3))
When n = 5, we choose with probability
	i	p(i)
	2	0.32
	3	0.26
	4	0.23
	5	0.20
and the corp's expected return for their pure strategy
(including the three bit benefit of preventing the run
as well as the cost to the runner of being guessed and of
playing the card) is
	i	e(i)
	2	2.6
	3	2.6
	4	2.6
	5	2.6
Thus, we have an optimal strategy for the runner under the new
conditions, under which it will cost the runner about 2.6 bits,
including the opportunity cost of missed runs, to play Inside
Job.  (There is some additional opportunity cost of playing a
Prep during the run, but this is difficult to estimate and
probably reasonably small in any case.)  Note how much the
probabilities have flattened out.  Indeed, as the value to the
corp of an unsuccessful run gets large, it swamps the expected
value of making the runner lose the guess bits, and the
runner's optimal strategy quickly converges on simply picking a
random value uniformly from the range 2..n .

A natural question to ask about strategy involves asymptotic
behavior:  what is the optimal strategy as the size of the
runner's bit pool n gets large?
	limit n->inf sum(i=2..n, 1/i)
diverges, so the more bits the runner has, the better off they
are.  Notice that the expected return for the corp as a function
of the runner's bit pool size n is
	e(n) = 1 + 1/(1/5 + 1/6 + ... + 1/(n+3))
Obviously, the expected corp return converges on 1.  But don't
be fooled: the divergence of this "harmonic series" is notoriously
slow.  When n = 100 (surely infinity from most corps' point
of view) we still have e(n)=1.32 .

Note that we picked a utility of 3 for a successful run fairly
arbitrarily.  Depending on the game situation, this utility
estimate could be wildly off.  E.g., you know there's a
Political Overthrow in a fort, and it's protected by
Filter-Filter-Colonel Failure.  Your icebreakers are Codecracker
and Raptor.  How much would you be willing to pay for a
Social?  On the other hand, a Filter protects a Holovid
Campaign, but you don't have any icebreakers yet.  How much
would you be willing to pay for a Social?

Fortunately, the calculations we have been performing
generalize quite easily to the case of a u-bit utility:
	p(i,n,u) = (1/(i+u)) / (1/(2+u) + 1/(3+u) + ... + 1/(n+u))
	e(n,u) = 1 + 1/(1/(2+u) + 1/(3+u) + ... + 1/(n+u))
One interesting way to pick the utility is to ask the card itself
how much it values a run at!  We do this by noting that the
card is presumably zero-sum, and thus when e(n,u) = u presumably
the corp expects to cost the runner exactly the amount that the
runner expects to benefit from the run.  Thus we need to solve
	1 + 1/(1/(2+u) + 1/(3+u) + ... + 1/(n+u)) = u
i.e.
	(1/(2+u) + 1/(3+u) + ... + 1/(n+u)) = 1/(u-1)
for u.  I don't see how to do this analytically, but numerically,
we find the following:
	n	u
	2	*
	3	4.5
	4	3.0
	5	2.4
	6	2.1
	7	2.0
	8	1.9
	9	1.8
	10	1.8
	...
	100	1.3
The * at n = 2 is because the equation for u is singular
at this value, and thus no solution exists.  Intuitively, this
is because the corp will never guess wrong in this case, and so
the runner is stupid to play the card!  The fact that the u
value is dependent on the size n of the runner bit pool is
rather strange, but points up the fact that as the runner has
more bits, his chances of using this card successfully just get
better and better, and so a run made using the card is worth
less and less.

Note that once the runner has 6 or more bits, the expected
return when playing optimally (at u=3) doesn't change much as
the bit pool increases.  However, what does change is the risk
of a disastrous outcome for the runner.  It is true that the
expected return for the corp if it always chooses n is exactly
the same as if it always chooses 2.  However, in the first
case, if the corp guesses right, the runner loses all their
bits.  This will happen very rarely, but the runner may be
risk-averse; willing to lose more on average to avoid the risk
of major losses.  Fortunately, limiting risk in this framework
is easily accomplished: the runner can treat the pool as though
it has only as many bits as the runner is willing to risk, and
calculate accordingly.  Again, the cost of this risk management
becomes very high at 4 bits or below, and I don't advise it ---
if losing 4 bits would be a disaster for the runner, the runner
shouldn't be playing a Social Engineering at that point.

