Sweet & Short Math Challenges #15: April 1st Spring Special



#26

Hi all,

    Where I am right now, April 1st has just begun, the very start of a brand-new month, so time for a new S&SMC, this time #15, an "April 1st Spring Special".

    What's so "special" about it ? Well, for one, you'll have
    5 individual, affordable mini-challenges to test your smarts and
    programming abilities against instead of a single,
    harder challenge.

    And secondly, this time you won't only be fighting to find
    your way through the challenges but you'll also experience
    the agony of a deadline, i.e., delivering the goods in a fixed,
    non-expandable amount of time. Namely this means you won't have
    a whole week before I post the solutions but just this very weekend.

    Next Monday I'll post my original solutions and comments,
    so if you want to try your hand at it before I spill them out,
    you'd better hurry up this one time, no excuses. :-)

    In all five cases, you must write a program for your chosen
    HP handheld (preferably, but other brands also acceptable as long
    as they're handheld but no PDAs!) to help you answer the challenge.
    Any language (RPN/RPL/BASIC/FORTH/C...) is acceptable as long as it
    actually runs in your handheld.

    Now for the Spring Special:

Take 1


    Given the coefficients a, b, c, d of the general
    4th-degree polynomial equation:
          x4 + a*x3 + b*x2 + c*x + d = 0
    you must write a program to compute the sum of the cubes
    of all its four roots (real and/or complex).

    If in RPN/RPL, your program must assume the coefficients have
    previously been entered into the stack as follows:

         T: d 
    Z: c
    Y: b
    X: a
    and upon running, it must stop with the sum of the cubes of the
    four roots in the display. You should try to make your program
    as short and fast as possible.

    For instance, given the equation:

         x4 - 2.006*x3 - 4*x2 - x + 4.44 = 0
    your program must output 35.144216216, which is the sum of
    the cubes of its four roots, namely:
         x1 = 0.848975027766
    x2 = 3.21371823523
    x3 = -1.0283466315 +0.754884455486*i
    x4 = -1.0283466315 -0.754884455486*i
    As this is S&SMC #15, I'll post my 15-step RPN solution for the HP-15C
    (which will also run in the HP-34C and other RPN models as well),
    plus a couple of solutions for the HP-71B, for good measure.

Take 2:


    Write a program to find all positive integers up to 10,000 such that
    they're simultaneously equal to the sum of the 1st powers of two
    consecutive integers and also to the positive difference of the
    2nd powers of those same integers.

    For instance, 153 is a solution because:

         153 = 761 + 771 
    = 772 - 762
    where 76 and 77 are consecutive integers.

    Your program must find out and output all solutions, and be as short and fast as possible.

Take 3:


    A cable is intended to tightly surround the Earth (assumed to be a perfect
    sphere) by the equator so that it touches the ground at all its points.
    Unfortunately, due to a slight manufacturing error, the cable, which should
    be some 40 million meters long, is actually a trifle 1 meter longer, so that
    regrettably, instead of fitting tightly as intended, there's some slack.

    To fix the problem, a man sits comfortably on a small cushion at ground level,
    picks the cable between his fingers, and raises his arm vertically upwards
    from the ground till the cable is perfectly tight again, with no slack
    whatsoever.

    Assuming that the Earth is a perfect sphere of radius exactly equal to
    6,400 Km, and that the cable doesn't stretch at all and has negligible radius,
    what's the name of the man's brother-in law ?

    Note:

      In order to help you find the answer, you are required to write
      a program to compute the required height to remove the slack for
      any given extra cable length (measurements in meters), and then run
      your program using the data for this particular case (1 extra meter).

Take 4:


    The subject of self-reproducing programs, i.e., programs that upon
    running do produce their own source code as their output, is highly
    fascinating and has tons of actual instances in every programming
    language known to man.