A final question is how to use our formulae in a game situation.
If one is playing online, it's relatively straightforward to
have a "Social Engineering bit picker" sitting in a window on
one's terminal, which will select bit value b(n,u) from the
appropriate probability distribution p(i,n,u).  If one is
playing head-to-head, it is possible to program a programmable
calculator, PDA, etc., to perform this same calculation.
Unfortunately, this may or may not be allowed by a tournament
director, and it may be cumbersome and unpleasant in any case.

Lacking a programmable calculator or PDA, the runner could use
a set of tables for d100, parameterized by n and u, to select
bit values.  These tables are included as Appendix A to this
article.  In fact, since it is highly desirable to screen the
die roll used for this selection from the runner, a very
reasonable approach is to glue the tables to the inside of a
cardboard screen, behind which the dice are rolled and the bits
selected.

What is really desired is an approximation or set of
approximations using just ordinary die rolls and no difficult
arithmetic, such that a selection reasonable approximating
optimality can be quickly and unobtrusively carried out in a
game situation.  Unfortunately, I see no obvious rules of thumb
for small n and/or u.  For u >= 4 and n >= 5, however, it
appears that the error induced by merely rolling a die and
guessing as many bits as it says (rerolling if it is out of
range) is fairly small.  We can quantify the cost of this
non-optimal strategy.  The expected return for the corp is
	c(i) = (1 + i + u) * (1/(n-1) - p(i))
where p(i) is the true probability that should be assigned to
the i-th bit.  It is clear that the corp's best bet is always
to take i=n, and that the largest error obtains when u=4.  It
turns out that the error gets larger as n increases (which
makes intuitive sense, since the runner is supposed to be
getting a better and better deal, and isn't) but stays
very close to 1/2 bit for a wide range of n and u values
above n >= 5 and u >= 4.  This is another interesting consequence
of the slow divergence of the harmonic series --- we know that
the error becomes infinite for large enough n, but when n=100,
we still have error less than 0.72!

One unfortunate effect of this strategy, however, is that the
optimal corp strategy is to guess the entire value of the bit
pool.  This might make a risk-averse runner very nervous,
especially since this corp strategy gets slightly better as the
size of the runner bit pool increases.  An obvious way to
counter this problem is to count die rolls of 1 as 2s: this
overweights the value 2 at the expense of the other values.  It
turns out that the optimal corp strategy (again for n >= 5 and
u >=4) is now to always guess 2.  Fortunately, the expected
corp gain over optimal strategy is still about 1/2 bit, and
falls rapidly off as n increases (for fixed u).

All of the above is pretty abstract.  Here's what I suggest
as a runner strategy for Social Engineering, given that neither
software nor tables are available:
	(1) Figure out how much the run is worth.  Call this
	    number u.  If u is less than 3 bits, don't play the
	    Social.
	(2) Figure out how many bits the pool will have after
	    playing the social.  Call this number n.
	(3) If n is more bits that the runner can safely lose,
	    reduce n to the number of bits the runner can safely lose.
	(4) If n is 4 or less and u < 5, don't play the Social.
	(5) If n is greater than 6:
		(5a) Obtain or synthesize an m-sided die,
		     with n > m.
		(5b) Reroll until the result r of the roll
		     satisfies 1 <= r <= n.
		(5c) If r is 1, make r be two.
		(5d) Risk r bits.
	(6) Otherwise:
		(6b) Obtain a 6-sided die.
		(6c) Reroll until the result r of the roll
		     satisfies 2 <= r <= n.
		(6d) Risk r bits.
When written out like this, the algorithm looks pretty complicated.
In point of fact, it's pretty straightforward once you get the hang
of it.