    For instance, in standard C you have:

          main(){char *c="main(){char *c=%c%s%c;printf(c,34,c,34);}";printf(c,34,c,34);}
    in RPL you have:
          \<<
    \<< "\<<" SWAP +
    "DUP EVAL" + OBJ\->
    \>>
    DUP EVAL
    \>>
    in Java you have:
          class S{public static void main(String[]a){String s="class
    S{public static void main(String[]a){String s=;char ='';
    System.out.println(s.substring(0,52)+c+s+c+s.substring
    (52,61)+c+s.substring(61));}}";char c='"';System.out.println
    (s.substring(0,52)+c+s+c+s.substring(52,61)+c+s.substring(61));}}
    and in Pascal you have:
          program s;const p='program s;const p=';a='a';aa=''';';aaa='a=''';aaaa
    ='''';aaaaa='begin write(p,aaaa,p,aa,aaa,a,aa,a,aaa,aaaa,aa,aa,a,a,
    aaa,aaa,aaaa,aa,a,a,a,aaa,aaaa,aaaa,aa,a,a,a,a,aaa,aaaaa,aa,aaaaa)
    end.';begin write(p,aaaa,p,aa,aaa,a,aa,a,aaa,aaaa,aa,aa,a,a,aaa,aaa,
    aaaa,aa,a,a,a,aaa,aaaa,aaaa,aa,a,a,a,a,aaa,aaaaa,aa,aaaaa)end.
    Now you must try your hand at writing a self-reproducing program for the
    HP-71B such that upon running produces as output its own source code.
    You may use not only statements and functions in the System ROMs but also
    in external ROMs, such as the Math ROM, the FORTH/Assembler ROM, or even
    the JPC ROM, to name a few. You should try to optimize for program size,
    the shorter the better.

Take 5:


    We're given a function f(x,y) recursively defined as follows:
         f(0,y) = y+1
    f(x,0) = f(x-1,1)
    f(x,y) = f(x-1,f(x,y-1))
    by applying this definition, it's straightforward to compute
    the function's exact value for various arguments, for instance
    we readily find the values:
         f(0,4) = 5
    f(1,2) = 4
    f(2,3) = 9
    f(3,1) = 13
    you're now asked to write a program that computes and outputs
    the exact value of f(4,2).

    I'll post my original 9-line solution for the HP-71B.

Caveats:


    That's all, the usual caveats apply, mainly:

    • Please refrain from posting just the solutions, actual code
      is mandatory. Googling solutions and posting them
      is pretty lame and only serves to spoil the challenge
      for others and makes blatantly public your unability
      to comply and your serious attitude problem and
      disrespect for rules.

    That said, let's see your answers before I post my solutions
    next Monday. You have more than 48 hours, so hurry up, you
    can do it ! :-)

Best regards from V.

Post-Edited 3 Apr 2006 solely to correct improper formatting (extremely long lines).

Edited: 3 Apr 2006, 5:48 a.m. after one or more responses were posted


#27

After trying a little program to solve 2, I stopped and thought. Looks like it is April 1st...

My solutions a bit later.

Arnaud


#28

Quote:
After trying a little program to solve 2, I stopped and thought.

So did I, but here is my 42S/41C program anyway:

LBL "SSMC152"
STO 00 ' Starting number in X
LBL 00
RCL 00
ENTER
ENTER
ENTER
1
STO+ 00
+
X<>Y
X^2
+/-
RCL 00
X^2
+
Rv
+
R^
X=Y?
VIEW ST X
GTO 00 ' Stopping at an upper limit is up to the user
END

Marcus

#29

A program you ask, a program you get:


solution to nr. 2:

10000
ENT^
1
LBL 1
R/S (or PSE)
2
+
x<=y?
GTO 1
R/S

I'm beginning to understand why you choose this day :-)



groeten,

Bram

#30

It appears Take 2 is the easiest of the bunch. Here's my HP 48 solution:

<< 1 9999 FOR i i 1 DISP 2 STEP >>

A second take on Take 2 doesn't work on the 48 because of RAM limitations---it may work on a 49:

<<'I' DUP 1 99999 2 SEQ >>

As for Take 4, I haven't touched a 71B in years, but I believe this should work:

10 PRINT

Or was that LIST? I'll have to go dig up my manuals...


#31

Duh! Not "Print", obviously.