One last obvious question is: what of the poor Corp?  Well,
the Runner has been careful to make any strategy yield
pretty similar results.  The Corp must simply pick randomly
among the available strategies, in order to prevent the Runner
guessing what's going to happen.  Here's the algorithm I suggest
for the Corp.
	(1) Figure out how many bits the pool will have after
	    playing the social.  Call this number n.
	(2) Figure out how many bits the runner can safely lose.
	    If this is less than n, set n to this value + 1.
	(3) If the runner is playing with a table or device,
	    use a die to select randomly from the range 2..n.
	(4) If n > 6, roll an m-sided die, and count any 1 result
	    as a n, until the result r is in range.  Guess r.
	(5) If n <= 6, roll a 6-sided die, and count any 1 result
	    as a 2, until the result r is in range.  Guess r.
This isn't perfect, but is good enough in my opinion.  There's
a little game the runner can play with n vs. n - 1 and a large
bit pool, but it hardly matters in this case anyhow.

So, how good a card is Social Engineering, anyhow?  Looks like
it's a little better deal than Inside Job in most cases: the
extra 0.5-1.5 bit expected penalty for choosing an arbitrary
piece of ice is pretty tasty in a lot of situations.  (On the
other hand, since Inside Job passes the 1st ice it encounters,
it can be useful in situations where the Corp can't afford to
rez more than 1 of the several pieces of ice on a fort.  Ce'st
La Vie.)  Hopefully, this exposition will help Runners and
Corps alike deal with Social Engineering, and encourage its use.

Appendix A: Guess Tables For Social Engineering

To use these tables:  Find the table for the u value you are
interested in, and go to the column for the n of the bit pool
you have and are willing to risk.  The e value, at the bottom,
tells how much, including opportunity cost, this Social run
will cost you.  If this is acceptable, then play the Social,
and roll d100.  Pick the row whose entry in your chosen column
is the largest value less than or equal to the die roll.  Look
to the left to see how many bits to risk.

u=0
b=  n=03 n=04 n=05 n=06 n=07 n=08 n=09 n=10 n=11 n=12 n=13 n=14 n=15
02----59---45---37---33---30---28---26---24---23---22---21---21---20
03----99---75---63---56---51---47---44---42---40---38---37---36---34
04---------99---83---73---67---62---58---55---52---50---48---47---45
05--------------99---87---79---73---69---65---62---60---57---55---54
06-------------------99---90---83---78---74---70---67---65---63---61
07------------------------99---91---86---81---77---74---72---69---67
08-----------------------------99---92---88---84---80---77---75---73
09----------------------------------99---93---89---85---82---80---77
10---------------------------------------99---94---90---87---84---82
11--------------------------------------------99---95---91---88---86
12-------------------------------------------------99---95---92---89
13------------------------------------------------------99---95---93
14-----------------------------------------------------------99---96
15----------------------------------------------------------------99
e=  2.20 1.92 1.78 1.69 1.63 1.58 1.55 1.52 1.50 1.48 1.46 1.44 1.43

u=1
b=  n=03 n=04 n=05 n=06 n=07 n=08 n=09 n=10 n=11 n=12 n=13 n=14 n=15
02----56---41---34---29---26---24---22---20---19---18---18---17---16
03----99---73---60---52---46---42---39---37---35---33---32---31---30
04---------99---81---70---63---57---53---50---47---45---43---42---40
05--------------99---85---77---70---65---61---58---55---53---51---49
06-------------------99---88---81---75---70---67---64---61---59---57
07------------------------99---90---84---79---74---71---68---65---63
08-----------------------------99---92---86---81---78---74---72---69
09----------------------------------99---93---88---84---80---77---74
10---------------------------------------99---93---89---85---82---79
11--------------------------------------------99---94---90---87---84
12-------------------------------------------------99---94---91---88
13------------------------------------------------------99---95---92
14-----------------------------------------------------------99---95
15----------------------------------------------------------------99
e=  2.71 2.28 2.05 1.92 1.82 1.75 1.70 1.66 1.62 1.60 1.57 1.55 1.53