10 LIST

or

10 PLIST

should work, assuming either of these commands are programmable. They are on many but not all BASIC dialects. Unfortunately I don't have a 71B or an emulator at hand so I can't try it out.


#32

10 LIST

works fine on my real 71B.

Marcus

#33

Well, the solution to problem 1 was already known to Newton (and I was aware of it before this challenge). But Newton didn't have a programmable calculator, so here's an RPL solution:


<< -> d c b a

<< '-a^3+3*a*b-3*c' EVAL >>

>>

I don't think it gets much shorter or faster than that.


#34

I guess these all might be April Fool's challenges?

For problem 2, the condition is that

x + (x + 1) = (x + 1)^2 - x^2

Expanding gives

2x + 1 = 2x + 1

which is always identically fulfilled. So, any number of the form 2x + 1 is a solution. Though we only allow integers..

Program:

<< 1 -> a
<< DO
a 1 DISP
a 2 + 'a' STO
UNTIL
a 9999 >
END
>>
>>


#35

Challenge 3

It's difficult to describe the nonprogramming side of the solution without a figure, but basically...

Having the rope pulled taut implies that it's tangent to the earth at the two points where it touches; ie., it's at 90 degrees to the earth's radius.

The part of the rope that is pulled tight is like a triangle. There is another triangle underneath it, made by the two radii. Call the angle by which the radii are separated 'y'.

Then it turns out that the horizontal distance of the base of either triangle (since they share that base) is

r * sin (y/2)

Then the length of the bit of wire that is pulled taut is

2 * r * tan (y/2)

and the total length of the wire is

r * (2 * pi - y + 2 * tan(y/2))

that has to equal

2 * pi * r + x

with x being the slack.

Simplifying that equation and solving for y is the first part of the program.

Then we have to find the height of the rope. The height of the first triangle is

r * cos(y/2)

The height of the second triangle is

r * sin(y/2) * tan(y/2)

The sum of those, minus the radius of the earth, is the height of the wire.


Putting it together, here's the program:


<< 'y' PURGE -> x
<< '-y+2*tan(y/2)=x/6400000'
EVAL 'y' x ROOT
'6400000*(cos(y/2)+sin(y/2)*tan(y/2)-1)'
EVAL
>>
>>

The answer, for x = 1 meter, is the surprisingly large 121.64 meters.

I don't know if I can guarantee I haven't made a mistake somewhere, but the first question is, is this reasonable? I think I can argue that it is.

If the person were to *pinch* the wire, so that all the slack would be in one little line above a point, then of course the height of that slack would be 0.5 meters. Much less than 121 meters. But as you release the "pinch", the area that's not touching the ground grows and grows, and the height of the lifted line can be much more than the original slack.

As for the "brother in law", I don't know. My first guesses would be that it would have something to do with Goliath, or Atlas, but, oh, well.


#36

Challenge 4

I don't have a 71B, so this shall remain unsolved by me.


Challenge 5

Sorry, I didn't do this one on the calculator; I did it by hand.

It's easy to verify that

f(0,y) = y + 1

f(1,y) = y + 2

f(2,y) = 2 * y + 3

f(3, y) = 8 * 2^y - 3

Then

f(4, 0) = f(3,1) = 13

f(4, 1) = f(3, f(4,0)) = f(3, 13) = 65533

f(4,2) = f(3, f(4,1)) = f(3, 65533) = 8 * 2 ^ 65533 - 3

Heh, now I have to find the *exact* value of that? That ain't happening. But this gives an approximate value in scientific notation (in Level 2 * 10 ^ Level 1 format)

<< 2 LOG 65536 * DUP FP DUP UNROT - >>

So I managed to fulfill the requirement of having a program after all.

The answer is

2.0035297704... * 10 ^ 19728

Edited: 1 Apr 2006, 12:52 a.m.


#37

Easy enough on a 49, were it not for the built-in limit of 9999 on integer powers..

So, all you have to to is write a 'IBIGPOW' program to compute a^b
(a and b integers), up to the limits of the available memory:

@ IBIGPOW
@ In : a b
@ Out: a^b
\<< \-> a b
\<< b 9999 IDIV2 a SWAP ^
IF OVER
THEN a 9999 ^ SWAP
1 4 ROLL
START OVER * NEXT
END
SWAP DROP
\>>
\>>
Now, I know Valentin asked for results, but you don't seriously want me to post the full number, do you? But the 49 is perfectly capable of computing it. After all, it's only 19729 digits.

Cheers, Werner


#38

Thanks! I really did learn something new. I didn't know the reason it refused to do it was because of an arbitrary (9999 limit on powers) restriction; I thought it might have been a memory restriction.

#39

Otherwise, I decided to try something else. I took emu49 and used the following program F with 2 and 4 on the stack.

Now, I will see what happens when I wake up tomorrow morning.


Program called F
<<
DUP2 ->STR " " + SWAP ->STR + 1 DISP
IF DUP 0 ==
THEN DROP 1 +
ELSE
IF OVER 0 ==
THEN SWAP DROP 1 SWAP 1 - F
ELSE DUP UNROT SWAP 1 - SWAP F SWAP 1 - F
END
END
>>

Now, don't try this on a real 49 if you want to keep your batteries.

Arnaud

#40

Now that I know how to do it, I couldn't resist: This should be the *exact* answer. In fact, this reveals that the last couple of digits I reported in the approximate answer were wrong. I should have known that, because 65536 * log(2) only gives seven places after the decimal (because the rest are taken up by the digits to the right of the decimal), so I should have only expected 6 or 7 digits of accuracy in the manitessa.

I'm also posting this because of how impressed I am by the size of numbers this calculator can deal with.


2003529930406846464979072351560255750447825475569751419265016973710
8940595563114530895061308809333481010382343429072631818229493821188
1266886950636476154702916504187191635158796634721944293092798208430
9104855990570159318959639524863372367203002916969592156108764948889
2540908059114570376752085002066715637023661263597471448071117748158
8091413574272096719015183628256061809145885269982614142503012339110
8273603843767876449043205960379124490905707560314035076162562476031
8637931264847037437829549756137709816046144133086921181024859591523
8019533103029216280016056867010565164675056803874152946384224484529
2537361442533614373729088303794601274724958414864915930647252015155
6939226281806916507963810641322753072671439981585088112926289011342
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0189920000241419637068135598404640394721940160695176901561197269823
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0168874458355579262584651247630871485663135289341661174906175266714
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0838890200741699262383002822862492841826712434057514241885699942723
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6143618601548909740241913509623043612196128165950518666022030715613
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7506072339445587895905719156733

#41

I also have been spending a couple of hours trying to find a suitable brother in law...
Still searching.

Arnaud

#42

Take 3

A little complicated, isn't it?

C = 2pi*r
C+1m = 2pi*r+1m = 2pi*(r+x) = 2pi*r+2pi*x
=> 1m = 2*pi*x
=> x = 1m / 2pi
C: Circumphere

r: Radius of the Earth (or of a tin can, doesn't matter at all)

x: value to increase the radius to get the additional length.

Lifting the wire on one end only needs a value of 2x.

Program in RPN:

PI
/
R/S
'Hulk Hogan'
AVIEW
END

Marcus

Edited: 2 Apr 2006, 11:24 a.m.

#43

Quote:
'-a^3+3*a*b-3*c'

It's interesting that the solution doesn't depend on d.


#44

It's not difficult to see (even without doing much algebra) why the sum of CUBES wouldn't depend on D, though.

Each coefficient of a polynomial in x^n + a * x^(n-1) + ... + d form (with leading coefficient of 1) is a symmetric polynomial of the roots.

You can see that by factoring the polynimal ( (x-x1)*(x-x2)*... etc.)

So, for the quartic case, if the roots are x1, x2, x3, x4...

a = -(x1 + x2 + x3 + x4)

and

d = x1 * x2 * x3 * x4


-a^3, by the multinomial theorem, contains a sum of the cubes. It also has some cross terms, which the next few terms in the answer get rid of. But a^3 ONLY has products of three terms taken at a time (again, by the multinomial theorem). So there's no need for d, which has 4 terms taken at a time.