u=2
b=  n=03 n=04 n=05 n=06 n=07 n=08 n=09 n=10 n=11 n=12 n=13 n=14 n=15
02----54---39---31---27---24---21---20---18---17---16---15---15---14
03----99---71---58---49---44---40---36---34---32---30---29---28---27
04---------99---80---68---60---55---50---47---44---42---40---38---37
05--------------99---84---75---68---63---58---55---52---50---48---46
06-------------------99---87---79---73---68---64---61---58---56---54
07------------------------99---89---82---77---72---69---66---63---60
08-----------------------------99---91---85---80---76---72---69---67
09----------------------------------99---92---87---82---78---75---72
10---------------------------------------99---93---88---84---81---78
11--------------------------------------------99---93---89---86---82
12-------------------------------------------------99---94---90---87
13------------------------------------------------------99---94---91
14-----------------------------------------------------------99---95
15----------------------------------------------------------------99
e=  3.22 2.62 2.32 2.13 2.00 1.91 1.84 1.79 1.74 1.71 1.67 1.65 1.62

u=3
b=  n=03 n=04 n=05 n=06 n=07 n=08 n=09 n=10 n=11 n=12 n=13 n=14 n=15
02----53---38---30---25---22---20---18---17---16---15---14---13---13
03----99---70---56---48---42---38---34---32---30---28---27---26---24
04---------99---79---67---59---53---48---45---42---40---38---36---35
05--------------99---84---74---66---61---56---53---50---47---45---43
06-------------------99---87---78---72---66---62---59---56---53---51
07------------------------99---89---81---76---71---67---64---61---58
08-----------------------------99---90---84---79---74---71---68---65
09----------------------------------99---91---86---81---77---74---71
10---------------------------------------99---92---87---83---79---76
11--------------------------------------------99---93---89---85---81
12-------------------------------------------------99---94---90---86
13------------------------------------------------------99---94---90
14-----------------------------------------------------------99---95
15----------------------------------------------------------------99
e=  3.73 2.96 2.58 2.34 2.18 2.07 1.98 1.91 1.86 1.81 1.77 1.74 1.71

u=4
b=  n=03 n=04 n=05 n=06 n=07 n=08 n=09 n=10 n=11 n=12 n=13 n=14 n=15
02----52---37---29---24---21---19---17---16---15---14---13---12---12
03----99---70---55---46---41---36---33---30---28---27---25---24---23
04---------99---78---66---57---51---47---43---40---38---36---34---33
05--------------99---83---73---65---59---55---51---48---46---44---42
06-------------------99---86---77---70---65---61---57---54---52---50
07------------------------99---88---81---75---70---66---62---59---57
08-----------------------------99---90---83---78---73---69---66---63
09----------------------------------99---91---85---80---76---73---69
10---------------------------------------99---92---87---82---78---75
11--------------------------------------------99---93---88---84---80
12-------------------------------------------------99---93---89---85
13------------------------------------------------------99---94---90
14-----------------------------------------------------------99---94
15----------------------------------------------------------------99
e=  4.23 3.30 2.83 2.55 2.36 2.22 2.12 2.03 1.97 1.91 1.86 1.83 1.79

u=5
b=  n=03 n=04 n=05 n=06 n=07 n=08 n=09 n=10 n=11 n=12 n=13 n=14 n=15
02----52---36---28---24---20---18---16---15---14---13---12---12---11
03----99---69---54---46---40---35---32---29---27---26---24---23---22
04---------99---78---65---57---50---46---42---39---37---35---33---32
05--------------99---83---72---64---58---54---50---47---44---42---40
06-------------------99---86---77---70---64---60---56---53---50---48
07------------------------99---88---80---74---69---65---61---58---55
08-----------------------------99---90---83---77---72---68---65---62
09----------------------------------99---91---85---80---75---72---68
10---------------------------------------99---92---86---82---78---74
11--------------------------------------------99---93---88---83---80
12-------------------------------------------------99---93---89---85
13------------------------------------------------------99---94---90
14-----------------------------------------------------------99---94
15----------------------------------------------------------------99
e=  4.73 3.64 3.09 2.75 2.53 2.37 2.25 2.15 2.07 2.01 1.96 1.91 1.87