If the problem was to find the sum of the FOURTH powers of the roots, then d would be needed.

#45

This is a variation of the old "how far is the horizon" problem.

Anyway, I get the formula:

'(R+r)*SIN(ACOS(R/(R+r)))-R*ACOS(R/(R+r))-S/2'

= 0

where R=6.4e6 meters, S, the slack = 1 meter, and we are solving for r, height above ground.

I didn't bother to write a program, but just keyed it in and solved for r; I got 121.645 meters (399 feet). I used my trusty 48GX and the Solver.

I don't know the brother-in-law's name but maybe our person was a princess sitting on a 40-story pile of mattresses.

- Michael

#46

Hi Valentin,

Take 5 is a battery drainer. I've written a small C program for my PB-2000C, recursive style:

/* SSMC15-5 */
main()
{
int x,y;
printf("x=");
scanf("%d",&x);
printf("y=");
scanf("%d",&y);
printf("f(%d,%d)=%d \n",
x,y,f(x,y));
}

int f(x,y)
int x,y;
{
gotoxy(0,0);printf("%5d %5d\n",x,y);
return x==0 ? y+1
: y==0 ? f(x-1,1)
: f(x-1,f(x,y-1));
}

I tried it for the solutions you gave:
     f(0,4) = 5
f(1,2) = 4
f(2,3) = 9
f(3,1) = 13

And the results were as expected. But entering x=4 and y=2 at the prompts seems to be an almost endless exercise, so no solution this time.

Marcus

Edited: 3 Apr 2006, 4:58 a.m.


#47

Hi Marcus,

this reminds me of the infamous Ackermann-function, which can be used as a benchmark for procedure calls.

Greetings, Klaus


#48

Hi Klaus,

you're perfectly right, it is the Ackermann-Péter function (See http://de.wikipedia.org/wiki/Ackermannfunktion or http://en.wikipedia.org/wiki/Ackermann_function).

The english article lists the results. Valentin is asking for
f(4,2) = 265536 - 3.

I don't see a chance to calculate it with my simple C program because arbitrary precision would be needed.

Marcus

Edited: 3 Apr 2006, 6:08 a.m.


#49

I posted an RPL version a bit higher which can work in arbitrary precision. I tried to run it on EMU49 see what was the first error I would get.
I had to stop before getting any error as my laptop was getting REALLY hot.
If someone has a very cool machine it would be good to run it without the first line see what happens...

Arnaud

#50

Hi all,

    Thank you very much to all of you who were interested in this Spring Special challenge and managed to find the time to try and solve its various parts despite the very short deadline, I certainly hope you enjoyed it. Mostly, you did unexpectedly well, essentially solving all of them five, and I must confess it never ceases to amaze me, the extreme ingenuity displayed in both actual solutions and close calls.

    These are my original solutions, plus comments:


Take 1:


    Though it might seem at first that you'd need to actually compute all four real and/or complex roots in order to be able to compute the sum of their cubes, actually this isn't so, because the asked value is a symmetric function of the roots (i.e.: it remains the same for every permutation of the roots) and so can be computed directly by using symmetric polynomials, a technique developed by Newton and Girard.

    For this particular Take, the asked sum S as a function of the coefficients is simply:

         S = -a3 + 3*a*b - 3*c
    For example, if we have the equation:
         x4 + x3 + 2*x2 + x + 1 = 0
    the coefficients are:
          a = 1, b = 2, c = 1, d = 1
    and the asked sum is:
          S = -13 + 3*1*2 - 3*1 = 2
    Let's check it. The roots are:
          x1 = i
    x2 = -i
    x3 = -1/2 + Sqrt(3)/2*i
    x4 = -1/2 - Sqrt(3)/2*i
    and the sum of their cubes comes indeed to 2. This is my HP-15C solution to compute the required sum, assuming the coefficientes are previously placed in the stack like this:
         T: d 
    Z: c
    Y: b
    X: a
    the corresponding program listing is:
         01 LBL A
    02 CHS
    03 X^2
    04 LASTX
    05 *
    06 X<>Y
    07 LASTX
    08 *
    09 GSB 0
    10 X<>Y
    11 LBL O
    12 3
    13 *
    14 -
    15 RTN
    My HP-71B solution (32 bytes) is:
         1 DESTROY ALL @ INPUT A,B,C @ DISP -A^3+3*A*B-3*C
    By the way, the HP-71B is powerful enough that even if you didn't realize that the sum could be obtained by using symmetric polynomials and tried for brute force instead (computing the four real and/or complex roots and then adding up their cubes), you can still make do with this 2-liner:
         1 DESTROY ALL @ OPTION BASE 0 @ DIM A(4) @ COMPLEX R(3),S @ MAT INPUT A
    2 MAT R=PROOT(A) @ S=0 @ FOR I=0 TO 3 @ S=S+R(I)^3 @ NEXT I @ DISP S
    For instance:
         >RUN
    A(0)? 1,-2,-3,4,-5 [ENTER] -> (14,0)
    i.e.: the sum of the cubes of the four roots is the complex value (14, 0), which is of course the real value 14


Take 2:



    This 2-liner (64 bytes) for the HP-71B will find all solutions quickly:
         1 FOR I=1 TO 10000 STEP 2 @ A=(I-1)/2 @ B=A+1 @ IF B*B-A*A=I THEN DISP I;
    2 NEXT I @ DISP "OK"
    Let's run it:
         >RUN
    1 3 5 7 9 11 [...] 9991 9993 9995 9997 9999
    though from an eminently practical point of view, it can actually be simplified to the following 1-line, 30-byte program:
        1 FOR I=1 TO 10000 STEP 2 @ DISP I; @ NEXT I @ DISP "OK"
    which outputs just about the same. Both programs can have their upper limits increased if desired. For the given range, there are 5000 solutions in all


Take 3:



    A little geometry will duly solve this. Let's call the Earth's radius r, the subtended angle between the point of maximum height and the point of tangency we'll call x, and let's say the length of the cable not in contact with the Earth is 2a. The corresponding circular arc length is 2rx. Let d be the extra length added to the cable. Then 2a = 2rx + d. Hence a = rx + d/2, and so a/r = x + d/2r.

    We also have, tan(x) = a/r. Therefore the equation to solve
    for the angle x is:

         tan(x) = x + d/2r
    Once we've got the angle x by solving it, we then have that the
    required height h over the ground is:
         h = r(sec(x) - 1)
    For our particular problem d = 1, r = 6,400,000, so
    we must solve:
         tan(x) = x + d/2r = x + 1/12,800,000
    which gives:
         x = 0.00616549902401 radians
    and from this, the height h is:
        h = r(sec(x) - 1) = 6,400,000*(sec(0.00616549902401) - 1)
    = 121.644736 m
    If the man can stretch his arm to a height in excess of 120 m (360 feet) while comfortably sitting at ground level, he must be
    Mr. Reed Richards, alias "Mr. Fantastic", so his brother-in-law's
    name should be Johnny Storm, alias "The Human Torch"
    , both members of the well-known Fantastic Four superhero team.

    This is my 20-step program for the HP-15C which will return the height for any given extra length. First store the constant 12,800,000 (2r) in R0 and set RAD mode:

         01 LBL A    
    02 STO1
    03 0
    04 SOLVE B
    05 COS
    06 1/X
    07 1
    08 -
    09 RCL*0
    10 2
    11 /
    12 RTN
    13 LBL B
    14 TAN
    15 LASTX
    16 -
    17 RCL 1
    18 RCL/0
    19 -
    20 RTN
    Running it, we can get for instance:
         For d =   1 m:    1 GSB A  ->  121.6448 m  (h)
    d = 0.1 m: 0.1 GSB A -> 26.2080 m
    d = 10 m: 10 GSB A -> 564.6400 m
    This is the 2-line, 65-byte HP-71B version:
         1 INPUT "D=";D @ R=6400000
    2 DISP "H=";R*(1/COS(FNROOT(0,1,TAN(FVAR)-FVAR-D/(2*R)))-1)
    Let's run it:
         >RUN
    D=1
    H= 121.644736
    As you may see, even a mere 0.1 meter (some 4 inches) of extra length can raise the highest point to a height of more than 26 meters (nearly 80 feet). Mr Fantastic would still be needed.