u=6
b=  n=03 n=04 n=05 n=06 n=07 n=08 n=09 n=10 n=11 n=12 n=13 n=14 n=15
02----51---36---28---23---20---17---16---14---13---12---12---11---10
03----99---69---54---45---39---34---31---28---26---25---23---22---21
04---------99---77---64---56---50---45---41---38---36---34---32---30
05--------------99---82---71---63---57---53---49---46---43---41---39
06-------------------99---85---76---69---63---59---55---52---49---47
07------------------------99---88---79---73---68---64---60---57---54
08-----------------------------99---89---82---76---72---67---64---61
09----------------------------------99---91---84---79---74---71---67
10---------------------------------------99---92---86---81---77---73
11--------------------------------------------99---92---87---83---79
12-------------------------------------------------99---93---88---84
13------------------------------------------------------99---94---89
14-----------------------------------------------------------99---94
15----------------------------------------------------------------99
e=  5.24 3.98 3.34 2.96 2.70 2.52 2.38 2.27 2.18 2.11 2.05 1.00 1.95

u=7
b=  n=03 n=04 n=05 n=06 n=07 n=08 n=09 n=10 n=11 n=12 n=13 n=14 n=15
02----51---35---27---23---19---17---15---14---13---12---11---10---10
03----99---68---53---44---38---34---30---28---26---24---22---21---20
04---------99---77---64---55---49---44---40---37---35---33---31---30
05--------------99---82---71---63---57---52---48---45---42---40---38
06-------------------99---85---75---68---63---58---54---51---48---46
07------------------------99---87---79---72---67---63---59---56---53
08-----------------------------99---89---82---76---71---67---63---60
09----------------------------------99---90---84---78---74---70---67
10---------------------------------------99---91---85---81---76---73
11--------------------------------------------99---92---87---82---78
12-------------------------------------------------99---93---88---84
13------------------------------------------------------99---93---89
14-----------------------------------------------------------99---94
15----------------------------------------------------------------99
e=  5.74 4.31 3.60 3.16 2.87 2.67 2.51 2.39 2.29 2.20 2.14 2.08 2.03

u=8
b=  n=03 n=04 n=05 n=06 n=07 n=08 n=09 n=10 n=11 n=12 n=13 n=14 n=15
02----51---35---27---22---19---17---15---14---12---12---11---10---10
03----99---68---53---44---38---33---30---27---25---23---22---21---20
04---------99---77---63---55---48---43---40---37---34---32---30---29
05--------------99---82---70---62---56---51---47---44---42---39---37
06-------------------99---85---75---68---62---57---53---50---48---45
07------------------------99---87---79---72---67---62---58---55---53
08-----------------------------99---89---81---75---70---66---63---59
09----------------------------------99---90---83---78---73---69---66
10---------------------------------------99---91---85---80---76---72
11--------------------------------------------99---92---87---82---78
12-------------------------------------------------99---93---88---83
13------------------------------------------------------99---93---89
14-----------------------------------------------------------99---94
15----------------------------------------------------------------99
e=  6.24 4.65 3.85 3.37 3.04 2.81 2.64 2.50 2.39 2.30 2.22 2.16 2.10