Take 4:



    This 1-line, 1-statement, 5-byte HP-71B program:
         1 LIST
    is surely the simplest one which fulfills the conditions. Upon running it, lo and behold:
         >RUN
    1 LIST
    it does produce its own source code as output !

    My, my, aren't we fortunate that HP-71B's BASIC is such a powerful programming language ?


Take 5:


    This recursively defined function is no other than the infamous Ackerman function, which among many interesting properties most of which are theoretically important, it has the unexpected characteristic that its value grows extremely quickly, even for very small input, which means a direct naive attempt will most likely overflow your handheld. This difficulty is easily overcome by doing a little research to see what we're trying to compute. We'll proceed to compute f(4,2) step by step:

    1. Let's find f(1, y):

      f(1,y) = f(0, f(1, y-1)) = f(1, y-1)+1

      and f(1,0) = f(0,1) = 1+1 = 2

      so we have the difference equation (not differential):

      f(1, y) = f(1, y-1)+1 with initial condition f(1,0) = 2

      whose exact solution is: f(1,y) = y + 2

    2. Let's find f(2, y):

      f(2, y) = f(1, f(2, y-1)) = f(2, y-1)+2

      and f(2,0) = f(1,1) = 1+2 = 3

      so we have the difference equation:

      f(2, y)= f(2, y-1)+2 with initial condition f(2,0) = 3

      whose exact solution is: f(2,y) = 2*y + 3

    3. Let's find f(3, y):

      f(3, y) = f(2, f(3, y-1)) = 2*f(3, y-1)+3

      and f(3,0) = f(2,1) = 2*1+3 = 5

      so we have the difference equation:

      f(3, y) = 2*f(3, y-1)+3 with initial condition f(3,0)=5

      whose exact solution is: f(3, y) = 2y+3-3

    4. Now for f(4,2):

      f(4,2) = f(3, f(4, 1))
      = f(3, f(3, f(4, 0))
      = f(3, f(3, f(3, 1)))
      = f(3, f(3, 13))
      = f(3, 65533)
      = 265536 - 3

      and our program must simply compute this value, exactly. This is my 9-line program for the HP-71B which does the job:

           1 DESTROY ALL @ OPTION BASE 0 @ N=65536 @ M=9 @ DIM A(1) @ K=10^M
      2 A(0)=1 @ P=0 @ L=9 @ R=2^L @ FOR J=0 TO N-L STEP L @ MAT A=(R)*A @ GOSUB 8
      3 IF NOT MOD(J,27) THEN DIM A(UBND(A,1)+1)
      4 NEXT J @ R=MOD(N,L) @ IF R THEN MAT A=(2^R)*A @ GOSUB 8
      5 A(0)=A(0)-3 @ J=0 @ PRINT @ PRINT STR$(A(P));
      6 FOR I=P-1 TO 0 STEP -1 @ J=J+1 @ IF J=7 THEN PRINT @ J=0
      7 A$=STR$(A(I)) @ PRINT RPT$("0",M-LEN(A$));A$; @ NEXT I @ END
      8 FOR I=0 TO P @ A(I+1)=A(I+1)+A(I) DIV K @ A(I)=MOD(A(I),K) @ NEXT I
      9 P=P+SGN(A(P+1)) @ RETURN
      You'll need 20 Kb RAM in your HP-71B or Emu71 to run it. After less than 25 minutes under Emu71 or just one week in a physical HP-71B, it outputs the exact, 19,729-digit result that follows.

      Best regards from V.

      >RUN

      20035299304068464649790723515602557504478254755697514192650169737108940595563114
      53089506130880933348101038234342907263181822949382118812668869506364761547029165
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Edited: 3 Apr 2006, 7:47 a.m.


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