u=9
b=  n=03 n=04 n=05 n=06 n=07 n=08 n=09 n=10 n=11 n=12 n=13 n=14 n=15
02----51---35---27---22---19---16---15---13---12---11---10---10---09
03----99---68---53---43---37---33---29---27---25---23---21---20---19
04---------99---76---63---54---48---43---39---36---34---31---30---28
05--------------99---81---70---62---55---51---47---44---41---39---37
06-------------------99---85---75---67---61---57---53---50---47---44
07------------------------99---87---78---72---66---62---58---55---52
08-----------------------------99---89---81---75---70---66---62---59
09----------------------------------99---90---83---78---73---69---65
10---------------------------------------99---91---85---80---75---72
11--------------------------------------------99---92---86---82---77
12-------------------------------------------------99---93---87---83
13------------------------------------------------------99---93---88
14-----------------------------------------------------------99---94
15----------------------------------------------------------------99
e=  6.74 4.98 4.09 3.57 3.21 2.96 2.77 2.62 2.50 2.40 2.31 2.24 2.18

u=10
b=  n=03 n=04 n=05 n=06 n=07 n=08 n=09 n=10 n=11 n=12 n=13 n=14 n=15
02----51---34---26---22---18---16---14---13---12---11---10---10---09
03----99---68---52---43---37---32---29---26---24---22---21---20---19
04---------99---76---63---54---47---42---39---36---33---31---29---28
05--------------99---81---70---61---55---50---46---43---40---38---36
06-------------------99---84---74---67---61---56---52---49---46---44
07------------------------99---87---78---71---66---61---57---54---51
08-----------------------------99---89---81---74---69---65---61---58
09----------------------------------99---90---83---77---72---68---65
10---------------------------------------99---91---85---79---75---71
11--------------------------------------------99---92---86---81---77
12-------------------------------------------------99---92---87---83
13------------------------------------------------------99---93---88
14-----------------------------------------------------------99---93
15----------------------------------------------------------------99
e=  7.24 5.32 4.35 3.77 3.38 3.10 2.89 2.73 2.60 2.49 2.40 2.32 2.26

Appendix B: Algorithms Used In This Article

Here are the algorithms I used in computing the numbers used in
the article, including the tables of Appendix A.  They are
written for Keith Packard's "nick" interactive calculator, but
should be fairly understandable without it.

	function norm(n,u) {
		auto i;
		auto s = 0;
		for (i = 2; i <= n; i++)
			s += 1/(i+u);
		return s;
	}

	function p(i,n,u) {
		return (1/(i+u)) / norm(n,u);
	}

	function ex(n,u) {
		return 1 + 1/norm(n,u);
	}

	function dd(i,n,u) {
		auto j;
		auto s = 0;
		for (j = 2; j <= i; j++)
			s += p(j,n,u);
		return s;
	}

	function d(i,n,u) {
		return floor(dd(i,n,u) * 100) - 1;
	}

	function er(i,n,u) {
		return (1 + i + u) * (1/(n-1) - p(i,n,u));
	}

	function er2(i,n,u) {
		auto k = 1 + i + u;
		if (i == 2)
			return k * (2/n - p(i,n,u));
		else
			return k * (1/n - p(i,n,u));
	}

	function normtest(n,u) {
		auto i;
		auto s = 0;
		for (i = 2; i <= n; i++)
			s += p(i,n,u);
		return s;
	}

	function printtab(u,nn) {
		auto i, n;
		printf("u=%d\n", u);
		printf("b= ");
		for (n = 3; n <= nn; n++)
			printf(" n=%02d", n);
		printf("\n");
		for (i = 2; i <= nn; i++) {
			printf("%02d-", i);
			for (n = 3; n <= nn; n++)
				if (i <= n)
					printf("---%02d", d(i,n,u));
				else
					printf("-----");
			printf("\n");
		}
		printf("e=  %4.2f", ex(3,u));
		for (n = 4; n <= nn; n++)
			printf(" %4.2f", ex(n,u));
		printf("\n");
	}

	function printtabs(nu,nn) {
		auto i;
		printtab(0,nn);
		for (i = 1; i <= nu; i++) {
			printf("\n");
			printtab(i,nn);
		}
	}

	/* printtabs(10,15) */